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Entropy of Phase Change

The enthalpy ...

Question

The enthalpy of vaporization of water at $100^{o} C$ is $40.63 k J m o l^{- 1}$ . It's entropy change for the vaporization would be :

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

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

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108.9 k J K^{- 1} m o l^{- 1}

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4063 k J K^{- 1} m o l^{- 1}

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Solution

The correct option is B $108.9 J K^{- 1} m o l^{- 1}$
We know that entropy change in phase transformation is given as:
$△ S_{v a p} = \frac{△ H_{v a p}}{T}$
According to given data
$△ S_{v a p} = \frac{40.63 \times 10^{3} J m o l^{- 1}}{373 K}$
so,
$△ S_{v a p} = 108.9 J K^{- 1} m o l^{- 1}$

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