This question is about thermodynamics. Consider the thermal decomposition of calcium carbonate shown below:
CaCO3(s)→CaO(s)+CO2(g) \text{CaCO}_3(\text{s}) \rightarrow \text{CaO}(\text{s}) + \text{CO}_2(\text{g}) CaCO3(s)→CaO(s)+CO2(g)The table below shows some thermodynamic data.
| 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) |
|---|---|---|---|
| ΔfH⊖ / kJ mol−1\Delta_f H^\ominus\ /\ \text{kJ mol}^{-1}ΔfH⊖ / kJ mol−1 | −1207-1207−1207 | −635-635−635 | −394-394−394 |
| S⊖ / J K−1mol−1S^\ominus\ /\ \text{J K}^{-1}\text{mol}^{-1}S⊖ / J K−1mol−1 | 939393 | 404040 | 214214214 |
Explain why the standard entropy value for carbon dioxide is greater than that for calcium carbonate.
State the temperature at which the standard entropy of calcium oxide is 0 J K−1mol−10\ \text{J K}^{-1}\text{mol}^{-1}0 J K−1mol−1.
Use the equation and the thermodynamic data provided to calculate the minimum temperature, in K\text{K}K, at which this reaction becomes feasible.
