For the reaction (1)and(2)
`A(g)hArrB(g)+C(g)`
` X(g) hArr2Y(g)`
Given , `K_(p1):K_(p2)=9:1`
If the degree of dissociation of A(g) and X(g) be same then the total pressure at equilibrium
(1) and (2) are in the ratio:

Options

A. (a) `3:1`
B. (b) `36:1`
C. (c) `1:1`
D. (d) `0.5:1`

Answer

To solve the problem step by step, we will analyze the two reactions given and derive the necessary relationships to find the total pressure at equilibrium for both reactions.

### Step 1: Write the reactions and initial conditions
1. **Reaction 1:** \( A(g) \rightleftharpoons B(g) + C(g) \)
2. **Reaction 2:** \( X(g) \rightleftharpoons 2Y(g) \)

Initially, we have:
- For Reaction 1: 1 mole of \( A \), 0 moles of \( B \), and 0 moles of \( C \).
- For Reaction 2: 1 mole of \( X \), 0 moles of \( Y \).

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

### Step 3: Calculate the moles at equilibrium
- For **Reaction 1**:
- Moles of \( A \) at equilibrium = \( 1 - \alpha \)
- Moles of \( B \) at equilibrium = \( \alpha \)
- Moles of \( C \) at equilibrium = \( \alpha \)
- Total moles at equilibrium = \( (1 - \alpha) + \alpha + \alpha = 1 + \alpha \)

- For **Reaction 2**:
- Moles of \( X \) at equilibrium = \( 1 - \alpha \)
- Moles of \( Y \) at equilibrium = \( 2\alpha \)
- Total moles at equilibrium = \( (1 - \alpha) + 2\alpha = 1 + \alpha \)

### Step 4: Write the expressions for \( K_{p} \)
- For **Reaction 1**:
\[ K_{p1} = \frac{P_B \cdot P_C}{P_A} = \frac{\left(\frac{\alpha}{1 + \alpha} P_1\right) \cdot \left(\frac{\alpha}{1 + \alpha} P_1\right)}{\frac{1 - \alpha}{1 + \alpha} P_1} \]
Simplifying this gives:
\[ K_{p1} = \frac{\alpha^2 P_1}{(1 - \alpha)(1 + \alpha)} \]

- For **Reaction 2**:
\[ K_{p2} = \frac{(P_Y)^2}{P_X} = \frac{\left(\frac{2\alpha}{1 + \alpha} P_2\right)^2}{\frac{1 - \alpha}{1 + \alpha} P_2} \]
Simplifying this gives:
\[ K_{p2} = \frac{4\alpha^2 P_2}{(1 - \alpha)(1 + \alpha)} \]

### Step 5: Use the ratio of \( K_{p} \)
Given that \( \frac{K_{p1}}{K_{p2}} = \frac{9}{1} \):
\[ \frac{\frac{\alpha^2 P_1}{(1 - \alpha)(1 + \alpha)}}{\frac{4\alpha^2 P_2}{(1 - \alpha)(1 + \alpha)}} = \frac{9}{1} \]
This simplifies to:
\[ \frac{P_1}{4 P_2} = 9 \implies P_1 = 36 P_2 \]

### Step 6: Find the total pressures at equilibrium
- Total pressure for Reaction 1:
\[ P_{total1} = P_1 \]
- Total pressure for Reaction 2:
\[ P_{total2} = P_2 \]

### Step 7: Ratio of total pressures
Thus, the ratio of total pressures at equilibrium for the two reactions is:
\[ \frac{P_{total1}}{P_{total2}} = \frac{P_1}{P_2} = 36:1 \]

### Final Answer
The total pressure at equilibrium for reactions (1) and (2) are in the ratio \( 36:1 \).

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