Question

The change in entropy for the fusion of $1$ mole of ice is [melting point of ice $= 273\, K$, molar enthalpy of fusion for ice $ =6.0\,kJ\,mol^{-1} $ ]:

• A. $ 11.73\,JK^{-1}\,mol^{-1} $
• B. $ 18.84\,JK^{-1}mol^{-1} $
• C. $ 21.97JK^{-1}mol^{-1} $
• D. $ 24.47\,JK^{-1}\,mol^{-1} $

Answer 1

Answer: (C)

Entropy change of fusion $\Delta S_{f}$
$\Delta S_{f}=\frac{\Delta H_{f}}{T} $
$\Delta H_{f}=6.0 \times 10^{3} J$,
$ T=273\, K$
$\Delta S_{f}=\frac{6000}{273}$
$=21.97 \,J / K \,mol$