Kinetics

[chiddler]

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A + B = C (fast)
C + A = D (slow)


Rate of reaction is given by the slow step. But do we write rate=k[C][A] even though C is an intermediate?

EK says so. I want to verify because this seems wrong since generally you can't have an initial concentration of intermediate.

 

[dsoz]

I would write it as rate =k[A]^2 since in order to make C, A and B need to combine (hence the A squared in the rate expression). has an effect on the overall rate, but A has twice the effect.

dsoz

 

[chiddler]

I would write it as rate =k[A]^2 since in order to make C, A and B need to combine (hence the A squared in the rate expression). has an effect on the overall rate, but A has twice the effect.

dsoz

I dont think A^2 is accurate because the first A is in the fast step and therefore has no significant effect on the rate.

 

[Melomare17]

i think for rate law you just write the rate as a function of the substrate concentrations of the slow step. even if its an intermediate i think u still would.

 

[MedPR]

A + B = C (fast)
C + A = D (slow)


Rate of reaction is given by the slow step. But do we write rate=k[C][A] even though C is an intermediate?

EK says so. I want to verify because this seems wrong since generally you can't have an initial concentration of intermediate.

Yes, rate=k[C][A] because even though C is an intermediate, the slow step is still dependent on its concentration.

It's kind of like how in a two lens system the first image serves as the second object. It's not really an object, but you count it as one for the sake of moving forward.

 

[chiddler]

Yes, rate=k[C][A] because even though C is an intermediate, the slow step is still dependent on its concentration.

It's kind of like how in a two lens system the first image serves as the second object. It's not really an object, but you count it as one for the sake of moving forward.

thanks that explains it well enough.

 

[sweet d]

A + B = C (fast)
C + A = D (slow)


Rate of reaction is given by the slow step. But do we write rate=k[C][A] even though C is an intermediate?

EK says so. I want to verify because this seems wrong since generally you can't have an initial concentration of intermediate.

I don't think that is correct either. The rate of the slow step is rate = k[C][A], but the rate of the overall reaction should not include any intermediates (they are often unstable and it is difficult, if not impossible, to determine the concentration).

Assuming the fast step is reversible, for the overall rate I get:

rate1 = k1[A] = k-1[C]
So [C] = (k1/k-1)[A]

rate2 = k2[C][A]

Substitute for [C]:
rate = (k1*k2/k-1)[A]^2, which could be rewritten as k[A]^2 if you want to combine the constants

Above you said that [A]^2 shouldn't be included because A is in the fast step and doesn't affect the rate. In this case, though, A is involved in producing an intermediate that does affect the rate, so [A] and affect the rate by proxy.

 

[chiddler]

I don't think that is correct either. The rate of the slow step is rate = k[C][A], but the rate of the overall reaction should not include any intermediates (they are often unstable and it is difficult, if not impossible, to determine the concentration).

Assuming the fast step is reversible, for the overall rate I get:

rate1 = k1[A] = k-1[C]
So [C] = (k1/k-1)[A]

rate2 = k2[C][A]

Substitute for [C]:
rate = (k1*k2/k-1)[A]^2, which could be rewritten as k[A]^2 if you want to combine the constants

Above you said that [A]^2 shouldn't be included because A is in the fast step and doesn't affect the rate. In this case, though, A is involved in producing an intermediate that does affect the rate, so [A] and affect the rate by proxy.

i like this and it makes a lot of sense.

the rate can still be expressed as rate = k[C][A], right? or is that entirely improper because C is an intermediate?

 

[sweet d]

As far as I know, the rate law for the overall reaction cannot include any intermediates.

 

[MedPR]

This seems like a pretty thorough explanation, but I can't be bothered to read it (sorry).

http://www.science.uwaterloo.ca/~cchieh/cact/c123/ratemech.html

 

[Buttafuoco]

Sweet D's explanation makes sense.

Here is a Yahoo Answers bleeb that covers this as well:

http://answers.yahoo.com/question/index?qid=20080610204445AA3TNBi

This is kind of a funny topic because I don't ever recall having to find a rate law like this through gen chem, pchem, or studying for the MCAT. Maybe I did and just forgot it though.

 

[MedPR]

TBR is a little more thorough than other prep companies. You might end up doing some low yield and maybe useless stuff, but you'll be prepared for it if it shows up on your MCAT.

 

[chiddler]

good finds. i'm learning a lot from those links.