- Email: [email protected]

7 Properties of Solutions We have seen in Chapter 6 that the thermodynamic properties of homogeneous pure substances dep

* Views 248*
* Downloads 4*
* File size 17MB*

7 Properties of Solutions We have seen in Chapter 6 that the thermodynamic properties of homogeneous pure substances depend only on the state of the system. The relationships developed for pure fluids are not applicable to solutions and need modification. The thermodynamic properties of solutions and heterogeneous systems consisting of more than one phase are influenced by the addition and removal of matter. The term solution includes homogeneous mixtures of two or more components in the gas, liquid or solid phase. The pressure, temperature and the amount of various constituents present determine the extensive state of a solution; and pressure, temperature and composition determine the intensive state. In this chapter, we discuss how the thermodynamic properties of a solution are determined and introduce certain concepts that are essential to the study of phase equilibria and chemical reaction equilibria.

7.1 PARTIAL MOLAR PROPERTIES The properties of a solution are not additive properties of its components. For example, the volume of a solution is not the sum of the volumes of the pure components constituting the solution. It means that when a substance becomes part of a solution it loses its identity. But it still contributes to the property of the solution as is evident from the fact that by changing the amount of substance, the solution property also changes. Thus we need a new set of concepts that enable us to apply thermodynamics to solutions of variable composition. In this connection, the concept of partial molar properties is of great use. The term partial molar property is used to designate the property of a component when it is in admixture with one or many components. To be more precise, the partial molar property of a particular component in a mixture measures the contribution of that component to the mixture property. If Mt is the total value of any extensive thermodynamic property of a solution, the partial molar property

of the component i in the solution is defined as

In Eq. (7.1), n is the total number of moles and M is the molar property of the solution. ni denotes the number of moles of component i in solution, so that n = Sni. In general, any partial molar property is the increase, in the property Mt of the solution resulting from the addition at constant temperature and pressure, of one mole of that substance to such a large quantity of the system that its composition remains virtually unchanged. It is an intensive property and its value depends only on the composition at the given temperature and pressure. The subscript njπi indicates that the number of moles of all components in the solution other than the number of moles of

i are kept constant.

7.1.1 Physical Meaning of Partial Molar Properties To understand the physical meaning of partial molar properties let us consider the partial molar volume, the simplest partial molar property to visualise. It is the contribution that a component in the solution makes to the total volume. Consider an open beaker containing a huge volume of water. Assume that one mole of water is added to it. The volume increases by 18 10–6 m3, which is the molar volume of pure water. If the same amount of water is added to a large amount of pure ethanol taken in the beaker, the increase in volume will be approximately 14 10–6 m3, which is the partial molar volume of water in pure ethanol. The difference in the increase in volumes can be explained thus: the volume occupied by a given number of water molecules depends on the molecules surrounding them. When water is mixed with a large volume of alcohol, there is so much alcohol present that each water molecule is surrounded by pure ethanol. Consequently, the packing of the molecules would be different from that in pure water, and the molecules occupy lesser volume. If one mole water is added to an equimolar mixture of alcohol and water, the increase in volume of the solution would be different from that resulted when the same quantity were added to pure alcohol. The partial molar properties of the components of a mixture vary with composition because the environment of each type of molecule changes as the composition changes. The intermolecular forces also get changed resulting in the changes in the thermodynamic properties of solutions with compositions. The variation of partial molar volumes with concentration is shown in Fig. 7.1 for ethanol (E) – water (W) system.

We have seen that the effective molar volume of water added to the ethanol–water solution, i.e. the

partial molar volume in the solution is less than the molar volume VW of pure water at the same temperature and pressure. To be specific, when pure water is added to an ethanol water solution of volume Vt and allowed sufficient time for heat exchange so that temperature remains the same as that before addition, the increase in volume of the solution DVt π DnWVW, where DnW is the moles of water added. The increase in volume is given by

In this process, a finite drop of water was added which may cause a finite change in composition. If were to represent a property of the solution, it must be based on data for the solution at this composition. For an infinitesimal amount of water added, Eq. (7.3) becomes

and it denotes the incremental change in mixture volume which occurs when a small quantity of component i is added at constant pressure and temperature. The amount of i added is so small that no detectable change in composition occurs. While the molar volume is always positive, the partial molar volume may even be negative. The partial molar volume of MgSO4 in water at infinite dilution (i.e. in the limit of zero concentration) is –1.4 10–6 m3/mol which means that the addition of one mole of MgSO4 to a large volume of water results in a decrease in volume of 1.4 10–6 m3. The contraction may be due to the breaking up and subsequent collapse of the open structure of water as the ions become hydrated. Though different from molar properties of the pure components, to get a physical picture of the concept of partial molar properties, we can treat them as the molar properties of the components in

solution. However, it is to be borne in mind that the components of a solution are intimately intermixed and cannot have individual properties of their own. The partial molar properties in fact, represent the contribution of individual components constituting the solution to the total solution property as described in the following section.

7.1.2 Partial Molar Properties and Properties of Solution Consider any thermodynamic extensive property (such as volume, free energy, heat capacity, etc.), its value for a homogeneous system being completely determined by the temperature, pressure and the amounts of various constituents present. Let M be the molar property of a solution and Mt be the total property. Then, Mt = nM, where n is n1 + n2 + n3 + . . . . Here, n1, n2, n3, . . . are the number of moles of the respective components 1, 2, 3, . . . of the system. The solution property is a function represented by Mt = f(P, T, n1, n2, . . ., ni, . . .)………(7.6) If there is a small change in the pressure, temperature and the amounts of various constituents, then

It is evident that the partial molar properties are not extensive properties, but are intensive properties of the solution. They depend, therefore, not upon the total amount of each constituent, but only upon the composition, or the relative amounts of the constituents. If we add several constituents simultaneously to a given solution at constant temperature and pressure, keeping the ratio of the various constituents constant, the partial molar properties are not changed. Then, the change in property

This equation along with Eq. (7.9), which can be written in the following form serves as the relationship between partial molar properties and total solution property.

We see that the partial molar property of any constituent may be regarded as the contribution of one mole of that constituent to the total value of the property under the specified conditions. In other words, the partial molar properties may be treated exactly as though they represent the molar properties of the components in the solution. EXAMPLE 7.1 Give an alternative derivation for Eqs. (7.12) and (7.14) starting from Eq. (7.9) Solution Equation (7.9) gives

n represents the total amount of various constituents and dn the changes in the total number of moles. One is free to choose any value for n as well as dn. In short, n and dn can be independently changed. For all possible values of n and dn, the above equation is to be satisfied. This is possible only if the

terms in brackets reduce to zero.

EXAMPLE 7.2 Will it be possible to prepare 0.1 m3 of alcohol-water solution by mixing 0.03 m3 alcohol with 0.07 m3 pure water? If not possible, what volume should have been mixed in order to prepare a mixture of the same strength and of the required volume? Density of ethanol and water are 789 and 997 kg/m3 respectively. The partial molar volumes of ethanol and water at the desired compositions are: Ethanol = 53.6

10–6 m3/mol; water = 18

10–6 m3/mol.

Solution Let us first find out the number of moles of ethanol and water mixed and their mole fractions in the resultant mixture. Moles of ethanol in the solution = (0.03 789 103)/46 = 514.57 mol Moles of water in the solution = (0.07 997 103)/18 = 3877.22 mol Mole fraction of ethanol desired = 514.57/(514.57 + 3877.22) = 0.1172 Mole fraction of water = 1 – 0.1172 = 0.8828 Actual volume of solution is 514.57 53.6 10–6 + 3877.22 18 10–6 = (0.02758 + 0.06979) = 0.09737 m3 That is, by mixing 0.03 m3 alcohol with 0.07 m3 water, we would get only 0.09737 m3 of solution. To prepare 0.1 m3 of solution the volumes to be mixed are: Ethanol = (0.03/0.09737) m3

0.1 = 0.03081 m3 and Water = (0.07/0.09737)

0.1 = 0.07189

EXAMPLE 7.3 A 30 per cent by mole methanol-water solution is to be prepared. How many cubic metres of pure methanol (molar volume, 40.727 10–6 m3/mol) and pure water (molar volume, 18.068 10–6 m3/mol) are to be mixed to prepare 2 m3 of the desired solution? The partial molar volumes of methanol and water in a 30 per cent solution are 38.632 and 17.765 10–6 m3/mol, respectively.

10–6 m3/mol

Solution Molar volume of the desired solution is V= m3/mol

= (0.3

38.632

10–6 + 0.7

Therefore, 2 m3 of the desired solution contains

17.765

10–6) = 24.0251

10–6

103 mol Number of moles of methanol in 2 m3 of solution = 83.2463

2/(24.0251

10–6) = 83.2463

103

0.3 = 24.9739

103

mol Number of moles of water in 2 m3 of solution = 83.2463 103 0.7 = 58.2724 103 mol Volume of methanol to be taken = 24.9739 103 40.727 10–6 m3 = 1.0171 m3 Volume of water to be taken = 58.2724 103 18.068 10–6 m3 = 1.0529 m3 EXAMPLE 7.4 Laboratory alcohol containing 96% alcohol and 4% water is to be diluted to a solution containing 56% alcohol and 44% water. All percentages are on weight basis. The partial specific volumes are as follows: In 96% alcohol solution, = 0.816 10–3 m3/kg, = 1.273 10–3 m3/kg. In 56% alcohol solution, density of water may be taken as 0.997

= 0.953 10–3 m3/kg, 103 kg/m3.

= 1.243

10–3 m3/kg. The

(a) How much water should be added to 2 10–3 m3 of the laboratory alcohol? (b) What is the volume of the dilute alcohol obtained? Solution Basis: 2

10–3 m3 laboratory alcohol.

1 kg laboratory alcohol (0.96 1.273 + 0.04 0.816) 10–3 = 1.255 10–3 m3 2 10–3 m3 of laboratory alcohol 2 10–3/(1.255 10–3) = 1.594 kg (a) Let the mass of water added be m kg. Taking an alcohol balance, we get 1.594 0.96 = (m + 1.594) 0.56 Thus mass of water added is, m = (1.594 0.96)/0.56 – 1.594 = 1.1386 kg and volume of water added is 1.1386/(0.997 103) = 1.142 10–3 m3 (b) Mass of dilute alcohol obtained = 1.594 + 1.1386 = 2.7326 kg Specific volume of 56% alcohol = (0.56 Therefore, Volume of dilute alcohol obtained = 1.115

1.243 + 0.44 10–3

0.953)

10–3 = 1.115

2.7326 = 3.0468

10–3 m3/kg

10–3 m3

7.1.3 Determination of Partial Molar Properties Method 1 (Analytical). If the volume of a solution is known as a function of its composition, the partial molar volume of a constituent may be found by partial differentiation with respect to the amount of that constituent.

Method 2 (Graphical). Let Vt, the volume of the solution containing a fixed amount of one of the constituents (say, n1) is known for several values of the amount of other constituent (say, n2). We may plot Vt against n2. See Fig. 7.2. The slope of the tangent to the curve is (∂Vt/∂n2)P,T,n1 which, by definition is , the partial molar volume of component 2. The volume of solution is assumed so large that no significant change in composition occurs when n2 is changed. This method has limitation of not yielding values of directly. Also, it is not advisable to use this method for determination of when n2 is large compared to n1. The method of tangent intercepts is free from such limitations and is therefore preferred for the determination of partial molar properties.

Method 3 (The tangent-intercept method). This is also a graphical method widely used for the determination of partial molar properties of both components in a binary solution. The molar volume V is plotted against mole fraction of one of the components (say, x2, the mole fraction of component 2). To determine the partial molar volumes, draw the tangent to the curve at the desired mole fraction. The intercept that this tangent makes with the vertical axis at x2 = 1 gives and the intercept on the vertical axis at x2 = 0 (or x1 = 1) gives

. In Fig. 7.3, BD =

and AC =

.

To prove this result, consider a binary solution containing n1 moles of component 1 and n2 moles of component 2. Let the total volume be Vt and let V be the molar volume. Then

In Fig. 7.3, the length BD = BE + ED, where BE is the slope of the tangent at P times the length PE. That is, BE = (1 – x2) (∂V/∂x2) and ED = V, the molar volume at the mole fraction x2. Thus BD = V + (1 – x2) (∂V/∂x2) which, by Eq. (7.22) is

. Similarly, the length

AC = FC – FA = V – x2 (∂V/∂x2) = The above methods are applicable for the determination of various other partial molar properties also. Of the various mixture properties, only the volume can be determined absolutely. For the determination of other properties like , etc., it becomes necessary to work with property changes on mixing (discussed later in this chapter) like DG, DH, etc. The method of tangent intercept for the determination of, say , requires the plot of DG per mole versus x2. EXAMPLE 7.5 At 300 K and 1 bar, the volumetric data for a liquid mixture of benzene and cyclohexane are represented by V = 109.4 10–6 – 16.8 10–6x – 2.64 10–6x2, where x is the mole fraction of benzene and V has the units of m3/mol. Find expressions for the partial molar volumes of benzene and cyclohexane. Solution The molar volume of the solution as a function of composition is given: V = 109.4 10–6 – 16.8 10–6x1 – 2.64 10–6x12………(7.23) where x1 = mole fraction of component 1 (in this case, benzene). By Eq. (7.20),

EXAMPLE 7.6 The enthalpy at 300 K and 1 bar of a binary liquid mixture is H = 400x1 + 600x2 + x1x2(40x1 + 20x2) where H is in J/mol. For the stated temperature and pressure, determine: (a) Expressions for and in terms of x1 (b) Numerical values for the pure component enthalpies H1 and H2 (c) Numerical values for the partial molar enthalpies at infinite dilution

.

Solution The molar enthalpy of the solution as function of concentrations of the constituents is given: H = 400x1 + 600x2 + x1x2(40x1 + 20x2)………(7.26) (a) Differentiating Eq. (7.26), we get

Since x1 = 1 – x2, dx1 = – dx2 and ∂x2/∂x1 = –1, the above equation simplifies to

EXAMPLE 7.7 The volume of an aqueous solution of NaCl at 298 K was measured for a series of molalities (moles of solute per kg of solvent) and it was found that the volume varies with molality according to the following expression. V = 1.003 10–3 + 0.1662 10–4m + 0.177 10–5m1.5 + 0.12 10–6m2 where m is the molality and V is in m3. Calculate the partial molar volumes of the components at m = 0.1 mol/kg. Solution The partial molar volume of NaCl:

Differentiating Eq. (7.29) with reference to m,

Partial molar volume of water = 18.05 Partial molar volume of NaCl = 17.48

10–6 m3/mol 10–6 m3/mol

7.2 CHEMICAL POTENTIAL The chemical potential, denoted by the symbol m, is a widely used thermodynamic property. It is used as an index of chemical equilibrium in the same manner as temperature and pressure are used as indices of thermal and mechanical equilibrium. The chemical potential mi of component i in a solution is the same as its partial molar free energy in the solution, component i in a solution can be defined as

. That is, chemical potential of a

………(7.30) The total free energy Gt of a solution is a function of pressure, temperature and number of moles of various components. Gt = f(P, T, n1, n2, . . ., ni, . . .)………(7.31) The total differential dGt is

We have shown that for a closed system, when there is no change in the amount of various constituents, dG = V dP – S dT (6.18) Considering the total properties of the system, dGt = Vt dP – St dT from which, it follows that

By reasoning analogous to that used in the derivation of Eq. (7.12), we have, at constant temperature and pressure, Gt = S mini For a binary solution, the molar free energy of the solution is G = x 1m 1 + x 2m 2 The chemical potential of a component is thus seen to be the contribution of that component to the Gibbs free energy of the solution. The chemical potential is an important property of solution extensively used in the study of phase and chemical equilibria.

7.2.1 Effect of Temperature and Pressure on Chemical Potential Effect of temperature. Consider Eqs. (7.30) and (7.34). Differentiate Eq. (7.30) with respect to temperature. Then

where is the partial molar entropy of the component i in the solution. This result, though gives the variation of chemical potential with temperature, can be put in a more useful form [compare with Eq. (6.73)] as follows: Since

Equation (7.41) predicts the effect of temperature on chemical potential. Effect of pressure. Equations (7.30) and (7.34) are further differentiated to develop equations that predict the effect of pressure on chemical potential. Differentiating Eq. (7.30) with respect to pressure, we obtain

The rate of change of chemical potential with pressure is thus equal to the partial molar volume of the constituent. EXAMPLE 7.8 Prove the alternative definition of chemical potential that mi = (∂U/∂ni)S,V,nj. Solution The internal energy of a system may be expressed as a function of thermodynamic state and moles of the components like the Gibb’s free energy. For the present purpose, it is convenient to express it as Ut = f(St, Vt, n1, n2, . . ., ni, . . .) which gives

But, we know that G = H – TS = U + PV – TS so that dG = dU + P dV + V dP – T dS – S dT

The change in the total free energy at constant temperature and pressure is therefore,

Equation (7.50) is an alternative definition of chemical potential. But it should be understood that (∂Ut/∂ni)S,V,njπi is not partial molar internal energy, for it refers to constant entropy and volume and not to constant temperature and pressure. Partial molar internal energy is not equal to chemical potential. EXAMPLE 7.9 Show that for an ideal gas,

Solution For a mixture of ideal gases,

7.3 FUGACITY IN SOLUTIONS The concept of fugacity was discussed in Chapter 6 with reference to pure substances. It was pointed out that fugacity is a useful concept in dealing with mixtures. For pure fluids, the definition of fugacity is provided by Eq. (6.118) and (6.122):

7.3.1 Fugacity in Gaseous Solutions The fugacity of a component i in a gaseous solution is given by Eq. (7.51). Equation (7.44) gives the effect of pressure on chemical potential.

For a mixture of ideal gases, we have the following simple equation of state:

PVt = (n1 + n2 + n3 + . . .) RT

which states that the fugacity of a component in a mixture of ideal gases is equal to the partial pressure of that component in the mixture. However, this is not true for real gases. Equation (7.57) provides the means for computing fugacities in the real gaseous solution. But this requires the evaluation of as a function of pressure, which in turn requires the knowledge of how the solution volume varies with composition at each pressure. These types of data are rarely available, and hence rigorous calculation of fugacities in gaseous mixtures using Eq. (7.57) is rarely done.

7.3.2 Lewis–Randall Rule As the calculation of fugacity in a mixture of gases through the general equation [Eq. (7.57)] is very difficult, we devise a model for mixtures known as the ideal solution model the fugacity of which can be easily evaluated. The fugacity in actual solution is then determined by taking into account the deviation of the actual solution from this ideal model behaviour. As an ideal gaseous solution we can consider a gas mixture formed without any volume change on mixing the components. A gas mixture that follows the Amagat’s law is an ideal gaseous solution. For such solutions, the volume of the mixture is a linear function of the mole numbers at a fixed temperature and pressure. That is, Vt = S niVi………(7.60) where Vi is the molar volume of pure i at the same temperature and pressure. For such ideal solutions,

Note that the right-hand side of Eq. (7.57) reduces to the same result as that given by Eq. (6.128) where the residual volume for the pure component is given by a = Vi – RT/P. That is, for pure components at a temperature T and pressure P,

Subtracting Eq. (7.62) from Eq. (7.63),

which is commonly known as Lewis–Randall rule or Lewis fugacity rule. It states that fugacity of a component in an ideal solution is directly proportional to the mole fraction of the component in the solution. In Eq. (7.65), is the fugacity of the species i in an ideal gaseous solution, and fi is the fugacity of pure i evaluated at the temperature and pressure of the mixture. Thus, we have now, for ideal gaseous solution and, for ideal (perfect) gases. For a gas mixture to behave as an ideal solution, it requires only that the molar volume in the pure state and the partial molar volume in the solution be the same, or

. For the mixture to be an ideal

gas it requires that , which means that the molar volumes of all the components are the same whether in the mixture or in the pure state. For ideal solutions, the volumes of components may differ from one another. In short, the concept of an ideal gaseous solution is less restrictive than that of a mixture of ideal gases. The Lewis–Randall rule is a simple equation and is therefore widely used for evaluating fugacities of components in gas mixtures. It allows the fugacity of a component in the mixture to be calculated without any information about the solution except its composition. However, it is not reliable because of the severe simplification inherent in Amagat’s law of additive volumes. But at high pressures it is often a very good assumption, because, at liquid like densities, fluids tend to mix with little or no change in volume (J.M. Prausnitz et al., 1986). Lewis fugacity rule is valid for systems where the intermolecular forces in the mixture are similar to those in the pure state. Thus, it can be said that this rule is valid

1. At low pressures when the gas phase behaves ideally 2. At any pressure if the component is present in excess 3. If the physical properties of the components are nearly the same 4. At moderate and high pressures, the Lewis–Randall rule will give incorrect results if the molecular properties of the components are widely different and the component under consideration is not present in excess.

7.3.3 Fugacities in Liquid Solutions Calculation of fugacity of a component in a liquid solution using Eq. (7.57) is not practical because the volumetric data at constant temperature and composition are rarely available. These data are required for the integration over the entire range of pressures from the ideal gas state to the pressure of the solution including the two-phase region. For calculation of fugacities in liquid solutions, another approach is used. We define an ideal solution whose fugacity can be easily calculated knowing its composition and measure the departure from ideal behaviour for the real solution. A quantitative measure of the deviation from ideality is provided by the function known as the activity coefficient which will be discussed in Section 7.6.

7.3.4 Ideal Solutions and Raoult’s Law A solution in which the partial molar volumes of the components are the same as their molar volumes in the pure state is called an ideal solution. There is no volume change when the components are mixed together to form an ideal solution. That is, for an ideal solution V = S xi = S xiVi, where V is the molar volume of the solution, Vi and are the molar volume and partial molar volume respectively of the component i, and xi is the mole fraction of component i in the solution. If a mixture of two liquids is to behave ideally, theoretical considerations reveal that the two types of molecules must be similar. The environment of any molecule and hence the force acting on it is then not appreciably different from that existing in the pure state. We have shown that for ideal gaseous solutions, the Lewis–Randall rule is applicable which states that fugacity of each constituent is directly proportional to the number of moles of the constituent in the solution. The Lewis–Randall rule is applicable to ideal liquid solutions also. It can be written as ………(7.66) where is the fugacity of component i in the solution, fi is the fugacity of i in the pure state, and xi is the mole fraction of component i in the solution. While the ideal solution model is adequate for many gas mixtures for reasonable temperature and pressure, the same is not true for the case of liquid solutions. Very few solutions follow Eq. (7.66) over the entire composition range. Ideal liquid solution behaviour is often approximated by solutions comprised of molecules not too different in size and chemical nature. Thus a mixture of isomers (e.g. ortho- , meta- and para-xylene), adjacent members of homologous series of organic compounds (e.g. n-hexane and n-heptane, ethanol and propanol, benzene and toluene, ethyl bromide and ethyl iodide) etc., are expected to form ideal solutions. Raoult’s Law. The criterion of phase equilibria permit us to replace the liquid phase fugacities

and fi with fugacities in the gas phase with which the liquid is in equilibrium. Thus, under equilibrium. Here superscripts V and L refer to the vapour phase and the liquid phase respectively. Thus, fugacity in Eq. (7.66) is equal to the fugacity of constituent i in the vapour phase. If the vapour phase is assumed to be ideal gas, which is true if the pressure is not too high, the vapour phase fugacity

is the same as partial pressure

of component i in the vapour. If the liquid phase is pure

i, the fugacity of pure i in the vapour phase can be replaced with the vapour pressure conditions the Lewis–Randall rule, Eq. (7.66), becomes

. Under these

………(7.67) This expression is known as Raoult’s Law. This is a simplified form of the Lewis–Randall rule. Whereas the Lewis–Randall rule is obeyed by all ideal solutions, the Raoult’s law is applicable to ideal solutions if the vapour phase with which it is in equilibrium is an ideal gas. Raoult’s law provides a very simple expression for calculating the fugacity of a component in the liquid mixture which is the same as the partial pressure of the component in the vapour. It says that the partial pressure is directly proportional to the mole fraction in the liquid solution. Ideal solutions which conform to Raoult’s law over the entire range of concentrations are rare. A frequently cited example for ideal solutions is mixtures of optical isomers of organic compounds. Raoult’s law applies as fair approximation to mixtures of hydrocarbons showing a reasonable similarity in molecular structure such as are encountered in petroleum industry. In most other cases Raoult’s law applies only over a limited concentration range.

7.4 HENRY’S LAW AND DILUTE SOLUTIONS Solutions conforming to Raoult’s law over the entire concentration range are rare as pointed out earlier. A solution, any of whose components does not obey Raoult’s law is designated as non-ideal solution. Even non-ideal solutions exhibit a common form of ideal behaviour over a limited concentration range where the fugacity (or, the partial pressure ) is directly proportional to the concentration in the liquid. This behaviour is exhibited by the constituent as its mole fraction approaches zero, and is generalised by Henry’s law.

Often, the solute portion of the non-ideal liquid solution can be assumed to follow Henry’s law. is the partial pressure of the solute over the solution, xi is its mole fraction in the solution and Ki is a proportionality constant known as Henry’s law constant. Ki may be greater or less than

, the

vapour pressure of the solute at the temperature and total pressure in question. When Ki and are equal, Henry’s law and Raoult’s law are identical. Henry’s law may be thought of as a general rule of which Raoult’s law is a special case. Henry’s law is obeyed in all solutions by the solute at extremely low concentrations. Essentially all liquids will obey Henry’s law close to mole fraction zero, but many will deviate from the law above 0.01–0.02 mole fraction. And almost all liquids

deviate above 0.1 mole fraction. But in some exceptional cases, Henry’s law is found to be obeyed quite well up to xi = 0.5. For ideal solutions, the partial fugacity (or partial pressure) of a component is proportional to its mole fraction. For a real solution it has been found experimentally that as the mole fraction of the component approaches unity, its fugacity approximates to the value for an ideal solution, though at lower mole fractions, the behaviour departs markedly from ideal behaviour. In Fig. 7.4, the fugacity curve becomes asymptotic to the straight line showing ideal behaviour as mole fraction approaches unity. In a dilute solution, the component present in larger proportions designated as solvent, obeys Raoult’s law even though it may depart from ideal solution behaviour in a more concentrated solution. As the mole fraction of the solute—the component present in smaller proportions—approaches zero, it will conform to the ideal behaviour predicted by Henry’s law. Thus, we can generalise by saying that the solute in a dilute solution obeys Henry’s law and the solvent obeys Raoult’s law. It can be shown that over the range of compositions where the solvent obeys Raoult’s law, the solute obeys Henry’s law (see Example 7.15).

7.4.1 Ideal Behaviour of Real Solutions The ideal behaviour exhibited by non-ideal solutions can be summarised by the following mathematical statements.

7.4.2 Henry’s Law and Gas Solubility Since the solubility of the gases in liquids is usually very low, the mole fraction of a gas in a saturated liquid solution is very small. The solute gas obeys Henry’s law and therefore its fugacity (or the partial pressure) would be directly proportional to its mole fraction, the proportionality constant being the Henry’s law constant [Eq. (7.69)]. In other words, the mole fraction or the solubility of the gas in the liquid is proportional to the partial pressure of the gas over the liquid as given by

where Ki is the Henry’s law constant. EXAMPLE 7.10 The Henry’s law constant for oxygen in water at 298 K is 4.4 104 bar. Estimate the solubility of oxygen in water at 298 K for a partial pressure of oxygen at 0.25 bar. Solution Equation (7.72) gives the solubility of a gas in liquid in terms of its mole fraction. Substituting the values Ki = 4.4 104 bar, and = 0.25 bar in Eq. (7.72) we get xi = 0.0568 10–4. For very dilute solutions, we can write

Therefore, the solubility of oxygen is 0.0568 be written as 0.0568 10–4 32 1/18 = 0.101

10–4 moles per mole of water. In mass units, it can

10–4 kg oxygen per kg water EXAMPLE 7.11 The partial pressure of acetone (A) and chloroform (B) were measured at 298 K and are reported below: xB

0

0.2

0.4

0.6

0.8

1.0

, bar

0.457

0.355

0.243

0.134

0.049

0

, bar

0

0.046

0.108

0.187

0.288

0.386

(a) Confirm that the mixture conforms to Raoult’s law for the component present in excess and Henry’s law for the minor component. (b) Determine the Henry’s law constants. Solution The partial pressures are plotted against mole fraction xA as shown in Fig. 7.5.

From the data given, it can be seen that = 0.457 and = 0.386 bar. The dotted line representing the ideal behaviour (Raoult’s law) of component A is drawn by joining the origin and (x = 1, p = 0.457) by a straight line. Raoult’s law for component B is also drawn. The dotted lines PA and QB represent the ideal behaviour. The Henry’s law line PR is drawn tangential to the curve versus xA as xA tends to 0 and the line QS is drawn tangential to the versus xA curve as xA tends to 1. (a) We see that the partial pressure curve for component A coincides with the Raoult’s law line in the region where mole fraction of component A approaches unity and in this region, the partial pressure of component B coincides with the Henry’s law line. Thus, in the region where Raoult’s law is obeyed by A, Henry’s law is obeyed by B, and vice versa. (b) The slopes of the Henry’s law line PR gives KA, the Henry’s law constant for A. KA = 0.23 bar. Similarly slope of QS is KB. KB = 0.217 bar.

7.5 ACTIVITY IN SOLUTIONS The activity with reference to pure substance was defined [see Eq. (6.145)] and the concept was discussed in Chapter 6. The activity of a component in a solution can be defined in a similar way. It is the ratio of fugacity of a component in the solution in a given condition to the fugacity of that component in the standard state. It is denoted by ai.

Since the fugacities are related to the chemical potential as

Dmi = mi – is the increase in the chemical potential of species i when it is brought into solution from its standard state. The concept of activity plays an important role in solution thermodynamics because activity can be related to compositions directly. For example, let the standard state for a substance be the pure component at the temperature and pressure of the solution. Then the activity of that component becomes equal to its mole fraction in the case of ideal solutions and is a strong function of mole fraction in the case of real solutions.

For ideal solutions as , the activity ai = xi. For real solutions, the activity can be shown to be equal to the product of activity coefficient and mole fraction. The activity coefficient is discussed later in this chapter. The term activity is a ratio without dimensions. It is a widely used function in solution thermodynamics, particularly in dealing with property changes of mixing. The relationship between property change of mixing and activity is discussed later in this chapter.

7.5.1 Selection of Standard States The numerical values of activity depend upon the choice of the standard state, this choice being based largely on experimental convenience and reproducibility. For all such standard states, the temperature is the same as the temperature of the solution under study and it is not a fixed value. Following are the commonly accepted standard states: Gases. Two standard states are common: 1. The pure component gas in its ideal state at 1 bar. At this state, the fugacity is unity if expressed in bar. The activity becomes

That is, the activity of a component in a mixture of gases is equal to its fugacity, numerically. If the mixture behaves as an ideal gas at the given conditions the activity and partial pressure are the same. This standard state is used in the study of chemical reaction equilibrium. 2. The pure component gas at the pressure of the system. With this choice the activity of each component in ideal gas solution becomes equal to its mole fraction.

This standard state becomes hypothetical at temperatures where the total pressure exceeds the saturation pressure of the component gas in the pure state. Vapour–liquid equilibrium studies conventionally use this standard state. Liquids. Two standard states are common for liquids also. 1. The pure component liquid at a pressure of 1 bar. This state is hypothetical if the vapour pressure of the pure liquid exceeds 1 bar. 2. The pure liquid at the pressure of the system. This state becomes hypothetical at temperatures above the critical or saturation temperature of the pure liquid. This standard state is used in vapour–liquid equilibrium studies. Solids. The standard state chosen for solid is usually the pure component in the solid state at a pressure of 1 bar.

7.6 ACTIVITY COEFFICIENTS We have already seen that the concept of ideal solution enables us to calculate the fugacity of a component in the liquid solution from the knowledge of its concentration in the solution and its fugacity in the pure state. The calculation of fugacity of a component in a real solution should take into account the degree of departure from ideal behaviour. Activity coefficients measure the extent to which the real solution departs from ideality. Activity coefficient of the component i in solution is denoted by gi and is defined by the following relationship.

where

is the fugacity in the standard state. For ideal solutions gi = 1, and we have

which is same as the Lewis–Randall rule [Eq. (7.66)] with the pure liquid at the system pressure as the standard state. Two types of ideal behaviour are observed; the first conforms to Lewis–Randall rule (or Raoult’s law) in which case

, the fugacity of the pure species at the system pressure and the second type

conforms to an ideal dilute solution behaviour (the Henry’s law), in which case , the Henry’s law constant. Depending upon the standard states on which they are based, the activity coefficients can take different numerical values. For the standard state in the sense of Lewis–Randall rule or Raoult’s law,

where ai is the activity of i in the solution. Equation (7.77) is, in fact, Lewis fugacity rule modified by

the factor gi to correct for deviation from ideality. This equation should reduce to Raoult’s law as x approaches unity and to Henry’s law as x approaches zero. For this to be possible, g must equal unity as mole fraction approaches unity (Raoult’s law region) and Ki/fi, as mole fraction, approaches zero (Henry’s law region). In terms of partial pressures, Eq. (7.77) may well be written as

Then the activity coefficient approaches unity as x approaches zero. In Eqs. (7.81) and (7.82), is the activity coefficient referred to infinite dilution. When activity coefficients are defined with reference to an ideal solution in the sense of Raoult’s law, then for each component i, gi 1…as…xi 1 On the other hand, if activity coefficients are defined with reference to an ideal dilute solution, then g1

1…as…x1 1…as…x2

1 (solvent) 0 (solute)

Activity coefficients with reference to ideal dilute solution would be useful when dealing with liquid mixtures that cannot exist over the entire composition range as happens, for example, in a liquid mixture containing gaseous solute. If the critical temperature of the solute is lower than the temperature of the mixture, then a liquid phase cannot exist as x2 1, and the relations based on an ideal mixture in the sense of Raoult’s law can be used only by introducing a hypothetical standard state for component 2. However, relations based on an ideal dilute solution eliminate this difficulty. Activity coefficients are very strong functions of concentration of solution. The variation of g with x over the entire range of composition is usually complex, but can often be roughly approximated in binary solutions by the empirical equations such as the one proposed by Porter:

where b is an empirical constant. These relationships apply best when the components are not too dissimilar in structure and polarity.

7.6.1 Effect of Pressure on Activity Coefficients The effect of pressure on fugacity was derived in Chapter 6 [Eq. (6.126)].

The molar volumes and Vi correspond to the particular phase under consideration. For liquid solutions, the effect of pressure on activity coefficients is negligible at pressures below atmospheric. For gaseous mixtures, activity coefficients are nearly unity at reduced pressures below 0.8.

7.6.2 Effect of Temperature on Activity Coefficients The effect of temperature on fugacity of a pure substance was given by Eq. (6.125) as

Equation (7.88) gives the effect of temperature on activity coefficients. The term ( – Hi) is the partial heat of mixing of component i from its pure state to the solution of given composition both in the same state of aggregation and pressure. For gaseous mixtures, this term is negligible at low pressures. EXAMPLE 7.12 The partial pressures of acetone (A) and chloroform (B) were measured at 298 K and are reported below: xA

0

0.2

0.4

0.6

0.8

1.0

, bar

0

0.049

0.134

0.243

0.355

0.457

, bar

0.386

0.288

0.187

0.108

0.046

0

Calculate the activity and activity coefficient of chloroform in acetone at 298 K, (a) Based on the standard state as per Lewis–Randall rule (b) Based on Henry’s law. Solution The Henry’s law constant was determined in Example 7.11. KB = 0.217 bar. The vapour pressure of pure chloroform, = 0.386 bar. The activity was defined by Eq. (7.73) and activity coefficient by Eq. (7.75). Combining these two we get, ai = gixi Based on the Lewis–Randall rule, the activity,

The activity coefficient based on the Lewis–Randall rule is

The above equations are used to calculate the activity and activity coefficients for different concentrations. A sample calculation is provided below for the second set where xA = 0.2, xB = 0.8,

= 0.049 bar,

= 0.288 bar, KB = 0.217 bar,

= 0.386 bar

The above calculations are repeated for other concentrations and the results are given below: xB

0

0.2

0.4

0.6

0.8

1.0

a

0

0.12

0.28

0.48

0.75

1.0

a

0

0.21

0.50

0.86

1.33

1.78

0.60

0.70

0.80

0.94

1.0

1.05

1.25

1.43

1.66

1.78

g g

1.0

EXAMPLE 7.13 The fugacity of component 1 in binary liquid mixture of components 1 and 2 at 298 K and 20 bar is given by

where is in bar and x1 is the mole fraction of component 1. Determine: (a) The fugacity f1 of pure component 1

(b) The fugacity coefficient f1 (c) The Henry’s law constant K1 (d) The activity coefficient g1. Solution (a) When the mole fraction approaches unity, the fugacity of a component in the solution becomes equal to the fugacity of the pure component. That is,

7.7 GIBBS–DUHEM EQUATIONS In a mixture, the partial molar properties of the components are related to one another by one of the most useful equations in thermodynamics, the Gibbs–Duhem equations. It tells us how the partial molar properties change with compositions at constant temperature and pressure. We have seen that at constant temperature and pressure, the property Mt of the solution is the sum of the partial molar properties of the constituents, each weighted according to the number of moles of the respective constituents.

Dividing throughout by n, the total number of moles in the solution, we get S xi dmi = 0………(7.96) Here xi is the mole fraction of component i in the solution and mi is the chemical potential of the component. Other forms of Gibbs–Duhem equation. Consider a binary solution made up of components 1 and 2 whose mole fractions in the solution are x1 and x2 respectively. Equation (7.96) can be written as x1 dm1 + x2 dm2 = 0………(7.97) where m1 and m2 are the chemical potentials of components 1 and 2 respectively. This can be rearranged as x1 dm1 = – x2 dm2 Dividing by dx1 and noting that dx1 = – dx2 in binary mixtures, the above result gives

Since

, the fugacity in the standard state, is independent of the composition of the solution,

The second terms on both sides of the above equation vanish, as they are equal to unity. Therefore,

As the activity coefficients directly measure the departure from the ideal solution behaviour, Eq. (7.101) is the most useful form of the Gibbs–Duhem equation. The various forms of Gibbs–Duhem equations are rigorous thermodynamic relations that are valid for conditions of constant temperature and pressure. They tell us that the partial molar proper-ties of a mixture cannot change independently; in a binary mixture, if the partial molar property of one of the component increases, the partial molar properties of the other should decrease. Gibbs–Duhem equations find wide applications in solution thermodynamics. These include: (a) In the absence of complete experimental data on the properties of the solution, Gibbs–Duhem

equations may be used to calculate additional properties. For example, if experimental data are available for the activity coefficient of one of the components in a binary solution over certain concentration range, the activity coefficient of the other component over the same composition range can be estimated using Gibbs–Duhem equations. This is particularly useful wherever the volatilities of the two components differ markedly. The measurements usually give the activity coefficient of the more volatile component whereas that of the less volatile component is calculated using Eq. (7.101). Thermodynamic properties of some high-boiling liquids (e.g. polymers) dissolved in a volatile liquid (say, benzene) can be computed by measuring the partial pressure of the latter in the solution. (b) Thermodynamic consistency of experimental data can be tested using Gibbs–Duhem equations. If the data on the partial molar property of each component measured directly in experiments satisfy the Gibbs–Duhem equation, it is likely that they are reliable, but if they do not satisfy the Gibbs–Duhem equation, it is certain that they are incorrect. (c) Gibbs–Duhem equations can be used for the calculation of partial pressure from isothermal total pressure data. Suppose that in an experimental investigation of vapour–liquid equilibrium, the total pressures are measured as a function of composition of one of the phases (usually the liquid phase) and the composition of the other phase is not measured. The Gibbs–Duhem equation facilitates the calculation of the composition of other phase thereby reducing the experimental work considerably. (d) Partial pressure data can be obtained from isobaric boiling point data using Gibbs–Duhem equations. The isobaric T-x data can be easily converted to x-y data. In the sections that follow the application of Gibbs-Duhem equation is illustrated in the derivation of the relationship between Henry’s law and Raoult’s law for a real solution (see Example 7.15), in proving the essential criterion that the vapour and liquid compositions are the same for an azeotropic mixture (see Example 8.11) etc. EXAMPLE 7.14 Show that in a binary solution, if the molar volume of one of the components increases with concentration, the molar volume of the other must decrease. Solution When Eq. (7.94) is written for one mole of the solution with M replaced by V, we get

It means that if is positive, must be negative. That is, if partial molar volume of component 1 increases the partial molar volume of component 2 must decrease. EXAMPLE 7.15 Prove that if Henry’s law is obeyed by component 1 in a binary solution over certain concentration range, Lewis–Randall rule (Raoult’s law) will be obeyed by component 2 over the same concentration range.

Solution Equation (7.99) gives Gibbs–Duhem equations in terms of fugacities

which is the Lewis–Randall rule for component 2. In Fig. 7.6, Henry’s law applies to component 1 in a binary system over the range 0 to . Lewis– Randall rule will be applicable to component 2 over the same composition range.

EXAMPLE 7.16 The activity coefficient of component 1 in a binary solution is given by

where a, b, c are constants independent of concentrations. Obtain an expression for g2 in terms of x1. Solution Using the Gibbs–Duhem equation [Eq. (7.101)], we get

where C is a constant of integration. Integrating and using the boundary condition that when x2 = 1 (or x1 = 0 ), g2 = 1 we get C = 0. Therefore, we get the required expression:

The above example illustrates how the activity coefficient of one of the species in a binary mixture can be evaluated if the activity coefficient of the other is known as an analytical equation in x. Now, suppose that g1 is determined experimentally and is reported as a function of x in a tabular form. How is g2 evaluated? Rearrange Gibbs–Duhem equation, Eq. (7.101) in the form

The integral in Eq. (7.106) is to be evaluated graphically. For this, plot a graph taking x1/x2 along the y-axis and ln g1 on the x-axis. The area under the curve from ln g1 at x1 = 0 to the ln g1 value at the desired concentration x1 will give the integral in Eq. (7.106). The negative of this is the value of ln g2 at x1.

7.8 PROPERTY CHANGES OF MIXING We know that the molar volume of an ideal solution is simply the average of the molar volumes of the pure components, each weighted according to its mole fraction. That is, V = S xiVi for ideal solutions. If such a relation could be written for all extensive thermodynamic properties of a solution, then M = S xiMi………(7.107) where M is the molar property of the solution, Mi and xi are the molar property of pure i and its mole fraction respectively. But Eq. (7.107) is not true even for ideal solutions when the property under consideration is entropy or entropy related functions like free energy. For non-ideal solutions, this equation cannot be used for the estimation of thermodynamic properties unless we apply a correction term DM, known as the property change of mixing. Thus, in general, when thermodynamic properties of a solution, whether ideal or real, are evaluated from the pure component properties the equation used should be

M = S xiMi + DM………(7.108) In Eq. (7.108), DM is the difference in the property of the solution M and sum of the properties of the pure components that make it up, all at the same temperature and pressure as the solution. Thus DM = M – S xiMi………(7.109) Replacing M in Eq. (7.108) by the molar volume V, V = S xiVi + DV where DV is the volume change on mixing. DV = 0, for ideal solutions. A more general definition of DM can be written as DM = M – S xi

………(7.110)

where is the molar property of pure i in a specified standard state. If the component exists in the pure form in the same state of aggregation as the solution and at the temperature and pressure as the solution, then = Mi. For example, if all components exist in the pure state as stable liquids at the temperature and pressure of the solution, = Vi and DV = V – S xiVi. Here, DV is the volume change of mixing when one mole of the solution is formed at constant temperature and pressure from the pure liquid constituents. Property change of mixing is a function of temperature and pressure like any other thermodynamic property of solution and its value depends on the standard state specified for the components. Comparison of Eq. (7.14), which relates the properties of the solution to the partial molar properties of the constituent species, with Eq. (7.108) yields DM = S xi

………(7.111)

The quantity can be treated as the change in the property of component i when one mole of pure i in its standard state is brought to the solution of given composition at the same temperature and pressure. Using Eq. (7.111), the volume change of mixing and free energy change of mixing can be written as DV = S xi

………(7.112)

DG = S xi

………(7.113)

7.8.1 Activity and Property Change of Mixing Free energy change of mixing, DG. Using the definition of fugacity, Eq. (6.118), the change in the free energy of a substance when it is brought from its standard state to the solution, can be written as

Volume change of mixing, DV. The partial molar free energy (the chemical potential ) varies with pressure as

are the partial molar volume and molar volume of component i in the standard state respectively. Replacing relations, we get

in Eq. (7.112) in terms of

using the preceding two

Enthalpy change of mixing, DH. The Gibbs–Helmholtz equation [see Eqs. (6.73) and (7.41)] relates the free energy of the substance in the pure state or in the solution to the corresponding enthalpies of the substance as

Substituting Eq. (7.114) into this, we get

Entropy change of mixing, DS. The partial molar entropy and molar entropy of component i in the standard state are related to the free energy of i in the solution and free energy of pure i as given below.

Equations (7.115), (7.117), (7.119) and (7.120) reveal that all property changes of mixing can be written in terms of activity of the components in solution.

7.8.2 Property Changes of Mixing for Ideal Solutions For ideal solutions fugacity is given by Lewis–Randall rule, . With reference to the pure component standard state, the activity is given as ai = /fi = xi. Replacing ai with xi in the results in the previous section and noting that the concentration is independent of pressure and temperature, we have the following results for the property changes of mixing of ideal solutions. DG = RT S xi ln xi………(7.121) DV = 0 DH = 0 DS = – R S xi ln xi………(7.122) Thus, we see that the volume change of mixing, and enthalpy change of mixing of ideal solution are zero. This is true for internal energy change and heat capacity change of mixing as well. But, the free energy change of mixing and entropy change of mixing are not zero. As an ideal gas is a special case of an ideal solution, the above equations are applicable for ideal gas mixtures also.

7.9 HEAT EFFECTS OF MIXING PROCESSES Since the energy of interaction between like molecules is different from that between unlike molecules, the energy of a solution is different from the sum of the energies of its constituents. This difference between the energy of the solution and the energy of the constituents leads to the absorption and evolution of heat during the mixing process. The heat of mixing, DH (or the enthalpy change of mixing), is the enthalpy change when pure species are mixed at constant pressure and temperature to form one mole (or unit mass) of solution. For binary mixtures, DH = H – (x1H1 + x2H2)………(7.123) Knowing the enthalpies of the pure constituents H1 and H2 and the heat of mixing at the given concentration, the enthalpy of the solution can be computed as H = (x1H1 + x2H2) + DH………(7.124) When solids or gases are dissolved in liquids, the accompanying enthalpy change is usually measured as heats of solution, which is defined as the enthalpy change when one mole of the solute dissolves in the liquid. Thus ………(7.125) where DHS is the heat of solution per mole of solute (component 1). When the constituents are all liquids and solutions of all proportions are possible, the heat effect is usually termed as heat of mixing. Figures 7.7 and 7.8 illustrate the two types of presentation of heat of mixing.

In Fig. 7.7, the heat of mixing of ethanol–water is shown from which it is clear that mixing process at lower temperatures and concentration of ethanol is exothermic and at higher temperatures and high

concentrations it is endothermic. In Fig. 7.8, heats of solution of various substances in water are plotted with moles of water per mole of solute as abscissa. Using the heat of mixing at one temperature and heat capacity data of pure species and the solution, the heat of mixing at any temperature can be calculated. The method of calculation is similar to the one employed for the calculation of standard heat of reaction at any temperature from the values at 298 K. EXAMPLE 7.17 The enthalpy change of mixing for a binary liquid solution at 298 K and 1 bar is given by the equation DH = x1x2(40x1 + 20x2), where DH is in J/mol and x1 and x2 are the mole fractions of components 1 and 2 respectively. The enthalpies of the pure liquids at the same temperature and pressure are 400 and 600 J/mol respectively. Determine numerical values of the partial molar enthalpies at infinite dilution

at 298 K and 1 bar.

Solution Refer Eqs. (7.20) and (7.21) and replace V by H.

EXAMPLE 7.18 At 300 K and 1 bar, the volumetric data for a liquid mixture of benzene and cyclohexane are represented by V = 109.4 10–6 – 16.8 10–6x – 2.64 10–6x2, where x is the mole fraction of benzene and V has the units of m3/mol. Determine the expression for volume change of mixing for the standard state based on Lewis–Randall rule. Solution The expressions for the partial molar volumes of benzene (1) and cyclohexane (2) were derived in Example 7.5. They are

EXAMPLE 7.19 A vessel is divided into two compartments. One contains 100 moles nitrogen at 298 K and 1 bar and the other contains 100 moles of oxygen at the same conditions. The barrier separating them is removed and the gases are allowed to reach equilibrium under adiabatic conditions. What is the change in entropy of the contents of the vessel? Solution As the gases are ideal, the temperature and pressure before and after mixing will be the same. For one mole of the mixture the change in entropy on its formation from pure components is

given by Eq. (7.122). DS = – R S xi ln xi Since xi = 0.5, DS = – R ln 0.5 = 8.314

ln 2 = 5.763 J/mol K

For 200 moles of the mixture, DS = 1152.57 J/K EXAMPLE 7.20 The heat of formation of LiCl is – 408.610 kJ/mol at 298 K. The heat of solution for 1 mol LiCl in 12 moles water is –33.614 kJ at 298 K. Calculate the heat of formation of LiCl in 12 moles water at 298 K. Solution The chemical reaction of formation of LiCl and the physical change of dissolution of LiCl in water can be represented by the following:

EXAMPLE 7.21 A container is divided into two compartments. One contains 3.0 moles hydrogen at 298 K and 1.0 bar and the other contains 1.0 mol nitrogen at 298 K and 3.0 bar. Calculate the free energy of mixing when the partition is removed. Solution Assume that the gases behave ideally. The volume occupied by hydrogen is (nRT)/P = 3RT Volume occupied by nitrogen is (nRT)/P = RT/3 Therefore, the total volume occupied after the partition is removed is 3RT + RT/3 = (10/3)RT The final pressure attained by the mixture is

It is assumed that the process is taking place in two steps. In the first, the individual gases are separately brought to the final pressure at constant temperature and in the second, the gases are mixed at constant pressure and temperature. For the first step, the change in free energy is due to change in the pressure and is equal to

EXAMPLE 7.22 Calculate the mean heat capacity of a 20 mole per cent solution of alcohol in water at 298 K, given the following: Heat capacity of water: 4.18 103 J/kg K; Heat capacity of ethanol: 2.18 103 J/kg K; Heat of mixing for 20 mole per cent ethanol–water at 298 K: – 758 J/mol; Heat of mixing for 20% (mole) ethanol–water at 323 K: – 415 J/mol. Assume that the heat capacities of pure liquids are constant between 298 and 523 K. Solution The enthalpy change when 0.8 moles of water and 0.2 moles of ethanol both at 323 K are mixed together is given by the heat of mixing at 323 K which is equal to – 415 J/mol of solution. 0.8 mol water at 323 K + 0.2 mol ethanol at 323 K 1.0 mole 20 per cent ethanol–water; DH = – 415 J/mol. This change can be assumed to be taking place in four steps as detailed below:

Step 1: 0.8 mol water is cooled from 323 K to 298 K. Let DH (1) be the enthalpy of cooling. Then DH(1) = 0.8

18

4.18

(298 – 323) = – 1504.8 J

Step 2: 0.2 mol ethanol is cooled from 323 K to 298 K. Let DH(2) be the enthalpy of cooling. DH(2) = 0.2 46 2.58 (298 – 323) = – 593.4 J Step 3: 0.8 mol water and 0.2 mol ethanol at 298 K are mixed together. Heat of mixing is DH(3) = –758 J/mol Step 4: 20 per cent ethanol–water solution is heated to 323 K. The enthalpy of heating is DH(4) = CPm(323 – 298)

where CPm is the mean specific heat of solution. DH = DH(1) + DH(2) + DH(3) + DH(4) – 415 = – 1504.8 – 593.4 – 758 + CPm (323 – 298) Thus the mean heat capacity of a 20 per cent solution is CPm = 97.65 J/mol K EXAMPLE 7.23 What temperature will be attained when a 20% mole ethanol–water mixture is adiabatically formed from the pure liquids at 298 K? Heat of mixing for 20% mole ethanol–water at 298 K: –758 J/mol. The mean heat capacity of a 20% mole solution of alcohol in water at 298 K: 97.65 J/mol K. Solution The adiabatic mixing process involves no change in enthalpy or DH = 0. It means the temperature of the solution will increase or decrease on mixing depending upon whether heat is absorbed or evolved during the process. Let T be the temperature attained by the solution on mixing. Then its enthalpy above 298 K is CP m (T – 298), where CPm is its mean heat capacity. The temperature T can be evaluated by assuming that the process is occuring in steps as shown in Fig. 7.10. DH = DHS + CPm(T – 298) 0 = – 758 + 97.65(T – 298) Therefore, T = 305.8 K.

7.10 EXCESS PROPERTIES The difference between the property of a real solution and that of an ideal solution is important in chemical thermodynamics, especially in the treatment of phase equilibria. The excess property, ME, is defined as the difference between an actual property and the property that would be calculated for the same temperature, pressure and composition by the equations for an ideal solution. ME = M – Mid………(7.131) M is the molar property of the solution and Mid is the property of an ideal solution under the same conditions. The excess property change of mixing is defined in a similar manner.

DME = DM – DMid………(7.132) DME is the excess property change of mixing, DM and DMid are the property changes of mixing for a real solution and an ideal solution respectively, both under the same conditions. As DM = M – S xi , DMid = Mid – S xi Equation (7.132) can be written as DME = M – Mid………(7.133) Compare Eq. (7.131) with Eq. (7.133). We see that DME = ME………(7.134) Equation (7.134) means that the excess property change of mixing and the excess property are the same. Let us consider the excess volume VE of a solution. VE = DVE = DV – DVid Since ideal solution involves no volume change of mixing, DVid = 0. Therefore, the excess volume of a solution and the volume change of mixing DV are the same. The same is true for some other extensive thermodynamic properties like enthalpy, internal energy, heat capacity, etc. Excess properties in these cases do not represent new thermodynamic properties. However, for entropy and entropy related functions, the excess properties are different from property changes of mixing and they represent new and useful quantities. Excess functions indicate the deviations from ideal solution behaviour and are easily related to activity coefficients. Excess functions may be positive or negative; when the excess Gibbs free energy of a solution is positive the solution is said to exhibit positive deviation from ideality, whereas if it is less than zero, the deviation from ideality is negative. The definition of partial molar excess functions is analogous to that of partial molar thermodynamic properties [see Eq. (7.1)].

Equation (7.136) says that the molar excess property ME of a solution is the average of the partial molar excess property of each component weighted according to its mole fractions.

7.10.1 Excess Gibbs Free Energy For phase equilibrium studies the most useful excess property is the partial molar excess Gibbs free energy which can be directly related to the activity coefficient. Excess Gibbs free energy is defined

as GE = G – Gid………(7.137) Using Eq. (7.136), we can write the excess Gibbs free energy as

In Eq. (7.140) the fugacity is related to xi,, i and

as

, so that Eq. (7.140) becomes

Dmi = RT ln xigi………(7.143) Substituting Eq. (7.142) and Eq. (7.143) into Eq. (7.139), the result is

EXAMPLE 7.24 The two-suffix-Margules equation is the simplest expression for excess Gibbs free

energy that is obeyed by chemically similar materials. GE = Ax1x2………(7.147) where A is an empirical constant independent of composition. Derive the expressions for the activity coefficients that result from this expression. Solution Write Eq. (7.146) for components 1 and 2. Then

SUMMARY A proper understanding of the thermodynamic properties of solutions is essential for the analysis of many chemical engineering problems such as the phase equilibria and chemical reaction equilibria. New concepts were found necessary to deal with the solutions, the concept of partial molar properties being the most important among them. The general definition of partial molar property was given by Eq. (7.1). The partial molar property of a substance in a solution is an intensive property strongly dependent on the concentration. It gives the increase, in the property of the solution resulting from the addition at constant temperature and pressure, of one mole of the substance, to such a large quantity of the system that its composition remains virtually unchanged. The method of tangent intercepts was found to be most suitable for the determination of partial molar properties. The partial molar free energy of a substance was designated as its chemical potential. It is the contribution that the component makes towards the total free energy of the solution. It is a widely used thermodynamic property and serves as an index of chemical equilibrium in the same manner as temperature and pressure are used as indices of thermal and mechanical equilibrium. The concept of fugacity introduced in Chapter 6 was extended to take care of mixtures through Eqs. (7.51) and (7.52). For the evaluation of fugacity of mixtures, data on the variation of the solution volume with composition at different pressures is necessary [Eq. (6.57)]. Since such data are scarce, the fugacity in the solution is to be evaluated by devising an ideal solution model and by measuring the extent to which the real solutions deviate from it. The Lewis–Randall rule allows the fugacity of a component in the mixture to be calculated without any information about the solution except its composition. It states that the fugacity of a component in an ideal solution is directly proportional to the mole fraction of the component in the solution [Eq. (7.65)]. For ideal liquid solutions, if the vapour phase with which it is in equilibrium is assumed to behave as an ideal gas, the Lewis–Randall rule may be simplified to the Raoult’s law [Eq. (7.67)]. The Raoult’s law states that the partial pressure of a component in the vapour phase is directly proportional to the mole fraction of that component in the liquid, which is in equilibrium with the vapour. Even for non-ideal solutions, the fugacity (or the partial pressure) was found to be directly proportional to the mole fraction in the liquid as the mole fraction approaches zero (Henry’s law). Often, the solute portion of a dilute nonideal liquid solution can be assumed to follow the Henry’s law and the solvent portion, the Raoult’s law. The activity of a component in a solution (Section 7.6) can be related to compositions directly and hence plays an important role in solution thermodynamics. Activity coefficients (Section 7.6) measure the extent to which real solutions deviate from ideal behaviour. Two types of ideal behaviour are observed; one conforming to the Lewis–Randall rule and the other conforming to the Henry’s law, thus giving rise to two types of activity coefficients. In Section 7.7, the Gibbs–Duhem equations were developed relating the partial molar properties of the components to one another. They tell us that the partial molar properties of a mixture cannot change independently; in a binary mixture, if the partial molar property of one of the component increases, that of the other should decrease. Various forms of the Gibbs–Duhem equations applicable for binary solutions were also developed. The property changes of mixing were defined (Section 7.8) as the difference in the property of the solution and the sum of the properties of the pure components constituting the solution, all at the same temperature and pressure as the solution. The property changes were presented in terms of the activity of the constituents as shown by Eq. (7.115) for the free energy, Eq. (7.117) for the volume, Eq.

(7.119) for the enthalpy and Eq. (7.120) for the entropy change on mixing. The difference between the energy of the solution and the energy of the constituents leads to absorption and evolution of heat during the mixing process. The heat effects of mixing process were dealt with in detail in Section 7.9. The excess property (Section 7.10) was defined as the difference between an actual property and the property that could be calculated for the same temperature, pressure and composition by the equations for an ideal solution. For phase equilibrium studies the most useful excess property is the partial molar excess Gibbs free energy which can be directly related to the activity coefficient. A simple relationship was shown to exist between activity coefficient and excess chemical potential which makes it possible to express the activity coefficient as a function of the composition.

REVIEW QUESTIONS 1. Distinguish between molar volume and partial molar volume. Does the partial molar volume of a substance vary with the concentration of the substance in the solution? 2. Express the partial molar property as the partial derivative of the total property of the solution. Is it an intensive property or an extensive property? 3. How are the partial molar volumes of the constituents of a binary mixture related to their mole fractions of the constituents and the molar volume of the solution? Explain how these equations are useful for the determination of partial molar volumes by the tangent-intercept method. 4. Define chemical potential. What is its physical significance? 5. Chemical potential can be equated to the partial derivatives of U, A, H or S under certain constraints. However, it cannot be treated as the partial molar internal energy, partial molar enthalpy, etc. Explain. 6. Show that the rate of change of chemical potential of a substance with pressure is equal to its partial molar volume in the solution. 7. What are the characteristics of an ideal solution? What is Lewis–Randall rule? 8. “The concept of an ideal gaseous solution is less restrictive than the concept of an ideal gas mixture.” Explain. 9. State Raoult’s law. Show that it is a simplified form of the Lewis–Randall rule. 10. State Henry’s law and show that the Raoult’s law is a special case of the Henry’s law. 11. Given the Henry’s law constant, how would you determine the solubility of a gas in a liquid? 12. Define activity and show that the activity and mole fraction in an ideal solution are identical. 13. The activities in a gas mixture may be numerically equal to the fugacities or the mole fractions in the mixture. Explain. 14. Define activity coefficient. How do you distinguish between the activity coefficient based on the Lewis–Randall rule and that based on the ideal dilute solution? 15. Do the activity coefficients vary with composition or not? What is the effect of temperature and pressure on the activity coefficient? 16. Discuss the Gibbs–Duhem equation and its various forms. What are the major fields of application of the Gibb’s Duhem equations? 17. What do you mean by property changes of mixing? How are these related to the activities of the

components in the mixture? 18. “All property changes of mixing are zero for ideal solutions”. Do you agree? Explain. 19. Define excess property. Under what circumstance the property change of mixing and the excess properties are identical? 20. How is the activity coefficient related to the excess free energy?

EXERCISES 7.1 Prove the following:

(a)

(b)

(c) 7.2 Discuss the method for the calculation of entropy of solutions. 7.3 Discuss the variable pressure and variable temperature modifications of Gibbs–Duhem equations. 7.4 Derive an expression for partial molar volumes using the following relation for the molar volume of the binary liquid mixture of components 1 and 2. V = x1V1 + x2V2 + x1x2[B + C(x2 – x1)] where x1 and x2 are the mole fractions and V1 and V2 are the molar volumes in the pure state. 7.5 Describe schematically an experimental technique for the determination of volume change and enthalpy change on mixing. 7.6 The activity coefficients in a binary mixture based on the Lewis–Randall rule standard state are given by Derive expressions for activity coefficients based on Henry’s law in terms of composition. 7.7 Show that the Henry’s law constant varies with pressure as

where is the partial molar volume of the solute at infinite dilution. 7.8 The enthalpy of a binary liquid mixture containing components 1 and 2 at 298 K and 1.0 bar is given by H = 400x1 + 600x2 + x1x2(40x1 + 4x2) where H is in J/mol. Determine

(a) Pure component enthalpies (b) Partial molar enthalpies. 7.9 The volume of a mixture of two organic liquids 1 and 2 is given by V = 110.0 – 17x1 – 2.5 where V is the volume in m3/mol at 1.0 bar and 300 K. Find the expressions for . 7.10 If the partial molar volumes of species 1 in a binary liquid solution at constant temperature and pressure is given by derive the equat3ion for . What equation for V is consistent with this? 7.11 The molar enthalpy of a binary mixture is given by H = x1(a1 + b1x1) + x2(a2 + b2x2) Derive an expression for . 7.12 Using the method of tangent intercepts plot the partial molar volume of HNO3 in aqueous solution at 293 K using the following data where w is the mass percentage of HNO3. w

2.162 10.98 20.80 30.00 39.20 51.68 62.64 71.57 82.33 93.4 99.60 10–3, kg/m3 1.01

r

1.06

1.12

1.18

1.24

1.32

1.38

1.42

1.46

1.49 1.51

7.13 On addition of chloroform to acetone at 298 K, the volume of the mixture varies with composition as follows: x

0

V

0.194 0.385 0.559 0.788 0.889 1.000

103, m3/kmol 73.99 75.29 76.50 77.55 79.08 79.82 80.67

where x is the mole fraction of chloroform. Determine the partial molar volumes of the components and plot against x. 7.14 The partial molar volumes of acetone and chloroform in a mixture in which mole fraction of acetone is 0.5307 are 74.166 10–6 m3/mol and 80.235 10–6 m3/mol respectively. What is the volume of 1 kg of the solution? 7.15 The volume of a solution formed from MgSO4 and 1.0 kg of water fits the expression V = 1.00121 10–3 + 34.69 10–6(m – 0.070)2 where m is the molality of the solution. Calculate the partial molar volume of the salt and solvent when m = 0.05 mol/kg. 7.16 Calculate the partial molar volumes of methanol and water in a 40 per cent (mol) methanol solution given the following data at 1 bar and 298 K. (x = mole fraction of methanol) x V

3 3 10 , m /mol

0

0.114

0.197

0.249

0.495

0.692

0.785

0.892

1.0

0.0181

0.0203

0.0219

0.023

0.0283

0.0329

0.0352

0.0379

0.0407

7.17 The standard enthalpy of formation of HCl (in kJ/mol) from the elements at 298 K are given

below: nw 1.0

2.0

3.0

4.0

5.0

6.0

8.0

10.0

50.0

100.0

92.66 119.0 141.67 149.73 156.96 158.81 161.16 162.42 166.22 166.79 205.9

Calculate the partial molar enthalpies of HCl and water in a solution containing 10 kmol HCl/m3 of solution. 7.18 The following table gives the molality and density of aqueous solutions of KCl at 298 K. Determine the partial molar volume of KCl at m = 0.3. m, mol/kg

0.0

0.1668

0.2740

0.3885

0.6840

0.9472

r, kg/m3

997.07

100.49

100.980

101.271

102.797

103.927

7.19 Calculate the concentration of nitrogen in water exposed to air at 298 K and 1 bar if Henry’s law constant for nitrogen in water is 8.68 104 bar at this temperature. Express the result in moles of nitrogen per kg water (Hint: Air is 79 per cent nitrogen by volume). 7.20 The partial pressure of methyl chloride in a mixture varies with its mole fraction at 298 K as detailed below: x

0.0005

0.0009

0.0019

0.0024

, bar

0.27

0.48

0.99

1.24

Estimate the Henry’s law constant of methyl chloride. 7.21 Two moles of hydrogen at 298 K and 2.0 bar and 4.0 moles of nitrogen at 298 K and 3.0 bar are mixed together. What is the free energy change on mixing and what would be the value had the pressures been identical initially? 7.22 Calculate the activity and activity coefficient of acetone based on Lewis–Randall rule and Henry’s law for the data given in Example 7.12. 7.23 The activity coefficient data for a binary solution at fixed temperature and pressure are correlated as Do these equations satisfy Gibbs–Duhem equations? 7.24 In a binary mixture, the activity coefficient g1 of component 1, in the entire range of composition, is given by where R, A and B are constants. Derive expression for the activity coefficient of component 2. 7.25 For a mixture of acetic acid and toluene containing 0.486 mole fraction toluene, the partial pressures of acetic acid and toluene are found to be 0.118 bar and 0.174 bar respectively at 343 K. The vapour pressures of pure components at this temperature are 0.269 bar and 0.181 bar for toluene and acetic acid respectively. The Henry’s law constant for acetic acid is 0.55 bar.

Calculate the activity and activity coefficient for acetic acid in the mixture (a) Based on Lewis–Randall rule (b) Based on Henry’s law. 7.26 Calculate the activity and activity coefficients for toluene for the conditions in Exercise 7.24 assuming pure liquid standard state. 7.27 Partial pressure of ammonia in aqueous solutions at 273 K varies with concentration as: x

0.05

0.10

0.15

0.50

1.00

, bar

0.0179

0.0358

0.062

1.334

4.293

Calculate (a) The activity coefficient of ammonia in 10 mole per cent solution using pure liquid standard state (b) The Henry’s law constant if the system obeys Henry’s law. 7.28 The activity coefficient of n-propyl alcohol in a mixture of water (A) and alcohol (B) at 298 K referred to the pure liquid standard state is given below: xB

0

0.01

0.02

0.05

0.10

0.20

gB

12.5

12.3

11.6

9.92

6.05

3.12

Find gA in the solution containing 10 per cent (mole) n-propyl alcohol. 7.29 The activity coefficient of thallium in amalgams at 293 K are given below. x2

0

0.00326

0.01675

0.04856

0.0986

0.168

0.2701

0.424

g2

1.0

1.042

1.231

1.776

2.811

4.321

6.196

7.707

Determine the activity coefficient of mercury (component 1) at various compositions. 7.30 A vessel is divided into two parts. One part contains 2 mol nitrogen gas at 353 K and 40 bar and the other contains 3 mol argon gas at 423 K and 15 bar. If the gases are allowed to mix adiabatically by removing the partition determine the change in entropy. Assume that the gases are ideal and CV is equal to 5/2 R for nitrogen and 3/2 R for argon. 7.31 A stream of nitrogen flowing at the rate of 7000 kg/h and a stream of hydrogen flowing at the rate of 1500 kg/h mix adiabatically in a steady flow process. If the gases are ideal and are at the same temperature and pressure, what is the rate of entropy increase in kJ/h K as a result of the process? 7.32 The molar volume of a binary liquid mixture is given by 90 10–3x1 + 50 10–3x2 + x1x2(6 10–3x1 + 9 10–3x2) Obtain expressions for

and show that they satisfy Gibbs–Duhem equations.

7.33 Water at a rate of 54 103 kg/h and Cu(NO3)2 ◊ 6H2O at a rate of 64.8 103 kg/h are mixed together in a tank. The solution is then passed through a heat exchanger to bring the temperature to 298 K, same as the temperature of the components before mixing. Determine the rate of heat transfer in the exchanger. The following data are available. Heat of formation at 298

K of Cu(NO3)2 is –302.9 kJ and that of Cu(NO3)2 ◊ 6H2O is –2110.8 kJ. The heat of solution

of Cu(NO3)2 ◊ nH2O at 298 K is – 47.84 kJ per mol salt and is independent of n. 7.34 If pure liquid H2SO4 is added to pure water both at 300 K to form a 20 per cent (weight) solution, what is the final temperature of the solution? The heat of solution of sulphuric acid in water is H2SO4 (21.8 H2O) = –70 103 kJ/kmol of sulphuric acid. Standard heat of formation of water = – 286 kJ/mol. Mean heat capacity of sulphuric acid may be taken from Chemical Engineer’s Handbook. 7.35 LiCl H2O (c) is dissolved isothermally in enough water to form a solution containing 5 mol of water per mole of LiCl. What is the heat effect? The following enthalpies of formation are given: LiCl (c) = – 409.05 kJ, LiCl◊H2O (c) = – 713.054 kJ LiCl (5H2O) = – 437.232 kJ, H2O (l) = – 286.03 kJ 7.36 Calculate the heat effects when 1.0 kmol of water is added to a solution containing 1.0 kmol sulphuric acid and 3.0 kmol of water. The process is isothermal and occurs at 298 K. Data: Heat of mixing for H2SO4 (3H2O) = – 49,000 kJ per kmol H2SO4. Heat of mixing for H2SO4 (4H2O) = – 54,100 kJ per kmol H2SO4. 7.37 A single effect evaporator is used to concentrate a 15% (weight) solution of LiCl in water to 40%. The feed enters the evaporator at 298 K at the rate of 2 kg/s. The normal boiling point of a 40% LiCl solution is 405 K and its specific heat is 2.72 kJ/kg K. For what heat transfer rate in kJ/h, should the evaporator be designed? The heat of solution of LiCl in water per mole of LiCl at 298 K are: DHS for LiCl(13.35 H2O) = – 33.8 kJ, for LiCl (3.53 H2O) = – 23.26 kJ. Enthalpy of superheated steam at 405 K, 1 bar = 2740.3 kJ/kg. Enthalpy of water at 298 K = 104.8 kJ/kg. Molecular weight of LiCl = 42.39. 7.38 The excess Gibbs free energy of solutions of methyl cyclohexane (MCH) and tetrahydrofuran (THF) at 303 K are correlated as GE = RTx(1 – x)[0.4587 – 0.1077(2x – 1) + 0.0191(2x – 1)2] where x is the mole fraction of methyl cyclohexane. Calculate the Gibbs free energy change on mixing when 1 mol MCH and 3 mol THF are mixed. 7.39 Derive the relation between the excess Gibbs free energy of a solution based on the Lewis– Randall rule and that based on the asymmetric treatment (Lewis–Randall rule for solvent and Henry’s law for solute) of solution ideality. 7.40 The excess enthalpy of a solution is given by HE = x1x2(40x1 + 20x2) J/mol Determine expressions for as functions of x1. 7.41 Given that ME = x1x2[A + B(x1 – x2) + C(x1 – x2)2]

derive expressions for . What are the limiting values for ME/x1x2 as x1 0 and x1 1? 7.42 The excess Gibbs free energy is given by GE/RT = – 3x1x2(0.4x1 + 0.5x2)

and

Find expressions for ln g1 and ln g2. 7.43 Do the following equations satisfy Gibbs–Duhem equations? Find expressions for GE/RT. 7.44 The excess volume (m3/kmol) of a binary liquid mixture is given by VE = 0.01 x1x2(3x1 + 5x2) at 298 K and 1 bar. Determine for an equimolar mixture of components 1 and 2 given that V1 = 0.12 m3/kmol and V2 = 0.15 m3/kmol.

8 Phase Equilibria A system is said to be in a state of equilibrium if it shows no tendency to depart from that state either by energy transfer through the mechanism of heat and work or by mass transfer across the phase boundary. Since a change of state is caused by a driving force, we can describe a system at equilibrium as one in which there are no driving forces for energy or mass transfer. That is, for a system in a state of equilibrium, all forces are in exact balance. It may be noted here that the state of equilibrium is different from a steady state condition. Under steady state there exist net fluxes for material or energy transfer across a plane surface placed anywhere in the system. Under equilibrium the net flux is zero. Transfer of material or energy across phase boundaries occurs till equilibrium is established between the phases. In our daily experience, we come across a number of processes in which materials are transferred from one phase to another. During breathing we take oxygen from the air through the lungs and dissolve it in the blood. During the preparation of tea or coffee we extract the soluble components in the powder into boiling water. Dilute aqueous solution of alcohol is concentrated by distillation in which a vapour rich in alcohol is produced from the boiling solution. The phase equilibrium thermodynamics is of fundamental importance in many branches of science, whether physical or biological. It is particularly important in chemical engineering, because majority of manufacturing processes involve transfer of mass between phases either during the preparation of the raw materials or during the purification of the finished products. Gas–liquid absorption, distillation, liquid–liquid extraction, leaching, adsorption, etc., are some of the important separation techniques employing mass transfer between phases. In addition to these, many industrial chemical reactions are carried out under conditions where more than one phase exist. A good foundation in phase equilibrium thermodynamics is essential for the analysis and design of these processes. In this chapter due emphasis is given to the development of the relationship between the various properties of the system such as pressure, temperature and composition when a state of equilibrium is attained between the various phases constituting the system. The temperature–pressure-composition relationships in multiphase system at equilibrium form the basis for the quantitative treatment of all separation processes. The two types of phase equilibrium problems that are frequently encountered are: 1. The determination of composition of phases which exist in equilibrium at a known temperature and pressure 2. The determination of conditions of temperature and pressure required to obtain equilibrium between phases of specified compositions. The present chapter tries to provide solutions to these problems.

8.1 CRITERIA OF PHASE EQUILIBRIUM Consider a homogeneous closed system in a state of internal equilibrium. The criteria of internal

thermal and mechanical equilibrium are that the temperature and pressure be uniform throughout the system. For a system to be in thermodynamic equilibrium, additional criteria are to be satisfied. Consider a closed system consisting of two phases of a binary solution, for example, the vapour and liquid phases of an alcohol–water solution. The requirement of uniformity of temperature and pressure does not preclude the possibility of transfer of mass between the phases. If the system is in thermodynamic equilibrium, mass transfer also should not occur. It means that additional criteria are necessary for establishing the state of thermodynamic equilibrium. A system can interact with the surroundings reversibly or irreversibly. In the reversible process, a state of equilibrium is maintained throughout the process. So it can be treated as a process connecting a series of equilibrium states. The driving forces are only infinitesimal in magnitude and the process can be reversed by an infinitesimal change (either increase or decrease) in the potential for the system or the surroundings. The irreversible process, in contrast, occurs with a finite driving force, and it can not be reversed by infinitesimal changes in the external conditions. However, all irreversible processes tend towards a state of equilibrium. We have shown in Chapter 4 under ‘Clausius inequality’,

In this equation, the equality sign refers to a reversible process which can be treated as a succession of equilibrium states and the inequality refers to the entropy change for a spontaneous process whose ultimate result would be an equilibrium state. The first law of thermodynamics expressed mathematically by Eq. (2.5) can be rewritten as dQ = dU + dW………(8.1) Substituting Eq. (8.1) into Eq. (4.44), we get T dS ≥ dU + dW dU T dS – dW………(8.2) dW in Eq. (8.2) may be replaced by P dV so that dU T dS – P dV………(8.3) Equation (8.3) is valid for cases where external pressure is the only force and the work is, therefore, the work of expansion only. By this, we exclude other effects like those due to gravitational and electromagnetic fields and surface and tensile forces. Equation (8.3) can be treated as the combined statement of the first and second law of thermodynamics applied to a closed system which interact with its surroundings through heat transfer and work of volume displacement. This equation is utilised for deriving the criteria of equilibrium under various sets of constraints, each set corresponding to a physically realistic or commonly encountered situation. These different criteria are discussed now. Constant U and V. An isolated system does not exchange mass, heat or work with the surroundings. In Eq. (8.1), d Q = 0, d W = 0 and hence d U = 0. A well-insulated vessel of constant volume would closely approximate this behaviour. Thus in Eq. (8.3) dU = 0 and dV = 0 so that

The entropy is constant in a reversible process and increases in a spontaneous process occurring in a

system of constant U and V. Since an irreversible process leads the system to an equilibrium state, the entropy is maximum at equilibrium when no further spontaneous processes are possible. Constant T and V. Helmholtz free energy is defined by Eq. (6.1). A = U – TS Rearranging Eq. (6.1), we get U = A + TS dU = dA + T dS + S dT Substitute this result in Eq. (8.3) and rearrange the resulting expression to the following form dA – P dV – S dT………(8.5) Under the restriction of constant temperature and volume, the latter implying no work, the equation simplifies to

Equation (8.6) means that the spontaneous process occurring at constant temperature and volume is accompanied by a decrease in the work function and consequently, in a state of thermodynamic equilibrium under these conditions the Helmholtz free energy or the work function is a minimum. Constant P and T. Equation (6.6) defines Gibbs free energy as G = H – TS Since H = U + PV we can write Eq. (6.6) as G = U + PV – TS Taking the differentials dG = dU + P dV + V dP – T dS – S dT rearranging these as dU = dG – P dV – V dP + T dS + S dT and combining this result with Eq. (8.3), we obtain dG V dP – S dT………(8.7) At constant temperature and pressure, Eq. (8.7) reduces to

Equation (8.8) means that the free energy either decreases or remains unaltered depending upon whether the process is spontaneous or reversible. It implies that for a system in equilibrium at a given temperature and pressure the free energy must be minimum. Since most chemical reactions and many physical changes are carried out under conditions of constant temperature and pressure, Eq. (8.8) is the commonly used criterion of thermodynamic equilibrium. It also provides a very convenient and simple test for the feasibility of a proposed process. No process is possible which results in an increase in the Gibbs free energy of the system, because according to Eq. (8.8) the Gibbs free energy always decreases in a spontaneous process and in the limit of the reversible process, the free energy doesn’t change at all.

In the equilibrium state, differential variations can occur in the system at constant temperature and pressure without producing any change in the Gibbs function. Thus, the equality in Eq. (8.8) can be used as the general criterion of equilibrium or as a thermodynamic statement that characterises the equilibrium state. dG = 0 (at constant T and P)………(8.9) To apply this criterion for phase equilibrium problems we need formulate an expression for dG as function of the number of moles of the components in various phases and set it equal to zero. This equation along with the mass conservation equations provides the solutions to phase equilibrium problems.

8.2 CRITERION OF STABILITY It can be shown that the criterion of equilibrium [Eq. (8.8)] can be used to formulate the criterion of stability for a binary mixture. When two pure liquids at a given temperature T and pressure P are mixed together, the resulting mixture should have a lower free energy at the same temperature and pressure. This is because the mixed state is an equilibrium state or stable state compared to the unmixed state. The molar free energy of the mixture is thus less than the sum of the molar free energies of the constituents for all possible concentrations. That is, G – S xiGi < 0………(8.10) The left-hand side in the above equation is the free energy change on mixing DG. Therefore, DG < 0………(8.11) When the free energy change on mixing DG is plotted against x1 —the mole fraction of constituent 1 in the binary mixture—the resulting curve is one of the two types shown in Fig. 8.1. The upper curve is for a binary mixture, which is miscible for the entire concentration range. Assume that the points A and B represent two binary mixtures of composition xA and xB respectively. Points on the dotted line AB represent the composition as well as DG of the mixture of two phases obtained when solutions represented by the points A and B are mixed together. Since the line AB is above the solid curve that represents the free energy of the miscible solution, the free energy of the mixture in the miscible state is the minimum and the mixture exists as a single homogeneous phase. However, this argument is not valid for the lower curve in Fig. 8.1. The dotted line MN represents the free energy of the two-phase mixture obtained when two binary mixtures of composition xM and xN, respectively, are mixed together. It lies below the DG curve of the homogeneous solution. Any point on the line MN represents the DG that would result for systems consisting of two phases of mole fraction xM and xN.. Thus, when the system moves from the solid curve to the dotted line MN, there is a decrease in the free energy. That is, the system attains stability when it moves from a homogeneous to a heterogeneous state. Therefore, for mixtures of composition between points M and N, the equilibrium or stable state consists of two immiscible phases. We see that the second derivative of DG with respect to x1 is always positive for stable liquid phase and if it becomes zero or negative, phase separation occurs. The criterion of stability is that at constant temperature and pressure the free energy change on mixing DG, its first and second derivatives are all continuous functions of the concentration x and

EXAMPLE 8.1 Show that for a stable liquid phase, the fugacity of each component in a binary mixture always increases with increase in concentration at constant temperature and pressure. Solution The excess free energy of mixing was defined in Chapter 7. It was shown there that DGE = DG – DGid DGE = GE = RT S xi ln gi DGid = RT S xi ln xi Combining these three equations we find that DG = RT S xi ln (gixi) = S xi ln (gixi) = x1 ln (g1x1) + x2 ln (g2x2) Differentiating this with respect to x1,

Differentiating Eq. (8.14) again with respect to x1 and noting that fugacities of pure components are independent of concentration, we get

Equation (8.12) reveals that the left-hand side of the above equation is greater than zero. Therefore,

The last two equations imply that fugacity of components in a stable solution always increase with increase in concentration.

8.3 PHASE EQUILIBRIA IN SINGLE-COMPONENT SYSTEMS Consider the thermodynamic equilibrium in a system consisting of two or more phases of a single substance. Though the individual phases can exchange mass with each other and are therefore open, the system as a whole is closed. As an example, we can treat the equilibrium between vapour and liquid phases of a single substance at a constant temperature and pressure. Applying the criterion of equilibrium [see Eqs. (8.8) and (8.9)] to this closed system, dG = 0

dGa + dGb = 0………(8.17) where the dGa and dGb are the changes in free energies of the phases ‘a’ and ‘b’ respectively. Since each phase is open, the change in its free energy may be due to the changes in temperature, pressure and the number of moles of the components that constitute the phase. Equation (7.35) expresses this mathematically as dG = V dP – S dT +

dni

Applying this equation to the phases ‘a’ and ‘b’, we can write dGa = Va dPa – Sa dTa + Ga dna, dGb = Vb dPb – Sb dTb + Gb dnb At constant temperature and pressure, dGa = Ga dna, dGb = Gb dnb………(8.18) As the system as a whole is closed, dna + dnb = 0, or dna = – dnb ………(8.19) Substituting Eqs. (8.18) and (8.19) into Eq. (8.17), we get (Ga – Gb)dna = 0………(8.20) Equation (8.20) means that Ga = Gb………(8.21) Whenever two phases of the same substance are in equilibrium under a given temperature and pressure, the molar free energy is the same in each phase. We can verify the above result easily by considering the example of boiling water. As long as both phases are present, an appreciable transfer of material from one phase to the other at constant temperature and pressure would not disturb the equilibrium. The change in the free energy for the equilibrium process (or reversible process) of evaporating a mole of liquid water is DG = DH – TDS………(8.22) As pressure is constant, DH = Q, and the process being reversible, Q = TDS. Equation (8.22) gives DG = 0 For vaporisation of 1 mol of liquid, DG = GV – GL, where GL and GV are the molar free energy of water in the liquid and vapour states at the given T and P. Therefore, under equilibrium GV = GL………(8.23) As the molar free energies are related to the fugacity of the substance by G = RT ln f + C Equation (8.21) can be expressed in terms of fugacity of the phases. fa = fb………(8.24) where fa and fb are the fugacities in phases a and b respectively. It is convenient to work with fugacities of substances as these have absolute values in contrast to free energies, which are usually

expressed as differences. The above conclusions can be extended to three phases, which is the maximum number of phases that can coexist under equilibrium in a system of one component. EXAMPLE 8.2 Using the criterion of phase equilibrium, show that the change in entropy during phase changes can be calculated from the latent heat of phase change and the absolute temperature as DS = DH/T. Solution Suppose that two phases a and b are in equilibrium. Using the definition of free energy [Eq. (6.6)], Ga = Ha – TSa, Gb = Hb – TSb Here, H and S denote the enthalpy and entropy of the substance. Substituting these results in Eq. (8.21), Ha – TSa = Hb – TSb This equation can be rearranged as,

The left-hand side of the above equation is the entropy change accompanying the phase change of one mole of the substance (DS), and the numerator on the right-hand side represents the enthalpy change for the phase change of one mole of the substance or the latent heat of phase change (DH). That is,

EXAMPLE 8.3 Deduce the Clapeyron equation using the criterion of equilibrium, Eq. (8.9). Solution In Chapter 6 we have derived the Clapeyron equation, Eq. (6.25), using Maxwell’s relations.

The criterion of equilibrium provides an alternate route for its derivation. Consider any two phases a and b of the same substance under equilibrium. Since Ga and Gb are both functions of temperature and pressure, and these functional relationships are different for different phases, the two phases can coexist only at such values of the temperature and pressure that Ga = Gb. If the temperature and pressure are altered infinitesimally without disturbing the equilibrium, the change in the free energy must be the same in each phase. dGa = dGb………(8.25) In a phase change there is no work other than the work of expansion, so that dG = V dP – S dT Using this in Eq. (8.25), Va dP – Sa dT = Vb dP – Sb dT………(8.26)

V and S are the molar volume and molar entropy of the fluid with the superscript representing the phase for which the properties correspond to. Equation (8.26) can be rearranged to the following form.

This relation gives the increase in pressure that is necessary to maintain the equilibrium between phases for a pure substance when the temperature is increased. By using the following simplifications Eq. (8.29) can be modified to yield the Clausius–Clapeyron equation applicable for vapour–liquid equilibria. 1. The latent heat of vaporisation is constant and independent of temperature. 2. The molar volume of liquid is negligible compared to that of vapour. 3. The vapour behaves as ideal gas. The Clausius–Clapeyron equation was derived [Eq. (6.28)] in Chapter 6 and is reproduced below.

where

and

are the vapour pressures at temperatures T1 and T2 respectively.

8.4 PHASE EQUILIBRIA IN MULTICOMPONENT SYSTEMS The criterion of equal molar free energy [Eq. (8.21)] is applicable for equilibrium between phases of a single component. This criterion needs modification when dealing with heterogeneous multicomponent systems. A heterogeneous closed system is made up of two or more phases with each phase behaving as open system within the overall closed system. Because each phase consists of two or more components in different proportions, it is necessary that the criterion of multicomponent phase equilibrium be developed in terms of partial molar free energies or the chemical potentials of the components. The criteria of thermal and mechanical equilibrium are, as discussed earlier, the uniformity of temperature and pressure. For the system to be in equilibrium with respect to mass transfer, the driving force for mass transfer—the chemical potential—must have uniform values for

each component in all phases. This criterion of internal equilibrium is derived in the following paragraphs. Consider a heterogeneous system consisting of p phases indicated by the letters, a, b, g, ..., p. The various components that constitute the system are 1, 2, 3, ..., C. The symbol denotes the chemical potential of component ‘i’ in phase ‘k’. Suppose that small amounts of various components are transferred from one phase to another, the system being in equilibrium and the temperature and pressure kept constant. Since the system as a whole is closed, the proposed transfer should satisfy the following criterion. dG = 0 (at constant T and P)………(8.9) The free energy change in a multicomponent system is given by Eq. (7.35) as dG = V dP – S dT + S mi dni At constant temperature and pressure, the above equation becomes dG = S mi dni Substitute this into Eq. (8.9) to get S mi dni = 0………(8.31) Let denote the increase in the number of moles of component i in phase k. Equation (8.31) may then be written as

The variations in the number of moles dni are independent of each other. However, they are subject to the constraints imposed by Eq. (8.33). For all possible variations This is possible only if

, Eq. (8.32) is to be satisfied.

Equation (8.34) means that when a system consisting of several components distributed between various phases is in thermodynamic equilibrium at a definite temperature and pressure, the chemical potential of each component is the same in all the phases. If they are different, the component for which such a difference exists will show a tendency to pass from the region of higher to the region of lower chemical potential. Thus the equality of chemical potential along with the requirement of uniformity of temperature and pressure serves as the general criterion of thermodynamic equilibrium in a closed heterogeneous multicomponent system. In short, we can write

where C is a constant, an alternative and equally general criterion of equilibrium can be written in terms of fugacities as

Fugacity is a more useful property than chemical potential for defining equilibrium since it can be expressed in absolute values, whereas chemical potential can be expressed only relative to some arbitrary reference state. Equation (8.36) is therefore widely used for the solution of phase equilibrium problems. EXAMPLE 8.4 Using the criterion of phase equilibrium show that the osmotic pressure over an ideal solution can be evaluated as

where xA is the mole fraction of solute and VB is the molar volume of the solvent. Solution Consider a vessel which is divided into two compartments by a semi-permeable membrane. Pure solvent (say, water) is taken in one of the compartments and a solution (say, sucrose in water) is taken in the other. Let T be the temperatures on both sides of the membrane and P be the pressure. While the membrane is impermeable to the flow of the solute, it permits the flow of solvent into the solution. This phenomenon of a solvent diffusing through a membrane which is permeable to it, but is

impermeable to the solute, is known as osmosis. Osmosis is caused by the difference in the chemical potentials of the solvent on the two sides of the membrane. At a given pressure, the chemical potential of a pure solvent is greater than that of the solvent in the solution. By increasing the pressure at the solution side of the membrane, the chemical potential of the solvent in the solution can be increased. When the pressure is increased to P keeping the temperature constant, the chemical potential of the solvent in the solution would become equal to that of the pure solvent at pressure P, and the diffusion would stop. If the pressure is increased above P , the direction of diffusion would be reversed. In that event, the solvent would diffuse from the solution to the pure solvent. This process is known as reverse osmosis. The excess pressure P – P to be applied over the solution at constant temperature to arrest the process of osmosis is known as the osmotic pressure. Thus, osmotic pressure is Posmotic = P – P Let the mole fraction of the solutes constituting the solution be represented by xA and the mole fraction of the solvent be represented by xB. Let denotes the chemical potential of the solvent at pressure P . Equation (7.51) relates the chemical potential of a component in a solution to its fugacity. Thus

Combining the preceding two equations, we get

In this equation, is the chemical potential of pure solvent at pressure P , fB is its fugacity and is its fugacity in the solution. Since the solution is ideal, the above equation may be simplified utilising the Lewis–Randall rule which relates fB and that result

for

the

chemical

potential

of

the

solvent

. Now we get the following in the

solution at

pressure P .

Since volume of a liquid is not affected by change in pressure, the integral in this equation can be easily determined in terms of the molar volume VB. Thus VB (P – P) = – RT ln xB Noting the definition of the osmotic pressure, the preceding equation may be written as

8.5 PHASE RULE FOR NON-REACTING SYSTEMS The essence of a phase equilibrium problem is to express quantitatively the relationship between the variables that describe the state of equilibrium of two or more homogeneous phases, which are free to interchange energy and matter. For a homogeneous phase at equilibrium the intensive properties are the same everywhere. In phase equilibrium studies, the intensive properties of interest are pressure, temperature, density and composition. The phase rule allows us to determine the number of degrees of freedom, denoted by F, which are the number of independent variables that must be arbitrarily fixed so as to establish uniquely the intensive state of the system. The phase rule was developed by J. Willard Gibbs (1875). The commonly used phase rule variables are the temperature, pressure and composition. By specifying these, the intensive state of the system at equilibrium is uniquely established. However, all these variables may not be independent. The phase rule gives the number of variables from this set, which when arbitrarily specified, would establish the remaining variables at fixed values. Thus, by degrees

of freedom, we mean the number of intensive properties that can be varied without changing the number of phases or the number of components in any phase. Consider a homogeneous phase of a pure substance like water contained in a beaker, or carbon dioxide gas confined in a cylinder. The state of the system is determined whenever two properties are set at definite values. Here, the number of degrees of freedom is two. This number is not necessarily two for more complex systems. For example, a mixture of steam and liquid water in equilibrium at 101.3 kPa can exist only at 373.15 K. It is impossible to change the temperature without also changing the pressure, if the liquid and vapour are to continue to exist in equilibrium. The number of degrees of freedom in this case is only one, for we can exercise independent control over only one variable, either the temperature or the pressure, and not both. It is evident that there is a connection between the number of degrees of freedom (F), the number of phases in equilibrium (p), and the number of components constituting the system (C). The phase rule gives the relationship between these three quantities. It states that F = C – p + 2………(8.37) The phase rule can be derived using the equilibrium criteria Eq. (8.35). Consider a heterogeneous system consisting of C components distributed between p phases. The composition of a phase containing C components is determined by specifying C – 1 concentration terms. This is because if the mole fractions of all but one of the components in a mixture are known, the mole fraction of the remaining component is the difference between one and the sum of the C – 1 concentration values. Since there are p phases, the number of concentration variables required to define the state of the system is p(C – 1). Recognising that the temperature and pressure are also to be specified, Total number of variables = p(C – 1) + 2………(8.38) If the system under consideration were not in internal equilibrium, the number of degrees of freedom would have been that given by Eq. (8.38). The fact that the system is in thermodynamic equilibrium implies that the system obeys the principle of equality of chemical potential [Eq. (8.35)]. Therefore, all concentrations are not independent. The concentrations of a component in the different phases are interrelated such that its chemical potential is the same in all phases. For any component there exist (p – 1) equations connecting the concentration in the p phases. Since there are C components, there are C(p – 1) independent equations which automatically fixes C(p – 1) of the possible variables in Eq. (8.38). Therefore, Number of variables that are not independent = C(p – 1)………(8.39) The number of degrees of freedom F, which is equal to the number of variables that are independent, and hence must be arbitrarily fixed to define the system completely is obtained as the difference between Eqs. (8.38) and Eq. (8.39). F = p(C – 1) + 2 – C(p – 1) F = C – p + 2………(8.40) Equation (8.40) is the phase rule derived by Gibbs. It may be noted that if a particular component is absent from any phase, the number of composition variables is reduced correspondingly. At the same time, there will be a similar decrease in the number of independent equations. The result is that Eq. (8.40) is valid even if all the components are not present in every phase.

The number C, which stands for the number of components, to be used in Eq. (8.40) is the least number of independent chemical compounds from which the system can be created. This becomes important in dealing with systems involving chemical reactions. For example, if a system consisting of methanol, hydrogen and carbon monoxide is at equilibrium, the number of components is only one as the chemical reaction CO + 2H2 CH3OH occurs in the system. The phase rule for the reacting system is discussed in Chapter 9. As an example of the application of the phase rule, consider the homogeneous system of a single substance. Here, C = 1, p = 1, and therefore, F = 2 (bivariant). Two variables, say, temperature and pressure are to be specified to define the state of the system. Consider a two-phase system, for example, pure liquid water in equilibrium with its vapour. Here, C = 1, p = 2 and F = 1 (univariant). One intensive property, either pressure or temperature but not both, may be varied freely. If, solid is also present in the system, p = 3 and F becomes zero (invariant). In this case, no intensive property can be varied, and the specification that three phases are in equilibrium, fixes the temperature and pressure. The triple point of water is at 273.16 K and 618 Pa and that of carbon dioxide is at 216.55 K and 523.9 kPa. For a binary liquid mixture of alcohol and water in equilibrium with its vapour, C = 2, p = 2 and F = 2. Two intensive properties may be freely varied. For example, both temperature and pressure may be varied freely over restricted ranges in composition of the phases, but all three variables cannot be independently changed. If the composition of the liquid and the pressure are specified, the boiling temperature and the equilibrium vapour composition are automatically fixed. A liquid mixture of benzene and toluene containing 39% benzene is in equilibrium with its vapour at 100 kPa. This determines the equilibrium temperature as 368 K and the composition of the vapour in equilibrium as 61.2% benzene.

8.6 DUHEM’S THEOREM The state of a system is completely determined when both the intensive as well as the extensive state of the system is fixed. The phase rule gives the number of independent intensive variables to be specified to define the intensive state of the system uniquely. It does not deal with the extensive state of the system. By extensive state we mean the amount (or the mass) of the various phases present and the total properties of the system. The Duhem’s theorem helps in establishing the extensive state of the system. It states that for any closed system formed initially from the given masses of prescribed chemical species, the equilibrium state is completely determined when any two independent variables are fixed. In addition to the phase rule variables, the mass of each phase should be known to define the state of the system completely. Therefore, the total number of variables = p(C – 1) + 2 + p, where the last term is the number of terms representing the mass of various phases. Since the system is closed, and is formed from specified amounts of the chemical species present, a material balance equation can be written for each of the C components present in the system. The total number of equations connecting the variables will thus be more than the corresponding number in the phase rule by C, the number of material balance equations. This is equal to C(p – 1) + C. The difference between the number of variables and the number of equations is therefore two indicating that only two independent variables need be fixed to define the state of the system completely. These may be either intensive or extensive. If F < 2, at least one extensive variable must be fixed for complete determination of the system. For

example, for water and water vapour in equilibrium, since the number of degrees of freedom is one, the intensive state of the system is specified by fixing either the pressure or the temperature. But the total properties can be evaluated only if the amount of liquid or vapour is also specified. However, for a binary mixture, say, water and alcohol, in vapour–liquid equilibrium, since the number of degrees of freedom is two, no additional specifications are needed to predict the amount of liquid and vapour present in equilibrium, provided, we know the amounts of the components from which the system is formed. Duhem’s theorem is applicable to reacting systems as well (see Chapter 9).

8.7 VAPOUR–LIQUID EQUILIBRIA The vapour-liquid equilibrium (VLE) data are essential for many engineering calculations, especially in the design and analysis of separation operations such as distillation, absorption, etc. Thermodynamics provides a system of equations relating the necessary experimental data and the unknown vapour-liquid equilibrium compositions, temperature and pressure. The conditions of equilibrium [Eq. (8.36)] require that the fugacity of a component in the liquid phase be equal to that in the vapour phase. That is,

where, represents the fugacity of component i in the solution and the superscripts V and L represent the vapour and liquid phases, respectively. Using this equation, the problem of determining the composition of the liquid and vapour phases in equilibrium is quite simple: it is necessary only to evaluate the compositions so that the fugacity of each component be the same in both phases. For example, for a binary mixture of ethanol and water in vapour-liquid equilibrium, at a definite temperature and pressure, the mole fractions in the liquid and vapour must be such that the fugacity of ethanol is the same in both phases. That is, . Here, is the fugacity of ethanol in the mixture. To evaluate quantitatively the equilibrium compositions, the fugacity of a component should be expressed in terms of its mole fraction in the mixture. Using the definition of activity coefficient, the fugacity of a substance in the vapour phase can be written in terms of its mole fraction yi in the mixture, the fugacity of pure i as a vapour at the system temperature and pressure.

If the stable state for i at T and P is not a vapour, evaluating requires the introduction of a hypothetical state. The use of the concept of fugacity coefficient helps to overcome this difficulty. The fugacity of a component in a gas mixture can be written as

where is the fugacity coefficient of i in the mixture. The fugacity coefficient may be evaluated from an equation of state for the mixture. For the liquid phase, the fugacity of a component can be expressed as the product of its mole fraction xi in the solution, the activity coefficient gLi and the fugacity of the component in the standard state.

Equations (8.45) and (8.46) are the fundamental relationships for estimating the vapour–liquid equilibrium by two different approaches—the equations of state approach and the activity coefficient model approach. Equation (8.45) forms the basis of estimating VLE by the equation of state approach, which will be discussed under Section 8.12.1. Equation (8.46) is the fundamental relationship in the study of vapour–liquid equilibrium based on activity coefficient model for the liquid phase fugacity. The liquid-phase activity coefficients in Eq. (8.46) can be estimated by any of the models described in the following sections. Estimation of the vapour–liquid equilibria using activity coefficients is useful when polar components are present in the system. For most substances at low pressures,

can be assumed to be unity. If the pressure is

very high, must be evaluated using an equation of state. Using the activity coefficient approach, the vapour–liquid equilibrium problems may be attacked, by dividing them into the following grouping for convenience. Case 1: Ideal gas-phase, ideal liquid solution. For mixtures of ideal gases, liquid solutions, at low pressures, gLi = 1, and the fugacity pure liquid

= 1. For ideal

is equal to the saturation pressure of

at the temperature of interest. Equation (8.46) becomes

Case 2: Low-pressure VLE problems. If the liquid phase is not an ideal solution so that g π 1, but the pressure is low enough that the assumption of ideal gas behaviour for the gas phase would not introduce any significant errors in practical calculations, Eq. (8.46) can be simplified as

Case 3: High-pressure VLE problems. In the general case where ideal behaviour cannot be assumed for the gas and liquid phases, the fugacity coefficient and the activity coefficient gLi

should first be determined for solving vapour–liquid equilibrium problems using Eq. (8.46). These are normally complex functions of temperature, pressure and compositions and can be written as

The fugacity in the reference state is the fugacity of pure i at the same T, P and state of aggregation as the mixture. To calculate this, it is convenient to determine first the fugacity of pure i in the liquid state at T under its equilibrium vapour pressure that

and then apply a correction term for the fact

. The fugacity of the liquid under its equilibrium vapour pressure is equal to the vapour

pressure times the fugacity coefficient . The fugacity coefficient is the ratio of the fugacity of the component i under its saturation conditions to the saturation pressure of the substance and tends to unity if the vapour behaves as an ideal gas. Using Eq. (6.31), we can write

The exponential in the above equation is known as the Poynting correction and it is approximately unity when pressure is low. Also at low pressures when the gas behaves ideally, the above equation reduces to Eq. (8.48).

and

8.8 PHASE DIAGRAMS FOR BINARY SOLUTIONS 8.8.1 Constant-pressure Equilibria Consider a binary system made up of components A and B. Component A is assumed to be more volatile than B, i.e. the vapour pressure of A is greater than that of B at any given temperature. For a binary liquid mixture in equilibrium with its vapour, according to the Gibbs Phase rule, the number of

degrees of freedom is two. When the pressure is fixed, only one variable, say liquid phase composition, can be changed independently and other properties such as the temperature and the vapour phase compositions get uniquely determined. Vapour–liquid equilibrium data at constant pressure are usually represented by means of either the temperature-composition diagrams (the T-x-y diagrams or the boiling point diagrams) or the distribution diagrams (x-y diagrams or equilibrium curves). Boiling-point diagram. The boiling point diagrams are plots of temperature as ordinate against composition of liquid and vapour as abscissa. The composition of liquid is usually indicated by the mole fraction of more volatile component in the liquid, x, and the composition of the vapour is indicated by the mole fraction of the more volatile component in the vapour, y. Therefore, the boiling point diagrams are also called T-x-y diagrams. The upper curve in Fig. 8.2 gives the temperature versus vapour composition (y), and is known as the ‘dew-point curve’. The lower curve in the figure is temperature versus liquid composition (x), also called the ‘bubble-point curve’. Below the bubblepoint curve the mixture is subcooled liquid and above the dew-point curve the mixture is superheated vapour. Between the bubble-point and dew-point curves the mixture cannot exist as a single phase, it spontaneously separates into saturated liquid and vapour phases that are in equilibrium.

To make these points clearer, consider a mixture whose temperature and composition ( x1) are such that it is represented by point A in Fig. 8.2. Since the point A lies below the bubble-point curve, the solution is entirely liquid. The mixture is taken in a closed container and the pressure over the system

is maintained at a constant value by a piston. The mixture is heated slowly so that its temperature increases along the vertical line passing through point A till point B on the bubble-point curve is reached. The temperature T1 corresponding to point B, is the bubble point of the original mixture. The first bubble of the vapour is produced at this temperature and it will have the composition (y1) represented by point C on the upper curve. The vapour is richer in the more volatile component. Therefore y1 > x1, and the dew-point curve lies above the bubble-point curve. The mixtures at points B and C are the liquid and vapour at equilibrium at the system pressure and temperature T1. Since both are at the same temperature, they can be joined by a horizontal line BC, known as a ‘tie line’. Further heating will result in the vaporisation of more liquid, and at temperature T2 the system will consist of saturated liquid represented by point D and saturated vapour represented by point E, which are in equilibrium. Since the vapour formed is not removed from the system, the overall composition of the combined mixture of liquid and vapour will be same as x1, the composition of the original mixture. However, the relative amounts of the liquid and the vapour change as the temperature is changed. These relative amounts are given by the ratio in which the point representing the combined mixture (in this case, point F) divides the tie line DE. By material balance consideration, it can be easily verified that

If heating is continued, eventually a temperature T3 is reached when almost all liquid is vaporised. The last drop of liquid getting vaporised at this temperature has a composition denoted by point G and the equilibrium vapour has the composition at H same as the original mixture. Temperature T3 is the dew point of the original mixture. The mixture temperature increases along the vertical line HJ on further heating. On cooling the superheated mixture at point J, the first drop of condensate appears when the temperature drops to T3, the dew point of the mixture and the composition of the liquid is given by point G. We have seen that the mixture at point A has vaporised over a temperature range from T1 (the bubble point) to T3 (the dew point), unlike a pure substance, which vaporises at a single temperature known as the boiling point of the substance. For a solution, the term ‘boiling point’ has no meaning, because, at a given pressure the temperature during vaporisation of a solution varies from the bubble point to the dew point. Equilibrium diagram. The vapour–liquid equilibrium data at constant pressure can be represented on a x versus y plot or an equilibrium distribution diagram. If the vapour composition is taken as the ordinate and the liquid composition is taken as the abscissa, a tie line such as line BC on the boiling point diagram gives rise to a point such as point P on the distribution diagram (Fig. 8.3). Since the vapour is richer in the more volatile component, the curve lies above the diagonal on which x = y.

A liquid–vapour equilibrium curve very close to the diagonal means that the composition of the vapour is not much different from the composition of the liquid with which it is in equilibrium; when the curve coincides with the diagonal, x and y are equal. Effect of pressure on VLE. On the boiling point diagram the temperatures corresponding to x = 0 and x = 1 are the boiling points of pure substances B and A respectively. The boiling points of pure substances increase with pressure. This is true for the bubble and dew points of a mixture of given composition. Consequently the boiling point diagrams at higher pressures will be above the boiling point diagrams at lower pressures as shown in Fig. 8.4. Since the relative volatility decreases as pressure is increased, the closed loop formed by the dew-point and bubble-point curves become narrow at high pressures. Figure 8.5 indicates the effect of pressure on the distribution diagram. In Fig. 8.4, P3 is the critical pressure for component A and above this pressure, the looped curves are shorter.

8.8.2 Constant-temperature Equilibria Vapour–liquid equilibrium data at constant temperature are represented by means of P-x-y diagrams; Fig. 8.6 shows a typical P-x-y diagram. The pressure at x = 0 is the vapour pressure of pure B(

) and the pressure at x = 1 is the vapour

pressure of pure A( ). Since component A is assumed to be more volatile, and therefore, the P-x-y diagram slopes upwards as shown in the figure. The P-y curve lies below the P-x curve so that for any given pressure, y > x. A solution lying above the P-x curve is in the liquid region and that lying below the P-y curve is in the vapour region. In between P-x and P-y curves the solution is a mixture of saturated liquid and vapour. A horizontal line such as AB connects the liquid and vapour phases in equilibrium and is therefore, a tie line. Assume that a liquid mixture whose conditions may be represented by the point C in Fig. 8.6, is taken in a closed container. When the pressure over this system is reduced at constant temperature, the first bubble of vapour forms at point D, and vaporisation goes to completion at point E. Further reduction in pressure leads to the production of superheated vapour represented by point F. The effect of temperature on P-x-y diagram is shown in Fig. 8.6(b). When the temperature is less than the critical temperature of both components, the looped curve such as the one shown at the bottom of Fig. 8.6(b) results. The other two curves refer to temperatures greater than the critical temperature of A.

8.9 VAPOUR–LIQUID EQUILIBRIA IN IDEAL SOLUTIONS It is possible to determine the vapour–liquid equilibrium (VLE) data of certain systems from the vapour pressures of pure components constituting the system. If the liquid phase is an ideal solution and the vapour behaves as an ideal gas, the VLE data can be estimated easily without resorting to direct experimentation. A solution conforming to the ideal behaviour has the following characteristics, all interrelated. 1. The components are chemically similar. The average intermolecular forces of attraction and repulsion in the pure state and in the solution are of approximately the same order of magnitude. 2. There is no volume change on mixing (DV = 0) or the volume of the solution varies linearly with composition. 3. There is neither absorption nor evolution of heat on mixing the constituents that form an ideal solution (DH = 0); that is, there is no temperature change on mixing. 4. The components in an ideal solution obey Raoult’s law, which states that the partial pressure in the vapour in equilibrium with a liquid is directly proportional to the concentration in the liquid. That is,

, where

is the partial pressure of component i and xi is its mole fraction in

the liquid. is the vapour pressure of pure i. This criterion also implies that the total vapour pressure over an ideal solution is a linear function of its composition. For an experimental test of an ideal solution, the last criterion is the safest one to use. For example, the solution formed by two chemically dissimilar materials like benzene and ethyl alcohol should definitely be non-ideal. It is found that for an equimolar mixture of benzene and ethyl alcohol, there is

no change in volume during mixing at room temperature. This peculiar behaviour is because of the fact that when this solution is formed from its constituents, there is increase in volume up to certain concentration and thereafter the volume decreases as shown in Fig. 8.7. When the solution volume is plotted against the composition, the curve will intersect the broken line representing the volume of an ideal solution at a particular concentration represented in the figure by point P.

If the volume were measured for the concentration of the solution corresponding to point P, no change in volume would be observed. This may be the case for enthalpy change of mixing also at some particular composition. The conclusion to be drawn is that negligible volume change or temperature change for one particular composition of the mixture is not a safe criterion of an ideal solution. If these are to be used as the tests for ideal behaviour, then these tests should be done for more than one concentration of the solution. In contrast, the criterion that the total vapour pressure over an ideal solution varies linearly with composition is safe and reliable. It should be understood that there exists no ideal solution in the strict sense of the word; but actual mixtures approach ideality as a limit. Ideality requires that the molecules of the constituents are similar in size, structure and chemical nature; only optical isomers of organic compounds meet these requirements. Thus a mixture of ortho- , meta- and para-xylene conforms very closely to the ideal solution behaviour. Practically, adjacent or nearly adjacent members of the homologous series of organic compounds can be expected to form ideal solutions. Thus mixtures of benzene and toluene, n-octane and n-hexane, ethyl alcohol and propyl alcohol, acetone and acetonitrile, paraffin hydrocarbons in paraffin oils, etc., can be treated as ideal solutions in engineering calculations. Consider an ideal binary solution made up of component 1 and component 2. We have shown in Chapter 7 that all ideal solutions obey Lewis–Randall rule.

Here

is the fugacity of the component i in the liquid and fi is the fugacity of pure i. Using the

criterion of equilibrium and noting that if pressure is not too high, the vapour would not depart too greatly from ideal gas behaviour, it is possible to write

Equation (8.51) shows that at a given temperature, the total pressure over an ideal solution is a linear function of composition thus establishing the fourth criteria given above. When the partial pressures and total pressure are plotted against mole fraction x1, we get according to Eq. (8.50) and Eq. (8.51) the straight lines shown in Fig. 8.8. The broken lines give the partial pressures and the continuous line gives the total pressure. The P-x-y diagram can be easily constructed. At any fixed temperature, the total pressure can be calculated using Eq. (8.51) for various x values ranging from 0 to 1. The corresponding equilibrium vapour phase compositions are obtained by applying Dalton’s law according to which the partial pressure in the vapour is equal to the mole fraction in the vapour (y) times the total pressure (P). That is

Thus Eq. (8.51) is used to calculate the total pressure at given x and Eq. (8.54) is used to calculate the corresponding equilibrium vapour phase composition y. The P-x-y diagram can now be plotted as shown in the Fig. 8.8.

To prepare the T-x-y diagrams at a given total pressure P we can again use Eqs. (8.51) and (8.54). Assume temperatures lying between the boiling points of pure liquids at the given pressure. For the temperature assumed, find the vapour pressures of the pure liquids and calculate x from Eq. (8.51). Use these in Eq. (8.54) and calculate the vapour composition y. Instead, if we attempt to find the equilibrium temperature for the solution of known concentration x, the temperature may be estimated by trial, such that the sum of the partial pressures is equal to the given total pressure. Once the temperature is thus known, the vapour phase composition is determined as before. The T-x curve is the lower curve in the figure and is called the bubble-point curve. The T-y curve is the upper curve and is called the dew-point curve. The y-x diagram is also prepared from the constant total pressure data. It can be constructed from the boiling point diagram by drawing horizontal tie lines. The intersections of these lines with the bubble-point curve give x and the intersections with the dew-point curve give y. Figure 8.10 shows a typical equilibrium diagram. There is an approximate method for the construction of the equilibrium diagram, and it is based on the assumption that the ratio of vapour pressures of the components is independent of temperature. This assumption may not introduce much error, as it is possible that the vapour pressures of both components vary with temperature and these variations are to the same extent that their ratio remains unaltered. Thus

which can be written in the following form.

Although Eq. (8.55) is not exact over a wide range of temperatures, the effect of variation in a is so small that an average a value can be used in Eq. (8.55) and the whole y-x data required for the preparation of the equilibrium curve can be evaluated. EXAMPLE 8.5 Prove that if Raoult’s law is valid for one constituent of a binary solution over the whole concentration range, it must also apply to the other constituent. Solution Assume that Raoult’s law is obeyed by component 1 in a binary mixture. Then

As pointed out earlier, Raoult’s law is obeyed by ideal solutions when the vapour phase behaves as an ideal gas whereas Lewis–Randall rule is obeyed by ideal solutions irrespective of whether the gas phase is ideal or not. So for component 1, we can write

This equation is sometimes referred to as Duhem–Margules equation. Comparing Eq. (8.56) with Eq. (8.57) we see that

This is Raoult’s law for component 2. The conclusion to be drawn from the above derivation is that if Raoult’s law is applicable to one of the constituents of a liquid mixture at all concentrations, it must be applicable to the other constituent as well. EXAMPLE 8.6 n-Heptane and toluene form ideal solution. At 373 K, their vapour pressures are 106 and 74 kPa respectively. Determine the composition of the liquid and vapour in equilibrium at 373 K and 101.3 kPa. Solution Refer Eq. (8.51). Then,

where x is the mole fraction of heptane in the liquid. On solving this, we get x = 0.853 From Eq. (8.54), y = 0.853 106/101.3 = 0.893 The liquid and the vapour at the given conditions contain respectively 85.3% (mol) and 89.3% (mol) heptane. EXAMPLE 8.7 An equimolar solution of benzene and toluene is totally evaporated at a constant temperature of 363 K. At this temperature, the vapour pressures of benzene and toluene are 135.4 and 54 kPa respectively. What are the pressures at the beginning and at the end of the vaporisation process? Solution Put x = 0.5 in Eq. (8.51). Then P = 94.7 kPa. This is the pressure at the beginning of vaporisation. Equation (8.51) can be written as

Put y = 0.5 in this. Thus, we get P = 77.2 kPa. (This is the pressure at the end of vaporisation). EXAMPLE 8.8 A mixture of A and B conforms closely to Raoult’s law. The pure component vapour pressures

in kPa at T K are given by

If the bubble point of a certain mixture of A and B is 349 K at a total pressure of 80 kPa, find the composition of the first vapour that forms. Solution At 349 K, the vapour pressures of the pure components are:

At a given temperature and pressure, the composition of the liquid and vapour phases in equilibrium is calculated using Eqs. (8.51) and (8.54), respectively. The composition of the vapour so calculated at the bubble-point temperature is the composition of the first vapour produced from a liquid on boiling. Using Eq. (8.51),

The vapour formed contains 92.5% A. EXAMPLE 8.9 The vapour pressures of acetone (1) and acetonitrile (2) can be evaluated by the Antoine equations

where T is in K and P is in kPa. Assuming that the solutions formed by these are ideal, calculate (a) x1 and y1 at 327 K and 65 kPa (b) T and y1 at 65 kPa and x1 = 0.4 (c) P and y1 at 327 K and x1 = 0.4 (d) T and x1 at 65 kPa and y1 = 0.4 (e) P and x1 at 327 K and y1 = 0.4 (f) The fraction of the system that is liquid and the composition of the liquid and vapour in equilibrium at 327 K and 65 kPa when the overall composition of the system is 70 mole per cent acetone. Solution (a) From the Antoine equations, at 327 K,

(b) Equation (8.51) can be written as

(c) At 327 K, we have = 85.12 kPa and = 39.31 kPa. Here x1 = 0.4. Using these values in Eq. (8.51), we get P = 57.63 kPa. Using Eq. (8.54) y1 = 0.4

85.12/57.63 = 0.5908

(d) Equation (8.51) can be written as

Assume a temperature and calculate the vapour pressures using Antoine equations. Substitute the vapour pressure values in the above equation. See whether the Left-hand side = 0.4. This is repeated till the left-hand side of the above equation becomes equal to 0.4. At T = 334 K,

= 107.91 kPa and

= 51.01 kPa.

(e) At 327 K, we have = 85.12 kPa and = 39.31 kPa. Here y1 = 0.4 Equation (8.54) relates y to x. When P in Eq. (8.54) is eliminated using Eq. (8.51) we get

(f) The composition of the vapour and liquid in equilibrium at P = 65 kPa and T = 327 K were determined in part (a). They are x1 = 0.5608 and y1 = 0.7344. Let f be the fraction of the mixture that is liquid. Then an acetone balance gives 1 0.7 = f 0.5608 + (1 – f) 0.7344 Solving this, we get f = 0.1982. That is, 19.82% (mol) of the given mixture is liquid. EXAMPLE 8.10 Mixtures of n-Heptane (A) and n-Octane (B) are expected to behave ideally. The total pressure over the system is 101.3 kPa. Using the vapour pressure data given below, (a) Construct the boiling point diagram and (b) The equilibrium diagram and (c) Deduce an equation for the equilibrium diagram using an arithmetic average a value. T, K

371.4

378

383

388

393

398.6

PSA, kPa

101.3

125.3

140.0

160.0

179.9

205.3

PSB, kPa

44.4

55.6

64.5

74.8

86.6

101.3

Solution Sample calculation: Consider the second set of data. T = 378 K; PSA = 125.3 kPa; PSB =

Solution Sample calculation: Consider the second set of data. T = 378 K; P A = 125.3 kPa; P B = 55.6 kPa. Using Eq. (8.51), 101.3 = 55.6 + xA(125.3 – 55.6) Therefore, xA = 0.656. Using Eq. (8.54), we see yA = 0.656

125.3/101.3 = 0.811

Relative volatility is a = PSA/PSB = 125.3/55.6 = 2.25 These calculations are repeated for other temperatures. The results are tabulated below: T, K

371.4

378

383

388

393

398.6

xA

1.000

0.656

0.487

0.312

0.157

0

yA

1.000

0.811

0.674

0.492

0.279

0

a

2.28

2.25

2.17

2.14

2.08

2.02

(a) Plot of T versus x and y gives the boiling point diagram (b) Plot of y against x gives the equilibrium diagram (c) The average of the last row gives a = 2.16. Use this value of a in Eq. (8.55) to get the equation for the equilibrium curve.

8.10 NON-IDEAL SOLUTIONS We have seen that the partial pressure of a component in an ideal solution varies linearly with concentration in the solution. If the solution behaves ideally, the different molecules should be chemically similar. In that case, the molecules of a particular substance, when brought into solution with other components, would not experience any difference in the environment surrounding them from that existed in their pure state. The intermolecular forces in the pure state of the substance and that in the solution would then be approximately of the same order of magnitude. Therefore, the fugacity (or the partial pressure) of a substance, which is a measure of the tendency of the substance to escape from the solution, is not affected by the properties of the other components in the solution. It depends only on the number of molecules of the substance present, or its concentration. In short, the components in an ideal solution obey Raoult’s law. But for non-ideal solutions, the partial pressures do not vary linearly with composition, as shown in Fig. 8.11 for the case of carbon disulphide– acetone system.

The non-ideal behaviour of liquid mixtures arises due to the dissimilarity among molecules. The dissimilarities arise from the difference in the molecular structure or from the difference in the molecular weight. The non-ideal behaviour of light hydrocarbons such as methane, ethylene, etc., in mixtures of heavier paraffin or crude oil is due to the difference in the molecular weights. In contrast, it is a type of intermolecular attraction called hydrogen bonding, that is responsible for the non-ideal behaviour resulting from the difference in the molecular structure. Molecules, which contain atoms such as oxygen, chlorine, fluorine or nitrogen, tend to be polar. When the electrons in the bonds between these atoms and hydrogen are not equally shared, a dipole is created. The electrons tend to be closer to the larger atoms, which become negatively charged compared to hydrogen which becomes the positive end of the dipole. In a solution of polar substances, the molecules tend to arrange themselves so that the charge deficiency of the hydrogen atoms is compensated by an intermolecular bond with a ‘donor’ or negatively charged atom. These hydrogen bonds have energies of the order of several kJ/mol. Because of hydrogen bonding, bimolecular complexes between like or unlike molecules are formed, and even chain-like or three-dimensional aggregates between a large number of molecules are sometimes formed. The formation or destruction of hydrogen bonding during mixing leads to very large heat effects and drastic changes in the thermodynamic properties. Non-ideal behaviour falls into one of the following two types: positive deviation from ideality and negative deviation from ideality. The positive deviation from ideality results when the actual partial pressure of each constituent is greater than it should be if Raoult’s law were obeyed. Solutions in which intermolecular forces between like molecules are stronger than those between unlike molecules, show appreciable positive deviation from ideality. On mixing the constituents which form a solution exhibiting positive deviation from ideality, there is an absorption of heat. This can be proved easily if we recognise the experimental observation that most solutions tend to exhibit ideal

behaviour as temperature is increased. For a solution showing positive deviation, for each component is greater than its mole fraction xi, and as temperature is increased it becomes equal to xi , because the solution tends to ideality as temperature is increased. It means that for a system of given composition for which deviation from Raoult’s law is positive, the ratio increasing temperature. That is

Comparing Eq. (8.58) with Eq. (8.59) we see that

/RT2 < 0, which means

decreases with

. The total

enthalpy of the solution is , whereas the enthalpy of the system before mixing is S niHi. Since the former is greater than the latter, there is absorption of heat during mixing. Examples of solutions showing positive deviation from ideality are oxygen–nitrogen, ethanol–ethyl ether, water–ethanol, carbon disulphide–acetone, benzene–cyclohexane, acetonitrile–benzene, n-hexane–nitroethane, etc. For solutions exhibiting negative deviation from ideal behaviour, the partial pressures are less than those given by Raoult’s law. By a derivation similar to the one presented in the preceding paragraph, it can be shown that when solutions showing negative deviation are formed from pure constituents there is evolution of heat. At the molecular level, appreciable negative deviation reflects stronger intermolecular forces between unlike than between like pairs of molecules. Examples are chloroform–ethyl ether, chloroform–benzene, hydrochloric acid–water, phenol–cyclohexanol, chloroform–acetone, etc. The general nature of the vapour pressure curves showing positive and negative deviation are shown in Fig. 8.12. Figures 8.12(a) and (b) refer to constant temperature conditions. The uppermost curves give the total vapour pressure as function of liquid composition. The corresponding curves, as a function of the vapour composition lie below it, so that the vapour is rich in the more volatile component.

8.10.1 Azeotropes Azeotropes are constant boiling mixtures. The word ‘azeotrope’ is derived from Greek word meaning ‘boiling without changing’. When an azeotrope is boiled, the resulting vapour will have the same composition as the liquid from which it is produced. Whereas, the equilibrium temperature of an ordinary solution varies from the bubble point to the dew point, the boiling point of an azeotrope remains constant till the entire liquid is vaporised. The azeotropes are formed by solution showing large positive or negative deviation from ideality. If the vapour pressures of the constituents of a solution are very close, then any appreciable positive deviation from ideality will lead to a maximum in the vapour pressure curve and negative deviations from ideality under the same conditions leads to a minimum in the vapour pressure curve. Even if an appreciable difference exists in the vapour pressures of the pure components, the chances for the occurrence of maximum or minimum in the vapour pressures should not be overruled if the deviation from ideal behaviour is quite high. At the composition at which there exists a maximum or minimum in the vapour pressure curve, a minimum or maximum, as the case may be, exists in the boiling point diagrams. The mixture is said to form an azeotrope at this composition under the given temperature and pressure and it will distil without change in composition, because the vapour produced has the same composition as the liquid. Minimum-boiling azeotropes. Solutions showing positive deviation from ideality in certain cases may lead to the formation of azeotropes of the minimum-boiling type. The P-x-y, T-x-y and x-y curves for the minimum-boiling azeotropes are shown in Fig. 8.13.

In the boiling point diagram, the liquid and vapour curves are tangent at point M, the point of azeotropism at this pressure. The temperature at M is the minimum temperature of boiling for the system. For all mixtures of composition less than M, the equilibrium vapour is richer than the liquid in the more volatile component. For all mixtures richer than M, the vapour is less rich than the liquid in the more volatile component. A mixture of composition M boils producing a vapour of identical compositions and consequently at a constant temperature and without change in composition. If solutions either at P or Q are boiled in an open vessel with continuous escape of vapours, the temperature and composition move along the lower curve away from M and towards the pure substances. Solutions like these cannot be distilled by usual distillation methods. One of the most important azeotropes in this category is ethanol–water which forms azeotrope at 89.4% (mol) ethanol at 351.4 K and 101.3 kPa. Other examples are benzene–ethanol (341.2 K, 55%

benzene), carbon disulphide–acetone (312.5 K, 61% carbon disulphide), isopropyl ether–isopropyl alcohol (345.1 K, 39.3% alcohol), all at 101.3 kPa. Maximum-boiling azeotropes. When the total pressure of the system at equilibrium is less than the ideal value, the system is said to exhibit negative deviation from ideality. When the difference in vapour pressures of the components is not too great, and in addition, the negative deviations are large, the curve for total pressure against composition passes through a minimum. This condition results in a maximum in the boiling temperature and a condition of azeotropism as at point M in Fig. 8.14.

The vapour is leaner in the more volatile component for liquids whose concentration is less than the azeotropic concentration. Solution on either side of the azeotrope, if boiled in an open vessel with escape of vapours will ultimately leave a residual liquid of the azeotropic composition in the vessel. Maximum-boiling azeotropes are less common than the minimum-boiling type. Hydrochloric acid–

water system forms an azeotrope at 11.1% (mol) HCl at 383 K and 101.3 kPa. Other examples are chloroform–acetone (337.7 K and 65.5% acetone), phenol–cyclohexanol (455.65 K, 90% phenol), all at 101.3 kPa. The plot of activity coefficients versus mole fraction in the liquid phase for ideal solutions, minimumboiling azeotrope and maximum-boiling azeotrope are given in Fig. 8.15.

Effect of pressure on azeotropes. The azeotropic composition shifts continuously with change in pressure or temperature. In some cases, changing the pressure may eliminate azeotropism from the system. Azeotropism disappears in ethanol–water system at pressures below 9.33 kPa. The table below illustrates the effect of pressure on this system. The last row gives the mole per cent of alcohol in the azeotrope. P, kPa

13.3

20.0

26.6

53.2

101.3

146.6

193.3

T, K

307.4

315.2

321

336

351.3

361

368.5

%(mol)

99.6

96.2

93.8

91.4

89.43

89.3

89.0

EXAMPLE 8.11 Prove that at the azeotropic composition, the vapour and liquid have the same composition. Solution Refer the Duhem–Margules equation, Eq. (8.57).

Assuming the vapour to behave ideally, it can be written as

For a maximum or minimum on the total pressure curve, (dP/dx1) is zero. Then Eq. (8.64) necessitates that either the terms in the parenthesis should be zero or is zero. If the latter term is zero the partial pressure would be unaffected by concentration changes, which is not true. Hence, at the point of azeotropism, we have, the following relation:

Since x2 = 1 – x1 and y2 = 1 – y1, the above result means that at azeotropic condition, x1 = y1 or the vapour composition and liquid composition are the same.

8.11 VAPOUR–LIQUID EQUILIBRIA (VLE) AT LOW PRESSURES It has been pointed out that if the pressure is not very high, the fundamental equation relating the compositions of the vapour and liquid under equilibrium is given by Eq. (8.47) as

yiP = gixi The activity coefficients gi vary with composition and temperature at a given pressure. In order to calculate the relationship between pressure, temperature and composition of the equilibrium phases for non-ideal solutions at low pressures where the vapour phase is assumed to behave as an ideal gas, we can utilise Eq. (8.47), provided that the equations relating the activity coefficients to compositions and temperature are available. Experimental vapour–liquid equilibrium data can be correlated using such equations and the empirical parameters in them are evaluated. Once an activity coefficient equation suitable for the given system is identified and the parameters evaluated, it can be used for vapour–liquid equilibrium calculations through Eq. (8.47). We have seen that the activity coefficients in a binary solution are not independent, and they are interrelated through the Gibbs– Duhem equations as

The minimum requirement to be met by an equation for activity coefficient is that it should conform to the restriction imposed by the above relationship. Among a number of equations between g and x that are available, as far as phase equilibrium problems are concerned, some equations have got wide acceptance. They are discussed in the following sections.

8.11.1 Activity Coefficient Equations Wohl’s three-suffix equations. The relationship between excess free energy and activity coefficient was discussed in Chapter 7. Most of the equations relating activity coefficient and concentration of the solution were derived from these excess free energy relationships. Wohl proposed, statistically, a general method for expressing excess free energy and provided some rough physical significance to the various parameters appearing in the equations. Wohl’s equation for excess free energy contained terms for compositions, effective molal volumes and effective volumetric fraction of the separate constituents of the solution. From these equations, the following empirical relations for activity coefficient could be written.

Equation (8.65) is known as Wohl’s three-suffix equation . It involves three parameters, A, B and (q1/q2) which are characteristics of the binary system. Margules equation. When the term (q1/q2) is unity in Eq. (8.65), we get the following expression, which is known as the Margules three-suffix equation.

The constant A in the above equation is the terminal value of ln g1 at x1 = 0 and the constant B is the terminal value of ln g2 at x2 = 0. The three-suffix Margules equation adequately represents the VLE data of systems like acetone–methanol, acetone–chloroform, chloroform–methanol, etc. When A = B in Eq. (8.66), the Margules equation takes the following simple form: ………(8.67) Equation (8.67) is called the Margules two-suffix equation. It represents sufficiently and accurately the activity coefficients of simple liquid mixtures, i.e. mixtures of molecules, which are similar in size, shape and chemical nature. The constant A may be positive or negative. While in general, the constant depends on temperature, for many systems it is a weak function of temperature. Vapour– liquid equilibrium data of argon–oxygen, benzene–cyclohexane, etc., are well represented by the Margules equation [Eq. (8.67)]. van Laar equation. Let (q1/q2) = (A/B) in Eq. (8.65). The resulting two-parameter equation is known as the van Laar equation. The van Laar equations can be written as

The constant A is the terminal value of ln g1 at x1 = 0 and B is the terminal value of ln g2 at x2 = 0. When A and B are equal, the van Laar equations simplify to the Margules equation, Eq. (8.67). The van Laar equation (8.68) may be rearranged to the following forms, which are very convenient for the evaluation of constants A and B.

Strictly speaking, van Laar equations are applicable only for solutions of relatively simple, preferably non-polar liquids. But empirically, it has been found that these are applicable for more complex mixtures. The van Laar equations are widely used for vapour–liquid equilibrium calculations because of their flexibility and mathematical simplicity. Activity coefficients in benzene– isooctane system, n-propanol–water system, etc., are accurately represented by the van Laar equations. The selection of a proper equation for VLE data correlation depends on the molecular complexity of the system and the precision of the experimental data. When an equation is selected that fits the experimental data well, the constants for the constant pressure conditions will be different from those applicable for constant temperature conditions. The effect of pressure on the constants is usually negligibly small, whereas the effect of temperature is appreciable and cannot be neglected. The van Laar constants vary with temperatures unless the temperature range involved is small. However, in vapour–liquid equilibrium calculations, the effect of temperature on the activity coefficient is usually

ignored (Prausnitz, 1985). The Margules three-suffix equation is suited for symmetrical systems, i.e. where the constants A and B are nearly the same. The van Laar equations can be used for unsymmetrical solutions, where the ratio A/B does not exceed 2. Though many systems follow van Laar equations, they cannot represent maxima or minima in the ln g curve. Margules three-suffix equation should be used in such cases. For choice of an appropriate equation, a rule of thumb usually employed is this: When the ratio of molar volumes is close to unity, the Margules equation is preferred. When the ratio is quite different from unity, as is the case when water is one of the constituents, the van Laar equations are found to be satisfactory. For example, the chloroform–ethyl alcohol system, which shows a maximum and a minimum on the ln g curves and whose ratio between pure component molar volumes is 1.38, is accurately represented by the Margules equation. For n-propanol–water system this ratio is 4.16 and the van Laar equations are found to represent the behaviour accurately. It is to be remembered that in equations having only two constants, determination of g1 and g2 at a single known composition permits the evaluation of the constants and the complete g curve. Equation (8.47) permits the evaluation of g1 and g2 when it is rearranged to the following form.

The data required are a single set of equilibrium vapour–liquid composition values and the vapour pressures of the pure components. When an azeotrope is formed, only the azeotropic composition need be known, because it represents the composition of both the liquid and the vapour phases. The activity coefficients can be evaluated by putting x = y in Eq. (8.70).

Wilson equation. All the activity coefficient equations discussed so far can be deduced from the original Wohl’s equation under proper simplifying assumptions. However, there are many equations that cannot be derived from the Wohl’s general equation. Among such equations, the Wilson equation, the NRTL equation and the UNIQUAC equation are important from practical point of view. All these are based on the concept of local compositions, which are different from the overall mixture compositions. Based on molecular considerations, Wilson (1964) proposed the following equations for activity coefficients in a binary mixture.

Wilson equations have two adjustable positive parameters L12 and L21. These are related to the pure component molar volumes and to the characteristic energy differences by

where V1 and V2 are the molar volumes of pure liquids and l’s are the energies of interaction between the molecules designated in the subscripts. The differences in the characteristic energies (aij) are assumed to be temperature independent and this introduces no serious error in practical calculations. Wilson equation provides a good representation of VLE of a variety of miscible mixtures. It is particularly suitable for solutions of polar or associating components like alcohols in non-polar solvents for which the Margules and van Laar equations are generally inadequate. Wilson equation suffers from two disadvantages, though not serious for many applications. Firstly, it is not suitable for systems showing maxima or minima on the ln g versus x curves. Secondly, it is not useful for systems exhibiting limited miscibility. The use of Wilson equation is therefore recommended only for liquid systems that are completely miscible, or for partially miscible systems in the region where only one liquid phase exists. Non-random two-liquid (NRTL) equation. The NRTL model, proposed by Renon and Prausnitz (1968), also is based on the local composition concept. The activity coefficients are

The constants b12 and b21 are similar to the constants representing characteristic energy differences appearing in the Wilson equation. These, as well as the constant a12 are independent of composition and temperature. The parameter a12 is related to the non-randomness in the mixture. If a1 2 is zero, the mixture is completely random and the NRTL equation reduces to the Margules equation. It is found from fitting of experimental data that a12 varies from about 0.20 to 0.47. In the absence of the experimental data, the value of a12 is arbitrarily set, a typical

choice being a12 = 0.3. When a12 is arbitrarily fixed, NRTL equation becomes a two-parameter model. NRTL equation is applicable to partially miscible as well as totally miscible systems. For moderately non-ideal systems, it offers no advantage over the van Laar and Margules equations. But, for strongly non-ideal solutions and especially partially miscible systems, the NRTL equations provide a good representation.

Universal quasi-chemical (UNIQUAC) equation. Abrams and Prausnitz (1975) extended the quasi-chemical theory of liquid mixtures to solutions containing molecules of different sizes. This extension is called the UNIQUAC theory. The UNIQUAC model consists of two parts—the combinatorial part, which describes the prominent entropic contribution and a residual part, which is due primarily to the intermolecular forces that are responsible for the enthalpy of mixing. The combinatorial part is determined by the sizes and shape of the molecules and requires only purecomponent data. The residual part depends on the intermolecular forces and involves two adjustable binary parameters. The UNIQUAC equations for activity coefficients are

z is the coordination number, r, q and q are pure-component molecular structure constants. The molecular size and surface area are given by r and q respectively. For fluids other than water or lower alcohols, q = q . For alcohols, the surface of interaction q is smaller than the geometric surface q. The adjustable binary parameters t12 and t21 are related to the characteristic energies Du as follows.

The UNIQUAC equation satisfies a large number of non-electrolyte mixtures containing non-polar fluids such as hydrocarbons, alcohols, nitriles, ketones, aldehydes, organic acids, etc., and water, including partially miscible mixtures. The main advantages of this equation are its wide applicability and simplicity arising primarily from the fact that there are only two adjustable parameters.

Universal functional activity coefficient (UNIFAC) method. In the UNIFAC method, the activity coefficients are estimated through group contributions. The liquid is treated as a solution of different structural groups from which the molecules are formed, rather than a solution of molecules themselves. This method is based on the UNIQUAC model where the activity coefficient is divided into two parts—the molecular size contribution (the combinatorial part) and the interaction contributions (the residual part). The combinatorial contribution can be estimated from pure-component properties and the size and shape of the molecules, whereas for the estimation of the second part, group areas and group contributions are needed. A large number of group interaction parameters are already reported. UNIFAC has been successfully used for the design of distillation columns, involving even azeotropic and extractive distillation (Prausnitz et al., 1986). EXAMPLE 8.12 Liquids A and B form an azeotrope containing 46.1 mole per cent A at 101.3 kPa and 345 K. At 345 K, the vapour pressure of A is 84.8 kPa and that of B is 78.2 kPa. Calculate the van Laar constants. Solution Let the material A be component 1 and B be component 2. The activity coefficients at the azeotropic concentration can be evaluated by Eq. (8.71)

EXAMPLE 8.13 The azeotrope of the ethanol–benzene system has a composition of 44.8% (mol) ethanol with a boiling point of 341.4 K at 101.3 kPa. At this temperature the vapour pressure of benzene is 68.9 kPa and the vapour pressure of ethanol is 67.4 kPa. What are the activity coefficients in a solution containing 10% alcohol? Solution Let benzene be component 1 and alcohol component 2. For the azeotrope

EXAMPLE 8.14 Water (1)–hydrazine (2) system forms an azeotrope containing 58.5% (mol) hydrazine at 393 K and 101.3 kPa. Calculate the equilibrium vapour composition for a solution containing 20% (mol) hydrazine. The relative volatility of water with reference to hydrazine is 1.6 and may be assumed to remain constant in the temperature range involved. The vapour pressure of hydrazine at 393 K is 124.76 kPa. Solution The vapour pressure of water at 393 K = 1.6 393 K = 1.6 124.76 = 199.62 kPa.

vapour pressure of hydrazine at

To evaluate the vapour compositions using these equations, we should know the vapour pressure values at the new equilibrium temperature. Taking the ratio of the last two equations, we get

The composition of the vapour in equilibrium with the liquid containing 20% hydrazine is 5.28% hydrazine and 94.72% water. EXAMPLE 8.15 At 318 K and 24.4 kPa, the composition of the system ethanol (1) and toluene (2) at equilibrium is x1 = 0.3 and y1 = 0.634. The saturation pressure at the given temperature for the pure components are (a) The liquid-phase activity coefficients (b) The value of GE/RT for the liquid phase

, respectively. Calculate:

Solution (a) At vapour–liquid equilibrium, the composition of the vapour and liquid phases are

related by Eq. (8.48)

Equation (8.48) may be used to evaluate the activity coefficients.

EXAMPLE 8.16 The activity coefficients in a mixture of components A and B at 313 K are given by

At 313 K, A and B form an azeotrope containing 49.4 mol percent A at a total pressure of 27 kPa. If the vapour pressures of pure A and pure B are 25.0 and 24.3 kPa, respectively, calculate the total pressure of the vapour at temperature 313 K in equilibrium with a liquid mixture containing 12.5 mol percent A. Solution At the azeotropic composition, Eq. (8.71) is applicable, so that the activity coefficients are:

EXAMPLE 8.17 Using van Laar constants and the vapour pressures of the pure substances how would you prove whether a given binary system forms an azeotrope or not? Solution If the mixture does not exhibit azeotropic behaviour, the ratio (y1/x1) will be greater than (y2/x2) for the entire concentration range 0 to 1. Denoting the ratio of (y1/x1) to (y2/x2) by a, then a > 1 for 0 < x < 1. However, if the mixture forms an azeotrope, then the value of a will be greater than 1 over some concentration range and will be less than 1 over the remaining portion. Since a varies continuously with x, a should have a value equal to 1 at some x which is the azeotropic composition. Writing Eq. (8.47) for both components and rearranging the result, we get

A and B are the van Laar constants. If the mixture forms an azeotrope, one of the above values will be greater than 1 and the other less than 1. EXAMPLE 8.18 For the binary system methanol (1) and benzene (2), the recommended values of the Wilson parameters at 341 K are L12 = 0.1751 and L21 = 0.3456. The vapour pressures of pure species are = 68.75 kPa and = 115.89 kPa. Show that the given system can form an azeotrope at 341 K. Assume that the vapour behaves like an ideal gas. Solution Wilson equations [Eq. (8.72)] provide the activity coefficients in a binary mixture as:

Equation (8.48) gives the relationship between activity coefficient and equilibrium phase compositions as

If the mixture forms an azeotrope, then the value of a will be greater than 1 over some concentration range and will be less than 1 over the remaining portion. Relative volatilities are calculated at x1 = 0 and x1 = 1. If the system forms an azeotrope, one of these values will be greater than unity and the other less than unity. At x1 = 0,

At x2 = 0,

Since the relative volatility at x1 = 0 is greater than unity, and that at x2 = 0 is less than unity, it is clear that the system forms an azeotrope. EXAMPLE 8.19 A stream of isopropanol–water mixture is flashed into a separation chamber at 353 K and 91.2 kPa. A particular analysis of the liquid product showed an isopropanol content of 4.7% (mol), a value that deviated from the norm. It is suspected that an air leak into the separator might have caused this. Do you agree? The vapour pressures of the pure propanol and water are 91.11 kPa and 47.36 kPa respectively, and the van Laar constants are A = 2.470 and B = 1.094. Solution For x1 = 0.047, x2 = 0.953, A = 2.47 and B = 1.094, Eq. (8.68) gives g1 = 7.388 and g2 = 1.011 The total pressure corresponding to this equilibrium composition is

This is less than the total pressure. This error must have been caused by an air-leak. EXAMPLE 8.20 Construct the P-x-y diagram for the cyclohexane (1)–benzene (2) system at 313 K given that at 313 K the vapour pressures are activity coefficients are given by

= 24.62 kPa and

Solution Assume x1 = 0.4, then x2 = 0.6. Therefore ln g1 = 0.458

0.62 = 0.16488; g1 = 1.1793.

Similarly, ln g2 = 0.458

0.42 = 0.0733; g2 = 1.0760

The total pressure is determined as:

= 24.41 kPa. The liquid-phase

The above calculation is repeated for various x1 values. The results are tabulated below: x1

0

0.2

0.4

0.6

0.8

1.0

g1

1.5809

1.3406

1.1793

1.0760

1.0185

1.000

g2

1.000

1.0185

1.0760

1.1793

1.3406

1.5809

P

24.41

26.49

27.37

27.41

26.61

24.62

y1

0

0.2492

0.4243

0.5799

0.7540

1.0

The results are plotted taking P on the y-axis and x1 and y1 on the x-axis. EXAMPLE 8.21 From vapour–liquid equilibrium measurements for ethanol–benzene system at 318 K and 40.25 kPa it is found that the vapour in equilibrium with a liquid containing 38.4% (mol) benzene contained 56.6% (mol) benzene. The system forms an azeotrope at 318 K. At this temperature, the vapour pressures of ethanol and benzene are 22.9 and 29.6 kPa respectively. Determine the composition and total pressure of the azeotrope. Assume that van Laar equation is applicable for the system. Solution Let benzene be component 1 and ethanol component 2. Using Eq. (8.47) the activity coefficients are determined.

For any liquid composition, the activity coefficients are calculated using these equations. If the mixture forms an azeotrope at any composition, then as per Eq. (8.71), the following relations also give the activity coefficients.

Thus, activity coefficients calculated using the van Laar equations should also satisfy the relation at the azeotropic composition. The azeotropic composition is obtained by trial assuming values of x. For x1 = 0.6, g1 = 1.3830 and g2 = 1.7806.

These values are so close that it can be assumed that the composition corresponds to an azeotrope. Thus, the liquid mixture forms an azeotrope containing 60% benzene which boils at 318 K and approximately 40.86 kPa, (mean of 40.937 and 40.775 kPa). EXAMPLE 8.22 The activity coefficients in a binary system are given by ln Show that if the system forms an azeotrope, then is given by

.

and the azeotropic composition

At the azeotropic composition, the P-x curve has a maximum or minimum. Therefore, the above result is equated to zero and after rearrangement, we get

Since the value of x1 lies between 0 and 1 it is essential that

. Noting that x1x2 for

a binary mixture is less than or equal to 0.25, the second root leads to A ≥ 2. This condition usually results in partial miscibility. Thus, condition for homogeneous azeotropy is and A < 2. EXAMPLE 8.23 The following values refer to the Wilson parameters for the system acetone(1)– water(2): a1 2 = 1225.31 J/mol, a21 = 6051.01 J/mol, V1 = 74.05 m3/mol. The vapour pressures are given by

10–6 m3/mol, V2 = 18.07

10–6

where P is in kPa and T is in K. Calculate the equilibrium pressure and composition of (a) Vapour in equilibrium with a liquid of composition x1 = 0.43 at 349 K. (b) The liquid in equilibrium with a vapour of concentration y1 = 0.8 at 349 K. Solution Using the Antoine equations, at 349 K, the vapour pressures are calculated as

(a) Now from Eq. (8.72), we obtain

The vapour composition is found out from the relation y1P = g1x1

(b) Since the liquid composition is not known, the activity coefficients cannot be calculated. Assume g1 = g2 = 1. The relation Pyi = xi can be written for component 1 and component 2, which on rearrangement gives

Activity coefficients can now be determined for this composition using Wilson equations.

Now the pressure is recalculated incorporating the activity coefficient values. The equation

This is same as the previous value calculated for P. Therefore, P = 164.48 kPa and x1 = 0.4568. EXAMPLE 8.24 The system methanol–methyl ethyl ketone forms an azeotrope containing 84.2% (mol) methanol at 337.5 K and 101.3 kPa. The vapour pressures of the pure species are given by the Antoine equation

Determine the parameters in the Wilson equation.

Solution The vapour pressures at 337.5 K are calculated using Antoine equations.

L12 is calculated by substituting the known values in the above equation.

8.12 VAPOUR-LIQUID EQUILIBRIA INVOLVING HIGH PRESSURES AND MULTICOMPONENT SYSTEMS 8.12.1 Equations of State Approach The equations of state approach is most useful for vapour–liquid equilibria at high pressures. In this method, the vapour and liquid compositions are related by Eq. (8.45):

The equations of state provide thermodynamic models for evaluating in Eq. (8.45) from volumetric properties. The fugacity coefficients for the vapour and liquid phases are to be obtained from appropriate equations of state. Equations of state widely used in engineering calculations were discussed in Chapter 3. The most generally used equations of state for VLE calculations are given in Table 8.1.

The fugacity coefficients in Eq. (8.45) can be calculated using a suitable equation of state. The relationship between fugacity coefficient and the volumetric properties can be written as

where Z is the compressibility factor. The equations of state approach has the advantage that the standard state fugacity need not be specified. Another advantage is that the continuity at the critical point is ensured as the properties in the gas and liquid phases are evaluated using the same model. The equations of state approach is more suitable to mixtures of nonpolar components. Also, the equations of state are widely used to predict VLE of light hydrocarbon mixtures.

8.12.2 Vaporisation Equilibrium Constants For high-pressure vapour–liquid equilibrium calculations, it is convenient to express the phase equilibrium relations in terms of vaporisation equilibrium constants or K factors. It is defined as the

ratio of mole fraction in the vapour phase y to that in the liquid x or K = (y/x). The equation of state form of the K-value is obtained from Eq. (8.45) as follows:

In Eq. (8.83a),

is the fugacity coefficient in the saturation state, Vi is the molar volume of pure i as

saturated liquid. are evaluated using an equation of state. Assuming the Poynting factor to be unity, Eq. (8.83a) can be rearranged as

which is nothing but Raoult’s law which is applicable for ideal solution where the vapour behaves as ideal gas. We see that for ideal solution, the vaporisation equilibrium constant is the ratio of vapour pressure to total pressure. The K-factors for ideal solutions depend only on the temperature and pressure and are readily correlated as a function of these two variables. For mixtures of light hydrocarbons, Eq. (8.83a) can be simplified using two assumptions:

1. Intermolecular forces are weak, so that vapour phase behaves as an ideal solution, so that where fi is the fugacity coefficient of the pure components.

,

2. Liquid phase behaves as an ideal solution so that =1.0. The resulting equation is very convenient as it involves properties of pure component only.

Equation (8.83e) reveals that for mixtures of light hydrocarbons, the vaporisation equilibrium constants are independent of the composition of the liquid and vapour phases in equilibrium and can be evaluated as pure component properties. Since can be determined from equations of state or generalised correlations, it is possible to provide correlations for K-values of substances as functions of temperature and pressure. DePriester nomographs [C.L. DePriester, Chem. Eng. Progr., Symposium Ser. 7, 49 (1953)] provide such correlations for many hydrocarbons. These nomographs are available in standard references such as Chemical Engineers Handbook. Figure 8.16 gives the K-factor for light hydrocarbons in the high temperature range.

8.12.3 Bubble-point Equilibria The bubble-point temperature is the one at which the first bubble of vapour is produced from the liquid on heating at constant pressure. At the bubble point the liquid has the same composition as the original mixture. Therefore, in problems where bubble-point temperature is to be determined, the xi

are known. Assume a temperature and get the Ki values at this temperature. Calculate yi using yi = Kixi. If the assumed temperature is correct then

S yi = S Kixi = 1………(8.84) Otherwise, repeat the calculations with another temperature. To find the bubble-point pressure , a similar procedure as above is adopted by assuming various values of pressure until S Kixi = 1.

8.12.4 Dew-point Equilibria The dew-point temperature is the one at which the first drop of condensate is formed on cooling a vapour at constant pressure. The vapour in equilibrium with the liquid at the dew point has the same composition as the original mixture. In order to find the dew-point temperature, a temperature is assumed arbitrarily and Ki is determined. Then,

Otherwise, repeat the calculation by assuming another temperature till this equation is satisfied. Determination of the dew-point pressure involves a similar procedure assuming pressure instead of temperature.

8.12.5 Flash Vaporisation The general flash vaporisation problem can be stated as: Given a mixture of known overall composition zi at temperature T and pressure P, what is the fraction that is vapour (V) and what are the composition of the liquid and vapour phases in equilibrium? The overall material balance for the system is F = V + L………(8.86) where F is the total number of moles of the initial mixture. The component-i balance for the system is Fzi = Vyi + Lxi………(8.87) Since yi = Kixi, it can be eliminated from Eq. (8.87) to get the following:

Equation (8.91) can also be utilised in an iterative procedure to estimate T, P or the fraction of the initial mixture that is vaporised. EXAMPLE 8.25 A mixture contains 45% (mol) methanol (A), 30% (mol) ethanol (B) and the rest n-propanol (C). Liquid solution may be assumed to be ideal and perfect gas law is valid for the vapour phase. Calculate at a total pressure of 101.3 kPa. (a) The bubble point and the vapour composition (b) The dew point and the liquid composition. The vapour pressures of the pure liquids are given below: Temperature, K 333 343 353 363 81.97 133.29 186.61 266.58 PSA, kPa 49.32 73.31 106.63 166.61 PSB, kPa 39.32 62.65 93.30 133.29 PSC, kPa

Solution The vapour pressures of the components are plotted against temperature so that interpolation of vapour pressure can be done easily. (a) If the vapour phase can be treated as an ideal gas and liquid phase, an ideal solution, the K-values can be written as Ki = yi/xi = . Equation (8.84) can be written as

Now temperatures are assumed till the above equality is satisfied. It is seen that at 344 K,

The bubble-point lies between 344 and 345 K. By interpolation, the bubble-point is obtained as 344.6 K. At this temperature the vapour pressures are obtained from the P vs T plots. 137.3 kPa,

= 76.20 kPa and

Component

xi

Methanol

0.45

Ethanol Propanol

=

= 65.40 kPa. Ki = /P

yi = Ki xi

137.30

1.355

0.610

0.30

76.20

0.752

0.226

0.25

65.40

0.646

0.162

S Ki xi

0.998

The equilibrium vapour contains 61% methanol, 22.6% ethanol and 16.2% propanol. (b) Equation (8.85) for the present case becomes

The dew-point temperature is to be determined by trial such that the above relation is satisfied. By trial, it can be seen that at 347.5 K, Component

yi

Methanol

0.45

Ethanol Propanol

Syi /Ki

= 153.28 kPa,

Ki = /P

xi = yi /Ki

153.28

1.5131

0.2974

0.30

85.25

0.8416

0.3565

0.25

73.31

0.7237

0.3454

= 85.25 kPa and

= 73.31 kPa.

0.9993

The values in the last column are the liquid composition at the dew point. Thus, liquid contains 29.7% methanol, 35.7% ethanol, and 34.5% propanol. EXAMPLE 8.26 A hydrocarbon mixture contains 25% (mol) propane, 40% (mol) n-butane and 35% (mol) n-pentane at 1447.14 kPa. Assume ideal solution behaviour and calculate (a) The bubble-point temperature and composition of the vapour (b) The dew-point temperature and the composition of the liquid (c) The temperature and the composition of the liquid and vapour in equilibrium when 45% (mol) of the initial mixture is vaporised. (The values of Ki can be obtained from Fig. 13.6 of Chemical Engineer’s Handbook, 5th ed.)

Solution (a) Assume temperature, say 355.4 K, and the Ki values are found out from the nomograph [Fig. 13.6(b) in Chemical Engineer’s Handbook]. The products of Ki and xi are calculated and their sum S xiKi is found out. The results for two temperatures 355.4 K and 366.5 K are shown below. T = 355.4 K

T = 366.5 K

xi

Ki

Ki xi

Ki

Ki xi

Propane

0.25

2.000

0.500

2.30

0.575

n-Butane

0.40

0.780

0.312

0.90

0.360

n-Pentane

0.35

0.330

0.116

0.40

0.140

Component

S Ki xi

0.928

1.075

The bubble-point temperature lies between 355.4 K and 366.5 K. By interpolation, the temperature is found out to be 361 K. The calculations are carried out at this temperature and the results are as follows: Component

xi

Ki

Ki xi

Propane

0.25

2.12

0.530

n-Butane

0.40

0.85

0.340

n-Pentane

0.35

0.37

0.130

S Ki xi

1.000

Since S xiKi is approximately 1.00, the bubble-point temperature is 361 K. The values in the last column are the mole fraction of various components in the vapour. At the bubble-point, the vapour contains 53% propane, 34% butane and 13% pentane. (b) At the dew-point temperature, S yi/Ki = 1. At 377.6 K, this value is 1.1598 and at 388.8 K it is 0.9677. T = 377.6 K

T = 388.8 K

Component

yi

Ki

yi /Ki

Ki

yi /Ki

Propane

0.25

2.6

0.0962

2.9

0.0862

n-Butane

0.40

1.1

0.3636

1.3

0.3077

n-Pentane

0.35

0.5

0.7000

0.61

0.5738

Syi /Ki

1.1598

0.9677

By interpolation, the dew-point temperature is found to be 387 K. The calculations for this temperature is given below.

Component

yi

yi /Ki

0.25

Ki 2.85

Propane n-Butane

0.40

1.25

0.3200

n-Pentane

0.35

0.59

0.5932

S yi /Ki

0.0877

1.0009

The last column in the above table is the liquid compositions. The equilibrium liquid at the dew point contains 8.77% propane, 32.0% butane and 59.32% pentane. (c) In the following calculations, temperature is assumed so as to satisfy Eq. (8.91). For a basis of 100 mol of the initial mixture, F = 100 mol, V = 45 mol and L = 55 mol. Equation (8.91) becomes

T = 366.5 K

T = 377.6 K

Component

zi

Ki

zi /[1 + L/(VKi )]

Ki

zi /[1 + L/(VKi )]

Propane

0.25

2.30

0.1632

2.6

0.1701

n-Butane

0.40

0.90

0.1696

1.1

0.1895

n-Pentane

0.35

0.40

0.0863

0.5

0.1016

S zi /[1 + L/(VKi )]

0.4191

0.4612

From the calculations given above, we see that the equilibrium temperature is between 366.5 K and 377.6 K. By interpolation, T = 374.6 K. T = 374.6 K Component

zi

Ki

zi /[1 + L/(VKi )]

Propane

0.25

2.50

0.1679

n-Butane

0.40

1.08

0.1876

n-Pentane

0.35

0.48

0.0987

S zi /[1 + L/(VKi )]

0.4542

Comparing Eqs. (8.90) and (8.91), we can see that

These are calculated using the values in the last column. Corresponding xi values are found out using the material balance [Eq. (8.87)]. Fzi = Vyi + Lxi The results of the calculation are given below:

Component

yi

xi

Propane

0.3697

0.1521

n-Butane

0.4130

0.3894

n-Pentane

0.2173

0.4586

8.13 CONSISTENCY TESTS FOR VLE DATA Many practical cases like distillation calculations are dependent on vapour–liquid equilibrium data and such data should be reasonably accurate if the results are to be reliable. As the VLE measurements are prone to inaccuracies, it is essential that we have some means for checking the consistency of the measured results. Thermodynamics provides tests for consistency of experimental VLE data. Almost all these tests are based on the Gibbs–Duhem equations written in terms of activity coefficients [Eq. (7.101)].

8.13.1 Using Slope of ln g Curves The Gibbs–Duhem equation in terms of activity coefficient [Eq. (7.101)] provides a very simple test for thermodynamic consistency.

Plot the logarithm of the activity coefficients against mole fraction x1 of component 1 in a binary solution as shown in Fig 8.17(a) and measure the slopes of the tangents drawn to the resulting curves at any selected composition x1. Equation (8.92) tells us that if the Gibbs–Duhem equation is to be satisfied, both slopes must have opposite sign. Otherwise, the data are inconsistent. If the slopes are of opposite sign, substitute the values in Eq. (8.92) and if it is satisfied reasonably well, then the data is consistent at the selected composition. For a complete test, the slopes determined at other compositions are substituted into Eq. (8.92) to see whether the equality is satisfied or not. In addition to the above observations, we can make the following generalisations with the help of Eq. (7.101). 1. If one of the ln g curves has a maximum (or minimum) at certain concentration, the other curve should have a minimum (or maximum) at the same composition. 2. If there is no maximum or minimum point, then both curves must be positive or both must be negative over the entire range. Or in other words, if one component has g values always greater than unity and has no maximum, the g values of the other component must likewise be greater than unity. This is a consequence of the fact that Raoult’s law is to be obeyed by the component

as its mole fraction tends to unity. From the above discussion, it is clear that Fig. 8.17(b) represents plots of consistent data whereas Figs. 8.17(c) and (d) are plots of thermodynamically inconsistent data. In Fig. 8.17(c), though there is maximum on one curve and minimum on the other, these are shown at different compositions. In Fig. 8.17(d), the slopes have the same sign and the data are thermodynamically inaccurate.

8.13.2 Using Data at the Mid-point

For testing the thermodynamic consistency of VLE data, the integrated forms of the Gibbs–Duhem equation like the van Laar or the Margules equation are sometimes found very convenient. Consider the van Laar equation Eq. (8.68). Put x1 = x2 = 0.5 and we see that

Case 1. Assume that A = B. Then ln g1 = B/4 and ln g2 = A/4 Case 2. Assume that A = 2B. Then ln g1 = (2/9)B and ln g2 = (2/9)A Case 3. Assume that A = 3B. Then ln g1 = (3/16)B and ln g2 = (3/16)A In all cases cited above, the ln g1 at the mid-point (i.e. at x1 = x2 = 0.5) is approximately one-fourth the van Laar constant B and ln g2 approximately one-fourth the constant A. We have already seen that the van Laar constant A is ln g1 as x1 tends to zero and B is ln g2 as x1 tends to one. Now we have a rough check on the consistency of VLE data for the mid-point. The lng value at this point will be approximately one-fourth the terminal value of the other ln g curve. That is the curve, which is highest at the end-point will be lowest at the mid-point and vice versa, as shown in Figs. 8.18(a) and (b). The mid-point values are approximately one-fourth the terminal values of the other curves in both the figures. However, Fig. 8.18(c) reveals an inconsistent data, as the curve which is highest at the endpoint is also the highest at the mid-point.

8.13.3 Redlich–Kister Method This method, also known as the zero area method can be applied to test the consistency of experimental data when the activity coefficient values over the entire concentration range is available. It is based on the excess free energy of mixing which is the difference between the free energy of mixing of a real solution and that of an ideal solution. Referring to Eqs. (7.134) and (7.145) the excess free energy of mixing can be written as DGE = RT S xi ln gi………(8.94) For a binary solution, it can be written as DGE = RT(x1 ln g1 + x2 ln g2) Differentiating this with respect to x1, we get

From the experimental values of activity coefficients, ln (g1/g2) values are calculated and plotted against x1 taken on the x-axis. The net area of the diagram should equal zero if the data are thermodynamically consistent. That is, the area above the x-axis will be equal to the area below it as shown in Fig. 8.19.

8.13.4 Using the Coexistence Equation The coexistence equation can be used for testing the consistency of vapour–liquid equilibrium data. If the vapour in equilibrium with a binary liquid mixture behaves as an ideal gas, Eq. (8.47) can be used to describe the equilibrium. Rearranging Eq. (8.47), the activity coefficients can be written as

Substitute this in the Gibbs–Duhem Equations [Eq. (8.92)] written in the following form

Equation (8.99) is known as the coexistence equation. It can be used to calculate any one of the three variables P, x or y if experimentally measured values of the other two variables are available. If all the three variables are experimentally determined, then Eq. (8.99) can be used to test the consistency

of the measured data.

8.13.5 Using the Partial Pressure Data At low pressures, the fugacity of a gas equals the pressure and therefore, the Gibbs–Duhem equation in terms of fugacity [Eq. (7.99)] can be rewritten as,

or

The partial pressures of both components are plotted against mole fraction x1 as in Fig. 8.20. The slopes,

are determined at any selected composition.

and

are calculated. Then, according to Eq. (8.100), the absolute values of these quantities should be the same if the data are thermodynamically consistent.

EXAMPLE 8.27 The following results were obtained by experimental VLE measurements on the system, ethanol (1)–benzene (2) at 101.3 kPa. Test whether the data are thermodynamically

consistent or not. x1

0.003

0.449

0.700

0.900

y1

0.432

0.449

0.520

0.719

, kPa

65.31

63.98

66.64

81.31

, kPa

68.64

68.64

69.31

72.24

Solution Assuming that the gas phase behaves ideally, the activity coefficients are calculated as

The values so calculated are listed below: x1

0.003

0.449

0.700

0.900

g1

223.3

1.58

1.13

0.99

g2

0.841

1.475

2.34

2.92

ln(g1/g2 )

5.58

0.068

– 0.734

– 1.082

ln (g1/g2) values are plotted against x1. The net area is found out. Since this is not equal to zero, the given experimental measurements do not satisfy the Redlich–Kister criterion [Eq. (8.97)] for consistency.

8.14 CALCULATION OF ACTIVITY COEFFICIENTS USING GIBBS–DUHEM EQUATION The activity coefficients in a binary solution are related to each other according to the Gibbs–Duhem equations as given below. See Eq. (7.101).

The integral can be evaluated analytically if g2 is expressed as a mathematical function of composition. To find the integral graphically, plot a graph with ( x2/x1) on the y-axis and ln g2 on the x-axis. The area under the curve from ln g2 at x2 = 0 to ln g2 at any desired composition x2 gives the value – ln g1 for the component 1. EXAMPLE 8.28 The activity coefficients for component 1 in a binary solution can be represented by , where a, b and c are concentration independent parameters. Derive an expression for ln g2. Solution Equation (8.101) may be written as

For the present case,

EXAMPLE 8.29 The following data gives the composition versus total pressure for the system chloroform (1)–ethyl alcohol (2) at 328 K are:

Vapour pressures of chloroform and acetone at 328 K are 82.35 and 37.30 kPa respectively. Estimate the constants in the Margules equation [Eq. (8.66)]. Solution The Margules equations are:

(Note: Equations (8.103) and (8.104) are known as Carlson and Colburn relations for activity coefficient.)

EXAMPLE 8.30 The following table gives the partial pressure of acetone versus liquid composition for acetone (1)–water (2) system at 333 K. x1

0

0.033

0.117

0.318

0.554

0.736

1.000

1, kPa

0

25.33

59.05

78.37

89.58

94.77

114.63

The vapour pressure of water at 333 K is 19.91 kPa. Calculate the partial pressure of water in the vapour phase. Solution Equation (8.100) is the Gibbs–Duhem equation in terms of partial pressures.

Using the given data calculate x1/[(1 – x1) ] and this is plotted against

. The area under the curve

between the limits 0 and gives the integral in Eq. (8.106) from which the partial pressure calculated. The results are given below: x1

0

0.033

0.117

0.318

0.554

0.736

1.000

, kPa

19.91

19.31

18.27

16.99

15.42

13.90

0

can be

8.15 VLE FOR SYSTEMS OF LIMITED MISCIBILITY 8.15.1 Partially Miscible Systems Figure 8.21 shows the temperature-composition diagram for a partially miscible system at constant total pressure. Points A and B indicate the boiling points of pure liquids A and B at this pressure. A binary system consisting of two liquid phases and a vapour phase is univariant according to the phase rule. By fixing the pressure, the system is completely defined. The states of the three phases in equilibrium lie on the horizontal line at T*, the three-phase equilibrium temperature. Points C and D represent the saturated liquid phases and point E the vapour in equilibrium with these liquids.

Below T* the system is entirely liquid. It may exist as a homogeneous system, or, as a heterogeneous system consisting of two saturated liquid phases, depending upon its overall composition. If the overall composition of the mixture lies within the region bounded by the curves FC and GD below temperature T*, the mixture cannot exist as a single phase. It separates into two saturated liquid phases, a A-rich phase (LA) represented by the curve GD and a B-rich phase (LB) represented by the curve FC. The compositions of these phases are determined by the intersection of a horizontal line corresponding to the temperature of the system and the curves GD and FC. Consider, for example, a liquid whose combined concentration and temperature are such that it is represented by the point M in Fig. 8.21(a). Since M lies in the two-liquid region (LB –LA), the mixture separates into a B-rich phase represented by point P, and a A-rich phase represented by point Q. The mutual solubility of A and B increases with increase in temperature as shown by the curves FC and GD in Fig. 8.21(a). The change in the solubility of liquid A in liquid B with temperature is along FC and that of B in A is along GD. To the right of the curve GD and to the left of the curve CF the mixture exists as homogeneous liquid, if the temperature is below the three-phase temperature. The higher the three-phase temperature the more the mutual solubility of liquids A and B. Suppose that the mixture at M in Fig. 8.21(a) is heated at constant pressure. Its temperature increases till T* is reached. At T*, the vapour of the composition corresponding to point E is formed. If additional quantities of liquid A or B are introduced into the system at this point, the relative amounts of the three phases in equilibrium will change, but the composition of these phases represented by points C, D and E must remain the same. The temperature T* remains constant until one of the liquid phases disappears. At temperatures above T*, the system can exist either as two phases (liquid and vapour) or as a single phase depending on the overall composition. The liquid phase that disappears depends on whether the combined mixture composition is to the left or right of point E. For the mixture at M,

the A-rich phase disappears. The system now consists of a saturated liquid (LB) in equilibrium with saturated vapour (V). When the temperature is increased the overall composition must lie on the vertical line MN, the relative amounts and composition of the liquid and the vapour phases change accordingly. Above the temperature corresponding to point R, the mixture is entirely vapour. If the initial mixture were to the right of point E, the liquid in equilibrium with the vapour would have been LA instead of LB. The three-phase equilibrium temperature increases with pressure as shown in Fig. 8.22 where temperature-composition diagrams are plotted at various pressures. The curves FC and GD eventually merge to a single point and the two liquid phases become identical. The temperature at which this occurs is known as the upper critical solution temperature (UCST). For pressures above this critical condition, the three-phase equilibrium conditions do not exist.

8.15.2 Immiscible Systems The phase diagram for a completely immiscible system is shown in Fig. 8.23. This diagram is a special case of Fig. 8.21 occurring when the two liquid phases LA and LB are the pure liquids A and B respectively. Consider a binary mixture of two immiscible liquids having an overall composition represented by point M. When this mixture is heated its temperature increases along the line MN. For all temperatures up to point N, each component exerts its full vapour pressure as its partial pressure in the vapour phase. When the temperature T* corresponding to point N is reached, the sum of the vapour pressures becomes equal to the surrounding pressure and the system cannot exist entirely in the liquid state and vapour is produced. T* is therefore known as the three-phase temperature.

The composition of the vapour in equilibrium with the pure liquids at the three-phase temperature is given by point E. Since there are three phases and two components present, according to the phase rule, the number of degrees of freedom is one as in the case of partially miscible systems. It means that, when pressure is fixed the system is completely defined. The three-phase temperature and the equilibrium vapour composition get automatically fixed by specifying the pressure in such a way that the sum of the vapour pressures equals the surrounding pressure.

On further addition of heat, the temperature remains constant at T* and more vapour of the same composition as given by point E is formed. This continues till one of the components disappears from the liquid and the system becomes a two-phase mixture either LA – V or LB – V depending upon the initial composition. Now let us consider the cooling of a vapour of initial composition and temperature indicated by point J. When the temperature is lowered to that corresponding to point K the vapour pressure of pure liquid B will be equal to the partial pressure of B in the vapour. Pure liquid B gets condensed and the vapour composition changes along the line KE. When the temperature T* is reached, the partial pressure of A in the vapour will be equal to the vapour pressure of A and at this condition, pure liquids A and B and the vapour are present in equilibrium. Further cooling results in the elimination of the vapour phase and the system now consists of two immiscible liquids. If the initial vapour were at point P, pure liquid A would have condensed out first, instead of B. EXAMPLE 8.31 A high boiling organic liquid is purified from non-volatile impurities by allowing it to mix with steam directly at a total pressure of 93.30 kPa. The vapour pressure data are given as follows:

Temperature, K Vapour pressure of water, kPa

353 47.98

373 101.3

Vapour pressure of liquid, kPa

2.67

5.33

Assume that water and the organic liquid are immiscible and the impurities do not affect the vaporisation characteristics. The vapour pressures vary linearly with temperature. Calculate under three-phase equilibrium (a) The equilibrium temperature and (b) The composition of the resulting vapour. Solution (a) At 353 K, sum of the vapour pressures is 50.65 kPa and at 373 K it is 106.63 kPa. Since the vapour pressures vary linearly, the temperature at which the sum of vapour pressures is 93.3 kPa is obtained by interpolation.

(b) At 368.2 K, the vapour pressure of water is 88.50 kPa and that of the liquid is 4.80 kPa. Since at three-phase equilibrium, the partial pressure is equal to the vapour pressure, the ratio of mole fractions of the components will be same as the ratio of vapour pressures. Let y be the mole fraction of water in the vapour. Then

The vapour contains 94.86% (mol) water vapour. EXAMPLE 8.32 Assuming that benzene is immiscible with water, prepare a temperaturecomposition diagram for benzene (1)–water (2) system at 101.3 kPa using the following vapour pressure data: T, K

323

333

343

348

353

363

373

, kPa

12.40

19.86

31.06

37.99

47.32

70.11

101.3

, kPa

35.85

51.85

72.91

85.31

100.50

135.42

179.14

The boiling point of pure benzene at 101.3 kPa is 353.1 K. Solution The three-phase temperature is first found out. At T*,

is calculated

for each given temperature, and this is plotted against temperature. T* is the temperature at which P is equal to 101.3 kPa. This is found out to be 342 K. The horizontal line CD in Fig. 8.24 is drawn at this temperature. The vapour pressures at this temperature are

=

71.18 kPa and = 30.12 kPa. The mole fraction of benzene in the vapour represented by point E in Fig. 8.24 is 71.18/101.3 = 0.70.

The dew-point curve BE is plotted by choosing a temperature lying between 373 K (boiling point of water) and 342 K (the three-phase temperature). For example, take T = 353 K. The partial pressure of water at the dew point is equal to the vapour pressure. For the dew-point temperature of 353 K, the partial pressure of water is 47.32 = (1 – y) P where y is the mole fraction of benzene in the vapour. We get, y = 0.5329. This calculation is repeated for various temperature and the entire curve BE is drawn. For getting the curve AE the procedure is the same. Here temperatures are assumed between 342 K and 353.1 K, the latter being the boiling point of pure benzene. On the curve AE, the partial pressure of benzene in the vapour equals its vapour pressure. For example, for a dew-point temperature of 348 K, 85.31 = P y, or y = 0.8422 The following table gives the results of a few such calculations: T, K

342

348

353

353.1

363

373

y (curve AE)

0.70

0.84

0.99

1.00

–

–

y (curve BE)

0.70

–

0.53

–

0.31

0

8.16 LIQUID–LIQUID EQUILIBRIUM DIAGRAMS 8.16.1 Binary Liquid–Liquid Equilibria When two liquids are only partially miscible, the equilibrium can be represented on rectangular coordinates as shown in Fig. 8.25. A dome-shaped region is formed by the mutual solubility curves and within the dome the mixture exists as two phases. The compositions of the equilibrium

phases lie at the ends of the horizontal line at a given temperature. For example, the mixture M in Fig. 8.25 will separate into two equilibrium phases A and B. The relative amounts of the phases are given by the inverse lever rule.

The point P gives the critical solution temperature. Outside the dome the mixture is homogeneous.

8.16.2 Ternary Equilibrium Diagrams Liquid–liquid equilibria involving three components are important in the analysis of extraction operations. The extraction process involves bringing a binary mixture of components A and C into intimate contact with a solvent B. The solvent B is either partially soluble in liquid A or is immiscible with it. The component C gets distributed in different proportions between the two insoluble phases known as the ‘raffinate’ and the ‘extract’. The A-rich phase is known as the raffinate and the B-rich phase is known as the extract. When the solvent added is only partially miscible with A, the extract and raffinate phases contain three components. The ternary liquid–liquid equilibrium diagrams are usually represented on equilateral triangular coordinates. On the equilateral triangle the length of the altitude is allowed to represent 100% composition and the length of the perpendiculars from any point to the bases represent the percentages of the three components. The apexes of the triangle represent the pure componen ts A, B, and C and points on the sides represent binary mixtures. Figure 8.26(a) shows the equilibrium diagram of type-I systems in which one pair is partially soluble. The pairs A-C and B-C are miscible in all proportions and the pair A-B is miscible only partially. Examples are water (A)–chloroform (B)–acetone (C), water (A)–benzene (B)–acetic acid (C), water (A)–methyl isobutyl ketone (B)–acetone (C), etc. Liquid C dissolves completely in A and B whereas A

and B dissolve only to a limited extent in each other. In the region below the mutual solubility curves the two liquid phases exist under equilibrium. The compositions of the equilibrium phases are obtained at the ends of the tie line passing through the point representing the overall composition of the mixture. For example, the mixture, whose combined composition is represented by point M separates into a raffinate R and an extract E at equilibrium. Thus, RE is a tie line for the system. The weight fraction of C in the raffinate is denoted by xR, and that in the extract is denoted by yE. Several tie lines can be drawn and each gives rise to a set of equilibrium xR and yE values, which can be used to plot the equilibrium diagram shown in Fig. 8.26(b).

The curve DRPF is the binodal solubility curve, which shows the change in the solubility of A-rich and B-rich phases upon addition of C at a fixed temperature. Any mixture outside this curve will be a homogeneous solution of a single liquid phase. There is one point on the binodal curve P, which will represent the last of the tie lines where the A-rich and B-rich phases become identical. It is known as the plait point. With increase in temperature the mutual solubilities of A and B increase and as a result the heterogeneous area shrinks. Above the critical solution temperature of the binary A-B they dissolve completely and the heterogeneous area vanishes completely. Extraction is not possible under this condition. The ternary equilibrium diagram for type II systems is shown in Fig. 8.27. In this type of systems, two pairs are partially soluble. Examples of type II systems are chlorobenzene (A)–water (B)–Methylethyl ketone (C) , n-heptane (A)–aniline (B)–methylcyclohexane(C), etc. Here A and C are completely miscible while A-B and B-C pairs show only limited solubility. Points D and F represent the mutual solubility of A and B and points H and G those of B and C at the prevailing temperature. Curves DRH and FEG are the ternary solubility curves. Mixtures such as at M inside the heterogeneous area form two liquid phases in equilibrium at E and R. As temperature is increased the mutual solubilities increase and above the critical solution temperature of the binary pair B-C, the system becomes identical to type I system.

SUMMARY The phase equilibrium thermodynamics is of fundamental importance in chemical engineering, because, majority of chemical process industries employ transfer of mass between phases either during the preparation of the raw materials or during the purification of the finished products. The major thrust of the present chapter was the development of the relationship between the various properties of the system such as pressure, temperature and composition when a state of equilibrium was attained between the different phases constituting the system. For a system to be in mechanical equilibrium, the pressure and temperature should be uniform throughout the system. Since, the uniformity of temperature and pressure do not eliminate the possibility of transfer of mass between the phases, to describe the state of thermodynamic equilibrium, additional criteria are developed (Section 8.1). They are: dSU, V ≥ 0, dAT, V

0, dGT, P

0

Since, most chemical reactions and physical changes are carried out at constant T and P, the last criterion formed the basis for phase equilibrium calculations. This criterion of equilibrium also led to the criterion of stability as given by Eq. (8.12). The criterion of stability requires that at constant temperature and pressure the free energy change on mixing DG, its first and second derivatives are all continuous functions of the concentration x, and the second derivative should be positive. For single-component systems in thermodynamic equilibrium under a given temperature and pressure, the molar free energy should be the same in each phase (Section 8.3). Its logical extension to multicomponent multi-phase systems reveals that if a system consisting of several components distributed between various phases is in thermodynamic equilibrium at a definite temperature and pressure, the chemical potential of each component will be the same in all the phases. Since absolute values of fugacities are known, it was found convenient to use fugacities in phase equilibrium calculations, rather than the chemical potentials. Accordingly, the general criterion of phase

equilibrium was expressed as the equality of fugacities [Eq. 8.36]. The Gibbs Phase rule follows from the criterion of equilibrium (Section 8.5). The phase rule allows us to determine the number of independent variables that must be arbitrarily fixed so as to establish uniquely the intensive state of the system. The Duhem’s theorem helps in establishing the extensive state of the system (Section 8.6). Vapour–liquid equilibrium problems essentially involve the calculation of the composition of the liquid and vapour phases such that the fugacities of the components are the same in both phases. To evaluate quantitatively the equilibrium compositions, the fugacity need be expressed in terms of the mole fractions in the mixture. The fundamental relationship for a general VLE problem was derived [Eq. (8.45)] and the various possible simplifications were described (Sections 8.7–8.12). For evaluating the liquid phase fugacity, the activity coefficients should be known as a function of the composition. Several equations were used for estimating the activity coefficients as function of composition of the liquid. The Wohl’s equations, the Margules equations and the van Laar equations, the local composition models for activity coefficients such as the Wilson equations, the NRTL equations and the UNIQUAC equations, and the UNIFAC group contribution model are some of the widely used activity coefficient equations. Thermodynamics provides tests for consistency of experimental VLE data (Section 8.13). Almost all these tests are based on the Gibbs–Duhem equations, the Redlich–Kister method [Eq. (8.97)] being the most reliable among them. The discussion on the vapour–liquid equilibrium for systems of limited miscibility (Section 8.15) and the liquid–liquid equilibrium (Section 8.16) would be helpful for the analysis of many important separation processes in chemical engineering.

REVIEW QUESTIONS 1. How would you state the criterion of equilibrium in terms of the entropy, the work function and the Gibbs free energy? 2. What do you know about the free energy change of mixing and its partial derivatives, for stable liquid phases? 3. Show that for equilibrium between phases of a pure substance, the fugacities in both phases should be equal. 4. How do you obtain the Clapeyron equation from the criterion of phase equilibrium? What simplifications are used in the derivation of the Clausius–Clapeyron equation? 5. For a heterogeneous multicomponent system, what is the general criterion of phase equilibrium? 6. What do you understand by the number of degrees of freedom? How is it determined using the phase rule for a non-reacting system? 7. State the Duhem’s theorem. What is its significance in establishing the state of the system? 8. What are the available degrees of freedom in the following non-reactive equilibrium systems? (a) Two partially miscible liquid phases, each containing the same three liquid phases. (b) A vapour phase containing ammonia in air and a liquid phase containing ammonia in water at a specified temperature. (c) A mixture of benzene and toluene undergoing a simple distillation operation. 9. Write down the equation for solving a general VLE problem. How does this equation get simplified for (a) ideal gas phase, ideal liquid phase and (b) low-pressure equilibrium?

10. What is Poynting correction? 11. Distinguish between the bubble-point and dew-point temperatures. 12. What is meant by a ‘tie line’? How does the tie line help in determining the amount of liquid and vapour in equilibrium? 13. Why does the boiling point diagram at a higher pressure lie above that at a lower pressure? 14. What are the salient features of an ideal liquid solution? How does the total pressure over an ideal solution vary with composition? 15. How would you calculate the constant pressure y-x data of a binary mixture using an average value of the relative volatility? 16. Component 1 in a binary non-ideal solution is found to obey the Raoult’s law over a certain concentration range. What do you know about the behaviour of component 2 over the same range? 17. What do you mean by positive and negative deviation from ideality? “A solution exhibiting positive deviation from ideality is formed accompanied by an absorption of heat and a solution exhibiting negative deviation from ideal behaviour is formed accompanied by an evolution of heat”. Explain. 18. What are azeotropes? With proper phase diagrams, distinguish between minimum and maximum boiling azeotropes. What is the effect of pressure on the azeotropic composition? 19. Discuss the suitability of different activity coefficient equations for VLE data correlation. 20. What is vaporisation equilibrium constant? How do you estimate the bubble-point temperature and the bubble-point pressure of a multicomponent system? 21. A multicomponent liquid mixture of known composition is flash vaporised at a given pressure and temperature. How would you estimate the fraction of the liquid vaporised? 22. How are the Gibbs–Duhem equations helpful in testing the consistency of the VLE data? 23. What is the zero area method for testing the consistency of VLE data? 24. What is coexistence equation? What are its major applications? 25. The activity coefficients of one of the components in a binary solution are known as function of concentration. How would you evaluate the activity coefficients of the other component as a function of composition? 26. What are the critical solution temperature and the three-phase temperature with reference to partially miscible liquid systems? 27. Why does immiscibility occur in liquid solutions? 28. How would you estimate the composition of the vapour phase in equilibrium with two immiscible liquid phases?

EXERCISES 8.1 Show that the following equations provide the criteria of equilibrium under certain constraints. (a) dUS,V = 0 (b) dSH,P = 0 (c) dHS,P = 0 8.2 For each of the following non-reactive equilibrium systems, determine the number of available

degrees of freedom. (a) Two miscible materials in vapour–liquid equilibrium with vapour composition specified at a given temperature and pressure. (b) A mixture of methane and air in contact with a solid adsorbent at atmospheric pressure and a specified temperature. (c) Liquid water in equilibrium with a mixture of water vapour and nitrogen. (d) Two partially miscible liquid phases and a vapour phase in equilibrium with them at a constant pressure. (e) A liquid mixture of benzene and toluene in equilibrium with its vapour at 1 bar. (f) A vapour phase consisting of ammonia and air and a liquid phase consisting of ammonia and water at a given temperature. (g) A liquid mixture of components A and C in equilibrium with a liquid solvent B in which only C is soluble at a given temperature and pressure. 8.3 Determine the mole fraction of methane, xi, dissolved in a light oil at 200 K and 20 bar. Henry’s law is valid for the liquid phase, and the gas phase may be assumed to be an ideal solution. At these conditions, Henry’s law constant for methane in oil = 200 bar, fugacity coefficient of pure methane gas = 0.90 and mole fraction of methane in the gas phase, y1 = 0.95. 8.4 The vapour pressures of benzene and toluene are given below.

Calculate the equilibrium data for the system at 101.3 kPa and formulate an equation for the equilibrium diagram in terms of average relative volatility. 8.5 At 303 K, the vapour pressures of benzene (A) and toluene (B) are 15.75 kPa and 4.89 kPa respectively. Determine the partial pressures and weight composition of the vapour in equilibrium with a liquid mixture consisting of equal weights of the two components. 8.6 An equimolar mixture of benzene and toluene is contained in a piston/cylinder arrangement at a temperature T. What is the maximum pressure below which the mixture exists as a vapour phase alone? At the given T, the vapour pressures of benzene and toluene are 203.9 kPa and 85.3 kPa, respectively. Assume that Raoult’s law is valid. 8.7 Two substances A and B are known to form ideal liquid solutions. A vapour mixture containing 50% (mol) A and 50% (mol) B is at 311 K and 101.3 kPa. This mixture is compressed isothermally until condensation occurs. At what pressure does condensation occur and what is the composition of the liquid that forms? The vapour pressures of A and B are 142 kPa and 122 kPa respectively. 8.8 Air is cooled to 80 K at 101.3 kPa. Calculate the composition of the liquid and vapour phases at this condition assuming that the mixture behaves ideally. The vapour pressure of nitrogen and oxygen at 80 K are 135.74 kPa and 30.04 kPa respectively. 8.9 The binary system, acetone (1)–acetonitrile (2) conforms closely to Raoult’s law. Using the

vapour pressure data given below plot the following (a) P-x1 and P-y1 curves at 323 K (b) T-x1 and T-y1 curves at 53.32 kPa

8.10 Assuming Raoult’s law to be valid for the system benzene (1)–ethyl benzene (2) and the vapour pressures are given by the Antoine equations

where P is in kPa and T is in K. Construct the following: (a) The P-x-y diagram at 373 K (b) The T-x-y diagram at 101.3 kPa. 8.11 A liquid mixture containing 65% (mol) benzene and 35% (mol) toluene is subjected to flash vaporisation at 363 K and 101.3 kPa. The vapour pressure of benzene at this temperature is 136.09 kPa and the vapour pressure of toluene is 54.21 kPa. Flash vaporisation is essentially an equilibrium stage operation. Calculate (a) The exit vapour composition (b) The exit liquid composition (c) The mole per cent of the feed that is vaporised. 8.12 For the system n-pentane (1)–n-heptane (2), the vapour pressures are given by the Antoine equation

Assuming that the solution formed is ideal, calculate: (a) The composition of the liquid and vapour in equilibrium at 95 kPa and 336.2 K. (b) The composition of the vapour in equilibrium with a liquid containing 34% (mol) pentane and the equilibrium temperature at P = 95 kPa.

(c) The total pressure and the vapour composition in equilibrium with a liquid of composition x1 = 0.44 at T = 333.2 K. 8.13 Using Gibbs–Duhem equations prove that if one constituent of a mixture exhibits positive deviation from ideal behaviour the other constituent also shows positive deviation. 8.14 Prove that a solution exhibiting negative deviation from ideal behaviour is formed with an evolution of heat. 8.15 Using van Laar method calculate the vapour–liquid equilibrium compositions for acetone (1)– chloroform (2) system at a pressure of 101.3 kPa. At this pressure, the system forms an azeotrope of composition 66.6% (mol) chloroform which boils at 337.7 K. The vapour pressures of the pure components are given below.

The normal boiling points of acetone and chloroform are respectively 329.5 K and 334.1 K. (Hint: The ratio of vapour pressures remains almost constant. Use the method employed in Example 8.13 for calculating y for arbitrarily chosen x values.) 8.16 Show that the van Laar equation and Margules equation are consistent with the Gibbs–Duhem equations. 8.17 The toluene–acetic acid mixture forms an azeotrope containing 62.7% (mol) toluene and having a minimum boiling point of 378.6 K at 101.3 kPa. The vapour pressure data are:

The normal boiling point of toluene and acetic acid are respectively 383.9 K and 391.7 K. (a) Calculate the van Laar constants A and B (b) Plot ln g1 and ln g2 as ordinate against mole fraction of toluene. 8.18 Under atmospheric pressure, the acetone–chloroform azeotrope boils at 337.8 K and contains 33.5% (mol) acetone. The vapour pressures of acetone and chloroform at 337.8 K are respectively 132.62 kPa and 113.96 kPa. (a) Calculate the composition of the vapour in equilibrium with a liquid analysing 11.1% (mol) acetone. How does it compare with the experimental value of 6.5% acetone in the vapour? (b) What is the total pressure at this condition? 8.19 Ethyl alcohol and hexane form an azeotrope at 33.2% (mol) ethanol. It boils at 331.9 K at 101.3 kPa. At 331.9 K, the vapour pressures are 44.25 kPa for ethanol and 72.24 kPa for hexane. Determine: (a) The van Laar constants (b) The vapour composition for a solution containing 50% (mol) hexane boiling at 331.9 K

(c) The total pressure for the conditions in part (b). 8.20 At atmospheric pressure, ethyl acetate and ethyl alcohol form an azeotrope containing 53.9% (mol) acetate boiling at 345 K. Determine: (a) The van Laar constants (b) The azeotropic composition and the total pressure if the mixture forms an azeotrope boiling at 329.5 K (c) The composition of the vapour in equilibrium with a liquid of composition 60% (mol) alcohol and boiling at 329.5 K

8.21 An organic liquid solution containing A (molecular weight 46) and B (molecular weight 78) form an azeotrope containing 52% by weight A at 333 K and 101.3 kPa. Vapour pressures of A and B are 69.31 kPa and 68 kPa respectively. Determine the van Laar constants. 8.22 For the acetone (1)–diethylamine (2) system the activity coefficients values as function of concentration are given below:

Using the above data estimate the van Laar constants for the system. (Hint: A = ln g1 as x1 0 and B = ln g2 as x2 0) 8.23 Find the van Laar constants for the binary system benzene (1)–ethanol (2) using the following data

8.24 The T-x-y data for the system metaxylene (1)–propionic acid (2) at 101.3 kPa is given below:

Does the system form an azeotrope? Give reasons. 8.25 For the conditions in Example 8.19, calculate (a) the equilibrium temperature and vapour composition for x1 = 0.32 and P = 101.3 kPa and (b) the equilibrium temperature and liquid composition for y1 = 0.57 and P = 101.3 kPa .

8.26 For isobutanol–water system, it is found from VLE measurements that the composition of the vapour and liquid in equilibrium at 101.3 kPa and 364.7 K are 28.6% (mol) and 1.4% (mol) i-butanol respectively. Vapour pressure of i-butanol at this temperature is 53.32 kPa. The activity coefficient of water may be taken as 1.0012. Compute the van Laar constants. 8.27 Ethanol–water mixture forms an azeotrope boiling at 351.4 K under a pressure of 101.3 kPa and its composition is 89.4% (mol) ethanol. The vapour pressures of ethanol and water at 351.4 K are 100 kPa and 44 kPa respectively. Using van Laar method and assuming that the ratio of vapour pressures remains constant calculate the composition of the vapour in equilibrium with a liquid containing 80% ethanol. 8.28 Two liquids P and Q form an azeotrope containing 58% (mol) P at 101.3 kPa pressure. At the azeotropic temperature the vapour pressures of P and Q are 200 kPa, 125.3 kPa respectively. Construct the equilibrium (y–x) diagram. List all the assumptions made. 8.29 For the binary system n-pentanol (1)–n-hexane (2), determine the activity coefficients at 313 K in an equimolal mixture. The Wilson parameters are given as follows: a12 = 7194.18 J/mol,……a21 = 697.52 J/mol V1 = 109.2 10–6 m3/mol,……V2 = 132.5 10–6 m3/mol 8.30 The Wilson parameters for acetone (1)–methanol (2) are obtained from the following values: a12 = – 712.51 J/mol,……a21 = 2487.71 J/mol V1 = 74.05 10–6 m3/mol,……V2 = 40.73 10–6 m3/mol The vapour pressures are given by the Antoine equations:

Calculate the total pressure and vapour composition in equilibrium with a liquid containing 31% (mol) acetone at 333 K. 8.31 For the 2-propanol (1)–water (2) system, the following Wilson parameters are reported. a12 = 1833.74 J/mol,……a21 = 5183.26 J/mol V1 = 76.92 10–6 m3/mol,……V2 = 18.07 10–6m3/mol The vapour pressures can be calculated by the Antoine equations, which are given below:

where T is in K and the vapour pressures are in kPa. Calculate: (a) Equilibrium pressure and vapour composition at T = 353.15 K and x1 = 0.25

(b) Equilibrium pressure and liquid composition for T = 353.15 K and y1 = 0.60 (c) Equilibrium temperature and vapour composition for P = 101.3 kPa and x1 = 0.85 (d) Equilibrium temperature and liquid composition for P = 101.3 kPa and y1 = 0.40. 8.32 A solution of hydrocarbons contains n-propane 5.0%, n-butane 30.0%, n-pentane 40.0% and n-hexane 25.0%. Compute the bubble point and the dew point at 350 kPa. The K-values can be taken from the DePriester nomographs. 8.33 A solution has the following composition in mol per cent: ethane 0.25%, propane 25.00%, isobutane 18.5%, n-butane 56.0% and isopentane 0.25%. For a pressure of 10 bars, calculate (a) The bubble point (b) The dew point (c) The composition of the liquid and vapour when 40% of the mixture is vaporised. 8.34 A stream of gas in a natural gasoline plant has the following composition by volume: ethane 10%, propane 14%, isobutane 19%, n-butane 54% and isopentane 3%. (a) Calculate the pressure necessary to condense this gas completely at 311 K. (b) For a condenser operating at the pressure in part (a), calculate the temperature at which condensation starts and the temperature at which 50% (mol) of the vapour gets condensed. Also, calculate the composition of the first liquid to condense and the composition of the liquid and vapour phases at 50% condensation. 8.35 Determine the composition of the vapour in equilibrium with the liquid and the pressure of the system at 313 K for a liquid mixture of 5% (mol) methane, 10% (mol) ethane, 30% (mol) propane, 25% (mol) isobutane, and 30% (mol) n-butane. Determine the pressure and composition of the liquid in equilibrium with a vapour mixture of the above composition. 8.36 Calculate the pressure at which condensation starts and the pressure at which condensation is complete when a vapour mixture of the following composition is subjected to condensation at a temperature of 300 K: 20% (mol) ethylene, 20% (mol) ethane, 40% (mol) propane and 20% (mol) n-butane. 8.37 A vapour mixture containing 15% ethane, 20% propane, 60% isobutane and the rest n-butane is subjected to partial condensation so that 75% of the vapour is condensed. If the condenser temperature is 300 K determine the pressure. 8.38 An equimolal mixture of propane (1) and n-butane (2) is partially condensed so that 50% (mol) of the mixture is in the liquid state at 311 K. Using the DePriester nomograph determine the following: (a) The pressure (b) The vapour and liquid compositions (c) The pressure at which condensation begins at the constant temperature of 311K. 8.39 Verify whether the following data are consistent.

8.40 Calculate the constants A and B in the van Laar equation from the following data. Check

whether the data are consistent.

8.41 The following data were reported for vapour–liquid equilibrium for ethanol–water system at 298 K. Test whether the data are thermodynamically consistent.

8.42 The following vapour-liquid equilibrium data were obtained for water (1)–nitric acid (2) system at 293 K.

Test the above data for thermodynamic consistency. 8.43 From the data activity coefficient versus mole fraction for the system acetone (1)– dichloroethylene (2) given below check their thermodynamic consistency.

8.44 The partial pressure of ether at 303 K for the ether (1)–acetone (2) system is given as follows:

The vapour pressure of pure acetone at 303 K is 37.72 kPa. (a) Calculate the activity coefficient of ether and plot the logarithm of the activity coefficient as

function of concentration. (b) Predict the partial pressure of acetone in the corresponding solution for which the activity coefficients of ether have been determined. 8.45 Vapour–liquid equilibrium data for the system methanol (1)–benzene (2) at 313 K are given below:

Use the area test to determine the thermodynamic consistency of the data. 8.46 The activity coefficient of thallium in amalgams at 293 K are given below:

Determine the activity coefficient of mercury as function of composition. 8.47 The following data refers to the system water (1)–n-propyl alcohol (2) at 298 K.

What is the activity coefficient of water in a 10% (mol) n-propyl alcohol solution? 8.48 At 323 K, the vapour pressures of pure ether and pure ethyl alcohol are 170.13 and 29.47 kPa respectively. The total pressures versus liquid composition data are given below:

Using Gibbs–Duhem equation compute from these data the partial pressures of ether and alcohol over liquid solutions of various compositions at 323 K. 8.49 The data given below refer to the boiling points of ethanol (1)–benzene (2) system at 100 kPa and the vapour pressures of pure ethanol and benzene at these temperatures.

Calculate the van Laar constants from these data assuming g to be independent of temperature. Also, find g1 and g2 from the van Laar equations. 8.50 The total pressure versus solution concentration data for the system dioxane (1)–water (2) at 353 K is given below:

The vapour pressures of pure water and dioxane at this temperature are 47.33 and 51.05 kPa. Calculate: (a) The van Laar constants (b) The constants in the Margules equation (c) The vapour composition in equilibrium with a liquid containing 60% water by weight and the total pressure over this solution using van Laar method. 8.51 Benzene (1)–cyclohexane (2) form an azeotrope at 0.525 mole fraction benzene at a temperature of 350.8 K and 101.3 kPa. At this temperature, the vapour pressure of benzene is 99.3 kPa and that of cyclohexane is 98 kPa. Using the van Laar model estimate the activity coefficients at x1 = 0.2 and 0.9. Using this activity coefficient information calculate the equilibrium pressure and the vapour compositions at 350.8 K for the two liquid compositions. 8.52 The azeotrope of the n-propanol–water system has a composition 56.83% (mol) water with a boiling point of 360.9 K at a pressure of 101.3 kPa. At this temperature, the vapour pressures of water and propanol are respectively 64.25 kPa and 69.71 kPa. Evaluate the activity coefficients for a solution containing 20% water through the van Laar equations. 8.53 The pressure exerted over the binary system ethanol–methylcyclohexane containing 40.5% (mol) ethanol at 308 K is 20.31 kPa. The vapour phase contained 54.7% (mol) ethanol. The vapour pressures at 308 K are 13.74 kPa for ethanol and 9.81 kPa for methylcyclohexane. What are the total pressure and composition of the vapour in equilibrium with a liquid containing 60% (mol) ethanol at 308 K? 8.54 A binary liquid mixture of components A and B containing 80% (mol) A is in equilibrium with a vapour containing 84.3% (mol) A at 101.3 kPa and 339 K. Estimate the pressure and composition of the vapour in equilibrium with a liquid containing 50% A at 339 K. The vapour pressures of A and B at this temperature are 106.6 kPa and 79.97 kPa respectively. 8.55 At 333 K, compounds A and B each has vapour pressures of 106.63 kPa. The mixture of A and B forms an azeotrope at 333 K and 133.29 kPa and has a composition of 50% A. (a) Calculate the equilibrium pressure and vapour composition over a liquid solution containing 25% A. (b) If A and B have equal latent heats of vaporisation, how do you expect the azeotropic

composition to respond to an increase in temperature? 8.56 At 353 K, compounds A and B each has vapour pressures of 93.30 kPa. At this temperature mixtures of A and B form azeotrope containing 50% (mol) A and exerts a pressure of 127.96 kPa. Calculate the equilibrium pressure and vapour composition at 353 K over a liquid solution containing 25% (mol) A. 8.57 For the binary mixture of A and B the activity coefficients are given by

The vapour pressures of A and B at 353 K are 119.96 kPa and 79.97 kPa respectively. Does an azeotrope exist at 353 K? If so, what is the azeotropic pressure and composition for A = 0.6? 8.58 It is proposed to purify benzene from small amounts of non-volatile impurities by subjecting it to distillation with saturated steam at 99.3 kPa. Calculate the temperature at which distillation will proceed and the weight of steam accompanying 1 kg benzene. The vapour pressure data is given in Example 8.28. 8.59 At 383 K, saturated solution of aniline in water contains 7.95% aniline by weight and a saturated solution of water in aniline contains 88.05% aniline by weight. The vapour pressures of pure aniline and of water at 383 K are 9.22 kPa and 143.10 kPa respectively. Construct the Px-y diagram for the mixture at 383 K. 8.60 Construct T-x-y diagram for the ether (1)–water (2) system at 101.3 kPa from the following data.

Assume that Raoult’s law is valid for ether in ether phase and for water in the water phase. 8.61 Dimethylaniline is distilled with steam at 90 kPa to free it from non-volatile impurities. Assuming it to be completely immiscible with water determine (a) The distillation temperature (b) The composition of the vapour produced. The vapour pressure data are following:

Plot of ln PS versus 1/T may be assumed linear. 8.62 A stream contains 30% (mol) toluene, 40% (mol) ethyl benzene and 30% (mol) water. Assuming that mixtures of ethylbenzene and toluene obey Raoult’s law and they are completely immiscible in water, calculate the following for a total pressure of 101.3 kPa: (a) The bubble-point temperature and the composition of the vapour (b) The dew-point temperature and the composition of the liquid. The vapour pressure data are given below:

8.63 n-Heptane (1) and water (2) are essentially immiscible as liquids. A vapour mixture containing 65% (mol) water at 373 K and 101.3 kPa is cooled slowly at constant pressure until condensation is complete. Construct a plot for the process showing temperature versus equilibrium mole fraction of heptane in the residual vapour. For n-heptane,

where P is in kPa and T in K. 8.64 Toluene (1) and water (2) are essentially immiscible in the liquid state. Determine the dewpoint temperature and the composition of the first drops of liquid formed when the vapour mixtures of these species containing (a) 23% (mol) toluene (b) 77% (mol) toluene at 101.3 kPa. What are the bubble-point temperature and the composition of the last drop of vapour in each case? The vapour pressure of toluene is

8.65 Components 1 and 2 are insoluble in the liquid phase. Estimate the dew-point temperature and the compositions of the first drop of liquid formed when vapour mixtures of components 1 and 2 containing (a) 75% (mol) component 1; (b) 25% (mol) component 1 are cooled at constant pressure of 101.33 kPa. The vapour pressures of the pure components in kPa are given against temperature in Kelvin in the following table.

8.66 An experimental determination of vapour–liquid equilibrium state of ether (1) and acetone (2) gave the following results: x1 = 0.3, y1 = 0.42, T = 313 K and P = 105 Pa The saturation vapour pressures of the pure components at 313 K are: ether = 1.21 105 Pa and acetone = 0.56 105 Pa. The vapour phase can be assumed ideal. (a) Calculate the liquid-phase activity coefficients. (b) What is the value of excess Gibbs free energy GE/RT for the liquid phase?

9 Chemical Reaction Equilibria The chemical process industries are concerned with the transformation of raw materials into useful products. Such transformation in most cases is achieved by means of chemical reactions. The design and operation of reaction equipment are therefore quite an important field in the chemical engineering profession. To be successful in this profession, the chemical engineer should be versatile with the thermodynamics and kinetics of chemical reactions. Thermodynamics predicts the equilibrium conversion that would be achieved in a chemical reaction and also the effect of operating conditions on it, whereas the kinetics deals with the rate or speed with which the desired conversion is attained in practice. Thermodynamic analysis can also give information about the feasibility of chemical reactions. The progress and extent of a chemical reaction are affected by changes in the reaction conditions like temperature, pressure, composition of the reactants, etc. For example, in the synthesis of methanol from carbon monoxide and hydrogen, the equilibrium conversion as well as the rate of reaction are affected by changing the pressure, temperature or the relative amounts of carbon monoxide and hydrogen in the reactant stream. The influence of these controllable variables on the thermodynamics of reaction, or to be specific, on the equilibrium conversion, in some situation may be in conflict with the influence of these variables on the kinetics of the reaction. This can be illustrated by considering the effect of temperature on the oxidation of sulphur dioxide to sulphur trioxide. The rate of this reaction increases with temperature and from the point of view of rate alone it is better to operate the reactor at as high a temperature as permissible. However, the equilibrium conversion to sulphur trioxide falls off sharply with increase in temperature. The conversion is above 90% at temperatures near 800 K, but it is only 50% at 950 K. It is clear that both the kinetics (the rate) and thermodynamics (the equilibrium) of the reaction must be considered in the choice of reaction conditions in the commercial process for any chemical reaction. The purpose of the present chapter is to identify the role of thermodynamics in the design and operation of chemical reaction systems. Equilibrium conversion of a reaction sets a limit and provides a goal by which we measure improvement in the process. It is impossible at a given set of conditions to attain a conversion that is better than the equilibrium value calculated from thermodynamic principles. Even if this conversion is not attainable in practice within a reasonable time, its knowledge is valuable because it represents the best that can be expected from the reaction. It tells us whether or not an experimental investigation of a proposed new process is worthwhile. There is no point in trying improvement in the process by improving the rate by introducing suitable catalysts, if thermodynamics predicts an equilibrium yield, of say, only 20% whereas a 50% yield is necessary for the process to be economically viable. The choice of an appropriate catalyst may give a better reaction rate, but it will not alter the equilibrium yield of the product. The emphasis in this chapter is on determining the conversion at equilibrium and on predicting the effect of controllable variables like temperature and pressure on the conversion.

9.1 REACTION STOICHIOMETRY

The generalised representation of a chemical reaction is given by

where A is the chemical symbol for the various species taking part in the reaction and n is the stoichiometric number. Consider the reaction 2A + 3B

L + 2M

This is a special case of the general form of Eq. (9.1), with nL = 1, nM = 2, nA = – 2, and nB = – 3. In the general form, this reaction may be represented as 0 = L + 2M – 2A – 3B The stoichiometric numbers are positive for products, negative for reactants and zero for inert species. The changes in the number of moles of various species taking part in the reaction are in direct proportion to their stoichiometric numbers. Let Dni denote the change in the number of moles of component i due to the reaction. For one mole of A disappearing in the reaction DnA = –1, DnB = –1.5, DnL = 0.5 and DnM = 1. We see that

For the thermodynamic analysis of chemical reactions the concept of ‘extent of reaction,’ also called ‘reaction coordinate’ is useful. It is denoted by e. The reaction coordinate measures the progress of a reaction and is defined as the degree to which a reaction has advanced. It has the advantage that the change in the extent of reaction de is the same for each component, whereas the changes in the number of moles are different for different species taking part in the reaction. The extent of reaction and the number of moles taking part in the reaction are related as

For the initial state of the system, that is, before the reaction, the value of e is zero. EXAMPLE 9.1 Derive the relationship between the mole fraction of the components taking part in the reaction and the extent of the reaction. Solution Let ni0 be the number of moles of the species initially present in the system and ni the number of moles present after the reaction. Then ni = ni0 + Dni where Dni is the change in the number of moles of i due to the reaction. Integration of Eq. (9.2) yields

EXAMPLE 9.2 A gas mixture containing 2 moles nitrogen, 7 moles hydrogen and 1 mole ammonia initially, is undergoing the following reaction: N2 + 3H2 2NH3 (a) Derive expressions for the mole fractions of various components in the reaction mixture in terms of the extent of reaction. (b) Explain how the conversion of limiting reactant is related to the extent of reaction. Solution (a) Equations (9.3) and (9.4) relate the mole fraction of various constituents in the system to the extent of reaction.

(b) The limiting reactant here is nitrogen. Let the fractional conversion of nitrogen be z. Then Moles of nitrogen in the reaction mixture is = Moles of nitrogen in the mixture in terms of the extent of reaction is = Comparing the two results, we see that

EXAMPLE 9.3 Derive the relationship between mole fraction of species in multiple reactions and the extent of reactions. Solution When two or more reactions occur simultaneously, the number of moles of each component changes because of several reactions. Equation (9.2) can be modified as

Let , the sum of the stoichiometric numbers in the jth reaction. Then the above equation can be written as

EXAMPLE 9.4 A gas mixture containing 3 mol CO2, 5 mol H2 and 1 mol water is undergoing the following reactions: CO2 + 3H2

CH3OH + H2O

CO2 + H2 CO + H2O Develop expressions for the mole fraction of the species in terms of the extent of reaction. Solution The total moles initially present, n0 = 3 + 5 + 1 = 9 For the first reaction, n1 = – 1 – 3 + 1 + 1 = – 2 For the second reaction, n2 = – 1 – 1 + 1 + 1 = 0 The mole fractions are calculated using Eq. (9.9)

9.2 CRITERIA OF CHEMICAL REACTION EQUILIBRIUM We have developed the criteria of phase equilibrium in Chapter 8. At constant temperature and pressure, the transfer of materials from one phase to another under equilibrium is found to occur with no change in the free energy. Stated mathematically, ………(9.10) Here is the total Gibbs free energy of the system at constant temperature and pressure. This criterion is quite general and is not restricted to physical transformations. When a chemical reaction occurs at equilibrium there is no change in the Gibbs free energy of the system, provided the change is taking place at constant temperature and pressure. Consider a closed system in which a chemical reaction represented by the following general equation has been allowed to reach a state of equilibrium at a given temperature and pressure. aA + bB lL + mM………(9.11) An infinitesimal change is allowed to occur in the system whereby the number of moles of various species change. The increments in the number of moles are dnA, dnB, dnL and dnM for components

A, B, L and M respectively. The free energy change for the process occurring at constant temperature and pressure is given by [see Eq. (7.36)]

where mi is the chemical potential of component i. For the reaction under consideration, Eq. (9.12) takes the form

where – a, – b, l and m are the stoichiometric numbers which are positive for products and negative for the reactants and e is the extent of reaction. In general, for an infinitesimal change in a reacting system, we can write Eq. (9.13) as

The left-hand side of Eq. (9.16) is the free energy change DG accompanying the complete reaction under equilibrium conditions. Hence, DG = 0 under equilibrium. The physical significance of the criterion of chemical equilibrium can now be examined. Consider a simple chemical reaction equilibrium: A D B. Let the extent of the reaction be e. The change in the number of moles of A = – de and the change in the number of moles of B = de. The change in free energy at constant temperature and pressure is found out by Eq. (9.14) dGt = (mB – mA) de………(9.17) This equation can be written in the following form.

Equation (9.18) gives the slope of the curve obtained when the Gibbs free energy is plotted against

extent of reaction as in Fig. 9.1.

The slopes given by Eq. (9.18) are not constant because the chemical potentials are functions of composition, which varies as the extent of reaction changes. Since the reaction proceeds in the direction of decreasing Gibbs free energy G, the forward reaction (A B) takes place if mA > mB and the backward reaction (A B) proceeds if mA < mB. When mA = mB, the slope of the curve is zero. This occurs at the minimum of the curve and corresponds to the position of chemical equilibrium. The composition of the reaction mixture at the point where the Gibbs free energy is the minimum is the equilibrium composition at the specified temperature and pressure. Thus the criterion of equilibrium, Eq. (9.10), means that differential displacement of chemical reaction can occur at the equilibrium state, but without changing the total Gibbs free energy. If the system is not in chemical equilibrium, the reaction occurring must be irreversible and the total Gibbs free energy must decrease at constant temperature and pressure.

9.3 EQUILIBRIUM CONSTANT Consider the chemical reaction given by Eq. (9.11) aA + bB lL + mM The equilibrium constant K or Ka for this reaction is defined in terms of the activities of the reactants and the products as

where ai is the activity of component i in the reaction mixture and ni is the stoichiometric number of i. Activities of the species appearing in Eq. (9.19) are raised to the respective stoichiometric numbers. Since the activity is defined as the ratio of the fugacity of the component in the solution to

the fugacity in the standard state,

Equation (9.19) can also be written as

Denoting

by Kf, we can write Eq. (9.23) as

K = Kf = KfKp………(9.24) This relationship is applicable for gaseous systems. If the gas mixture behaves as an ideal gas, then Kf = 1 and Eq. (9.24) leads to K = Kf = Kp………(9.25) The numerical value of the equilibrium constant depends upon the form of the stoichiometric equation. Consider the decomposition of water vapour into hydrogen and oxygen as represented by the following equation: 2H2O

2H2 + O2

The equilibrium constant K

for this reaction is calculated as

Thus, it is seen that K≤ = (K )1/2. The form of the stoichiometric equation should be specified along with the numerical values of the equilibrium constant.

9.4 EQUILIBRIUM CONSTANT AND STANDARD FREE ENERGY CHANGE The criterion of equilibrium, Eq. (9.15), can be written for the general chemical reaction represented by Eq. (9.11) as (lmL + mmM) – (amA + bmB) = 0………(9.16) The chemical potential of a component in the equilibrium state of the reaction mixture is related to its fugacity in that state as given below

Equation (9.29) can be put into the following form:

The left-hand side gives the standard free energy change DG0, the free energy change accompanying the reaction when each of the reactants and the products is in its standard state. Using the definition of the equilibrium constant [Eq. (9.19)], the above equation is written as

Thus the equilibrium constant is determined by the standard free energy change and the temperature. The standard free energy change depends on the temperature, the specification of standard state for each component and the number of moles involved in the stoichiometric equation under consideration. The numerical values of the equilibrium constant will be of no significance unless accompanied by the specifications for these three factors. However, it is independent of pressure at equilibrium. The effect of the reaction stoichiometry on the equilibrium constant has already been discussed. The choice of standard state is being dealt with in the following section.

9.4.1 Choice of Standard State Though the choice of standard state in Eq. (9.31) is arbitrary and is left to our convenience, certain conventions are followed in this choice. The choice of pure component standard state will be convenient in many situations, as this requires only the specification of temperature and pressure for defining the state completely. The temperature in the standard state is the same as that of the reaction. If the standard state chosen for a substance is a solution, the composition must also be specified. For gases, as has been pointed out earlier, the standard state chosen is the pure component at the temperature of the reaction and at unit fugacity. Fugacity will be unity at 1 bar (or 1 atm) if the gas behaves as an ideal gas at this condition. For ideal gases, therefore, the standard state pressure approaches 1 bar and DG0 can be easily evaluated at this pressure. By this choice, K = Kf and Eq. (9.31) becomes DG0 = – RT ln Kf………(9.32) The standard state of unit fugacity may not be convenient for reactions involving solids, liquids or solutions. By convention, the standard state chosen for solids and liquids is the pure solid or liquid as the case may be, at a pressure of 1 bar (or 1 atm), the temperature being the same as the temperature of the reaction.

9.4.2 Feasibility of a Reaction From the values of standard free energy change, we can formulate an approximate criterion for the feasibility of a chemical reaction, which will be useful in preliminary exploratory work. It would be

worthwhile to have some idea about whether or not the equilibrium is favourable, before we search for catalysts and other conditions necessary to cause the reaction. If the reaction is not thermodynamically feasible, there is no point in pursuing a long and expensive experimental investigation on improving the rate of reaction. Any reaction starting with pure reactants uncontaminated with any of the products will have a tendency to proceed to some extent, though this may be infinitesimally small. It is the value of the equilibrium constant, which, in turn, is related to the standard free energy of the reaction that gives the necessary information on the thermodynamic possibility of the reaction. Even the decomposition of water vapour to hydrogen and oxygen will proceed to some extent under atmospheric temperature and pressure. From the value of the standard free energy change, DG0, the equilibrium constant for the reaction

at 298 K is found to be about 1 10–40. This means that the extent of decomposition of water vapour is infinitesimally small at equilibrium and the reaction is not thermodynamically feasible. I f DG0 for a reaction is zero, then K = 1, the reaction proceeds to a considerable extent before equilibrium is reached. If DG0 is negative, then K > 1, the reaction is quite favourable. But the situation becomes less favourable as DG0 increases in the positive direction. It should be borne in mind that many reactions with positive values of DG0 are certainly feasible from the standpoint of industrial operation. For example, the methanol synthesis reaction with DG0 = 46,200 kJ/kmol at 600 K is found to be feasible. This reaction is carried out at high pressure to overcome the unfavourable free energy change. In short, there is no well-defined demarcation to separate favourable and unfavourable reactions. The following guide may be useful as an approximate criterion for ascertaining the feasibility of chemical reactions: DG0 < 0, the reaction is promising. 0 < DG0 < 40,000 kJ/kmol, the reaction may or may not be possible and needs further study. DG0 > 40,000 kJ/kmol, the reaction is very unfavourable. EXAMPLE 9.5 Device a series of hypothetical steps for carrying out the gas-phase reaction aA + bB lL + mM when the reactants and the products are at their standard state. Show that the free energy changes calculated for these series of steps add up to give the same result as the one provided by Eq. (9.31). Solution The free energy change accompanying the process in which the reactants at their standard state are converted to products also at their standard state may be calculated via any convenient path. Let us assume the following computational path for carrying out the reaction, which is represented in Fig. 9.2.

Step 1: The reactants are initially in their pure form and are at their standard state of unit fugacity and at the temperature of the reaction. Then they are compressed to a fugacity of the reaction mixture at equilibrium. The free energy change for this process is DG1.

Step 2: The pure reactants are introduced to the reaction system through membranes permeable only to single species. Since the fugacities of the components before and after this step are the same, the free energy change DG2 for this process is zero. DG2 = 0. Step 3: The introduction of the reactants disturbs the state of equilibrium prevailing in the reaction system. To bring the system back to the equilibrium condition the forward reaction occurs at the given temperature and pressure. According to the criterion of equilibrium, this reaction proceeds without any change in the free energy of the system. Therefore, DG3 = 0. Step 4: The product gases are separated by means of membranes into pure components at the reaction temperature and pressure. As in step 2, the free energy change in this process is zero. That is, DG4 = 0. Step 5: The pure components with fugacities equal to . The free energy change for this step,

are expanded to standard state fugacities

which is same as Eq. (9.31).

9.5 EFFECT OF TEMPERATURE ON EQUILIBRIUM CONSTANT The effect of the operating variables on equilibrium can be qualitatively explained by means of the Le Chatelier’s principle, which states that a system at equilibrium when subjected to a disturbance, responds in a way that tends to minimise the effect of that disturbance. An increase in temperature will shift the equilibrium state in the direction of absorption of heat. That is, the equilibrium will shift in the endothermic direction if the temperature is raised, for then, energy is absorbed as heat. In a similar way, the equilibrium can be expected to shift in the exothermic direction if the temperature is lowered, for then the reduction in temperature is opposed. Thus, an endothermic reaction is favoured by an increase in temperature and an exothermic reaction is favoured by a decrease in temperature. Or stated in another way, increased temperature favours the reactants in exothermic reactions and the products in endothermic reactions. The effect of temperature on equilibrium constant is quantitatively expressed by van’t Hoff equation, which is developed below. The relationship of equilibrium constant to the standard free energy of reaction is given by Eq. (9.31). The standard state is identified by specifying a definite pressure (or fugacity), but the temperature is always the same as that of the reaction mixture at equilibrium. DG0 and hence K will vary with this temperature. For a single species the effect of temperature on its free energy is predicted by Gibbs–Helmholtz equation [Eq. (6.73)].

Note that the partial derivative notation is dropped from the above equation, as the standard free energy, by virtue of the definition of the standard state, depends on the equilibrium temperature, but not on the equilibrium pressure. Multiplying the above equation by i and summing over all species present in the system, we get

Substitute Eq. (9.31) into Eq. (9.35), we get

Equation (9.36), known as van’t Hoff equation, predicts the effect of temperature on the equilibrium constant and hence on the equilibrium yield. DH0 in Eq. (9.36) is the standard heat of reaction. It is apparent that if DH0 is negative, i.e. if the reaction is exothermic, the equilibrium constant decreases as the reaction temperature increases. Alternatively, for an endothermic reaction, the equilibrium constant will increase with increase in temperature. If DH0, the standard heat of reaction, is constant, Eq. (9.36) on integration yields

K and K1 are the equilibrium constant values at temperatures T and T1 respectively. Equation (9.37) may be used to evaluate the equilibrium constant with good results over small temperature ranges.

The equation is exact if DH0 is independent of temperature. A reasonably accurate method of interpolation or extrapolation of equilibrium constant is provided by plotting ln K versus reciprocal of temperature, which leads to a straight line according to Eq. (9.37). The variation of the standard heat of reaction with temperature may be taken into account if the molal heat capacities of the various species taking part in the reaction are known as functions of temperature. Suppose that the specific heats at constant pressure are expressed as a power function in T. CP = a + bT + gT2………(9.38) Then the effect of temperature on the standard heat of reaction may be developed as follows: Since heat of reaction is the enthalpy change between the given initial and final states, it may be evaluated by devising any convenient path between these terminal states for which the enthalpy changes are readily available. Assume that the standard heat at temperature T1, calculate the standard heat at temperature T.

, is known and it is desired to

The actual reaction occurring at temperature T for which the heat of reaction is as occurring along the three paths as depicted in Fig. 9.3.

may be treated

1. The reactants are cooled from temperature T to T1. The enthalpy change for this step is

2. The reaction is allowed to occur at temperature T1. The enthalpy change is

3. The temperature of the products is raised from T1 to T in this step. The enthalpy change is

The standard heat of reaction at temperature T, is obtained by adding the preceding three equations.

Equation (9.43) may be expanded to yield the following result.

The constant DH in the above equation can be evaluated if the heat of reaction at a single temperature is known. Equation (9.46) can then be used for the evaluation of the heat of reaction at any temperature T. Substitute Eq. (9.46) into Eq. (9.36) and integrate the resulting expression. The result is

A in Eq. (9.47) is a constant of integration, which may be evaluated from the knowledge of the equilibrium constant at one temperature. Equation (9.31) relates the equilibrium constant to the standard free energy change. Using this relationship, we get

9.5.1 Evaluation of Equilibrium Constants Equation (9.47) can be used for the evaluation of the equilibrium constant, provided, we know the dependence of heat capacities on temperature and we also have enough information for the evaluation of the constants DH and A. Assuming that the heat capacity data are available, the general methods used for the evaluation of the constants DH and A are listed below. M ethod 1. K may be calculated from the experimentally measured composition of the equilibrium mixture using Eq. (9.19). If K values are thus known at two different temperatures, they may be substituted into Eq. (9.47). The resulting two equations are solved for the constants DH and A. M ethod 2. Standard heat of reaction at one temperature and one value for the equilibrium constant that is determined by direct experimental measurements are available. The former is used in Eq. (9.46) for the evaluation of the constant DH and the latter in Eq. (9.47) for evaluating the constant A. M ethod 3. This method involves no direct experimental measurements for the equilibrium constant and therefore this is the most convenient and most widely used method. The method makes use of thermal data only, usually in the form of standard heat of reaction DH0, and a standard free energy change of reaction DG0. Then the constants DH and A are evaluated using Eq. (9.46) and Eq. (9.48) respectively. DH0 for a reaction may be evaluated from the standard heat of formation, , that are tabulated for most of the compounds. The standard free energy of a reaction can be estimated from the values of standard free energy of formation, respective stoichiometric numbers as

of the various species participating in the reaction and their

That is, the standard free energy of a reaction is the algebraic sum of the free energies of formation of the products minus the algebraic sum of the free energies of formation of the reactants. When an element enters into a reaction, its standard free energy of formation may be taken to be zero.

9.5.2 Giauque Functions Data for calculation of standard free energy of reactions are sometimes tabulated as Giauque functions. These are Gibbs free energy functions that vary very slowly with temperature. Two such functions are in general use—the first is referred to 0 K and the second referred to 298 K. These are written as

where are respectively the free energy in the standard state at temperature T, the enthalpy in the standard state at T and the enthalpy in the standard state at 298 K. Because only standard state properties are involved, these functions depend only on temperature. This temperature dependence is found to be very weak which makes these functions suitable for tabular interpolation. Using the definition of free energy, we can show that

The difference in enthalpy values, the terms in brackets in Eq. (9.53), needed for applying Eq. (9.53) also are listed in tables along with 0. The standard free energy change of a reaction may be calculated from the Gibbs free energy functions. Equation (9.51) can be rearranged as

The Gibbs free energy at the standard state for each of the species taking part in the reaction as given by Eq. (9.54) or Eq. (9.55) multiplied by the respective stoichiometric numbers add together to give the standard free energy of the reaction.

Note that the enthalpy of a compound in the standard state,

, is the same as its standard enthalpy of

formation, . The standard free energy of a reaction determined using Eqs. (9.56) or (9.57) may be used in Eq. (9.31) to calculate the equilibrium constant. EXAMPLE 9.6 Calculate the equilibrium constant at 298 K of the reaction N2O4 (g) 2NO2 (g) given that the standard free energies of formation at 298 K are 97,540 J/mol for N2O4 and 51,310 J/mol for NO2. Solution Using Eq. (9.50) for the dissociation of N2O4, DG0 = 2 51,310 – 97,540 = 5080 J/mol From Eq. (9.31), DG0 = – RT ln K which gives

Therefore, K = 0.1287. EXAMPLE 9.7 The standard heat of formation and standard free energy of formation of ammonia at 298 K are –46,100 J/mol and –16,500 J/mol respectively. Calculate the equilibrium constant for the reaction N2 (g) + 3H2 (g) 2NH3 (g) at 500 K assuming that the standard heat of reaction is constant in the temperature range 298 to 500 K. Solution The standard free energy of reaction is estimated from Eq. (9.50).

The standard heat of reaction at 298 K = 2 – 46,100 = – 92,200 J/mol. This is assumed constant within the temperature range involved. Now use Eq. (9.37) to evaluate the equilibrium constant.

Therefore, the equilibrium constant at 500 K, K = 0.18 EXAMPLE 9.8 n-Butane is isomerised to i-butane by the action of catalyst at moderate temperatures. It is found that the equilibrium is attained at the following compositions. Temperature, K

Mol %, n-butane

317

31.00

391

43.00

Assuming that activities are equal to the mole fractions, calculate the standard free energy of the reaction at 317 K and 391 K and average value of heat of reaction over this temperature range. Solution Since activities are equal to mole fractions, K = yib/ynb, where yib is the mole fraction of ibutane and ynb the mole fraction of n-butane in the equilibrium mixture. Therefore,

Assuming that the heat of reaction is independent of temperature we can use Eq. (9.37) for calculating it.

EXAMPLE 9.9 Estimate the standard free energy change and equilibrium constant at 700 K for the reaction N2 (g) + 3H2 (g) 2NH3 (g) given that the standard heat of formation and standard free energy of formation of ammonia at 298 K to be – 46,100 J/mol and –16,500 J/mol respectively. The specific heat (J/mol K) data are given below as function of temperature (K): CP = 27.27 + 4.93 10–3T for N2 CP = 27.01 + 3.51 10–3T for H2 CP = 29.75 + 25.11 10–3T for NH3 Solution The standard heat of reaction and standard free energy of reaction at 298 K were estimated in Example 9.7. DH0 = – 92,200 J/mol; DG0 = – 33,000 J/mol Also, Da = 2

29.75 – 27.27 – 3

Db = (2 25.11 – 4.93 – 3 Equation (9.46) gives

27.01 = – 48.8 3.51) 10–3 = 34.76

10–3

– 92,200 = DH

– 48.8T + 17.38

= DH

– 48.8

= DH

– 1.3

298 + 17.38 104

Therefore, DH

= – 7.9201

10–3T2 10–3 (298)2

104. Equation (9.48) gives

– DaT ln T – T2 – ART 104 + 48.8 298 ln 298 – 17.38

– 33,000 = DH

= – 7.9201 = 2105 – 2477.57 A

Therefore, A = 14.169. Substitute DH

10–3

2982 – A

8.314

298

and A into Eq. (9.47) and Eq. (9.48), we get

EXAMPLE 9.10 Evaluate the equilibrium constant at 600 K for the reaction CO (g) + 2H2 (g)

CH3OH (g) given that the Gibbs free energy function

for CO, H2 and methanol at 600 K are respectively –203.81, –136.39 and –249.83 J/mol K. The heats of formation at 298 K of CO (g) and CH3OH (g) at 298 K are –110,500 J/mol and –200,700 J/mol. Solution The standard free energy of formation at 600 K is evaluated by means of Eq. (9.57).

EXAMPLE 9.11 Calculate the equilibrium constant for the reaction N2 (g) + 3H2 (g) 2NH3 (g) at 500 K, given that the free energy function

at 500 K for nitrogen, hydrogen and ammonia are respectively –177.5, –116.9 and –176.9 J/mol K. The function for nitrogen, hydrogen and ammonia are respectively 8669, 8468 and 9920 J/mol. The free energy of formation of ammonia at 298 K is –46,100 J/mol. Solution Equation (9.53) gives

Equation (9.57) can be used to evaluate

.

9.6 EFFECT OF PRESSURE ON EQUILIBRIUM 9.6.1 Effect of Pressure on Equilibrium Constant We have shown that the equilibrium constant K is related to the standard free energy change by the equation, DG0 = – RT ln K, where K is defined by Eq. (9.19) as

The equilibrium constant defined above is independent of the pressure. By Eq. (9.31), the equilibrium constant is known if the standard free energy of the reaction and the reaction temperature are known. The standard free energy of a reaction is determined by the free energies of the substances in their standard states. The standard states are defined by specifying a pressure and are in no way affected by the reaction pressure. That is, the standard free energy of a reaction, and hence, the equilibrium constant are not affected by changes in the equilibrium pressure.

9.6.2 Effect of Pressure on Equilibrium Composition Though the equilibrium constant is unaffected by pressure, it does affect the equilibrium composition in gas-phase reactions. This effect is explained qualitatively by Le Chatelier’s principle. Consider for example, the equilibrium in the gas-phase reaction A 2B. When pressure is applied to this system, it responds in such a way as to minimise the effect of the increase in pressure. This is achieved by decreasing the number of moles in the system, which in turn is achieved by the reaction A 2B. Thus, increase in pressure decreases the number of B molecules and increases the number of A molecules. By the same reasoning we can deduce that in the case of the reaction equilibrium for N2 + 3H2 2NH3 formation of ammonia will be favoured by an increase in pressure as there is a reduction in the number of moles due to this reaction. It should be remembered that when the composition of the system changes in this manner in response to increase or decrease in pressure, it does so without changing the equilibrium constant. Except at very high pressures, properties of solids, liquids or solutions are not affected appreciably by pressure. Therefore, the equilibrium concentrations in reactions involving solids, liquids or solutions are not affected significantly by changes in pressure. To predict the effect of pressure quantitatively, the relationship between equilibrium constant and

equilibrium composition must be established. Equation (9.19) defines the equilibrium constant as a function of activities of the species in the reacting system. The activities of the components are affected by changes in pressure, temperature and composition. As K is independent of pressure, and activities are not, it requires that the activities of the components change with pressure in such a way that the complex function of activities, which we have defined as equilibrium constant, remains unaltered. The equilibrium constant written in terms of activities, K, and the equilibrium constant Kf, which is written in terms of the fugacities of the components were shown to be equal for gaseous systems employing ideal-gas standard state through Eq. (9.21).

Since, K is independent of pressure, the variation in the Pn term in the above equations must be balanced by a corresponding change in the value for Ky. The change in Ky means the change in the equilibrium compositions. If there is a decrease in the number of moles during the reaction as in the case of ammonia synthesis reaction, n will be negative. An increase in pressure in this case will decrease Pn and as a result, Ky and the equilibrium yield would increase. On the other hand, if the reaction results in an increase in the number of moles, n will be positive and the equilibrium yield would decrease with increase in pressure. The above observations are in agreement with the Le Chatelier’s principle. In addition, Eq. (9.62) can be used to explain the effect of pressure on reactions where n is zero, which cannot be explained by Le Chatelier’s principle. One would expect pressure to have no effect on reaction such

as the water-gas shift reaction CO (g) + H2O (g)

CO2 (g) + H2 (g) because there is no change in the number of moles during the reaction. The effect of pressure on the equilibrium composition in this case can be explained by the effect of pressure on Kf. Kf measures the deviation from ideal-gas behaviour, and its value may change with change in pressure. If Kf decreases in any reaction, then Ky and the equilibrium yield would increase even when n is zero. The effect of pressure on Kf can be calculated from fugacity coefficients. It is seen that when the compressibility of the products is greater than the compressibility of the reactants, Kf decreases with pressure, thereby increasing the conversion. EXAMPLE 9.12 Industrial grade methanol can be produced according to the reaction

For this reaction, kJ. If an equimolar mixture of CO and H2 is fed to a reactor maintained at 400 K and 10 bar, determine the fraction of CO that is converted into CH3OH at equilibrium. Assume that the reaction mixture behaves like an ideal gas. Solution Basis: 1 mol CO and 1 mol hydrogen in the reaction mixture. Let e be the extent of reaction. The mole fractions of the components under equilibrium are:

EXAMPLE 9.13 A compound M polymerises in the gas phase at low pressure to Mn, where n > 1. (a) Show that the mole fraction of the polymer at equilibrium increases with increase in pressure at constant temperature

(b) The mole fraction of the polymer in the equilibrium mixture at 300 K is 0.15 at 1 bar and 0.367 at 2 bar. Find the value of n. Solution (a) The reaction is nM Mn. There is a decrease in the number of moles during the forward reaction. The increase in pressure therefore favours the polymerisation reaction and as a result, the mole fraction of the polymer at equilibrium increases with pressure. (b) From Eq. (9.62), Ky = (K/Kf)P–n. Assuming ideal gas behaviour, Ky = KP–n. Here, n = 1 – n and at 1 bar, Ky = mole fraction of Mn/(mole fraction of M)n = 0.15/0.85n Ky at 2 bar = 0.367/0.633n Therefore, 0.15/0.85n = KPn–1 = K 0.367/0.633n = KPn–1 = K 2n–1 Dividing the second equation by the first,

On solving, we get n = 4. EXAMPLE 9.14 In the synthesis of ammonia, stoichiometric amounts of nitrogen and hydrogen are sent to a reactor where the following reaction occurs N2 + 3H2

2NH3

The equilibrium constant for the reaction at 675 K may be taken equal to 2 10–4. (a) Determine the per cent conversion of nitrogen to ammonia at 675 K and 20 bar. (b) What would be the conversion at 675 K and 200 bar? Solution Basis: 1 mol nitrogen and 3 mol hydrogen are in the reactant mixture. Let e be the extent of reaction. Then the number of moles of various species at equilibrium are calculated using Eq. (9.3) as ni = ni0 + nie. Thus the moles of nitrogen, hydrogen and ammonia at equilibrium are, respectively, 1 – e, 3 – 3e and 2e. Total moles at equilibrium is = 4 – 2e. The mole fractions of nitrogen, hydrogen and ammonia are, respectively,

Therefore, e = 0.5375. So, conversion of nitrogen = 53.75%. We see that the increase in pressure favours the formation of ammonia as this reaction is accompanied by a decrease in the number of moles.

9.7 OTHER FACTORS AFFECTING EQUILIBRIUM CONVERSION The reaction conditions that influence the extent of reaction under equilibrium are the temperature, pressure, presence of inert materials, presence of excess of reactants and presence of the products of

the reaction in the initial mixture. The effects of temperature and pressure on the equilibrium composition have already been discussed in the previous sections. Here, we discuss the effects of other factors. Rearrange Eq. (9.62) to the following form.

where i is any species taking part in the reaction and ni is the number of moles of i. N represents the total number of moles in the reaction mixture, and if any inert material is present in the system, N includes nI moles of inert material also. N = S ni + nI Combining Eqs. (9.65) and (9.66) we obtain

Any changes in the reaction conditions that results in an increase in the right-hand side of Eq. (9.67) leads to an improved conversion.

9.7.1 Presence of Inert Materials Diluting the reaction mixture with an inert material will increase N in Eq. (9.67). This will result in an increased conversion, if n is positive. That is, if the reaction proceeds with an increase in the number of moles, presence of inerts in the system will increase the equilibrium yield. The effect, as we see, is just the opposite to the effect of increased pressure in such reactions. The presence of inerts will decrease conversion if the reaction is accompanied by a decrease in the number of moles; and the inerts present in the system will have no influence on the degree of completion if n is zero, that is, if there is no change in the number of moles during a reaction. EXAMPLE 9.15 A mixture of 1 mol CO, and 1 mol water vapour is undergoing the water-gas shift reaction at a temperature of 1100 K and a pressure of 1 bar. CO (g) + H2O (g)

CO2 (g) + H2 (g) The equilibrium constant for the reaction is K = 1. Assume that the gas mixture behaves as ideal gas. Calculate

(a) The fractional dissociation of steam (b) The fractional dissociation of steam if the reactant stream is diluted with 2 mol nitrogen. Solution The mole fractions of the species at equilibrium are related to the equilibrium constant which is given by

where n is the sum of stoichiometric numbers. Here, n = 1 + 1 – 1 – 1 = 0. As the gas mixture behaves as an ideal gas, Kf = 1. Equation (9.65) gives Ky = K = 1. Ky is related to the mole fractions of various components as . Equation (9.4) gives the relationship between mole fractions and extent of reaction at equilibrium as

(a) The mole fractions of the constituents in the equilibrium mixture are expressed in terms of the extent of reaction as given in the table below:

Solving the above, e = 0.5. Conversion of steam z is obtained from Eq. (9.5)

This means that 50% of steam is converted in the reaction. (b)

e = 0.5. The conversion of water vapour is 50%. The conversion remains the same as that resulted when the reactant stream contained only the stoichiometric quantities of CO and H2O vapour. This is because n = 0 or the reaction produces no change in the number of moles. EXAMPLE 9.16 Ammonia synthesis reaction is represented by N2 + 3H2 2NH3 The reactant stream consists of 1 mol N2, 3 mol H2 and 2 mol argon. The temperature and pressure of the reaction are 675 K and 20 bar. The equilibrium constant for the reaction is 2 10–4. Determine how the conversion of nitrogen is affected by the presence of argon. Solution The total number of moles of the initial mixture, n0 = 1 + 3 + 2 = 6.

Solving we get, e = 0.1022. Thus, it is seen that the conversion of nitrogen decreases to 10.22% in the presence of argon, from a value of 14.48% (Example 9.14) achieved in the absence of argon. EXAMPLE9.17 The reaction takes place in the gas phase at 2975 K and 2025 kPa. The reaction mixture initially comprises 15 mol percent oxygen, 77 mol percent nitrogen and the rest inerts. The standard Gibbs free energy change for the reaction is 113.83 kJ/mol at this temperature. Assuming ideal gas behaviour, calculate the partial pressures of all species at equilibrium. How is the conversion of oxygen affected when the initial mixture were free of inerts? Solution Basis: 15 mol oxygen, 77 mol nitrogen and 8 mol inert in the reaction mixture. Let e be the extent of reaction. The mole fractions of the components under equilibrium are: Oxygen: (15 – e)/100, Nitrogen: (77 – e)/100, NO: 2e/100

Partial pressures are obtained by multiplying the mole fractions by the total pressure. The values are given below: O2: 271.6 kPa, N2: 1527.1 kPa, NO: 64.4 kPa, Inerts: 162 kPa If the initial mixture were free of inerts, the mole fractions of the components under equilibrium will be Oxygen: (15 – e)/92, Nitrogen: (77 – e)/92, NO: 2e/92 and the equilibrium constant will be given by the same expression as before:

9.7.2 Presence of Excess of Reactants When the reactants are not present in stoichiometric proportions, increasing the number of moles of

the excess reactant will result in increase in the number of moles of the products and improved conversion of the limiting reactant at equilibrium. This is evident from the left-hand side of Eq. (9.67), the value of which increases when the number of moles of the excess reactant is increased. Therefore, it is desirable to supply all the reactants except the limiting reactant in excess of the stoichiometric requirement, in order to increase the conversion with respect to the limiting reactant. EXAMPLE 9.18 One mole steam undergoes the water-gas shift reaction at a temperature of 1100 K and a pressure of 1 bar. CO (g) + H2O (g)

CO2 (g) + H2 (g) The equilibrium constant for the reaction is K = 1. Assuming ideal gas behaviour, calculate the fractional dissociation of steam in the following cases and discuss the effect of the presence of excess reactant on the extent of reaction. (a) CO supplied is 100% in excess of the stoichiometric requirement. (b) CO supplied is only 50% of the theoretical requirement. Solution Basis: 1 mol water vapour present in the reactant stream. (a)

(b)

We see that the equilibrium conversion of water vapour was 50% when the reactants were in stoichiometric proportions, it increases to 66.7% when CO was present 100% in excess and falls to 33.3% when CO becomes the limiting reactant. EXAMPLE9.19 Ethanol is produced by the vapour phase hydration of ethylene according to the reaction:

The reactor operates at 400 K and 2 bar and the feed is a gas mixture of ethylene and steam in the ratio 1:3. The equilibrium constant is 0.25. Estimate the composition (mol %) of the equilibrium mixture. Assume ideal gas behaviour. How is the conversion of ethylene affected when the initial reactant stream contains stoichiometric quantities of the reactants? Solution Basis: 1 mole of ethylene and 3 moles of N2 in the reactant stream. K = KyPv. Here, n = –1. Therefore, 0.25 = Ky2–1. Hence, Ky = 2 K = 0.5 Let e be the extent of reaction. Then the mole fraction in the equilibrium mixture are: Ethylene: (1 – e)/(4 – e), Steam: (3 – e)/(4 – e), Ethanol: e/(4 – e) Using these values, we get

Equilibrium mixture contains 19.6% ethylene, 73.2% steam and 7.2% ethanol. Let the initial mixture contain 1 mol ethylene and 1 mol steam and let e be the extent of reaction. Then the mole fraction in the equilibrium mixture are: Ethylene: (1 – e)/(2 – e), Steam: (1 – e)/(2 – e), Ethanol: e/(2 – e) Using these values, we get

Solving this, we get e = 0.183. That is, conversion of ethylene is 18.3%. The conversion was 26.8% in the first case when excess of water vapour was present in the reactant stream.

9.7.3 Presence of Products If the initial reaction mixture contained any of the products of the reaction, then the number of moles of that product formed by the reaction so as to establish equilibrium will decrease as indicated by Eq. (9.67). Therefore, the addition of the products to the original reactant stream decreases the equilibrium conversion. EXAMPLE 9.20 A gas mixture which contained 1 mol CO, 1 mol water vapour and 1 mol CO2 is undergoing the following reaction at a temperature of 1100 K and a pressure of 1 bar. CO (g) + H2O (g)

CO2 (g) + H2 (g) The equilibrium constant for the reaction is K = 1. Assume that the gas mixture behaves as ideal gas. Calculate the fractional dissociation of steam and discuss the effect of the presence of the products on the equilibrium conversion. Solution The mole fractions of the different species in the equilibrium mixture is expressed in terms of the extent of reaction as below:

Solving the above equation, we get e = 0.333, which means that the conversion of water vapour gets reduced to 33.3% due to the presence of CO2, the product of the reaction in the reactant stream. EXAMPLE 9.21 A gas mixture containing 25% CO, 55% H2 and 20% inert gas is to be used for methanol synthesis. The gases issue from the catalyst chamber in chemical equilibrium with respect to the reaction CO (g) + 2H2 (g)

CH3OH (g) at a pressure of 300 bar and temperature of 625 K. Assume that the equilibrium mixture forms an ideal solution and Kf and Kf are 4.9 10–5 and 0.35 respectively. What is the per cent conversion of CO? Solution Basis: 100 moles of initial gas mixture. Let e be the extent of reaction at equilibrium. n0 = 100. n=–1–2+1=–2 The mole fractions in the equilibrium mixture are calculated using Eq. (9.4)

For gases, K = Kf. Therefore,

Therefore, 61% of CO gets converted. EXAMPLE 9.22 A gas mixture consisting of 60% H2, 20% N2 and the rest inert gas is passed over a suitable catalyst for the production of ammonia.

The equilibrium constant Kp = 1.25 10–2. The pressure is maintained at 50 bar. Assume ideal gas behaviour for the gas mixture. Determine the composition of the gases leaving the reactor. Solution Basis: 100 moles of the reactant gases.

Solving this, we get e = 8.71. Mole fraction of nitrogen is obtained as [20 – (e/2)]/(100 – e) = 0.1714 and mole fraction of hydrogen is [60 – (3e/2)]/(100 – e) = 0.5141 Mole fraction of ammonia: e/(100 – e) = 0.0954 Mole fraction of inert gas: 1 – 0.1714 – 0.5141 – 0.0954 = 0.2191 Analysis of exit gases from the reactor: N2 = 17.14%, H2 = 51.41%, NH3 = 9.54% and inert gas = 21.91%

9.8 LIQUID-PHASE REACTIONS The equilibrium constant as defined by Eq. (9.19) is applicable for all chemical reactions. For liquid-phase reactions, the evaluation of equilibrium constant using this equation requires a relationship between activity and composition. Since, activity is the ratio of the fugacity to the fugacity in the standard state, such a relationship can be established once the standard state is specified. The standard state for liquid-phase reactions may be the pure liquid at 1 bar and the reaction temperature. The fugacity in this state is not much different from the fugacity of pure liquid at the pressure and temperature of the reaction fi. This is because, pressure has very negligible effect on the properties of liquids. With this choice, the equilibrium constant becomes

The fugacity of a component in the solution is related to the fugacity in the pure state by = gixifi, where i is the activity coefficient in the solution. Using this in Eq. (9.68), we get

K is an equilibrium constant in terms of activity coefficients. Accurate values of activity coefficients are rarely available and in practical calculations we set K = 1. This is equivalent to assuming that the solution is ideal and ai = xi. The components present in large proportions obey Lewis–Randall rule and for them the activity and the mole fraction in the solution are the same. Even if Lewis–Randall rule is not applicable, the assumption that K = 1 is not a very serious limitation as the function denoted by K may become nearly unity even if the individual activity coefficients are

not. Thus

For components present in low concentration, the standard state of the solute is usually the fictitious or hypothetical state which would exist if the solute obeyed Henry’s law over a concentration range extending up to a molality of unity. This hypothetical state is illustrated in Fig. 9.4.

The fugacity and molality (mol/kg solvent) are related as

where Ki is the Henry’s law constant and mi is the molality. Using the hypothetical standard state, it can be shown that the standard state fugacity is equal to the Henry’s law constant and the activity and the molality are equal. That is ai = mi………(9.72) With this choice for the standard state, a very simple relationship exists between the activity and the concentration for cases where Henry’s law is applicable.

9.9 HETEROGENEOUS REACTION EQUILIBRIA In the study of heterogeneous reaction equilibria presented in this section, we are concerned with a gas phase that is in equilibrium with a liquid or a solid phase. When the heterogeneous system is in equilibrium we would have to consider the equilibrium with respect to chemical reactions in the gas phase as well as the phase equilibria between the components in the gas phase and the liquid or the solid phase as the case may be.

9.9.1 Reactions in Solutions Consider the reaction between a gas A and liquid B to produce a solution C.

A (g) + B (l) C (aq) The equilibrium in this reaction can be studied in different ways: 1. The reaction may be assumed to take place entirely within the gas phase and the equilibrium constant for the reaction is evaluated using the standard state for gases, i.e. the ideal gas standard state at a pressure of 1 bar and the reaction temperature. The resulting equations are coupled with equations for material transfer between phases to maintain equilibrium. 2. The reaction is assumed to occur in the liquid phase with simultaneous transfer of material between phases to maintain equilibrium. The calculations of reaction equilibria are based on the liquid standard state. 3. The third method involves the use of mixed standard states. Thus, the standard state for A is the pure ideal gas at 1 bar, that for B is pure liquid at 1 bar, and for C it is the solute in an ideal 1molal aqueous solution. The equilibrium constant in this case may be evaluated as

All the above methods give the same results for equilibrium compositions, but the values for equilibrium constant depend on the choice of the standard state.

9.9.2 Equilibria involving Pure Solids and Liquids When a pure liquid or a pure solid is involved in a heterogeneous reaction with gases, its activity may be taken as unity provided the pressure of the system is not much different from the standard state. Activity as we know, is defined as the ratio of the fugacity to the fugacity in the standard state. The fugacity in the standard state is almost equal to that in the equilibrium state, as these two states differ only in their pressures and not in their temperatures. Pressure, unless extremely high, has only a negligible effect on the properties of liquids and solids. Where the standard state for solids and liquids is taken at 1 bar or at low equilibrium vapour pressures, the activities of pure solids and pure liquids may be taken as unity at moderate pressures. Therefore, the composition of the gaseous phase at equilibrium is not affected by the presence of the solid or liquid.

9.9.3 Pressures of Decomposition Many solid compounds decompose to give another solid and a gas, as in the calcination of calcium carbonate to CO2 and lime. CaCO3 (s) CaO (s) + CO2 (g) The equilibrium constant for this reaction is

The activities of the solid components present at equilibrium are close to unity provided the pressure is moderate and both solids are present at equilibrium. Since the standard state for gases is the idealgas state at 1 bar, the standard state fugacity is equal to unity and therefore, the activity of CO2 in the equilibrium mixture is equal to its fugacity,

. But fugacity of a component is equal to its partial

pressure at low pressures and, therefore, Eq. (9.74) reduces to

This is the equilibrium partial pressure exerted by CO2 and its value depends only on temperature. If the partial pressure is lowered below this equilibrium value, CaCO3 will decompose and will eventually disappear. On the other hand, if the pressure on the system is kept above the equilibrium partial pressure, CaO will combine with CO2 resulting in the formation of CaCO3. For a general solid decomposition reaction represented by aA (s) lL (s) + mM (g) the above treatment can be generalised as

In the above equation, m is the stoichiometric coefficient; DH0 and DS0 are the standard heat of reaction and standard entropy of reaction respectively. EXAMPLE 9.23 Ethylene gas reacts with water forming aqueous solution of ethanol. C2H4 (g) + H2O (l) C2H5OH (aq) Equilibrium measurements at 530 K and 85 bar showed that the aqueous phase contained 1.5% (mole) ethanol and 95.0% (mole) water. The vapour phase analysed 48% ethylene. The fugacity coefficient for ethylene is estimated to be 0.9. Evaluate the equilibrium constant. Solution Equation (9.73) may be used for evaluating K.

The standard state for aqueous solution is 1 molal solution; for water, it is pure liquid water at 1 bar; and for gaseous ethylene, it is the pure ethylene at 1 bar. The molality of aqueous solution = moles ethanol/kg water = 1.5/(95.0 18 10–3) = 0.8772 mol/kg water

EXAMPLE 9.24 Calculate the decomposition pressure of limestone at 1000 K. CaCO3 (s) CaO (s) + CO2 (g) The standard free energy of this reaction as function of temperature is DG0 = 1.8856 105 – 243.42T + 11.8478T ln T – 3.1045 10–3T2 + 1.7271 – 4.1784 105/T

10–6 T3

Also calculate the decomposition temperature at 1 bar. Solution From Eq. (9.75), the decomposition pressure is

= K, where K can be calculated by

EXAMPLE 9.25 Solid calcium oxalate dissociates at high temperatures into solid calcium carbonate and carbon monoxide:

Solution By Eq. (9.75), Therefore,

EXAMPLE 9.26 Iron oxide is reduced to iron by passing over it a mixture of 20% CO and 80% N2 at 1200 K and 1 bar. FeO (s) + CO (g)

Fe (s) + CO2 (g)

The equilibrium constant for this reaction is 0.403. Assuming that equilibrium is attained, calculate the weight of metallic iron produced per 100 m3 of gas admitted at 1200 K and 1 atm. Gas mixture may be assumed to behave as ideal gas. Solution Basis: 100 mol of gas entering. The activities of solid components can be taken to be unity.

9.10 SIMULTANEOUS REACTIONS With a given set of reactants many reactions may be possible. When we consider the equilibrium yield of methanol in the reaction CO + 2H2

CH3OH………(9.78) by the methods already discussed, we are in fact ignoring the presence of intermediate product, formaldehyde in the reaction mixture. The above reaction proceeds in two steps in series as: CO + H2 HCHO + H2

HCHO………(9.79) CH3OH………(9.80)

For the thermodynamic analysis of a reaction that proceeds in two or more steps, the presence of intermediate products can sometimes be ignored on the assumption that they are very unstable and their concentrations at equilibrium are negligible in comparison with that of the main product. The above assumption is implicit in treating the equilibrium mixture in the methanol synthesis as consisting of only CO, H2 and CH3OH. In this case, this assumption is a valid one as formaldehyde is very unstable, but in many other situations, the presence of intermediate products in the reaction mixture at equilibrium also should be taken into account as explained below: The free energy change for a reaction is equal to the sum of the free energy changes in the individual step reactions. Thus,

are the free energy changes in the two step reactions that occur and DG0 is the standard free energy change in the overall reaction. Since DG0 = – RT ln K, the above equation gives where

K = K1K2 K1 and K2 are the equilibrium constants for the individual steps and K is the equilibrium constant for the combined reaction. For a given value of K, an infinite number of combinations of K1 and K2 are possible such that K = K1K2. For example, let us take K = 10–4 and consider the cases where (a) 10– 4 = 10–10 106, (b) 10– 4 = 10–2 10–2, and (c) 10–4 = 106 10–10. For case (a), the concentration of intermediate products at equilibrium would be negligible and correct result would be obtained by considering only the overall reaction. For case (b), there would be considerable amounts of intermediates at equilibrium and their presence cannot be ignored. For case (c), the equilibrium mixture would be mostly intermediates. The use of an overall equilibrium constant for the calculation of equilibrium compositions is limited to cases where the intermediate products are not present in significant quantities. In addition to the formation of intermediate products, which subsequently reacts to form the final desired products, many side reactions may also occur within the system. For example, starting with the pairs CO and H2 some of the possible reactions are: CO + H2

HCHO

CO + 2H2

CH3OH

CO + 3H2

CH4 + H2O

2CO + 5H2 3CO + 6H2

C2H6 + 2H2O

C3H7OH + 2H2O In dealing with methanol synthesis, it was assumed that the side reactions proceeded at a negligible rate in comparison with the steps involved in the synthesis reaction. Theoretically, when the equilibrium yield of a particular component is to be determined, we should consider simultaneous equilibria in all possible reactions between the substances involved. However, for practical calculations, it is possible to reduce the number of reactions that are to be considered. In the general case when all intermediates and final products must be considered, it is necessary that the equilibrium equations of all reactions must be satisfied by the compositions of the system at equilibrium. Determination of the equilibrium compositions involves simultaneous solution of r equilibrium equations where r is the number of independent reactions that can be written. After determining the number of independent reactions as explained later, the equilibrium constant is evaluated for each reaction by

Here the suffix j is used to represent the jth reaction under consideration. The above equation is written for all r independent reactions. Assuming the equilibrium mixture to behave as ideal gases, these lead to r equations relating the composition to the pressure and the equilibrium constant.

Let the equilibrium constants be K1 and K2 for the reactions indicated by Eqs. (9.84) and (9.85) respectively and let the corresponding extent of reaction be e1 for reaction (9.84) and e2 for reaction (9.85). The initial reactant mixture is assumed to consist of 1 mol A and x mol B. The mole fractions in simultaneous reactions can be calculated using Eq. (9.9).

The mole fractions of various components are yA = (1 – ae1)/[1 + x + (l + m1 – a – b1)e1 + (m2 + n – l – b2)e2] yB = (x – b1e1 – b2e2)/[1 + x + (l + m1 – a – b1)e1 + (m2 + n – l – b2)e2] yL = l(e1 – e2)/[1 + x + (l + m1 – a – b1)e1 + (m2 + n – l – b2)e2] yM = (m1e1 + m2e2)/[1 + x + (l + m1 – a – b1)e1 + (m2 + n – l – b2)e2] yN = ne2/[1 + x + (l + m1 – a – b1)e1 + (m2 + n – l – b2)e2] These are substituted into the following equilibrium relations.

These two equations are solved simultaneously to obtain the variables e1 and e2. Equation (9.9) can now be utilised to evaluate the equilibrium compositions. EXAMPLE 9.27 Five moles of steam reacts with one mole methane according to the following reaction at 850 K and 1 bar. CH4 + H2O

CO + 3H2; K1 = 0.574………(9.88)

CO2 + H2; K2 = 2.21………(9.89) Calculate the composition at equilibrium assuming ideal gas behaviour. CO + H2O

Solution Number of moles of a component at equilibrium = ni = ni0 + Le t e1 and e2 be the extent of reactions for reaction (9.88) and (9.89) respectively. Thus at equilibrium the number of moles are, CH4: 1 – e1,……H2O: 5 – e1 – e2,……CO: e1 – e2,……H2: 3e1 + e2,……CO2: e2 Total number of moles at equilibrium = 6 + 2e1. The mole fractions of various components in the equilibrium mixture are:

Note that K1 = 0.574, K2 = 2.21 and P = 1 bar. The resulting equations are solved for e1 and e2. Assume a value for e1 and calculate e2 by each equation. These two e2 values are plotted against e1. This is repeated for various assumed e1 values. The intersection of the two curves gives the solution. e1 = 0.9124; and e2 = 0.623. The mole fractions are evaluated by supplying the values of e1 and e2. The results are: CH4: 0.0112,……H2O: 0.4415,……CO: 0.0357,……H2: 0.4307……and……CO2: 0.0804

9.11 PHASE RULE FOR REACTING SYSTEMS We have used the criteria of phase equilibrium to develop the phase rule for non-reacting systems in

Chapter 8. F=C–p+2 The criterion of phase equilibrium is valid even when chemical reactions occur within the system. However, the phase rule needs modification for it to be applicable for reacting systems. This is because for each independent reaction occurring, an additional constraint is imposed on the system through Eq. (9.15) or (9.31). Thus, the number of degrees of freedom will be reduced by one for each independent chemical reaction. If r independent reactions occur in the system, then the phase rule becomes F = C – p – r + 2………(9.90) For example, consider a system containing five components distributed between two phases. If the number of independent chemical reactions occurring is one, then the number of degrees of freedom will be 5 – 2 – 1 + 2 = 4. Assuming that the reaction occurring is a gas-phase isomerisation reaction involving two of the components (say, A and B), we can write the equilibrium relationship as

This in fact is a relationship between T, yA and yB. Only two of these three variables are therefore independent. As the degree of freedom is 4 it means that in addition to these three variables, two more variables are to be specified to define the intensive state of the system uniquely. O.A. Hougen, et al., define the number of independent reactions that must be considered as the least number that includes every reactant and product present to an appreciable extent in all phases of the equilibrium system, and accounts for the formation of each product from the original reactants. It can be determined as follows: 1. For each chemical compound present in the system, equation for its formation reaction from its elements is written. 2. The elements that are not present in the system are eliminated by properly combining the equations written in step 1. The number of equations, r, that results from the above procedure is equal to the number of independent chemical reactions occurring. EXAMPLE 9.28 Determine the number of degrees of freedom in a gaseous system consisting of CO, CO2, H2, H2O and CH4 in chemical equilibrium. Solution The number of independent chemical reactions occurring in the system is first determined. The formation reactions for each of the compounds are written:

The elements C and O2 are not present in the system. C is eliminated first, from Eqs. (9.91), (9.92) and (9.94). Combining Eq. (9.91) with Eq. (9.92) we get

The equations that remain after this elimination process are Eqs. (9.97) and (9.98) which represent the independent chemical reactions occurring in the system. Therefore, r = 2. Equation (9.90) gives the degrees of freedom as F = C – p – r + 2. Here C = 5; p = 1 and therefore, F = 4.

SUMMARY Thermodynamics of chemical reactions is mainly concerned with the prediction of the equilibrium conversion attainable in a chemical reaction and the effect of operating conditions on the degree of completion of the reaction. The criterion of chemical equilibrium requires that for a chemical reaction occurring at equilibrium, there should be no change in the Gibbs free energy of the system at constant temperature and pressure. If the system is not in chemical equilibrium, the reaction occurring must be irreversible and the total Gibbs free energy must decrease at constant temperature and pressure (Section 9.2). The equilibrium constant K for a reaction was defined in terms of the activities of the reactants and the products as

where ai is the activity of component i in the reaction mixture and i is the stoichiometric number of i (Section 9.3). The equilibrium constant was related to the standard free energy change by Eq. (9.31). Thus the numerical value of the equilibrium constant depends upon the temperature, the form of the stoichiometric equation and the definition of the standard state for each component. However, it is independent of the pressure at equilibrium (Section 9.4). Equation (9.31) also provided an approximate criterion for feasibility of reactions. If DG0 for a reaction is zero, then K = 1, the reaction proceeds to a considerable extent before equilibrium is reached. If DG0 is negative, then K > 1, the reaction is quite favourable. The effect of temperature on the equilibrium constant was quantitatively expressed by

van’t Hoff equation [Eq. (9.36)]. For an exothermic reaction, the equilibrium constant decreases as the reaction temperature increases and for an endothermic reaction, the equilibrium constant will increase with increase in temperature (Section 9.5). Three methods for the evaluation of equilibrium constant were discussed; the one which made use of thermal data in the form of standard heat of reaction DH0, and a standard free energy change of reaction DG0 at a given temperature was found to be the most convenient and widely used. The usefulness of the Giauque functions for tabulation of standard free energy of reactions and calculation of the equilibrium constant was also established. The equilibrium constant is independent of pressure whereas the composition at equilibrium varies with pressure as evident from Eq. (9.62). If there is a decrease in the number of moles during the reaction, the equilibrium yield would increase with increase in the pressure, whereas if the reaction results in an increase in the number of moles, the equilibrium yield would decrease with increase in pressure. It was also shown that the effect of the presence of inert gas in the reactant stream on the equilibrium conversion was just the opposite of the effect of pressure (Section 9.7). For liquid-phase reactions, the equilibrium constant may be written as K = KgKx. K is an equilibrium constant in terms of activity coefficients, which may be assumed, equal to unity. For components present in low concentration, the activity and the molality are equal (Section 9.8). Under heterogeneous equilibrium (Section 9.9), a brief discussion on the reaction between a gas and liquid resulting in the formation of a solution and reaction equilibria in which a solid or liquid reacted with a gas, were provided. Also, it was seen that for reactions in which solid compounds decomposed to give another solid and a gas, the equilibrium constant was equal to the partial pressure of the gas. If the partial pressure was lowered below this equilibrium value the solid would decompose and if the pressure on the system was maintained above this value, the formation of solid was favoured. For simultaneous reactions in which all intermediate and final products in the equilibrium mixture were to be considered for determining the composition, equilibrium equations were written for all the independent reactions and these were solved simultaneously (Section 9.10).

REVIEW QUESTIONS 1. What do you mean by the ‘extent of reaction’? How is it related to the mole fraction of the species in the reaction mixture? 2. What is the criterion of chemical reaction equilibria? 3. Define equilibrium constant K of a chemical reaction. How is it related to Kf and KP? 4. Does the numerical value of the equilibrium constant depend on the form of the stoichiometric equation? 5. How is the equilibrium constant K related to the standard free energy change? Does K vary with pressure? 6. What is the effect of temperature on the equilibrium constant? Using van’t Hoff equation predict the effect of increasing the temperature on endothermic and exothermic reactions. 7. How would you predict the feasibility of a reaction from the value of the standard free energy change? 8. How would the equilibrium yield in a gaseous chemical reaction be affected by increasing the pressure, if there is a decrease in the number of moles during the reaction? How would you

explain the effect of pressure on reactions such as the water–gas shift reaction, where there is no change in the number of moles? 9. How would the equilibrium yield of ammonia be affected if argon is present in the synthesis gas fed to the ammonia converter? 10. Explain how the equilibrium constant for liquid-phase reactions is evaluated. 11. Show that the equilibrium constant in the decomposition of calcium carbonate into CO2 and lime is equal to the partial pressure of carbon dioxide. Explain how would you estimate the decomposition pressure? What would happen if the CO2 pressure is reduced below this value? 12. A reaction proceeds in two steps. The equilibrium constants for the individual steps are K1 and K2. What would be the equilibrium constant for the overall reaction? 13. What do you mean by the number of independent reactions in a chemically reacting system? How would you determine it? 14. What is phase rule as applicable to a reacting system?

EXERCISES 9.1 Water vapour decomposes according to the following reaction:

Derive expressions for the mole fraction of each species in terms of the extent of reaction assuming that the system contained n0 moles of water vapour initially. 9.2 The following reaction occurs in a mixture consisting of 2 mol methane, 1 mol water, 1 mol carbon monoxide and 4 mol hydrogen initially. CH4 + H2O CO + 3H2 Deduce expression relating the mole fractions of various species to the extent of reaction. 9.3 A system consisting of 2 mol methane and 3 mol water is undergoing the following reaction CH4 + H2O

CO + 3H2

CH4 + 2H2O CO2 + 4H2 Derive expressions for mole fractions in terms of the extent of reactions. 9.4 The following gas-phase reactions occur in a mixture initially containing 3 mol ethylene and 2 mol oxygen.

Derive expressions for mole fractions in terms of extent of reactions. 9.5 Calculate the equilibrium constant at 298 K of the reaction N2 + 3H2 free energy of formation of ammonia at 298 K is –16,500 J/mol.

2NH3, given that the

9.6 Calculate the standard free energy change at 298 K in the gas-phase alkylation of isobutane with ethylene to form neohexane. C4H10 (g) + C2H4 (g) C6H14 (g) The free energies of formation at 298 K are –21,000 J/mol, 68,460 J/mol and –9,950 J/mol for isobutane, ethylene and neohexane respectively. 9.7 Calculate the equilibrium constant at 673 K and 1 bar for the reaction N2 (g) + 3H2 (g) 2NH3 (g) assuming that the heat of reaction remains constant in the temperature range involved. Take the standard heat of formation and standard free energy of formation of ammonia at 298 K to be – 46,110 J/mol and –16,450 J/mol respectively. 9.8 Is the following reaction promising at 600 K? NaOH (s) + CO (g) HCOONa (s) The free energy of formation, the heat of formation and the specific heat of the components are given below:

9.9 Methanol is produced by the following reaction: CO (g) + 2H2 (g)

CH3OH (g) The standard heat of formation of CO (g) and CH3OH (g) at 298 K are –110,500 J/mol and –200,700 J/mol respectively. The standard free energies of formation are –137,200 J/mol and – 162,000 J/mol respectively. (a) Calculate the standard free energy change and determine whether the reaction is feasible at 298 K. (b) Determine the equilibrium constant at 400 K assuming that the heat of reaction is constant. (c) Derive an expression for the standard free energy of reaction as function of temperature if the specific heats of the components are: CP = 3.376R + 0.557 10–3RT – 0.031 105RT–2 for CO CP = 3.249R + 0.422 10–3RT + 0.083 105RT–2 for H2 CP = 2.211R + 12.216 10–3RT – 3.450 10–6RT2 for CH3OH (d) Use the equation obtained in part (c) to calculate the equilibrium constant at 400 K and compare with the result in part (b). 9.10 Calculate the equilibrium constant at 298 K for the reaction

9.11 The standard free energy change for the reaction C4H8 (g) C4H6 (g) + H2 (g) is given by the relation

(a) Over what range of temperature is the reaction promising from a thermodynamic viewpoint? (b) For reaction of pure butene at 800 K, calculate the equilibrium conversion for operation at 1 bar and 5 bar. (c) Repeat part (b) if the feed consists of 50% (mol) butene and the rest inerts. 9.12 Calculate the equilibrium constant for the vapour-phase hydration of ethylene to ethanol at 600 K C2H4 + H2O C2H5OH The following data are available:

9.13 The equilibrium constant at 420 K for the vapour-phase hydration of ethylene to ethanol according to the reaction C2H4 + H2O C2H5OH is 6.8 10–2 and standard heat of reaction at 298 K is –45.95 are as follows:

103 J. The specific heat data

Formulate general relationships for estimating the equilibrium constant and standard free energy change as functions of temperature. 9.14 For the vapour-phase hydration of ethylene to ethanol according to C2H4 + H2O C2H5OH the equilibrium constants were measured at temperature 420 K and 600 K. They are 6.8 10–2 and 1.9 10–3 respectively. The specific heat (J/mol K) data are:

Develop general expressions for the equilibrium constant and standard free energy change as functions of temperature. 9.15 The water–gas shift reaction CO (g) + H2O (g)

CO2 (g) + H2 (g)

takes place at 373 K. The equilibrium constant KP for this reaction at 537 K = 9.8 10–4. The heats of formation at 298 K are: CO = –110,525 J/mol, CO2 = –393,509 J/mol, H2O = –241,818 J/mol. Calculate the equilibrium constant at 1000 K. 9.16 Calculate the fraction of pure ethane that would dehydrogenate at 750 K and 5 atm, if the following reaction goes to equilibrium. DG0 for the reaction at 750 K is 42.576 kJ. Assume ideal behaviour. 9.17 Ethanol can be prepared by the following vapour-phase reaction from ethylene:

The value of DG0 for the above reaction at 1 bar and 398 K is 5040 J. Calculate the conversion obtained if an isothermal reactor operating at 398 K and 2 bar is fed with a mixture containing 50 mol percent ethylene and 50 mol percent steam. Assume that equilibrium is reached at the exit of the reactor and the gases behave ideally. 9.18 A gaseous mixture containing 30% CO, 50% H2 and the rest inert gas is sent to a reaction

chamber for methanol synthesis. The following reaction occurs at 635 K and 310 bar. CO (g) + 2H2 (g)

CH3OH (g) Assuming that the gas mixture behaves as an ideal solution calculate the per cent conversion of CO given that Kf = 5 10–5 and Kf = 0.35. 9.19 Estimate the maximum conversion of ethylene to alcohol by vapour phase hydration at 523 K and 34 bar. C2H4 (g) + H2O (g) C2H5OH (g) The equilibrium constant varies with temperature as ln K = 4760/T – 1.558 ln T + 2.22 10–3T – 0.29

10–6T2 – 5.56 The steam–ethylene ratio in the initial mixture is 5.0. The fugacity coefficients for ethylene, ethanol and water vapour are 0.98, 0.84 and 0.91. 9.20 Ethanol is manufactured by the vapour-phase hydration of ethylene to ethanol according to the reaction, C2H4 (g) + H2O (g) C2H5OH (g) Starting with a gas mixture containing 25% ethylene and 75% steam, determine the composition of the products if the reaction were carried out at 400 K and 1 bar. The standard free energy of reaction at 400 K is 4548.3 J. 9.21 What would be the equilibrium yield of ethanol at 1 bar and 373 K in the following reaction? C2H4 (g) + H2O (g) C2H5OH (g) The reactant stream consists of an equimolar mixture of steam and ethylene. The standard free energy change may be taken as = 1264 J/mol. 9.22 Calculate the equilibrium percentage conversion of nitrogen to ammonia at 700 K and 300 bar, if the gas enters the converter with a composition of 75% (mol) hydrogen and 25% (mol) nitrogen. For the reaction

equilibrium constant may be taken as K = 9.1 10–3. Assume that Kf = 0.72. 9.23 The gases from the pyrites burner of a contact sulphuric acid plant have the following composition: SO2 = 7.80%, O2 = 10.80% and N2 = 81.40%. This is then passed into a converter where the SO2 is converted to SO3. The temperature and pressure in the converter are 775 K and 1 bar. The equilibrium constant for the reaction

may be taken as K = 85. Calculate the composition of gases leaving the converter. 9.24 One mol carbon at 298 K reacts with 2 mol oxygen at 298 K to form an equilibrium mixture of CO2, CO and O2 at 3000 K and 1 bar. If the equilibrium constant K = 0.328, determine the

equilibrium composition. 9.25 One mol carbon at 298 K and 1 bar reacts with 1 mol oxygen at 298 K and 1 bar to form an equilibrium mixture of CO2, CO and O2 at 3000 K and 1 bar in a steady flow process. Determine the equilibrium composition and heat transfer for this process if the equilibrium constant K = 0.328. Standard heat of formation are 393.509 kJ/mol for CO2, 110.525 kJ/mol for CO. The mean heat capacity of products = 45 J/mol K. 9.26 Pure N2O4 at a low temperature is diluted with air and heated to 298 K and 1 bar. The following reaction occurs N2O4 (g) 2NO2 (g) If the mole fraction of N2O4 in the N2O4–air mixture before dissociation begins is 0.2, calculate the extent of decomposition and mole fraction of NO2 and N2O4 present at equilibrium. The standard free energy change for the reaction at 298 K = 4644.7 J/mol. 9.27 Methanol is manufactured according to the reaction CO (g) + 2H2 (g)

CH3OH (g) The reaction is carried out at 400 K and 1 bar. The standard heat of reaction at this condition is – 9.4538 104 J and the equilibrium constant is 1.52. Analysis of the equilibrium vapour product from the reactor shows 40% hydrogen. Equilibrium gas mixture can be treated as an ideal gas. (a) Determine the concentrations of CO and CH3OH in the product. (b) If the reaction occurred at 500 K and 1 bar starting with the same feed as in part (a) would you expect the concentration of hydrogen in the equilibrium mixture to be greater or less than 40% mole? Why? 9.28 Determine the maximum percentage of ethane that may get dehydrogenated to ethylene at 750 K and 5 bar according to the reaction C2H6 (g)

C2H4 (g) + H2 (g)

The standard free energy of reaction at 750 K is 4.2593 104 J. 9.29 Hydrogen cyanide can be produced by the gas-phase nitrogenation of acetylene according to the reaction N2 (g) + C2H2 (g) 2HCN (g) The feed to the reactor consists of an equimolar mixture of acetylene and nitrogen. The temperature of the reaction is 575 K. At this temperature, the standard free energy of reaction is 3.0181 104 J. Determine the percentage of cyanide in the reaction mixture if (a) The pressure is 1 bar (b) The pressure is 200 bar. The fugacity coefficients for HCN, C2H2 and N2 may be taken as 0.607, 0.942 and 1.07 respectively. 9.30 For the synthesis of ammonia according to the reaction

a mixture consisting of 0.5 mol N2, and 1.5 mol H2 is send to the reactor. The equilibrium mixture behaves as ideal gas. Show that the extent of reaction e is given by e = 1 – (1 + 1.299 KP)–1/2 9.31 For the reaction

in equilibrium at 775 K what pressure is required for a 90 per cent conversion of SO2 if the initial mixture is equimolar in the reactants. Assume ideal gases. Take the free energy of the reaction at 775 K to be –2.8626 104 J. 9.32 1-butene is dehydrogenated to 1,3-butadiene according to the reaction C4H8 (g) C4H6 (g) + H2 (g) Determine the extent of reaction at equilibrium at 900 K and 1 bar with (a) 1 mol butene as the reactant (b) a reactant mixture consisting of 1 mol butene and 10 mol steam. The following free energy functions and heat of formation data are available:

9.33 An experimental investigation on the effect of temperature on the reaction A (g) + B (g) C (g) gave the following equilibrium compositions at 373 K and 473 K. The pressure was maintained at 1 bar. At 373 K, yA = 0.414, yB = 0.414 and yC = 0.172; At 473 K, yA = 0.179, yB = 0.179 and yC = 0.642. What will be the equilibrium composition at 423 K and 10 bar if equimolar quantities of the reactants are used? 9.34 Determine the ranges of temperature and pressure for which the equilibrium conversion is at least 10% in the following reaction: CO (g) + 2H2 (g)

CH3OH (g) Assume that stoichiometric quantities of reactants are used. The standard free energy of formation of methanol and CO are respectively –1.626 105 J/mol and –1.374 105 J/mol at 298 K. The standard heat of formation at 298 K are –2.013 105 J/mol and –1.106 105 J/mol. Heat of reaction may be assumed to remain constant.

9.35 The equilibrium constant for the oxidation of SO2 to SO3 according to the reaction

where T is in K and K is in (bar)–1/2. A feed mixture containing 12% SO2, 9% O2 and 79% N2 is reacted at 749 K and 1 bar. Calculate the fractional conversion of SO2. 9.36 Ethanol is produced by vapour-phase hydration of ethylene: C2H4 (g) + H2O (g)

C2H5OH (g)

9.37 Acetic acid is esterified in the liquid phase with ethanol at 373 K and 1 bar to produce ethyl acetate and water according to the reaction CH3COOH (l) + C2H5OH (l) CH3COOC2H5 (l) + H2O (l) The feed consists of 1 mol each of acetic acid and ethanol, estimate the mole fraction of ethyl acetate in the reacting mixture at equilibrium. The standard heat of formation and standard free energy of formation at 298 K are given below:

Assume that the heat of reaction is independent of temperature and the liquid mixture behaves as ideal solution. 9.38 The esterification of ethanol with acetic acid occurs in an aqueous solution as follows: C2H5OH (aq) + CH3COOH (aq) CH3COOC2H5 (aq) + H2O (l) The free energies of formation of acetic acid, ethanol and ethyl acetate in a hypothetical 1 molal solution at 298 K are –3.9645 105 J, –1.8053 105 J and –3.3296 105 J respectively. The free energy of formation of water at 298 K is –2.3735 105 J. What is the equilibrium constant? Starting with a dilute equimolar mixture of ethanol and acetic acid, calculate the extent of reaction and the molalities of ethyl acetate and acetic acid in the equilibrium solution. Assume dilute solution behaviour. 9.39 Carbon dioxide is reduced by graphite according to the equation C (s) + CO2 (g)

2CO (g)

Assuming that equilibrium is attained at 1000 K and 1 bar, calculate the degree of completion of reduction of CO2. The following data are available:

9.40 Carbon dioxide is reduced by graphite according to the equation C (s) + CO2 (g)

2CO (g)

Calculate the effect of pressure on the degree of completion of pure CO2 at 1000 K assuming total pressures of 1, 2 and 3 bar. Gas mixture may be treated as ideal gas and an equilibrium constant value of K = 1.778 may be assumed. 9.41 Calculate the decomposition pressure of limestone at 1000 K given that

9.42 Ammonium chloride decomposes upon heating to yield a gas mixture of ammonia and hydrochloric acid. At what temperature does ammonium chloride exert a decomposition pressure of 1 bar? The standard heat of formation and the standard free energy of formation are as follows:

9.43 The following decomposition reaction occurs at 373 K in the liquid phase. A B+C The equilibrium constant based on pure liquid standard state is 2. The vapour pressures are PA = 5 bar, PB = 20 bar and PC = 2 bar. Assume that all vapours are ideal, liquid B is immiscible with A–C liquid mixture and the A–C mixture is ideal. Calculate the equilibrium pressure and the composition of the liquid and vapour phases. 9.44 The equilibrium constant for the following reaction is found to be 2. A (l) B (l) + C (l) The vapour pressures are PA = 5 bar, PB = 20 bar and PC = 2 bar. A and C form ideal solution and B is immiscible with either A and C or their mixtures. The system consisted of pure A initially. Find the pressure below which only a gas phase exists. 9.45 Mixtures of CO and CO2 are to be processed at temperatures between 900 K and 1000 K. Determine the conditions under which solid carbon might deposit according to the reaction CO2 (g) + C (s) 2CO (g) The equilibrium constants for this reaction are 0.178 at 900 K and 1.58 at 1000 K. (Hint: The activity of solid carbon is less than unity if carbon is not present in the system.) 9.46 Acetylene is catalytically hydrogenated to ethylene at 1500 K and 1 bar. Starting with an equimolar mixture of acetylene and hydrogen what will be the mole fractions at equilibrium? Assume ideal gases. C2H2

2C + H2; K = 5.2

C2H4; K = 0.1923 9.47 What would be the equilibrium conversion of ethyl alcohol to butadiene at 700 K and 1 bar given the following reactions? C2H5OH C2H4 + H2O; DG0 = – 45,427 J/mol C2H5OH CH3CHO + H2; DG0 = – 15,114 J/mol C2H4 + CH3CHO C4H6 + H2O; DG0 = – 5,778 J/mol 9.48 The feed to a reactor consists of an equimolar mixture of A and B. Determine the equilibrium composition of the mixture if the following gas-phase reaction occurs at 1000 K and 1 bar. 2C + 2H2

A+B A + 2B

C + D; K = 0.4 C + E; K = 0.5067

9.49 The following reactions occur at 1500 K and 10 bar. A+B

C + D; K = 2.67

A + C 2E; K = 3.20 The initial mixture consists of 2 mol A and 1 mol B, determine the composition at equilibrium assuming ideal gas behaviour. 9.50 Determine the number of degrees of freedom in a gaseous system consisting of NH3, NO2, NO, H2O, O2 and N2. 9.51 Determine the number of degrees of freedom in a gaseous system consisting of H2O, HCl, O2 and Cl2.

APPENDIX

A

Table for Thermodynamic Properties of Saturated Steam

APPENDIX

B

Table for Thermodynamic Properties of Superheated Steam

APPENDIX

C

Aptitude Test in Chemical Engineering Thermodynamics Given below are some questions which will help the students to check their knowledge in the subject of Chemical Engineering Thermodynamics. These questions are prepared for the benefit of students appearing for various competitive examinations. In fact, some of these are taken from the GATE (Graduate Aptitude Test in Engineering) question papers. C.1 Write the most appropriate answer to the following multiple-choice questions: 1. The major limitation of the first law of thermodynamics is that it does not consider A. Heat as a form of energy B. Rate of change of a process C. Direction of change D. Spontaneous processes 2. All spontaneous processes are A. Reversible B. Irreversible C. Reversible adiabatic D. Adiabatic 3. Which one of the following may be treated as a statement of the second law of thermodynamics A. Heat and work are energy in transit B. It is impossible to convert mechanical work to heat with 100% efficiency C. Absolute zero of temperature cannot be attained in practice D. Energy of an isolated system is conserved 4. CP = CV for a fluid A. Which is compressible B. Whose volume coefficient of expansion is negligible C. Which is homogeneous D. Under normal temperature and pressure 5. Compared to an ordinary vapour compression refrigerator, the COP of an expansion engine vapour compression cycle is A. Greater B. Lesser C. Equal D. No generalisation possible 6. Entropy change of a system is zero in A. Reversible process

B. Adiabatic process C. Reversible adiabatic process D. Isothermal process 7. The heat capacity at constant pressure of a single component system consisting of liquid and vapour phases in equilibrium is A. Infinite B. Zero C. Positive D. Negative 8. The ratio of the fugacity to the pressure is known as A. Activity B. Activity coefficient C. Fugacity coefficient D. Acentric factor 9. Which one of the following is not a property of an ideal gas? A. Internal energy U is a function of temperature alone B. Enthalpy H is a function of temperature alone C. Entropy S is a function of temperature alone D. Heat capacities CP and CV are functions of temperature alone 10. For ideal gases, DH = dT is valid A. For constant volume process B. For constant pressure process C. Irrespective of the nature of the process D. The statement is never true 11. The equation of state for a certain gas is given by P(V – b) = RT, where b is a positive constant. The Joule–Thomson coefficient of this gas would be A. Positive B. Negative C. Zero D. Positive within the inversion points 12. A three-stage compressor is used to compress a gas at 1 bar to a final pressure of 125 bar. For minimum work, the pressure ratios in each stage should be A. 25 B. 5 C. 41.67 D. None of the above 13. The ordinary vapour compression cycle for refrigeration is less efficient than the Carnot cycle, because in the former, A. Evaporation process is non-isothermal B. A two-phase mixture is to be compressed

C. Vapour leaving the compressor is superheated D. Expansion process results in liquefaction 14. The main feature of an absorption refrigeration unit is A. The absence of compression step B. The absence of expansion step C. The absence of condensation step D. None of the above 15. The shaft work done by the fluid in a reversible flow process in which there are no changes in the kinetic, potential and surface energies, is given by A. B. C. D. 16. The reversible work of expansion in a flow process under isothermal condition is equal to A. –(DA)T B. –(DG)T C. –(DU)T D. –(DH)T 17. The decrease in enthalpy accompanying a reversible expansion measures the shaft work in the case of A. A non-flow isothermal process B. An isothermal flow process C. An isentropic non-flow process D. An isentropic flow process 18. For a reversible process occurring at constant temperature and pressure, the decrease in Gibbs free energy measures A. The maximum reversible work B. The maximum reversible work, other than the electrical work C. The maximum reversible work, other than the wok of expansion D. The heat supplied 19. CP = CV when A. B.

C. D. 20. At the triple point of water, the number of degrees of freedom is A. zero B. one C. two D. three 21. The canonical variables for H are: A. P and T B. V and T C. P and S D. V and S 22. Which one of the following is incorrect? A. dU = T dS – P dV B. dH = T dS – V dP C. dA = – S dT – P dV D. dG = – S dT + V dP 23. The volume coefficient of expansion b of an ideal gas equals A. 1/T B. 1/P C. T D. P 24. Fugacity has the same dimensions as that of A. Gibbs free energy B. Pressure C. Temperature D. Fugacity is dimensionless 25. The change in free energy when a real gas undergoes an isothermal change in state is A. DG = RT ln (V2/V1) B. DG = RT ln (P2/P1) C. DG = RT ln (f2/f1) D. DG = RT ln (g2/g1) 26. The difference between the heat supplied and the work extracted in a steady flow process in which the kinetic and potential energy changes are negligible, is equal to A. The change in internal energy B. The change in enthalpy C. The change in the work function

D. The change in the Gibbs free energy 27. Compressibility factor Z of a gas is A. The ratio of fugacity in the given state to fugacity in the standard state B. The ratio of actual volume to the volume of the gas if it were ideal C. The change in volume with temperature at constant pressure D. The difference between actual volume and ideal gas volume 28. The temperature at which a transition occurs from a compressibility factor less than 1.00 to that greater than 1.00 is known as A. The critical temperature B. The critical solution temperature C. The inversion point D. The Boyle point 29. The coefficient of compressibility k is defined as A. B. C. D. 30. For any equation of state to be valid, at the critical point the critical isotherm should have A. A maximum B. A minimum C. A point of inflection D. Negative slope 31. At constant temperature and pressure, the decrease in Gibbs free energy is a measure of A. The maximum work B. The maximum net work C. The unavailable energy D. The loss in capacity to do work 32. In thermodynamics, a phase means A. A closed system B. An open system C. A homogeneous system D. A heterogeneous system 33. The net change in a state function is zero for A. A reversible process B. An Irreversible process

C. A cyclic process D. A non-cyclic process 34. The third law of thermodynamics deals with A. Chemical reactions B. Quantitative equivalence between heat and work C. Rate of change of a process D. Absolute entropy of perfect crystalline substances 35. DG = DA for a process occurring at A. Constant pressure B. Constant volume C. Constant pressure and constant temperature D. Constant pressure and constant volume 36. As pressure approaches zero, fugacity coefficient value tends to A. Pressure B. Zero C. Unity D. Infinity 37. For a gas obeying the van der Waals equation of state, at the critical temperature, A. Both (∂P/∂V)T and (∂2P/∂V2)T are zero B. The first derivative is zero, while the second derivative is non-zero C. The second derivative is zero while the first derivative is non-zero D. Both the derivatives are non-zero (1991) 38. For an ideal gas, the slope of the pressure-volume curve at a given point will be A. Steeper for an isothermal than for an adiabatic process B. Steeper for an adiabatic than for an isothermal process C. Identical for both the processes D. Of opposite sign (1991) 39. The shape of T-S diagram for Carnot cycle is A. A rectangle B. A rhombus C. A trapezoid D. A circle (1991) 40. During Joule–Thomson expansion of gases, A. Enthalpy remains constant ]B. Entropy remains constant C. Temperature remains constant D. None of the above (1992)

41. For a single-component, two-phase mixture, the number of independent variable properties are A. Two B. One C. Zero D. Three (1992) 42. Ideal gas law is applicable at A. Low T, low P B. High T, high P C. Low T, high P D. High T, low P (1994) 43. The second law of thermodynamics states that A. The energy change of a system undergoing any reversible process is zero B. It is not possible to transfer heat from a lower temperature to a higher temperature C. The total energy of the system and the surroundings remain constant D. None of the above (1994) 44. A solid is transformed into vapour without changing into the liquid phase A. At the triple point B. At the boiling point C. Below the triple point D. Always (1995) 45. At the inversion point, the Joule–Thomson coefficient is A. Positive B. Negative C. Zero D. Cannot be generalised (1995) 46. The kinetic energy of a gas molecule is zero at A. 0∞C B. 273∞C C. 100∞C D. –273∞C (1995) 47. Assuming that CO2 obeys the perfect gas law, the density of CO2 in kg/m3 at 536 K and 202.6 kPa is A. 1 B. 2

C. 3 D. 4 (1995) 48. A closed system is cooled reversibly from 373 K to 323 K. If no work is done on the system A. Its internal energy (U) decreases and its entropy (S) increases B. U and S both decrease C. U decreases but S is constant D. U is constant but S decreases (1995) 49. The equation dU = T dS – P dV is applicable to infinitesimal changes occurring in A. An open system of constant composition B. A closed system of constant composition C. An open system with changes in composition D. A closed system with changes in composition (1996) 50. A system undergoes a change from a given initial state to a given final state either by an irreversible process or by a reversible process. Then A. DSI is always > DSR B. DSI is sometimes > DSR C. DSI is always < DSR D. DSI is always = DSR where DSI and DSR are the entropy changes of the system for the irreversible and reversible processes, respectively. (1997) 51. The change in Gibbs free energy for vaporisation of a pure substance is A. Positive B. Negative C. Zero D. May be positive or negative (1997) 52. A change in state involving a decrease in entropy can be spontaneous only if A. It is exothermic B. It is isenthalpic C. It takes place isothermally D. It takes place at constant volume (1998) 53. A Carnot cycle consists of the following steps: A. Two isothermals and two isentropics B. Two isobarics and two isothermals C. Two isochorics and two isobarics

D. Two isothermals and two isochorics (1998) 54. It is desired to bring about certain change in the state of a system by performing work on the system under adiabatic conditions A. The amount of work needed is path-dependent B. Work alone cannot bring about such a change of state C. The amount of work needed is independent of path D. More information is needed to conclude anything about the path-dependence or otherwise of the work needed. (1998) 55. Chemical potential is A. An extensive property B. An intensive property C. A path property D. A reference property 56. According to the phase rule, the triple point of a pure substance is A. Invariant B. Univariant C. Bivariant D. None of the above 57. Which one of the following is incorrect with reference to partial molar properties? A. They are intensive properties B. They are always positive C. They represent the contribution of individual components to the total solution property D. They vary with composition of the solution 58. All but one of the following represent the chemical potential of component i in solution. Find the odd man out.

A.

B.

C.

D. 59. Which one of the following statements is not valid for an ideal solution? A. There is no volume change on mixing B. There is no enthalpy change on mixing

C. There is no entropy change on mixing D. Fugacity is directly proportional to concentration. 60. For the standard state of pure component at the solution pressure, the activity of a component in an ideal solution is equal to A. its fugacity in the solution B. its mole fraction in the solution C. its partial pressure D. its chemical potential 61. Which one of the following is the correct form of Gibbs–Duhem equation for a binary solution? A. B. C. D. 62. Which one of the following is true for the excess property ME? A. ME = M – S xiMi B. ME = M – S xi C. ME = M – Mid D. ME = DM 63. One of the following statements is incorrect for a multicomponent system consisting of two phases in thermodynamic equilibrium. Identify it. A. The temperatures of both phases are the same B. The pressure is uniform throughout C. The concentrations of a component in both phases are equal D. The chemical potentials of a component in both phases are equal 64. When an ideal binary solution is boiled at constant pressure in a closed container, A. The boiling temperature remains constant at some value between the bubble point and the dew point, till the entire liquid is vaporised B. The boiling temperature varies between the bubble point and the dew point of the solution C. The boiling occurs at a constant temperature known as the bubble point D. The boiling occurs either at the bubble point or at the dew point 65. The value of activity coefficient for an ideal solution is A. One B. Zero C. Equal to Henry’s law constant

D. Equal to the vapour pressure 66. A solution exhibiting positive deviation from ideality A. Always forms a minimum boiling azeotrope B. Always forms a maximum boiling azeotrope C. Has a total vapour pressure that is less than that predicted by Raoult’s law D. When formed from its constituents there is an absorption of heat 67. Which one of the following statements is true with reference to the minimum boiling azeotropes? A. There is a minimum on the vapour-pressure curve B. The solution exhibits positive deviation from ideality C. The dew point is greater than the bubble point D. The activity coefficients are less than unity 68. The vaporisation equilibrium constant (K-factor) depends upon A. Temperature only B. Pressure only C. Temperature and pressure only D. Temperature, pressure and concentration 69. The vapour–liquid equilibrium data are thermodynamically inconsistent if A. The slopes of the ln g1 vs x1 and ln g2 vs x1 curves have opposite signs B. When plotted against x1, ln g2 and ln g1 curves pass through a maximum at the same composition C. Both g1 and g2 are greater than unity D. ln g1 vs x1 curve has a maximum and ln g2 vs x1 curve has a minimum at a particular x1. 70. A mixture of two immiscible liquids A and B is in equilibrium with its vapour at temperature T and pressure P. The vapour pressures of pure A and pure B are respectively. The relation applicable to the system is A. B. C. D. T > TA + TB, where TA and TB are boiling points of pure A and pure B respectively 71. The mutual solubility of two partially miscible liquids increases with temperature. At what temperature do the two liquid phases become identical? A. At the critical point B. At the three-phase temperature C. At the upper critical solution temperature D. At the dew point 72. Benzene and water may be considered immiscible. A mixture of benzene (20 g) and water (80 g) is taken in a vessel and boiled. It boils at 101.3 kPa and 342 K. At this temperature vapour pressure of benzene is 71.18 kPa and that of water is 30.12 kPa. What is the concentration of benzene in the vapour in mass per cent?

A. 70% B. 91% C. 20% D. 80% 73. The necessary and sufficient condition for equilibrium between two phases is A. Concentration of each component should be same in the two phases B. The temperature of each phase should be the same C. The pressure should be same in the two phases D. The chemical potential of each component should be same in the two phases (1992) 74. For a system in equilibrium, at a given temperature and pressure, A. The entropy must be a minimum B. The enthalpy must be a minimum C. The internal energy must be a minimum D. The Gibbs free energy must be a minimum (1991) 75. To obtain the integrated form of Clausius–Clapeyron equation

from the exact Clapeyron equation, it is assumed that: A. The volume of the liquid phase is negligible compared to that of the vapour phase B. The vapour phase behaves as an ideal gas C. The heat of vaporisation is independent of temperature D. All the above are applicable (1991) 76. One mole of a binary mixture of a given composition is flash vaporised at a fixed P and T. If Raoult’s law is obeyed, then changing the feed composition would affect A. The product composition but not the fraction vaporised B. The product composition as well as the fraction vaporised C. The fraction vaporised but not the product composition D. Neither the product composition nor the fraction vaporised (1997) 77. The molar excess free energy, GE, for a binary liquid mixture at T and P is given by GE/RT = Ax1x2, where A is constant. The corresponding equation for ln g1, where g1 is the activity coefficient of component 1, is A. B. Ax1 C. Ax2 D.

(1997) 78. A liquid mixture contains 30% o-xylene, 60% p-xylene and 10% m-xylene (all percentages in w/w). Which of the following statements would be true for this mixture? A. The mixture exhibits an azeotrope at 101.3 kPa B. The composition of the mixture in per cent by volume is: o-xylene 30, p-xylene 60 and mxylene 10 C. The composition of the mixture in mole per cent is: o-xylene 30, p-xylene 60 and m-xylene 10 D. The mixture contains optical isomers (1998) 79. The theoretical minimum work required to separate one mole of a liquid mixture at 1 bar, containing 50 mole per cent each of n-heptane and n-octane into pure compounds each at 1 bar is A. –2RT ln 0.5 B. –RT ln 0.5 C. 0.5RT D. 2RT (1996) 80. If the heat of solution of an ideal gas in a liquid is negative, then its solubility at a given partial pressure varies with temperature as A. Solubility increases as temperature increases B. Solubility decreases as temperature increases C. Solubility is independent of temperature D. Solubility increases or decreases with temperature depending on the Gibbs free energy of the solution (1998) 81. For evaluation of heat effects, all thermochemical equations can be treated as algebraic equations. This is a consequence of A. Le Chatlier’s principle B. Third law of thermodynamics C. Hess’s law D. Principle of corresponding states 82. For a chemical reaction occurring at equilibrium under constant temperature and pressure, the change in Gibbs free energy is A. Maximum B. Minimum C. Zero D. None of the above 83. The equilibrium constant for the reaction N2 + 3H2 = 2NH3 is 0.1084. Under the same conditions, the equilibrium constant for the reaction N2 + H2 = NH3 is A. 0.1084

B. 0.3292 C. 0.0118 D. 0.0542 84. The equilibrium constant is independent of A. The pressure at equilibrium B. The temperature at equilibrium C. The number of moles involved in the stoichiometric equation for the reaction D. The temperature and pressure at the equilibrium 85. For a highly favourable chemical reaction, the standard free energy change is A. Zero B. Unity C. Positive D. Negative 86. For an exothermic reaction, the increase in temperature results in A. Increase of K B. Decrease of K C. No change of K D. None of the above 87. For the reaction N2 + 3H2 2NH3, the increase in pressure results in A. Increase of K B. Increase in the concentration of ammonia at equilibrium C. Decrease of K D. Decrease in the concentration of ammonia at equilibrium 88. For the equilibrium yield in a gas-phase reaction, diluting the reaction mixture with an inert gas A. Has the same effect as that of an increase in pressure B. Has the same effect as that of a decrease in pressure C. Has no correlation with a change in pressure D. Always produces unfavourable results 89. Which one of the following statements is true for ammonia synthesis reaction? A. Increase in temperature increases K B. Increase in pressure decreases the conversion C. Presence of argon in the reactant stream decreases conversion D. Increase in pressure increases K 90. The number of degrees of freedom for a system prepared by partially decomposing CaCO3 into an evacuated space is A. 0 B. 1 C. 2 D. 3 91. An exothermic gas-phase reaction proceeds according to the equation

3A + 2B 2R The equilibrium conversion for this reaction: A. Increases with an increase in temperature B. Decreases on dilution with an inert gas C. Decreases with an increase in pressure D. Is affected by the presence of a catalyst (1990) 92. The reaction A (l) = R (g) + S (g) is allowed to reach equilibrium conditions in an autoclave. At equilibrium there are two phases, one pure liquid phase of A and the other a vapour phase of A, R and S. Initially, A alone is present. The number of degrees of freedom are A. 1 B. 2 C. 3 D. 0 (1996) 93. Given 3H2 + CO CH4 + H2O, Kp = 101.84 and 4H2 + CO2 CH4 + 2H2O, Kp = 101.17 the Kp for the reaction CO + H2O A. 103.01

CO2 + H2 is

B. 10–0.67 C. 10–3.01 D. 100.67 (1996) 94. Which of the following is true for virial equation of state? A. Virial coefficients are universal constants B. Virial coefficient B represents three-body interactions C. Virial coefficients are functions of temperature only D. For some gases, virial equations and ideal gas equation are the same (1999) 95. A gas mixture of three components is brought in contact with a dispersion of an organic phase in water. The degrees of freedom of the system are A. 4 B. 3 C. 5 D. 6 (1999)

96. Maxwell’s equation corresponding to the identity, dH = T dS + V dP + S mi dni, is A. B. C. D. (1999) 97. In a binary liquid solution of components A and B, if component A exhibits positive deviation from Raoult’s law then component B A. exhibits positive deviation from Raoult’s law B. exhibits negative deviation from Raoult’s law C. obeys Raoult’s law D. may exhibit either positive or negative deviation from Raoult’s law (2000) 98. Assume that benzene is insoluble in water. The normal boiling points of benzene and water are 353.3 K and 373.2 K, respectively. At a pressure of 1 atm, the boiling point of a mixture of benzene and water is A. 353.3 K B. less than 353.3 K C. 373.2 K D. greater than 353.3 K but less than 373.2 K (2000) 99. On a P-V diagram of an ideal gas, suppose a reversible adiabatic line intersects a reversible isothermal line at point A. Then at point A, the slope of the reversible adiabatic line (∂P/∂V)S and the slope of the reversible isothermal line (∂P/∂V)T are related as A. B. C. D.

where g = CP/CV. (2000) 100. The thermal efficiency of a reversible heat engine operating between two given thermal reservoirs is 0.4. The device is used either as a refrigerator or as a heat pump between the same reservoirs. The coefficient of performance as a refrigerator (COP)R and the coefficient of performance as a heat pump (COP)HP are A. (COP)R = (COP)HP = 0.6 B. (COP)R = 2.5; (COP)HP = 1.5 C. (COP)R = 1.5; (COP)HP = 2.5 D. (COP)R = (COP)HP = 2.5 (2000) 101. At a given temperature, K1, K2 and K3 are the equilibrium constants for the following reactions 1, 2, 3 respectively:

Then K1, K2 and K3 are related as A. K3 = K1K2 B. K3 = (K1K2)0.5 C. K3 = (K1 + K2)/2 D. K3 = (K1K2)2 (2000) 102. A reasonably general expression for vapour–liquid equilibrium at low to moderate pressures is

where fi is a vapour fugacity coefficient, gi is the liquid activity coefficient and is the fugacity of pure component i. The Ki value (yi = Kixi) is therefore, in general, a function of A. temperature only B. temperature and pressure only C. temperature, pressure and liquid composition xi only D. temperature, pressure, liquid composition xi, and vapour composition yi (2001) 103. High pressure steam is expanded adiabatically and reversibly through a well insulated turbine which produces some shaft work. If the enthalpy change and entropy change across the turbine are represented by DH and DS respectively, for this process:

A. DH = 0 and DS = 0 B. DH 0 and DS = 0 C. DH 0 and DS 0 D. DH = 0 and DS 0 (2001) 104. For the case of a fuel gas undergoing combustion with air, if the air/fuel ratio is in-creased, the adiabatic flame temperature will A. increase B. decrease C. increase or decrease depending on the fuel type D. not change (2001) 105. The Maxwell relation derived from the differential equation for the Helmholtz free energy (DA) is A. B. C. D. (2001) 106. At 373 K, water and methyl cyclohexane both have a vapour pressure of 1.0 atm. The latent heats of vaporization are 40.63 kJ/kmol for water and 31.55 kJ/kmol for cyclohexane. The vapour pressure of water at 423 K is 4.69 atm. The vapour pressure of methyl-cyclohexane at 423 K is expected to be A. Significantly less than 4.69 atm B. Nearly equal to 4.69 atm C. Significantly more than 4.69 atm D. Indeterminate due to lack of data (2001) 107. Air enters an adiabatic compressor at 300 K. The exit temperature for a compression ratio of 3, assuming air to be an ideal gas (g = Cp/CV = 7/5) and the process to be reversible, is A. 300(32/7) B. 300(33/5) C. 300(33/7) D. 300(35/7) (2001)

108. The extent of reaction is A. different for reactants and products B. dimensionless C. dependent on the stoichiometric coefficients D. all of the above (2002) 109. An exothermic reaction takes place in an adiabatic reactor. The product temperature the reactor feed temperature. A. is always equal to B. is always greater than C. is always less than D. may be greater or less than (2002) 110. The number of degrees of freedom for an azeotropic mixture of ethanol and water in vapourliquid equilibrium is A. 3 B. 1 C. 2 D. 0 (2002) 111. The partial molar enthalpy of a component in an ideal binary gas mixture of composition z, at a temperature T and pressure P, is a function only of A. T B. T and P C. T, P and z D. T and z (2002) 112. Which of the following identities can most easily be used to verify steam table data for superheated steam? A. B. C. D. (2002) 113. Steam undergoes isentropic expansion in a turbine from 5000 kPa and 673 K (entropy = 6.65 kJ/kg K) to 150 kPa (entropy of saturated liquid = 1.4336 kJ/kg K, entropy of saturated vapour =

7.2234 kJ/kg K). The exit condition of steam is A. superheated vapour B. partially condensed vapour with quality of 0.9 C. saturated vapour D. partially condensed vapour with quality of 0.1 (2002) 114. A rigid vessel, containing three moles of nitrogen gas at 303 K is heated to 523 K. Assume the average heat capacities of nitrogen to be CP = 29.1 J/mol K and CV = 20.8 J/mol K. The heat required, neglecting the heat capacity of the vessel, is A. 13728 J B. 19206 J C. 4576 J D. 12712 J (2002) 115. One cubic metre of an ideal gas at 500 K and 1000 kPa expands reversibly to 5 times its volume in an insulated container. If the specific heat capacity (at constant pressure) of the gas is 21 J/mol K, the final temperature will be A. 35 K B. 174 K C. 274 K D. 154 K (2002) 116. Ammonia is produced by the following reaction: N2 + 3H2 2NH3 In a commercial process for ammonia production, the feed to an adiabatic reactor con-tains 1 kmol/s of nitrogen and stoichiometric amounts of hydrogen at 700 K. Assume the feed and product streams to be ideal gas mixtures. The heat of reaction at 700 K for the above reaction is calculated to be – 94.2 kJ/mol. The mean molar heat capacity in the range of 700–800 K are 0.03, 0.0289 and 0.0492 kJ/mol K for nitrogen, hydrogen and ammonia respectively. What is the maximum allowable conversion in the reactor, if the adiabatic temperature rise across the reactor should not exceed 100 K? A. 87.9% B. 12.1% C. 25.8% D. 74.2% (2002) 117. In Joule’s experiments, an insulated container contains 20 kg of water initially at 25°C. It is stirred by an agitator, which is made to turn by a slowly falling body weighing 40 kg through a height of 4 m. The process is repeated 500 times. The acceleration due to gravity is 9.8 m/s2. Neglecting the heat capacity of the agitator, the temperature of water in (°C) is A. 40.5

B. 34.4 C. 26.8 D. 25 (2003) 118. One mole of nitrogen at 8 bar and 600 K is contained in a piston-cylinder assembly. It is brought to 1 bar isothermally against a resisting pressure of 1 bar. The work done (in joules) by the gas is A. 30554 B. 10373 C. 4988.4 D. 4364.9 (2003) 119. For water at 573 K, it has a vapour pressure of 8592.7 kPa and fugacity 6738.9 kPa. Under these conditions, one mole of water in liquid phase has a volume of 25.28 cm3, and that in vapour phase 391.1 cm3. The fugacity of water (in kPa) at 9000 kPa is A. 6738.9 B. 6753.5 C. 7058.3 D. 9000 (2003) 120. The heat capacity of air can be approximately expressed as CP = 26.693 + 7.365 10–3T where CP is in J/mol K and T is in K. The heat given off by one mole of air when cooled at 1 atmospheric pressure from 500°C to –100°C is A. 10.73 kJ B. 16.15 kJ C. 18.11 kJ D. 18.33 kJ (2003) 121. A solid metallic block weighing 5 kg has an initial temperature of 500°C; 40 kg of water initially at 25°C is contained in a perfectly insulated tank. The metallic block is brought into contact with water. Both of them come to equilibrium. The specific heat of the block material is 0.4 kJ/kg K. Ignoring the effect of expansion and contraction, and also the heat capacity of the tank, the total entropy change in kJ/kg K is A. –1.87 B. 0.0 C. 1.26 D. 3.91 (2003) 122. The following heat engine produces a power of 100,000 kW. The heat engine operates

between 800 K and 300 K. It has a thermal efficiency equal to 50% of that of the Carnot engine for the same temperatures. The rate at which heat is absorbed from the hot reservoir is A. 100,000 kW B. 160,000 kW C. 200,000 kW D. 320,000 kW (2003) 123. A steam turbine operates with a superheated steam flowing at 1 kg/s. The steam is supplied at 41 bar and 500°C, and discharges at 1.01325 bar and 100°C.

The maximum power output (in kW) will be A. 644.0 B. 767.9 C. 871.3 D. 3024.8 (2003) 124. At 60°C, the vapour pressures of methanol and water are 84.562 kPa and 19.953 kPa respectively. An aqueous solution of methanol at 60°C exerts a pressure of 39.223 kPa; the liquid phase and the vapour phase mole fractions of methanol are 0.1686 and 0.5714 respectively. The activity coefficient of methanol is A. 1.572 B. 1.9398 C. 3.389 D. 4.238 (2003) 125. One kilogram of saturated steam at 373 K and 1.01325 bar is contained in a rigid walled vessel. It has a volume of 1.673 m3. It is cooled to 371 K; the saturation pressure is 0.943 bar. One kilogram of water vapour under these conditions has a volume of 1.789 m3. The amount of water vapour condensed in kilograms is A. 0.0 B. 0.065 C. 0.1 D. 1.0 (2003)

126. One kilogram of saturated steam at 373 K and 1.01325 bar is contained in a rigid walled vessel. It has a volume of 1.673 m3. It is cooled to 371 K; the saturation pressure is 0.943 bar. One kilogram of water vapour under these conditions has a volume of 1.789 m3. The latent heat of condensation in kJ/kg under these conditions is A. 40732 B. 2676 C. 2263 D. 540 (2003) 127. For an ideal gas mixture undergoing a reversible gaseous phase chemical reaction, the equilibrium constant A. is independent of pressure B. increases with pressure C. decreases with pressure D. increases/decreases with pressure depending on the stoichiometric coefficients of the reaction (2004) 128. As pressure approaches zero, the ratio of fugacity to pressure (f/P) for a gas approaches A. zero B. unity C. infinity D. an indeterminate value. (2004) 129. A perfectly insulated container of volume V is divided into two equal halves by a partition. One side is under vacuum and the other side contains one mole of an ideal gas (with constant heat capacity) at 298 K. If the partition is broken, the final temperature of the gas in the container A. will be greater than 298 K B. will be 298 K C. will be less than 298 K D. cannot be determined (2004) 130. One mole of methane at 298 K undergoes complete combustion in a stoichiometric amount of air also at 298 K. Both the reactants and products are in the gas phase. CH4 + 2CO2 CO2 + 2H2O DH0298 = –730 kJ/mol If the average specific heat of all the gases/vapours is 40 J/mol K, the maximum temperature rise (in K) of the exhaust gases would be approximately A. 1225 B. 1335 C. 1525 D. 1735

(2004) 131. A vessel of volume 1000 m3 contains air which is saturated with water vapour. The total pressure and temperature are 100 kPa and 293 K respectively. Assuming that the vapour pressure of water at 293 K is 2.34 kPa, the amount of water vapour in kilograms in the vessel is approximately A. 17 B. 20 C. 25 D. 34 (2004) 132. A car tyre of volume 0.057 m3 is inflated to 300 kPa and 300 K. After the car is driven for 10 hours, the pressure in the tyre increases to 330 kPa. Assume air is an ideal gas and CV for air is 21 J/mol K. The change in internal energy of air in tyre in J/mol is A. 380 B. 630 C. 760 D. 880 (2004) 133. A gas obeys P(V – b) = RT. The work obtained from reversible isothermal expansion of one mole of this gas from an initial volume Vi to a final volume Vf is A. B. C. D. (2004) 134. A cyclic engine exchanges heat with two reservoirs maintained at 100°C and 300°C, respectively. The maximum work (in J) that can be obtained from 1000 J of heat extracted from the hot reservoir is A. 349 B. 651 C. 667 D. 1000 (2004) 135. The vapour pressure of water is given by Psat = A –

, where A is a constant, Psat is

vapour pressure in atm, and T is temperature in K. The vapour pressure of water (in atm) at 50°C is approximately A. 0.07 B. 0.09 C. 0.11 D. 0.13 (2004) 136. At standard conditions,

The standard free energy of formation of NO in kJ/mol is A. 15 B. 30 C. 85 D. 170 (2004) 137. The boiling points for pure water and toluene are 100°C and 110.6°C respectively. Toluene and water are completely immiscible in each other. A well-agitated equimolar mixture of toluene and water is prepared. The temperature at which the above mixture will exert a pressure of one standard atm is A. less than 100°C B. 100°C C. between 100 and 110.6°C D. 110.6°C (2004) 138. The boiling points for pure water and toluene are 100°C and 110.6°C respectively. Toluene and water are completely immiscible in each other. A well-agitated equimolar mixture of toluene and water is prepared. At a total pressure of one standard atm exerted by the vapours of water and toluene, the mole fraction of water xw in the vapour phase satisfies A. 0 < xw < 0.5 B. xw = 0.5 C. 0.5 < xw < 1.0 D. xw = 1.0 (2004) 139. In the van der Waal equation of state, what are the criteria applied at the critical point to determine the parameters a and b?

A.

B.

C.

D. (2005) 140. Which one of the following statements is true? A. Heat can be fully converted into work. B. Work cannot be fully converted to heat. C. The efficiency of a heat engine increases as the temperature of the heat source is increased while keeping the temperature of the heat sink fixed. D. A cyclic process can be devised whose sole effect is to transfer heat from a lower temperature to a higher temperature. (2005) 141. A Carnot heat engine cycle is working with an ideal gas. The work performed by the gas during the adiabatic expansion and compression steps, W1 and W2 respectively, are related as A. |W1| > |W2| B. |W1| < |W2| C. W1 = W2 D. W1 = –W2 (2005) 142. The van Laar activity coefficient model for a binary mixture is given in the form

Given g1 = 1.40, g2 = 1.25, x1 = 0.25, x2 = 0.75, determine the constants A* and B*. A. A* = 0.5, B* = 0.3 B. A* = 3, B* = 0.5

C. A* = 0.333, B* = 0.2 D. A* = 2, B* = 0.333 (2005) 143. A liquid mixture of benzene and toluene is in equilibrium with its vapour at 101.3 kPa and 373 K. The vapour pressures of benzene and toluene at 373 K are 156 and 63 kPa respectively. Assuming that the system obeys Raoult’s law, the mole fraction of benzene in the liquid phase is A. 0.65 B. 0.41 C. 0.065 D. 0.04 (2005) 144. A frictionless cylinder piston assembly contains an ideal gas. Initially at pressure (P1) = 100 kPa, temperature (T1) = 500 K and volume (V1) = 700 10–6 m3. This system is supplied with 100 J of heat and pressure is maintained constant at 100 kPa. The enthalpy variation is given by h (J/mol) = 30000 + 50T, where T is the temperature in K, and the universal gas constant R = 8.314 J/mol K. The final volume of the gas (V2) in m3 is A. 700 10–6 B. 866.32 C. 934.29 D. 1000.23

10–6 10–6 10–6

(2005) 145. A frictionless cylinder piston assembly contains an ideal gas. Initially pressure (P1) = 100 kPa, temperature (T1) = 500 K and volume (V1) = 700 10–6 m3. This system is supplied with 100 J of heat and pressure is maintained constant at 100 kPa. The enthalpy variation is given by h (J/mol) = 30000 + 50T, where T is the temperature in K, and the universal gas constant R = 8.314 J/mol K. The change in internal energy of the gas is A. 0 B. 100 C. 23.43 D. 83.37 (2005) 146. Heat and work are A. intensive properties B. extensive properties C. point functions D. path functions 147. A frictionless piston-cylinder device contains a gas initially at 0.8 MPa and 0.015 m3. It expands quasi-statically at constant temperature to a final volume of 0.030 m3. The work output

(in kJ) during the process will be A. 8.32 B. 12.0 C. 554.67 D. 8320.00 148. The contents of a well-insulated tank are heated by a resistor of 23 W in which 10 A current is flowing. Consider the tank along with its contents as a thermodynamic system. The work done by the system and the heat transfer to the system are positive. The rates of heat Q, work W and change in internal energy DU during the process in kW are A. Q = 0, W = –2.3, DU = +2.3 B. Q = +2.3, W = 0, DU = +2.3 C. Q = +2.3, W = 0, DU = –2.3 D. Q = 0, W = +2.3, DU = –2.3 149. A compressor undergoes a reversible steady-flow process. The gas at inlet and outlet of the compressor is designated as state 1 and state 2, respectively. Potential and kinetic energy changes are to be ignored. The following notations are used: V specific volume and P pressure of the gas The specific work required to be supplied to the compressor for this gas compression process is A. B. C. V1(P2 – P1) D. P2(V1 – V2)

150. A gas contained in a cylinder is compressed, the work required for compression being 5000 kJ. During the process heat interaction of 2000 kJ causes the surroundings to be heated. The change in internal energy of gas during the process is A. –7000 kJ B. –3000 kJ C. 3000 kJ D. 7000 kJ 151. A mono-atomic ideal gas (g = 1.67, molecular weight = 40) is compressed adiabatically from 0.1 MPa, 300 K to 0.2 MPa. The universal gas constant is 8.314 kJ/kmol K. The work of compression of the gas (in kJ/kg) is A. 29.7 B. 19.9 C. 13.3 D. 0 152. A gas expands in a frictionless piston-cylinder arrangement. The expansion process is very

slow and is resisted by an ambient pressure of 100 kPa. During the expansion process, the pressure of the system (gas) remains constant at 300 kPa. The change in volume of the gas is 0.01 m3. The maximum amount of work that could be utilized from the above process is A. zero B. 1 kJ C. 2 kJ D. 3 kJ 153. One kilogram water at room temperature is brought into contact with a high temperature thermal reservoir. The entropy change of the universe is A. equal to entropy change of the reservoir B. equal to entropy change of water C. equal to zero D. always positive 154. If a closed system is undergoing an irreversible process, the entropy of the system A. must increase B. always remains constant C. must decrease D. can increase, decrease or remain constant 155. Two moles of oxygen are mixed adiabatically with another 2 mol of oxygen in a mixing chamber, so that the final total pressure and temperature of the mixture become equal to that of the individual constituents at their initial states. The universal gas constant is given as R. The change in entropy due to mixing per mole of oxygen is given by A. –R ln 2 B. zero C. R ln 2 D. R ln 4 156. Availability of a system at any given state is A. a property of the system B. the maximum work obtainable as the system goes to dead state C. the total energy of the system D. the maximum useful work obtainable as the system goes to dead state 157. Consider the following two processes: I. A heat source at 1200 K loses 2500 kJ of heat to sink at 800 K II. A heat source at 800 K loses 2000 kJ of heat to sink at 500 K which of the following statements is true? A. Process I is more irreversible than Process II B. Process II is more irreversible than Process I C. Irreversibility associated in both the processes is equal D. Both the processes are reversible 158. An irreversible heat engine extracts heat from a high temperature source at a rate of 100 kW and rejects heat to a sink at a rate of 50 kW. The entire work output of the heat engine is

used to drive a reversible heat pump operating between a set of independent isothermal heat reservoirs at 17°C and 75°C. The rate (in kW) at which the heat pump delivers heat to its high temperature sink is A. 50 B. 250 C. 300 D. 360 Common data for Questions 159 and 160. In an experimental set-up, air flows between two stations P and Q adiabatically. The direction of flow depends on the pressure and temperature conditions maintained at P and Q. The conditions at station P are 150 kPa and 350 K. The temperature at station Q is 300 K. The following are the properties and relations pertaining to air: Specific heat at constant pressure CP = 1.005 kJ/kg K, Specific heat at constant volume CV = 0.718 kJ/kg K, Universal gas constant R = 0.287 kJ/kg K, Enthalpy H = CpT, Internal energy U = CVT. 159. If the air has to flow from station P to station Q, the maximum possible value of pressure in kPa at station Q is close to A. 50 B. 87 C. 128 D. 150 160. If the pressure at station Q is 50 kPa, the change in entropy (SQ – SP) in kJ/kg K is A. – 0.155 B. 0 C. 0.160 D. 0.355 161. A cyclic device operates between three thermal reservoirs, as shown in the figure. Heat is transferred to/from the cyclic device. It is assumed that heat transfer between each thermal reservoir and cyclic device takes place across negligible temperature difference. Interactions between the cyclic device and the respective thermal reservoirs that are shown in the figure are all in the form of heat transfer.

The cyclic device can be:

A. A reversible heat engine B. A reversible heat pump or a reversible refrigerator C. An irreversible heat engine D. An irreversible heat pump or an irreversible refrigerator 162. A heat transformer is a device that transfers a part of the heat supplied to it at an intermediate temperature, to a high temperature reservoir while rejecting the remaining part to a low temperature heat sink. In such a heat transformer, 100 kJ of heat is supplied at 350 K. The maximum amount of heat in kJ that can be transferred to 400 K, when the rest is rejected to a heat sink at 300 K, is A. 12.50 B. 14.29 C. 33.33 D. 57.14 163. An ideal gas of mass m and temperature T1 undergoes a reversible isothermal process from an initial pressure P1 to final pressure P2. The heat loss during the process is Q. The entropy change DS of the gas is A. B. C. D. zero Common data for Questions 164 and 165. Nitrogen gas (molecular weight 28) is enclosed in a cylinder by a piston, at the initial condition of 2 bar, 298 K and 1 m3. In a particular process, the gas slowly expands under isothermal condition, until the volume becomes 2 m3. Heat exchange occurs with the atmosphere at 298 K during this process. 164. The work interaction for the nitrogen gas is A. 200 kJ B. 138.6 kJ C. 2 kJ D. –200 kJ 165. The entropy change for the universe during the process in kJ/K is A. 0.4652 B. 0.0067 C. 0 D. – 0.6711 166. A Carnot cycle is having an efficiency of 0.75. If the temperature of the high temperature reservoir is 727°C, what is the temperature of the low temperature reservoir? A. 23°C

B. –23°C C. 0°C D. 250°C 167. A cyclic heat engine does 50 kJ of work per cycle. If the efficiency of the heat engine is 75%, the heat rejected per cycle is A.

kJ

B.

kJ

C.

kJ

D.

kJ

168. A solar collector receiving solar radiation at the rate of 0.6 kW/m2 transforms it to the internal energy of a fluid at an overall efficiency of 50%. The fluid heated to 350 K is used to run a heat engine which rejects heat at 313 K. If the heat engine is to deliver 2.5 kW power, the minimum area of the solar collector required would be A. 8.33 m2 B. 16.66 m2 C. 39.68 m2 D. 79.36 m2 169. Considering the relationship TdS = d U + P d V between the entropy S, internal energy U, pressure P, temperature T and volume V, which of the following statements is correct? A. It is applicable only for a reversible process B. For an irreversible process, TdS > dU + PdV C. It is valid only for an ideal gas D. It combines first and second laws for a reversible process 170. A balloon containing an ideal gas is initially kept in an evacuated and insulated room. The balloon ruptures and the gas fills up the entire room. Which one of the following statements is true at the end of the above process? A. The internal energy of the gas decreases from its initial value but the enthalpy remains constant. B. The internal energy of the gas increases from its initial value but the enthalpy remains constant. C. Both internal energy and enthalpy of the gas remain constant. D. The internal energy and enthalpy of the gas increase. 171. In a steady-state flow process taking place in a device with a single inlet and single outlet, the

work done per unit mass flow rate is given by and P is the pressure. The expression for W given above is A. valid only if the process is both reversible and adiabatic B. valid only if the process is both reversible and isothermal C. valid for any reversible process

, where V is the specific volume

D. incorrect; it must be 172. The following four figures have been drawn to represent a fictitious thermodynamic cycle on the P–V and T–S planes:

According to the first law of thermodynamics, equal areas are enclosed by A. Figs. 1 and 2 B. Figs. 1 and 3 C. Figs. 1 and 4 D. Figs. 2 and 3 173. A P–V diagram has been obtained from a test on a reciprocating compressor. Which of the following represents that diagram?

Common data for Questions 174 and 175 A football was inflated to a gauge pressure of 1 bar when the ambient temperature was 15°C. When the game started next day, the air temperature at the stadium was 5°C. Assume that the volume of the football remains constant at 2.5 10–3 m3. Take ratio of specific heats to be 1.4. 174. The amount of heat lost by the air in the football and the gauge pressure of air in the football at the stadium respectively equal A. 30.6 J, 1.94 bar B. 21.8 J, 0.93 bar C. 61.1 J, 1.94 bar D. 43.7 J, 0.93 bar 175. Gauge pressure of air to which the ball must have been originally inflated so that it would equal 1-bar gauge at the stadium is A. 2.23 bar B. 1.94 bar C. 1.07 bar D. 1.00 bar 176. A positive value of Joule Thomson coefficient of a fluid means A. Temperature drops during throttling B. Temperature remains constant during throttling C. Temperature rises during throttling D. None of the above

Common data for Questions 177, 178 and 179 In the figure shown, the system is a pure substance kept in a piston-cylinder arrangement. The system is initially a two-phase mixture containing 1 kg of liquid and 0.03 kg of vapour at a pressure of 100 kPa. Initially the piston rests on a set of stops as shown in the figure. A pressure of 200 kPa is required to exactly balance the weight of the piston and the outside atmospheric pressure. Heat transfer takes place into the system until its volume increases by 50%. Heat transfer to the system occurs in such a manner that the piston, when allowed to move, does so in a very slow quasi-static/quasi-equilibrium process. The thermal reservoir from which heat is transferred to the system has a temperature of 400°C. Average temperature of the system boundary can be taken as 175°C. The heat transfer to the system is 1 kJ, during which its entropy increases by 10 J/K. Specific volumes of liquid (Vl) and vapour (Vg) phases, as well as values of saturation temperatures, are given in the table below.

177. At the end of the process, which one of the following situations will be true? A. Superheated vapour will be left in the system B. No vapour will be left in the system C. A liquid + vapour mixture will be left in the system D. The mixture will exist at a dry saturated vapour state 178. The work done by the system during the process is A. 0.1 kJ B. 0.2 kJ C. 0.3 kJ D. 0.4 kJ 179. The net entropy generation (considering the system and the thermal reservoir together) during the process is closest to A. 7.5 J/K B. 7.7 J/K

C. 8.5 J/K D. 10 J/K 180. A gas having a negative Joule Thomson coefficient (m < 0), when throttled will A. become cooler B. become warmer C. remain at the same temperature D. either be cooler or warmer depending on the type of gas 181. An ideal Brayton cycle operating between the pressure limits of 1 bar and 6 bar has minimum and maximum temperatures of 300 K and 1500 K. The ratio of the specific heats of the working fluid is 1.4. The approximate final temperatures in kelvin at the end of the compression and expansion processes are, respectively A. 500 and 900 B. 900 and 500 C. 500 and 500 D. 900 and 900 182. The values of enthalpy of steam at the inlet and outlet of a steam turbine in a Rankine cycle are 2800 kJ/kg and 1800 kJ/kg respectively. Neglecting pump work, the specific steam consumption in kg/kWh is A. 3.6 B. 0.36 C. 0.06 D. 0.01 Statements for linked answer Questions 183 and 184. The temperature and pressure of air in a large reservoir are 400 K and 3 bar, respectively. A converging-diverging nozzle of exit area 0.005 m2 is fitted to the reservoir as shown in the figure. The static pressure of air at the exit section, for isentropic flow through the nozzle, is 50 kPa. The characteristic gas constant and the ratio of specific heats of air are 0.287 kJ/kg K and 1.4, respectively.

183. The density of air in kg/m3 at the nozzle exit is A. 0.560 B. 0.600 C. 0.727 D. 0.800

184. The mass flow rate of air through the nozzle in kg/s is A. 1.30 B. 1.77 C. 1.85 D. 2.06 185. A cyclic process is represented on P–V diagram as shown below:

186. Nitrogen at an initial state of 10 bar, 1 m3 and 300 K is expanded isothermally to a final

volume of 2 m3. The P-V-T relation is , where a > 0. The final pressure. A. will be slightly less than 5 bar B. will be slightly more than 5 bar C. will be exactly 5 bar D. cannot be ascertained in the absence of the value of a Common data for Questions 187 and 188 The following table of properties was printed out for saturated liquid and saturated vapour of ammonia. The titles for only the first two columns are available. All that we know is that the other columns (columns 3 to 8) contain data on specific properties, namely, internal energy (kJ/kg), enthalpy (kJ/kg) and entropy (kJ/kg K).

187. The specific enthalpy data are in columns A. 3 and 7 B. 3 and 8 C. 5 and 7 D. 5 and 8 188. When the saturated liquid at 40°C is throttled to –20°C, the quality at exit will be A. 0.189 B. 0.212 C. 0.231 D. 0.788 189. A single acting two-stage compressor with complete intercooling delivers air at 16 bar. Assuming an intake state of 1 bar at 15°C, the pressure ratio per stage is A. 16 B. 8 C. 4 D. 2 190. A small steam whistle (perfectly insulated and doing no shaft work) causes a drop of 0.8 kJ/kg in the enthalpy of steam from entry to exit. If the kinetic energy of the steam at entry is negligible, the velocity of the steam at exit is A. 4 m/s B. 40 m/s C. 80 m/s

D. 120 m/s 191. When an ideal gas with constant specific heats is throttled adiabatically, with negligible changes in kinetic and potential energies A. DH = 0, DT = 0 B. DH > 0, DT = 0 C. DH > 0, DS > 0 D. DH = 0, DS > 0 where H, T and S represent respectively, enthalpy, temperature and entropy. Common data for Questions 192 and 193. Air enters an adiabatic nozzle at 300 kPa, 500 K with a velocity of 10 m/s. It leaves the nozzle at 100 kPa with a velocity of 180 m/s. The inlet area is 80 cm2. The specific heat CP of air is 1008 J/kg K. 192. The exit temperature of air is A. 516 K B. 532 K C. 484 K D. 468 K 193. The exit area of the nozzle in cm2 is A. 90.1 B. 56.3 C. 4.4 D. 12.9 194. A heat engine operates at 75% of the maximum possible efficiency. The ratio of the heat source temperature (in K) to the heat sink temperature (in K) is 5/3. The fraction of the heat supplied that is converted to work is A. 0.2 B. 0.3 C. 0.4 D. 0.6 (2006) 195. For the isentropic expansion of an ideal gas from the initial conditions P1, T1 to the final conditions P2, T2, which one of the following relations is valid? (g = CP/CV) A. (P1/P2) = (T2/T1)g B. (P1/P2) = (T1/T2)g/(g–1) C. (P1/P2) = (T1/T2) D. (P1/P2) = (T1/T2)(g–1)/g (2006)

196. Match the following:

A. (a)-(ii), (b)-(i), (c)-(i), (d)-(i) B. (a)-(ii), (b)-(i), (c)-(ii), (d)-(ii) C. (a)-(ii), (b)-(ii), (c)-(i), (d)-(i) D. (a)-(ii), (b)-(i), (c)-(ii), (d)-(i) 197. For a reversible exothermic gas phase reaction, A + B increase with A. increase in pressure and increase in temperature B. decrease in pressure and increase in temperature C. increase in pressure and decrease in temperature D. decrease in pressure and decrease in temperature

(2006) C, the equilibrium conversion will

(2006) 198. For a binary mixture of A and B at 400 K and 1 atm, which one of the following equilibrium states deviates significantly from ideality? , where is vapour pressure of A in atm, T = temperature, K, is partial pressure of A in atm, xA is mole fraction of A in liquid and yA is mole fraction of A in vapour. A. xA = 0.5; yA = 0.25 B. xA = 0.5; = 0.25 C. xA = 0.5; = 0.5 D. xA = 0.6; yA = 0.3 (2006) 199. Pure A at 473 K is fed to a steady-state adiabatic continuous reactor at the rate of 100 kg/h, where it undergoes an exothermic reaction to give its isomer B. The product stream is at temperature 773 K. The heat of reaction is 21 kJ/mol of A and the specific heat of the reaction mixture is constant at 35 J/mol K. The conversion in the reactor is A. 25% B. 50% C. 75% D. 100% (2006) 200. The molar density of water vapour at the normal boiling point of water is 33 mol/m3. The compressibility factor under these conditions is close to which one of the following? R = 8.314

J/mol K. A. 0.75 B. 1 C. 1.25 D. 1.5 (2006) 201. If TA and TB are the boiling points of pure A and pure B respectively, and TAB is that of a non-homogeneous immiscible mixture of A and B, then A. TAB < TA and TB B. TAB > TA and TB C. TA > TAB > TB D. TB > TAB > TA (2007) 202. The state of an ideal gas is changed from (T1, P1) to (T2, P2) in a constant volume process. To calculate the change in enthalpy, DH, all of the following properties/variables are required. A. CV, P1, P2 B. CP, T1, T2 C. CP, T1, T2, P1, P2 D. CV, P1, P2, T1, T2 (2007) 203. The change in entropy of the system DSsys, undergoing a cyclic irreversible process, is A. greater than zero B. equal to zero C. less than zero D. equal to the DSsurroundings (2007) 204. Parameters a and b in the van der Waals and other cubic equations of state represent A. a-molecular weight, b-molecular polarity B. a-molecular size, b-molecular attraction C. a-molecular size, b-molecular speed D. a-molecular attraction, b-molecular size (2007) 205. For the two paths as shown in the figure, one reversible and one irreversible, to change the state of the system from a to b,

A. DU, Q, W are the same B. DU are the same C. Q, W are the same D. DU, Q are different (2007) 206. For a pure substance, the Maxwell’s relation obtained from the fundamental property relation dU = TdS – PdV is A. (∂T/∂V)S = – (∂P/∂S)V B. (∂P/∂T)V = (∂S/∂V)T C. (∂T/∂P)S = (∂V/∂S)P D. (∂V/∂T)P = – (∂S/∂P)T (2007) 207. Which of the following represents the Carnot cycle (ideal engine)?

(2007) 208. Two kilograms of steam in a piston-cylinder device at 400 kPa and 448 K undergoes a mechanically reversible, isothermal compression to a final pressure such that the steam becomes just saturated. What is the work W, required for the process? Data: T = 448 K, P = 400 kPa, V = 0.503 m3/kg, U = 2606 kJ/kg, S = 7.055 kJ/kg K T = 448 K, saturated vapour, V = 0.216 m3/kg, U = 2579 kJ/kg, S = 6.622 kJ/kg K A. 0 kJ B. 230 kJ C. 334 kJ D. 388 kJ (2007) 209. Vapour-phase hydration of C2H4 to ethanol by the following reaction

attains equilibrium at 400 K and 3 bar. The standard Gibbs free energy change of reaction at these condition is DG0 = 4000 J/mol. For two moles of an equimolar feed of ethylene and steam, the equation in terms of the extent of reaction e (in moles) at equilibrium is A. B. C. D. (2007) 210. A methanol-water vapour liquid system is at equilibrium at 333 K and 60 kPa. The mole fraction of methanol in liquid is 0.5 and in vapour is 0.8. Vapour pressures of methanol and water at 333 K are 85 kPa and 20 kPa, respectively. Assuming vapour phase to be an ideal gas mixture, what is the activity coefficient of water in the liquid phase? A. 0.3 B. 1.2 C. 1.6 D. 7.5 (2007) 211. For conditions in Question 210, what is the excess Gibbs free energy (GE, J/mol) of the liquid mixture? A. 9.7

B. 388 C. 422 D. 3227 (2007) 212. A perfectly insulated cylinder of volume 0.6 m3 is initially divided into two parts by a thin, frictionless piston, as shown in the figure. The smaller part of volume 0.2 m3 has ideal gas at 6 bar pressure and 373 K. The other part is evacuated.

At certain instant of time t, the stopper is removed and the piston moves out freely to the other end. The final temperature is A. 124 K B. 240 K C. 306 K D. 373 K (2007) 213. The cylinder insulation is removed and the piston is pushed back to restore the system to the initial state. If this is to be achieved only by doing work on the system (no heat addition, only heat removal allowed), what is the minimum work required? A. 3.4 kJ B. 107 kJ C. 132 kJ D. 240 kJ (2007) 214. For a Carnot refrigerator operating between 40°C and 25°C, the coefficient of performance is A. 1 B. 1.67 C. 19.88 D. 39.74 (2008) 215. The work done by one mole of a van der Waals fluid undergoing reversible isothermal expansion from initial volume Vi to final volume Vf is A.

B.

C. D. (2008) 216. The standard Gibbs free energy change and enthalpy change at 298 K for the liquid phase reaction CH3COOH(l) + C2H5OH(l) CH3COOC2H5(l) + H2O(l) are given as DG0 = – 4650 J/mol and DH0 = – 3640 J/mol. If the solution is ideal and enthalpy change is assumed to be constant, the equilibrium constant at 368 K is A. 0.65 B. 4.94 C. 6.54 D. 8.65 (2008) 217. A binary mixture containing species 1 and 2 forms an azeotrope at 378.6 K and 1.013 bar. The liquid phase mole fraction of component 1 (xl) of this azeotrope is 0.62. At 378.6 K, the pure component vapour pressures for species 1 and 2 are 0.878 bar and 0.665 bar, respectively. Assume that the vapour phase is an ideal gas mixture. The van Laar constants, A and B, are given by the expressions

The activity coefficients g1 and g2 under these conditions are A. 0.88, 0.66 B. 1.15, 1.52 B. 1.52, 1.15 D. 1.52, 0.88 (2008) 218. The van Laar constants A and B for conditions in Question 217 are: A. 0.92, 0.87 B. 1.00, 1.21 C. 1.12, 1.00 D. 1.52, 1.15 (2008) 219. An ideal gas at temperature T1 and pressure P1 is compressed isothermally to pressure P2 (>

P1) in a closed system. Which one of the following is true for internal energy (U) and Gibbs free energy G of the gas at the two states? A. U1 = U2, G1 > G2 B. U1 = U2, G1 < G2 C. U1 > U2, G1 = G2 D. U1 < U2, G1 = G2 (2009) 220. For a binary mixture at constant temperature and pressure, which one of the following relations between activity coefficient gi and mole fraction xi is thermodynamically consistent? A. B. C. D. (2009) 221. An ideal gas with molar heat capacity CP = 5/2 R (where R = 8.314 J/mol K) is compressed adiabatically from 1 bar and 300 K to pressure P2 in a closed system. The final temperature after compression is 600 K and the mechanical efficiency of compression is 50%. The work required for compression in (kJ/mol) is A. 3.74 B. 6.24 C. 7.48 D. 12.48 (2009) 222. In the above problem, the pressure P2 (in bar) is A. 23/4 B. 25/4 C. 23/2 D. 25/2 (2009) 223. A new linear temperature scale, denoted by °S, has been developed, where the freezing point of water is 200°S and the boiling point is 400°S. On this scale, 500°S corresponds, in degree Celsius, to A. 100°C B. 125°C C. 150°C

D. 300°C (2010) 224. An equimolar mixture of species 1 and 2 is in equilibrium with its vapour at 400 K. At this temperature, the vapour pressures of the species are Raoult’s law is valid, the value of y1 is

kPa and

kPa. Assuming

A. 0.30 B. 0.41 C. 0.50 D. 0.60 (2010) 225. A saturated liquid at 1500 kPa and 500 K, with an enthalpy of 750 kJ/kg, is throttled to a liquid-vapour mixture at 150 kPa and 300 K. At the exit conditions, the enthalpy of the saturated liquid is 500 kJ/kg and the enthalpy of the saturated vapour is 2500 kJ/kg. The percentage of the original liquid, which vaporises, is A. 87.5% B. 67% C. 12.5% D. 10% (2010) 226. At constant temperature and pressure, the molar density of a binary mixture is given by r = 1 + x2, where x2 is the mole fraction of component 2. The partial molar volume at infinite dilution of component 1, is A. 0.75 B. 1.0 C. 2.0 D. 4.0 (2010) 227. Minimum work (W) required to separate a binary gas mixture at a temperature T0 and pressure P0 is

where, y1 and y2 are mole fractions, fpure, 1, and fpure, 2 are fugacities of pure species at T0 and P0 and are fugacities of species in the mixture at To, Po and y1. If the mixture is ideal, then W is A. 0 B. W = – RT0[y1 ln y1 + y2 ln y2]

C. W = RT0[y1 ln y1 + y2 ln y2] D. W = RT0 (2011) 228. The partial molar enthalpies of mixing (in J/mol) for benzene (component 1) and cyclohexane (component 2) at 300 K and 1 bar are given by , where x1 and x2 are the mole fractions. When 1 mol of benzene is added to 2 mol of cyclohexane, the enthalpy change (in J) is A. 3600 B. 2400 C. 2000 D. 800 (2011) 229. One mol of methane is contained in a leak-proof piston-cylinder assembly at 8 bar and 1000 K. The gas undergoes isothermal expansion to 4 bar under reversible conditions. Methane can be considered as an ideal gas under these conditions. The value of universal gas constant is 8.314 J/mol K. The heat transferred (in kJ) during the process is A. 11.52 B. 5.76 C. 4.15 D. 2.38 (2011) 230. Consider a binary mixture of methyl ethyl ketone (component 1) and toluene (component 2). At 323 K, the activity coefficients g1 and g2 are given by

where x1 and x2 are the mole fractions in the liquid mixture, and Y1 and Y2 are parameters independent of composition. At the same temperature, the infinite dilution activity coefficients, are given by . The vapour pressures of methyl ethyl ketone and toluene at 323 K are 36.9 and 12.3 kPa, respectively. Assuming that the vapour phase is ideal, the equilibrium pressure (in kPa) of a liquid mixture containing 90 mol % toluene is A. 19 B. 18 C. 16 D. 15 (2011) 231. In a throttling process, the pressure of an ideal gas reduces by 50%. If CP and CV are the heat capacities at constant pressure and constant volume, respectively (g = CP/CV), the specific volume will change by a factor of A. 2

B. 21/g C. 2g–1/g D. 0.5 (2012) 232. If the temperature of saturated water is increased infinitesimally at constant entropy, the resulting state of water will be A. liquid B. liquid-vapour coexistence C. saturated vapour D. solid (2012) 233. In a parallel flow heat exchanger operating under steady state, hot liquid enters at a temperature Th, in and leaves at a temperature Th, out. Cold liquid enters at a temperature Tc, in and leaves at a temperature Tc, out. Neglect any heat loss from the heat exchanger to the surrounding. If Th, in > > Tc, in, then for a given time interval, which one of the following statements is true? A. Entropy gained by the cold stream is greater than the entropy lost by the hot stream B. Entropy gained by the cold stream is equal to the entropy lost by the hot stream C. Entropy gained by the cold stream is less than the entropy lost by the hot stream D. Entropy gained by the cold stream is zero. (2012) 234. For an exothermic reversible reaction, which one of the following correctly describes the dependence of the equilibrium constant (K) with temperature (T) and pressure (P)? A. K is independent of T and P. B. K increases with an increase in T and P. C. K increases with T and decreases with P. D. K decreases with an increase in T and is independent of P. (2012) 235. An insulated, evacuated container is connected to a supply line of an ideal gas at pressure PS, temperature TS and specific volume VS. The container is filled with the gas until the pressure in the container reaches PS. There is no heat transfer between the supply line to the container, and kinetic and potential energies are negligible. If CP and CV are the heat capacities at constant pressure and constant volume, respectively (g = CP/CV), then the final temperature of the gas in the container is A. gTS B. TS C. (g – 1) TS D. (g – 1)TS/g

(2012) 236. Consider a binary liquid mixture at constant temperature T and pressure P. If the enthalpy change of mixing DH = 5x1x2, where xl and x2 are mole fraction of species 1 and 2, respectively, and the entropy change of mixing DS = – R(x1 ln x1 + x2 ln x2) with R = 8.314 J/mol K, then the minimum value of the Gibbs free energy change of mixing at 300 K occurs when A. x1 = 0 B. x1 = 0.2 C. x1 = 0.4 D. x1 = 0.5 (2012) C.2 Fill in the blanks: 1. properties of a system do not depend on the quantity of matter contained in it. 2. An open system exchanges with the surroundings. 3. The maximum efficiency of a heat engine depends only on the between which it operates and is independent of the nature of the cyclic process. 4. Gibbs free energy is defined as . 5. Mollier diagram is a plot of versus . 6. The efficiency of a Carnot engine working between 1000 K and 300 K is . 7. In the statement (DS)total ≥ 0 , the inequality refers to process. 8. P, V, T and S are properties whereas U, H, G, A are properties. 9. The principle of corresponding states may be stated thus: “At same TR and PR all gases have the same .” 10. A gaseous phase may be termed a vapour, if it can be condensed by . 11. A refrigerator of capacity 2 tons is working on ammonia at 273 K. The heat of vaporisation of ammonia is 1260 kJ/kg. The circulation rate of ammonia under this condition is approximately kg/h. 12. Gibbs–Helmholtz equation relates the change in with changes in . 13. A system from which finite quantities of heat can be removed without affecting its temperature is called . 14. The maximum velocity attained by a fluid in a pipe of uniform cross-section is equal to the in the fluid. 15. The maximum velocity attainable in a convergent nozzle is equal to and it is attained when the equals the critical value. 16. The ratio of the intake volume to the displacement volume in a single-stage compressor is called the . 17. The ratio of the velocity of flow to the sonic velocity is designated as . 18. The decrease in is a measure of the maximum work obtainable in an isothermal process.

19. As pressure tends to zero, fugacity of a pure gas becomes equal to its . 20. The ratio of the fugacity to the fugacity in the standard state is called . 21. Isothermal mixing of pure gases always produces a decrease in the . Hence work has to be done the system for separating a mixture of gases into its components. (1990) 22. The maximum work obtainable from a closed system under isothermal expansion is given by ; For one mole of an ideal gas expanding isothermally to twice its volume this is equal to . (1990) 23. The phase rule is given as . (1994) 24. Raoult’s law states that the of a component over an ideal solution is directly proportional to its mole fraction in the solution. 25. In a dilute solution, the obeys Henry’s law and the obeys Raoult’s law. 26. When the Henry’s law constant is equal to , Henry’s law becomes identical to Raoult’s law. 27. The activity coefficient (gi ) in a solution is related to the chemical potential as = . 28. The phase rule indicates the number of variables needed to specify the intensive state of the system whereas the indicates those needed to specify the extensive state of the system. 29. A mixture exists as a superheated vapour above its temperature. 30. If the intermolecular forces between unlike molecules are than those between like molecules, the solution will exhibit negative deviation from ideality. 31. The constant boiling mixtures are called . 32. The vaporisation equilibrium constant Ki is defined as Ki = . 33. Among three liquids A, B and C, the A-B binary is partially miscible whereas A-C and B-C binaries are totally soluble. On the binodal curve, the A-rich and B-rich phases in equilibrium become identical in properties at the of the system. 34. The number of degrees of freedom for a system consisting of two miscible non-reacting species which exists as an azeotrope in vapour–liquid equilibrium is . 35. The equilibrium state for a closed system is the state for which the total is a minimum at constant temperature and pressure. 36. A binary hydrocarbon liquid mixture of A and B (KA = 1.5) containing 60% (mol) A is flash vaporised. If 40% of the feed is vaporised, the mole fraction of A in the liquid product is . (1990) 37. A system of unit mass at equilibrium consists of two phases a and b of extent x and (1 – x) respectively. Write down expressions for the pressure and the specific enthalpy of the system as a whole in terms of the properties Pa, Pb, Ha and Hb of the individual phases: (a) P = , (b) H =

. (1990)

38. The heats of formation of CO (g), H2O (g), and CO2 (g) are respectively –110.525 kJ, –393.509 kJ and –241.818 kJ. The heat of reaction for CO (g) + H2O (g)

CO2 (g) + H2 (g)

is kJ. 39. reactions are favoured by increase in temperature. 40. The following data on heats of combustion at 298 K are given:

Heats of formation of CO2 (g) and H2O (l) are –390 and –280 kJ/kmol respectively. (a) The heat of formation of gaseous n-heptane at 298 K is . (b) The heat of formation of gaseous ethyl alcohol at 298 K is . (1990) 41. The heat absorbed for isothermal reaction C4H10 (g) C2H4 (g) + C2H6 (g) at 298 K and 101.3 kPa is . Standard heat of combustion in kJ/kmol are: C4H10 (g) = – 2873.5, C2H4 (g) = – 1411.9 and C2H6 (g) = – 1561.0

(1991) 42. The heat of formation of a compound is defined as the heat of reaction leading to the formation of the compound from its . (1994) C.3 Say, whether the following statements are TRUE or FALSE. Give correct statements to the false ones. 1. Internal energy is a state function whereas entropy is a path function. 2. Heat capacity and specific heat are extensive properties whereas volume and temperature are intensive properties. 3. Heat and work are not properties of a system; they are properties of a process. 4. At constant pressure, the change in enthalpy DH = . 5. Entropy cannot have absolute values; they are always expressed as a difference. 6. Energy of the universe is conserved whereas entropy is increasing. 7. So long as the process is reversible, the value of is the same for the change of the gas from any given state to another. 8. Heat involved in any process can be expressed as dQ = T dS. 9. Enthalpy and entropy of an ideal gas are functions of temperature alone. 10. If DS refers to the entropy change between the same initial and final states of the system for two

different processes, one reversible (R) and the other irreversible (I), then DSI = DSR. 11. The heat capacities CP and CV of an ideal gas are independent of temperature. 12. The second law of thermodynamics states that heat cannot be completely converted to work. 13. For any process, the second law of thermodynamics requires that the entropy change of the system is either zero or positive. 14. The entropy change of a chemical reaction is calculated as the ratio of the heat of reaction to the temperature of the reaction. 15. When water freezes to form ice, the atoms arrange themselves in a highly ordered manner. Since the increasing order is associated with the decreasing entropy, we must conclude that entropy of the universe decreases as a result of this process. 16. Entropy of a rotating flywheel is the same as that of the flywheel at rest. 17. For given operating temperatures all heat engines have the same efficiency regardless of the nature of the working substance. 18. Real gases behave ideally at high pressures and temperatures. 19. A reversible adiabatic process is essentially isenthalpic. 20. The heat capacity at constant pressure and constant volume of all gases are related as CP – CV = R. 21. For an ideal gas, the activity and fugacity are numerically equal. 22. For an ideal solution, all property changes of mixing are zero. 23. Raoult’s law is applicable to all ideal liquid solutions. 24. On the P-T diagram of a pure substance, the vaporisation curve and the fusion curve extend up to infinity. 25. The change in internal energy of an ideal gas is DU = irrespective of the nature of the process. 26. For gases, the Joule–Thomson coefficient is always positive. 27. Work required for isothermal compression is less than that of adiabatic compression. 28. The clearance has no effect on the work of compression in a single-stage compressor. 29. The reversible work of expansion in a non-flow process under isentropic condition is equal to –(DU)S. 30. For an ideal gas, the fugacity and pressure are equal. 31. The excess volume and the volume change on mixing are the same. 32. For a multicomponent system, equilibrium between two phases is established when the concentrations in both the phases are uniform. 33. For a solution at a given pressure, the vapour phase can exist in equilibrium with the liquid phase only at its bubble point. 34. In an ideal binary solution, component A obeys Raoult’s law and component B obeys Henry’s law. 35. Maximum boiling azeotropes may be formed if the solution exhibits very large positive deviation from ideality.

36. Azeotropic composition can be shifted by changing the pressure. 37. For a chemically reacting system at equilibrium at constant temperature and pressure, the Gibbs free energy is maximum. 38. The numerical value of the equilibrium constant depends upon the stoichiometric equation. 39. If there is decrease in the total number of moles during a gas-phase chemical reaction, the increase in pressure decreases the formation of products. 40. The equilibrium conversion in a gaseous reaction which produces no change in the number of moles (e.g., the water-gas shift reaction) is not affected by the change in pressure. Answers C.1 1. C 8. C 15. D 22. B 29. A 36. C 43. B 50. D 57. B 64. B 71. C 78. C 85. D 92. A 99. C 106. A 113. B 120. C 127. A 134. A 141. D 148. A 155. B 162. D 169. D 176. A 183. C 190. B

2. B 9. C 16. B 23. A 30. C 37. A 44. C 51. C 58. C 65. A 72. B 79. B 86. B 93. D 100. C 107. A 114. A 121. C 128. B 135. D 142. B 149. B 156. D 163. B 170. C 177. A 184. D 191. A

3. C 10. C 17. D 24. B 31. B 38. B 45. C 52. A 59. C 66. D 73. D 80. B 87. B 94. D 101. A 108. B 115. B 122. D 129. B 136. C 143. B 150. C 157. B 164. B 171. C 178. D 185. C 192. C

4. B 11. B 18. C 25. C 32. C 39. A 46. D 53. A 60. B 67. B 74. D 81. C 88. B 95. A 102. C 109. B 116. B 123. C 130. D 137. A 144. B 151. A 158. C 165. A 172. A 179. C 186. B 193. D

5. A 12. B 19. B 26. B 33. C 40. A 47. B 54. C 61. A 68. D 75. D 82. C 89. C 96. D 103. B 110. B 117. B 124. A 131. A 138. C 145. D 152. C 159. B 166. B 173. D 180. B 187. B 194. B

6. C 13. C 20. A 27. B 34. D 41. B 48. B 55. B 62. C 69. B 76. C 83. B 90. B 97. A 104. B 111. D 118. B 125. B 132. B 139. A 146. D 153. D 160. C 167. A 174. D 181. A 188. B 195. B

7. A 14. A 21. C 28. D 35. D 42. D 49. B 56. A 63. C 70. A 77. A 84. A 91. B 98. B 105. D 112. B 119. B 126. C 133. D 140. C 147. A 154. D 161. A 168. D 175. C 182. A 189. C 196. D

197. C 204. D 211. C 218. C 225. C 232. A

198. C 205. B 212. D 219. B 226. A 233. A

199. B 206. A 213. C 220. D 227. B 234. D

200. B 207. C 214. C 221. C 228. D 235. A

201. A 208. C 215. D 222. D 229. B 236. D

C.2 1. Intensive 2. Mass and energy 3. Temperature 4. G = H – TS 5. Enthalpy, entropy 6. 70% 7. Irreversible 8. Reference, energy 9. Z (Compressibility factor) 10. Compression at constant temperature 11. 20 kg/h 12. G/T with T 13. Heat reservoir 14. Sonic velocity 15. Sonic velocity, Pressure ratio 16. Theoretical volumetric efficiency 17. Mach number 18. Helmholtz free energy 19. Pressure 20. Fugacity coefficient 21. Gibbs free energy, on 22. P dV, RT ln 2 23. F = C – p + 2 24. Partial pressure (fugacity) 25. Solute, solvent 26. Vapour pressure 27. RT ln gi 28. Duhem’s theorem 29. Dew point 30. Stronger (greater) 31. Azeotropes

202. B 209. D 216. B 223. C 230. C

203. B 210. B 217. B 224. D 231. A

32. yi/xi 33. Plait point 34. One 35. Gibbs free energy 36. 0.46 37. (a) Pa = Pb (b) xHa + (1 – x)Hb 38. 262.22 39. Endothermic 40. (a) – 120 kJ/kmol (b) – 210 kJ/kmol 41. 99.4 kJ 42. Constituent elements C.3 1. False. Both are state functions. 2. False. Heat capacity and volume are extensive properties whereas specific heat and temperature are intensive properties. 3. True. 4. True. 5. False. Entropy can have absolute values. 6. True. 7. False. The value of P dV is dependent on the path followed. 8. False. Heat involved in a reversible process can be expressed as dQ = T dS. 9. False. Enthalpy of an ideal gas is a function of temperature only. Entropy depends on pressure as well. 10. True. 11. False. The heat capacities CP and CV of an ideal gas are dependent on temperature only. 12. False. The second law of thermodynamics states that heat cannot be completely converted into work continuously (or in a cyclic process). 13. False. For any process, the second law of thermodynamics requires that the entropy change of the system and the surroundings together is either zero or positive. 14. False. The entropy change of a chemical reaction is to be computed as the sum of the absolute entropies of the products minus the sum of the absolute entropies of the reactants. 15. False. When water freezes to form ice, greater disorder may result in the surroundings due to transfer of heat with a consequent increase in the total entropy. 16. True. 17. False. For given operating temperatures all reversible heat engines (Carnot engines) have the same efficiency regardless of the nature of the working substance. 18. False. Real gases behave ideally at low pressures and or high temperatures. 19. False. A reversible adiabatic process is essentially isentropic. 20. False. The heat capacity at constant pressure and constant volume of ideal gases are related as

CP – CV = R. 21. True. 22. False. For an ideal solution, property changes of mixing are not zero for entropy and entropyrelated functions such as the free energy. 23. False. Raoult’s law is applicable to all ideal liquid solutions provided the vapour phase is an ideal gas. 24. False. On the P-T diagram of a pure substance, the vaporisation curve lies between the triple point and the critical point whereas the fusion curve extends up to infinity. 25. True. 26. False. For gases the Joule–Thomson coefficient may be positive, zero or negative. 27. True. 28. True. 29. True. 30. True. 31. True. 32. False. For a multicomponent system in equilibrium, the chemical potentials in both phases are uniform. 33. False. For a solution at a given pressure, the vapour phase can exist in equilibrium for a range of temperatures lying between the bubble point and the dew point. 34. False. In an ideal binary solution both components obey Raoult’s law. 35. False. Minimum boiling azeotropes may be formed if the solution exhibits very large positive deviation from ideality. 36. True. 37. False. The Gibbs free energy is minimum. 38. True. 39. False. If there is decrease in the total number of moles during a gas-phase chemical reaction, the increase in pressure favours the formation of products. 40. False. If the compressibility of the components are affected by the change in pressure, the equilibrium conversion also will be affected.

References 1. Abrams, D. and Prausnitz, J.M., AIChE J., 21, 116, 1975. 2. Atkins, P.W., Physical Chemistry, 4th ed., ELBS-Oxford, Oxford University Press, 1990. 3. Benedict, M., Webb, G. and Rubin, L., J. Chem. Phys., 8, 334, 1940. 4. Callen, H.B., Thermodynamics, John Wiley, New York, 1960. 5. Daubert, T.E., Chemical Engineering Thermodynamics, McGraw-Hill, New York, 1985. 6. Denbigh, K., The Principles of Chemical Equilibrium, 4th ed., Cambridge, New York, 1981. 7. Dodge, B.F., Chemical Engineering Thermodynamics, McGraw-Hill, New York, 1944. 8. Fredenslund, Aa., Jones, R.L. and Prausnitz, J.M., AIChE J., 21, 1086, 1975. 9. Glasstone, S., Thermodynamics for Chemists, Van Nostrand, New York, 1958. 10. Hill, L., Statistical Mechanics, McGraw-Hill, New York, 1956. 11. Hougen, O.A., Watson, K.M. and Ragatz, R.A., Chemical Process Principles, Part II, 2nd ed., John Wiley, New York, 1960. 12. Karapetyants, M.Kh., Chemical Thermodynamics, Mir Publishers, Moscow, 1978. 13. Kirkwood, I.J. and Oppenheim, I.,Chemical Thermodynamics, McGraw-Hill, New York, 1961. 14. Kyle, B.G., Chemical and Process Thermodynamics, 2nd ed., Prentice-Hall of India, New Delhi, 1994. 15. Lewis, G.N., Randall, M., Pitzer, K.S. and Brewer, L., Thermodynamics, McGraw-Hill, New York, 1981. 16. Modell, M. and Reid, R.C.,Thermodynamics and Its Applications, Prentice Hall, New Jersey, 1974. 17. Peng, D.Y. and Robinson, D.B., Ind. Eng. Chem. Fundam., 15, 59, 1976. 18. Perry, J.H. and Chilton, C.H. (Eds.),Chemical Engineers’ Handbook, 5th ed., McGraw-Hill, Tokyo, 1973. 19. Prausnitz, J.M., Lichtenthaler, R.N. and Azevedo, E.G.,Molecular Thermodynamics of Fluid Phase Equilibria, 2nd ed., Prentice Hall, New Jersey, 1986. 20. Rao, Y.V.C., An Introduction to Thermodynamics, Wiley Eastern, New Delhi, 1993. 21. Redlich, O. and. Kwong, J.N.S., Chem. Rev., 44, 233, 1949. 22. Renon, H. and Prausnitz, J.M., AIChE J., 14, 135, 1968. 23. Reynolds, W.C. and Perkins, H.C., Engineering Thermodynamics, McGraw-Hill, Tokyo, 1977. 24. Saad, M.A., Thermodynamics for Engineers, Prentice-Hall of India, New Delhi, 1969. 25. Sandler, S.I., Chemical and Engineering Thermodynamics, John Wiley, New York, 1977. 26. Smith, J.M. and Van Ness, H.C.,Introduction to Chemical Engineering Thermodynamics, 4th ed., McGraw-Hill, New York, 1987. 27. Smith, N.O., Chemical Thermodynamics: A Problems Approach, Reinhold, New York, 1967. 28. Soave, G., Chem. Engg. Sci., 27, 1197, 1972. 29. Sonntag, R.E. and Van Wylen, G.J., Introduction to Thermodynamics, John Wiley,

New York, 1971. 30. Treybal, R.E., Mass-Transfer Operations, 3rd ed., McGraw-Hill, New York, 1981. 31. Wark, K., Thermodynamics, 2nd ed., McGraw-Hill, New York, 1971. 32. Weber, H.E. and Meissner, H.C., Thermodynamics for Chemical Engineers, 2nd ed., John Wiley, New York, 1957. 33. Wilson, E.D. and Ries, H.C., Principles of Chemical Engineering Thermodynamics, McGraw-Hill, New York, 1956.

Answers to Exercises CHAPTER 1 1.1 89.55 kg; 878.51 N 1.2 102.14 N 1.3 698 mm 1.4 1.6783 bar 1.5 3.02 bar 1.6 1.5453 105 N/m2 1.7 17.845 m, 25 kJ 1.8 1.667 102 kJ 1.9 7.355 kJ; 61.29 W 1.10 16.8 W 1.11 20.39 m 1.12 (a) 4.164 103 N (b) 1.3254 (c) 2.082 (d) 490.5 J

105 N/m2 103 J

1.13 2.156 105 J 1.14 17.15 m/s 1.15 1.146 104 kJ 1.16 6.073

104 J

CHAPTER 2 2.1 40.75 kJ 2.2 400 K, 500 kPa 2.3 (a) 183.94 kJ (b) 1839.4 kJ (c) 1655.46 kJ (d) 38.36 m/s (e) 1839.4 kJ 2.4 (a) 981 J (b) 981 J (c) 0.234 K 2.5 (a) 14 m/s (b) 0.78 K 2.6 18,467 kJ

2.7 4466 kJ 2.8 195,814 kJ 2.9 2423.9 kJ/kg; 2214 kJ/kg 2.10 (a) 3.572 kJ (b) 10.97 min 2.11 Q = 914.86 kJ, W = 139.35 kJ 2.12 227 V 2.13 Q = 1.246 106 kJ 2.14 Q = 58.4 104 J 2.15 435.5 HP 2.16 35 kJ 2.17 70 kJ 2.18 2453 kJ/kg 2.19 389.9 K 2.20 (a) 78.9 m/s (b) 72.7 kPa 2.21 870.9 kJ/kg, 5.74 kg/s 2.22 55.5 kW 2.23 3.68 MW 2.24 75% 2.25 141.3 MW 2.26 1.722 K 2.27 100 kJ 2.28 (a) 22,480 kJ (b) 8730 kJ 2.29 166.67 kJ 2.30 (a) V = 4.65584 m3, P = 1 bar, T = 56 K DU = 4656 kJ, DH = 6518 kJ, W = 0, Q = 4656 kJ (b) V = 23.2792 m3, P = 5 bar, T = 1400 K, DU = 22,380.3 kJ, DH = 32,592.0 kJ, W = 0, Q = 22,380.3 kJ 2.31 DU = 2168.6 kJ, DH = 3000 kJ, W = – 125.32 kJ, Q = 2043.28 kJ

CHAPTER 3 3.1 (a) 38.301 (b) 0 (c) 4.31 bar

103 J

(d) – 47.817 103 J 3.2 (a) 24.0685 kJ

(b) – 17.695 kJ (c) 6.374 kJ 3.3 (a) 2282 kJ (b) 2206 kJ (c) 0 3.4 (a) 472.5 kJ (b) – 4863 kJ (c) 0 (d) 4863 kJ (e) 5981 kJ 3.5 (a) DU = 997.68 kJ/kmol; DH = 1662.8 kJ/kmol; W = – 997.68 kJ/kmol; Q = 0 (b) DU = – 997.68 kJ/kmol; DH = – 1662.8 kJ/kmol; W = – 665.12 kJ/kmol; Q = – 1662.8 kJ/kmol (c) DU = 0; DH = 0; W = 1490 kJ/kmol; Q = 1490 kJ/kmol (d) DU = 0; DH = 0; W = – 172.8 kJ/kmol; Q = – 172.8 kJ/kmol 3.6 DU = – 2329 kJ/kmol; DH = – 3252 kJ/kmol; W = 3395 kJ/kmol; Q = 1066 kJ/kmol 3.7 (a) 0.373 kg (b) 0.304 kg 3.8 418.5 K, 7.65 bar 3.9 (a) 4605 kJ (b) 132.9 K, 2376 kJ (c) 900 kJ 3.10 3.66 10–4 m3/mol 3.11 (a) 3

10–3 m3/mol 10–3 m3/mol

(b) 2.98 3.12 23.84 bar 3.13 (a) 65.54 bar (b) 57.23 bar (c) 57.87 bar

10–4 m3/mol 3.15 7.134 10–5 m3/mol (liquid), 1.712 10–3 m3/mol (vapour) 3.14 1.8

10–3 m3/mol, 0.8862 3.17 (a) 4.157 10–5 m3/mol 3.16 3.485

(b) 6.44

10–5 m3/mol 10–5 m3/mol

(c) 5.3 3.19 – 110.6 kJ/mol 3.20 – 1207.69 kJ/mol 3.21 – 1655.07 kJ/mol 3.22 48.70 kJ/mol 3.23 – 42.62 kJ 3.24 – 395.2 kJ/mol 3.25 – 799.3 kJ/mol

3.26 = – 3.8235 104 – 31.82 T + 1.776 3.27 – 224.673 kJ 3.28 – 207.2 103 kJ

10–2T2 – 3.108

10–6 T3

3.29 32,528.5 kJ 3.30 1360 K 3.31 1216 K 3.32 – 311.627 kJ/mol 3.33 – 3283.5 kJ 3.34 2.7989 104 kJ 3.35 2089.5 K 3.36 – 114.408 kJ/mol 3.38 (a) – 103.2 kJ/mol (b) = – 75,964 – 62.71T + 4.496 3.39 2055.8 K 3.40 3.766 106 kJ 3.41 – 3535.50 kJ 3.42 2141.9 K

CHAPTER 4 4.1 329.84 kJ 4.2 1.448

109 J/s

4.3 (a) 2.1502 (b) 18.1%

106 kJ/h

10–2T2 – 9.561

10–6T3 + 11.224

104(1/T)

4.4 (a) 2.9736 kJ (b) 3.9736 kJ 4.5 Unacceptable 4.6 Unacceptable 4.7 Unrealistic 4.8 (a) –19.1437 J/mol K (b) 30.63 J/K (c) 11.49 J/K 4.9 79.91 J/K 4.10 (a) 6.66 kJ/K, – 4.6204 kJ/K, 2.04 kJ/K (b) 6.66 kJ/K, – 4.925 kJ/K, 1.735 kJ/K (c) 6.66 kJ/K, – 4.89 kJ/K, 1.77 kJ/K (d) 6.66 kJ/K, – 6.66 kJ/K, 0 4.11 18.531 kJ/kmol K, Possible 4.12 4579.6 kJ/kmol 4.13 7.1 105 kJ 4.14 0.0508 kJ 4.16 31.28 kJ/kmol K, 111.9 kJ/kmol K 4.17 – 2.46 kJ/kmol K 4.18 (a) – 1.99 kJ/K (b) 3.44 kJ/K (c) 1.45 kJ/K 4.19 (a) 0.37 (b) 0.56 4.20 (a) 0 (b) 19.5 kJ 4.21 5966 kJ/kmol 4.22 Yes 4.23 10.65 J/K 4.24 1.07 kJ/K 4.25 (a) 5.46 kJ/kmol K, – 4.48 kJ/kmol K, 0.98 kJ/kmol K (b) 8.6 kJ/kmol K, – 7.9 kJ/kmol K, 0.70 kJ/kmol K 4.26 348.5 K 4.27 64.6% 4.28 21.995 103 kJ/kmol 4.29 13.38 kJ/kmol K 4.30 76.1% 4.31 (a) – 1375.35 kJ/K (b) 326.45 kJ/K (c) 1.16887 105 kJ 4.32 (a) 397.7 kW

(b) 329.2 kW 4.33 8.62 J/g K

CHAPTER 5 5.1 13.54 m/s 5.2 0.9948 kg/s 5.3 350 K 5.4 722.51 kW 5.5 6.877 105 kJ 5.6 144.1 kJ/h 5.7 567 K 5.8 6558 kg 5.9 759 K 5.10 0.9159 kg 5.11 229 K, 3.82 bar 5.12 420 K 5.13 (a) 865.8 K (b) 18.4 kg 5.14 (a) 45.61 kg (b) –104594 kJ 5.15 0.0744, 9018 kJ 5.16 343.3 K 5.17 109.3 HP 105 N/m2 5.19 4.316 kW, 0.8662 105 N/m2, 9.66 105 N/m2 5.18 1481.3 W, 2.95

5.20 (a) 402.64 m/s (b) 5.751 10–4 m2 5.21 1.14 kg/s 5.22 2.19 5.23 61.04 K 5.24 813.8 m/s 5.25 11.33 5.26 (a) 0.5457, 542.96 m/s (b) 28.323 kPa 5.27 429.4 K, 3.65P2 5.29 541 kW, 3060.97 kJ/kg, 7.3544 kJ/kg K 5.30 (a) 114.91 kW (b) 0.49 m3/s

5.31 31.84 kW, 0.2963 m3/s, 393 K 5.32 71.8 kJ/kg, 339 K 5.33 14 5.34 1.11 kJ/s, 2.5 kJ/s 5.35 2.525 kg/s, 280 kJ/s, 4 5.36 419.32 kg/h 5.37 (a) 0.5079 kg/min (b) 0.311 kW (c) 2.857 5.38 (a) 3.42 HP (b) 63.3 kg/h (c) 1731.1 kg/h (d) 6.9 5.39 11.39 kg/h, 5.03 5.40 (a) 162.2 kg/h (b) 3.5% (c) 131.55 kg/h (d) 4.11, 5.0, 3.76 5.41 4.78 kW, 8.9214 103 kg/h 5.42 (a) 3 (b) 4 kJ/s (c) 231 K 5.43 6976.7 kg/h, 57.67 ton 5.44 9.73 kW 5.45 0.084 (winter), 0.048 (summer) 5.46 (a) 2% liquefied (b) 23% liquefied 5.47 (a) 0.048, 4.92 kg/h (b) 0.006, 0.65 kg/h 5.48 (a) 0.054 (b) 230 K 5.49 (a) 13.3 kg/h (b) 58.87 kW 5.50 40% 5.51 (a) 2.9245 kJ/kg (b) 578.17 kJ/kg (c) 55.4 kJ/kg (d) 2867.4 kJ/kg (e) 2211.6 kJ/kg 5.52 (a) 38.83% (b) 37.22%

5.53 38.42% 5.54 54.14% 5.55 (a) 34.95% (b) 29.68% (c) 4232 kg/h 5.56 34.25% 5.57 34.81% 5.58 (a) 29.72% (b) 4000 kg/h 5.59 (a) 39.66% (b) 29.61% (c) 3.0505

105 kg/h 105 kW, 1.9016

(d) 2.7016 105 kW 5.60 39.12% 5.61 25.45%, 0.941 (at 700 kPa), 33.34%, 0.8627 (at 3500 kPa), 36.16%, 0.771 (at 7000 kPa), 38.22%, 0.6944 (at 14,000 kPa) 5.62 32.42%, 0.7835 (at 573 K), 32.99%, 0.8128 (at 623 K), 36.69%, 0.920 (at 873 K), 37.64%, 0.9361 (at 923 K) 5.63 (a) 510.34 kJ/kg (b) 222.15 kJ/kg (c) 56.47% 5.64 1129.4 kJ/kg 5.65 (a) 7.233 (b) 54.68% (c) 909.24 kJ/kg 5.66 6.13 5.67 (a) 13.94 (b) 1.578 (c) 17.874 kJ (d) 61.49% 5.68 (a) 80 kPa, 310 K (step 1), 2009.5 kPa, 778.69 K (step 2), 2368.9 kPa, 917.96 K (step 3), 94.3 kPa, 365.45 K (step 4) (b) 300.93 kJ (c) 60.2% 5.69 43.2%

5.70 (a) 9 (b) 2167.4 kPa (c) 49.68% (d) 78.618 kW 5.71 (a) 13.94 (b) 1.64 (c) 366.81 kJ/kg (d) 61.13% 5.72 Compression ratio = 2.69 Expansion ratio = 2.408 5.73 (a) 3101.3 K (b) 3.3 (c) 2172.5 kJ/kg (d) 55.74% 5.74 19.16 5.75 (a) 40.62% (b) 543.27 kJ/kg (c) 1.22 kJ/kg K 5.76 (a) 1828 kJ/kg (b) 1040.6 kJ/kg (c) 56.93% 5.77 (a) 1203.26 kJ/kg (b) 1976.9 kJ/kg 5.78 (a) 220 kJ/kg (b) 372.2 kJ (c) 22.2% 5.79 (a) 100 kPa (b) 244.6 kJ (c) 585.2 kJ (d) 44.79% 5.80 (a) 40.06% (b) 24.7% 5.81 (a) 1416 kW (b) 3616 kW (c) 32.72% (d) 421 kPa 5.82 (a) 24,714.4 kJ/s (b) 0.3973 (c) 66.54 kg/s 5.83 (a) 300 K, 100 kPa (point 1), 445.8 K, 400 kPa (point 2), 900 K, 390 kPa(point 3), 618.6 K, 105 kPa (point 4)

(b) 29.85%

CHAPTER 6 6.11 m = – b/CP, Temperature increases 6.12 z = (1 + 2aP)/(1 + aP) 6.13 RT ln f = PV – RT – RT ln

6.16 DV = 1.378 6.17 377.24 K

10–5 m3/kg

6.18 33.51 103 kJ/kmol, PS = a exp (b/T) 6.19 393.43 K 6.20 (a) 1832.4 kJ/kg (b) 2145.4 kJ/kg 6.21 (b) – 0.0982 (kJ/kmol) (bar)–1 6.22 0.1451 kJ/kmol K 6.23 DH = 51,666.14 kJ/kmol DU = 47,731.74 kJ/kmol DS = 131.72 kJ/kmol K 6.24 DH = 5796 kJ/kmol DU = 4133.20 kJ/kmol DS = – 23.50 kJ/kmol K DG = – 8754 kJ/kmol 6.25 H = 421.48 kJ/kg, S = 1.22 kJ/kg K 6.26 8.434 J/mol K 6.27 59.65 J/mol K 6.28 97.34% vapour 6.29

= – 2.681

10–3 kJ/kmol bar

= 8.997 kJ/kmol bar m = – 0. 0652 K (bar)–1 6.31 73.1 bar 6.32 (a) 48.9 bar, 406.9 bar (b) 48.78 bar, 408.8 bar 6.33 50.41 bar 6.34 1156 bar

6.35 0.2% 6.36 9.94 bar, 15.79 bar 6.37 0.9954 bar 6.38 50.71 bar 6.39 (a) ln f = A ln P + BP + CP2/2 + DP3/3 (b) 301.02 bar 6.40 (a) ln f = BP/RT + (C – B2) P2/(2R2T2) (b) 0.9991 bar 6.41 f = (a – 1) ln P + BP + CP2/2 93.44 bar

CHAPTER 7

7.12 103 (m3/kmol) are given in brackets against x : 0.0667 (10.332), 0.16(13.86), 0.30 (19.78), 0.3634 (22.16), 0.84 (36.83) 7.14 0.8866 10–3 m3/kg 7.15

= 18.023 7.16

10–6 m3/mol (salt) 10–6 m3/mol (solvent)

= – 1.3873 = 0.0389

10–6 m3/mol (methanol) 10–6 m3/mol (water)

= 0.0175 7.17 – 22 kJ/kmol (HCl), – 9.1 kJ/kmol (water) 7.18 0.02903 m3/mol

7.19 5.056 10–4 mol/kg (water) 7.20 533 bar

7.21 – 9.74 kJ, – 9.46 kJ 7.22 gA and are shown in brackets against mole fraction x. 0 (–, 1.00), 0.2 (0.5361, 1.0652), 0.4 (0.7330, 1.4565), 0.6 (0.8862, 1.7609), 0.8 (0.9710, 1.9273), 1.0 (1.0, 1.987) 7.23 Yes 7.24 7.25 (a) 1.27, 0.652 (b) 0.4174, 0.2145 7.26 1.331 7.27 (a) 0.0834 (b) 0.36 bar 7.28 0.953 7.29 0.8939 7.30 34.2589 J/K 7.31 – 4675.3 kJ/kmol K h 7.32 7.33 9.661 106 kJ 7.34 340.62 K 7.35 – 10.217 kJ/mol 7.36 5100 kJ to be removed 7.37 1.2914 109 kJ/h 7.38 – 1.17 kJ/mol

7.39 7.40 7.41

+ and ME/x1x2 are respectively, A – B + C, 0, A – B + C (for x1

0) and

0, A + B + C, A + B + C (for x1

1)

7.42

7.43 GE/RT = x1x2 [A + B(x1 – x2)] 7.44 = 0.1275 m3/kmol = 0.1625 m3/kmol

CHAPTER 8 8.2 (a) 2, overdefined (b) 1 (c) 2 (d) 0 (e) 1 (f) 2 (g) 1 8.3 0.0855 8.4 x and y are given in brackets against T 353.1 (1, 1), 358 (0.78, 0.90), 363 (0.581, 0.777), 368 (0.411, 0.632), 373 (0.258, 0.456), 378 (0.130, 0.261), 383 (0.017, 0.039), 383.6 (0, 0) y = 2.45 x/(1 + 1.45 x) 8.5 8.5239 kPa, 76.3% benzene 8.6 120.3 kPa 8.7 131.24 kPa, 46.2% A 8.8 N2: 67.4% (liquid), 90.34% (vapour) 8.9 (a) x1 and y1 are given in brackets against P in kPa: 33.79 (0, 0), 43.426 (0.2, 0.3775), 53.062 (0.4, 0.6179), 62.698 (0.6, 0.7844), 72.334 (0.8, 0.9066), 81.79 (1, 1) (b) x1 and y1 are given in brackets against T in K: 311.45 (1, 1), 315 (0.787, 0.902), 319 (0.581, 0.773), 323 (0.405, 0.623), 327 (0.254, 0.449), 331 (0.123, 0.250), 335.33 (0, 0) 8.10 (a) x1 and y1 are given in brackets against P in kPa: 34.02(0, 0), 63.03(0.2, 0.568), 92.04(0.4, 0.778), 121.04(0.6, 0.888), 150.05(0.8, 0.995), 179.06(1, 1) (b) x1 and y1 are given in brackets against T in K: 353.3(1, 1), 363.3(0.686, 0.925), 373.3(0.458, 0.816), 383.3(0.287, 0.666), 393.3(0.156, 0.464), 403.3(0.053, 0.198), 409.4(0, 0) 8.11 (a) 77.3% benzene (b) 57.5% benzene 37.9% benzene 8.12 (a) x1 = 0.3138, y1 = 0.7730 (b) 334.4 K, 79.6% pentane (c) 110.25 kPa, 85.7% pentane 8.15 x1 and y1 are given in brackets against T in K: 334.4 (0, 0), 336.5 (0.2, 0.165), 337.7 (0.334, 0.334), 337 (0.4, 0.4277), 335 (0.6, 0.6858), 331.5 (0.8, 0.8750), 329.5 (1, 1)

8.17 (a) A = 1.0624, B = 1.0217 (b) ln g1 and ln g2 are shown in brackets against x1: 0 (1.0624, 0), 0.2 (0.6299, 0.0435), 0.4 (0.3706, 0.1713), 0.6 (0.1621, 0.3793), 0.8 (0.0399, 0.6640), 1.0 (0, 1.0217) 8.18 (a) 7.19% acetone (b) 108.35 kPa 8.19 (a) A = 1.6625, B = 2.7475 (b) 67.81% hexane (c) 101.56 kPa 8.20 (a) A = 0.8940, B = 0.8426 (b) 60.3% acetate, 54.72 kPa (c) 48.64% acetate 8.21 A = 0.9376, B = 3.0119 8.22 A = 0.1365, B = 0.1122 8.23 A = 1.7492, B = 1.4446 8.24 Yes 8.25 (a) 335.53 K, 80.9% acetone (b) 351.96 K, 4.07% acetone 8.26 A = 3.8297, B = 2.3540 8.27 83.16% alcohol 8.28 y1 is shown in brackets against x1: 0 (0), 0.2 (0.1355), 0.4 (0.3299), 0.6 (0.6123), 0.8 (0.9212), 1.0 (1.0) 8.29 g1 = 1.3551, g2 = 1.682 8.30 108.4 kPa, 43.46% acetone 8.31 (a) 91.50 kPa, 53.77% propanol (b) 96.75 kPa, 43.91% propanol (c) 353.84 K, 81.5% propanol (d) 360.615 K, 6.38% propanol 8.32 329.7 K, 356.9 K 8.33 (a) 330 K (b) 340.6 K (c) Mole % in the liquid and vapour are given in brackets: Ethane (0.48, 0.09), Propane (36.3, 17.45), Isobutane (18.18, 18.74), n-Butane (44.98, 63.35), Isopentane (0.13, 0.38) 8.34 (a) 930.3 kPa (b) 337.4 K, Composition of condensate: Ethane (1.8%), Propane (6.2%), Isobutane (17.3%),nButane (67.1%), Isopentane (7.50%) 330.2 K, Composition of the liquid and vapour: Ethane (3.22%, 16.81%), Propane (9.31%, 18.7%), Isobutane (19.55%, 18.46%), n-Butane (63.38%, 44.57%), Isopentane (4.54%, 1.46%) 8.35 2205 kPa, Methane: 41.91, Ethane: 20.25, Propane: 21.96, Isobutane: 8.57, n-Butane: 7.31 706 kPa, Methane: 0.2, Ethane: 1.91, Propane: 16.14, Isobutane: 30.88, n-Butane: 50.88

8.36 861.4 kPa, 2446 kPa 8.37 758 kPa 8.38 (a) 717 kPa (b) Propane: 63.93%, n-Butane: 36.07% (c) Propane: 36.07%, n-Butane: 63.93% 8.39 Consistent 8.40 A = 0.685, B = 0.785, consistent 8.41 Inconsistent 8.42 Consistent 8.43 Consistent 8.44 (a) g1 given in brackets against x1: 1.0 (1.00), 0.87 (1.016), 0.50 (1.2074), 0.30 (1.3778) (b) (kPa) in brackets against x1: 1.0 (0), 0.87 (7.45), 0.50 (20.05), 0.30 (31.63), 0 (37.72) 8.45 Inconsistent 8.46 g2 in brackets against x1: 0 (1.0), 0.0033 (0.9999), 0.0168 (0.9982), 0.0486 (0.9854), 0.0986 (0.9746), 0.168 (0.9313), 0.2701 (0.8535), 0.424 (0.7688) 8.47 1.0493 8.48 (kPa) in brackets against x1: 0.065 (24.20), 0.14 (47.90), 0.211 (66.67), 0.293 (84.56), 0.383 (100.31), 0.483 (114.08), 0.587 (125.48), 0.713 (137.19), 0.854 (150.67) 8.49 A = 1.7405, B = 1.4012. g1 and g2 are in brackets against x1: 0 (5.7, 0), 0.04 (4.8231, 1.0034), 0.11 (3.6990, 1.0251), 0.28 (2.2064, 1.1603), 0.43 (1.5902, 1.3880), 0.61(1.2226, 1.8417), 0.80 (1.0561, 2.6406), 0.89 (1.0144, 3.1869), 0.94 (1.004, 3.5522), 1.00 (1.0, 4.06) 8.50 (a) 1.7951, 1.4679 (b) 1.7951, 1.4679 (c) 65.21% water, 65.82 kPa 8.51 x1 = 0.20, g1 = 1.0720, g2 = 1.0059, P = 100.15 kPa, y1 = 0.2126 x1 = 0.90, g1 = 1.0007, g2 = 1.0815, P = 100.03 kPa, y1 = 0.894 8.52 g1 = 1.0288, g2 = 2.6100 8.53 20.22 kPa, 53.57% ethanol 8.54 93.3 kPa, 57.1% A 8.55 (a) 128.6 kPa, 34.25% A (b) No change 8.56 123.21 kPa, 38.54% A 8.57 Azeotrope exists at

121.8672 kPa, y1 = 0.8379 8.58 342 K, 0.0975 kg

8.59 Mole fractions of water in the liquid and vapour against pressure (kPa): Aniline-rich phase: 9.22 (0, 0), 42.475 (0.1, 0.805), 75.73 (0.2, 0.903), 108.99 (0.3, 0.941), 142.24 (0.4, 0.961), 146.23 (0.412, 0.963) Water-rich phase: 146.23 (0.984, 0.963), 145.06 (0.990, 0.977), 144.67 (0.992, 0.981), 144.08 (0.995, 0.988), 143.49 (0.998, 0.995), 143.10 (1.00, 1.00) 8.60 Mole fractions of ether in the liquid and vapour against temperature in K: Water-rich phase: 307 (0.0123, 0.9295), 313 (0.00914, 0.9277), 323 (0.0056, 0.8796), 333 (0.0034, 0.8047), 343 (0.0018, 0.6945), 353 (0.00098, 0.5334), 363 (0.00037, 0.3092) 8.61 (a) 369.2 K, (b) 4.78% aniline 8.62 (a) 361.6 K, 21.9% toluene, 13.1% ethylbenzene, 65.0% water (b) 387.8 K, 26.8% toluene, 73.2% ethyl benzene 8.63 Mole fraction of heptane in the vapour against temperature in K: 361.53 (0.35), 359.45 (0.40), 357.23 (0.45), 354.83 (0.50), 353.82 (0.52), 352.40 (0.548) 8.64 (a) 366 K, pure water (b) 374.9 K, pure toluene Last drop of vapour contains 44.36% toluene 8.65 (a) 388.8 K, pure component 1 (b) 365.5 K, pure component 2 8.66 (a) g1 = 1.1570, g2 = 1.4796 (b) 0.318

CHAPTER 9

9.5 6.09 105 9.6 – 57,350 J/mol 9.7 5.7498 10–4 9.8 Yes 9.9 (a) – 24,800 J/mol, feasible (b) 2.067 (c) DG0 = – 7.53305 104 + 6.12 3T2 + 4.7805 10–6T3 (d) 1.7297 9.10 14.3692 9.11 (a) Above 812.4 K, quite favourable Below 550 K, unfavourable (b) 75.42%, 45.69%

104(T)–1 + 63.710T ln T – 181.11T – 44.958

10–

(c) 81.94%, 55% 9.12 1.0506 10–3 9.13 DG0 = – 4.35946 104 + 13.003 T lnT – 1.8564 10–2T2 + 2.43835 10–6T3 + 54.972T

+ 2.2329 10–3T – 1.5640 lnT – 6.612 9.14 DG0 = – 3.92016 104 + 13.003T lnT – 1.8564 + 44.51T

2.2329

10–2T2 + 2.43835

10–6T3

10–3T – 1.5640 lnT – 5.3538 10–5

9.15 1.37 9.16 40.7% 9.17 16.6% 9.18 51.35% 9.19 15.05% 9.20 Ethylene: 21.9%, Steam: 74.0%, Ethanol: 4.1% 9.21 22.51% 9.22 58.92% 9.23 7.77% SO3, 0.34% SO2, 7.34% O2, 84.55% N2 9.24 32.05% CO2, 14.36% CO, 53.59% O2 9.25 46.3% CO2, 35.8% CO, 17.9% O2 Q = – 121.859 kJ/mol C burned 9.26 e = 0.07243, yNO2 = 0.135 yN2O4 = 0.119 9.27 (a) CO = 48.26%, CH3OH = 11.74% (b) Concentration will increase. 9.28 3.285% 9.29 (a) 2.08% (b) 3.4% 9.31 11.58 bar 9.32 (a) 0.4976 (b) 0.8251 9.33 11.93% A, 11.93% B, 76.14% C 9.34 Pressure in bar is given in brackets against temperature in K:

298 (0.0032), 400 (0.342), 500 (5.23) 600 (32.22), 700 (118.06) 9.35 0.4192 9.36 (a) 14.781 (b) Below about 345 K, the reaction is highly favourable. But above 420 K, unfavourable (c) 0.85%, 4.04% (d) 1.13%, 5.38% (e) 0.847% 9.37 0.1028 9.38 0.003733, 2.074 10–4 mol/kg, 0.0553 mol/kg 9.39 81.09% 9.40 1 bar (55.47%), 2 bar (42.64%), 3 bar (35.92%) 9.41 0.0481 bar 9.42 617.87 K 9.43 23 bar, yA = 0.072, yB = 0.87, yC = 0.058 9.44 4.1 bar 9.45 T (K) and P (bar) given in brackets against mole fraction of CO resulting in deposition of C: 0.3422 (900, 1), 0.3434 (900, 5), 0.1248 (900, 10), 0.6946 (1000, 1), 0.4259 (1000, 5), 0.3263 (1000, 10) 9.46 Acetylene: 0.4142, Hydrogen: 0.4142, Ethylene: 0.1716 9.47 8% 9.48 A: 31.58%, B: 26.32%, C: 21.05%, D: 15.79%, E: 5.26% 9.49 A: 23.54%, B: 5.53%, C: 12.47%, D: 27.81%, E: 30.65% 9.50 4 9.51 4

Index Absolute temperature, 12–14, 97–99 Absorption refrigeration, 163–165 Acentric factor, 64 Activity, 255–256, 431, 461–462 effect of temperature and pressure on, 256 and equilibrium constant, 431–433 in solutions, 296 standard state for, 255, 296–297 Activity coefficient, 292, 297–300,401–402, 460 calculation of, 401 and excess chemical potential, 319 effect of pressure on, 299 effect of temperature on, 299–300 equations for, 368–374 and Gibbs–Duhem equation, 303–304 in testing consistency of VLE data, 395–400 Adiabatic flame temperature, 77 mixing, 105 reaction temperature, 77 Adiabatic process, 55–57 work of compression, 144 work of expansion, ideal gas, 55–57 Air refrigeration cycle, 161–162 Air–standard cycles, 180–182, 183–185 Availability, 90, 93 Azeotropes, 364–368, 370–371 Benedict–Webb–Rubin equation, 64–65 Bernoulli equation, 129 Boiling point diagram, 347–348, 403–406 Brayton cycle, 189–190 Bubble point, 347, 349, 389 Canonical variables, 211–212 Carlson–Colburn relation, 402 Carnot cycle, 95–97 for power plants, 171–172 for refrigeration, 152–154 Carnot efficiency, 97–98 Carnot’s principle, 95–97 Chemical equilibrium (see Chemical reaction equilibrium) Chemical potential, 284–287, 338–339, 430, 433 as criteria of equilibrium, 338–339 effect of T and P on, 285–286 excess, 318–319 Chemical reactions: adiabatic, 77 entropy change of, 107 equilibrium constant for, 431–436, 462, 466–467 equilibrium conversion, 2, 425–426, 428, 446–448, 451, 454–455 extent of, 426, 428

feasibility of, 434–435 heat effects of, 69–77 independent, 467, 469 liquid-phase, 459–460 reaction coordinates (see also Extent of), 426 simultaneous, 465–467 in solutions, 461 stoichiometry of, 426 Chemical reaction equilibrium, 425–470 constants, 431–436, 440 criteria for, 429–431 effect of excess reactant on, 454 effect of inerts on, 451 effect of pressure on, 446–448 effect of products on, 457 effect of temperature on, 436–440 in heterogeneous reactions, 461–462 in homogeneous gas phase reactions, 446–448 in liquid phase reactions, 459–461 in multiple reactions, 465–468 standard state for, 434 Clapeyron equation, 213–214, 337–338 Claude process, 168 Clausius inequality, 108–110, 330 statement, 91–92 Clausius–Clapeyron equation, 214, 338 Clearance, 147–149 effect of, 147–149 volume, 147–149 Closed cycle gas turbine system, 188 heat engine system, 180 Closed system, 3, 26, 112, 343 Coefficient of compressibility, 207, 227, 239 of volume expansion, 207, 227, 239 Coefficient of performance (COP) of heat pump, 15, 91, 65 of refrigeration cycles, 15, 152, 153, 156, 162 Coexistence equation, 398–399 Combustion, standard heat of, 70 Compressibility, isothermal (see Coefficient of compressibility) Compressibility factor, 65, 67, 247, 257–258 Compression ratio, 184–185 Compressors, 143–149 multi-stage, 145–146 theoretical volumetric efficiency of, 147–149 work of isentropic, 144 work of isothermal, 145 Conservation of energy, 24, 127 of mass, 126–127 Consistency test of VLE, 395–400 Continuity equation, 126–127 Control volume, 126–127 Convergent–divergent nozzle, 138–139 Corresponding states, principle of, 68–69

Criterion of equilibrium, 330–332, 429–431 of stability, 332–333 Critical point, 50 pressure ratio, nozzle, 139 properties, 50, 62, 68 solution temperature, 405, 408 Cut-off ratio, 184 Cycle efficiency, 169 Cycles Carnot, 95, 152–154, 172 gas turbine, 188–190 internal combustion engine, 180–187 refrigeration, 151–165 steam power plant, 170–178 Dalton’s law, 353 Degrees of freedom, 11, 342–344, 469 Departure functions, 256–259 DePriester nomographs, 389 Derived properties, 206–207 Dew point temperature, 347–348, 389 Diesel cycle, 183–185 Differentials of energy properties, 210 Differential expressions for enthalpy, 210, 217–218 for entropy, 216–217 for internal energy, 210, 217–218 Diffuser, 139, 142 Dilute solutions, 293 Distillation, 1, 395 Displacement volume, 148 Dual cycle, 187–188 Duhem–Margules equation, 356 Duhem’s theorem, 343–344 Efficiency Carnot, 96–100, 172 compressor, 144, 148–149 ejector, 143 gas liquefaction, 169 gas turbine, 190 of internal combustion engine, 182, 184–185, 188 of steam power plant, 171–172, 175, 178 theoretical volumetric, 147–149 Endothermic reactions, 70, 437–438 Energy, 6, 8, 24, 34–36 availability of, 90, 93–94 balance, 35, 127–129 conservation of, 24, 127 degradation of, 90, 91, 93–95 equation, 127–129 internal, 24–27, 34–35, 41, 51, 93 kinetic, 8, 24–28, 35–36 potential, 8, 24–27, 35–36 properties, 206

Engines Carnot, 171–172 heat, 14, 180 internal combustion, 180–187 Enthalpy, 31–33 calculation of, 32–34, 220–226 change of chemical reaction, 69–70 change of combustion, 70 change of formation, 70–71 change of mixing, 309–312, 351 change of phase change, 213 change, constant pressure process, 32, 41 change, constant volume process, 32 change, steady–state flow process, 36 departure, 257–259, 263 differential equation for, 210, 217–218 effect of T and P on, 220–222 ideal gas, 55, 218–219 partial molar, 281–284, 300 residual (see Enthalpy departure) Enthalpy–entropy diagram, 260–261 Entrance work, 35 Entropy, 84–86, 92–97, 99–109 absolute, 119 calculation of, 103–107, 216, 220–226 change as criterion of equilibrium, 330–331 change of mixing, 105–106, 310–311 departure, 257–259, 263–264 differential equation for, 216–217 effect of T and P on, 220–222 and heat, 93–94 heat capacity relations, 215–216 for ideal gas, 103–104, 106–107 and irreversibility, 115–116 of isolated systems, 111–112 and nature of processes, 94–95 partial molar, 286, 310 principle of increase of, 112 and probability, 118–119 residual (see Entropy departure) statistical explanation of, 118–119 and temperature, 94 Equality of chemical potentials, 339 Equations of state, 51, 60–65, 250 approach for VLE, 345, 386 Benedict–Webb–Rubin, 64 limiting conditions for, 61 Peng–Robinson, 64, 387 Redlich–Kwong, 63, 387 Redlich–Kwong–Soave, 64, 387 van der Waals, 61–62 Virial, 65 Equilibrium, 1, 3, 329–332 chemical reaction, 425–479 constant for vaporisation, 363 constant, 431–436 criteria of, 330–332, 429–431 liquid–liquid, 408–410

mechanical, 10, 330 phase, 329–410 state, 1, 3, 329, 331–332, 436 thermal, 10–12, 330 vapour–liquid, 344–406 Equilibrium constant, 431–436 and activity, 431–433 effect of pressure on, 446 effect of reaction stoichiometry on, 432–433 effect of temperature on, 436–440 and free energy change, 433–434 evaluation of, 440 for multiple reactions, 466–467 Equilibrium conversion, 1, 2, 425–426, 428 effect of excess reactants on, 454 effect of inerts on, 451 effect of pressure on, 446–448 effect of products on, 457 in multiple reactions, 465–467 Equilibrium yield (see Equilibrium conversion) Exact differential equation, 209–210, 237 Excess properties, 317–319, 333 Exothermic reactions, 69, 436–437 Expansion engine vapour compression, 156–157 Extensive properties, 4, 206–207, 273, 276 Extent of reaction, 426–428, 430, 467–468 Extract, 409 Extraction, 409 Fanning friction factor, 135 Feasibility of process, 332, 434–435 First law of thermodynamics, 24–29 for cyclic process, 25 for flow process, 34–35, 127–129 for non–flow process, 26–31 limitations of, 89–90 Flame temperature, 77 Flash calculation, 390–391, 392–395 Flow energy, 35, 127 Flow in pipes, 135–136 Flow processes in compressors, 143–149 continuity equation of, 126–127 energy balance, 127–129 in ejectors, 142–143 mechanical energy balance for, 129 in nozzles, 137–140 in pipes, 135–137 in throttling, 143–144 total energy balance for, 128 Force, 5 Formation Gibbs free energy of, 440 heat of, 70–71, 442 Free energy Gibbs, 206–209 Helmholtz, 206–209

Free expansion, 16, 167–168 Friction factor, 135 Fugacity coefficient, 246–248, 289, 344–346, 386, 432, 448 generalised chart for, 248–249 Fugacity, pure fluids, 244–255 calculation of, 247–253 effect of pressure on, 246–247 effect of temperature on, 246 of gases, 244–246 of liquids and solids, 254 of van der Waals gas, 251–252 standard state for, 245 Fugacity, solutions, 288–293, 333–335, 355–356, 362 as criterion of equilibrium, 340 and equilibrium constant, 431–433, 436 in gaseous mixtures, 288–291 of ideal solutions, 290–293 in liquid mixtures, 291–293, 334–335, 352–353, 356, 363, 460 in VLE, 344–345 Function path, 4, 102–103 state, 4, 25–26, 101–102 Fundamental property relations, 210–211, 238 Fusion curve, 50–51 Gas ideal, 12, 51–57 liquefaction, 166–169 real, 60–65 solubility, Henry’s law for, 294 turbines, 188–190 Gas turbine cycle, 188–190 Generalised charts compressibility factor, 68–69, 258 enthalpy departure, 259 entropy departure, 259 fugacity coefficient, 248–249 Generalised correlation for enthalpy departure, 258–259 entropy departure, 258–259 Giauque functions, 440–442 Gibbs–Duhem equation, 302–304, 334, 368, 395–396, 398–399 Gibbs free energy, 206–209 at equilibrium, 331–332, 335–336, 428–430 change as criteria of equilibrium, 330–331, 429–431 change of mixing, 308, 311, 315, 332 change on reaction, 429–431, 433–435 differential equation for, 210 effect of temperature on, 235–236 excess, 318–319, 333, 369, 398 of formation, 440 functions (see Giauque functions) partial molar, 284 (see also Chemical potential) Gibbs–Helmholtz equation, 235–236, 437 Gibbs paradox, 107 Gross heating value, 70

Heat, 4, 7–8, 25, 35, 40–41, 89–90, 93 of combustion, 70–71 and entropy, 93–94 of formation, 70–71 latent, 213, 337 of mixing, 300, 311–312 quality of, 89–90 of reaction, 69–74 reservoir, 14, 94, 96 sign convention for, 25 of solution, 311–312 specific, 40 of vaporisation, 214 Heat capacity, 4, 40–41, 53, 215 constant pressure, 40–41, 53, 239 constant volume, 40–41, 53, 239 effect of pressure and volume on, 229–230 entropy change and, 215 of ideal gas, 53–54, 231 mean, 315–316 ratio of, 229 relationship between, 226–227, 243 Heat engines, 14, 91, 180 efficiency, 14, 91 Heat pump, 15, 91, 165 COP of, 15, 165 Heat of reaction, effect of T on, 72–74 Helmholtz free energy, 206–208 as criterion of equilibrium, 331 differential equations for, 210 Henry’s law, 293–294, 295–296, 298 for activity coefficients, 298 and gas solubility, 294 and Lewis–Randall rule, 293–294, 305 for standard state, 298–301, 460–461 Hess’s law, 71 HS diagram, 260–261 HT diagram, 260, 262–263 construction of, 262–263 Ideal gas, 12, 51–57 adiabatic process for, 55–57 as Carnot cycle fluid, 98–99 constant pressure process for, 53–54 constant volume process for, 52–53 enthalpy of, 51, 218–219 equation, 12, 51 heat capacity of, 53–54, 231 internal energy of, 51, 218–219 isothermal process for, 54–55 Joule–Thomson coefficient of, 52, 143, 235 polytropic process for, 57 property changes of mixing for, 310–311 Ideal solution, 290–293, 351–357 and Lewis–Randall rule, 290–292 phase equilibrium for, 351–355 property changes of mixing, 310–311

and Raoult’s law, 292–293, 298, 351–357 and standard states, 298 Immiscible systems, 332–333 Independent chemical reactions, 469 Intensive property, 4, 206, 274, 276, 341–342 Internal combustion engines, 180–187 Internal energy, 25–27, 30, 34–35, 41 change in constant volume process, 32, 41 change in non–flow process, 26–27 differential equation for, 218–219 effect of T and P on, 220–222 Invariant system, 343 Inversion temperature, 143 Irreversible process, 16–19, 94–95, 102–103, 108–110, 115–116, 208, 330–331 Isenthalpic process, 52 Isentropic expansion, liquefaction, 168–169 flow, 136 Isolated system, 3, 94, 112, 330 Isothermal compressibility, 207, 227, 239 compression, work of, 144–145 expansion, work of, 54–55 mixing, ideal gases, 106–107 process, 54–55 Jacobians, 236–238 thermodynamic relations using, 238–244 Jet pumps, 142 Joules experiment, 24 Joule–Thomson coefficient, 52, 143, 166–167, 233–235, 244 expansion, 143 inversion curve, 233–234 inversion temperature, 143 liquefaction, 166–168 Kelvin–Planck statement, second law, 91–92 Kelvin temperature, 12–14 Kinetic energy, 8, 25–26, 35–36 K–values for VLE, 386–390 DePriester nomographs, 389–390 Latent heat (see Enthalpy change of phase change) Law of conservation of energy, 24, 127 LeChatlier’s principle, 436, 446–447 Lewis fugacity rule (see Lewis–Randall rule) Lewis–Randall rule, 290–292, 341, 356 for activity coefficient, 298 and Henry’s law, 293–294, 305 standard state based on, 300–301, 313 Limiting conditions for equations of state, 51 Linde process, air liquefaction, 167–168 Liquefaction process, 166–169 by isentropic expansion, 168–169 by throttling, 166–168 by vaporisation of liquid, 166

Liquid–liquid equilibrium, 408–410 Liquids, fugacity of, 254, 291–293 Liquid–vapour equilibrium (see Vapour–liquid equilibrium) Lost work, 116 Mach number, 138 Margules equation, 319, 369–370 Maximum net work, 209 velocity for flow, 136–137 work, 18, 207–208 Maximum–boiling azeotrope, 365–366 Maxwell’s equations, 211–213, 215, 238–239 Mechanical energy balance, 128–129 Minimum–boiling azeotrope, 364–365 Mnemonic diagrams, 212 Molality, 460–461 Mollier diagrams, 260–261 Multistage compression, 145–147 with interstage cooling, 146 work requirement, 146 Negative deviation, 318, 363–365 Net heating value, 70 Newton’s second law, 5 Non–ideal solutions, 293–294, 307, 361–367 ideal behaviour of, 293–294 negative deviation in, 318, 363–365 positive deviation in, 318, 363–365 Nozzles, 137–140 convergent–divergent, 138–140 critical pressure ratio in, 139 relation between A and u, 137–138 throat velocity for, 138–140 NRTL equation, 372 Number of degrees of freedom, 11, 342–344 Open cycle, 180 gas turbine power plant, 188 Open systems, 3 Ordinary vapour compression cycle, 154–156 Osmosis, 340 Osmotic pressure, 340–341 Otto cycle, 180–182 Partial heat of mixing, 300 pressure, 288, 290 Partially miscible systems, 403–405 Partial molar property, 273–281 determination of, 279–281 physical meaning of, 274–275 and properties of solutions, 276–277 Path functions, 4, 102 Peng–Robinson equation, 64, 387 Perpetual motion machine, 24

Phase, 3 change and entropy change, 103 diagrams, 346–351, 353–355, 362–366, 404–406, 408–410 equilibrium, 329–410 rule, 11, 341–343, 469 Phase equilibrium, 329–410 criterion of, 292, 330–332 in multicomponent systems, 338–340 in single component systems, 335–336 PH diagrams, 155, 259–260 Plait point, 409 Polytropic process, 57 Positive deviation from ideality, 318, 363–365 Potential energy, 8, 25–27, 35–36 Power–plant cycles, 170–179 Poynting correction, 346, 388 Practical efficiency, 169 Pressure, 5–6 critical, 50 of decomposition, 462 drop, 135 osmotic, 340–341 partial, 288, 290, 353–354, 362–364, 432, 462 reduced, 68–69 Pressure ratio critical, 139 turbine, 190 Principle of corresponding states, 68–69 Probability and entropy, 118–119 Properties, 4 critical, 50, 62 derived, 206–207 energy, 206 excess, 317–319 extensive, 4 intensive, 4 partial molar, 273–281 path, 4, 102 reduced, 68–69 reference, 206 residual, 257–259 Property changes on mixing, 296, 307–311 and excess property changes, 317 for ideal solutions, 310–311 PT diagram, 50–51, 61 PV diagram for IC engines, 181, 184, 187 PV isotherm, 49–50, 61, 65 PVT behaviour of fluids, 49–51 Raffinate, 409 Rankine cycle, 171–172 Raoult’s law, 292–293, 295–296, 298, 305, 351–355, 362–363 Reaction coordinate, 426 Redlich–Kister consistency test, 398 Redlich–Kwong equation, 63, 387 Redlich–Kwong–Soave equation, 64, 387 Reduced properties, 68–69

Reference properties, 206 Refrigerant, choice of, 159–160 Refrigeration cycle, 151–165 absorption, 163–165 air, 161–162 capacity of, 152 Carnot cycle for, 152–153 COP of, 152, 153, 156, 162 vapour compression cycle, 154–156 Regenerative cycle, 177–178 Reheat cycle, 174–175 Relative volatility, 354–355 Residual properties (see Departure functions) Reverse osmosis, 340 Reversible process, 16–19, 95, 103, 108–110, 207–209, 213, 330 work, 17–19, 207–209 Saturated phases, 50 Saturation pressure, 50 temperature, 50 Second law of thermodynamics, 90–92, 111–112 Shaft work, 34–36, 122–129, 144–145 Solubility of gas in liquid, Henry’s law, 294 Sonic velocity, 137 Specific heat (see also Heat capacity), 4, 40–41, 215 Stability criteria, 332–333 Standard Gibbs free energy change, 433–437, 441 heat of combustion, 70–71 heat of formation, 70–71, 442 states, 69–70, 245, 255, 296–301, 308 vapour power cycle (see Rankine cycle) Standard heat of reaction, 69–71, 438–439 effect of temperature on, 72–74 State, 3 equilibrium, 3, 329, 331, 332, 436 functions, 4, 25–26, 101–102, 207 steady, 10, 34, 329 Steady–state flow process, 34–36 Steam–power plants, 170–179 Rankine cycle for, 171–173 regenerative cycle for, 177–179 reheat cycle for, 174–176 Steam tables, 483–490 Stoichiometric numbers, 72, 426 reaction, 426 Sublimation, 51 Surroundings, 3, 26, 112 System, 2 adiabatic, 112 closed, 3, 26, 112 heterogeneous, 3 homogeneous, 3 isolated, 3, 25, 84, 112–113, 330

open, 3 Tangent–intercept method, 280–281 Temperature, 11–14 absolute, 12–14, 97–99 adiabatic reaction, 77 critical solution, 409 critical, 50 and entropy, 94 flame, 77 Joule–Thomson inversion, 143 of reaction, 77 reduced, 68–69 saturation, 50 three–phase equilibrium, 404–406 upper critical solution, 405 Temperature scale, 12–14, 97–99 ideal gas, 12–14, 98–99 thermodynamic, 97–99 Ternary equilibrium diagrams, 408–410 TH diagram, 260 construction of, 262–263 Theoretical volumetric efficiency, 147–149 Thermal efficiency of gas turbines, 190 of IC engines, 182, 184–185, 188 of steam power plants, 171–172, 175, 178 Thermodynamic consistency, 395–400 efficiency, 169 equilibrium, 11, 330–331, 339 temperature, 13–14, 97–98 Thermodynamic diagrams, 259–264 construction of, 262–264 types of, 259–262 Thermodynamics limitations of, 1–2 scope of, 1–2 third law of, 118–119 zeroth law of, 11–12 Three–phase equilibrium temperature, 404–406 Throttling process, 143, 233, 259–260 Tie line, 347, 354, 410 Ton of refrigeration, 152 Total energy balance, 128 Triangular diagrams, 409–410 Triple point, 13, 51, 97, 343 TS diagram, 260–261 construction of, 263–264 for IC engines, 181, 184, 187 for liquefaction, 167–168 for refrigeration, 153, 155–157, 159, 161 for steam power plants, 172, 175, 178 UNIFAC equation, 374 UNIQUAC equation, 371–373 Unit

operations, 1–2 processes, 1 Univariant system, 343 Upper critical solution temperature, 405 van der Waals equation, 61–62 van Laar equation, 370–371 van’t Hoff equation, 437 Vaporisation curves, 50–51 Vaporisation equilibrium constant (see K–values for VLE) Vapour–compression refrigeration cycle, 154–157 expansion engine, 156–157 ordinary, 154–156 Vapour–liquid equilibrium, 344–406 activity coefficient equations for, 369–374 activity coefficient model approach for, 345 basic equations for, 344–345 boiling point diagrams, 347–348 consistency tests for, 395–400 diagrams, 346–350 effect of pressure on, 349 equations of state approach for, 345, 386 equilibrium diagram, 348 flash calculations, 390–391 fugacity of components under, 344–346 immiscible systems, 405–406 K–values for, 387–390 partially miscible systems, 403–404 P–x–y diagrams, 350 Virial equation, 65–66 Volume change of mixing, 308–309, 313–314, 317, 351–352 coefficient of expansion, 207, 227, 239 critical, 50 excess, 317 partial molar, 274–275, 280–284 reduced, 68 residual, 248–250 Wilson equation, 371–372 Wohl’s equation, 369 Work, 4, 6–8, 25, 34–35, 89–90 of adiabatic compression, 144–145 of adiabatic expansion, 56–57 function (see Helmholtz free energy) of isothermal compression, 144–145 of isothermal expansion, 54–55 lost, 116 requirement for compressors, 144–145 reversible, 17–19 shaft, 31–33, 127–129, 144–149 sign convention for, 25 x–y diagrams, 348–349 x–y–P diagrams, 349, 350, 353–354 x–y–T diagrams (see Boiling point diagram)

Zero area method, 398 Zeroth law of thermodynamics, 11–12