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 \(K_{p}=4.34 \times 10^{-3} \mathrm{at} 300^{\circ} \mathrm{C}\) . Pure \(\mathrm{NH}_{3}\) is placed in a \(1.00-\mathrm{L}\) 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?

To calculate the equilibrium concentrations, we need to use the given equilibrium constant (\(K_p\)) and the stoichiometry of the reaction. Given that there are 1.05 g NH3 in the equilibrium mixture, we can calculate the moles of NH3 using its molar mass (17 g/mol): Moles of NH3 = mass/(molar mass) = \( \frac{1.05 \, g}{17 \frac{g}{mol}} = 0.0618 \, mol\) Since the reaction takes place in a 1.00 L flask, the concentration of NH3 at equilibrium is: \[ [\mathrm{NH}_3]_{eq} = \frac{0.0618 \, mol}{1.00 \, L} = 0.0618 \, M \] Now we can set up an ICE table for the reaction: Initial | Change | Equilibrium N2 0 -x x H2 0 -3x 3x NH3 0.0618 2x 0.0618 - 2x Since we know that: \[ K_p = \frac{[NH_3]_{eq}^2}{[N_2]_{eq} [H_2]_{eq}^3} = 4.34 \times 10^{-3} \], we can plug in the equilibrium concentrations from our ICE table: \[ 4.34 \times 10^{-3} = \frac{(0.0618 - 2x)^2}{x(3x)^3} \]

Solving the equation for x, we get: \[ x = 1.062 \times 10^{-3} \] Now, we can calculate the moles of N2 and H2 in the equilibrium mixture: Moles of N2 = x = 1.062 \times 10^{-3} mol Moles of H2 = 3x = 3.186 \times 10^{-3} mol Next, we can convert these moles into masses: Mass of N2 = moles × molar mass = (1.062 \times 10^{-3} mol) × (28 g/mol) = 0.0297 g Mass of H2 = moles × molar mass = (3.186 \times 10^{-3} mol) × (2 g/mol) = 0.00637 g Thus, the masses of N2 and H2 in the equilibrium mixture are 0.0297 g and 0.00637 g, respectively.

Since equilibrium is established, we know that: Initial moles of NH3 - 2x = moles of NH3 at equilibrium Initial moles of NH3 = 0.0618 mol + 2x = 0.0618 + (2 × 1.062 × 10^{-3}) mol = 0.0640 mol Now, we can calculate the initial mass of ammonia: Initial mass of NH3 = Initial moles of NH3 × molar mass = (0.0640 mol) × (17 g/mol) = 1.09 g The initial mass of ammonia placed in the vessel is 1.09 g.

At equilibrium, the partial pressures of N2, H2, and NH3 can be calculated using their equilibrium concentrations and the gas constant R(0.0821 L atm/mol K) and temperature (300°C = 573 K). Partial pressure of N2 = \( P_{N_2} = \frac{n_{N_2} RT}{V} = \frac{(1.062 \times 10^{-3} mol)(0.0821 \frac{L \cdot atm}{mol \cdot K})(573 K)}{1.00 L} = 0.0441\, atm\) Partial pressure of H2 = \( P_{H_2} = \frac{n_{H_2} RT}{V} = \frac{(3.186 \times 10^{-3} mol)(0.0821 \frac{L \cdot atm}{mol \cdot K})(573 K)}{1.00 L} = 0.132\, atm\) Partial pressure of NH3 = \( P_{NH_3} = \frac{n_{NH_3} RT}{V} = \frac{(0.0618 \, mol)(0.0821 \frac{L \cdot atm}{mol \cdot K})(573 K)}{1.00 L} = 2.57\, atm\) Now, we can find the total pressure in the vessel by adding the partial pressures: Total Pressure = \( P_{N_2} + P_{H_2} + P_{NH_3} = 0.0441 \, atm + 0.132 \, atm + 2.57 \, atm = 2.75 \, atm\) The total pressure in the vessel at equilibrium is 2.75 atm.