For the reaction
`NH_(2)COONH_(4)(g)hArr 2NH_(3)(g)+CO_(2)(g)` the equilibrium constant `K_(p)=2.92xx10^(-5)atm^(3)`. The total pressure of the gaseous products when 1 mole of reactant is heated, will be

Options

A. 0.0194 atm
B. 0.0388 atm
C. 0.058 atm
D. 0.0667 atm

Answer

To solve the problem, we need to analyze the given reaction and the equilibrium constant. The reaction is:

\[ \text{NH}_2\text{COONH}_4 (g) \rightleftharpoons 2 \text{NH}_3 (g) + \text{CO}_2 (g) \]

The equilibrium constant \( K_p \) is given as \( 2.92 \times 10^{-5} \, \text{atm}^3 \).

### Step-by-Step Solution:

1. **Identify Initial Conditions:**
- We start with 1 mole of the reactant \( \text{NH}_2\text{COONH}_4 \).
- At the beginning (time \( t = 0 \)), we have:
- Moles of \( \text{NH}_2\text{COONH}_4 = 1 \)
- Moles of \( \text{NH}_3 = 0 \)
- Moles of \( \text{CO}_2 = 0 \)

2. **Define Change in Moles:**
- Let \( x \) be the number of moles of \( \text{NH}_2\text{COONH}_4 \) that decomposes at equilibrium.
- At equilibrium, we will have:
- Moles of \( \text{NH}_2\text{COONH}_4 = 1 - x \)
- Moles of \( \text{NH}_3 = 2x \)
- Moles of \( \text{CO}_2 = x \)

3. **Write the Expression for \( K_p \):**
- The expression for the equilibrium constant \( K_p \) is given by:
\[ K_p = \frac{(P_{\text{NH}_3})^2 \cdot (P_{\text{CO}_2})}{1} = (2p)^2 \cdot (p) \]
where \( P_{\text{NH}_3} = 2p \) and \( P_{\text{CO}_2} = p \).

4. **Substitute Values into \( K_p \):**
- Substituting into the equation, we have:
\[ K_p = (2p)^2 \cdot p = 4p^2 \cdot p = 4p^3 \]
- Setting this equal to the given \( K_p \):
\[ 4p^3 = 2.92 \times 10^{-5} \]

5. **Solve for \( p \):**
- Rearranging gives:
\[ p^3 = \frac{2.92 \times 10^{-5}}{4} = 7.3 \times 10^{-6} \]
- Taking the cube root:
\[ p = (7.3 \times 10^{-6})^{1/3} \approx 1.94 \times 10^{-2} \, \text{atm} \]

6. **Calculate Total Pressure:**
- The total pressure \( P_{\text{total}} \) at equilibrium is the sum of the partial pressures:
\[ P_{\text{total}} = P_{\text{NH}_3} + P_{\text{CO}_2} = 2p + p = 3p \]
- Substituting the value of \( p \):
\[ P_{\text{total}} = 3 \times 1.94 \times 10^{-2} \approx 0.0582 \, \text{atm} \]

### Final Answer:
The total pressure of the gaseous products when 1 mole of reactant is heated will be approximately \( 0.0582 \, \text{atm} \).