Question

The value of Ke for the reaction, $2A\rightleftharpoons B+C$ is $2\times {10}^{-3}$. At a given time, if the composition of reaction mixture is $\left[A\right]=\left[B\right]=\left[C\right]=3\times {10}^{-3}M$.

Which is true?

• A. The reaction will proceed in forward direction
• B. The reaction will proceed in backward direction
• C. The reaction will proceed in any direction
• D. None of the above

Answer 1

The correct option is B The reaction will proceed in backward direction

Reaction is $2A\rightleftharpoons B+C$

Reaction quotient will be $Q_c=\frac{\left[B\right]\left[C\right]}{{\left[A\right]}^2}$

Now $\left[A\right]=\left[B\right]=\left[C\right]=3\times {10}^{-3}$

$Q_c=1$

But equilibrium constant is $K_c=2\times {10}^{-3}$.

$K_c<Q_c$

The concentrations of products at the given time are more than equilibrium concentrations of products. so to decrease the concentrations of products, the backward reaction should occur.

Hence, the correct option is $\text{B}$