Question 19.10: Calculating ΔG° and K at Various Temperatures a. What is ΔG°...

a. From Tables 6.2 and 19.1, you have

\begin{aligned} & \mathrm{CaCO}_{3}(s) &\rightleftharpoons &\mathrm{CaO}(s)&+\mathrm{CO}_{2}(g) \\ & \Delta H_{f}^{\circ}:-1206.9 \quad &&-635.1 \quad &-393.5 \mathrm{~kJ} \\ & S^{\circ}: & 92.9 & 38.2 & 213.7 \mathrm{~J} / \mathrm{K} \end{aligned}

You calculate \Delta H^{\circ} and \Delta S^{\circ} from these values.

\begin{gathered} \Delta H^{\circ}=[(-635.1-393.5)-(-1206.9)] \mathrm{kJ}=178.3 \mathrm{~kJ} \\ \Delta S^{\circ}=[(38.2+213.7)-(92.9)] \mathrm{J} / \mathrm{K}=159.0 \mathrm{~J} / \mathrm{K} \end{gathered}

Now you substitute \Delta H^{\circ}, \Delta S^{\circ}(=0.1590 \mathrm{~kJ} / \mathrm{K}), and T(=1273 \mathrm{~K}) into the equation for \Delta G_{T}^{\circ}.

\Delta G_{T}^{\circ}=\Delta H^{\circ}-T \Delta S^{\circ}=178.3 \mathrm{~kJ}-(1273 \mathrm{~K})(0.1590 \mathrm{~kJ} / \mathrm{K})=\mathbf{- 2 4 . 1} \mathbf{k J}

Because \Delta G_{T}^{\circ} is negative, the reaction should be spontaneous at 1000^{\circ} \mathrm{C} and 1 \mathrm{~atm}

b. Substitute the value of \Delta G^{\circ} at 1273 \mathrm{~K}, which equals -24.1 \times 10^{3} \mathrm{~J}, into the equation relating \ln K and \Delta G^{\circ}.

\begin{aligned} \ln K & =\frac{\Delta G^{\circ}}{-R T}=\frac{-24.1 \times 10^{3}}{-8.31 \times 1273}=2.278 \\ K & =K_{p}=e^{2.278}=\mathbf{9 . 7 6} \end{aligned}