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Enthalpy

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Question

Calculate the enthalpy of vaporisation per mole for ethanol. Given, $Δ S = 109.8 J K^{- 1} {m o l}^{- 1}$ and boiling point of ethanol is ${78.5}^{o} C$ .

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Solution

We know that
$Δ S_{v a p o u r} = \frac{Δ H_{v a p o u r}}{T_{b p}}$
Given, $Δ S_{v a p o u r} = 109.8 J K^{- 1} {m o l}^{- 1}$
$T_{b p} = 78.5 + 273 = 351.5 K$
Substituting these values in above equation, we get
$109.8 = \frac{Δ H_{v a p o u r}}{351.5}$
$Δ H_{v a p o u r} = 38594 J {m o l}^{- 1} = 38.594 k J {m o l}^{- 1}$

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