Question 19.6: Interpreting the Sign of ΔG° Calculate ΔH° and ΔG° for the r...
Interpreting the Sign of \Delta G^{\circ}
Calculate \Delta H^{\circ} and \Delta G^{\circ} for the reaction
2 \mathrm{KClO}_{3}(s) \longrightarrow 2 \mathrm{KCl}(s)+3 \mathrm{O}_{2}(g)
Interpret the signs obtained for \Delta H^{\circ} and \Delta G^{\circ}. Values of \Delta H_{f}^{\circ} (in \mathrm{kJ} / \mathrm{mol} ) are as follows: \mathrm{KClO}_{3}(s),-397.7 ; \mathrm{KCl}(s),-436.7. Similarly, values of \Delta G_{f}^{\circ} (in \mathrm{kJ} / \mathrm{mol}) are as follows: \mathrm{KClO}_{3}(s),-296.3 ; \mathrm{KCl}(s),-408.8. Note that \mathrm{O}_{2}(g) is the reference form of the element, so \Delta H_{f}^{\circ}=\Delta G_{f}^{\circ}=0 for it.
PROBLEM STRATEGY
Calculate \Delta H^{\circ} and \Delta G^{\circ}; then interpret the signs of these quantities. The interpretation of the sign of \Delta H was discussed in Chapter 6. As long as \Delta G^{\circ} is not close to zero, you can interpret its sign as follows: a negative \Delta G^{\circ} means the reactants tend to go mostly to products; a positive \Delta G^{\circ} means that the reaction is nonspontaneous as written. When the value of \Delta G^{\circ} is close to zero, the reaction gives an equilibrium mixture.
The problem is set up as follows:
\begin{aligned} & 2 \mathrm{KClO}_{3}(s) &\longrightarrow & \qquad 2 \mathrm{KCl}(s)&+3 \mathrm{O}_{2}(g) \\ & \Delta H_{f}^{\circ}: 2 \times(-397.7)&& \quad 2 \times(-436.7) \quad &0 \mathrm{~kJ} \\ & \Delta G_{f}^{\circ}: 2 \times(-296.3) &&\quad 2 \times(-408.8) &\quad 0 \mathrm{~kJ} \end{aligned}
Then,
\begin{aligned} \Delta H^{\circ} & =[2 \times(-436.7)-2 \times(-397.7)] \mathrm{kJ}=-\mathbf{7 8 . 0} \mathbf{k J} \\ \Delta G^{\circ} & =[2 \times(-408.8)-2 \times(-296.3)] \mathrm{kJ}=-\mathbf{2 2 5 . 0} \mathbf{k J} \end{aligned}
The reaction is exothermic, liberating 78.0 \mathrm{~kJ} of heat. The large negative value for \Delta G^{\circ} indicates that the equilibrium composition is mostly potassium chloride and oxygen.