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Chemistry Question on Spontaneity

The change in entropy for the fusion of 11 mole of ice is [mp of ice = 273 K273\, K, molar enthalpy of fusion for ice =6.0 kJ mol−1= 6.0\,kJ\,mol^{-1} ]

A

11.73 JK−1 mol−111.73\, JK ^{-1} \,mol ^{-1}

B

18.84 JK−1 mol−118.84\, JK ^{-1} \,mol ^{-1}

C

21.97 JK−1 mol−121.97\, JK ^{-1} \,mol ^{-1}

D

24.47 JK−1 mol−124.47\, JK ^{-1} \,mol ^{-1}

Answer

21.97 JK−1 mol−121.97\, JK ^{-1} \,mol ^{-1}

Explanation

Solution

Entropy change of fusion ΔSf=ΔHfT\Delta S_{f}=\frac{\Delta H_{f}}{T} ΔHf=60×103J,mol−1,T=273 K\Delta H_{f}=60 \times 10^{3} J , mol ^{-1}, T=273\, K ΔSf=60×103273\Delta S_{f}=\frac{60 \times 10^{3}}{273} =21.97 JK−1mol−1=21.97\, J K ^{-1} mol ^{-1}