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 −3.58×103K.What is the enthalpy of vaporization of the substance?29.8 kJ/mol2.32×1023kJ/mol0.431 kJ/mol294 kJ/mol

Answer 1

Answer:

ΔHvap = 431 J/mol

Explanation:

The relation between the vapor pressure and the absolute temperature is given by the Clausius-Clapeyron equation.

$lnp=\frac{-\Delta Hvap}{R} .\frac{1}{T} +C$

where,

p: vapor pressure

ΔHvap: enthalpy of vaporization

T: absolute temperature

C: constant

We can see that this corresponds to a linear equation with slope -ΔHvap/R and intercept C. If the slope is -3.58 × 10³ K,

-3.58 × 10³ K = -ΔHvap/R

3.58 × 10³ K = ΔHvap/(8.314 J/K.mol)

ΔHvap = 431 J/mol