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Exercise · Q29

Q.The thermal decomposition of calcium carbonate, CaCO3(s)→CaO(s)+CO2(g)\text{CaCO}_3(s) \rightarrow \text{CaO}(s) + \text{CO}_2(g), has ΔH∘=+178 kJ mol−1\Delta H^\circ = +178\ \text{kJ mol}^{-1} and ΔS∘=+160 J K−1mol−1\Delta S^\circ = +160\ \text{J K}^{-1}\text{mol}^{-1}. Calculate the minimum temperature above which this decomposition becomes spontaneous, assuming ΔH∘\Delta H^\circ and ΔS∘\Delta S^\circ do not vary with temperature.

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The decomposition becomes spontaneous once ΔG∘\Delta G^\circ turns negative. The threshold (minimum) temperature at which this first happens is where ΔG∘=0\Delta G^\circ = 0 exactly, i.e. where ΔH∘=TΔS∘\Delta H^\circ = T\Delta S^\circ, so T=ΔH∘ΔS∘T = \dfrac{\Delta H^\circ}{\Delta S^\circ}. Substituting the given values (converting ΔH∘\Delta H^\circ to joules to match the units of ΔS∘\Delta S^\circ in J K−1mol−1\text{J K}^{-1}\text{mol}^{-1}): T=178,000 J mol−1160 J K−1mol−1=1112.5 KT = \dfrac{178{,}000\ \text{J mol}^{-1}}{160\ \text{J K}^{-1}\text{mol}^{-1}} = 1112.5\ \text{K}. Below 1112.5 K1112.5\ \text{K}, ΔG∘>0\Delta G^\circ > 0 and calcium carbonate does not decompose spontaneously; above 1112.5 K1112.5\ \text{K} (≈839.4∘C\approx 839.4^\circ\text{C}), ΔG∘<0\Delta G^\circ < 0 and decompositio …

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