`NH_4COONH_2 (s) hArr 2NH_3(g)+CO_(2)(g)`
If equilibrium pressure is 1 atm for the above reaction, `K_p` will be :(volume and temperature are constant)

Options

A. `1/4`
B. `4/9`
C. `9/4`
D. `4/1`

Explanation

To solve the problem, we need 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 1 atm, we can follow these steps:

### Step 1: Determine the total number of moles at equilibrium
From the balanced equation, we see that:
- 1 mole of \( NH_4COONH_2 \) decomposes to produce 2 moles of \( NH_3 \) and 1 mole of \( CO_2 \).
- Therefore, the total number of moles at equilibrium is \( 2 + 1 = 3 \) moles.

### Step 2: Relate total pressure to partial pressures
Since the total pressure at equilibrium is given as 1 atm, we can express the partial pressures in terms of a variable \( P \):
- Let the partial pressure of \( NH_3 \) be \( 2P \) (since there are 2 moles of \( NH_3 \)).
- Let the partial pressure of \( CO_2 \) be \( P \) (since there is 1 mole of \( CO_2 \)).
- The total pressure is then \( 2P + P = 3P \).

### Step 3: Solve for \( P \)
Given that the total pressure is 1 atm:
\[ 3P = 1 \text{ atm} \implies P = \frac{1}{3} \text{ atm} \]

### Step 4: 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 considered to be 1. Thus, we can simplify the expression to:
\[ K_p = (P_{NH_3})^2 \cdot P_{CO_2} \]

### Step 5: Substitute the values of the partial pressures
Substituting the values we found:
- \( P_{NH_3} = 2P = 2 \times \frac{1}{3} = \frac{2}{3} \text{ atm} \)
- \( P_{CO_2} = P = \frac{1}{3} \text{ atm} \)

Now substituting these into the \( K_p \) expression:
\[ K_p = \left(\frac{2}{3}\right)^2 \cdot \left(\frac{1}{3}\right) \]
\[ K_p = \frac{4}{9} \cdot \frac{1}{3} = \frac{4}{27} \]

### Final Answer
Thus, the value of \( K_p \) is:
\[ K_p = \frac{4}{27} \]