Step-by-Step Solution

1. Understand the relationship between activation energies and enthalpy change.

The enthalpy change of a reaction, $\Delta H$, is related to the activation energies of the forward ($E_{a,f}$) and backward ($E_{a,b}$) reactions by the following equation: $\Delta H = E_{a,f} - E_{a,b}$ We are given that the activation energy of the backward reaction exceeds that of the forward reaction by $2RT$. $E_{a,b} - E_{a,f} = 2RT$ Therefore, the enthalpy change is: $\Delta H = E_{a,f} - E_{a,b} = -(E_{a,b} - E_{a,f}) = -2RT$

2. Relate rate constants to the equilibrium constant.

The equilibrium constant ($K$) for a reversible reaction is the ratio of the forward rate constant ($k_f$) to the backward rate constant ($k_b$): $K = \frac{k_f}{k_b}$

3. Use the Arrhenius equation.

The Arrhenius equation relates the rate constant ($k$), the pre-exponential factor ($A$), and the activation energy ($E_a$) as follows: $k = A e^{-E_a/RT}$ Writing this for the forward and backward reactions: $k_f = A_f e^{-E_{a,f}/RT}$ $k_b = A_b e^{-E_{a,b}/RT}$ Substituting these into the expression for the equilibrium constant: $K = \frac{A_f e^{-E_{a,f}/RT}}{A_b e^{-E_{a,b}/RT}} = \left( \frac{A_f}{A_b} \right) e^{(-E_{a,f} + E_{a,b})/RT} = \left( \frac{A_f}{A_b} \right) e^{(E_{a,b} - E_{a,f})/RT}$

4. Substitute the given values into the expression for K.

We are given:

• The pre-exponential factor of the forward reaction is 4 times that of the reverse reaction: $A_f = 4A_b$, which means $\frac{A_f}{A_b} = 4$.
• The difference in activation energies: $E_{a,b} - E_{a,f} = 2RT$.

Substituting these values into the equation for $K$: $K = (4) e^{(2RT)/RT} = 4e^2$

5. Calculate the standard Gibbs free energy change ($\Delta G^\circ$).

The standard Gibbs free energy change is related to the equilibrium constant by the equation: $\Delta G^\circ = -RT \ln K$ Substituting the value of $K$ we found: $\Delta G^\circ = -RT \ln(4e^2)$ Using the properties of logarithms, $\ln(xy) = \ln x + \ln y$ and $\ln(x^n) = n\ln x$: $\Delta G^\circ = -RT (\ln 4 + \ln e^2)$ $\Delta G^\circ = -RT (\ln(2^2) + 2 \ln e)$ $\Delta G^\circ = -RT (2 \ln 2 + 2)$ $\Delta G^\circ = -2RT (\ln 2 + 1)$

6. Substitute the numerical values.

We are given:

• $T = 300K$
• $RT = 2500 \, J\,mo{l^{ - 1}}$
• $\ln(2) = 0.7$

Substitute these values into the equation for $\Delta G^\circ$: $\Delta G^\circ = -2(2500) (0.7 + 1)$ $\Delta G^\circ = -5000 (1.7)$ $\Delta G^\circ = -8500 \, J\,mo{l^{ - 1}}$

7. Find the absolute value of $\Delta G^\circ$.

The question asks for the absolute value of $\Delta G^\circ$: $|\Delta G^\circ| = |-8500| = 8500 \, J\,mo{l^{ - 1}}$

The final answer is 8500.