For the reaction CaCO_{3(s)} to CaO_{(s)} + C | vedclass

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(C) For a reaction to be spontaneous,the Gibbs free energy change $\Delta G$ must be negative $(\Delta G < 0)$.Given $\Delta G = \Delta H - T \cdot \D

Vedclass · Accepted Answer

$1114 \, K$

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(C) For a reaction to be spontaneous,the Gibbs free energy change $\Delta G$ must be negative $(\Delta G Given $\Delta G = \Delta H - T \cdot \Delta S$.At equilibrium,$\Delta G = 0$,so $T = \frac{\Delta H}{\Delta S}$.Converting $\Delta H$ to Joules: $\Delta H = 178.3 	imes 1000 \, J = 178300 \, J$.$T = \frac{178300 \, J}{160 \, J \cdot K^{-1}} = 1114.375 \, K \approx 1114 \, K$.Since the reaction is endothermic $(\Delta H > 0)$ and entropy increases $(\Delta S > 0)$,the reaction becomes spontaneous at temperatures higher than the equilibrium temperature.Therefore,the reaction becomes spontaneous at $T > 1114 \, K$.

For the reaction $CaCO_{3(s)} \to CaO_{(s)} + CO_{2(g)}$,given $\Delta H^o = 178.3 \, kJ$ and $\Delta S^o = 160 \, J \cdot K^{-1}$. At what temperature will the reaction become spontaneous?

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