Ammonium carbamate decomposes as :
`NH_(2)COONH_(4) (s) rarr 2NH_(3)(g) + CO_(2)(g)`
For the reaction, `K_(P) = 2.9 xx 10^(-5) atm^(3)` If we start with 1 mole of the compound, the total pressure at equilibrium would be

Options

A. 0.766 atm
B. 0.0582 atm
C. 0.0388 atm
D. 0.0194 atm

Explanation

To solve the problem, we will follow these steps:

### Step 1: Write the balanced chemical equation
The decomposition of ammonium carbamate is given by:
\[ \text{NH}_2\text{COONH}_4 (s) \rightleftharpoons 2\text{NH}_3 (g) + \text{CO}_2 (g) \]

### Step 2: Define the initial conditions
We start with 1 mole of ammonium carbamate (NH₂COONH₄), which is a solid and does not contribute to the equilibrium expression. Therefore, at the start (t = 0):
- Moles of NH₂COONH₄ = 1
- Moles of NH₃ = 0
- Moles of CO₂ = 0

### Step 3: Set up the change in moles at equilibrium
Let \( x \) be the amount of ammonium carbamate that decomposes at equilibrium. Thus, at equilibrium:
- Moles of NH₂COONH₄ = \( 1 - x \) (but this does not affect Kp)
- Moles of NH₃ = \( 2x \)
- Moles of CO₂ = \( x \)

### Step 4: Calculate the total moles at equilibrium
The total number of moles of gases at equilibrium is:
\[ \text{Total moles} = 2x + x = 3x \]

### Step 5: Express partial pressures in terms of total pressure (P)
The partial pressure of NH₃ is given by:
\[ P_{\text{NH}_3} = \frac{2x}{3x} \cdot P = \frac{2}{3} P \]

The partial pressure of CO₂ is given by:
\[ P_{\text{CO}_2} = \frac{x}{3x} \cdot P = \frac{1}{3} P \]

### Step 6: Write the expression for Kp
The equilibrium constant \( K_p \) is given by:
\[ K_p = \frac{(P_{\text{NH}_3})^2 \cdot (P_{\text{CO}_2})}{1} \]
Substituting the expressions for partial pressures:
\[ K_p = \left(\frac{2}{3} P\right)^2 \cdot \left(\frac{1}{3} P\right) \]
\[ K_p = \frac{4}{9} P^2 \cdot \frac{1}{3} P = \frac{4}{27} P^3 \]

### Step 7: Set Kp equal to the given value and solve for P
Given \( K_p = 2.9 \times 10^{-5} \):
\[ \frac{4}{27} P^3 = 2.9 \times 10^{-5} \]
Multiplying both sides by 27:
\[ 4 P^3 = 27 \times 2.9 \times 10^{-5} \]
\[ 4 P^3 = 7.83 \times 10^{-4} \]
Now, divide by 4:
\[ P^3 = \frac{7.83 \times 10^{-4}}{4} = 1.9575 \times 10^{-4} \]

### Step 8: Take the cube root to find P
Taking the cube root:
\[ P = \sqrt[3]{1.9575 \times 10^{-4}} \approx 0.0582 \, \text{atm} \]

### Final Answer
The total pressure at equilibrium is approximately \( 0.0582 \, \text{atm} \).

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