[beheada]

In drawing an energy diagram with the reactants and products at approximately the same level, but with an intermediate which is at a more stable energy level (i.e. lower), does the reaction favor the intermediate since it is more stable than both the reactant and product, or does the reaction favor neither the reactant, intermediate, or product?

And in literal terms, if it does favor the intermediate, what does that tell you about the intermediate? That the intermediate will tend to form more than the product? Can't seem to get the answer out of my book.

[english]

Are you referring to thermodynamic stability or kinetic stability?

I'm assuming you are referring to thermodynamic stability.  It seems that the intermediate is more stable than both reactants and products, following your description.

This would simply mean that the intermediate will form more; however, this seems quite odd because normally intermediates are quite unstable.

Given the description this is what I am getting though.

This makes me ask the question to anyone more knowledgeable:  Is this a hypothetical reaction or one that can actually happen this way, and can you give an example of one that follows such a reaction diagram?

[beheada]

It is a thermodynamic energy diagram... and I'm assuming it is hypothetical because it is a question on my homework (probably made up by the professor). The exact wording for the diagram is thus:

"Sketch a reaction energy diagram for a two-step reaction, where the second step is slowest. Assume that the reactants are higher in energy than the intermediates. Assume that the intermediates are lower in energy than the products. Assume that the reactants and products are of the same energy.

a) Overall reaction is at? Equilibrium
b) Will the equilibrium constant most likely be at? 1
c) Does the reaction favor reactants, products, intermediates, neither? Intermediates or neither... not sure
d) Gibbs Free Energy change = approximately 0
e) If a more stable intermediate was formed, would the rate of the overall rxn be faster? NOPE, because according to Hammond's Postulate, the species each transition state resembles is either the product or the reactant, and thus the changing of stability of the intermediate would not affect either transition state.


I'd draw my diagram, but you get the picture. I'm just stumped as to whether the reaction would be favoring the intermediates, since technically they are the most stable.... I know they're more STABLE, but does that mean that they are FAVORED in the reaction? I posted the other questions in case someone felt like correcting me if I made any errors.

[english]

By definition, whatever is thermodynamically stable is favored.  Stability means favorability, simply because the more stable it is, the more feasible it is for it to form.

If you are saying that the intermediate is more stable, and your K=1, then this obviously means that your G° change is 0, which means that the products and reactants are at the same energy, which is higher than that of the intermediate.

The intermediate should form more readily then.  Your answer makes sense.

[english]

e) If a more stable intermediate was formed, would the rate of the overall rxn be faster? NOPE, because according to Hammond's Postulate, the species each transition state resembles is either the product or the reactant, and thus the changing of stability of the intermediate would not affect either transition state.

Hammond's Postulate does not necessarily apply to the reactants and products of the reaction, such that the transition state cannot be similar in energy to the intermediate.  The transition state, in this case there are 2, will resemble the species to which it is closer in energy. 

If we were to cut this diagram in two, and focus just on reactant to intermediate, and intermediate to product, respectively, the transition state of each could easily be seen as resembling the reactant and product.

My point is that transition states will not necessarily resemble reactant and product, although in this problem that is the case.

[beheada]

Right. I worded that a bit wrong. What I meant was according to H's P... the transition state resembles the closest species to it and in THIS case, since there are two transition states, the first one favors the reactants, the other favors the product, so adding stability to the intermediate is moot.

[english]

Right. I worded that a bit wrong. What I meant was according to H's P... the transition state resembles the closest species to it and in THIS case, since there are two transition states, the first one favors the reactants, the other favors the product, so adding stability to the intermediate is moot.

Right.  Well you are right from what you've said.  Don't doubt yourself.