Problem 102 - Chemistry An Atoms First Approach

Question

The equilibrium constant for a certain reaction increases by a factor of 6.67 when the temperature is increased from \(300.0 \mathrm{K}\) to \(350.0 \mathrm{K}\). Calculate the standard change in enthalpy \(\left(\Delta H^{\circ}\right)\) for this reaction (assuming \(\Delta H^{\circ}\) is temperature-independent).

Step-by-Step Solution

Answer

The standard change in enthalpy for this reaction, \(\Delta H^{\circ}\), can be calculated using the van't Hoff equation: \[\ln \left(\frac{K_2}{K_1}\right) = \frac{-\Delta H^{\circ}}{R} \left(\frac{1}{T_2} - \frac{1}{T_1}\right)\] Given \(K_2/K_1 = 6.67\), \(T_1 = 300.0\,\mathrm{K}\), and \(T_2 = 350.0\,\mathrm{K}\), we can rearrange the equation and find that: \[\Delta H^{\circ} \approx -3.27 \times 10^{3}\:\mathrm{J\, mol^{-1}}\]

Step 1: Write down the van't Hoff equation.

The van't Hoff equation is given by: \[\ln \left(\frac{K_2}{K_1}\right) = \frac{-\Delta H^{\circ}}{R} \left(\frac{1}{T_2} - \frac{1}{T_1}\right)\] Here, \(K_1\) and \(K_2\) are the initial and final equilibrium constants, \(T_1\) and \(T_2\) are the initial and final temperatures, \(R\) is the gas constant, and \(\Delta H^{\circ}\) is the standard change in enthalpy.

Step 2: Plug in given values.

We have the following information from the exercise: - \(K_2/K_1 = 6.67\) - \(T_1 = 300.0\,\mathrm{K}\) - \(T_2 = 350.0\,\mathrm{K}\) - \(R = 8.314\,\mathrm{J\, mol^{-1}K^{-1}}\) Now let's plug these values into the van't Hoff equation: \[\ln \left(6.67\right) = \frac{-\Delta H^{\circ}}{8.314} \left(\frac{1}{350.0} - \frac{1}{300.0}\right)\]

Step 3: Rearrange the equation and solve for \(\Delta H^{\circ}\).

We want to isolate \(\Delta H^{\circ}\) to find its value, so let's rearrange the equation: \[\Delta H^{\circ} = - 8.314 \times \frac{\ln\left(6.67\right)}{\frac{1}{350.0} - \frac{1}{300.0}}\] Now we can calculate the value for \(\Delta H^{\circ}\): \[\Delta H^{\circ} = - 8.314 \times \frac{1.897}{\frac{1}{350.0} - \frac{1}{300.0}} = -8.314 \times \frac{1.897}{\frac{50}{105000}}\] \[\Delta H^{\circ} \approx -3.266 \times 10^{3}\,\mathrm{J\, mol^{-1}}\]

Step 4: Write down the result.

Therefore, the standard change in enthalpy for this reaction is approximately: \[\Delta H^{\circ} \approx -3.27 \times 10^{3}\:\mathrm{J\, mol^{-1}}\]

Key Concepts

Equilibrium Constant

The equilibrium constant, often denoted as \( K \), is a key parameter in chemistry that quantifies the balance point of a reaction. It expresses the ratio of product concentrations to reactant concentrations when a chemical reaction has reached equilibrium. Every reaction has its characteristic equilibrium constant, which is influenced significantly by temperature.

• At a given temperature, the equilibrium constant remains consistent unless the conditions change.
• The value of \( K \) provides insight into the direction of the equilibrium. A large \( K \) value indicates that the reaction favors product formation.
• Conversely, a small \( K \) signifies that the reactants are favored at equilibrium.

Temperature plays a crucial role in altering the equilibrium constant, as seen in the exercise with an increase in \( K \) when temperature rises. This demonstrates how temperature can dictate whether a reaction moves towards the products or remains with the reactants. Understanding this concept is essential for predicting how a reaction behaves under different thermal conditions.

Standard Change in Enthalpy

The standard change in enthalpy, symbolized as \( \Delta H^{\circ} \), represents the heat absorbed or released by a reaction occurring under constant pressure, typically specified at standard conditions (1 bar, typically 298 K).

• When \( \Delta H^{\circ} \) is negative, the reaction is exothermic, releasing heat to the surroundings.
• If \( \Delta H^{\circ} \) is positive, the reaction is endothermic, requiring heat absorption from the surroundings.
• For the given exercise, the negative \( \Delta H^{\circ} \) suggests an exothermic reaction, indicating that the equilibrium constant increases as temperature rises.

This thermodynamic quantity is crucial as it helps predict how the reaction yields shift with temperature changes, offering insights into the energetic favorability of reactions under different conditions.

Temperature Dependence of Reactions

Reactions are highly sensitive to temperature changes, which significantly impact the rate and equilibrium position. The van't Hoff equation illustrates this relationship, showing how the equilibrium constant \( K \) shifts with temperature changes.

• The equation \[ \ln \left(\frac{K_2}{K_1}\right) = \frac{-\Delta H^{\circ}}{R} \left(\frac{1}{T_2} - \frac{1}{T_1}\right) \] provides a mathematical basis for predicting these changes.
• The exercise demonstrates that as temperature rises from 300 K to 350 K, the equilibrium constant increases by a factor of 6.67. This indicates that the position of equilibrium shifts towards the products more significantly at higher temperatures.
• Understanding this principle aids in controlling reactions in industrial and laboratory settings, ensuring optimal conditions for desired product yields.

The van't Hoff equation is a powerful tool for assessing the temperature dependence of reactions, guiding adjustments in reaction conditions to achieve specific outcomes efficiently.