Question

Chloroform, CHCl3, has a vapor pressure of 197 mmHg at 23.0 °C, and 448 mmHg at 45.0 °C. Estimate its heat of vaporization and normal boiling point.

Answer 1

(1) we have to use the integral form of the Clausius-Clapeyron equation:
ln(p₁/p₂) = (ΔH_vap/R)∙{(1/T₂) - (1/T₁)}
p₁ and p₂ are vapor pressure at thermodynamic temperatures T₁ and T₂ respectively. R is the universal gas constant, ΔH_vap the heat of vaporization.

T1=23+273= 296k T2= 45+273=318K

Hence,
ΔH_vap = R ∙ ln(p₁/p₂) / {(1/T₂) - (1/T₁)}
= 8.3145 J∙K⁻¹∙mol⁻¹ ∙ ln(448mm/197mm) / { (1/296 K)-(1/318K) )}
= 29169 J∙mol⁻¹

(2)now we use use ∆Hvap and either condition to calculate the normal boiling point

using the first condition listed,

P2= 197 mmHg
T2=296K
Then
P1 = 760 mmHg (normal atmospheric pressure)
T1 = unknown (normal boiling point)

ln(p₁/p₂) = (ΔH_vap/R)∙{(1/T₂) - (1/T₁)}

ln(760mm/197mm)=(29169/8.314 ) {(1/296) - (1/T₁)}

so 3.84 x 10^-4= 1/296 -1/T1

1/T1= 2.993 x 10^-3

T1= 334 k

so T1= 334-273 = 61.1 ^oC

so boiling point of chloroform=61.1 ^oC