For the decomposition reaction `NH_(2)COONH_(4)(s) hArr2NH_(3)(g)+CO_(2)(g)`, the `K_(p)=3.2xx10^(-5)atm^(3)`. The total pressure of gases at equilibrium when `1.0` mol of `NH_(2)COONH_(4)(s)` was taken to start with will be:

Options

A. `0.25atm`
B. `0.12atm`
C. `0.04atm`
D. `0.06atm`

Explanation

To solve the problem, we need to analyze the decomposition reaction and apply the equilibrium constant expression for the gases produced. Here’s a step-by-step solution:

### Step 1: Write the Reaction
The decomposition reaction is given as:
\[ \text{NH}_2\text{COONH}_4 (s) \rightleftharpoons 2 \text{NH}_3 (g) + \text{CO}_2 (g) \]

### Step 2: Identify the Change in Moles of Gas
From the reaction, we see that 1 mole of solid ammonium carbamate decomposes to produce 2 moles of ammonia gas and 1 mole of carbon dioxide gas. Therefore, the total moles of gas produced are:
\[ 2 \text{ moles of } \text{NH}_3 + 1 \text{ mole of } \text{CO}_2 = 3 \text{ moles of gas} \]

### Step 3: Set Up the Equilibrium Expression
The equilibrium constant \( K_p \) for the reaction is given as:
\[ K_p = \frac{(P_{\text{NH}_3})^2 \cdot (P_{\text{CO}_2})}{1} \]
Where \( P_{\text{NH}_3} \) is the partial pressure of ammonia and \( P_{\text{CO}_2} \) is the partial pressure of carbon dioxide.

### Step 4: Define Partial Pressures
Let \( P \) be the partial pressure of \( \text{CO}_2 \) at equilibrium. Then, the partial pressure of \( \text{NH}_3 \) will be \( 2P \) (since there are 2 moles of \( \text{NH}_3 \) for every mole of \( \text{CO}_2 \)).

### Step 5: Substitute into the Equilibrium Expression
Substituting the partial pressures into the \( K_p \) expression:
\[ K_p = \frac{(2P)^2 \cdot P}{1} = 4P^3 \]

### Step 6: Solve for \( P \)
We know that \( K_p = 3.2 \times 10^{-5} \, \text{atm}^3 \):
\[ 4P^3 = 3.2 \times 10^{-5} \]
\[ P^3 = \frac{3.2 \times 10^{-5}}{4} = 0.8 \times 10^{-5} = 8.0 \times 10^{-6} \]
\[ P = (8.0 \times 10^{-6})^{1/3} \]

Calculating \( P \):
\[ P \approx 0.0193 \, \text{atm} \]

### Step 7: Calculate Total Pressure at Equilibrium
The total pressure \( P_{\text{total}} \) at equilibrium is given by:
\[ P_{\text{total}} = P_{\text{NH}_3} + P_{\text{CO}_2} = 2P + P = 3P \]
Substituting the value of \( P \):
\[ P_{\text{total}} = 3 \times 0.0193 \, \text{atm} = 0.0579 \, \text{atm} \]

### Step 8: Round the Total Pressure
Rounding \( 0.0579 \, \text{atm} \) gives approximately:
\[ P_{\text{total}} \approx 0.0582 \, \text{atm} \]

### Final Answer
The total pressure of gases at equilibrium is approximately:
\[ \boxed{0.06 \, \text{atm}} \]