For the decomposition reaction
`NH_2COONH_4 (s) hArr 2NH_3(g) +CO_2(g)`
The `K_p=2.9 xx 10^(-5) atm^3`. The total pressure of gases at equilibrium when 1 mole of `NH_2COONH_4` (s) was taken to start with would be

Options

A. 0.0194 atm
B. 0.0388 atm
C. 0.0582 atm
D. 0.0766 atm

Explanation

To solve the problem, we need to analyze the decomposition reaction of ammonium formate (NH2COONH4) into ammonia (NH3) and carbon dioxide (CO2). The equilibrium constant \( K_p \) is given, and we need to find the total pressure of gases at equilibrium when starting with 1 mole of the solid reactant.

### Step-by-Step Solution:

1. **Write the balanced chemical equation:**
\[ \text{NH}_2\text{COONH}_4 (s) \rightleftharpoons 2 \text{NH}_3 (g) + \text{CO}_2 (g) \]

2. **Set up the initial conditions:**
- Initially, we have 1 mole of NH2COONH4 (solid), and no products:
\[ \text{Initial: } [\text{NH}_2\text{COONH}_4] = 1 \text{ mole}, \quad [\text{NH}_3] = 0, \quad [\text{CO}_2] = 0 \]

3. **Define the change in moles at equilibrium:**
- Let \( x \) be the amount of NH2COONH4 that decomposes. At equilibrium:
\[ [\text{NH}_2\text{COONH}_4] = 1 - x, \quad [\text{NH}_3] = 2x, \quad [\text{CO}_2] = x \]

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

5. **Express the total pressure in terms of \( x \):**
- Let \( P \) be the total pressure at equilibrium. The partial pressures can be expressed as:
\[ P_{\text{NH}_3} = \frac{2}{3}P, \quad P_{\text{CO}_2} = \frac{1}{3}P \]

6. **Write the expression for \( K_p \):**
- The equilibrium constant \( K_p \) is given by:
\[ K_p = \frac{(P_{\text{NH}_3})^2 \cdot (P_{\text{CO}_2})}{1} = \left(\frac{2}{3}P\right)^2 \cdot \left(\frac{1}{3}P\right) \]
- Simplifying this gives:
\[ K_p = \frac{4}{9}P^2 \cdot \frac{1}{3}P = \frac{4}{27}P^3 \]

7. **Substitute the given \( K_p \) value and solve for \( P \):**
- We know \( K_p = 2.9 \times 10^{-5} \):
\[ \frac{4}{27}P^3 = 2.9 \times 10^{-5} \]
- Rearranging gives:
\[ P^3 = \frac{2.9 \times 10^{-5} \cdot 27}{4} \]
- Calculate \( P^3 \):
\[ P^3 = \frac{2.9 \times 27 \times 10^{-5}}{4} = \frac{78.3 \times 10^{-5}}{4} = 19.575 \times 10^{-6} \]
- Taking the cube root:
\[ P = \sqrt[3]{19.575 \times 10^{-6}} \approx 0.0580 \text{ atm} \]

8. **Final answer:**
- The total pressure of gases at equilibrium is approximately:
\[ P \approx 0.0580 \text{ atm} \]