(i) X(g) $$\rightleftharpoons$$ Y(g) + Z(g) K$$_{p1}$$ = 3
(ii) A(g) $$\rightleftharpoons$$ 2B(g) K$$_{p2}$$ = 1
If the degree of dissociation and initial concentration of both the reactants X(g) and A(g) are equal, then the ratio of the total pressure at equilibrium $$\frac{p_1}{p_2}$$ is equal to x : 1. The value of x is (Nearest integer)

We are given two equilibrium reactions with the same degree of dissociation $$\alpha$$ and same initial concentration:

(i) $$X(g) \rightleftharpoons Y(g) + Z(g)$$, $$K_{p1} = 3$$

(ii) $$A(g) \rightleftharpoons 2B(g)$$, $$K_{p2} = 1$$

For reaction (i), starting with 1 mole of X at pressure $$P_1$$, at equilibrium: moles of X = $$1 - \alpha$$, moles of Y = $$\alpha$$, moles of Z = $$\alpha$$, total moles = $$1 + \alpha$$. The mole fractions give:

$$K_{p1} = \frac{\left(\frac{\alpha}{1+\alpha}\right)^2}{\frac{1-\alpha}{1+\alpha}} \cdot P_1 = \frac{\alpha^2 P_1}{1-\alpha^2} = 3 \quad \cdots (1)$$

For reaction (ii), starting with 1 mole of A at pressure $$P_2$$, at equilibrium: moles of A = $$1 - \alpha$$, moles of B = $$2\alpha$$, total moles = $$1 + \alpha$$. Similarly:

$$K_{p2} = \frac{\left(\frac{2\alpha}{1+\alpha}\right)^2}{\frac{1-\alpha}{1+\alpha}} \cdot P_2 = \frac{4\alpha^2 P_2}{1-\alpha^2} = 1 \quad \cdots (2)$$

Now, dividing equation (1) by equation (2), the $$\alpha^2$$ and $$(1-\alpha^2)$$ terms cancel completely because $$\alpha$$ is the same for both reactions:

$$\frac{K_{p1}}{K_{p2}} = \frac{P_1}{4P_2}$$

Substituting the given values:

$$\frac{3}{1} = \frac{P_1}{4P_2}$$

$$P_1 = 12 P_2$$

Hence, $$\frac{P_1}{P_2} = 12$$, so the answer is $$x = 12$$.