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 \(K_{p}=4.34 \times 10^{-3} \mathrm{at} 300^{\circ} \mathrm{C}\) . 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?

Short answer

The masses of \(\mathrm{N_2}\) and \(\mathrm{H_2}\) in the equilibrium mixture are 0.0297 g and 0.00637 g, respectively. The initial mass of ammonia placed in the vessel is 1.09 g. The total pressure in the vessel at equilibrium is 2.75 atm.

Step-by-step solution

Calculate the equilibrium concentrations of all components

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: \[ 4.34 \times 10^{-3} = \frac{(0.0618 - 2x)^2}{x(3x)^3} \]

Solve for x and find masses of N2 and H2

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.

Find the initial mass of ammonia

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.

Calculate the total pressure in the vessel

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.

Key concepts

Equilibrium Constant

In the context of chemical reactions, the equilibrium constant (K_p) plays a crucial role in describing the balance of reactants and products in their gaseous forms. It provides a measure of the position of equilibrium; when equilibrium is reached, the rates of the forward and reverse reactions are equal, and the concentrations of the reactants and products remain constant.
For the given reaction \( \mathrm{N}_2(g) + 3 \mathrm{H}_2(g) \rightleftharpoons 2 \mathrm{NH}_3(g) \), the equilibrium constant, K_p= 4.34 × 10^-3 at 300°C, helps us analyze the system under equilibrium conditions. The expression of K_p is given by:

• \( K_p = \frac{[\mathrm{NH}_3]^2}{[\mathrm{N}_2][\mathrm{H}_2]^3} \)

This equation indicates that the value of K_p relates to the partial pressures of the gases involved when the system reaches equilibrium.
A smaller K_p value, like the one mentioned, indicates that at equilibrium, the concentration of products will be smaller relative to the concentration of reactants.

ICE Table

The ICE Table is a tool utilized to systematically determine the concentrations or pressures of species in a chemical reaction at equilibrium. ICE stands for Initial, Change, and Equilibrium, representing three stages of concentration or pressure we expect through the course of the reaction. This structured approach helps in simplifying the calculations required to find equilibrium concentrations.
To set up an ICE Table, start by listing the initial concentrations or pressures of all reactants and products. Usually, these start at zero for products if they are initially absent. The 'Change' row captures the shifts in concentration based on the stoichiometry of the reaction, and finally, the 'Equilibrium' row combines the Initial and Change rows to show what's at equilibrium.
In our example, the Initial concentration of NH_3 is known, and through the ICE table, we can determine how the concentrations or pressures of N_2 and H_2 develop as NH_3 is decomposed into them. Solving the corresponding equilibrium equation with correct stoichiometric shifts gives us the desired equilibrium concentrations.

Partial Pressure

Partial pressure is a concept from gas law, reflecting the pressure exerted by a single type of gas in a mixture of gases. In equilibrium problems, this becomes particularly important when dealing with gases because it helps in understanding and calculating the equilibrium condition via K_p.
Each gas in a mixture contributes to the total pressure proportionally to the number of moles it has in the vessel. For a reaction such as \(\mathrm{N}_2(g)+3 \mathrm{H}_2(g)\rightleftharpoons 2 \mathrm{NH}_3(g)\), the partial pressures of N_2, H_2, and NH_3 will be dictated by their mole fractions and the overall temperature and volume of the system.
To calculate the partial pressure, we employ the ideal gas law formula, \(P = \frac{{nRT}}{V}\), where R is the gas constant and T the temperature in Kelvin, adapting for pressure contribution by each component. The summation of all partial pressures in the vessel will yield the total pressure experienced by the system.

Stoichiometry

Stoichiometry is the area of chemistry that deals with the relationships between reactants and products in a chemical reaction. It leverages the balanced chemical equations to predict the amount of reactants consumed and products formed. This integral concept helps translate theoretical chemical reactions into quantifiable amounts.
In our reaction,\(\mathrm{N}_2(g) + 3 \mathrm{H}_2(g) \rightleftharpoons 2 \mathrm{NH}_3(g)\), the stoichiometric coefficients depict the ratio of molecules participating in the reaction. This means for every 1 mole of \(\mathrm{N}_2\)gas that reacts, 3 moles of \(\mathrm{H}_2\) are required to produce 2 moles of \(\mathrm{NH}_3\).
By knowing one component's amount in moles, you can easily determine the others using these ratios. This is especially important in calculating the changes in concentrations for constructing ICE Tables and determining the exact shifts taking place in the equilibrium reaction. Stoichiometry thereby provides the essential quantitative framework for any equilibrium calculation.