Exercises

Exercise 1. What is the enthalpy of vaporization of a compound that has a vapor pressure of 154 torr at -20.0°C K and 0.350 torr at -82.2°C?

\(\displaystyle ΔH_{vap}=ln\biggl(\frac{154\;torr}{0.350\;torr}\biggr)\times\frac{8.314\frac{J}{mol⋅K}}{\biggl(\frac{1}{190.8\;K}-\frac{1}{253\;K}\biggr)}=\mathbf{39274\;J/mol = 39.3 kJ/mol}\)

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Exercise 2. A liquid has an enthalpy of vaporization of 45.6 kJ/mol. Its vapor pressure at 78.0 °C is 224 mmHg. What is the normal boiling point of the liquid?

At the normal boiling point, the pressure is 760 mmHg. We can solve the Clausius Clapeyron Equation for 1/T₂. P₂ = 760 mmHg, T₂ = ?, P₁ = 224 mmHg, and T₁ = 351 K. Convert R to kg, R = 8.314 x 10^-3 kg/(mol⋅K)

\(\displaystyle \frac{1}{T_2}=\frac{ln\biggl(\frac{P_1}{P_2}\biggr)}{\frac{ΔH_{vap}}{R}}+\frac{1}{T_1}=\frac{ln\biggl(\frac{224\;mmHg}{760\;mmHg}\biggr)}{\frac{45.6\;kJ/mol}{8.314\times 10^{-3}\frac{kJ}{mol⋅K}}}+\frac{1}{351\;K}=0.00263/K\)
 
Take the reciprocal. 1/T₂ = 0.00263/K, and 1/0.00263 = 380 K = 107°C

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Exercise 3. A liquid has a vapor pressure of 14.5 mmHg at 25.0°C and 150.2 mmHg at 150.0°C. What is the enthalpy of vaporization for the liquid?

Solve the Clausius-Clapeyron Equation for the enthalpy of vaporization, ΔH_vap.

\(\displaystyle ΔH_{vap}=ln\biggl(\frac{14.5\;mmHg}{150.2\;mmHg}\biggr)\times\frac{8.314\;J}{mol⋅K}{\biggl(\frac{1}{423\;K}-\frac{1}{298\;K}\biggr)}\)
 

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