Kinetics!

[chiddler]

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Step 1: H2O2 + I (-) --> H2O + IO (-) (fast)

Step 2: H2O2 + IO (-) --> H2O + O2 + I (-) (slow)

Sum of reactions: 2 H2O2 --> 2H2O + O2

what is the rate law here? IO- is an intermediate and therefore cannot be part of the rate law.

So is it

rate=k [H2O2]^2 * [I-] ?

 

[Postictal Raiden]

Step 1: H2O2 + I (-) --> H2O + IO (-) (fast)

Step 2: H2O2 + IO (-) --> H2O + O2 + I (-) (slow)

Sum of reactions: 2 H2O2 --> 2H2O + O2

what is the rate law here? IO- is an intermediate and therefore cannot be part of the rate law.

So is it

rate=k [H2O2]^2 * [I-] ?

Seems that I- is a catalyst. I say the rate law should be
rate=k [H2O2]
since the rate law can be determined using the reactants of the rate determining step.

 

[Endure]

Step 1: H2O2 + I (-) --> H2O + IO (-) (fast)

Step 2: H2O2 + IO (-) --> H2O + O2 + I (-) (slow)

Sum of reactions: 2 H2O2 --> 2H2O + O2

what is the rate law here? IO- is an intermediate and therefore cannot be part of the rate law.

So is it

rate=k [H2O2]^2 * [I-] ?

I would agree with your answer, yes (assuming water is excluded from the expression).

In any complex reaction where a fast step precedes the slow step, assume that the fast step reaches dynamic equilibrium quickly, equate the forward and reverse reaction rates, and solve for the intermediate. After substitution into your original rate law, this will give you a rate law that includes the reactants in the net reaction.

Seems that I- is a catalyst. I say the rate law should be
rate=k [H2O2]
since the rate law can be determined using the reactants of the rate determining step.

Since I- is the catalyst, it's logical to assume that it would be present in the rate law. The rate of the reaction depends on the slow step, but the law you provided doesn't say anything about what happens to the rate of the reaction when the concentration of I- is increased. This would alter your reaction order, rate constant, and ability to predict the half-life.

 

[slz1900]

You can rewrite [H2O2][IO] so that it doesn't include an intermediate. You have to figure out what [IO] is equal to.

Look at the first equation:

H2O2 + I (-) --> H2O + IO (-)

[H2O2] = [H2O][IO]

[IO] = [H2O2]/[H2O]

Plug that back into the rate equation and you get:

k = [H2O2]^2/[H2O]

I don't think you can exclude water here since it isn't just acting as a solvent. Increasing its concentration will slow your reaction, and removing it will drive the reaction forward. Why would you leave it out?

 

[Postictal Raiden]

I would agree with your answer, yes (assuming water is excluded from the expression).

In any complex reaction where a fast step precedes the slow step, assume that the fast step reaches dynamic equilibrium quickly, equate the forward and reverse reaction rates, and solve for the intermediate. After substitution into your original rate law, this will give you a rate law that includes the reactants in the net reaction.



Since I- is the catalyst, it's logical to assume that it would be present in the rate law. The rate of the reaction depends on the slow step, but the law you provided doesn't say anything about what happens to the rate of the reaction when the concentration of I- is increased. This would alter your reaction order, rate constant, and ability to predict the half-life.

Oops, sorry. I totally ignored the fact that the slow step was NOT the first step. In this case, yes, the rate law should still be the rate determining step, but the intermediate must be replaced by its algebraic equivalent from the previous step.

However, shouldn't we also account for the different rate constants, k, that are involved in each step? In the rate law shouldn't we have something like k2*k1/k-1 instead of k?

 

[ouroboros39]

Fun question actually, we covered this type of problem in the Ochem class I took recently. You can make 1 one of two assumptions: if you assume water is negligible (as it might be in the very start of the reaction) the rate would be

k1[H2O2][I-]

after a while water would build and the rate would change to something closer to this

Keq*k2[H2O2]^2*[I-]/[H2O]

I dont think they wanted that one so the first is probably what they are looking for. Where did you get this questions by the way?

 

[Endure]

The rate law can be expressed using the species present in the rate determining step, but it can't be compared directly to experimental data in that form. Assuming equilibrium and solving for the reaction intermediate allows you to (a) write a rate law according to the reactants in the net equation and (b) find the experimentally observed rate constant, k.

This mechanism doesn't allow you to do that, meaning the OP probably mixed up which elementary reaction was slow. If we go by this assumption, then the rate law becomes:

Rate = k[H2O2][I-]

This makes sense to me. It also includes the catalyst (which, according to the Arrhenius equation, affects k by either lowering the activation energy or increasing the frequency factor).

Look at the first equation:

H2O2 + I (-) --> H2O + IO (-)

[H2O2] = [H2O][IO]

[IO] = [H2O2]/[H2O]

Plug that back into the rate equation and you get:

k = [H2O2]^2/[H2O]

The rate law should never be equivalent to k unless it's zero order.

Oops, sorry. I totally ignored the fact that the slow step was NOT the first step. In this case, yes, the rate law should still be the rate determining step, but the intermediate must be replaced by its algebraic equivalent from the previous step.

However, shouldn't we also account for the different rate constants, k, that are involved in each step? In the rate law shouldn't we have something like k2*k1/k-1 instead of k?

Yep. The resulting rate constant is the experimentally observed rate constant.

 

[slz1900]

The rate law can be expressed using the species present in the rate determining step, but it can't be compared directly to experimental data in that form. Assuming equilibrium and solving for the reaction intermediate allows you to (a) write a rate law according to the reactants in the net equation and (b) find the experimentally observed rate constant, k.

This mechanism doesn't allow you to do that, meaning the OP probably mixed up which elementary reaction was slow. If we go by this assumption, then the rate law becomes:

Rate = k[H2O2][I-]

This makes sense to me. It also includes the catalyst (which, according to the Arrhenius equation, affects k by either lowering the activation energy or increasing the frequency factor).



The rate law should never be equivalent to k unless it's zero order.



Yep. The resulting rate constant is the experimentally observed rate constant.

Sorry, meant to just type k[H2O2]^2/[H2O] I am fairly certain that water needs to be in there. I could be wrong, but unlike an equilibrium constant where varying water doesn't usually affect the direction of the reaction, it certainly affects the rate.