For a reversible reaction `A underset(K_(2))overset(K_(1))(hArr) `B rate constant `K_(1)` (forward) = `10^(15)e^(-(200)/(T))` and `K_(2)` (backward) = `10^(12)e^(-(200)/(T))`. What is the value of `(-Delta G^(@))/(2.303 RT)` ?

Answer

To solve the problem, we need to determine the value of \(-\Delta G^\circ / (2.303 RT)\) for the given reversible reaction \(A \rightleftharpoons B\) with the provided rate constants for the forward and backward reactions.

### Step-by-Step Solution:

1. **Identify the Rate Constants**:
- The forward rate constant \(K_1\) is given by:
\[ K_1 = 10^{15} e^{-\frac{200}{T}} \]
- The backward rate constant \(K_2\) is given by:
\[ K_2 = 10^{12} e^{-\frac{200}{T}} \]

2. **Determine the Equilibrium Constant \(K_c\)**:
- The equilibrium constant \(K_c\) for the reaction can be expressed as the ratio of the forward rate constant to the backward rate constant:
\[ K_c = \frac{K_1}{K_2} \]

3. **Substitute the Values of \(K_1\) and \(K_2\)**:
- Substitute the expressions for \(K_1\) and \(K_2\) into the equation for \(K_c\):
\[ K_c = \frac{10^{15} e^{-\frac{200}{T}}}{10^{12} e^{-\frac{200}{T}}} \]

4. **Simplify the Expression**:
- The exponential terms \(e^{-\frac{200}{T}}\) cancel out:
\[ K_c = \frac{10^{15}}{10^{12}} = 10^{15 - 12} = 10^{3} \]

5. **Relate \(-\Delta G^\circ\) to \(K_c\)**:
- From thermodynamics, we know that:
\[ -\Delta G^\circ = 2.303 RT \log K_c \]
- Thus, we can express \(-\Delta G^\circ / (2.303 RT)\) as:
\[ -\frac{\Delta G^\circ}{2.303 RT} = \log K_c \]

6. **Calculate \(\log K_c\)**:
- Since \(K_c = 10^{3}\), we can find:
\[ \log K_c = \log(10^{3}) = 3 \]

7. **Final Result**:
- Therefore, the value of \(-\Delta G^\circ / (2.303 RT)\) is:
\[ -\frac{\Delta G^\circ}{2.303 RT} = 3 \]

### Conclusion:
The final answer is:
\[ -\frac{\Delta G^\circ}{2.303 RT} = 3 \]