Calculation of Standard Gibbs Free Energy from Kinetic Parameters

To determine the absolute value of the standard Gibbs free energy change, $|\Delta G^\ominus|$, we use the relationship between the equilibrium constant ($K_{\text{eq}}$) and the rate constants of the forward ($k_f$) and backward ($k_b$) reactions.

For the reversible reaction: $\text{A(g)} + \text{B(g)} \rightleftharpoons \text{AB(g)}$

The rate constants are given by the Arrhenius equation: $k_f = A_f e^{-E_{a,f}/RT}$ $k_b = A_b e^{-E_{a,b}/RT}$

The equilibrium constant is the ratio of the forward and backward rate constants: $K_{\text{eq}} = \frac{k_f}{k_b} = \left(\frac{A_f}{A_b}\right) e^{-(E_{a,f} - E_{a,b})/RT} = \left(\frac{A_f}{A_b}\right) e^{(E_{a,b} - E_{a,f})/RT}$

From the problem statement:

• The pre-exponential factor of the forward reaction is $4$ times that of the reverse reaction: $\frac{A_f}{A_b} = 4$
• The activation energy of the backward reaction exceeds that of the forward reaction by $2RT$: $E_{a,b} - E_{a,f} = 2RT$

Substituting these values into the expression for $K_{\text{eq}}$: $K_{\text{eq}} = 4 \cdot e^{(2RT)/RT} = 4 e^2$

The standard Gibbs free energy change is related to the equilibrium constant by: $\Delta G^\ominus = -RT \ln K_{\text{eq}}$

Taking the absolute value: $|\Delta G^\ominus| = RT \ln(K_{\text{eq}}) = RT \ln(4 e^2)$

Using logarithmic properties: $\ln(4 e^2) = \ln(2^2) + \ln(e^2) = 2\ln(2) + 2$

Given that $\ln(2) = 0.7$ and $RT = 2500\text{ J mol}^{-1}$: $\ln(4 e^2) = 2(0.7) + 2 = 1.4 + 2 = 3.4$

Now, calculate $|\Delta G^\ominus|$: $|\Delta G^\ominus| = 2500 \times 3.4 = 8500\text{ J mol}^{-1}$

8500