`NH_4COONH_2 (s) hArr 2NH_3(g)+CO_(2)(g)`
If equilibrium pressure is 4 atm for the above reaction, `K_p` will be :(volume and temperature are constant)

Options

A. 2.57
B. 7.11
C. 1.525
D. 2.25

Explanation

To find the equilibrium constant \( K_p \) for the reaction:

\[ NH_4COONH_2 (s) \rightleftharpoons 2NH_3(g) + CO_2(g) \]

given that the equilibrium pressure is 4 atm, we can follow these steps:

### Step 1: Determine the total pressure and individual partial pressures
At equilibrium, we know the total pressure is given as 4 atm. The reaction produces 2 moles of \( NH_3 \) and 1 mole of \( CO_2 \).

Let the partial pressure of \( NH_4COONH_2 \) be \( P \) (which is solid and does not contribute to \( K_p \)), the partial pressure of \( NH_3 \) be \( 2P \) (since there are 2 moles), and the partial pressure of \( CO_2 \) be \( P \) (since there is 1 mole).

Thus, the total pressure \( P_{total} \) can be expressed as:
\[ P_{total} = P_{NH_3} + P_{CO_2} = 2P + P = 3P \]

Given that \( P_{total} = 4 \, \text{atm} \):
\[ 3P = 4 \implies P = \frac{4}{3} \, \text{atm} \]

### Step 2: Calculate the partial pressures of the gases
Now we can find the individual partial pressures:
- \( P_{NH_3} = 2P = 2 \times \frac{4}{3} = \frac{8}{3} \, \text{atm} \)
- \( P_{CO_2} = P = \frac{4}{3} \, \text{atm} \)

### Step 3: Write the expression for \( K_p \)
The expression for the equilibrium constant \( K_p \) for the reaction is given by:
\[ K_p = \frac{(P_{NH_3})^2 \cdot (P_{CO_2})}{(P_{NH_4COONH_2})} \]
Since \( NH_4COONH_2 \) is a solid, its activity is 1 and does not appear in the expression. Thus:
\[ K_p = (P_{NH_3})^2 \cdot P_{CO_2} \]

### Step 4: Substitute the values into the \( K_p \) expression
Substituting the values we found:
\[ K_p = \left(\frac{8}{3}\right)^2 \cdot \left(\frac{4}{3}\right) \]

Calculating \( K_p \):
\[ K_p = \frac{64}{9} \cdot \frac{4}{3} = \frac{256}{27} \]

### Step 5: Final Calculation
Now, we can calculate the numerical value:
\[ K_p \approx 9.481 \text{ (approximately)} \]

### Conclusion
Thus, the final value of \( K_p \) is approximately:
\[ K_p \approx 9.481 \]