The molar enthalpy of vaporization of benzene at is boiling point (353 K) is 7.4 kcal/mol. The molar — Thermodynamics and Thermochemistry Chemistry Question

💡 Solution & Explanation

$\Delta H = \Delta U + \Delta n_g RT \implies \Delta U = \Delta H - \Delta n_g RT$. For vaporization, $\Delta n_g = 1$. Using $R \approx 1.987 \times 10^{-3}\text{ kcal/(K mol)}$, $\Delta U = 7.4 - (1 \times 1.987 \times 10^{-3} \times 353) = 7.4 - 0.701 = 6.699\text{ kcal/mol}$, mapping closest to 6.694 kcal/mol (using exactly $R=8.314/4.184$).