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. 81
B. 9
C. 6561
D. 40.5

Explanation

To solve the problem, we need to determine 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. **Identify the given reaction and its equilibrium constant:**
The first reaction is:
\[ 2XY \rightleftharpoons X_2 + Y_2 \]
with an equilibrium constant \( K_c = 81 \).

2. **Write the new reaction:**
The second reaction we need to find the equilibrium constant for is:
\[ XY \rightleftharpoons \frac{1}{2} X_2 + \frac{1}{2} Y_2 \]

3. **Relate the new reaction to the original reaction:**
The new reaction is derived from the original reaction by dividing the entire equation by 2.

4. **Apply the equilibrium constant relationship:**
When a balanced equation is divided by a factor, the equilibrium constant for the new reaction is related to the original equilibrium constant raised to the power of that factor. In this case, since we divided the reaction by 2, we take the square root of the original equilibrium constant:
\[ K_c' = (K_c)^{1/2} \]

5. **Calculate the new equilibrium constant:**
Substitute the value of \( K_c \):
\[ K_c' = (81)^{1/2} = 9 \]

6. **Conclusion:**
The equilibrium constant for the reaction \( XY \rightleftharpoons \frac{1}{2} X_2 + \frac{1}{2} Y_2 \) is \( 9 \).

### Final Answer:
The value of the equilibrium constant for the reaction is \( 9 \).

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