For the decomposition of the compound, represented as NH$$_2$$COONH$$_4$$(s) $$\rightleftharpoons$$ 2NH$$_3$$(g) + CO$$_2$$(g), the K$$_p$$ = $$2.9 \times 10^{-5}$$ atm$$^3$$. If the reaction is started with 1 mole of the compound, the total pressure at equilibrium would be:

We are given the decomposition reaction: NH2COONH4(s) ⇌ 2NH3(g) + CO2(g) with Kp = $$2.9 \times 10^{-5}$$ atm3. We start with 1 mole of NH2COONH4(s). Since the solid does not affect the equilibrium constant, we focus on the gaseous products.

Let the number of moles of NH2COONH4 that decompose be $$x$$. At equilibrium:

• Moles of NH3 produced = $$2x$$
• Moles of CO2 produced = $$x$$
• Total moles of gas, $$n_{\text{total}} = 2x + x = 3x$$

Let the total pressure at equilibrium be $$P$$ atm. The partial pressures are:

• Partial pressure of NH3, $$P_{\text{NH}_3} = \left( \frac{2x}{3x} \right) P = \frac{2}{3}P$$
• Partial pressure of CO2, $$P_{\text{CO}_2} = \left( \frac{x}{3x} \right) P = \frac{1}{3}P$$

The equilibrium constant Kp is given by:

$$K_p = (P_{\text{NH}_3})^2 \times (P_{\text{CO}_2})$$

Substituting the partial pressures:

$$K_p = \left( \frac{2}{3}P \right)^2 \times \left( \frac{1}{3}P \right) = \frac{4}{9}P^2 \times \frac{1}{3}P = \frac{4}{27}P^3$$

Given that Kp = $$2.9 \times 10^{-5}$$ atm3, we set up the equation:

$$\frac{4}{27}P^3 = 2.9 \times 10^{-5}$$

Solving for $$P^3$$:

$$P^3 = 2.9 \times 10^{-5} \times \frac{27}{4}$$

First, compute $$\frac{27}{4} = 6.75$$:

$$P^3 = 2.9 \times 10^{-5} \times 6.75$$

Now, multiply 2.9 by 6.75:

$$2.9 \times 6.75 = 2.9 \times (6 + 0.75) = (2.9 \times 6) + (2.9 \times 0.75) = 17.4 + 2.175 = 19.575$$

So:

$$P^3 = 19.575 \times 10^{-5} = 1.9575 \times 10^{-4}$$

Take the cube root of both sides to find $$P$$:

$$P = \sqrt[3]{1.9575 \times 10^{-4}}$$

We can write this as:

$$P = (1.9575 \times 10^{-4})^{1/3} = (1.9575)^{1/3} \times (10^{-4})^{1/3} = (1.9575)^{1/3} \times 10^{-4/3}$$

Compute $$10^{-4/3} = 10^{-1.3333} \approx 0.046416$$ (since $$10^{-4/3} = 1 / 10^{4/3}$$ and $$10^{4/3} \approx 21.5443$$, so $$1 / 21.5443 \approx 0.046416$$).

Now, compute the cube root of 1.9575. We know that $$1.25^3 = 1.25 \times 1.25 = 1.5625$$, then $$1.5625 \times 1.25 = 1.953125$$, which is close to 1.9575. Adjusting slightly, $$1.251^3 = 1.251 \times 1.251 = 1.565001$$, then $$1.565001 \times 1.251 \approx 1.957816251$$, which is very close to 1.9575. So $$(1.9575)^{1/3} \approx 1.251$$.

Thus:

$$P \approx 1.251 \times 0.046416 \approx 0.05806$$

So:

$$P \approx 0.05806 \text{ atm} = 5.806 \times 10^{-2} \text{ atm}$$

Comparing with the options:

• A: $$7.66 \times 10^{-2}$$ atm
• B: $$38.8 \times 10^{-2}$$ atm
• C: $$5.82 \times 10^{-2}$$ atm
• D: $$1.94 \times 10^{-2}$$ atm

The value $$5.806 \times 10^{-2}$$ atm is closest to option C ($$5.82 \times 10^{-2}$$ atm). To verify, let's check with option C:

If $$P = 5.82 \times 10^{-2} = 0.0582$$ atm, then:

$$P^3 = (0.0582)^3 = 0.0582 \times 0.0582 = 0.00338724, \text{ then } 0.00338724 \times 0.0582 \approx 0.00019713768 = 1.9713768 \times 10^{-4}$$

Then:

$$\frac{4}{27} \times 1.9713768 \times 10^{-4} = \frac{4 \times 1.9713768 \times 10^{-4}}{27} = \frac{7.8855072 \times 10^{-4}}{27} \approx 2.920558 \times 10^{-5} \approx 2.92 \times 10^{-5}$$

This is very close to the given Kp = $$2.9 \times 10^{-5}$$ (difference due to rounding).

Hence, the correct answer is Option C.