In a reversible reaction, two substances are in equilibrium. If the concentration of each one is reduced to half, the equilibrium constant will be

Let us consider equilibrium for a reversible reaction as given below.

                    $\text{A}\rightleftharpoons \text{B}$

The initial concentrations of A and B to be $x$  and$y$ .

If we reduce the concentration of each species to one half , then we will get

 $\frac{x}{2}$ and $\frac{y}{2}$ for A and B respectively.

Now, we will find the equilibrium constant (${\text{K}}_{\text{c}}$ ) for each condition.

Case 1: Equilibrium constant, if the concentrations of A and B are$x$  &$y$ .

From the equilibrium constant expression,

                              ${\text{K}}_{\text{c}}=\frac{\left[\text{conc}\text{. of products}\right]}{\left[\text{conc}\text{.of reactants}\right]}=\frac{\left[\text{B}\right]}{\left[\text{A}\right]}$

                   Then   ${\text{K}}_{\text{c}}=\frac{y}{x}$

Case 2: Equilibrium constant, if the concentrations of A and B

           are$\frac{x}{2}$  &$\frac{y}{2}$ .  From the equilibrium constant expression,

                              ${\text{K}}_{\text{c}}=\frac{\left[\text{conc}\text{. of products}\right]}{\left[\text{conc}\text{.of reactants}\right]}=\frac{\left[\text{B}\right]}{\left[\text{A}\right]}$

                 Then     ${\text{K}}_{\text{c}}=\frac{(y/2)}{(x/2)}=\frac{y}{x}$

Therefore, in the above two cases the equilibrium constant remains same.