Question

The equilibrium constant $K_{C}$ of the reaction, $2 A \rightleftharpoons B+C$ is $0.5$ at $25^{\circ} C$ and $1 \,atm$. The reaction will proceed in the backward direction, when concentrations $[A],[B]$ and $[C]$ are, respectively

• A. $10^ {-3},10^{- 2}$ and $10^{-2}$ M
• B. $10^{-1} , 10^{-2}$ and $10^{-2}$ M
• C. $10^{-2} , 10^{-2}$ and $10^{-3}$ M
• D. $10^{-2} , 10^{-3}$and $10^{-3}$ M

Answer 1

Answer: (A)

(a) For the reaction,
$2 A \rightleftharpoons B+C $
$Q_{C}=\frac{[B][C]}{[A]^{2}}$
(Given, $ K_{c}=0.5$)
If $Q_{C}>K_{C}$ then the reaction will proceed in backward direction.
The value of $Q_{C}$ in the given options can be calculated as follows:
(i) $Q_{C}=\frac{\left[10^{-2}\right]\left[10^{-2}\right]}{\left[10^{-3}\right]^{2}}=\frac{10^{-4}}{10^{-6}}=10^{2}$
As $Q_{C}>K_{C}$ the reaction will proceed in backward direction.
(ii) $Q_{C}=\frac{\left[10^{-2}\right]\left[10^{-2}\right]}{\left[10^{-1}\right]^{2}}=\frac{10^{-4}}{10^{-2}}=10^{-2}$
As $Q_{C} < K_{C}$, the reaction will proceed in forward direction.
(iii) $Q_{C}=\frac{\left[10^{-2}\right]\left[10^{-3}\right]}{\left[10^{-2}\right]^{2}}=\frac{10^{-5}}{10^{-4}}=10^{-1}$
Here, $Q_{C} < K_{C}$, the reaction will proceed in forward direction.
(iv) $Q_{C}=\frac{\left[10^{-3}\right]\left[10^{-3}\right]}{\left[10^{-2}\right]^{2}}=\frac{10^{-6}}{10^{-4}}=10^{-2}$
The reaction will proceed in forward direction as $Q_{C} < K_{C}$