The equilibrium constant `K_(p_(1))` and `K_(p_(2))` for the reactions `XhArr2Y` and `ZhArrP+Q`, respectively are in the ratio of `1:9`. If the degree of dissociation of `X` and `Z` be equal, then the ratio of total pressure at these equilibrium is:

Options

A. `1:36`
B. `1:1`
C. `1:3`
D. `1:9`

Answer

To solve the problem step by step, we will analyze the two reactions and their equilibrium constants, given that the degree of dissociation for both reactions is equal.

### Step 1: Define the Reactions and Initial Conditions
1. **First Reaction**: \( X \rightleftharpoons 2Y \)
- Initial moles of \( X = 1 \)
- Initial moles of \( Y = 0 \)

2. **Second Reaction**: \( Z \rightleftharpoons P + Q \)
- Initial moles of \( Z = 1 \)
- Initial moles of \( P = 0 \)
- Initial moles of \( Q = 0 \)

### Step 2: Degree of Dissociation
Let the degree of dissociation for both reactions be \( \alpha \).

- For the first reaction:
- Moles of \( X \) at equilibrium = \( 1 - \alpha \)
- Moles of \( Y \) at equilibrium = \( 2\alpha \)

- For the second reaction:
- Moles of \( Z \) at equilibrium = \( 1 - \alpha \)
- Moles of \( P \) at equilibrium = \( \alpha \)
- Moles of \( Q \) at equilibrium = \( \alpha \)

### Step 3: Calculate Total Moles at Equilibrium
- For the first reaction:
\[ \text{Total moles} = (1 - \alpha) + 2\alpha = 1 + \alpha \]

- For the second reaction:
\[ \text{Total moles} = (1 - \alpha) + \alpha + \alpha = 1 + \alpha \]

### Step 4: Calculate Equilibrium Constants
1. **Equilibrium Constant for the First Reaction** \( K_{p1} \):
\[ K_{p1} = \frac{(P_Y)^2}{P_X} = \frac{(2\alpha P_1)^2}{(1 - \alpha) P_1} = \frac{4\alpha^2 P_1^2}{(1 - \alpha) P_1} = \frac{4\alpha^2 P_1}{(1 - \alpha)} \]

2. **Equilibrium Constant for the Second Reaction** \( K_{p2} \):
\[ K_{p2} = \frac{P_P \cdot P_Q}{P_Z} = \frac{(\alpha P_2)(\alpha P_2)}{(1 - \alpha) P_2} = \frac{\alpha^2 P_2^2}{(1 - \alpha) P_2} = \frac{\alpha^2 P_2}{(1 - \alpha)} \]

### Step 5: Set Up the Ratio of Equilibrium Constants
Given that \( K_{p1} : K_{p2} = 1 : 9 \):
\[ \frac{K_{p1}}{K_{p2}} = \frac{4\alpha^2 P_1}{(1 - \alpha)} \cdot \frac{(1 - \alpha)}{\alpha^2 P_2} = \frac{4 P_1}{P_2} = \frac{1}{9} \]

### Step 6: Solve for the Ratio of Total Pressures
From the equation:
\[ 4 P_1 = \frac{1}{9} P_2 \]
This implies:
\[ \frac{P_1}{P_2} = \frac{1}{36} \]

### Conclusion
The ratio of total pressures at equilibrium for the two reactions is:
\[ \frac{P_1}{P_2} = \frac{1}{36} \]