The combustion step has a standard entropy change of at 298 K

CHEMISTRY · Single correct · A Level · Chemistry · Thermodynamics · Thermodynamics

Problem

The combustion step CO(g)+12 O2(g)→CO2(g)\text{CO}(g)+\tfrac{1}{2}\,\text{O}_2(g)\rightarrow \text{CO}_2(g) has a standard entropy change of ΔS∘=−86.4 J K−1\Delta S^{\circ}=-86.4\,\text{J K}^{-1} at 298 K. Given S∘(O2)=205.0 J K−1 mol−1S^{\circ}(\text{O}_2)=205.0\,\text{J K}^{-1}\,\text{mol}^{-1} and S∘(CO2)=213.7 J K−1 mol−1S^{\circ}(\text{CO}_2)=213.7\,\text{J K}^{-1}\,\text{mol}^{-1}, determine the standard molar entropy S∘(CO)S^{\circ}(\text{CO}) at 298 K. (a) 154  J K−1 mol−1154\;\text{J K}^{-1}\,\text{mol}^{-1} (b) 198  J K−1 mol−1198\;\text{J K}^{-1}\,\text{mol}^{-1} (c) 95  J K−1 mol−195\;\text{J K}^{-1}\,\text{mol}^{-1} (d) 230  J K−1 mol−1230\;\text{J K}^{-1}\,\text{mol}^{-1}

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