[ajax0604]

For this equilibrium system, 2HI  H₂ + I₂ ΔH = -ve, I would like to sketch a rate vs time graph for both the forward and reverse reaction rates if the temperature was decreased. Would it look like this? (I have attached a picture)
I understand that the rate of both forward and reverse reactions will decrease if temperature decreased but would it result in a sudden drop in the rates as I have drawn on my graph or would it be more gradual? Also, would the rate of the reverse reaction drop more initially? Thank you.

[mjc123]

I understand that the rate of both forward and reverse reactions will decrease if temperature decreased but would it result in a sudden drop in the rates as I have drawn on my graph or would it be more gradual?

Depends how sudden the temperature change is; how quickly thermal equilibrium is established at the new temperature.

Also, would the rate of the reverse reaction drop more initially?

Which reaction has the higher activation energy; forward or reverse?
Which rate constant is more sensitive to temperature change, the one with higher or lower activation energy?

[ajax0604]

The question doesn't specify how sudden the temperature change is. if the temperature change is sudden, can I assume that the rates would drop sharply as I have shown on my sketch?
Since it's an exothermic reaction, the reverse reaction would have higher activation energy. I think the reverse reaction will be more sensitive to temperature changes. So the reaction rate for the reverse reaction will drop more than the forward reaction rate initially.
Is the second part of my graph correct where the rate of the reverse reaction catches up to the forward one?

[mjc123]

Yes, the reverse reaction rate will initially be slower, so the equilibrium will shift to the right. As the concentrations change, the forward reaction will become slower and the reverse reaction faster, until the new equilibrium situation is reached.

[ajax0604]

Thank you. Sorry I have one more question. If the temperature was lowered gradually, would the graph look something like this?

[mjc123]

Yes, something like that