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Step-by-Step Solution
1. Write Down the Two Reactions and Their Equilibrium Constants
Reaction (i):
X(g) \rightleftharpoons Y(g) + Z(g), \quad K_{p1} = 3
Reaction (ii):
A(g) \rightleftharpoons 2B(g), \quad K_{p2} = 1
Both reagents ( X and A ) are assumed to start with the same initial amount (say 1 mole each) and have the same degree of dissociation, denoted by \alpha .
2. Set Up Equilibrium Amounts for Reaction (i)
Initial moles of X : 1
At equilibrium:
Moles of X = 1 - \alpha
Moles of Y = \alpha
Moles of Z = \alpha
Total moles at equilibrium for Reaction (i) =
n_{1} = (1 - \alpha) + \alpha + \alpha = 1 + \alpha.
Let the total pressure in the container for Reaction (i) be p_1 . The partial pressures at equilibrium are then:
P_X = \frac{(1 - \alpha)\,p_1}{1 + \alpha}, \quad
P_Y = \frac{\alpha\,p_1}{1 + \alpha}, \quad
P_Z = \frac{\alpha\,p_1}{1 + \alpha}
\,.
Because K_{p1} = 3 , given by
K_{p1} = \frac{P_Y \cdot P_Z}{P_X},
we substitute these expressions:
K_{p1}
= \frac{\bigl(\frac{\alpha\,p_1}{1 + \alpha}\bigr)\,\bigl(\frac{\alpha\,p_1}{1 + \alpha}\bigr)}{\frac{(1 - \alpha)\,p_1}{1 + \alpha}}
= \frac{\alpha^2\,p_1^2}{(1 + \alpha)^2}
\times \frac{1 + \alpha}{(1 - \alpha)\,p_1}
= \frac{\alpha^2\,p_1}{(1 - \alpha)(1 + \alpha)}.
Set this equal to 3:
\frac{\alpha^2\,p_1}{(1 - \alpha^2)} = 3
\quad \Longrightarrow \quad
p_1 = 3\,\frac{(1 - \alpha^2)}{\alpha^2}.
3. Set Up Equilibrium Amounts for Reaction (ii)
Initial moles of A : 1
At equilibrium:
Moles of A = 1 - \alpha
Moles of B = 2\alpha
Total moles at equilibrium for Reaction (ii) =
n_{2} = (1 - \alpha) + 2\alpha = 1 + \alpha.
Let the total pressure in the container for Reaction (ii) be p_2 . Then the partial pressures are:
P_A = \frac{(1 - \alpha)\,p_2}{1 + \alpha}, \quad
P_B = \frac{2\alpha\,p_2}{1 + \alpha}.
Because K_{p2} = 1 , given by
K_{p2} = \frac{(P_B)^2}{P_A},
we substitute these expressions:
K_{p2}
= \frac{\Bigl(\frac{2\alpha\,p_2}{1 + \alpha}\Bigr)^2}{\frac{(1 - \alpha)\,p_2}{1 + \alpha}}
= \frac{4\,\alpha^2\,p_2^2}{(1 + \alpha)^2}
\times \frac{1 + \alpha}{(1 - \alpha)\,p_2}
= \frac{4\,\alpha^2\,p_2}{(1 - \alpha^2)}.
Set this equal to 1:
\frac{4\,\alpha^2\,p_2}{1 - \alpha^2} = 1
\quad \Longrightarrow \quad
p_2 = \frac{1 - \alpha^2}{4\,\alpha^2}.
4. Find the Ratio of Total Pressures, p_1 / p_2
We now divide p_1 by p_2 :
\frac{p_1}{p_2}
= \frac{\,3\,\frac{(1 - \alpha^2)}{\alpha^2}\,}{\,\frac{1 - \alpha^2}{4\,\alpha^2}\,}
= 3 \times \frac{(1 - \alpha^2)}{\alpha^2} \times
\frac{4\,\alpha^2}{1 - \alpha^2}
= 3 \times 4
= 12.
Hence,
p_1 : p_2
= 12 : 1.
Therefore, x = 12 .
5. Final Answer
The ratio of the total pressures at equilibrium for the two reactions, under the given condition, is 12 : 1.
Hence,
\boxed{x = 12}.
