`NH_4COONH_2 (s) hArr 2NH_3(g)+CO_(2)(g)`
If equilibrium pressure is 9 atm for the above reaction, `K_p` will be :(volume and temperature are constant)

Options

A. 9
B. 27
C. 36
D. 12

Explanation

To solve the problem, we need to determine the equilibrium constant \( K_p \) for the reaction given the equilibrium pressure.

The reaction is:
\[ \text{NH}_4\text{COONH}_2 (s) \rightleftharpoons 2 \text{NH}_3(g) + \text{CO}_2(g) \]

### Step-by-Step Solution:

1. **Identify the Reaction Components:**
- Reactant: \( \text{NH}_4\text{COONH}_2 \) (solid, does not contribute to \( K_p \))
- Products: \( 2 \text{NH}_3(g) \) and \( \text{CO}_2(g) \)

2. **Understand the Equilibrium Pressure:**
- The total equilibrium pressure is given as \( 9 \, \text{atm} \).

3. **Determine the Partial Pressures:**
- Let the partial pressure of \( \text{NH}_4\text{COONH}_2 \) (solid) be negligible since it does not affect \( K_p \).
- Let \( P \) be the partial pressure of \( \text{CO}_2 \).
- The partial pressure of \( \text{NH}_3 \) is \( 2P \) (since there are 2 moles of \( \text{NH}_3 \)).
- Thus, the total pressure at equilibrium can be expressed as:
\[ P_{\text{total}} = P_{\text{NH}_3} + P_{\text{CO}_2} = 2P + P = 3P \]

4. **Set Up the Equation for Total Pressure:**
- Given that the total pressure is \( 9 \, \text{atm} \):
\[ 3P = 9 \]
- Solving for \( P \):
\[ P = \frac{9}{3} = 3 \, \text{atm} \]

5. **Calculate the Partial Pressures:**
- Partial pressure of \( \text{NH}_3 \):
\[ P_{\text{NH}_3} = 2P = 2 \times 3 = 6 \, \text{atm} \]
- Partial pressure of \( \text{CO}_2 \):
\[ P_{\text{CO}_2} = P = 3 \, \text{atm} \]

6. **Write the Expression for \( K_p \):**
- The expression for \( K_p \) is given by:
\[ K_p = \frac{(P_{\text{NH}_3})^2 \cdot P_{\text{CO}_2}}{P_{\text{NH}_4\text{COONH}_2}} = \frac{(6)^2 \cdot (3)}{1} = 36 \]

7. **Final Result:**
- Therefore, the value of \( K_p \) is:
\[ K_p = 36 \]