Given that the equilibrium constant k for the vaporization of a liquid is 3.2 × 10^{-3} at 0 °C and 8.1 × 10^{-3} at 50 °C, what is the enthalpy of vaporization (ΔH_{vap}) in kJ/mol? (Assume R = 8.314 J/mol·K)

• A. 19 kJ/mol
• B. 8.2 kJ/mol
• C. 12 kJ/mol
• D. 25 kJ/mol

Answer: (C)

Verified step by step guidance

1. Identify the relationship between the equilibrium constant and temperature using the Van't Hoff equation: \(\ln\left(\frac{K_2}{K_1}\right) = -\frac{\Delta H_{vap}}{R} \left(\frac{1}{T_2} - \frac{1}{T_1}\right)\), where \(K_1\) and \(K_2\) are the equilibrium constants at temperatures \(T_1\) and \(T_2\) respectively, \(R\) is the gas constant, and \(\Delta H_{vap}\) is the enthalpy of vaporization.
2. Convert the given temperatures from Celsius to Kelvin by adding 273.15: \(T_1 = 0 + 273.15 = 273.15\,K\) and \(T_2 = 50 + 273.15 = 323.15\,K\).
3. Substitute the known values into the Van't Hoff equation: \(K_1 = 3.2 \times 10^{-3}\), \(K_2 = 8.1 \times 10^{-3}\), \(R = 8.314\,J/(mol\cdot K)\), \(T_1 = 273.15\,K\), and \(T_2 = 323.15\,K\).
4. Calculate the natural logarithm of the ratio of equilibrium constants: \(\ln\left(\frac{K_2}{K_1}\right)\), and the difference in the reciprocals of the temperatures: \(\left(\frac{1}{T_2} - \frac{1}{T_1}\right)\).
5. Rearrange the Van't Hoff equation to solve for \(\Delta H_{vap}\): \(\Delta H_{vap} = -R \times \frac{\ln\left(\frac{K_2}{K_1}\right)}{\left(\frac{1}{T_2} - \frac{1}{T_1}\right)}\). Calculate this value and convert the units from joules per mole to kilojoules per mole by dividing by 1000.