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 find the value of \( x \) in the given reaction \( A + B \rightleftharpoons C + D \), we can use the relationship between the equilibrium constant \( K \) and the rate constants of the forward and reverse reactions.

### Step-by-Step Solution:

1. **Write the expression for the equilibrium constant \( K \)**:
\[ K = \frac{k_f}{k_b} \]
where \( k_f \) is the rate constant for the forward reaction and \( k_b \) is the rate constant for the reverse reaction.

2. **Substitute the known values into the equation**:
We know that:
- \( K = 100 \)
- \( k_f = 4 \times 10^5 \)
- \( k_b = x \times 10^3 \)

Therefore, we can write:
\[ 100 = \frac{4 \times 10^5}{x \times 10^3} \]

3. **Rearrange the equation to solve for \( x \)**:
Multiply both sides by \( x \times 10^3 \):
\[ 100 \times (x \times 10^3) = 4 \times 10^5 \]
Simplifying gives:
\[ 100x \times 10^3 = 4 \times 10^5 \]

4. **Divide both sides by \( 100 \times 10^3 \)**:
\[ x = \frac{4 \times 10^5}{100 \times 10^3} \]

5. **Calculate the value of \( x \)**:
Simplifying the right-hand side:
\[ x = \frac{4 \times 10^5}{10^5} = \frac{4}{100} = 0.04 \]
Therefore, we can express \( x \) in terms of \( 10^3 \):
\[ x = 4 \]

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