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 -3.46 * 10 3 K. Find the enthalpy of vaporization of the substance.

Answer 1

Answer:

28.7664 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.

The graph of ln P and 1/T gives a slope of - ΔHvap/ R and intercept of c.

Given :

Slope = -3.46×10³ K

So,

- ΔHvap/ R = -3.46×10³ K

ΔHvap = 3.46×10³ K × 8.314×10⁻³ kJ /mol K = 28.7664 kJ /mol