The standard free energies of formation () at for substance in its pure liquid and perfect gaseous s — Thermodynamics and Thermochemistry Chemistry Question
Question
The standard free energies of formation ($\Delta_f G^\circ$) at $500 \text{ K}$ for substance $S$ in its pure liquid and perfect gaseous states are precisely $+100.7 \text{ kcal mol}^{-1}$ and $+103.0 \text{ kcal mol}^{-1}$, respectively. Calculate the equilibrium vapour pressure of liquid $S$ at $500 \text{ K}$. (Assume perfect gas behavior and use $R = 2.0 \text{ cal K}^{-1} \text{mol}^{-1}$)
💡 Solution & Explanation
The equilibrium vaporization process is $S(l) \rightleftharpoons S(g)$. The standard Gibbs free energy change for this reaction is $\Delta_r G^\circ = \Delta_f G^\circ(g) - \Delta_f G^\circ(l) = 103.0 - 100.7 = +2.3 \text{ kcal mol}^{-1} = +2300 \text{ cal mol}^{-1}$. At phase equilibrium, $\Delta G^\circ = -RT \ln K_p$. For vaporization, $K_p = P_{vap}$. Thus, $2300 = -(2.0 \text{ cal K}^{-1} \text{mol}^{-1})(500 \text{ K}) \ln(P_{vap}) \Rightarrow 2300 = -1000 \ln(P_{vap}) \Rightarrow \ln(P_{vap}) = -2.3$. Converting from natural log to base 10 (since $\ln 10 \approx 2.303$), this gives $P_{vap} \approx 10^{-1} = 0.1 \text{ atm}$.