A particular compound has an enthalpy of vaporization of 28400 J/mol.At 274 K it has a vapor pressure of 122 mmHg.What is its vapor pressure at 311 K? (R = 8.31 J/(K· mol))

Question

Quizplus · Accepted Answer

First, we need to calculate the heat required to vaporize one mole of the compound:

ΔH_vap = 28,400 J/mol

Next, we need to use the Clausius-Clapeyron equation to relate the vapor pressure of the compound at two different temperatures:

ln(P₂/P₁) = -ΔH_vap/R (1/T₂ - 1/T₁)

We can rearrange this equation to solve for P₂:

P₂ = P₁ * e^(-ΔH_vap/R * (1/T₂ - 1/T₁))

We plug in the given values:

P₁ = 122 mmHg
T₁ = 274 K
T₂ = 311 K
ΔH_vap = 28,400 J/mol
R = 8.31 J/(K·mol)

P₂ = 122 mmHg * e^(-28,400 J/mol / (8.31 J/(K·mol)) * (1/311 K - 1/274 K))

P₂ = 538 mmHg

Therefore, the vapor pressure at 311 K is 538 mmHg, which corresponds to choice D.

A Particular Compound Has an Enthalpy of Vaporization of 28400