For the standard reaction at , the standard Gibbs free energy change () is . At a given instant, the — Chemical Equilibrium Chemistry Question
Question
For the standard reaction $2A(g) \rightleftharpoons B(g) + C(g)$ at $300\text{ K}$, the standard Gibbs free energy change ($\Delta G^\circ$) is $+2494.2\text{ J mol}^{-1}$. At a given instant, the reaction mixture composition is $[A] = 0.5\text{ M}$, $[B] = 2.0\text{ M}$, and $[C] = 0.5\text{ M}$. In which direction will the reaction proceed? (Use $R = 8.314\text{ J K}^{-1}\text{mol}^{-1}$ and $e \approx 2.718$)
💡 Solution & Explanation
First, find $K_c$ using $\Delta G^\circ = -RT \ln K_c$. $2494.2 = -(8.314 \times 300) \ln K_c \Rightarrow 2494.2 = -2494.2 \ln K_c \Rightarrow \ln K_c = -1 \Rightarrow K_c = 1/e \approx 0.368$. Next, calculate the reaction quotient $Q_c = \frac{[B][C]}{[A]^2} = \frac{2.0 \times 0.5}{(0.5)^2} = \frac{1.0}{0.25} = 4.0$. Since $Q_c$ (4.0) is much greater than $K_c$ (0.368), the system shifts backward to reach equilibrium.