Ethanol boils at ${78.4}^{o} C$ and the enthalpy of vaporisation of ethanol is $42.4 k J m o l^{- 1}$ . What will be the entropy of vaporisation of ethanol:

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

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

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

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

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Solution

The correct option is C $120.66 J K^{- 1} m o l^{- 1}$
We are given that,
$△ H_{v a p} = 42.4 k J m o l^{- 1}, T = 273 + 78.4 = 351.4 K$

We know that,
$△ S_{v a p} = \frac{△ H_{v a p}}{T}$
putting the values,
$\Rightarrow \frac{42.4 \times 10^{3}}{351.4} = 120.66 J K^{- 1} m o l^{- 1}$

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