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

Options

A. 1.7
B. 2.7
C. 8.7
D. 16

Explanation

To solve the equilibrium reaction \( NH_4COONH_2 (s) \rightleftharpoons 2NH_3(g) + CO_2(g) \) and find the equilibrium constant \( K_p \) when the equilibrium pressure is given as 2 atm, we can follow these steps:

### Step 1: Understand the Reaction
The reaction shows that 1 mole of ammonium carbamate (\( NH_4COONH_2 \)) decomposes into 2 moles of ammonia (\( NH_3 \)) and 1 mole of carbon dioxide (\( CO_2 \)).

### Step 2: Identify Moles of Gases at Equilibrium
At equilibrium, we have:
- 2 moles of \( NH_3 \)
- 1 mole of \( CO_2 \)

Thus, the total number of moles of gas produced is \( 2 + 1 = 3 \) moles.

### Step 3: Relate Total Pressure to Partial Pressures
Given that the total equilibrium pressure is 2 atm, we can express this in terms of the partial pressures of the gases:
- Let the partial pressure of \( CO_2 \) be \( P \).
- The partial pressure of \( NH_3 \) will be \( 2P \) (since there are 2 moles of \( NH_3 \)).

### Step 4: Write the Total Pressure Equation
The total pressure at equilibrium can be expressed as:
\[ P_{total} = P_{NH_3} + P_{CO_2} = 2P + P = 3P \]
Since the total pressure is given as 2 atm, we can set up the equation:
\[ 3P = 2 \text{ atm} \]

### Step 5: Solve for \( P \)
From the equation \( 3P = 2 \):
\[ P = \frac{2}{3} \text{ atm} \]

### Step 6: Calculate Partial Pressures
Now we can find the partial pressures:
- \( P_{NH_3} = 2P = 2 \times \frac{2}{3} = \frac{4}{3} \text{ atm} \)
- \( P_{CO_2} = P = \frac{2}{3} \text{ atm} \)

### Step 7: Write the Expression for \( K_p \)
The equilibrium constant \( K_p \) for the reaction is given by:
\[ K_p = \frac{(P_{NH_3})^2 \cdot P_{CO_2}}{P_{NH_4COONH_2}} \]
Since \( NH_4COONH_2 \) is a solid, its activity is considered to be 1. Therefore:
\[ K_p = (P_{NH_3})^2 \cdot P_{CO_2} \]

### Step 8: Substitute the Values
Substituting the values of the partial pressures:
\[ K_p = \left(\frac{4}{3}\right)^2 \cdot \left(\frac{2}{3}\right) \]
Calculating this:
\[ K_p = \frac{16}{9} \cdot \frac{2}{3} = \frac{32}{27} \]

### Step 9: Final Calculation
Calculating \( \frac{32}{27} \) gives approximately \( 1.185 \).

### Conclusion
Thus, the value of \( K_p \) for the reaction is approximately \( 1.185 \).