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Entropy and Spontaneity (Additional Higher Level) · Paper 2 — Free Chemistry SL Practice Questions

1ChallengeSAQ-LEntropy and the second law of thermodynamics7 marksPaper 2~11 min

Data

- ΔH⊖=+178 kJ mol−1\Delta H^{\ominus} = +178\,\text{kJ mol}^{-1} - ΔSsystem⊖=+161 J K−1mol−1\Delta S^{\ominus}_{system} = +161\,\text{J K}^{-1}\text{mol}^{-1} Equations: ΔG⊖=ΔH⊖−TΔS⊖\Delta G^{\ominus} = \Delta H^{\ominus} - T\Delta S^{\ominus}; ΔStotal=ΔSsystem+ΔSsurroundings\quad \Delta S_{total} = \Delta S_{system} + \Delta S_{surroundings}; ΔSsurroundings=−ΔH⊖T\quad \Delta S_{surroundings} = -\dfrac{\Delta H^{\ominus}}{T}
This question is about the spontaneity of thermal decomposition of calcium carbonate, a key reaction in the cement industry. CaCO3(s)→CaO(s)+CO2(g)CaCO_3(s) \rightarrow CaO(s) + CO_2(g)
(a)
Calculate the temperature, in K, above which the decomposition of calcium carbonate becomes spontaneous. [3 marks]
(b)
Explain why ΔSsystem⊖\Delta S^{\ominus}_{system} is positive for this reaction. [2 marks]
(c)
Calculate the total entropy change of the universe, ΔStotal\Delta S_{total}, at 298 K298\,\text{K}, and hence evaluate whether this consistent with the reaction being non-spontaneous at this temperature. [2 marks]
diagram

Solutions

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