For the standard gaseous reaction , the standard Gibbs free energy change at is exactly . Assuming t — Chemical Equilibrium Chemistry Question
Question
For the standard gaseous reaction $X(g) \rightleftharpoons Y(g) + Z(g)$, the standard Gibbs free energy change $\Delta G^\circ$ at $300\text{ K}$ is exactly $2494.2\text{ J mol}^{-1}$. Assuming the universal gas constant $R = 8.314\text{ J K}^{-1}\text{mol}^{-1}$, what is the exact integer value of $\ln\left(\frac{1}{K_{eq}}\right)$ for this system at $300\text{ K}$?
💡 Solution & Explanation
The standard thermodynamic relationship is $\Delta G^\circ = -RT \ln K_{eq}$. Using logarithmic properties, $-\ln K_{eq} = \ln(K_{eq}^{-1}) = \ln(1/K_{eq})$. Thus, $\Delta G^\circ = RT \ln(1/K_{eq})$. Substituting the given values: $2494.2 = (8.314 \times 300) \times \ln(1/K_{eq})$. Since $8.314 \times 300 = 2494.2$, the equation simplifies to $2494.2 = 2494.2 \times \ln(1/K_{eq})$, which gives $\ln(1/K_{eq}) = 1$.