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 $3M$.

$K_c$ is:

• A. $K_c=1/12$
• B. $K_c=1/10$
• C. $K_c=1/11$
• D. $K_c=1/13$

Answer 1

The correct option is A $K_c=1/12$

$P+2Q\rightleftharpoons R$

Let $2x$ be the concentration of R at first equilibrium.

$P+2Q\rightleftharpoons R$

Concentration at first equilibrium are: $P=3,Q=4,R=2x$

When volume is doubled concentration becomes half i.e $P=1.5,Q=2,R=x$ Concentration at second equilibrium is $P=2,Q=3,R=x-0.5$

The expression for the equilibrium constant is $K_c=\frac{\left[R\right]}{\left[P\right][Q{]}^2}$

For two equilibria, $\frac{2x}{3\times 4^2}=\frac{x-0.5}{2\times 3^2}$

Thus $x=2and{K}_c=\frac{1}{12}$