For `A(g) hArr 2B(g)`, equilibrium constant at total equilibrium pressure `p_(1)` is
`K_(p1)& ` for `C(g)hArrD(g)+E(g)`.
equilibrium constant at total equilibrium pressure `p_(2)` is `K_(p2)`. If degree of dissciation of `A & C` are same, then the ratio `K_(p1) //K_(p2)` , if `2p_(1)=p_(2)`, is `:`

Options

A. 2
B. `1//8`
C. `1//2`
D. 8

Explanation

To solve the problem, we need to find the ratio \( \frac{K_{p1}}{K_{p2}} \) given the reactions and conditions provided. Let's break down the solution step by step.

### Step 1: Analyze the first reaction \( A(g) \rightleftharpoons 2B(g) \)

1. **Initial Moles**: Assume at time \( t = 0 \):
- Moles of \( A = 1 \)
- Moles of \( B = 0 \)

2. **At Equilibrium**: Let the degree of dissociation of \( A \) be \( \alpha \).
- Moles of \( A = 1 - \alpha \)
- Moles of \( B = 2\alpha \)

3. **Total Moles at Equilibrium**:
\[ n_{total} = (1 - \alpha) + 2\alpha = 1 + \alpha \]

4. **Expression for \( K_{p1} \)**:
\[ K_{p1} = \frac{(P_B)^2}{P_A} = \frac{(2\alpha \cdot \frac{P_1}{1 + \alpha})^2}{(1 - \alpha) \cdot \frac{P_1}{1 + \alpha}} \]
Simplifying this gives:
\[ K_{p1} = \frac{4\alpha^2 P_1}{(1 - \alpha)(1 + \alpha)} \]

### Step 2: Analyze the second reaction \( C(g) \rightleftharpoons D(g) + E(g) \)

1. **Initial Moles**: Assume at time \( t = 0 \):
- Moles of \( C = 1 \)
- Moles of \( D = 0 \)
- Moles of \( E = 0 \)

2. **At Equilibrium**: Let the degree of dissociation of \( C \) be \( \alpha \).
- Moles of \( C = 1 - \alpha \)
- Moles of \( D = \alpha \)
- Moles of \( E = \alpha \)

3. **Total Moles at Equilibrium**:
\[ n_{total} = (1 - \alpha) + \alpha + \alpha = 1 + \alpha \]

4. **Expression for \( K_{p2} \)**:
\[ K_{p2} = \frac{(P_D)(P_E)}{P_C} = \frac{(\alpha \cdot \frac{P_2}{1 + \alpha})(\alpha \cdot \frac{P_2}{1 + \alpha})}{(1 - \alpha) \cdot \frac{P_2}{1 + \alpha}} \]
Simplifying this gives:
\[ K_{p2} = \frac{\alpha^2 P_2}{(1 - \alpha)(1 + \alpha)} \]

### Step 3: Find the ratio \( \frac{K_{p1}}{K_{p2}} \)

1. **Setting up the ratio**:
\[ \frac{K_{p1}}{K_{p2}} = \frac{\frac{4\alpha^2 P_1}{(1 - \alpha)(1 + \alpha)}}{\frac{\alpha^2 P_2}{(1 - \alpha)(1 + \alpha)}} \]
The terms \( (1 - \alpha)(1 + \alpha) \) cancel out:
\[ \frac{K_{p1}}{K_{p2}} = \frac{4P_1}{P_2} \]

2. **Using the condition \( 2P_1 = P_2 \)**:
\[ \frac{K_{p1}}{K_{p2}} = \frac{4P_1}{2P_1} = 2 \]

### Final Answer
Thus, the ratio \( \frac{K_{p1}}{K_{p2}} \) is \( 2 \).