Problem 64 - Chemistry: The Central Science

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?

Step-by-Step Solution

Answer

The masses of N2 and H2 at equilibrium are 0.0375 g and 0.00805 g, respectively. The initial mass of ammonia placed in the vessel was 1.102 g. The total pressure in the vessel is 0.2074 atm.

Step 1: Calculate the equilibrium concentrations

To calculate the equilibrium concentrations, we must convert the given mass of NH3 into moles. Since we know the volume of the container, we can then determine the molar concentration at equilibrium. Molar mass of NH3 = 14 (N) + (3 × 1) (H) = 17 g/mol Moles of NH3 = (1.05 g) / (17 g/mol) = 0.06176 mol Equilibrium concentration of NH3 = [NH3] = moles / volume = (0.06176 mol) / (1 L) = 0.06176 M

Step 2: Relate the equilibrium concentrations to K_p

The given equilibrium constant, K_p, can be expressed in terms of the equilibrium concentrations of each substance: K_p = [NH3]^2 / ([N2][H2]^3) We know the equilibrium constant and the equilibrium concentration of NH3, so we can treat the concentrations of N2 and H2 at equilibrium as variables (x and y, respectively): 4.34 × 10^(-3) = (0.06176)^2 / (x * (y)^3)

Step 3: Calculate how much NH3 dissociated

Given the stoichiometry of the reaction (1 mole N2 reacts with 3 moles H2 to produce 2 moles NH3), we can calculate the number of moles of NH3 that dissociated: Moles of NH3 dissociated = 2x Moles of N2 formed = x Moles of H2 formed = 3x Since we are dealing with molar concentrations, we can rewrite these relationships in terms of concentrations: [NH3] initial - 2x = 0.06176 M [N2] initial + x = x [H2] initial + 3x = 3x

Step 4: Determine the initial mass of NH3

Now, we can return to the relationship between the molar concentrations and K_p: 4.34 × 10^(-3) = (0.06176 - 2x)^2 / (x * (3x)^3) Solving for x: x = 0.00134 M This x value represents the equilibrium concentration of N2. We can also calculate the equilibrium concentration of H2: [H2] = 3x = 3 * (0.00134 M) = 0.00402 M Now we can find the masses of N2 and H2 at equilibrium: Mass of N2 = (0.00134 mol) * (28 g/mol) = 0.0375 g Mass of H2 = (0.00402 mol) * (2 g/mol) = 0.00805 g To find the initial mass of NH3: From the difference in moles of NH3 between initial and final state, we can calculate the initial moles of NH3: Initial moles of NH3 = 0.06176 mol + 2x = 0.06484 mol Initial mass of NH3 = (0.06484 mol) * (17 g/mol) = 1.102 g

Step 5: Calculate the total pressure in the vessel

To find the total pressure, we need to find the total moles of all gases and use the ideal gas law: Total moles of gas = moles of N2 + moles of H2 + moles of NH3 Total moles of gas = 0.00134 + 0.00402 + 0.06176 = 0.06712 mol Now, using the ideal gas law (PV=nRT), we can find the total pressure: P = nRT/V P = (0.06712 mol) * (8.314 J/(mol·K)) * (273 + 300° C) / (1 L) P = 21015.814 J/L Since 1 J = 1 Pa·L, we can convert the pressure to atm: P = 21015.814 Pa * (1 atm/101325 Pa) = 0.2074 atm So, the total pressure in the vessel is 0.2074 atm.

Key Concepts

Equilibrium Constant

In the realm of chemical equilibrium, the equilibrium constant, often represented by the symbol \( K \), is pivotal. It provides insight into the balance of products and reactants in a reaction at equilibrium. For the given reaction \( N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g) \), the equilibrium constant \( K_p \) is set at \( 4.34 \times 10^{-3} \) at 300°C.

This particular value of \( K_p \) allows us to determine the extent of the reaction's progress towards products or reactants. A small \( K_p \) value, like we see here, suggests that at equilibrium, the concentration of products (\( NH_3 \)) is small compared to reactants (\( N_2 \) and \( H_2 \)).

The expression for \( K_p \) for the reaction is formulated as follows:

• \( K_p = \frac{[NH_3]^2}{[N_2][H_2]^3} \)

In this equation, square brackets denote the partial pressures or concentrations of the respective gases. The powers of 2 and 3 in the denominator and numerator stem from the stoichiometric coefficients in the balanced chemical equation.

Stoichiometry

Stoichiometry involves the calculation of reactants and products in chemical reactions, and it is crucial to understanding chemical equilibrium. In the exercise, we see this concept in action through the balanced equation for ammonia synthesis:

• \( N_2 + 3H_2 \rightarrow 2NH_3 \)

This balanced equation tells us a lot about the reaction. For every molecule of \( N_2 \) that reacts, three molecules of \( H_2 \) are required to form two molecules of \( NH_3 \).

In the solution provided, stoichiometry helps us determine the amount of \( NH_3 \) that dissociates into \( N_2 \) and \( H_2 \) at equilibrium. The change in moles from the initial amount to the equilibrium state is calculated by using stoichiometric ratios:

• 1 mole of \( N_2 \) forms 2 moles of \( NH_3 \)
• 3 moles of \( H_2 \) form 2 moles of \( NH_3 \)

These relationships allow us to calculate the equilibrium concentrations and subsequently the masses of each component in the mixture.

Ideal Gas Law

The Ideal Gas Law is an essential tool in chemistry that relates the pressure, volume, and temperature of a gas to the number of moles of the gas. It is represented by the equation \( PV = nRT \). In this exercise, it helps calculate the total pressure in the reaction vessel once equilibrium is established.

After determining the total number of moles of gases (sum of \( N_2 \), \( H_2 \), and \( NH_3 \)) in the vessel, the Ideal Gas Law allows us to find the pressure by substituting the known values:

• \( P \) is the pressure in atm
• \( V \) is the volume in liters
• \( n \) is the total moles of gas
• \( R \) is the gas constant, 8.314 J/(mol·K)
• \( T \) is the temperature in Kelvin

Using \( P = \frac{nRT}{V} \), we calculated the total pressure to be 0.2074 atm, by converting moles to an ideal gas's pressure at the given temperature and volume.