Question

The vapor pressure of the liquid SO2 is measure at different temperatures. the following vapor pressure data are obtained.Temp (K) Pressure mmHg241.3 271.2259.3 641.8Calculate the enthalpy of vaporization(delta H vap) in KJ/mol for this liquid.

Answer 1

24.895 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 J /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` = 271.2 mmHg

`P_2` = 641.8 mmHg

`T_1` = 241.3 K

`T_2` = 259.3 K

So,

`\ln \:\left(\:(271.2)/(641.8)\right)\:=\:(\Delta \:H_(vap))/(8.314)\:\left(\:(1)/(259.3)-\:(1)/(241.3)\:\right)`

`\Delta \:H_(vap)=\ln \left((271.2)/(641.8)\right)(8.314)/(\left((1)/(259.3)-\:(1)/(241.3)\right))\ J/mol`

`\Delta \:H_(vap)=\left(-(520199.41426)/(18)\right)\left(\ln \left(271.2\right)-\ln \left(641.8\right)\right)\ J/mol`

ΔHvap = 24895.015 J/mol = 24.895 kJ/mol ( 1 J = 0.001 kJ )