For the decomposition of the compound, represented as
`NH_(2)COONH_(4)(s)hArr 2NH_(3)(g)+CO_(2)(g)` the `_K(p)=2.9xx10^(-5)" atm"^(3)`. If the reaction is started with 1 mol of the compound, the total pressure equilibrium would be:

Options

A. `7.66xx10^(-2)" atm"`
B. `38.8xx10^(-2)" atm"`
C. `5.82xx10^(-2)" atm"`
D. `1.94xx10^(-2)" atm"`

Answer

To solve the problem, we need to determine the total pressure at equilibrium for the decomposition reaction of the compound \( NH_2COONH_4(s) \) into \( 2NH_3(g) + CO_2(g) \). We are given the equilibrium constant \( K_p = 2.9 \times 10^{-5} \, \text{atm}^3 \) and that we start with 1 mole of the solid compound.

### Step-by-Step Solution:

1. **Write the balanced reaction:**
\[ NH_2COONH_4(s) \rightleftharpoons 2NH_3(g) + CO_2(g) \]

2. **Set up the initial conditions:**
- Initial moles of \( NH_2COONH_4 = 1 \) mol (solid, does not contribute to pressure)
- Initial moles of \( NH_3 = 0 \)
- Initial moles of \( CO_2 = 0 \)

3. **Define the change in moles at equilibrium:**
Let \( x \) be the amount of \( NH_2COONH_4 \) that decomposes.
- At equilibrium:
- Moles of \( NH_2COONH_4 = 1 - x \)
- Moles of \( NH_3 = 2x \)
- Moles of \( CO_2 = x \)

4. **Calculate total moles at equilibrium:**
The total moles of gas at equilibrium:
\[ \text{Total moles} = 2x + x = 3x \]

5. **Express the partial pressures:**
The partial pressures of the gases at equilibrium can be expressed as:
- \( P_{NH_3} = \frac{2x}{3x} P_{total} = \frac{2}{3} P_{total} \)
- \( P_{CO_2} = \frac{x}{3x} P_{total} = \frac{1}{3} P_{total} \)

6. **Write the expression for \( K_p \):**
\[ K_p = \frac{(P_{NH_3})^2 \cdot (P_{CO_2})}{1} = \left(\frac{2}{3} P_{total}\right)^2 \cdot \left(\frac{1}{3} P_{total}\right) \]
Simplifying this gives:
\[ K_p = \frac{4}{27} P_{total}^3 \]

7. **Substitute \( K_p \) and solve for \( P_{total} \):**
Set the expression equal to the given \( K_p \):
\[ 2.9 \times 10^{-5} = \frac{4}{27} P_{total}^3 \]
Rearranging gives:
\[ P_{total}^3 = \frac{2.9 \times 10^{-5} \times 27}{4} \]
\[ P_{total}^3 = \frac{7.83 \times 10^{-5}}{4} = 1.9575 \times 10^{-5} \]

8. **Calculate \( P_{total} \):**
\[ P_{total} = (1.9575 \times 10^{-5})^{1/3} \]
Calculating this gives:
\[ P_{total} \approx 5.80 \times 10^{-2} \, \text{atm} \]

### Final Answer:
The total pressure at equilibrium is approximately \( 5.80 \times 10^{-2} \, \text{atm} \).