Question 19.4: Calculating ΔG° from ΔH° and ΔS° What is the standard free-e...
Calculating ΔG° from ΔH° and ΔS°
What is the standard free-energy change, \Delta G^{\circ}, for the following reaction at 25^{\circ} \mathrm{C} ?
\mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \longrightarrow 2 \mathrm{NH}_{3}(g)
Use values of \Delta H_{f}^{\circ} and S^{\circ}, from Tables 6.2 and 19.1 .
PROBLEM STRATEGY
Calculate \Delta H^{\circ} and \Delta S^{\circ}, then substitute these values into \Delta H^{\circ}-T \Delta S^{\circ} to obtain \Delta G^{\circ}.
Write the balanced equation and place below each formula the values of \Delta H_{f}^{\circ} and S^{\circ} multiplied by stoichiometric coefficients.
\begin{array}{cccc} \mathrm{N}_{2}(g)+ & 3 \mathrm{H}_{2}(g) & & 2 \mathrm{NH}_{3}(g) \\ \Delta H_{f}^{\circ}: 0 & 0 & & 2 \times(-45.9) \mathrm{kJ} \\ S^{\circ}: 191.6 & 3 \times 130.6 & & 2 \times 192.7 \mathrm{~J} / \mathrm{K} \end{array}
You calculate \Delta H^{\circ} and \Delta S^{\circ} by taking values for products and subtracting values for reactants.
\begin{aligned} \Delta H^{\circ} & =\sum n \Delta H_{f}^{\circ}(\text { products })-\sum m \Delta H_{f}^{\circ}(\text { reactants }) \\ & =[2 \times(-45.9)-0] \mathrm{kJ}=-91.8 \mathrm{~kJ} \\ \Delta S^{\circ} & =\sum n S^{\circ}(\text { products })-\sum m S^{\circ}(\text { reactants }) \\ & =[2 \times 192.7-(191.6+3 \times 130.6)] \mathrm{J} / \mathrm{K}=-198.0 \mathrm{~J} / \mathrm{K} \end{aligned}
You now substitute into the equation for \Delta G^{\circ} in terms of \Delta H^{\circ} and \Delta S^{\circ}. Note that you substitute \Delta S^{\circ} in units of \mathrm{kJ} / \mathrm{K}.
\Delta G^{\circ}=\Delta H^{\circ}-T \Delta S^{\circ}=-91.8 \mathrm{~kJ}-(298 \mathrm{~K})(-0.1980 \mathrm{~kJ} / \mathrm{K})=-\mathbf{3 2 . 8} \mathbf{k J}