Thermodynamics vs. Kinetics

[amikhchi]

---

Members don't see this ad.

I know most books will swear by the fact that these two are separate and I get that, and I usually don't fall for the trick-questions giving you a thermo stat and a kinetic answer.

But can someone explain what this means in a simpler way, I tried asking a teacher if they are separate than why does temperature affect both (i'm sure that doesnt' violate the law, but my teacher had a hard time explaining that)

Changing temperature changes your rate law (by changing the k right?)

and changing temperature will alter your Gibbs-free-energy right?

Side q:

If you have an exothermic rxn, heat is considered a product, if you add heat i know it will stress the reverse rxn b/c of le chatlier; will this alter the Keq or does the ratio of products/reactants stay the same?

 

[Bernoull]

I know most books will swear by the fact that these two are separate and I get that, and I usually don't fall for the trick-questions giving you a thermo stat and a kinetic answer.

But can someone explain what this means in a simpler way, I tried asking a teacher if they are separate than why does temperature affect both (i'm sure that doesnt' violate the law, but my teacher had a hard time explaining that)

Changing temperature changes your rate law (by changing the k right?)

and changing temperature will alter your Gibbs-free-energy right?

Side q:

If you have an exothermic rxn, heat is considered a product, if you add heat i know it will stress the reverse rxn b/c of le chatlier; will this alter the Keq or does the ratio of products/reactants stay the same?

From the Arrhenius Eq and the delta (Gibbs) equations, I can see how temperature affects both. To explain it conceptually, I'll leave that to someone else. Chemistry is not my strongest suit.

k= zpe(RT/Ea) where z=collision frequency, p= steric factor, Ea =activation energy, T = temp in K

From here, its clear that k is directly proportional to temperature, in fact reaction rate doubles per 10 degree (C) increase in temperature. Just bear in mind that this is a generalisation.

For Gibbs free energy (change) = delta (H) - delta(S)T
where H is enthalpy, S is entropy and T is temp (K)
We all know a spontaneous rxn has negative delta(G), therefore as temperature increases, the (- delta(S)T) portion of the Eq becomes more negative and ceteris paribus, the spontaneity of the rxn increases!!

I hope this helped.