The thermodynamic equilibrium constant () for a reversible chemical process is precisely 10 at 300 K — Thermodynamics and Thermochemistry Chemistry Question
Question
The thermodynamic equilibrium constant ($K_{eq}$) for a reversible chemical process is precisely 10 at 300 K. Calculate the standard Gibbs free energy change ($\Delta G^\circ$) for this reaction. ($R = 8.314 \text{ J K}^{-1}\text{mol}^{-1}, \ln 10 = 2.303$)
Answer: A
💡 Solution & Explanation
The relation connecting standard free energy change and the equilibrium constant is $\Delta G^\circ = -RT \ln K_{eq} = -2.303 RT \log K_{eq}$. Substituting values: $\Delta G^\circ = -2.303 \times 8.314 \times 300 \times \log(10) = -5744.14 \text{ J/mol}$.
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