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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.

1 Answer

4 votes

Answer:

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 )

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