`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

Options

A. `4`
B. `27`
C. `4//27`
D. `1//27`

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 3 atm, we can follow these steps:

### Step 1: Identify the reaction and the equilibrium condition
The reaction involves a solid reactant and gaseous products. The equilibrium pressure of the gaseous products is given as 3 atm.

### Step 2: Set up the initial conditions
Assume we start with \( x \) moles of the solid reactant \( NH_4COONH_2 \). Since it is a solid, it does not affect the equilibrium expression. Let \( \alpha \) be the amount of the solid that reacts at equilibrium.

### Step 3: Determine the moles of products at equilibrium
From the stoichiometry of the reaction:
- For every \( 1 \) mole of \( NH_4COONH_2 \) that reacts, \( 2 \) moles of \( NH_3 \) and \( 1 \) mole of \( CO_2 \) are produced.
- Therefore, if \( \alpha \) moles of \( NH_4COONH_2 \) react, we will have:
- Moles of \( NH_3 = 2\alpha \)
- Moles of \( CO_2 = \alpha \)

### Step 4: Calculate the total moles of gas at equilibrium
The total moles of gas at equilibrium will be:
\[ \text{Total moles} = 2\alpha + \alpha = 3\alpha \]

### Step 5: Determine the partial pressures of the products
Using the total pressure \( P = 3 \) atm:
- The partial pressure of \( NH_3 \):
\[ P_{NH_3} = \frac{2\alpha}{3\alpha} \times P = \frac{2}{3} \times 3 = 2 \text{ atm} \]
- The partial pressure of \( CO_2 \):
\[ P_{CO_2} = \frac{\alpha}{3\alpha} \times P = \frac{1}{3} \times 3 = 1 \text{ atm} \]

### Step 6: Write the expression for \( K_p \)
The equilibrium constant \( K_p \) for the reaction is given by:
\[ K_p = \frac{(P_{NH_3})^2 \cdot P_{CO_2}}{1} = (P_{NH_3})^2 \cdot P_{CO_2} \]

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

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