The reaction $$A(g) \rightleftharpoons B(g) + C(g)$$ was initiated with the amount `$$a$$` of $$A(g)$$. At equilibrium it is found that the amount of $$A(g)$$ remaining is $$(a - x)$$ at a total pressure of $$p$$.
The equilibrium constant $$K_p$$ of the reaction can be calculated from the expression:

For the reaction

$$A(g)\rightleftharpoons B(g)+C(g),$$

let the initial number of moles of (A) be (a), and let (x) moles dissociate at equilibrium.

The equilibrium moles are therefore:

• (A = a-x)
• (B = x)
• (C = x)

The total number of moles at equilibrium is

$$(a-x)+x+x=a+x.$$

The mole fractions of the species are

$$x_A=\frac{a-x}{a+x},\qquad x_B=\frac{x}{a+x},\qquad x_C=\frac{x}{a+x}.$$

If the total pressure at equilibrium is (p), then the partial pressures are

$$P_A=\frac{a-x}{a+x},p,\qquad P_B=\frac{x}{a+x},p,\qquad P_C=\frac{x}{a+x},p.$$

The equilibrium constant in terms of partial pressures is

$$K_p=\frac{P_B\times P_C}{P_A}.$$

Substituting the expressions for the partial pressures,

$$K_p=\frac{\left(\frac{x}{a+x}p\right)\left(\frac{x}{a+x}p\right)}{\left(\frac{a-x}{a+x}p\right)}.$$

Simplifying,

$$K_p=\frac{\frac{x^2}{(a+x)^2}p^2}{\frac{a-x}{a+x}p}$$

$$=\frac{x^2}{(a+x)^2}\times\frac{a+x}{a-x}\times p$$

$$=\frac{x^2}{(a+x)(a-x)}\times p$$

$$=\frac{x^2}{a^2-x^2}\times p.$$

Hence, the correct expression for the equilibrium constant is

$$K_p=\frac{x^2}{a^2-x^2}\times p.$$