Using the Clausius-Clapeyron equation, what is the enthalpy of vaporization (ΔHvap) for the reaction A(l) ⇌ B(g) if the equilibrium constant K is 3.2 × 10^(-3) at 0 °C and 8.1 × 10^(-3) at 50 °C?

• A. ΔHvap = 31.2 kJ/mol
• B. ΔHvap = 24.5 kJ/mol
• C. ΔHvap = 18.7 kJ/mol
• D. ΔHvap = 42.0 kJ/mol

Answer: (A)

Verified step by step guidance

1. Understand the Clausius-Clapeyron equation, which is used to relate the change in vapor pressure with temperature to the enthalpy of vaporization. The equation is: $d^{\ln }\left(P\right)/d^T=-\frac{\mathrm{\Delta H}}{}$_vapRT2, where P is the pressure, T is the temperature, R is the gas constant, and ΔHvap is the enthalpy of vaporization.
2. Convert the equilibrium constants to pressures using the relation between K and P for the reaction A(l) ⇌ B(g). Since K is proportional to the pressure of B(g), you can use the given values of K as proxies for pressure at the respective temperatures.
3. Use the natural logarithm of the equilibrium constants at the two temperatures to find the change in ln(P) with respect to temperature. Calculate $ln(\frac{K}{}$₂K₁), where K1 and K2 are the equilibrium constants at 0 °C and 50 °C respectively.
4. Determine the change in temperature in Kelvin. Convert the temperatures from Celsius to Kelvin by adding 273.15 to each temperature. Calculate the difference: $T$₂-T₁.
5. Rearrange the Clausius-Clapeyron equation to solve for ΔHvap: $\mathrm{\Delta H}$_vap=-R×ln(K₂K₁)T₂-T₁. Use the gas constant R = 8.314 J/(mol·K) to calculate ΔHvap.