Question

The vapor pressure of a substance is measured over a range of temperatures. A plot of the natural log of the vapor pressure versus the inverse of the temperature (in Kelvin) produces a straight line with a slope of −3.55×103 K . Find the enthalpy of vaporization of the substance.

Answer 1

The Clausius-Clapeyron equation is: P_vap = A exp (- H_vap / R T) taking logarithm both sides, so the equation becomes : ln(P_vap) = -H_vap/RT + A

This is a linear equation of the form: y = mx + c . where, y = ln( P_vap ) , x = 1/T and slope,m =-H_vap/R . So,the logarithm of the vapor pressure varies linearly with the reciprocal of the temperature.

A plot of the logarithm of the vapor pressure versus the reciprocals of the temperature will be a straight line with a slope of - H_vap /R. The value for H_vap can be calculated from slope.

slope,m =-H_vap/R

so. H_vap = slope X (-R)

Given:  slope = −3.55×10³ K and R = 8.314 JK^-1mol^-1

H_vap = slope X (-R) = (−3.55×10³ K)( -8.314 JK^-1mol^-1) = 29.5147X 10³ Jmol^-1

H_vap = 29.5147 kJmol^-1   ( 1kJ = 10³J)

The enthalpy of vaporization of the substance = 29.5147 kJmol^-1