Question

The equilibrium constant $K_C$ of the reaction, $2A\rightleftharpoons B+C$ is $0.5$ at $25^{\circ }C$ and $1atm$. 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,
$2A\rightleftharpoons B+C$
$Q_C=\frac{[B][C]}{[A{]}^2}$
(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 $Q_C<K_C$