If the equilibrium constant for the reaction $2 X Y ⇌ X_{2} + Y_{2} i s 81$ ,
what is the value of equilibrium constant for the reaction ?
$X Y ⇌ \frac{1}{2} X_{2} + \frac{1}{2} Y_{2}$ :

81

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9

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6561

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40.5

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Solution

The correct option is B $9$
$2 X Y ⇌ X_{2} + Y_{2}, K_{c} = 81$
$X Y ⇌ \frac{1}{2} X_{2} + \frac{1}{2} Y_{2}$
As the molecules in the first reaction are twice that of the second reaction
The equilibrium constant for the second equation can be calculated as
$K_{c}^{'} = {\sqrt{K}}_{c}$ $K_{c}^{'} = \sqrt{8} 1 = 9$

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