Question

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 \(K_{p}=\) \(4.34 \times 10^{-3}\) 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?

Short answer

The masses of N2 and H2 in the equilibrium mixture are approximately 0.865 g and 0.185 g, respectively. The initial mass of NH3 placed in the vessel is approximately 1.58 g. The total pressure in the vessel is approximately 8.93 atm.

Step-by-step solution

(a) Calculating equilibrium amounts of N2 and H2

1. Write down the balanced chemical equation: N2(g) + 3 H2(g) ↔ 2 NH3(g) 2. Calculate the moles of NH3 at equilibrium: We are given that the equilibrium mixture contains 1.05 g of NH3. To find the moles of NH3, we can use the molar mass of NH3, which is approximately 17 g/mol. \[\text{moles of} \hspace{2mm}NH_{3} = \frac{1.05 \hspace{2mm}g}{17 \hspace{2mm}g/mol} \approx 0.0618 \hspace{2mm}mol\] 3. Define the change in moles for the reaction: Let the change in moles of N2 and H2 be x and 3x, respectively. For every x moles of N2 and 3x moles H2 consumed, 2x moles of NH3 are produced at equilibrium. 4. Calculate the equilibrium moles of N2 and H2: Since we know the moles of NH3 at equilibrium, we can use the stoichiometry of the equation to find x and then calculate the moles of N2 and H2 at equilibrium. \[0.0618 \hspace{2mm}mol \hspace{2mm}NH_{3} = 2x\] \[x = \frac{0.0618}{2} = 0.0309\] The equilibrium moles of N2 and H2 are: \[\text{N}_{2} \hspace{2mm}mol=\hspace{2mm}0.0309, \hspace{2mm}\text{H}_{2} \hspace{2mm}mol=\hspace{2mm}3 \times 0.0309 = 0.0927\] 5. Convert moles to mass: Now, we can find the masses of N2 and H2 in the equilibrium mixture by multiplying the moles by their respective molar masses: Mass of N2 = 0.0309 mol × 28 g/mol ≈ 0.865 g Mass of H2 = 0.0927 mol × 2 g/mol ≈ 0.185 g So, the masses of N2 and H2 in the equilibrium mixture are approximately 0.865 g and 0.185 g, respectively.

(b) Calculating the initial mass of NH3

1. Calculate moles of NH3 initially present: Initial moles of NH3 = Moles at equilibrium + Change in moles Initial moles of NH3 = 0.0618 + 0.0309 = 0.0927 mol 2. Convert moles to mass: Initial mass = Initial moles × Molar mass Initial mass = 0.0927 mol × 17 g/mol ≈ 1.58 g The initial mass of NH3 placed in the vessel is approximately 1.58 g.

(c) Calculating the total pressure in the vessel

1.To find the total pressure, we will use Kp given in the exercise and the stoichiometry of the reaction: \[K_{p} = \frac{(P_{NH_{3}})^{2}}{P_{N_{2}} \times (P_{H_{2}})^{3}}\] 2. Calculate partial pressures using moles and volume: Since we have the moles of each substance at equilibrium and the volume of the flask, we can calculate their partial pressures using the ideal gas law. \(P = \frac{nRT}{V}\) (Note: T = 300°C + 273.15 = 573.15 K, R = 0.08206 L·atm/mol·K) Partial pressures of N2, H2, and NH3 at equilibrium are: \(P_{N_{2}} = \frac{0.0309 \times 0.08206\times 573.15}{1} \approx 1.49 \hspace{2mm}atm\) \(P_{H_{2}} = \frac{0.0927 \times 0.08206\times 573.15}{1} \approx 4.47 \hspace{2mm}atm\) \(P_{NH_{3}} = \frac{0.0618 \times 0.08206\times 573.15}{1} \approx 2.97 \hspace{2mm}atm\) 3. Solve for total pressure (Pt): Using Kp value: \(4.34 \times 10^{-3} = \frac{(P_{NH_{3}})^{2}}{P_{N_{2}} \times (P_{H_{2}})^{3}}\) (Note: We will use the calculated partial pressures.) \(4.34 \times 10^{-3} = \frac{(2.97)^{2}}{1.49 \times (4.47)^{3}}\) Now, we can solve for the total pressure (Pt), knowing that it is the sum of partial pressures: \(Pt = P_{N_{2}} + P_{H_{2}} + P_{NH_{3}} = 1.49 + 4.47 + 2.97 \approx 8.93 \hspace{2mm}atm\) The total pressure in the vessel is approximately 8.93 atm.