Le Chatelier's Principle- Temperature

[Ashley Han 1F]

Hello,

I understand how a change in volume and moles affects an equilibrium reaction, but I'm confused on how temperature affects the reaction. I was hoping that someone could explain how an increase or decrease in temperature affects both an exothermic and endothermic reaction at equilibrium.

Thank you so much!

[K Honeychurch 1K]

Think of energy (in this case heat) as a reactant or product. In an exothermic reaction heat is a product, so if you increase the temperature (more heat) the reaction will favor the reactants and K will decrease. In an endothermic reaction heat is a reactant, so if you increase the temperature (more heat) the reaction will favor the products and K will increase.

[Vasudev Tadimeti 3B]

If you increase the temperature in an endothermic reaction, the the reaction will favor the products, while in an exothermic reaction, increasing the temperature will favor the reactants. If heat is given off, the reaction will be exothermic, and if absorbed then the reaction will be endothermic. So look if the change in heat is positive or negative. Hope this helps!

[John Colarusso 1F]

Raising the temperature for an endothermic reaction will favor the formation of products, so K will shift right. The reasoning for this is that, because the reaction requires heat, it will use the excess heat to further the reaction.

Raising the temperature for an exothermic reaction will favor the formation of reactants, so K will shift left. Because the reverse reaction of an exothermic reaction is endothermic, and endothermic reactions thrive under increased temperatures, as shown above, the reverse reaction is favored, so the formation of reactants is favored.

[504420311]

Its also important to remember that because a change in volume or moles just shifts the equilibrium in one direction it does not change the K value, but because temperature drives a reaction in one direction completely based on adding or subtracting heat and being endothermic or exothermic, the K value will change when temperature does.

[katia makhovik 4F]

what if it doesn't specify whether a reaction is endothermic or exothermic? Then how would you know whether the conditions favor high or low temp? For example, on the fall 2013 Q5D, when Br2(g) is in equilibrium with 2Br(g), and it asks what physical conditions favor the production of bromine atoms, how would you know that you want High temp?

[EJunChang1C]

Br2(g) to 2Br(g) will be a bond-breaking reaction which is an endothermic reaction. So to produce more bromine atoms (products), temperature should increase shifting the reaction to the right.