`ArArr2B,K_(p),CrArrD+E,K_(p)` . If degrees of dissociation of A and C are same and `K_(p)=2K_(p)`, then the ratio of total presure `p//p=`?

Options

A. `(1)/(2)`
B. `(1)/(3)`
C. `(1)/(4)`
D. 2

Answer

To solve the problem, we will analyze the two reactions and their equilibrium constants step by step.

### Step 1: Define the Reactions
We have two reactions:
1. \( A \rightleftharpoons 2B \) with equilibrium constant \( K_p \)
2. \( C \rightleftharpoons D + E \) with equilibrium constant \( K'_p \)

We are given that the degrees of dissociation of \( A \) and \( C \) are the same, denoted as \( \alpha \), and that \( K_p = 2K'_p \).

### Step 2: Set Up Initial Pressures
Let the initial pressure of \( A \) be \( P \) and the initial pressure of \( C \) be \( P' \). Initially, the pressures of \( B \), \( D \), and \( E \) are zero.

### Step 3: Calculate Equilibrium Pressures for Reaction 1
For the reaction \( A \rightleftharpoons 2B \):
- Initial pressure of \( A = P \)
- Change in pressure due to dissociation = \( -P\alpha \) for \( A \) and \( +2P\alpha \) for \( B \)

At equilibrium:
- Pressure of \( A = P(1 - \alpha) \)
- Pressure of \( B = 2P\alpha \)

### Step 4: Write the Expression for \( K_p \) for Reaction 1
The expression for \( K_p \) for the first reaction is:
\[ K_p = \frac{(2P\alpha)^2}{P(1 - \alpha)} = \frac{4P^2\alpha^2}{P(1 - \alpha)} = \frac{4P\alpha^2}{1 - \alpha} \]

### Step 5: Calculate Equilibrium Pressures for Reaction 2
For the reaction \( C \rightleftharpoons D + E \):
- Initial pressure of \( C = P' \)
- Change in pressure due to dissociation = \( -P'\alpha \) for \( C \) and \( +P'\alpha \) for \( D \) and \( E \)

At equilibrium:
- Pressure of \( C = P'(1 - \alpha) \)
- Pressures of \( D \) and \( E = P'\alpha \)

### Step 6: Write the Expression for \( K'_p \) for Reaction 2
The expression for \( K'_p \) for the second reaction is:
\[ K'_p = \frac{(P'\alpha)(P'\alpha)}{P'(1 - \alpha)} = \frac{P'^2\alpha^2}{P'(1 - \alpha)} = \frac{P'\alpha^2}{1 - \alpha} \]

### Step 7: Relate \( K_p \) and \( K'_p \)
Given that \( K_p = 2K'_p \):
\[ \frac{4P\alpha^2}{1 - \alpha} = 2 \cdot \frac{P'\alpha^2}{1 - \alpha} \]
Cancelling \( \alpha^2 \) and \( (1 - \alpha) \) from both sides (assuming \( \alpha \neq 0 \) and \( \alpha \neq 1 \)):
\[ 4P = 2P' \implies P' = 2P \]

### Step 8: Calculate the Total Pressures
Total pressure for the first reaction:
\[ P_{total} = P(1 + \alpha) \]
Total pressure for the second reaction:
\[ P'_{total} = P'(1 + \alpha) = 2P(1 + \alpha) \]

### Step 9: Find the Ratio of Total Pressures
Now, we find the ratio of total pressures:
\[ \frac{P_{total}}{P'_{total}} = \frac{P(1 + \alpha)}{2P(1 + \alpha)} = \frac{1}{2} \]

### Step 10: Final Ratio of Total Pressures
Thus, the ratio of total pressures \( \frac{p}{p'} \) is:
\[ \frac{p}{p'} = \frac{1}{2} \]