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 ^{o}C

so boiling point of chloroform=61.1 ^{o}C

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°C.

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