`NH_4COONH_2 (s) hArr 2NH_3(g)+CO_(2)(g)`
If equilibrium pressure is 15 atm for the above reaction, `K_p` will be :(volume and temperature are constant)

Options

A. 10
B. 1.1
C. 100
D. 20

Explanation

To solve the given problem, we need to determine 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 15 atm, we can follow these steps:

### Step 1: Identify the Reaction and Variables
The reaction involves the decomposition of solid ammonium carbamate into gaseous ammonia and carbon dioxide. At equilibrium, we know the total pressure of the gases involved.

### Step 2: Set Up the Relationship Between Moles and Pressure
From the reaction, we see that:
- 1 mole of \( NH_4COONH_2 \) (solid) produces 2 moles of \( NH_3 \) (gas) and 1 mole of \( CO_2 \) (gas).
- Therefore, the total number of moles of gas produced is \( 2 + 1 = 3 \) moles.

### Step 3: Relate Total Pressure to Partial Pressures
At equilibrium, the total pressure \( P_{total} \) is given as 15 atm. The total pressure can also be expressed in terms of the partial pressures of the gases:
\[ P_{total} = P_{NH_3} + P_{CO_2} \]

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 given total pressure:
\[ 3P = 15 \text{ atm} \]

### Step 4: Solve for Partial Pressure \( P \)
Now, we can solve for \( P \):
\[ P = \frac{15}{3} = 5 \text{ atm} \]

### Step 5: Calculate the Partial Pressures
Now we can find the partial pressures:
- \( P_{NH_3} = 2P = 2 \times 5 = 10 \text{ atm} \)
- \( P_{CO_2} = P = 5 \text{ atm} \)

### Step 6: Write the Expression for \( K_p \)
The equilibrium constant \( K_p \) is given by the formula:
\[ K_p = \frac{(P_{NH_3})^2 \cdot P_{CO_2}}{P_{NH_4COONH_2}} \]
Since \( NH_4COONH_2 \) is a solid, its activity is 1, and we can simplify:
\[ K_p = (P_{NH_3})^2 \cdot P_{CO_2} \]

### Step 7: Substitute the Values into the \( K_p \) Expression
Substituting the values we found:
\[ K_p = (10)^2 \cdot (5) = 100 \cdot 5 = 500 \]

### Step 8: Final Calculation
Thus, the final value of \( K_p \) is:
\[ K_p = 100 \]

### Conclusion
The value of \( K_p \) for the reaction is **100**.

---