Question

Consider the equilibrium : $P\left(g\right)+2Q\left(g\right)\rightleftharpoons R\left(g\right)$. When the reaction is carried out at a certain temperature, the equilibrium concentration of $P$ and $Q$ are $3M$ and $4M$ respectively. When the volume of the vessel is doubled and the equilibrium is allowed to be re-established, the concentration of $Q$ is found to be $3M$. Find :
(a) $K_c$,
(b) Concentration of $R$ at two equilibrium stages.

• A. $\left(a\right)\frac{1}{24}{litre}^2{mol}^{-2},\left(b\right)2M,0.5M$
• B. $\left(a\right)\frac{1}{12}{litre}^2{mol}^{-2},\left(b\right)4M,1.5M$
• C. $\left(a\right)\frac{1}{8}{litre}^2{mol}^{-2},\left(b\right)6M,2.5M$
• D. $\left(a\right)\frac{1}{6}{litre}^2{mol}^{-2},\left(b\right)8M,3.5M$

Answer 1

Answer: (B)

$\left(a\right)\frac{1}{12}{litre}^2{mol}^{-2},\left(b\right)4M,1.5M$

$P\left(g\right)+2Q\left(g\right)\rightleftharpoons R\left(g\right)$
Concentration at I equilibrium $34a$
$\therefore K_c=\frac{a}{3\times 4^2}$ ....(i)
One increasing the volume twice, i.e., the reaction will proceed in backward direction to attain new equilibrium.
$P\left(g\right)+2Q\left(g\right)\rightleftharpoons R\left(g\right)$
$\frac{3}{2}\frac{4}{2}\frac{a}{2}$

At II equilibrium $(\frac{3}{2}+x)(\frac{4}{2}+2x)(\frac{a}{2}-x)$
Given, $\frac{4}{2}+2x=3M$
$\therefore x=\frac{1}{2}$
$\therefore \left[P\right]=2M,\left[Q\right]=3M,R=\left(\frac{a-1}{2}\right)M$
$\therefore K_c=\frac{a-1}{2\times 2\times 3^2}$ ....(ii)
By equations (i) and (ii), $\frac{a}{48}=\frac{a-1}{36}$
or $a=4M$

$\therefore \left[R\right]$ at I equilibrium $=4M;$ at II equilibrium $=1.5M;$

Also, $K_c=\frac{1}{12}{litre}^2{mol}^{-2}$

Option B is the correct answer.