`NH_4COONH_2 (s) hArr 2NH_3(g)+CO_(2)(g)`
If equilibrium pressure is 6 atm for the above reaction, `K_p` will be :(volume and temperature are constant)

Options

A. 16
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 6 atm, we can follow these steps:

### Step 1: Understand the Reaction
The reaction shows that 1 mole of ammonium carbamate decomposes into 2 moles of ammonia and 1 mole of carbon dioxide.

### Step 2: Determine the Total Moles at Equilibrium
At equilibrium, we have:
- 1 mole of \( NH_4COONH_2 \) (which is a solid and does not contribute to the pressure)
- 2 moles of \( NH_3 \)
- 1 mole of \( CO_2 \)

This gives a total of \( 2 + 1 = 3 \) moles of gas.

### Step 3: Relate Total Pressure to Partial Pressures
The total pressure at equilibrium is given as 6 atm. Since the total pressure is the sum of the partial pressures of the gases, we can express this as:

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

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, we can write:

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

Setting this equal to the total pressure:

\[ 3P = 6 \text{ atm} \]

### Step 4: Solve for Partial Pressures
From the equation \( 3P = 6 \), we can solve for \( P \):

\[ P = \frac{6}{3} = 2 \text{ atm} \]

Now we have:
- \( P_{NH_3} = 2P = 4 \text{ atm} \)
- \( P_{CO_2} = P = 2 \text{ atm} \)

### Step 5: Write the Expression for \( K_p \)
The equilibrium constant \( K_p \) is given by the expression:

\[ K_p = \frac{(P_{NH_3})^2 \cdot (P_{CO_2})}{(P_{NH_4COONH_2})} \]

Since \( NH_4COONH_2 \) is a solid, its activity is 1 and does not appear in the expression. Thus, we have:

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

### Step 6: Calculate \( K_p \)
Now we can calculate \( K_p \):

\[ K_p = \frac{16 \cdot 2}{1} = 32 \]

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

\[ \boxed{32} \]