Question

The reaction $A(g) + B(g) \rightleftharpoons 2C(g)$ occurs by mixing $3$ moles of A and $1$ mole in the one-litre container. If $\alpha $ of B is $\dfrac{1}{3}$ , then ${K_C}$ for this reaction is
(A) $0.12$
(B) $0.25$
(C) $0.50$
(D) $0.75$

Answer 1

Hint: The equilibrium constant is calculated by the concentration of products divided by the concentration of reactants at equilibrium. The initial number of moles is given and the concentration at equilibrium needs to be found out. Then the equilibrium constant will be the ratio of the concentrations at equilibrium.

Complete step by step answer:
We know that the equilibrium constant ${K_C}$ is the ratio of the concentration of products and reactants at equilibrium.
The reaction is given as follows
$A(g) + B(g) \rightleftharpoons 2C(g)$

Conc	A	B	C
at t=0	3	1	0
at equilibrium	$3 - \alpha $	$1 - \alpha $	$2\alpha $
	$3 - \dfrac{1}{3}$	$1 - \dfrac{1}{3}$	$\dfrac{2}{3}$

We knew that concentration is the number of moles per unit volume
$ \Rightarrow c = \dfrac{n}{V}$ as the volume is given is one litre so the concentration is numerically equal to the number of moles.
Now we have been given
$\alpha = \dfrac{1}{3}$
${K_C} = \dfrac{{{{[C]}^2}}}{{[A][B]}}$
Here in the above expression of ${K_C}$ the stoichiometric coefficient is raised to the power of concentration.
$ \Rightarrow {K_C} = \dfrac{{{{(\dfrac{2}{3})}^{^2}}}}{{(\dfrac{2}{3})(\dfrac{8}{3})}}$
The concentration here used is at the equilibrium at
$\alpha = \dfrac{1}{3}$
$ \Rightarrow {K_C} = \dfrac{1}{4}$
$ \Rightarrow {K_C} = 0.25$
Therefore the value of the equilibrium constant ${K_C}$ is $0.25$
Hence the correct answer is option B.

Note:
The concentration here is numerically equal to the number of moles as the volume here mentioned is one litre otherwise it would be the number of moles divided by the volume of the container. Here $\alpha $ is the degree of dissociation. And here one mole of B reacts with three moles of A to form two moles of C then similarly $\alpha $ mol of A will react and $\alpha $ mole of B will react to form the product.