The standard Gibbs free energy change () for a reversible chemical synthesis at exactly is measured — Thermodynamics and Thermochemistry Chemistry Question
Question
The standard Gibbs free energy change ($\Delta G^\circ$) for a reversible chemical synthesis at exactly $300 \text{ K}$ is measured as $-11488.28 \text{ J mol}^{-1}$. Calculate the thermodynamic equilibrium constant ($K_{eq}$) for this synthesis reaction. (Use the universal gas constant $R = 8.314 \text{ J K}^{-1} \text{mol}^{-1}$ and assume $\ln 10 = 2.303$).
💡 Solution & Explanation
The equilibrium constant is fundamentally related to the standard free energy change via the isotherm equation: $\Delta G^\circ = -RT \ln K_{eq}$. Using base-10 logarithms, the equation transforms to $\Delta G^\circ = -2.303 RT \log K_{eq}$. Substituting the provided parameters into the formula: $-11488.28 \text{ J mol}^{-1} = -2.303 \times 8.314 \text{ J K}^{-1}\text{mol}^{-1} \times 300 \text{ K} \times \log K_{eq}$. The product $2.303 \times 8.314 \times 300 = 5744.14$. So, $-11488.28 = -5744.14 \log K_{eq}$. Dividing both sides gives $\log K_{eq} = 2$. Therefore, $K_{eq} = 10^2 = 100$.