`NH_(4)COONH_(2)(s)hArr2NH_(3)(g)+CO_(2)(g)` If equilibrium pressure is 3 atm for the above reaction, then `K_(p)` for the reaction is

Answer

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

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

given that the equilibrium pressure is 3 atm, we can follow these steps:

### Step 1: Write the balanced chemical equation
The balanced equation for the dissociation of ammonium carbamate is:

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

### Step 2: Identify the initial and equilibrium conditions
Initially, we have:
- Moles of \( \text{NH}_4\text{COONH}_2 \) = \( X \) (solid, does not contribute to \( K_p \))
- Moles of \( \text{NH}_3 \) = 0
- Moles of \( \text{CO}_2 \) = 0

At equilibrium, let \( \alpha \) be the amount that dissociates:
- Moles of \( \text{NH}_4\text{COONH}_2 \) = \( X - \alpha \) (but it is solid, so we ignore it for \( K_p \))
- Moles of \( \text{NH}_3 \) = \( 2\alpha \)
- Moles of \( \text{CO}_2 \) = \( \alpha \)

### Step 3: Calculate total moles of gas at equilibrium
The total number of moles of gas at equilibrium is:
\[ \text{Total moles} = 2\alpha + \alpha = 3\alpha \]

### Step 4: Calculate partial pressures
The total pressure at equilibrium is given as \( P = 3 \) atm.

The partial pressure of \( \text{NH}_3 \) is:
\[ P_{\text{NH}_3} = \frac{\text{moles of } \text{NH}_3}{\text{total moles}} \times P = \frac{2\alpha}{3\alpha} \times P = \frac{2}{3} P \]

The partial pressure of \( \text{CO}_2 \) is:
\[ P_{\text{CO}_2} = \frac{\text{moles of } \text{CO}_2}{\text{total moles}} \times P = \frac{\alpha}{3\alpha} \times P = \frac{1}{3} P \]

### Step 5: Substitute the equilibrium pressure
Substituting \( P = 3 \) atm into the equations for partial pressures:
\[ P_{\text{NH}_3} = \frac{2}{3} \times 3 = 2 \text{ atm} \]
\[ P_{\text{CO}_2} = \frac{1}{3} \times 3 = 1 \text{ atm} \]

### Step 6: Write the expression for \( K_p \)
The expression for \( K_p \) is given by:
\[ K_p = \frac{(P_{\text{NH}_3})^2 \cdot (P_{\text{CO}_2})}{1} = (P_{\text{NH}_3})^2 \cdot (P_{\text{CO}_2}) \]

### Step 7: Substitute the partial pressures into the \( K_p \) expression
Substituting the values we found:
\[ K_p = (2 \text{ atm})^2 \cdot (1 \text{ atm}) = 4 \text{ atm}^2 \cdot 1 \text{ atm} = 4 \]

### Final Answer
Thus, the value of \( K_p \) for the reaction is:
\[ \boxed{4} \]