For which of the following reaction the degree of dissociation `(alpha)` and equilibrium constant `(K_p)` are
related as `K_p =(4alpha^2P)/((1-alpha^2))` ?

Options

A. `N_2O_4(g) hArr2NO_2(g)`
B. `H_2(g) +I_2(g) hArr 2HI(g)`
C. `N_2(g)+3H_2(g) hArr 2NH_3(g)`
D. `PCI_3(g)+CI_2(g) hArr PCI_5(g)`

Answer

To solve the problem, we need to determine for which of the given reactions the degree of dissociation (α) and the equilibrium constant (K_p) are related by the equation:

\[ K_p = \frac{4 \alpha^2 P}{1 - \alpha^2} \]

We will analyze each reaction step by step.

### Step 1: Analyze Reaction 1
**Reaction:** \( N_2O_4(g) \rightleftharpoons 2 NO_2(g) \)

1. **Initial Moles:**
- At \( t = 0 \): 1 mole of \( N_2O_4 \), 0 moles of \( NO_2 \).

2. **At Equilibrium:**
- Moles of \( N_2O_4 \) = \( 1 - \alpha \)
- Moles of \( NO_2 \) = \( 2\alpha \)

3. **Total Moles at Equilibrium:**
- Total moles = \( (1 - \alpha) + 2\alpha = 1 + \alpha \)

4. **K_p Expression:**
- \( K_p = \frac{(P_{NO_2})^2}{P_{N_2O_4}} \)
- \( P_{NO_2} = \frac{2\alpha}{1 + \alpha} P \)
- \( P_{N_2O_4} = \frac{1 - \alpha}{1 + \alpha} P \)

5. **Substituting into K_p:**
\[ K_p = \frac{\left(\frac{2\alpha}{1 + \alpha} P\right)^2}{\frac{1 - \alpha}{1 + \alpha} P} \]
\[ K_p = \frac{4\alpha^2 P^2}{(1 + \alpha)^2} \cdot \frac{1 + \alpha}{1 - \alpha} \]
\[ K_p = \frac{4\alpha^2 P}{1 - \alpha^2} \]

This matches the given relationship.

### Step 2: Analyze Reaction 2
**Reaction:** \( H_2(g) + I_2(g) \rightleftharpoons 2 HI(g) \)

1. **Initial Moles:**
- At \( t = 0 \): 1 mole of \( H_2 \), 1 mole of \( I_2 \), 0 moles of \( HI \).

2. **At Equilibrium:**
- Moles of \( H_2 \) = \( 1 - \alpha \)
- Moles of \( I_2 \) = \( 1 - \alpha \)
- Moles of \( HI \) = \( 2\alpha \)

3. **Total Moles at Equilibrium:**
- Total moles = \( (1 - \alpha) + (1 - \alpha) + 2\alpha = 2 \)

4. **K_p Expression:**
- \( K_p = \frac{(P_{HI})^2}{P_{H_2} P_{I_2}} \)
- \( P_{HI} = \frac{2\alpha}{2} P = \alpha P \)
- \( P_{H_2} = \frac{1 - \alpha}{2} P \)
- \( P_{I_2} = \frac{1 - \alpha}{2} P \)

5. **Substituting into K_p:**
\[ K_p = \frac{(\alpha P)^2}{\left(\frac{1 - \alpha}{2} P\right)\left(\frac{1 - \alpha}{2} P\right)} \]
\[ K_p = \frac{4\alpha^2 P^2}{(1 - \alpha)^2 P^2} \]
\[ K_p = \frac{4\alpha^2}{(1 - \alpha)^2} \]

This does not match the given relationship.

### Step 3: Analyze Reaction 3
**Reaction:** \( N_2(g) + 3 H_2(g) \rightleftharpoons 2 NH_3(g) \)

1. **Initial Moles:**
- At \( t = 0 \): 1 mole of \( N_2 \), 3 moles of \( H_2 \), 0 moles of \( NH_3 \).

2. **At Equilibrium:**
- Moles of \( N_2 \) = \( 1 - \alpha \)
- Moles of \( H_2 \) = \( 3 - 3\alpha \)
- Moles of \( NH_3 \) = \( 2\alpha \)

3. **Total Moles at Equilibrium:**
- Total moles = \( (1 - \alpha) + (3 - 3\alpha) + 2\alpha = 4 - 2\alpha \)

4. **K_p Expression:**
- \( K_p = \frac{(P_{NH_3})^2}{P_{N_2} (P_{H_2})^3} \)
- \( P_{NH_3} = \frac{2\alpha}{4 - 2\alpha} P \)
- \( P_{N_2} = \frac{1 - \alpha}{4 - 2\alpha} P \)
- \( P_{H_2} = \frac{3 - 3\alpha}{4 - 2\alpha} P \)

5. **Substituting into K_p:**
\[ K_p = \frac{\left(\frac{2\alpha}{4 - 2\alpha} P\right)^2}{\left(\frac{1 - \alpha}{4 - 2\alpha} P\right)\left(\frac{3 - 3\alpha}{4 - 2\alpha} P\right)^3} \]
This expression will not simplify to the required form.

### Step 4: Analyze Reaction 4
**Reaction:** \( PCl_3(g) + Cl_2(g) \rightleftharpoons PCl_5(g) \)

1. **Initial Moles:**
- At \( t = 0 \): 1 mole of \( PCl_3 \), 1 mole of \( Cl_2 \), 0 moles of \( PCl_5 \).

2. **At Equilibrium:**
- Moles of \( PCl_3 \) = \( 1 - \alpha \)
- Moles of \( Cl_2 \) = \( 1 - \alpha \)
- Moles of \( PCl_5 \) = \( \alpha \)

3. **Total Moles at Equilibrium:**
- Total moles = \( (1 - \alpha) + (1 - \alpha) + \alpha = 2 - \alpha \)

4. **K_p Expression:**
- \( K_p = \frac{P_{PCl_5}}{P_{PCl_3} P_{Cl_2}} \)
- \( P_{PCl_5} = \frac{\alpha}{2 - \alpha} P \)
- \( P_{PCl_3} = \frac{1 - \alpha}{2 - \alpha} P \)
- \( P_{Cl_2} = \frac{1 - \alpha}{2 - \alpha} P \)

5. **Substituting into K_p:**
\[ K_p = \frac{\frac{\alpha}{2 - \alpha} P}{\left(\frac{1 - \alpha}{2 - \alpha} P\right)\left(\frac{1 - \alpha}{2 - \alpha} P\right)} \]
This will not yield the required form.

### Conclusion
After analyzing all four reactions, we find that only the first reaction \( N_2O_4(g) \rightleftharpoons 2 NO_2(g) \) satisfies the relationship \( K_p = \frac{4 \alpha^2 P}{1 - \alpha^2} \).

### Final Answer
The correct reaction is:
**Option 1: \( N_2O_4(g) \rightleftharpoons 2 NO_2(g) \)**

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