Given vapour pressure of Benzaldehyde is 400 torr at 154°C and its normal BP is 179° C, what is its molar enthalpy of vaporisation, ${ΔH}_{vap}$ ?

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56.4 KJ mol^–1

10.3 KJ mol^–1

67.9 KJ mol^–1

41.2 KJ mol^–1

answer is C.

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Detailed Solution

$\begin{array}{l} \log (\frac{P_{2}}{P_{1}}) = \frac{ΔH}{2 .303 R} [\frac{1}{T_{1}} - \frac{1}{T_{2}}] \\ \Rightarrow \log (\frac{700}{400}) = \frac{ΔH}{2 .303 R} [\frac{1}{427} - \frac{1}{452}] \\ \Rightarrow ΔH = 41 .2 KJ / mole \end{array}$

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