The compounds A and B are mixed in equimolar proportion to form the products, A+B leftharpoons C+D At equilibrium, one-third of A and B are consumed. The equilibrium constant for the reaction is

Question

The compounds A and B are mixed in equimolar proportion to form the products, A+B leftharpoons C+D At equilibrium, one-third of A and B are consumed. The equilibrium constant for the reaction is (A) 0.5 (B) 4.0 (C) 2.5 (D) 0.25

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<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/chemistry/eecf8c7e5a0c3ece5c26377c09a07345-.png" /> K=([C][D]/[A][B])=((1/3) × (1/3)/(2)3 × (2)3=0.25

Q. The compounds $A$ and $B$ are mixed in equimolar proportion to form the products, $A + B ⇌ C + D$
At equilibrium, one-third of $A$ and $B$ are consumed. The equilibrium constant for the reaction is

0.5

4.0

2.5

0.25

$K = \frac{[ C ] [ D ]}{[ A ] [ B ]} = \frac{\frac{1}{3} \times \frac{1}{3}}{\frac{2}{3} \times \frac{2}{3}} = 0.25$

Q. The compounds A and B are mixed in equimolar proportion to form the products, A+B⇌C+D At equilibrium, one-third of A and B are consumed. The equilibrium constant for the reaction is

0.5

4.0

2.5

0.25

K=[A][B][C][D]​=32​×32​31​×31​​=0.25

Q. The compounds $A$ and $B$ are mixed in equimolar proportion to form the products, $A + B ⇌ C + D$
At equilibrium, one-third of $A$ and $B$ are consumed. The equilibrium constant for the reaction is

$K = \frac{[ C ] [ D ]}{[ A ] [ B ]} = \frac{\frac{1}{3} \times \frac{1}{3}}{\frac{2}{3} \times \frac{2}{3}} = 0.25$