Question

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

Calculate the enthalpy of vaporization (?Hvap) in kJ/mol for this liquid.

P1 = 234.2 mmHg T1 = 217.9 K
P2 = 522.6 mmHg T2 = 232.4 K

Answer 1

`\Delta \:H_(vap)=23.3054\ kJ/mol`

Step-by-step explanation:

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

`\ln P = \frac{-\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( (P_1)/(P_2) \right) = (\Delta H_(vap))/(R) \left( (1)/(T_2)- (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(\:(234.2\ mmHg)/(522.6\ mmHg)\right)\:=\:(\Delta \:H_(vap))/(8.314* 10^(-3)\ kJ /mol K)\:\left(\:(1)/(232.4\ K)-\:(1)/(217.9\ K)\:\right)`

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

`\Delta \:H_(vap)=-(10^(-3)* \:8.314\ln \left((234.2)/(522.6)\right))/((14.5)/(50639.96))\ kJ/mol`

`\Delta \:H_(vap)=-(421020.62744\ln \left((234.2)/(522.6)\right))/(14500)\ kJ/mol`

`\Delta \:H_(vap)=23.3054\ kJ/mol`