91Ó°ÊÓ

The equilibrium constant, denoted as \( K \), is a fundamental concept in understanding chemical reactions at equilibrium. It is a numerical value that describes the ratio of the concentrations of products to reactants at equilibrium for a reversible chemical reaction. In the reaction \( 2 \mathrm{NO}_{2}(\mathrm{g}) \rightleftharpoons \mathrm{N}_{2}\mathrm{O}_{4}(\mathrm{g}) \), the equilibrium constant changes with temperature. At \( 0^{\circ} \text{C} \), the equilibrium constant \( K \) is 58, indicating a greater concentration of \( \mathrm{N}_{2}\mathrm{O}_{4} \) relative to \( \mathrm{NO}_{2} \). In contrast, at \( 100^{\circ} \text{C} \), \( K \) drops to 0.065, suggesting a predominance of \( \mathrm{NO}_{2} \).

• Equilibrium constants vary with temperature, impacting reaction direction.
• High \( K \) value: favors product formation.
• Low \( K \) value: favors reactant formation.

Understanding \( K \) helps predict how a reaction responds to temperature changes, as illustrated by the Van't Hoff equation, which links \( K \) with enthalpy change and temperature.

Enthalpy change, denoted as \( \Delta_{\mathrm{rxn}} H^{\circ} \), represents the heat absorbed or released during a reaction at constant pressure. In this exercise, we calculate the standard enthalpy change using the Van't Hoff equation, which relates changes in equilibrium constant with temperature.

• The reaction in consideration has a \( \Delta_{\mathrm{rxn}} H^{\circ} \) of approximately 18.8 kJ/mol.
• This positive value indicates the reaction absorbs heat, hence it is endothermic.

Calculating the enthalpy change involves several steps:

• Assessing the natural log change in equilibrium constant \( \ln\left(\frac{K_2}{K_1}\right) \)
• Computing the reciprocal temperature change \( \left(\frac{1}{T_2} - \frac{1}{T_1}\right) \)
• Substituting these into the Van't Hoff equation to find \( \Delta_{\mathrm{rxn}} H^{\circ} \)

The positive \( \Delta_{\mathrm{rxn}} H^{\circ} \) signifies the reaction's requirement for heat input to progress, aligning with the shift back to reactants at higher temperatures.

Endothermic reactions are those that absorb heat from their surroundings. This is a crucial concept in thermodynamics, and it explains why certain reactions are more product-favorable at lower temperatures. For the reaction \( 2 \mathrm{NO}_{2}(\mathrm{g}) \rightleftharpoons \mathrm{N}_{2}\mathrm{O}_{4}(\mathrm{g}) \), we've determined that \( \Delta_{\mathrm{rxn}} H^{\circ} \) is positive, confirming it as endothermic.

• Endothermic processes pull heat in, making products less stable than reactants at higher temperatures.
• Such reactions often result in a decrease in equilibrium constant with increasing temperature.

In our example, as temperature rises, the reaction shifts towards reactants \( \mathrm{NO}_{2} \) while reducing the equilibrium constant. This is typical for endothermic reactions, where the need for heat favors the reverse reaction at higher temperatures. Understanding the endothermic nature of chemical processes is essential for predicting how they respond to temperature variations and applying this knowledge to control reaction conditions in practical applications.