When heated , ammonium carbamate decomposes as follows `NH_4 COONH_2 (s) hArr 2NH_3(g)+CO-2(g)` At a certain temperature , the equilibrium pressure of the system is 0.318 atm , `K_p` for the reaction is

Options

A. 0.128
B. 1.146
C. `4.76xx10^(-3)`
D. `2.24xx10^(-2)`

Answer

To find the equilibrium constant \( K_p \) for the reaction

\[ NH_4 COONH_2 (s) \rightleftharpoons 2NH_3(g) + CO_2(g) \]

given that the equilibrium pressure of the system is 0.318 atm, we can follow these steps:

### Step 1: Understand the Reaction
The decomposition of ammonium carbamate produces 2 moles of ammonia gas and 1 mole of carbon dioxide gas.

### Step 2: Set Up the Equilibrium Expression
The equilibrium constant \( K_p \) for the reaction can be expressed in terms of the partial pressures of the gases:

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

Since ammonium carbamate is a solid, its concentration does not appear in the expression.

### Step 3: Determine Partial Pressures
Let the partial pressure of \( NH_3 \) be \( P_{NH_3} \) and the partial pressure of \( CO_2 \) be \( P_{CO_2} \).

From the stoichiometry of the reaction:
- For every 2 moles of \( NH_3 \), there is 1 mole of \( CO_2 \).
- Therefore, if we denote the partial pressure of \( CO_2 \) as \( P \), then the partial pressure of \( NH_3 \) will be \( 2P \).

### Step 4: Total Pressure
The total pressure at equilibrium is given as 0.318 atm:

\[ P_{total} = P_{NH_3} + P_{CO_2} = 2P + P = 3P \]

Setting this equal to the total pressure:

\[ 3P = 0.318 \text{ atm} \]

### Step 5: Solve for \( P \)
To find \( P \):

\[ P = \frac{0.318}{3} = 0.106 \text{ atm} \]

### Step 6: Calculate Partial Pressures
Now we can find the partial pressures:
- \( P_{CO_2} = P = 0.106 \text{ atm} \)
- \( P_{NH_3} = 2P = 2 \times 0.106 = 0.212 \text{ atm} \)

### Step 7: Substitute into the \( K_p \) Expression
Now substitute these values into the \( K_p \) expression:

\[ K_p = \frac{(P_{NH_3})^2 \cdot (P_{CO_2})}{1} = \frac{(0.212)^2 \cdot (0.106)}{1} \]

Calculating this gives:

\[ K_p = (0.212)^2 \cdot (0.106) = 0.044944 \cdot 0.106 \approx 0.00476 \text{ atm}^3 \]

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

\[ K_p \approx 4.76 \times 10^{-3} \text{ atm}^3 \]

---