Question 19.1: Calculating the Entropy Change for a Phase Transition The he...
Calculating the Entropy Change for a Phase Transition
The heat of vaporization, \Delta H_{\text {vap }}, of carbon tetrachloride, \mathrm{CCl}_{4}, at 25^{\circ} \mathrm{C} is 39.4 \mathrm{~kJ} / \mathrm{mol}.
\mathrm{CCl}_{4}(l) \longrightarrow \mathrm{CCl}_{4}(g) ; \Delta H_{\text {vap }}=43.0 \mathrm{~kJ} / \mathrm{mol}
If 1 \mathrm{~mol} of liquid carbon tetrachloride at 25^{\circ} \mathrm{C} has an entropy of 216 \mathrm{~J} / \mathrm{K}, what is the entropy of 1 \mathrm{~mol} of the vapor in equilibrium with the liquid at this temperature?
PROBLEM STRATEGY
The entropy change for this equilibrium vaporization is \Delta H_{v a p} / T. The entropy of the vapor equals the entropy of the liquid plus the entropy change.
When liquid \mathrm{CCl}_{4} evaporates, it absorbs heat: \Delta H_{\text {vap }}=39.4 \mathrm{~kJ} / \mathrm{mol}(39.4 \times 10^{3} \mathrm{~J} / \mathrm{mol} ) at 25^{\circ} \mathrm{C}, or 298 \mathrm{~K}. The entropy change, \Delta S, is
\Delta S=\frac{\Delta H_{v a p}}{T}=\frac{39.4 \times 10^{3} \mathrm{~J} / \mathrm{mol}}{298 \mathrm{~K}}=132 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K})
In other words, 1 \mathrm{~mol} of carbon tetrachloride increases in entropy by 132 \mathrm{~J} / \mathrm{K} when it vaporizes. The entropy of 1 \mathrm{~mol} of the vapor equals the entropy of 1 \mathrm{~mol} of liquid (216 \mathrm{~J} / \mathrm{K}) plus 132 \mathrm{~J} / \mathrm{K}.
\text { Entropy of vapor }=(216+132) \mathrm{J} /(\mathrm{mol} \cdot \mathrm{K})=\mathbf{3 4 8} \mathbf{J} /(\mathbf{m o l} \cdot \mathbf{K})
Note: This is the entropy at the equilibrium vapor pressure, which differs from S^{\circ} (entropy at 1 \mathrm{~atm} ).