Question

In an experiment the equilibrium constant for the reaction $A+B\rightleftharpoons C+D$ is $K$ when the initially concentration of $A$ and $B$ each is $0.1mol{L}^{-1}$. Under the similar conditions in an another experiment if the initial concentration of $A$ and $B$ are take $2$ and $3mol{L}^{-1}$ respectively then the value of equilibrium constant will be:

• A. $\frac{K}{2}$
• B. $\frac{K}{736}$
• C. $K^2$
• D. $\frac{1}{K}$

Answer 1

The correct option is B $\frac{K}{736}$

$A+B\to C+D$

Initial 0.1 0.1 0 0

equilibrium 0.1-x 0.1-x x x

$K=\frac{x^2}{{(0.1-x)}^2}$

For second reaction

$A+B\to C+D$

Initial 2 3 0 0

equilibrium 2-x 3-x x x

$K_2=\frac{x^2}{(2-x)(3-x)}$

lets lake x=0.01

$K=\frac{1}{81}$

$K_2=\frac{1}{59501}$

$\frac{K}{K_2}=\frac{1}{81}\times \frac{59501}{1}=736$

so $K_2=\frac{K}{736}$