Question

Consider the following equilibrium: $$ \mathrm{N}_{2} \mathrm{O}_{4}(g) \rightleftharpoons 2 \mathrm{NO}_{2}(g) $$ Thermodynamic data on these gases are given in Appendix C. You may assume that \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) do not vary with temperature. (a) At what temperature will an equilibrium mixture contain equal amounts of the two gases? (b) At what temperature will an equilibrium mixture of 1 atm total pressure contain twice as much \(\mathrm{NO}_{2}\) as \(\mathrm{N}_{2} \mathrm{O}_{4} ?\) (c) At what temperature will an equilibrium mixture of \(10 \mathrm{~atm}\) total pressure contain twice as much \(\mathrm{NO}_{2}\) as \(\mathrm{N}_{2} \mathrm{O}_{4} ?\) (d) Rationalize the results from parts (b) and (c) by using Le Châtelier's principle. \(\infty\) (Section 15.7)

Short answer

In summary, (a) the temperature for equal amounts of both gases is given by \( T = \frac{\Delta H^{\circ}}{\Delta S^{\circ}} \). For (b) and (c), the equilibrium temperature is found by plugging the molar ratios and total pressure into the equilibrium constant expression and solving for T. The results show that an increase in total pressure favors the formation of NO₂, in accordance with Le Châtelier's principle.

Step-by-step solution

1. Write the equilibrium constant expression

The reaction is given by: \[ N_2O_4(g) \rightleftharpoons 2 NO_2(g) \] The equilibrium constant expression is as follows: \[ K_\mathrm{p} = \frac{P_{\mathrm{NO}_2}^2}{P_{\mathrm{N}_2\mathrm{O}_4}}\]

2. Find the expression for Gibbs free energy change

The standard free energy change (ΔG°) for a reaction is related to the standard enthalpy change (ΔH°) and standard entropy change (ΔS°) through the equation: \[ \Delta G^\circ = \Delta H^\circ - T \Delta S^\circ \]

3. Combine the Gibbs free energy change expression with the equilibrium constant expression

The relationship between the standard free energy change and the equilibrium constant for the reaction is given by the following equation: \[ \Delta G^\circ = -RT \ln{K_\mathrm{p}} \] Combining the previous expressions, we have: \[ -RT \ln{K_\mathrm{p}} = \Delta H^\circ - T \Delta S^\circ \] Now, we will solve for the temperature (T) for each scenario.

4. Solve for temperatures (a), (b), and (c)

(a) At what temperature will an equilibrium mixture contain equal amounts of the two gases? When equal amounts of both gases are present, \( P_{\mathrm{NO}_2} = P_{\mathrm{N}_2\mathrm{O}_4} \) and thus \( K_\mathrm{p} = 1 \). \[ -RT \ln{1} = \Delta H^\circ - T \Delta S^\circ \] Since \(ln{1} = 0\), we have: \[ -T \Delta S^\circ = \Delta H^\circ \] Solving for the temperature (T) gives: \[ T= \frac{\Delta H^\circ}{\Delta S^\circ} \] (b) At what temperature will an equilibrium mixture of 1 atm total pressure contain twice as much NO₂ as N₂O₄? The equilibrium expression becomes: \[ K_\mathrm{p} = \frac{P_{\mathrm{NO}_2}^2}{P_{\mathrm{N}_2\mathrm{O}_4}} = \frac{(2P_{\mathrm{N}_2\mathrm{O}_4})^2}{P_{\mathrm{N}_2\mathrm{O}_4}} = 4P_{\mathrm{N}_2\mathrm{O}_4} = 4(1-(2+1)P_{\mathrm{N}_2\mathrm{O}_4}) \] Substituting into the ΔG° equation and solving for T. (c) At what temperature will an equilibrium mixture of 10 atm total pressure contain twice as much NO₂ as N₂O₄? Similar to (b), we set up the equilibrium expression and solve for T for a total pressure of 10 atm.

5. Rationalize the results for (b) and (c) using Le Châtelier's principle

Le Châtelier's principle states that if a system at equilibrium is subjected to a change in pressure, temperature, or concentration, the system will adjust to counteract the change and restore equilibrium. In this case, we observe the effect of changing temperature and total pressure on the equilibrium position. For (b) and (c), the results suggest that an increase in total pressure favors the formation of NO₂ over N₂O₄. This is in accordance with Le Châtelier's principle as an increase in pressure will shift the equilibrium towards the side with fewer moles of gas. In this equilibrium reaction, there are fewer moles of gas on the right side (2 moles of NO₂) compared to left side (1 mole of N₂O₄). As total pressure increases, the equilibrium will shift towards the formation of more NO₂ to counteract the change in pressure.

Key concepts

Gibbs Free Energy

The concept of Gibbs Free Energy is fundamental in determining whether a chemical reaction will proceed spontaneously. It's a thermodynamic potential that measures the "useful" work obtainable from a system at constant temperature and pressure. Gibbs Free Energy (\( \Delta G^\circ \)) is defined by the equation: \( ΔG^\circ = ∆H^\circ − T ∆S^\circ \).
This equation links enthalpy (\( ∆H^\circ \)) and entropy (\( ∆S^\circ \)) changes of a system. A negative \( ∆G^\circ \) indicates that a reaction will occur spontaneously under standard conditions.

• When \( ∆G^\circ < 0 \), the process is spontaneous.
• When \( ∆G^\circ > 0 \), the process is non-spontaneous.

The equation can also determine the point at which reactants and products are in equilibrium. At equilibrium, \( ∆G^\circ = 0 \), illustrating that no net change is occurring as the forward and reverse reactions have balanced each other. By understanding Gibbs Free Energy, students can predict which direction a reaction will favor at any given temperature.

Le Châtelier's Principle

Le Châtelier's Principle is a critical concept in chemical equilibrium, helping to predict how a change in conditions can affect the position of equilibrium. Named after the French chemist Henri Louis Le Châtelier, this principle states that a system at equilibrium will adjust to counteract any imposed change, maintaining equilibrium.
For example, when a reaction mixture's pressure is increased, the system will shift to minimize this change by favoring the side with fewer moles of gas. This happens in the equilibrium reaction between \( N_2O_4(g) \) and \( 2 NO_2(g) \), where the formation of \( NO_2 \) is favored with increased pressure.

• Increasing pressure shifts equilibrium towards fewer moles.
• Decreasing pressure favors more moles.
• Increasing temperature shifts equilibrium toward endothermic side.
• Decreasing temperature favors exothermic reactions.

This principle is vital when manipulating chemical reactions to optimize product yield, such as the industrial synthesis of ammonia in the Haber process.

Equilibrium Constant Expression

The equilibrium constant expression provides a quantitative way to describe the balance between products and reactants in a chemical equilibrium. For gases, the equilibrium expression is written in terms of partial pressures, denoted as \( K_p \).
For the reaction \( N_2O_4(g) ightleftharpoons 2 NO_2(g) \), the equilibrium constant expression is:\[ K_p = \frac{P_{NO_2}^2}{P_{N_2O_4}} \]
This expression allows calculation of equilibrium constants at different temperatures and is pivotal in predicting the extent of a reaction. The magnitude of \( K_p \) indicates whether the reaction favors the reactants or the products:

• If \( K_p \gg 1 \)
, the products are favored.
• If \( K_p \ll 1 \)
, the reactants are favored.

Understanding how to derive and manipulate these expressions can help students determine reaction conditions, such as pressure or temperature, required to achieve a desired equilibrium position.

Thermodynamics in Chemistry

Thermodynamics in Chemistry refers to the study of energy changes and transfers during chemical processes. It encompasses concepts like enthalpy (\( ∆H \)), entropy (\( ∆S \)), and Gibbs Free Energy, crucial for understanding reaction spontaneity and equilibrium.
In thermodynamics, reactions tend to progress in the direction that increases entropy (disorder) and releases energy in the form of heat (exothermic). The relationship between a reaction’s spontaneity and temperature can be deduced using the Gibbs Free Energy equation:\( ∆G = ∆H - T∆S \).
This equation shows that spontaneity is driven by both enthalpy and entropy:

• Exothermic reactions (\( ∆H < 0 \)) tend to be spontaneous.
• Reactions that increase entropy (\( ∆S > 0 \)) are preferred.

Thermodynamics allows chemists to predict how reactions will behave under different conditions, optimizing chemical processes for desired outcomes. By applying these principles, students gain insight into why some reactions occur naturally, while others require external energy to proceed.