Question
How does temperature affect the equilibrium constant in the reaction N2O4(g) --> 2NO2(g)?
Answers
GPT-4o mini
The equilibrium constant for a given reaction is affected by temperature because it is derived from the Gibbs free energy change for that reaction. For the reaction you've mentioned, \( \text{N}_2\text{O}_4(g) \rightleftharpoons 2 \text{NO}_2(g) \), we need to consider the nature of the reaction to understand how temperature affects the equilibrium constant (\( K \)).
This reaction is an endothermic process, meaning that it absorbs heat. In general, the relationship between temperature and the equilibrium constant can be analyzed using the van 't Hoff equation, which describes how the equilibrium constant changes with temperature:
\[
\ln \left( \frac{K_2}{K_1} \right) = -\frac{\Delta H^\circ}{R} \left( \frac{1}{T_2} - \frac{1}{T_1} \right)
\]
Where:
- \( K_1 \) and \( K_2 \) are the equilibrium constants at temperatures \( T_1 \) and \( T_2 \), respectively.
- \( \Delta H^\circ \) is the standard enthalpy change of the reaction.
- \( R \) is the universal gas constant.
- \( T \) is the temperature in Kelvin.
For endothermic reactions like the formation of NO2 from N2O4 (\( \Delta H^\circ > 0 \)), increasing the temperature will result in an increase in the equilibrium constant \( K \). This is because more heat will drive the reaction toward the products (NO2) to absorb the added heat.
Conversely, if the temperature is decreased, the equilibrium constant \( K \) will decrease, favoring the formation of the reactants (N2O4).
In summary, for the reaction \( \text{N}_2\text{O}_4(g) \rightleftharpoons 2 \text{NO}_2(g) \):
- **Increase in temperature**: \( K \) increases (favoring products)
- **Decrease in temperature**: \( K \) decreases (favoring reactants)
This reaction is an endothermic process, meaning that it absorbs heat. In general, the relationship between temperature and the equilibrium constant can be analyzed using the van 't Hoff equation, which describes how the equilibrium constant changes with temperature:
\[
\ln \left( \frac{K_2}{K_1} \right) = -\frac{\Delta H^\circ}{R} \left( \frac{1}{T_2} - \frac{1}{T_1} \right)
\]
Where:
- \( K_1 \) and \( K_2 \) are the equilibrium constants at temperatures \( T_1 \) and \( T_2 \), respectively.
- \( \Delta H^\circ \) is the standard enthalpy change of the reaction.
- \( R \) is the universal gas constant.
- \( T \) is the temperature in Kelvin.
For endothermic reactions like the formation of NO2 from N2O4 (\( \Delta H^\circ > 0 \)), increasing the temperature will result in an increase in the equilibrium constant \( K \). This is because more heat will drive the reaction toward the products (NO2) to absorb the added heat.
Conversely, if the temperature is decreased, the equilibrium constant \( K \) will decrease, favoring the formation of the reactants (N2O4).
In summary, for the reaction \( \text{N}_2\text{O}_4(g) \rightleftharpoons 2 \text{NO}_2(g) \):
- **Increase in temperature**: \( K \) increases (favoring products)
- **Decrease in temperature**: \( K \) decreases (favoring reactants)