Question

A given compound in liquid state has a vapor pressure of $190torr$ at ${27}^{\circ }C$ and $1atm$ at ${127}^{\circ }C$.
Calculate $\Delta H_{vap}$ for the compound
Given $\ln \left(2\right)=0.693$

• A. $13.83kJmo{l}^{-1}$
• B. $5.24kJmo{l}^{-1}$
• C. $6.66kJmo{l}^{-1}$
• D. $29.11kJmo{l}^{-1}$

Answer 1

The correct option is A $13.83kJmo{l}^{-1}$

Effect of temperature on the vapour pressure of a liquid is given by Clausius – Clapeyron equation.

$\ln \left(\frac{P_2}{P_1}\right)=\frac{\Delta H}{R}(\frac{1}{T_1}-\frac{1}{T_2})$

where,

$\Delta H\text{is the enthalpy of vapourisation}{P}_1\text{is the vapour pressure of the liquid at temperature}{T}_1{P}_2\text{is the vapour pressure of the liquid at temperature}{T}_2$
Here,
$P_1=190torr,T_1=300K{P}_2=1atm=760torr,T_2=400K$

Putting values in the equation:

$\ln \left(\frac{760}{190}\right)=\frac{\Delta H_{vap}}{8.314Jmo{l}^{-1}{K}^{-1}}(\frac{1}{300}-\frac{1}{400})$

$\ln \left(4\right)\times 8.314\times 1200=\Delta H_{vap}$

$\Delta H_{vap}=13830.7Jmo{l}^{-1}\Delta H_{vap}=13.8307kJmo{l}^{-1}$