temperature dependence of reactions where entropy goes down

[qtpai]

So, the rate constant is determined by the following equation:

k = Ae^(-Eact/RT)

Gibbs free energy is given by the following equation:

G = dH - TdS

If dS is negative for a reaction, then increasing temperature should make the G more positive, i.e. should make the reaction less favorable.

This leads to question 1:
Even as the reaction is becoming less favorable, should the reaction rate still increase according to the equation that determines the rate constant?

Similarly:

With Le Chatelier's Principle, for the following reaction:

Reactants -> Products + Heat,

increasing temperature shifts the reaction to the right from equilibrium.

Question 2: But should the rate constant still increase with an increase in temperature when reactants are at their initial concentration?

This seems counterintuitive to me. Looking at it from the first angle, I can see how an increase in temperature could increase the rate of reaction, even as it would make Gibbs free energy more positive, by decreasing the activation energy. Right?

But with respect to question 2, I can't see how adding heat, even if added when the reaction is still spontaneous, should cause a faster initial reaction rate.

(EDIT: The reason I ask this is that part of the explanation in TBR, GChem, Chapter 9, for Problem 72, says:

"Because the rate constant varies directly with temperature, it is always true that as the temperature is increased, the reaction rate increases.")

Please clarify. Thank you.

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[SaintJude]

With Le Chatelier's Principle, for the following reaction:

Reactants -> Products + Heat,

increasing temperature shifts the reaction to the right from equilibrium.

You're saying a lot right now & I don't have time (&intellect) to tackle it all. But I did want to make a quick note.

Increasing temp, will shift the reaction above to the left so forward reaction will actually be less favorable

 

[milski]

It all revolves around what SaintJude said. For the reaction to happen you need a negative ΔG. T is always positive and if ΔS is negative your only way to have ΔG negative is a negative ΔH. Negative ΔH means that the reaction has to be exothermic:

Reactants <-> Products + Heat

Adding heat to this reaction will move the equilibrium further left, which is also what you get from the &#916;G equation - higher T will make it less negative and the reaction will be less favorable.

Your conclusions about the kinetics of the reaction are correct. If you increase the temperature the reaction will happen faster even though the equilibrium will not be so far right in that case. How fast the reaction happens and where the equilibrium is are two very different things. You'll see that more in organic chemistry but for certain reactions you can control the products that you get based on the temperature and pushing the equilibrium to one or the other side.

 

[qtpai]

It all revolves around what SaintJude said. For the reaction to happen you need a negative &#916;G. T is always positive and if &#916;S is negative your only way to have &#916;G negative is a negative &#916;H. Negative &#916;H means that the reaction has to be exothermic:

Reactants <-> Products + Heat

Adding heat to this reaction will move the equilibrium further left, which is also what you get from the &#916;G equation - higher T will make it less negative and the reaction will be less favorable.

Your conclusions about the kinetics of the reaction are correct. If you increase the temperature the reaction will happen faster even though the equilibrium will not be so far right in that case. How fast the reaction happens and where the equilibrium is are two very different things. You'll see that more in organic chemistry but for certain reactions you can control the products that you get based on the temperature and pushing the equilibrium to one or the other side.

For an endothermic reaction, K can be lower at higher temperature, but the reaction rate still faster. That's very interesting.

And I suppose the same goes for a reaction where entropy goes down: even though Gibbs free energy becomes more positive (but is still negative), the reaction is less favorable (and therefore a lower K), but the reaction rate in the forward direction is still faster.

Equilibrium being different from reaction rate. I just wanted to make sure my thinking was right. Still seems a bit odd to me, but that's what the math and concepts say. So I'll keep thinking through it.

Thanks.