Problem 69 - Chemistry The Molecular Science

Question

Nitrogen, oxygen, and nitrogen monoxide are in equilibrium in a container of fixed volume. \(\mathrm{N}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NO}(\mathrm{g}) \quad \Delta_{\mathrm{r}} H^{\circ}=180.5 \mathrm{~kJ} / \mathrm{mol}\) How will each of these changes affect the indicated quantities? Write "increase," "decrease," or "no change." \begin{tabular}{lccc} \hline Change & {\(\left[\mathrm{N}_{2}\right]\)} & {\([\mathrm{NO}]\)} & \(K_{c}\) & \(K_{p}\) \\ \hline Some NO is added to the & & & \\ container. \\ The temperature of the gases & & & \\ in the container is decreased. & & & \\ The pressure of \(\mathrm{N}_{2}\) is & & & \\ decreased. \end{tabular}

Step-by-Step Solution

Answer

Adding NO: [N2] increases, [NO] decreases, no change in Kc and Kp. Decreasing temperature: [N2] increases, [NO] decreases, Kc and Kp decrease. Decreasing N2 pressure: no change in [N2], [NO] decreases, no change in Kc and Kp.

Step 1: Analyze the Effect of Adding NO

When NO is added to the system, the equilibrium shifts to the left according to Le Chatelier's principle to counteract the change. This means that the concentrations of \([\mathrm{N}_2]\) will increase to compensate for the added \([\mathrm{NO}]\). The concentration of \([\mathrm{NO}]\) will initially increase but will decrease towards equilibrium. Since the system is compensating for the stress by re-establishing equilibrium, both \(K_c\) and \(K_p\) remain unchanged.

Step 2: Analyze the Effect of Decreasing Temperature

The given reaction is endothermic with \(\Delta_{r}H^{\circ} = 180.5 \mathrm{~kJ/mol}\), meaning it absorbs heat. Decreasing the temperature shifts the equilibrium to the left, favoring the formation of \(\mathrm{N}_2\) and \(\mathrm{O}_2\), leading to an increase in \([\mathrm{N}_2]\) and a decrease in \([\mathrm{NO}]\). Because the equilibrium shifts to the reactant side in an endothermic process, both \(K_c\) and \(K_p\) decrease.

Step 3: Analyze the Effect of Decreasing \(\mathrm{N}_2\) Pressure

Decreasing the pressure of \(\mathrm{N}_2\) reduces its concentration in the equilibrium mixture, causing the equilibrium to shift towards the left to produce more \(\mathrm{N}_2\) and \(\mathrm{O}_2\) to counter the decrease. Therefore, \([\mathrm{N}_2]\) stays relatively constant due to re-establishing equilibrium, but \([\mathrm{NO}]\) decreases as it is consumed to form more reactants. Like before, both \(K_c\) and \(K_p\) remain unchanged because pressure changes alone do not affect the equilibrium constant unless temperature changes.

Key Concepts

Chemical Equilibrium

In chemical reactions, the term "chemical equilibrium" refers to a state where the reactants and products maintain constant concentrations over time. It's a balance achieved in a closed system, where the rate of the forward reaction equals the rate of the reverse reaction.
This balance does not necessarily mean that the quantities of reactants and products are equal; instead, they remain stable.

• At equilibrium, the mixture appears static because the macroscopic properties like concentration, color, and pressure remain unchanged.
• Le Chatelier's principle helps predict how an equilibrium system responds to changes in concentration, temperature, or pressure.
• The system will adjust itself in a way that alleviates the stress and re-establishes equilibrium.

This concept is crucial for understanding how equilibrium reacts to external changes, such as adding more of a substance or changing the system's temperature, as it directly influences the direction in which the reaction "shifts."

Endothermic Reactions

Endothermic reactions are those that absorb heat to proceed. In the context of an equilibrium system, the reaction rate can change depending on the temperature.
An endothermic reaction, like the one in the given exercise, involving nitrogen monoxide, involves an increase in the system's energy content when heat is added.

• If the temperature increases, the equilibrium shifts to the right, favoring the endothermic direction which produces more heat-consuming products.
• Conversely, decreasing the temperature shifts the equilibrium to the left, thereby favoring the reactants.

For the example reaction, a decrease in temperature results in more nitrogen and oxygen being formed since the system attempts to produce more heat.
The effect of temperature changes on equilibrium emphasizes the critical role of thermal energy in chemical reactions.

Equilibrium Constant (Kc and Kp)

In chemical equilibria, the equilibrium constants denoted as \(K_c\) and \(K_p\) are essential parameters.\(K_c\) and \(K_p\) describe the ratio of product concentrations to reactant concentrations at equilibrium, with \(K_c\) used for concentrations and \(K_p\) for partial pressures.

• \(K_c\) is calculated using molarity (mol/L) for reactions in solution.
• \(K_p\) is used for gaseous reactions, considering the partial pressures measured in atm or other pressure units.
• The values of \(K_c\) and \(K_p\) remain constant at a given temperature, meaning they are unaffected by changes in concentration or pressure alone.

This constancy underscores one essential aspect of equilibria: while a system can shift to accommodate stress (like changes in reactant quantities or pressure), these constants remain unchanged as long as the temperature is constant.
However, temperature shifts can alter the value of the equilibrium constant, reflecting a new balance point between reactants and products.