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Question: If the equilibrium constant for the reaction. \(0.05\) \(1 \times 10^{- 14}\) is 81, What is the v...

If the equilibrium constant for the reaction.

0.050.05 1×10−141 \times 10^{- 14} is 81,

What is the value of equilibrium constant for the reaction.

XY 1×10−21 \times 10^{- 2}

A

81

B

9

C

6561

D

1×10−71 \times 10^{- 7}

Answer

9

Explanation

Solution

: 2XY X2+Y2;Kc=81X_{2} + Y_{2};K_{c} = 81

XY 12X2+12Y2;Kc′=?\frac{1}{2}X_{2} + \frac{1}{2}Y_{2};K_{c}^{'} = ?

Kc′=Kc=81=9K_{c}^{'} = \sqrt{K_{c}} = \sqrt{81} = 9