Question 19.3: Calculating ΔS° for a Reaction Calculate the change of entro...
Calculating \Delta S^{\circ} for a Reaction
Calculate the change of entropy, \Delta S^{\circ}, at 25^{\circ} \mathrm{C} for the reaction in which urea is formed from \mathrm{NH}_{3} and \mathrm{CO}_{2}.
2 \mathrm{NH}_{3}(g)+\mathrm{CO}_{2}(g) \longrightarrow \mathrm{NH}_{2} \mathrm{CONH}_{2}(a q)+\mathrm{H}_{2} \mathrm{O}(l)
The standard entropy of \mathrm{NH}_{2} \mathrm{CONH}_{2}(a q) is 174 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K}). See Table 19.1 for other values.
PROBLEM STRATEGY
The calculation is similar to that used to obtain \Delta H^{\circ} from \Delta H_{f}^{\circ} values.
It is convenient to put the standard entropy values multiplied by stoichiometric coefficients below the formulas in the balanced equation.
\begin{array}{rccc} 2 \mathrm{NH}_{3}(g)&+\mathrm{CO}_{2}(g) & \longrightarrow & \mathrm{NH}_{2} \mathrm{CONH}_{2}(a q)&+\mathrm{H}_{2} \mathrm{O}(l) \\ S^{\circ}: 2 \times 193 & 214 && 174 & 70 \end{array}
You calculate the entropy change by subtracting the entropy of the reactants from the entropy of the products.
\begin{aligned} \Delta S^{\circ} & \left.\left.=\sum n S^{\circ} \text { (products }\right)-\sum m S^{\circ} \text { (reactants }\right) \\ & =[(174+70)-(2 \times 193+214)] \mathrm{J} / \mathrm{K}=-356 \mathrm{~J} / \mathrm{K} \end{aligned}
Note that the sign of \Delta S^{\circ} agrees with the solution of Example 19.2b.