Question

The homogeneous reaction is carried out in a 2-litre container at a particular temperature by taking 1 mole each of A, B, C and D respectively. If ${k_c}$ for the reaction is $\dfrac{1}{4}$ then the equilibrium concentration of C is:
  \[A\left( g \right) + B\left( g \right) \rightleftharpoons C\left( g \right) + D\left( g \right)\]
A. $\dfrac{1}{3}M$
B. $\dfrac{2}{3}M$
C. $\dfrac{4}{3}M$
D. $\dfrac{1}{2}M$

Answer 1

Hint: The equilibrium constant ${k_c}$ is calculated by the ratio of products to reactants at equilibrium.

Complete step by step answer:
 \[A\left( g \right) + B\left( g \right) \rightleftharpoons C\left( g \right) + D\left( g \right)\]
 This reaction is homogeneous reaction
Equilibrium constant $ = {k_c} = \dfrac{{{{[C]}^C}{{[D]}^D}}}{{{{[A]}^A}{{[B]}^B}}}$
Where, [A] = concentration of reactant A
[B] = concentration of reactant B
[C] = concentration of product C
[D] = concentration of product D
If initial concentrations of [A] and [B] are 1 then initial concentration of [C] and [D] are 0. And if concentration of [C] and [D] are ‘x’, then concentration of [A] and [B] becomes \[\left( {1 - x} \right)\]
So, $[A] = \dfrac{{1 - x}}{2}$ which becomes (0.5 – 0.5x) , similarly of [B]
So, $[C] = \dfrac{{1 + x}}{2}$ which becomes (0.5 – 0.5x), similarly of [D]
 ${k_c} = \dfrac{{[0.5 + 0.5x][0.5 + 0.5x]}}{{[0.5 - 0.5x][0.5 - 0.5x]}}$
 $\dfrac{1}{4} = \dfrac{{[0.5 + 0.5x][0.5 + 0.5x]}}{{[0.5 - 0.5x][0.5 - 0.5x]}}$
 $x = 0.33 = \dfrac{1}{3}$
Now, substituting the value of x for [C]
 $[C] = \dfrac{{1 + x}}{2} = \dfrac{{1 + \dfrac{1}{3}}}{2} = \dfrac{2}{3}M$

So, the correct answer is option (B)

Note: When we calculate Equilibrium constant ${k_c}$ , then pure solids and liquids are not taken into account because their concentration does not change during the reaction. In a homogeneous reaction, substances are in the same phases.