Question

The value of $K_c$ for the reaction $2A\rightleftharpoons B+C$ is $2\times {10}^{-3}$.
At a given time, the composition of reaction mixture is $\left[A\right]=\left[B\right]=\left[C\right]=3\times {10}^{-4}M$.
In which direction the reaction will proceed?

Answer 1

$\text{Step 1: Reaction quotient}$

For a general reaction $aA+bB\rightleftharpoons cC+dD$

$Q_c=\frac{\left[C{]}^c\right[D{]}^d}{\left[A{]}^a\right[B{]}^b}$

• If $Q_c>K_c$ , the reaction proceeds in the direction of reactants (backward reaction).
• If $Q_c<K_c$ , the reaction proceeds in the direction of the products (forward reaction).
• If $Q_c=K_c$ , the reaction mixture is at equilibrium already.

$\text{Step 2: Value of}{Q}_c$

For the reaction $2A\rightleftharpoons B+C;K_c=2\times {10}^{-3}$

Therefore, $Q_c=\frac{\left[B\right]\left[C\right]}{[A{]}^2}$

As $\left[A\right]=\left[B\right]=\left[C\right]=3\times {10}^{-4}M$

$\Rightarrow Q_c=\frac{(3\times {10}^{-4})(3\times {10}^{-4})}{(3\times {10}^{-4}{)}^2}=1$

$\text{Step 3: Direction of reaction}$

$\because Q_c>K_c$ , reaction proceeds in the backward direction.

$\text{Final answer}:$

$Q_c=1$ and reaction proceeds in the backward direction.