Question

The change in entropy for the fusion of 1 mole of ice is [ $ \text{mp}\,\text{of}\,\text{ice}\,\,\text{=}\,\,\text{273}\,\text{K,} $ , molar enthalpy of fusion for ice $ =\,\,6.0\,\text{kJ}\,\text{mo}{{\text{l}}^{-1}} $ ]

• A. $ 11.73\,J{{K}^{-1}}\,mo{{l}^{-1}} $
• B. $ 18.84\,J{{K}^{-1}}\,mo{{l}^{-1}} $
• C. $ 21.97\,J{{K}^{-1}}\,mo{{l}^{-1}} $
• D. $ 24.47\,J{{K}^{-1}}\,mo{{l}^{-1}} $

Answer 1

Answer: (C)

Entropy change of fusion
$ \Delta {{S}_{f}}=\frac{\Delta {{H}_{f}}}{T} $
$ \Delta {{H}_{f}}=60\times {{10}^{3}}\,J\,mo{{l}^{-1}},T=273\,K $
$ \Delta {{S}_{f}}=\frac{60\times {{10}^{3}}}{273} $
$ =21.97\,J\,K{{\,}^{-1}}\,mo{{l}^{-1}} $