Question

As shown in Table \(15.2, K_{p}\) for the equilibrium $$\mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \rightleftharpoons 2 \mathrm{NH}_{3}(g)$$ is \(4.39 \times 10^{-9}\) at \(450^{\circ} \mathrm{C}\). For each of the mixtures listed here, indicate whether the mixture is at equilibrium at \(450^{\circ} \mathrm{C}\). If it is not at equilibrium, indicate the direction (toward product or toward reactants) in which the mixture must shift to achieve equilibrium. (a) \(9.93 \mathrm{MPa} \mathrm{NH}_{3}, 4.56 \mathrm{MPa} \mathrm{N}_{2}, 5.57 \mathrm{MPa} \mathrm{H}_{2}\) (b) \(5.78 \mathrm{MPa} \mathrm{NH}_{3}, 14.49 \mathrm{MPa} \mathrm{N}_{2},\) no \(\mathrm{H}_{2}\) (c) \(1.32 \mathrm{MPa} \mathrm{NH}_{3}, 2.74 \mathrm{MPa} \mathrm{N}_{2}, 8.31 \mathrm{Mpa} \mathrm{H}_{2}\)

Short answer

For the given mixtures at \(450^{\circ}C\), none of them are at equilibrium: (a) the mixture must shift towards the reactants (N2 and H2), (b) the mixture requires the addition of H2 and will then shift towards the products (NH3), (c) the mixture must shift towards the products (NH3).

Step-by-step solution

Calculate Qp for each mixture

The reaction quotient, Qp, is given by the following expression for the given reaction: \[ Q_p = \frac{(P_{NH_3})^2}{(P_{N_2})(P_{H_2})^3} \] Calculate Qp for each given mixture using the pressure (P) values provided: (a) For the mixture \( 9.93 \mathrm{MPa} \mathrm{NH}_{3}, 4.56 \mathrm{MPa} \mathrm{N}_{2}, 5.57 \mathrm{MPa} \mathrm{H}_{2} \): \[ Q_{p(a)} = \frac{ (9.93)^2 }{ (4.56)(5.57)^3 } \] (b) For the mixture \( 5.78 \mathrm{MPa} \mathrm{NH}_{3}, 14.49 \mathrm{MPa} \mathrm{N}_{2}, \) no H2: Since there is no H2 in this mixture, the denominator of the Qp expression will be zero, resulting in an undefined value for Qp. We can conclude that this mixture is not at equilibrium since it requires some amount of H2 for the reaction to take place. (c) For the mixture \( 1.32 \mathrm{MPa} \mathrm{NH}_{3}, 2.74 \mathrm{MPa} \mathrm{N}_{2}, 8.31 \mathrm{Mpa} \mathrm{H}_{2} \): \[ Q_{p(c)} = \frac{ (1.32)^2 }{ (2.74)(8.31)^3 } \]

Compare Qp values with Kp

Now, compare the Qp values we obtained in Step 1 with the given Kp value: (a) \( Q_{p(a)} = 0.224 \) (approx.) Since \( Q_{p(a)} > K_p \), the mixture must shift towards the reactants (N2 and H2). (b) The mixture is not at equilibrium since there is no H2 present. Adding H2 will cause the reaction to proceed towards the formation of NH3. (c) \( Q_{p(c)} = 4.00 \times 10^{-9} \) (approx.) Since \( Q_{p(c)} < K_p \), the mixture must shift towards the products (NH3).

Summary

To summarize, all three mixtures are not at equilibrium at \( 450^{\circ}C \). Mixture (a) will shift towards the reactants (N2 and H2), mixture (b) will shift towards the products (NH3) after addition of H2, and mixture (c) will also shift towards the products (NH3).

Key concepts

Reaction Quotient

The reaction quotient, denoted as \( Q \), is a snapshot of the equilibrium state's progress at any given moment. It uses the anologue of the equilibrium constant equation but with the current concentrations or pressures of reactants and products. This helps predict the direction a reaction must shift to reach equilibrium. For the given reaction:

• \( \mathrm{N}_{2}(g) + 3 \mathrm{H}_{2}(g) \rightleftharpoons 2 \mathrm{NH}_{3}(g) \)

The reaction quotient formula is:\[ Q_p = \frac{(P_{NH_3})^2}{(P_{N_2})(P_{H_2})^3} \]Where \( P \) represents the partial pressures of the gases involved. By calculating \( Q \), you're essentially creating a snapshot of where your mixture stands relative to equilibrium. If \( Q \) equals the equilibrium constant \( K \), the reaction is at equilibrium. If \( Q > K \), the reaction will shift towards the reactants, and if \( Q < K \), it will shift towards the products. This intuitive concept helps chemists understand and predict how systems respond under different conditions.

Equilibrium Constant

The equilibrium constant, symbolized as \( K \), is a crucial value depicting the ratio of product concentrations to reactant concentrations at equilibrium. For gas-phase reactions, like the one under study, the pressure-based equilibrium constant is written as \( K_p \). It is defined at a specific temperature and is a measure of the tendency of a reaction to proceed to completion.For the reaction \( \mathrm{N}_{2}(g) + 3 \mathrm{H}_{2}(g) \rightleftharpoons 2 \mathrm{NH}_{3}(g) \), the given \( K_p \) is \( 4.39 \times 10^{-9} \) at \( 450^{\circ} \mathrm{C} \). A high \( K_p \) value signifies a reaction that favors the formation of products at equilibrium, whereas a low \( K_p \) indicates stronger preference for reactants.

• If \( Q_p = K_p \), the system is at equilibrium, and no shift occurs.
• If \( Q_p eq K_p \), it predicts the shift necessary to reach equilibrium.

Understanding \( K \) and its comparison with \( Q \) is essential for solving equilibrium problems, allowing chemists to adjust conditions to favor desired products.

Le Chatelier's Principle

Le Chatelier's Principle provides a framework to predict how a chemical system at equilibrium responds to external changes. When a system at equilibrium experiences a change in pressure, concentration, or temperature, it will adjust to counteract that change and re-establish equilibrium.Applying this principle to our exercise:

• If a reaction mixture has \( Q > K \), it has too much product. The principle suggests shifting towards reactants to reduce the product amount.
• Conversely, if \( Q < K \), the product concentration is too low, so the reaction shifts towards forming more products.

Let's consider our second mixture that lacks \( H_2 \). According to Le Chatelier, by adding \( H_2 \), the system will shift to the right, producing more \( NH_3 \) to maintain equilibrium.Le Chatelier’s Principle is a powerful tool for chemists to manipulate reaction conditions, ensuring maximum yield under given circumstances.