For the reaction 2A ⇌ B + C, K_c = 4 × 10^–3 . At a given time, the composition of reaction mixture is: [A] = [B] = [C] = 2 × 10^–3 M.

Then, which of the following is correct?

The correct option is: Reaction has a tendency to go in backward direction.

To determine the direction in which the reaction will proceed, we need to calculate the reaction quotient ($Q_c$) and compare it with the equilibrium constant ($K_c$).

For the given reversible reaction:
$2A\rightleftharpoons B+C$

The expression for the reaction quotient ($Q_c$) is:
$Q_c=\frac{\left[B\right]\left[C\right]}{{\left[A\right]}^2}$

Given the concentrations at the specified time:
$\left[A\right]=\left[B\right]=\left[C\right]=2\times {10}^{-3}\text{M}$

Substitute these concentration values into the reaction quotient expression:
$Q_c=\frac{(2\times {10}^{-3})\times (2\times {10}^{-3})}{{(2\times {10}^{-3})}^2}$

Simplifying the expression:
$Q_c=\frac{4\times {10}^{-6}}{4\times {10}^{-6}}=1$

The given equilibrium constant is:
$K_c=4\times {10}^{-3}=0.004$

Comparing $Q_c$ with $K_c$:
$Q_c>K_c$
(since 1 > 0.004)

Because the reaction quotient is greater than the equilibrium constant, the concentration of products is higher than the equilibrium value. Therefore, the reaction will favor the reverse pathway, meaning it has a tendency to go in the backward direction to establish equilibrium.