If the equilibrium constant for the reaction,
`2XYhArrX_(2)+Y_(2)` is `81`,
what is the value of equilibrium constant for the reaction
`XYhArr(1)/(2)X_(2)+(1)/(2)Y_(2)`

Options

A. (a) 81
B. (b) 9
C. (c) 6561
D. (d) 40.5

Explanation

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

\[ XY \rightleftharpoons \frac{1}{2}X_2 + \frac{1}{2}Y_2 \]

given that the equilibrium constant for the reaction:

\[ 2XY \rightleftharpoons X_2 + Y_2 \]

is \( K_c = 81 \).

### Step-by-Step Solution:

1. **Write the expression for the equilibrium constant of the first reaction:**
The equilibrium constant \( K_c \) for the reaction \( 2XY \rightleftharpoons X_2 + Y_2 \) is given by:
\[ K_c = \frac{[X_2][Y_2]}{[XY]^2} \]
We know from the problem statement that \( K_c = 81 \).

2. **Write the expression for the equilibrium constant of the second reaction:**
The equilibrium constant \( K_c' \) for the reaction \( XY \rightleftharpoons \frac{1}{2}X_2 + \frac{1}{2}Y_2 \) can be expressed as:
\[ K_c' = \frac{[X_2]^{1/2}[Y_2]^{1/2}}{[XY]^1} \]

3. **Relate the two equilibrium constants:**
Notice that the second reaction is derived from the first reaction by halving all coefficients. When the coefficients in a balanced equation are multiplied by a factor, the equilibrium constant is raised to the power of that factor. Here, we are halving the coefficients (multiplying by \( \frac{1}{2} \)):
\[ K_c' = K_c^{1/2} \]

4. **Substitute the value of \( K_c \):**
Since \( K_c = 81 \), we can substitute this into the equation:
\[ K_c' = (81)^{1/2} \]

5. **Calculate \( K_c' \):**
Now we calculate the square root of 81:
\[ K_c' = 9 \]

### Final Answer:
The value of the equilibrium constant \( K_c' \) for the reaction \( XY \rightleftharpoons \frac{1}{2}X_2 + \frac{1}{2}Y_2 \) is \( 9 \).