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 find the equilibrium constant for the reaction \( C \rightleftharpoons A + \frac{1}{2}B \) in terms of the equilibrium constant \( K \) for the reaction \( 2A + B \rightleftharpoons 2C \), where \( K = x \), we can follow these steps:

### Step 1: Write the given reaction and its equilibrium expression
The given reaction is:
\[ 2A + B \rightleftharpoons 2C \]

The equilibrium constant \( K \) for this reaction is given by:
\[ K = \frac{[C]^2}{[A]^2[B]} \]
Given that \( K = x \), we can write:
\[ x = \frac{[C]^2}{[A]^2[B]} \]

### Step 2: Write the reverse reaction and its equilibrium expression
Now, we need to consider the reverse reaction:
\[ C \rightleftharpoons A + \frac{1}{2}B \]

The equilibrium constant \( K' \) for this reaction can be expressed as:
\[ K' = \frac{[A][B]^{1/2}}{[C]} \]

### Step 3: Relate the two equilibrium constants
To relate \( K' \) to \( K \), we can use the fact that reversing a reaction inverts the equilibrium constant. Additionally, when we change the coefficients of a reaction, the equilibrium constant is raised to the power of the coefficient.

For the reaction \( 2A + B \rightleftharpoons 2C \), when we reverse it, we get:
\[ C \rightleftharpoons A + \frac{1}{2}B \]
This is equivalent to taking the square root of the original equilibrium constant because the coefficients are halved. Therefore:
\[ K' = \frac{1}{\sqrt{x}} \]

### Step 4: Final expression for the equilibrium constant
Thus, the equilibrium constant for the reaction \( C \rightleftharpoons A + \frac{1}{2}B \) is:
\[ K' = \frac{1}{\sqrt{x}} \]

### Conclusion
The equilibrium constant for the reaction \( C \rightleftharpoons A + \frac{1}{2}B \) in terms of \( x \) is:
\[ \boxed{\frac{1}{\sqrt{x}}} \]