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

Options

A. 27
B. 4
C. 3
D. 9

Explanation

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

\[ NH_4COONH_4(s) \rightleftharpoons 2NH_3(g) + CO_2(g) \]

given that the equilibrium pressure is \( 3 \, \text{atm} \), we can follow these steps:

### Step 1: Identify the reaction components
The reaction involves:
- Reactant: \( NH_4COONH_4 \) (solid)
- Products: \( 2NH_3 \) (gas) and \( CO_2 \) (gas)

### Step 2: Understand the equilibrium expression
The equilibrium constant \( K_p \) is expressed in terms of the partial pressures of the gaseous products. The general form for \( K_p \) is:

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

Since the reactant is a solid, it does not appear in the expression for \( K_p \).

### Step 3: Define partial pressures
Let \( P \) be the total pressure at equilibrium, which is given as \( 3 \, \text{atm} \).

From the stoichiometry of the reaction:
- The partial pressure of \( NH_3 \) is \( P_{NH_3} = 2 \alpha \)
- The partial pressure of \( CO_2 \) is \( P_{CO_2} = \alpha \)

Where \( \alpha \) is the change in moles of \( NH_4COONH_4 \) that decomposes.

### Step 4: Calculate total pressure
The total number of moles of gas at equilibrium is:

\[ P_{total} = P_{NH_3} + P_{CO_2} = 2\alpha + \alpha = 3\alpha \]

Given that the total pressure \( P_{total} = 3 \, \text{atm} \):

\[ 3\alpha = 3 \implies \alpha = 1 \]

### Step 5: Calculate partial pressures
Now substituting \( \alpha \) back into the expressions for partial pressures:

\[ P_{NH_3} = 2\alpha = 2(1) = 2 \, \text{atm} \]
\[ P_{CO_2} = \alpha = 1 \, \text{atm} \]

### Step 6: Substitute into the \( K_p \) expression
Now, substituting the partial pressures into the \( K_p \) expression:

\[ K_p = \frac{(P_{NH_3})^2 \cdot (P_{CO_2})}{1} = (2)^2 \cdot (1) = 4 \]

### Conclusion
Thus, the value of \( K_p \) for the reaction is:

\[ \boxed{4} \]

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