For reaction `2A+B hArr 2C, K=x`
Equilibrium constant for `C hArr A+1//2 B` will be

Options

A. x
B. `x/2`
C. `1/sqrtx`
D. `sqrtx`

Answer

To solve the problem, we need to find the equilibrium constant for the reaction:

\[ C \rightleftharpoons A + \frac{1}{2} B \]

given that the equilibrium constant for the reaction:

\[ 2A + B \rightleftharpoons 2C \]

is \( K = x \).

### Step-by-Step Solution:

1. **Write the equilibrium expression for the first reaction:**
The equilibrium constant \( K \) for the reaction \( 2A + B \rightleftharpoons 2C \) is given by:
\[ K = \frac{[C]^2}{[A]^2[B]} \]
Since we know \( K = x \), we can write:
\[ x = \frac{[C]^2}{[A]^2[B]} \]

2. **Write the equilibrium expression for the second reaction:**
For the reaction \( C \rightleftharpoons A + \frac{1}{2} B \), the equilibrium constant \( K' \) is given by:
\[ K' = \frac{[A][B]^{1/2}}{[C]} \]

3. **Relate the two reactions:**
The second reaction can be derived from the first reaction by reversing it and adjusting the stoichiometry. When we reverse the first reaction, we get:
\[ 2C \rightleftharpoons 2A + B \]
The equilibrium constant for this reversed reaction is:
\[ K_{reverse} = \frac{1}{K} = \frac{1}{x} \]

4. **Adjust for stoichiometry:**
The reaction \( 2C \rightleftharpoons 2A + B \) can be divided by 2 to match the stoichiometry of the second reaction:
\[ C \rightleftharpoons A + \frac{1}{2} B \]
When we divide the reaction by 2, the equilibrium constant for the new reaction is:
\[ K' = \sqrt{K_{reverse}} = \sqrt{\frac{1}{x}} = \frac{1}{\sqrt{x}} \]

5. **Final Result:**
Thus, the equilibrium constant \( K' \) for the reaction \( C \rightleftharpoons A + \frac{1}{2} B \) is:
\[ K' = \frac{1}{\sqrt{x}} \]

### Conclusion:
The equilibrium constant for the reaction \( C \rightleftharpoons A + \frac{1}{2} B \) is \( \frac{1}{\sqrt{x}} \).