Chapter 11, Problem 11.1P

PRACTICE 11.1 The boiling point of ethanol is 78.4 °C, and the enthalpy change for theconversion of liquid to vapor is $\Delta H_{\text{vap}}=\text{38}.\text{56 kJ}/\text{mol}$ . What is the entropy change for vaporization, $\Delta S_{\text{vap}}$ , in $\text{J/(K•mol)}$ ?

Interpretation Introduction

Interpretation:The boiling point of ethanol is 78.4 $℃$, and the enthalpy change for the conversion of liquid to vapor is $\Delta H_{vap}=38.56kJ/mol.$

Concept introductionWhen a liquid undergoes vaporization, its entropy increases. Thus, the change in entropy is always positive for vaporization. This is due to the change of phase from liquid to vapor

The relation between change in Gibbs free energy, change in enthalpy and change in entropy is as follows:

  $\Delta G_{\text{vap}}=\Delta H_{\text{vap}}-T_{\text{vap}}\Delta S_{\text{vap}}$

In the vaporization of liquid, the liquid and vapor phases exist in equilibrium thus, the change in Gibbs free energy is zero.

Thus,

  $\Delta G_{\text{vap}}=\Delta H_{\text{vap}}-T_{\text{vap}}\Delta S_{\text{vap}}=0$

Or,

  $\Delta S_{\text{vap}}=\frac{\Delta H_{\text{vap}}}{T_{\text{vap}}}$

Given:The enthalpy change for vaporization, $\Delta H_{vap}$is 38.56 kJ/mol and boiling point of ethanol is $78.4{}^{°}\text{C}$.

Answer to Problem 11.1P

Solution:The value of $\Delta S_{vap}$is $109.7J/molK.$

Explanation of Solution

Converting the boiling point from degree Celsius to Kelvin as follows:

$\begin{array}{l}T=(78.4+273.15)K\\ =351.55\text{}\text{}K\end{array}$

The enthalpy change for the vaporization is 38.56 kJ/mol. Thus, the entropy change for the vaporization can be calculated as follows:

  $\Delta S_{\text{vap}}=\frac{\Delta H_{\text{vap}}}{T_{\text{vap}}}$

Putting the values,

$\begin{array}{c}\Delta S_{vap}=\frac{(38.56kJ/mol)\times \left(\frac{{10}^{3}J}{1kJ}\right)}{351.55K}\\ =109.7J/molK\end{array}$

Conclusion

Therefore, the value of $\Delta S_{vap}$is $109.7J/molK.$