The equilibrium constant `K_(p_(2))` and `K_(p_(2))` for the reactions `A hArr 2B` and `P hArr Q+R`, respectively, are in the ratio of `2:3`. If the degree of dissociation of A and P are equal, the ratio of the total pressure at equilibrium is,

Options

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

Explanation

To solve the problem, we need to analyze the two equilibrium reactions and their respective equilibrium constants. Let's break down the solution step by step.

### Step 1: Write the reactions and their equilibrium expressions

1. **Reactions**:
- Reaction 1: \( A \rightleftharpoons 2B \)
- Reaction 2: \( P \rightleftharpoons Q + R \)

2. **Equilibrium Constants**:
- For Reaction 1:
\[ K_{p1} = \frac{(P_B)^2}{P_A} \]
- For Reaction 2:
\[ K_{p2} = \frac{P_Q \cdot P_R}{P_P} \]

### Step 2: Define initial conditions and degree of dissociation

Assume we start with 1 mole of A and P each.

- For Reaction 1:
- Initial moles: \( A = 1, B = 0 \)
- At equilibrium: \( A = 1 - \alpha \) and \( B = 2\alpha \)

- For Reaction 2:
- Initial moles: \( P = 1, Q = 0, R = 0 \)
- At equilibrium: \( P = 1 - \alpha \), \( Q = \alpha \), \( R = \alpha \)

### Step 3: Calculate total moles at equilibrium

- For Reaction 1:
- Total moles = \( (1 - \alpha) + 2\alpha = 1 + \alpha \)

- For Reaction 2:
- Total moles = \( (1 - \alpha) + \alpha + \alpha = 1 + \alpha \)

### Step 4: Calculate partial pressures using Dalton's Law

Using Dalton's Law, the partial pressure of each component is given by:
\[ P_i = \text{mole fraction} \times P_{total} \]

- For Reaction 1:
- \( P_B = \frac{2\alpha}{1+\alpha} P_1 \)
- \( P_A = \frac{1 - \alpha}{1 + \alpha} P_1 \)

- For Reaction 2:
- \( P_Q = \frac{\alpha}{1 + \alpha} P_2 \)
- \( P_R = \frac{\alpha}{1 + \alpha} P_2 \)
- \( P_P = \frac{1 - \alpha}{1 + \alpha} P_2 \)

### Step 5: Substitute into equilibrium constant expressions

Substituting the partial pressures into the equilibrium constant expressions:

1. For \( K_{p1} \):
\[ K_{p1} = \frac{\left(\frac{2\alpha}{1+\alpha} P_1\right)^2}{\frac{1 - \alpha}{1 + \alpha} P_1} = \frac{4\alpha^2 P_1}{(1 - \alpha)(1 + \alpha)} \]

2. For \( K_{p2} \):
\[ K_{p2} = \frac{\left(\frac{\alpha}{1+\alpha} P_2\right) \left(\frac{\alpha}{1+\alpha} P_2\right)}{\frac{1 - \alpha}{1 + \alpha} P_2} = \frac{\alpha^2 P_2}{(1 - \alpha)(1 + \alpha)} \]

### Step 6: Set up the ratio of equilibrium constants

Given \( \frac{K_{p1}}{K_{p2}} = \frac{2}{3} \):
\[ \frac{4\alpha^2 P_1}{(1 - \alpha)(1 + \alpha)} \div \frac{\alpha^2 P_2}{(1 - \alpha)(1 + \alpha)} = \frac{2}{3} \]
This simplifies to:
\[ \frac{4 P_1}{P_2} = \frac{2}{3} \]

### Step 7: Solve for the ratio of total pressures

Cross-multiplying gives:
\[ 4 P_1 = \frac{2}{3} P_2 \]
Thus,
\[ \frac{P_1}{P_2} = \frac{2}{12} = \frac{1}{6} \]

### Final Answer

The ratio of the total pressure at equilibrium is:
\[ \frac{P_1}{P_2} = \frac{1}{6} \]