Consider the following reversible reaction :
`A(g)+B(g) Leftrightarrow AB(g)`.
The activation energy of the backward reaction exceeds that of the forward reaction by 2RT `("in J mol"^(-1))`. If the pre-exponential factor of the forward reaction is four times that of the reverse reaction, the absolute value of `triangleG^@ ("in J mol"^(-1))` for the reaction at 300K is...... `("Given ln 2=0.7, RT=2500 J mol"^(-1)" at 300K and G is the Gibbs energy")`

Explanation

To solve the problem step by step, we will analyze the given reaction and the relationships between the activation energies, pre-exponential factors, and Gibbs free energy.

### Step 1: Understand the Reaction and Given Data
We have the reversible reaction:
\[ A(g) + B(g) \leftrightarrow AB(g) \]

Given:
- The activation energy of the backward reaction exceeds that of the forward reaction by \( 2RT \).
- The pre-exponential factor of the forward reaction is four times that of the reverse reaction.
- \( RT = 2500 \, \text{J mol}^{-1} \) at \( 300 \, \text{K} \).
- \( \ln 2 = 0.7 \).

### Step 2: Define Activation Energies
Let:
- \( E_{a,f} \) = Activation energy of the forward reaction
- \( E_{a,b} \) = Activation energy of the backward reaction

From the problem, we know:
\[ E_{a,b} = E_{a,f} + 2RT \]

### Step 3: Define Pre-exponential Factors
Let:
- \( A_f \) = Pre-exponential factor of the forward reaction
- \( A_b \) = Pre-exponential factor of the backward reaction

Given:
\[ A_f = 4A_b \]

### Step 4: Write the Rate Constants
Using the Arrhenius equation:
\[ k_f = A_f e^{-E_{a,f}/RT} \]
\[ k_b = A_b e^{-E_{a,b}/RT} \]

### Step 5: Find the Equilibrium Constant
The equilibrium constant \( K \) is given by:
\[ K = \frac{k_f}{k_b} \]

Substituting the expressions for \( k_f \) and \( k_b \):
\[ K = \frac{A_f e^{-E_{a,f}/RT}}{A_b e^{-E_{a,b}/RT}} \]

### Step 6: Substitute \( A_f \) and \( A_b \)
Using \( A_f = 4A_b \):
\[ K = \frac{4A_b e^{-E_{a,f}/RT}}{A_b e^{-E_{a,b}/RT}} \]
\[ K = 4 e^{-(E_{a,f} - E_{a,b})/RT} \]

Using \( E_{a,b} = E_{a,f} + 2RT \):
\[ K = 4 e^{-(E_{a,f} - (E_{a,f} + 2RT))/RT} \]
\[ K = 4 e^{2} \]

### Step 7: Calculate \( \Delta G^\circ \)
The relationship between Gibbs free energy and the equilibrium constant is:
\[ \Delta G^\circ = -RT \ln K \]

Substituting \( K = 4e^2 \):
\[ \Delta G^\circ = -RT \ln(4e^2) \]
\[ \Delta G^\circ = -RT (\ln 4 + 2) \]

### Step 8: Substitute Values
Using \( RT = 2500 \, \text{J mol}^{-1} \) and \( \ln 4 = 2 \ln 2 = 2 \times 0.7 = 1.4 \):
\[ \Delta G^\circ = -2500 \times (1.4 + 2) \]
\[ \Delta G^\circ = -2500 \times 3.4 \]
\[ \Delta G^\circ = -8500 \, \text{J mol}^{-1} \]

### Final Answer
The absolute value of \( \Delta G^\circ \) is:
\[ |\Delta G^\circ| = 8500 \, \text{J mol}^{-1} \]

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