The equilibrium constant ( K ) of the reaction `A + B rarr C + D` at 297 K is 100. If the rate constant of the forward reaction is `4 xx 10^(5)` , the rate constant of the reverse reaction is `x xx 10^(3)`. Find the value of x.

Answer

To solve the problem, we need to find the value of \( x \) in the rate constant of the reverse reaction given the equilibrium constant \( K \) and the rate constant of the forward reaction.

### Step-by-Step Solution:

1. **Identify the Given Data:**
- The equilibrium constant \( K \) for the reaction \( A + B \rightleftharpoons C + D \) is given as \( 100 \).
- The rate constant for the forward reaction \( K_F \) is given as \( 4 \times 10^5 \).
- The rate constant for the reverse reaction \( K_B \) is expressed as \( x \times 10^3 \).

2. **Write the Relationship of Equilibrium Constant and Rate Constants:**
- At equilibrium, the relationship between the rate constants and the equilibrium constant is given by:
\[ K = \frac{K_F}{K_B} \]
- Rearranging this gives:
\[ K_B = \frac{K_F}{K} \]

3. **Substitute the Known Values:**
- Substitute \( K_F \) and \( K \) into the equation:
\[ K_B = \frac{4 \times 10^5}{100} \]

4. **Calculate \( K_B \):**
- Performing the division:
\[ K_B = \frac{4 \times 10^5}{10^2} = 4 \times 10^{5-2} = 4 \times 10^3 \]

5. **Relate \( K_B \) to \( x \):**
- Since \( K_B = x \times 10^3 \), we can equate:
\[ 4 \times 10^3 = x \times 10^3 \]

6. **Solve for \( x \):**
- Dividing both sides by \( 10^3 \):
\[ x = 4 \]

### Final Answer:
The value of \( x \) is \( 4 \).