The thermal decomposition of calcium carbonate is represented by the equation:
CaCO3(s)⟶CaO(s)+CO2(g)ΔH⊖=+178 kJ mol−1 \text{CaCO}_3(\text{s}) \longrightarrow \text{CaO}(\text{s}) + \text{CO}_2(\text{g}) \quad \Delta H^{\ominus} = +178 \text{ kJ mol}^{-1} CaCO3(s)⟶CaO(s)+CO2(g)ΔH⊖=+178 kJ mol−1Some entropy data are shown in the table below:
| Substance | CaCO3(s)\text{CaCO}_3(\text{s})CaCO3(s) | CaO(s)\text{CaO}(\text{s})CaO(s) | CO2(g)\text{CO}_2(\text{g})CO2(g) |
|---|---|---|---|
| S⊖ / J K−1mol−1S^{\ominus} \text{ / J K}^{-1} \text{mol}^{-1}S⊖ / J K−1mol−1 | 92.992.992.9 | 39.739.739.7 | 214214214 |
Use the equation and the entropy data to calculate the Gibbs free-energy change (ΔG\Delta GΔG) for this reaction at 950 ∘C950\,^{\circ}\text{C}950∘C. Give your answer to the appropriate number of significant figures.
Use your answer to explain whether this reaction is feasible.
