Ammonium carbamate decomposes as `NH_(2)COONH_(4)(s) hArr 2NH_(3)(g) +CO_(2)(g)` . The value of `K_(p)` for the reaction is `2.9 xx 10^(-5) atm^(3)`. If we start the reaction with 1 mole of the compound, the total pressure at equilibrium would be

Options

A. 0.0766 atm
B. 0.0194 atm
C. 0.194 atm
D. 0.0582 atm

Explanation

To solve the problem, we need to determine the total pressure at equilibrium for the decomposition of ammonium carbamate. Let's break it down step by step.

### 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: Identify initial conditions
We start with 1 mole of solid ammonium carbamate:
- Initial moles of NH2COONH4 = 1 mole
- Initial moles of NH3 = 0
- Initial moles of CO2 = 0

### Step 3: Define change in moles at equilibrium
Let \( x \) be the amount of ammonium carbamate that decomposes. At equilibrium:
- Moles of NH2COONH4 = \( 1 - x \)
- Moles of NH3 = \( 2x \) (since 2 moles of NH3 are produced for every mole of ammonium carbamate decomposed)
- Moles of CO2 = \( x \)

### Step 4: Calculate total moles at equilibrium
The total moles of gas at equilibrium will be:
\[ \text{Total moles} = 2x + x = 3x \]

### Step 5: Express partial pressures in terms of \( x \)
Using the ideal gas law, we can express the partial pressures in terms of \( x \):
- Partial pressure of NH3, \( P_{\text{NH3}} = \frac{2x}{3x} P = \frac{2}{3} P \)
- Partial pressure of CO2, \( P_{\text{CO2}} = \frac{x}{3x} P = \frac{1}{3} P \)

### Step 6: Write the expression for \( K_p \)
The equilibrium constant \( K_p \) is given by:
\[ K_p = \frac{(P_{\text{NH3}})^2 \cdot (P_{\text{CO2}})}{1} \]
Substituting the expressions for partial pressures:
\[ K_p = \frac{\left(\frac{2}{3} P\right)^2 \cdot \left(\frac{1}{3} P\right)}{1} \]
\[ K_p = \frac{4}{9} P^2 \cdot \frac{1}{3} P = \frac{4}{27} P^3 \]

### Step 7: Set up the equation using the given \( K_p \)
We know that \( K_p = 2.9 \times 10^{-5} \, \text{atm}^3 \):
\[ \frac{4}{27} P^3 = 2.9 \times 10^{-5} \]

### Step 8: Solve for \( P^3 \)
Rearranging gives:
\[ P^3 = \frac{2.9 \times 10^{-5} \cdot 27}{4} \]
\[ P^3 = \frac{7.83 \times 10^{-4}}{4} \]
\[ P^3 = 1.9575 \times 10^{-4} \]

### Step 9: Calculate \( P \)
Taking the cube root:
\[ P = \sqrt[3]{1.9575 \times 10^{-4}} \]
\[ P \approx 0.0581 \, \text{atm} \]

### Step 10: Calculate total pressure at equilibrium
The total pressure at equilibrium is:
\[ P_{\text{total}} = 3P \]
\[ P_{\text{total}} = 3 \times 0.0581 \]
\[ P_{\text{total}} \approx 0.1743 \, \text{atm} \]

### Final Answer
The total pressure at equilibrium is approximately \( 0.0581 \, \text{atm} \).