The standard free energies of formation at 500 K for substance S in its pure liquid and gaseous stat — Thermodynamics and Thermochemistry Chemistry Question

💡 Solution & Explanation

The process is $S(l) \rightleftharpoons S(g)$. $\Delta_r G^\circ = 103.0 - 100.7 = +2.3 \text{ kcal/mol} = +2300 \text{ cal/mol}$. At phase equilibrium, $\Delta G^\circ = -2.303 RT \log P_{vap}$. $2300 = -2.303(2)(500) \log P_{vap} \Rightarrow 2300 = -2303 \log P_{vap} \Rightarrow \log P_{vap} \approx -1 \Rightarrow P_{vap} = 0.1$ atm.