[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

[Mocking_bird]

(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.