Question

Consider the following reaction at \(25.0^{\circ} \mathrm{C}\) : $$2 \mathrm{NO}_{2}(g) \rightleftharpoons \mathrm{N}_{2} \mathrm{O}_{4}(g)$$ The values of \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) are \(-58.03 \mathrm{~kJ} / \mathrm{mol}\) and \(-176.6 \mathrm{~J} / \mathrm{K}\). mol, respectively. Calculate the value of \(K\) at \(25.0^{\circ} \mathrm{C}\). Assuming \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) are temperature independent, estimate the value of \(K\) at \(100.0^{\circ} \mathrm{C}\).

Short answer

The equilibrium constant (\(K\)) at \(25.0^{\circ}C\) (298.15 K) is approximately \(2.576\), and at \(100.0^{\circ}C\) (373.15 K) is approximately \(1.176\).

Step-by-step solution

(Step 1: Convert temperatures to Kelvin)

Convert the given temperatures to Kelvin, as the Van't Hoff equation requires temperatures to be in Kelvin: $$T_1 = 25.0^{\circ}C + 273.15K = 298.15 \, K$$ $$T_2 = 100.0^{\circ}C + 273.15K = 373.15 \, K$$

(Step 2: Calculate the initial equilibrium constant)

Use the Van't Hoff equation to calculate the equilibrium constant (\(K_1\)) for \(25.0^{\circ}C\) (\(298.15K\)), with the given values of \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\): $$\ln K_1 = -\frac{-58.03 \, kJ/mol}{8.314 \, J/mol \cdot K} \cdot \frac{1}{298.15 \, K} + \frac{-176.6 \, J/K}{8.314 \, J/mol \cdot K}$$ $$\ln K_1 = -\frac{-58.03 \times 10^3 \, J/mol}{8.314 \, J/mol \cdot K} \cdot \frac{1}{298.15 \, K} + \frac{-176.6 \, J/K}{8.314 \, J/mol \cdot K}$$ After calculating the values, we get: $$\ln K_1 = 0.94662$$ Convert the natural logarithm value to the actual value of \(K_1\): $$K_1 = e^{0.94662} \approx 2.576$$

(Step 3: Calculate the equilibrium constant at the new temperature)

Use the Van't Hoff equation again to calculate the equilibrium constant (\(K_2\)) for \(100.0^{\circ}C\) (\(373.15K\)), with the same values of \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\): $$\ln K_2 = -\frac{-58.03 \, kJ/mol}{8.314 \, J/mol \cdot K} \cdot \frac{1}{373.15 \, K} + \frac{-176.6 \, J/K}{8.314 \, J/mol \cdot K}$$ $$\ln K_2 = -\frac{-58.03 \times 10^3 \, J/mol}{8.314 \, J/mol \cdot K} \cdot \frac{1}{373.15 \, K} + \frac{-176.6 \, J/K}{8.314 \, J/mol \cdot K}$$ After calculating the values, we get: $$\ln K_2 = 0.16241$$ Convert the natural logarithm value to the actual value of \(K_2\): $$K_2 = e^{0.16241} \approx 1.176$$ So, the values of \(K\) at \(25.0^{\circ}C\) and \(100.0^{\circ}C\) are approximately \(2.576\) and \(1.176\), respectively.

Key concepts

Equilibrium Constant

In the context of chemical reactions, the equilibrium constant, often represented as \( K \), is a crucial value that helps us understand the balance of reactants and products at equilibrium. This constant is affected by temperature and can be used to predict the direction of the reaction. For the given reaction: \[ 2 \text{NO}_2(g) \rightleftharpoons \text{N}_2\text{O}_4(g) \], the equilibrium constant can be calculated using thermodynamic data such as the change in enthalpy \( \Delta H^\circ \) and the change in entropy \( \Delta S^\circ \). At a specific temperature, the Van't Hoff equation links these values with the equilibrium constant. It is given by: \[ \ln K = -\frac{\Delta H^\circ}{R} \cdot \frac{1}{T} + \frac{\Delta S^\circ}{R} \] where:

• \( R \) is the universal gas constant \( (8.314 \, \text{J/mol} \cdot \text{K}) \)
• \( T \) is the temperature in Kelvin

Understanding the equilibrium constant is key as it tells us to what extent a reaction will proceed and how conditions like temperature can shift the balance of this equilibrium.

Temperature Conversion

When working with thermochemical equations and calculations, it's essential to express temperature in Kelvin rather than Celsius. The Kelvin scale is an absolute thermodynamic temperature scale that eliminates negative values and aligns directly with the precision needed in scientific calculations. To convert from Celsius to Kelvin, the formula is straightforward: \[ T_{K} = T_{C} + 273.15 \] where:

• \( T_{C} \) is the temperature in degrees Celsius
• \( T_{K} \) is the temperature in Kelvin

For instance, if you have a temperature of \( 25.0^{\circ}C \), adding \( 273.15 \) gives you \( 298.15 \, K \). Similarly, for \( 100.0^{\circ}C \), it converts to \( 373.15 \, K \). Always ensure temperatures are in Kelvin when applying the Van't Hoff equation for consistency and accuracy in results.

Gibbs Free Energy Change

Gibbs free energy change, denoted by \( \Delta G^\circ \), is another thermodynamic parameter crucial for understanding chemical reactions. It merges the concepts of enthalpy and entropy, providing insight into the spontaneity of a reaction.The formula for Gibbs free energy change at constant temperature is: \[ \Delta G^\circ = \Delta H^\circ - T \times \Delta S^\circ \] where:

• \( \Delta H^\circ \) is the standard enthalpy change
• \( T \) is the temperature in Kelvin
• \( \Delta S^\circ \) is the standard entropy change

If \( \Delta G^\circ \) is negative, it indicates that the reaction can occur spontaneously. In contrast, a positive \( \Delta G^\circ \) suggests that the reaction is non-spontaneous under standard conditions. By relating \( \Delta G^\circ \) to the equilibrium constant \( K \) through the relation: \[ \Delta G^\circ = -RT \ln K \], we can see the interplay between temperature and equilibrium position, as well as how changes in thermodynamic properties influence reaction spontaneity.