π Le Chatelier's Principle: Temperature vs. Pressure
Le Chatelier's Principle states that if a change of condition (like temperature, pressure, or concentration) is applied to a system in equilibrium, the system will shift in a direction that relieves the stress.
π‘οΈ Temperature Effects on Equilibrium
Temperature changes affect reactions differently depending on whether they are endothermic (heat absorbed) or exothermic (heat released). Think of heat as a reactant or product.
- π₯ Endothermic Reactions: π§ͺ If you increase the temperature in an endothermic reaction (heat is a reactant), the equilibrium will shift to the products side. Conversely, decreasing the temperature shifts the equilibrium towards the reactants. For example, consider the reaction: $N_2O_4(g) \rightleftharpoons 2NO_2(g)$ (endothermic). Adding heat will favor the formation of $NO_2$.
- βοΈ Exothermic Reactions: π‘ If you increase the temperature in an exothermic reaction (heat is a product), the equilibrium will shift to the reactants side. Decreasing the temperature shifts the equilibrium towards the products. For example: $2SO_2(g) + O_2(g) \rightleftharpoons 2SO_3(g)$ (exothermic). Adding heat will favor the formation of $SO_2$ and $O_2$.
π¨ Pressure Effects on Equilibrium
Pressure changes primarily affect reactions involving gases. The key is to look at the number of moles of gas on each side of the equation.
- βοΈ Increase in Pressure: βοΈ Increasing the pressure will shift the equilibrium towards the side with fewer moles of gas. This reduces the volume and thus relieves the pressure. For example: $N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g)$. Increasing pressure favors the formation of $NH_3$ because there are 4 moles of gas on the reactant side and 2 moles on the product side.
- π Decrease in Pressure: π Decreasing the pressure will shift the equilibrium towards the side with more moles of gas. For example, in the same reaction: $N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g)$, decreasing pressure favors the formation of $N_2$ and $H_2$.
- π― No Change: π’ If the number of moles of gas is the same on both sides of the equation, changes in pressure will have virtually no effect on the equilibrium. Example: $H_2(g) + I_2(g) \rightleftharpoons 2HI(g)$.
π Temperature and Pressure: Side-by-Side Comparison
| Feature |
Temperature |
Pressure |
| Applicability |
All reactions (endothermic & exothermic) |
Primarily reactions involving gases |
| Effect on Equilibrium |
Shifts based on whether reaction is endothermic or exothermic |
Shifts towards side with fewer moles of gas when pressure increases |
| Key Consideration |
Heat as a reactant (endothermic) or product (exothermic) |
Number of moles of gas on each side of the equation |
| No Effect |
Adding or removing catalyst |
Equal number of gas moles on both sides |
π Key Takeaways
- π― Temperature: π₯ Affects all equilibria by shifting towards products in endothermic reactions when heated and towards reactants in exothermic reactions when heated.
- π‘ Pressure: π¨ Primarily affects gaseous equilibria by shifting towards fewer gas moles when pressure increases.
- π§ͺ Moles of Gas: π’ Pressure changes are ineffective if there are equal gas moles on both sides of the reaction.