Question

The vapor pressure of the liquid NH3 is measured at different temperatures. The following vapor pressure data are obtained.

Answer 1

Answer:

$\Delta \:H_{vap}=23.3054\ kJ/mol$

Explanation:

The expression for Clausius-Clapeyron Equation is shown below as:

$\ln P = \dfrac{-\Delta{H_{vap}}}{RT} + c$

Where,  

P is the vapor pressure

ΔHvap  is the Enthalpy of Vaporization

R is the gas constant (8.314×10⁻³ kJ /mol K)

c is the constant.

For two situations and phases, the equation becomes:

$\ln \left( \dfrac{P_1}{P_2} \right) = \dfrac{\Delta H_{vap}}{R} \left( \dfrac{1}{T_2}- \dfrac{1}{T_1} \right)$

Given:

$P_1$ = 234.2 mmHg

$P_2$ = 522.6 mmHg

$T_1$ = 217.9 K

$T_2$ = 232.4 K

So,  

$\ln \:\left(\:\frac{234.2\ mmHg}{522.6\ mmHg}\right)\:=\:\frac{\Delta \:H_{vap}}{8.314\times 10^{-3}\ kJ /mol K}\:\left(\:\frac{1}{232.4\ K}-\:\frac{1}{217.9\ K}\:\right)$

$\Delta \:H_{vap}=\frac{\left\{\ln \left(\:\frac{234.2}{522.6}\right)\:\times \:8.314\times 10^{-3}\right\}}{\left(\:\frac{1}{232.4}-\:\frac{1}{217.9}\:\right)}\ kJ/mol$

$\Delta \:H_{vap}=-\frac{10^{-3}\times \:8.314\ln \left(\frac{234.2}{522.6}\right)}{\frac{14.5}{50639.96}}\ kJ/mol$

$\Delta \:H_{vap}=-\frac{421020.62744\ln \left(\frac{234.2}{522.6}\right)}{14500}\ kJ/mol$

$\Delta \:H_{vap}=23.3054\ kJ/mol$