The equilibrium constant for a reaction
`A+2B hArr 2C` is `40`. The equilibrium constant for reaction `C hArr B+1//2 A` is

Options

A. `1//40`
B. `(1//40)^(1//2)`
C. `(1//40)^(2)`
D. `40`

Explanation

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

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

given that the equilibrium constant for the reaction:

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

is \( K_1 = 40 \).

### Step-by-Step Solution:

1. **Write the equilibrium expression for the first reaction:**
The equilibrium constant \( K_1 \) for the reaction \( A + 2B \rightleftharpoons 2C \) is given by:
\[ K_1 = \frac{[C]^2}{[A][B]^2} \]
Given \( K_1 = 40 \), we have:
\[ \frac{[C]^2}{[A][B]^2} = 40 \]

2. **Write the equilibrium expression for the second reaction:**
For the reaction \( C \rightleftharpoons B + \frac{1}{2} A \), the equilibrium constant \( K_2 \) can be expressed as:
\[ K_2 = \frac{[B]^{1}[A]^{\frac{1}{2}}}{[C]^{1}} = \frac{[B] \sqrt{[A]}}{[C]} \]

3. **Relate \( K_2 \) to \( K_1 \):**
To find \( K_2 \), we can manipulate the first reaction. The second reaction is essentially the reverse of the first reaction, but with different stoichiometry.

To find \( K_2 \), we can use the relationship between the equilibrium constants when reactions are reversed or modified:
- Reversing a reaction inverts the equilibrium constant.
- Halving the coefficients of a balanced equation takes the square root of the equilibrium constant.

Since the first reaction is:
\[ A + 2B \rightleftharpoons 2C \]
and we want to find \( C \rightleftharpoons B + \frac{1}{2} A \), we first reverse the first reaction:
\[ 2C \rightleftharpoons A + 2B \]
The equilibrium constant for this reversed reaction is:
\[ K' = \frac{1}{K_1} = \frac{1}{40} \]

4. **Adjust for stoichiometry:**
Now, we need to adjust for the fact that the coefficients in the new reaction are halved:
\[ C \rightleftharpoons B + \frac{1}{2} A \]
Therefore, we take the square root of \( K' \):
\[ K_2 = \sqrt{K'} = \sqrt{\frac{1}{40}} = \frac{1}{\sqrt{40}} = \frac{1}{\sqrt{40}} = \frac{1}{\sqrt{40}} \approx 0.158 \]

### Final Answer:
Thus, the equilibrium constant for the reaction \( C \rightleftharpoons B + \frac{1}{2} A \) is:
\[ K_2 = \frac{1}{\sqrt{40}} \]