As shown in Table \(15.2,\) the equilibrium constant for the reaction $\mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \rightleftharpoons 2 \mathrm{NH}_{3}(g)\( is \)K_{p}=4.23 \times 10^{-7}\( at \)300^{\circ} \mathrm{C}\(. Pure \)\mathrm{NH}_{3}$ is placed in a 1.00-L flask and allowed to reach equilibrium at this temperature. There are $1.05 \mathrm{~g} \mathrm{NH}_{3}$ in the equilibrium mixture. (a) What are the masses of \(\mathrm{N}_{2}\) and \(\mathrm{H}_{2}\) in the equilibrium mixture? (b) What was the initial mass of ammonia placed in the vessel? (c) What is the total pressure in the vessel?

From the given mass of ammonia (1.05 g), we can calculate the moles of \(\mathrm{NH}_{3}\) using its molar mass. Moles of \(\mathrm{NH}_{3}\) = $$\frac{1.05 \ \text{g}}{17.03 \ \frac{\text{g}}{\mathrm{mol}}} = 0.0616 \ \mathrm{mol}$$

We can find the partial pressure of \(\mathrm{NH}_{3}\) in the equilibrium mixture by considering its mole fraction: Partial pressure of \(\mathrm{NH}_{3}\), \(P_{\mathrm{NH}_{3}} = \frac{0.0616 \ \mathrm{mol}}{1.00 \ \text{L}} = 0.0616 \ \mathrm{atm}\)

The expression for the equilibrium constant (\(K_p\)) considering the partial pressures of each reactant and product is: $$K_p = \frac{P_{\mathrm{NH}_{3}}^2}{P_{\mathrm{N}_{2}} \times P_{\mathrm{H}_{2}}^3}$$ Since \(K_p = 4.23 \times 10^{-7}\), we can write, $$4.23 \times 10^{-7} = \frac{P_{\mathrm{NH}_{3}}^2}{P_{\mathrm{N}_{2}} \times P_{\mathrm{H}_{2}}^3}$$

From the balanced chemical equation, we know that 1 mole of \(\mathrm{N}_{2}\) reacts with 3 moles of \(\mathrm{H}_{2}\) to produce 2 moles of \(\mathrm{NH}_{3}\). Therefore, we can create a table of partial pressures at equilibrium: | | N2 | 3H2 | 2NH3 | |---------------|----|-----|-------| | Initial | 0 | 0 | 0.0616 | | Change | x | 3x | -2x | | Equilibrium | x | 3x | 0.0616-2x | |-| Thus, $$K_p = \frac{(0.0616-2x)^2}{x \times (3x)^3}$$

Now we can solve for x: $$4.23 \times 10^{-7} = \frac{(0.0616-2x)^2}{27x^4}$$ By solving the equation for \(x\), we find that \(x = 0.00456\). From the table in Step 4, we know the partial pressures of \(\mathrm{N}_{2}\) and \(\mathrm{H}_{2}\) at equilibrium are \(x\) and \(3x\) respectively. Now, we can determine the masses of \(\mathrm{N}_{2}\) and \(\mathrm{H}_{2}\): Mass of N2 = Partial pressure of \(\mathrm{N}_{2}\) × Molar mass of \(\mathrm{N}_{2}\) = \(0.00456 \ \text{atm} \times \frac{28.02 \ \mathrm{g}}{\mathrm{mol}} = 0.128 \ \text{g}\) Mass of H2 = Partial pressure of \(\mathrm{H}_{2}\) × Molar mass of \(\mathrm{H}_{2}\) = \(3(0.00456) \ \text{atm} \times \frac{2 \ \mathrm{g}}{\mathrm{mol}} = 0.0274 \ \text{g}\) So, the masses of \(\mathrm{N}_{2}\) and \(\mathrm{H}_{2}\) in the equilibrium mixture are approximately 0.128 g and 0.0274 g, respectively. (Answer for part a)

The change in ammonia's partial pressure is -2x, which can be converted to mass by multiplying with the molar mass of ammonia: Change in mass of ammonia = \(-2x \times 17.03 \ \frac{\text{g}}{\mathrm{mol}} = -0.155 \ \text{g}\) The initial mass of \(\mathrm{NH}_{3}\) can be obtained by summing the change in mass with the final mass of \(\mathrm{NH}_{3}\): Initial mass of ammonia = \(1.05 \ \text{g} - 0.155 \ \text{g} = 0.895 \ \text{g}\) Thus, the initial mass of ammonia placed in the vessel was approximately 0.895 g. (Answer for part b)

The total pressure in the vessel can be found by summing the partial pressures of \(\mathrm{N}_{2}\), \(\mathrm{H}_{2}\), and \(\mathrm{NH}_{3}\): Total pressure = \(P_{\mathrm{N}_{2}} + P_{\mathrm{H}_{2}} + P_{\mathrm{NH}_{3}} = x + 3x + 0.0616 \ \text{atm} = 0.0752 \ \text{atm}\) So, the total pressure in the vessel is approximately 0.0752 atm. (Answer for part c)