Example 16.5: Determination of \(ΔS°\)
Calculate the standard entropy change for the following process:
\(\ce{H2O(g)}\)\(\ce{->}\)\(\ce{H2O(l)}\)\(\ce{ }\)
Solution
The value of the standard entropy change at room temperature,
\(ΔS°_{\mathrm{298}}\), is the difference between the standard entropy of the product,
\(\ce{H2O(l)}\), and the standard entropy of the reactant,
\(\ce{H2O(g)}\).
\(ΔS°_{\mathrm{298}} = S°_{\mathrm{\ce{H2O(l)},298}} - S°_{\mathrm{\ce{H2O(g)},298}}\)
\(S°_{\mathrm{\ce{H2O(g)},298}}\) \(= 188.8\ \frac{\mathrm{J}}{\mathrm{mol}\ \mathrm{K}}\)
\(ΔS°_{\mathrm{298}}\) \(= S°_{\mathrm{\ce{H2O(l)},298}} - S°_{\mathrm{\ce{H2O(g)},298}}\)
\(\ \ \ =70.0\ \frac{\mathrm{J}}{\mathrm{mol}\ \mathrm{K}} - 188.8\ \frac{\mathrm{J}}{\mathrm{mol}\ \mathrm{K}}\)
\(\ \ \ =-118.8\ \frac{\mathrm{J}}{\mathrm{mol}\ \mathrm{K}}\)
The value for
\(ΔS°_{\mathrm{298}}\) is negative, as expected for this phase transition (condensation), which the previous section discussed.