Problem 60 - Chemistry: An Atoms-Focused Approach

Question

At \(650 \mathrm{K},\) the valuc of \(K_{\mathrm{p}}\) for the ammonia synthesis reaction $$ \mathrm{N}_{2}(\mathrm{g})+3 \mathrm{H}_{2}(\mathrm{g}) \rightleftharpoons 2 \mathrm{NH}_{3}(g) $$ is \(4.3 \times 10^{-4} .\) If a vessel contains a reaction mixture in which \(\left[\mathrm{N}_{2}\right]=0.010 M,\left[\mathrm{H}_{2}\right]=0.030 M,\) and \(\left[\mathrm{NH}_{3}\right]=0.00020 \mathrm{M},\) will more ammonia form?

Step-by-Step Solution

Answer

Answer: No, more ammonia will not form. The reaction will proceed in the reverse direction to reach equilibrium since Q > Kp (0.00148 > \(4.3 × 10^{-4}\)).

Step 1: Write the expression for equilibrium constant (Kp) and reaction quotient (Q)

The expression for Kp and Q for the given reaction can be written as: $$ K_{p} = \frac{[\mathrm{NH}_3]^2}{[\mathrm{N}_2][\mathrm{H}_2]^3} \quad \text{and} \quad Q = \frac{[\mathrm{NH}_3]^2}{[\mathrm{N}_2][\mathrm{H}_2]^3} $$

Step 2: Calculate the reaction quotient (Q) using the given concentrations

Substitute the given concentrations into the Q expression and calculate its value: $$ Q = \frac{(0.00020)^2}{(0.010)(0.030)^3} = 0.00148 $$

Step 3: Compare Q with Kp to determine the direction of the reaction

We have Q = 0.00148 and Kp = \(4.3 \times 10^{-4}\). So to determine the direction of the reaction: - If Q > Kp, the reaction will proceed in the reverse direction to reach equilibrium - If Q < Kp, the reaction will proceed in the forward direction to reach equilibrium - If Q = Kp, the reaction is already at equilibrium In our case, Q > Kp (0.00148 > \(4.3 \times 10^{-4}\)), so the reaction will proceed in the reverse direction to reach equilibrium, meaning that more ammonia (NH3) will not form.

Key Concepts

Equilibrium Constant

The Equilibrium Constant (\(K\)) is a critical concept in chemical reactions. It gives us a way to quantify the position of equilibrium in a chemical reaction at a given temperature. Generally, \(K\) is determined by the ratio of the concentration or partial pressure of products to reactants, each raised to the power of their stoichiometric coefficients in the balanced equation of the reaction.
For the ammonia synthesis reaction: \(\mathrm{N}_2(\mathrm{g}) + 3\mathrm{H}_2(\mathrm{g}) \rightleftharpoons 2\mathrm{NH}_3(\mathrm{g})\), the equilibrium constant \(K_p\) is expressed as:

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

At specific conditions, like 650 K for this reaction, \(K_p\) allows us to predict whether the reactants or products are favored at equilibrium. In this exercise, \(K_p\) is given the value \(4.3 \times 10^{-4}\), indicating the position of equilibrium for ammonia synthesis under the given temperature.

Ammonia Synthesis

Ammonia Synthesis is a crucial industrial process that produces ammonia (\(\mathrm{NH}_3\)), an essential ingredient in fertilizers and various chemical products. This process is typically carried out via the Haber-Bosch method, which involves the direct reaction of nitrogen (\(\mathrm{N}_2\)) from the air with hydrogen (\(\mathrm{H}_2\)).
The balanced chemical equation for this reaction is:

• \(\mathrm{N}_2(\mathrm{g}) + 3\mathrm{H}_2(\mathrm{g}) \rightleftharpoons 2\mathrm{NH}_3(\mathrm{g})\)

According to the equation, one mole of nitrogen reacts with three moles of hydrogen to produce two moles of ammonia. The reaction is exothermic, meaning it releases heat. A challenge of ammonia synthesis lies in achieving an efficient yield due to its dependency on factors such as temperature, pressure, and catalyst effectiveness. In practice, high temperatures and pressures are often used to drive the formation of ammonia, although these conditions can also lead to a less favorable equilibrium constant.

Chemical Equilibrium

Chemical Equilibrium is a state in a chemical reaction where the rates of the forward and reverse reactions are equal, causing no net change in the concentrations of reactants and products. At equilibrium, a mixture will contain amounts of reactants and products that remain constant over time.
However, equilibrium does not imply that the reaction stops altogether. Instead, both forward and backward reactions continue to occur, but at the same rate, thus stabilizing the concentration of species involved.

• The reaction quotient (\(Q\)) is a measure used to predict the direction in which a reaction must proceed to reach equilibrium. It is calculated using the same expression as the equilibrium constant (\(K\)), but with the concentrations or pressures from any given condition, not necessarily at equilibrium.
• In this exercise, by calculating \(Q\) and comparing it with \(K_p\), we determine that \(Q > K_p\), indicating that the reaction mixture contains more products than it would at equilibrium, prompting a shift towards reactants.

Le Chatelier's Principle

Le Chatelier's Principle provides insight into how a chemical equilibrium responds to changes in the system. Essentially, if a dynamic equilibrium is disturbed by changing the conditions, the position of equilibrium moves to counteract the change.
This principle helps predict how conditions such as concentration, pressure, and temperature affect the equilibrium. For example, if the concentration of a reactant is increased, the system shifts to produce more products to re-establish equilibrium.

• In the context of this ammonia synthesis exercise, Le Chatelier's Principle explains the backward reaction observed: since \(Q > K_p\), the reaction shifts towards the reactants to decrease the concentration of ammonia and establish equilibrium.
• Changes in temperature can further be explained by this principle, especially in reactions like ammonia synthesis, where applying heat (exothermic reactions) may shift equilibrium backward to absorb excess heat.

By understanding this principle, chemists can optimize conditions to maximize the yield of desired products in industrial processes.