Question

The normal boiling point of benzene is 80.1°C. What is its enthalpy of vaporization if the vapor pressure at 26.1°C is 100 torr?

Answer 1

The answer is, 33.0 kJ/mol

Answer 2

The heat of vaporization of benzene is required.

The heat of vaporization of benzene is 33009 J/kg.

$T_0$ = Normal boiling point = 80.1+273.15 K

$T_B$ = Boiling point at given pressure = 26.1+273.15 K

$R$ = Gas constant = 8.314 J/mol K

$P$ = Pressure at given $T_B$ = 100 torr

$\Delta H$ = Heat of vaporization

From the Clausius–Clapeyron equation

$\dfrac{1}{T_B}=\dfrac{1}{T_0}-\dfrac{R\ln(\dfrac{P}{P_0})}{\Delta H}\\\Rightarrow \Delta H=\dfrac{R\ln\dfrac{P}{P_0}}{\dfrac{1}{T_0}-\dfrac{1}{T_B}}\\\Rightarrow \Delta H=\dfrac{8.314\times \ln\left(\frac{100}{760}\right)}{\frac{1}{80.1+273.15}-\frac{1}{26.1+273.15}}\\\Rightarrow \Delta H=33008.99\ \text{J/kg}$

The heat of vaporization of benzene is 33009 J/kg.

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