Question

For reactions carried out under standard-state conditions, Equation 18.10 takes the form \(\Delta G^{\circ}=\Delta H^{\circ}-T \Delta S^{\circ} .\) (a) Assuming \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) are independent of temperature, derive the equation: $$ \ln \frac{K_{2}}{K_{1}}=\frac{\Delta H^{\circ}}{R}\left(\frac{T_{2}-T_{1}}{T_{1} T_{2}}\right) $$ where \(K_{1}\) and \(K_{2}\) are the equilibrium constants at \(T_{1}\) and \(T_{2},\) respectively (b) Given that at \(25^{\circ} \mathrm{C} K_{c}\) is \(4.63 \times 10^{-3}\) for the reaction: $$ \mathrm{N}_{2} \mathrm{O}_{4}(g) \rightleftharpoons 2 \mathrm{NO}_{2}(g) \quad \Delta H^{\circ}=58.0 \mathrm{~kJ} / \mathrm{mol} $$ calculate the equilibrium constant at \(65^{\circ} \mathrm{C}\).

Short answer

The equilibrium constant at 65°C is approximately 0.086.

Step-by-step solution

Derive the Van't Hoff Equation - Part 1

Start with the Gibbs free energy equation under standard state conditions, \(\Delta G^{\circ} = \Delta H^{\circ} - T \Delta S^{\circ}\). The relationship between \( \Delta G^{\circ} \) and the equilibrium constant \( K \) is given by \( \Delta G^{\circ} = -RT \ln K \). Equating the two expressions for \( \Delta G^{\circ} \), we have \( -RT \ln K = \Delta H^{\circ} - T \Delta S^{\circ} \).

Derive the Van't Hoff Equation - Part 2

Rearrange the equation \(-RT \ln K = \Delta H^{\circ} - T \Delta S^{\circ}\) to express \(\ln K\):\[ \ln K = -\frac{\Delta H^{\circ}}{RT} + \frac{\Delta S^{\circ}}{R} \]This equation can be used to find \(K\) at different temperatures if \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) are constant.

Derive the Van't Hoff Equation - Part 3

For two states at temperatures \(T_1\) and \(T_2\), the equilibrium constants are \(K_1\) and \(K_2\) respectively. Using the expression \(\ln K\) for the two temperatures:\[ \ln K_1 = -\frac{\Delta H^{\circ}}{RT_1} + \frac{\Delta S^{\circ}}{R} \]\[ \ln K_2 = -\frac{\Delta H^{\circ}}{RT_2} + \frac{\Delta S^{\circ}}{R} \]Subtracting these two equations gives:\[ \ln \frac{K_2}{K_1} = \frac{\Delta H^{\circ}}{R} \left( \frac{1}{T_1} - \frac{1}{T_2} \right) \]Recognizing \(\frac{1}{T_1} - \frac{1}{T_2}\) as \(\frac{T_2 - T_1}{T_1 T_2}\), this results in: \[ \ln \frac{K_2}{K_1} = \frac{\Delta H^{\circ}}{R} \left( \frac{T_2 - T_1}{T_1 T_2} \right) \]

Apply the Van't Hoff Equation - Part 4

Given \(K_1 = 4.63 \times 10^{-3}\) at \(T_1 = 25^{\circ} \text{C} = 298 \text{ K}\), and \(T_2 = 65^{\circ} \text{C} = 338 \text{ K}\), use the derived equation to find \(K_2\).Convert \(\Delta H^{\circ} = 58.0 \text{ kJ/mol} = 58000 \text{ J/mol}\), and use \(R = 8.314 \text{ J/molK}\):\[ \ln \frac{K_2}{4.63 \times 10^{-3}} = \frac{58000}{8.314} \times \frac{338 - 298}{298 \times 338} \]Calculate this value.

Solve for K2 - Part 5

Calculate the right side of the equation from Part 4:\[ \frac{58000}{8.314} \approx 6974 \]\\[ T_2 - T_1 = 40 \text{ K}, \quad 298 \times 338 = 100924 \text{ K}^2\]\[ \ln \frac{K_2}{4.63\times10^{-3}} \approx \frac{6974 \times 40}{100924} \approx 2.768 \]Then, solve for \(K_2\):\[ \ln K_2 = \ln(4.63 \times 10^{-3}) + 2.768 \]\(\ln(4.63 \times 10^{-3}) \approx -5.223\)\[ \ln K_2 \approx -5.223 + 2.768 = -2.455 \]\(K_2 = e^{-2.455} \approx 0.086\).

Key concepts

Equilibrium Constant

The equilibrium constant, represented as \( K \), is a crucial concept in chemical reactions that tells us about the balance between reactants and products at equilibrium. In essence, it provides a snapshot of where a reaction's equilibrium lies. If \( K \) is larger than 1, the products are favored at equilibrium; conversely, if \( K \) is smaller than 1, the reactants are favored.

Understanding \( K \) helps us predict how changes in conditions, such as temperature, might shift a reaction. The equilibrium constant is inherently tied to Gibbs free energy and can be expressed using the formula \( \Delta G^{\circ} = -RT \ln K \). This connection allows us to use measurements of \( K \) to calculate the energy changes in a reaction, facilitating deeper insights into the chemical process.

Through the Van’t Hoff equation, we can compare equilibrium constants at different temperatures, which is particularly valuable for understanding how heat influences reactions. The equation \[ \ln \frac{K_2}{K_1} = \frac{\Delta H^{\circ}}{R} \left( \frac{T_2 - T_1}{T_1 T_2} \right) \] allows us to see the effect of temperature changes on \( K \) without directly measuring Gibbs free energy. This is incredibly useful for chemists looking to optimize reactions.

Gibbs Free Energy

Gibbs free energy, denoted as \( \Delta G \), is a measure of the maximum reversible work a thermodynamic system can perform at constant temperature and pressure. It is a fundamental thermodynamic potential that helps predict the behavior of reactions. At equilibrium, \( \Delta G = 0 \), and any movement from equilibrium conditions implies a change in the system's free energy.

To express the change in Gibbs free energy at standard conditions, we use the equation \( \Delta G^{\circ} = \Delta H^{\circ} - T \Delta S^{\circ} \). This relationship shows how enthalpy (\( \Delta H \)) and entropy (\( \Delta S \)) contribute to the free energy of the system.

When analyzing a reaction's spontaneity, \( \Delta G^{\circ} \) is key:

• If \( \Delta G^{\circ} < 0 \), the reaction is spontaneous.
• If \( \Delta G^{\circ} > 0 \), the reaction is non-spontaneous.

A negative \( \Delta G^{\circ} \) indicates that the reaction can proceed without any external input of energy, favoring product formation. Gibbs free energy changes also link directly to equilibrium constants, thus entwining with the Van’t Hoff equation to provide insight into how reaction dynamics engage with temperature variations.

Enthalpy

Enthalpy, symbolized as \( \Delta H \), refers to the total heat content of a system. It’s a thermodynamic property that signifies the energy transferred between a system and its surroundings as heat when the system undergoes a process at constant pressure.

The change in enthalpy, \( \Delta H \), relates to the heat absorbed or released:

• When \( \Delta H \) is positive, the reaction is endothermic, absorbing heat.
• When \( \Delta H \) is negative, the reaction is exothermic, releasing heat.

Enthalpy changes are crucial for assessing how energy flows through chemical reactions. Together with the Van’t Hoff equation, \( \Delta H^{\circ} \) indicates how temperature affects reaction equilibria.

By combining \( \Delta H^{\circ} \) with entropy and temperature, we can determine the Gibbs free energy change. This holistic approach helps predict a reaction's spontaneity and stability under different conditions. Remember, while enthalpy tells part of the story, it is the combination with entropy and temperature (in Gibbs free energy) that gives the full thermodynamic picture.

Entropy

Entropy, expressed as \( \Delta S \), is a measure of disorder or randomness in a system. In thermodynamics, it reflects how energy is dispersed or spread out in a process. A high entropy (positive \( \Delta S \)) indicates a greater degree of unpredictability or randomness.

In chemical reactions, entropy is crucial for understanding how systems evolve. It's part of the larger equation of Gibbs free energy, represented as \( \Delta G^{\circ} = \Delta H^{\circ} - T \Delta S^{\circ} \). Here, entropy shows its power in determining the direction and feasibility of a reaction.

Entropy increases when:

• Gases form from solids or liquids.
• A substance is mixed with another, creating a solution.
• The complexity of molecules increases.

The concept of entropy ties deeply into both equilibrium and the underlying energy dynamics of reactions. Whereas enthalpy reflects heat changes, entropy shifts based on the system’s order. This combination ultimately affects \( \Delta G^{\circ} \) and thus the position of equilibrium at various temperatures, illustrating the intricate balance of energy and matter transitions.