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Step-by-Step Solution
Step 1: Write down the given reactions and define symbols
We have two equilibria:
1) $X \leftrightharpoons 2Y$, with equilibrium constant $K_{P1}$.
2) $Z \leftrightharpoons P + Q$, with equilibrium constant $K_{P2}$.
They are given to be in the ratio
$$\frac{K_{P1}}{K_{P2}} = \frac{1}{9}.$$
Further, the degree of dissociation (denoted as $\alpha$) is the same for both $X$ and $Z$. Our goal is to find the ratio of total pressures $P_{T1} : P_{T2}$ at these equilibria.
Step 2: Set up the initial moles and moles at equilibrium for the first reaction
Let the initial moles of $X$ be $a$, and at the start, there are no moles of $Y$. Define $\alpha$ as the degree of dissociation of $X$. The equilibrium situation for the reaction
$$X \leftrightharpoons 2Y$$
is shown in the table below:
SpeciesInitial molesMoles at equilibrium
$X$$a$$a(1 - \alpha)$
$Y$$0$$2a \alpha$
Hence, the total number of moles at equilibrium for the first reaction is
$$a(1 - \alpha) + 2a\alpha = a \bigl(1 - \alpha + 2\alpha \bigr) = a(1 + \alpha).$$
Step 3: Express $K_{P1}$ in terms of partial pressures
The expression for $K_{P1}$ can be written as:
$$
K_{P1} = \frac{(P_Y)^2}{P_X},
$$
adjusted for any total pressure dependence. Let $P_{T1}$ be the total pressure for this system. In terms of mole fractions and total pressure,
Mole fraction of $X$ at equilibrium:
$$x_X = \frac{a(1 - \alpha)}{a(1 + \alpha)} = \frac{1 - \alpha}{1 + \alpha}.$$
Mole fraction of $Y$ at equilibrium:
$$x_Y = \frac{2a\alpha}{a(1 + \alpha)} = \frac{2\alpha}{1 + \alpha}.$$
Thus, partial pressures become $P_X = x_X \, P_{T1}$ and $P_Y = x_Y \, P_{T1}$. However, since $\Delta n$ (change in moles) for $X \leftrightharpoons 2Y$ is $+1$, one finds that an appropriate factor of $\left(\tfrac{P_{T1}}{\text{total moles}}\right)^{\Delta n}$ is also used in detailed derivations. Eventually, one arrives at a form proportional to:
$$
K_{P1} \propto \frac{(2a\alpha)^2 \, P_{T1}}{a(1 - \alpha)\, a(1 + \alpha)}.
$$
Step 4: Set up the initial moles and moles at equilibrium for the second reaction
Let the initial moles of $Z$ be $b$. At the start, there are no moles of $P$ and $Q$. Define the same degree of dissociation $\alpha$. The equilibrium situation for the reaction
$$Z \leftrightharpoons P + Q$$
is shown in the table below:
SpeciesInitial molesMoles at equilibrium
$Z$$b$$b(1 - \alpha)$
$P$$0$$b\alpha$
$Q$$0$$b\alpha$
Hence, the total number of moles at equilibrium for the second reaction is
$$b(1 - \alpha) + b\alpha + b\alpha = b(1 + \alpha).$$
Step 5: Express $K_{P2}$ in terms of partial pressures
Similarly, for $Z \leftrightharpoons P + Q$, we have one mole of $Z$ forming two moles ($P$ and $Q$). The change in moles is also $+1$. One obtains a relation of the form:
$$
K_{P2} \propto \frac{(b\alpha)\,(b\alpha)\,P_{T2}}{b(1 - \alpha)\,b(1 + \alpha)}.
$$
Step 6: Form the ratio $K_{P1} / K_{P2}$ and substitute the given ratio
Upon dividing the two expressions and simplifying (while noting $\alpha$ is the same for both reactions), nearly all $\alpha$-dependent factors simplify in the final step, giving us:
$$
\frac{K_{P1}}{K_{P2}} = \frac{4\,P_{T1}}{P_{T2}}.
$$
We are given
$$
\frac{K_{P1}}{K_{P2}} = \frac{1}{9}.
$$
Thus,
$$
\frac{1}{9} = \frac{4\,P_{T1}}{P_{T2}} \quad \Rightarrow \quad \frac{P_{T1}}{P_{T2}} = \frac{1}{36}.
$$
Step 7: Conclude the desired ratio of total pressures
Therefore, the ratio of total pressures is:
$$
P_{T1} : P_{T2} = 1 : 36.
$$
