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

The enthalpy of vaporization of ethanol is 38.7 kJ/mol at its boiling point $\left({78}^{\circ }C\right)$. Determine when 1.00 mole of ethanol is vaporized at ${78}^{\circ }C$ and 1.00atm.
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Donald Cheek

Here we have the reaction:
${C}_{2}{H}_{5}OH\left(1\right)⇒{C}_{2}{H}_{5}OH\left(g\right)$
The values are:
$\mathrm{△}{H}_{vap}=38.7k\frac{J}{mol}$
$T={78}^{\circ }C$
Firstly we will calculate $\mathrm{△}{S}_{sys}$:
$\mathrm{△}{S}_{sys}=\frac{\mathrm{△}{H}_{vap}}{T}$
$=\frac{38.7k\frac{J}{mol}}{78+273K}$
$=+110.2\frac{J}{mol}K$
Now, we will calculate $\mathrm{△}{S}_{surr}$:
$\mathrm{△}{S}_{surr}=-\frac{\mathrm{△}{H}_{vap}}{T}$
$=-\frac{38.7k\frac{J}{mol}}{78+273K}$
$=-110.2\frac{J}{mol}K$
Therefore, at the end we can calculate $\mathrm{△}{S}_{univ}$ as the sum of $\mathrm{△}{S}_{sys}$ and $\mathrm{△}{S}_{surr}$:
$\mathrm{△}{S}_{univ}=\mathrm{△}{S}_{sys}+\mathrm{△}{S}_{surr}$
$=+110.2\frac{J}{mol}K+\left(-110.2\frac{J}{mol}K\right)$
$=0$

Foreckije

Step 1
Given:
Enthalpy of vaporization of ethanol $=38.7k\frac{J}{mol}$
Boiling point $\left(T\right)={78}^{\circ }C=351K$
Step 2 Calculation
Vaporisation of ethanol is given by,
${C}_{2}{H}_{5}OH\left(1\right)⇒{C}_{2}{H}_{5}OH\left(g\right)$
$\mathrm{△}{H}_{vap}=38.7k\frac{J}{mol}$
Step 3
Calculation for entropy of change system $\left(\mathrm{△}{S}_{sys}\right)$:
$\mathrm{△}{S}_{sys}=+\frac{\mathrm{△}{H}_{vap}}{T}$
$\mathrm{△}{S}_{sys}=\frac{38.7×{10}^{3}\frac{J}{mol}}{351K}$
$\mathrm{△}{S}_{sys}=+110.26\frac{J}{mol}.K$
Calculation for entropy change of surroundings $\left(\mathrm{△}{S}_{surr}\right)$:
$\mathrm{△}{S}_{surr}=-\frac{\mathrm{△}{H}_{vap}}{T}$
$\mathrm{△}{S}_{surr}=\frac{-38.7×{10}^{3}\frac{J}{mol}}{351K}$
$\mathrm{△}{S}_{surr}=-110.26\frac{J}{mol}.K$
Step 4
Calculation for entropy change of universe $\left(\mathrm{△}{S}_{univ}\right)$:
$\mathrm{△}{S}_{univ}=\mathrm{△}{S}_{sys}+\mathrm{△}{S}_{surr}$
$\mathrm{△}{S}_{univ}=+110.26\frac{J}{mol}.K-110.26\frac{J}{mol}.K$
$\mathrm{△}{S}_{univ}=0$