`NH_4COONH_2 (s) hArr 2NH_3(g)+CO_(2)(g)`
If equilibrium pressure is 5 atm for the above reaction, `K_p` will be :(volume and temperature are constant)

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 5 atm, we can follow these steps:

### Step 1: Understand the Reaction
The reaction involves the decomposition of solid ammonium carbamate into gaseous ammonia and carbon dioxide. At equilibrium, we have:

- 2 moles of \( NH_3 \) (g)
- 1 mole of \( CO_2 \) (g)

### Step 2: Calculate Total Moles of Gas
At equilibrium, the total number of moles of gas produced is:

\[ n_{total} = 2 \text{ (from } NH_3\text{)} + 1 \text{ (from } CO_2\text{)} = 3 \text{ moles} \]

### Step 3: Relate Total Pressure to Partial Pressures
Given that the total equilibrium pressure \( P_{total} = 5 \, \text{atm} \), we can relate the total pressure to the partial pressures of the gases:

\[ P_{total} = P_{NH_3} + P_{CO_2} \]

Let \( P \) be the partial pressure of \( CO_2 \). Then, the partial pressure of \( NH_3 \) will be \( 2P \) (since there are 2 moles of \( NH_3 \)). Therefore:

\[ P_{total} = 2P + P = 3P \]

### Step 4: Solve for Partial Pressure
From the total pressure equation:

\[ 3P = 5 \implies P = \frac{5}{3} \, \text{atm} \]

### Step 5: Calculate Partial Pressures
Now we can find the partial pressures:

- \( P_{NH_3} = 2P = 2 \times \frac{5}{3} = \frac{10}{3} \, \text{atm} \)
- \( P_{CO_2} = P = \frac{5}{3} \, \text{atm} \)

### Step 6: Write the Expression for \( K_p \)
The expression for \( K_p \) 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 taken as 1. Thus, we have:

\[ K_p = (P_{NH_3})^2 \cdot P_{CO_2} \]

### Step 7: Substitute the Values
Now substituting the values we calculated:

\[ K_p = \left(\frac{10}{3}\right)^2 \cdot \left(\frac{5}{3}\right) \]

Calculating this gives:

\[ K_p = \frac{100}{9} \cdot \frac{5}{3} = \frac{500}{27} \]

### Final Answer
Thus, the value of \( K_p \) is:

\[ K_p = \frac{500}{27} \approx 18.52 \]