You are Here: Home

Announcements Posted on
Uni student? You could win a Samsung Galaxy S7 or £500 of travel vouchers by taking this quick survey 05-05-2016
Talking about ISA/EMPA specifics is against our guidelines - read more here 28-04-2016
1. Apologies if this comes across a silly question, but I'm having trouble getting my head around this concept.

There is a reversible reaction, exothermic in one direction and endothermic in the opposite, e.g.
2SO₂ + O₂ ↔ 2SO₃ ∆H = -197mol-1

I know that when temperature is increased, position of equilibrium moves in the direction that decreases temperature.

What I don't understand is why, when the temperature of this system is increased, does it move in the endothermic direction? To me it makes more sense to move in the exothermic direction as that would give out heat where the endothermic direction would take in heat?

I'm clearly missing something here, so if anybody could explain this to me I'd much appreciate it, thanks
2. (Original post by pjanoo)
Apologies if this comes across a silly question, but I'm having trouble getting my head around this concept.

There is a reversible reaction, exothermic in one direction and endothermic in the opposite, e.g.
2SO₂ + O₂ ↔ 2SO₃ ∆H = -197mol-1

I know that when temperature is increased, position of equilibrium moves in the direction that decreases temperature.

What I don't understand is why, when the temperature of this system is increased, does it move in the endothermic direction? To me it makes more sense to move in the exothermic direction as that would give out heat where the endothermic direction would take in heat?

I'm clearly missing something here, so if anybody could explain this to me I'd much appreciate it, thanks

I was confused at exactly the same thing as you, but the way my teacher described it helped:

Imagine you're doing this reaction in a sealed container - its the temperature of the container which is effecting the reaction. So if the temperature is increased, it will move in the endothermic direction to absorb the heat from the container and decrease the temperature.

## Register

Thanks for posting! You just need to create an account in order to submit the post
1. this can't be left blank
2. this can't be left blank
3. this can't be left blank

6 characters or longer with both numbers and letters is safer

4. this can't be left empty
1. Oops, you need to agree to our Ts&Cs to register

Updated: March 14, 2012
TSR Support Team

We have a brilliant team of more than 60 Support Team members looking after discussions on The Student Room, helping to make it a fun, safe and useful place to hang out.

This forum is supported by:
Today on TSR

Don't be late