1. The Relationship:
* For exothermic reactions (ΔH < 0): Increasing temperature shifts the equilibrium to the left, favoring the reactants. This is because the system wants to relieve the stress of added heat by consuming some of it, which occurs by shifting back towards the reactants. Consequently, Keq decreases with increasing temperature.
* For endothermic reactions (ΔH > 0): Increasing temperature shifts the equilibrium to the right, favoring the products. The system absorbs heat to relieve the stress, which means it favors the reaction that produces heat, the forward reaction. Therefore, Keq increases with increasing temperature.
2. Van't Hoff Equation:
The relationship between temperature and Keq is quantified by the Van't Hoff equation:
```
ln(K2/K1) = -ΔH°/R * (1/T2 - 1/T1)
```
where:
* K1 and K2 are the equilibrium constants at temperatures T1 and T2, respectively.
* ΔH° is the standard enthalpy change of the reaction.
* R is the ideal gas constant.
3. Key Points:
* The change in Keq with temperature is directly related to the enthalpy change (ΔH°) of the reaction.
* A large enthalpy change results in a more significant change in Keq with temperature.
* The Van't Hoff equation is a powerful tool for predicting how temperature affects the equilibrium of a reaction.
In summary:
* Exothermic reactions: Higher temperature favors reactants, smaller Keq.
* Endothermic reactions: Higher temperature favors products, larger Keq.
Example:
Consider the Haber process for ammonia synthesis:
N2(g) + 3H2(g) ⇌ 2NH3(g) ΔH < 0 (exothermic)
Increasing temperature will shift the equilibrium to the left, favoring the reactants (N2 and H2). This means the yield of ammonia (NH3) will decrease at higher temperatures.
Important Note: The effect of temperature on Keq is only one factor that can influence the outcome of a reaction. Other factors like pressure, concentration, and catalysts can also play a significant role.
