Le Chatelier's Principle- Temperature

Moderators: Chem_Mod, Chem_Admin

Ashley Han 1F
Posts: 12
Joined: Fri Sep 26, 2014 2:02 pm

Le Chatelier's Principle- Temperature

Postby Ashley Han 1F » Tue Dec 09, 2014 12:13 am

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
Posts: 27
Joined: Fri Sep 26, 2014 2:02 pm

Re: Le Chatelier's Principle- Temperature

Postby K Honeychurch 1K » Tue Dec 09, 2014 12:19 am

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
Posts: 10
Joined: Fri Sep 26, 2014 2:02 pm

Re: Le Chatelier's Principle- Temperature

Postby Vasudev Tadimeti 3B » Tue Dec 09, 2014 10:32 am

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
Posts: 8
Joined: Fri Sep 26, 2014 2:02 pm

Re: Le Chatelier's Principle- Temperature

Postby John Colarusso 1F » Tue Dec 09, 2014 9:58 pm

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
Posts: 9
Joined: Fri Sep 26, 2014 2:02 pm

Re: Le Chatelier's Principle- Temperature

Postby 504420311 » Thu Dec 11, 2014 11:06 am

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
Posts: 13
Joined: Fri Sep 26, 2014 2:02 pm

Re: Le Chatelier's Principle- Temperature

Postby katia makhovik 4F » Fri Dec 12, 2014 1:37 am

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
Posts: 11
Joined: Fri Sep 26, 2014 2:02 pm

Re: Le Chatelier's Principle- Temperature

Postby EJunChang1C » Fri Dec 12, 2014 2:41 am

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.


Return to “Applying Le Chatelier's Principle to Changes in Chemical & Physical Conditions”

Who is online

Users browsing this forum: No registered users and 1 guest