The standard free energies of formation at 500 K for substance S in its pure liquid and gaseous stat — Thermodynamics and Thermochemistry Chemistry Question
Question
The standard free energies of formation at 500 K for substance S in its pure liquid and gaseous states are +100.7 kcal/mol and +103.0 kcal/mol respectively. Calculate the equilibrium vapour pressure of liquid S at 500 K. ($R = 2$ cal/K-mol, $\ln 10 = 2.303$)
Answer: A
💡 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.
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