Ammonium carbamate decomposes as :
`NH_(2)COONH_(4) 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. 0.388 atm
D. 0.0194 atm

Answer

To solve the problem of finding the total pressure at equilibrium for the decomposition of ammonium carbamate, we can follow these steps:

### Step 1: Write the balanced equation and identify the initial conditions.
The decomposition of ammonium carbamate can be represented as:
\[ \text{NH}_2\text{COONH}_4 (s) \rightarrow 2 \text{NH}_3 (g) + \text{CO}_2 (g) \]

Initially, we have:
- Moles of NH2COONH4 = 1 mole (solid, does not contribute to pressure)
- Moles of NH3 = 0
- Moles of CO2 = 0

### Step 2: Define the change in moles at equilibrium.
Let \( x \) be the amount of ammonium carbamate that decomposes. At equilibrium, we will have:
- Moles of NH3 = \( 2x \)
- Moles of CO2 = \( x \)

Since we started with 1 mole of ammonium carbamate, the amount remaining will be:
- Moles of NH2COONH4 = \( 1 - x \)

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

### Step 4: Write the expression for the equilibrium constant \( K_p \).
The expression for \( K_p \) is given by:
\[ K_p = \frac{(P_{\text{NH}_3})^2 \cdot (P_{\text{CO}_2})}{1} \]

Where:
- \( P_{\text{NH}_3} = \frac{2P}{3} \)
- \( P_{\text{CO}_2} = \frac{P}{3} \)

### Step 5: Substitute the pressures into the \( K_p \) expression.
Substituting the pressures into the \( K_p \) expression gives:
\[ K_p = \left(\frac{2P}{3}\right)^2 \cdot \left(\frac{P}{3}\right) \]
\[ K_p = \frac{4P^2}{9} \cdot \frac{P}{3} = \frac{4P^3}{27} \]

### Step 6: Set up the equation using the given \( K_p \) value.
We know that:
\[ K_p = 2.9 \times 10^{-5} \]
So, we can set up the equation:
\[ 2.9 \times 10^{-5} = \frac{4P^3}{27} \]

### Step 7: Solve for \( P^3 \).
Rearranging gives:
\[ P^3 = \frac{2.9 \times 10^{-5} \cdot 27}{4} \]
\[ P^3 = 1.9575 \times 10^{-4} \]

### Step 8: Calculate \( P \).
Taking the cube root:
\[ P = \left(1.9575 \times 10^{-4}\right)^{1/3} \]
Calculating this gives:
\[ P \approx 5.82 \times 10^{-2} \text{ atm} \]

### Final Answer:
The total pressure at equilibrium is approximately:
\[ P \approx 0.0582 \text{ atm} \]

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