Calculate the enthalpy of vaporisation (in $k J / m o l$ ) for ethanol. Given, entropy change for the process is $109 J K^{- 1} m o l^{- 1}$ and boiling point of ethanol is $78.5^{o} C .$

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Solution

We know that,
$△ S_{v a p o u r i s a t i o n} = \frac{△ H_{v a p o u r i s a t i o n}}{T_{b}}$

Given: $△ S_{v a p o u r} = 109 J K^{- 1} m o l^{- 1}$
$T_{b} = 78.5 + 273 = 351.5 K$

Substituting these values in above equation, we get
$109 = \frac{△ H_{v a p}}{351.5}$
$△ H_{v a p} = 38313.5 J m o l^{- 1}$
$= 38.31 k J m o l^{- 1}$

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