When heated , ammonium carbamate decomate decompoes as follows :
`NH_(4)COOH_(2)(s) gives 2NH_(3)(g)+CO_(2)(g)`
At a certain temperature , the equilibrium pressure of the system is `0.318atm`. `Kp` for the reaction is:

Options

A. `0.128`
B. `0.426`
C. `4.76xx10^(-3)`
D. none of these

Answer

To find the equilibrium constant \( K_p \) for the decomposition of ammonium carbamate, we can follow these steps:

### Step-by-Step Solution:

1. **Write the Balanced Chemical Equation**:
The decomposition of ammonium carbamate can be represented as:
\[ \text{NH}_4\text{CO}_2\text{NH}_2(s) \rightleftharpoons 2\text{NH}_3(g) + \text{CO}_2(g) \]

2. **Identify the Equilibrium Pressures**:
At equilibrium, the total pressure of the system is given as \( P_{total} = 0.318 \, \text{atm} \).

3. **Define the Changes in Moles**:
Let \( x \) be the amount of ammonium carbamate that decomposes. The changes in moles can be represented as:
- Moles of \( \text{NH}_3 \) produced = \( 2x \)
- Moles of \( \text{CO}_2 \) produced = \( x \)
- Total moles at equilibrium = \( 2x + x = 3x \)

4. **Calculate the Mole Fractions**:
The mole fractions of the gases at equilibrium can be calculated as:
- Mole fraction of \( \text{NH}_3 \) = \( \frac{2x}{3x} = \frac{2}{3} \)
- Mole fraction of \( \text{CO}_2 \) = \( \frac{x}{3x} = \frac{1}{3} \)

5. **Calculate the Partial Pressures**:
The partial pressures of the gases can be calculated using the mole fractions and the total pressure:
- Partial pressure of \( \text{NH}_3 \):
\[ P_{\text{NH}_3} = \left(\frac{2}{3}\right) \times 0.318 \, \text{atm} = 0.212 \, \text{atm} \]
- Partial pressure of \( \text{CO}_2 \):
\[ P_{\text{CO}_2} = \left(\frac{1}{3}\right) \times 0.318 \, \text{atm} = 0.106 \, \text{atm} \]

6. **Calculate \( K_p \)**:
The expression for \( K_p \) for the reaction is given by:
\[ K_p = \frac{(P_{\text{NH}_3})^2 \cdot (P_{\text{CO}_2})}{1} \]
Substituting the values:
\[ K_p = (0.212)^2 \cdot (0.106) = 0.045024 \cdot 0.106 = 0.004771 \, \text{atm}^3 \]

7. **Final Result**:
Thus, we can express \( K_p \) in scientific notation:
\[ K_p \approx 4.77 \times 10^{-3} \, \text{atm}^3 \]