In the reaction A+B C+D, initial concentration of A and B both are equal. At equilibrium, the concentration of D is twice as that of A. What will be the equilibrium constant of the reaction?

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4/9

9/4

1/9

answer is D.

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Detailed Solution

Let the initial concentration be ‘a’. A+B→C+aa0 At equilibrium a−x a−x x Given: x=2 (a−x)3x=2ax=23a⇒ a3 a3 2a3 2a3Keq=[C][D][A][B]=2a32a3a3a3=4

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