Enthalpy change for the process,
$H_{2} O (s) ⇌ H_{2} O (l)$
is $6.01 k J m o l^{- 1} .$ The entropy change for 1 mole of ice at its melting point will be

12 J K^{- 1} m o l^{- 1}

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22 J K^{- 1} m o l^{- 1}

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100 J K^{- 1} m o l^{- 1}

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30 J K^{- 1} m o l^{- 1}

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Solution

The correct option is B $22 J K^{- 1} m o l^{- 1}$
The entropy change in a phase transformation (fusion or melting here) is given as:
$△ S_{f u s i o n} = \frac{△ H_{f u s i o n}}{T_{f u s i o n}}$
melting point of ice = $273 K$
so
$△ S_{f u s i o n} = \frac{6.01 \times 1000}{273}$

$= 22 J K^{- 1} m o l^{- 1}$

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