The enthalpy of vaporization of ethanol is 38.7 kJ/mol at

Katherine Walls 2021-12-19 Answered
The enthalpy of vaporization of ethanol is 38.7 kJ/mol at its boiling point (78C). Determine Ssys,Ssurr,and Suniv when 1.00 mole of ethanol is vaporized at 78C and 1.00atm.
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Expert Answer

Donald Cheek
Answered 2021-12-20 Author has 41 answers

Here we have the reaction:
C2H5OH(1)C2H5OH(g)
The values are:
Hvap=38.7kJmol
T=78C
Firstly we will calculate Ssys:
Ssys=HvapT
=38.7kJmol78+273K
=+110.2JmolK
Now, we will calculate Ssurr:
Ssurr=HvapT
=38.7kJmol78+273K
=110.2JmolK
Therefore, at the end we can calculate Suniv as the sum of Ssys and Ssurr:
Suniv=Ssys+Ssurr
=+110.2JmolK+(110.2JmolK)
=0

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Foreckije
Answered 2021-12-21 Author has 32 answers

Step 1
Given:
Enthalpy of vaporization of ethanol =38.7kJmol
Boiling point (T)=78C=351K
Step 2 Calculation
Vaporisation of ethanol is given by,
C2H5OH(1)C2H5OH(g)
Hvap=38.7kJmol
Step 3
Calculation for entropy of change system (Ssys):
Ssys=+HvapT
Ssys=38.7×103Jmol351K
Ssys=+110.26Jmol.K
Calculation for entropy change of surroundings (Ssurr):
Ssurr=HvapT
Ssurr=38.7×103Jmol351K
Ssurr=110.26Jmol.K
Step 4
Calculation for entropy change of universe (Suniv):
Suniv=Ssys+Ssurr
Suniv=+110.26Jmol.K110.26Jmol.K
Suniv=0

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