Rate law when the slow step is NOT the 1st step in a reaction

[EECStoMed]

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So according to EK, when the slow step is not the 1st step in an overall reaction, we must take in account the intermediates that are consumed in the slow step of the reaction. On the MCAT do we need to know this? Here's the example:

1. NO2 + NO2 -> NO3 + NO slow step
2. NO3 + CO -> NO2 + CO2 fast step

rate law: k2[NO2]^2

but if the speeds are reversed, then it's this:

1. NO2 + NO2 -> NO3 + NO fast step
2. NO3 + CO -> NO2 + CO2 slow step

rate law NOT: k2[NO3][CO] but IS: k2k1/k-1 [NO3][CO] k-1 = reverse reaction of step 1

 

[axp107]

For some reason... I REALLY don't think we need to know that.
It's not covered in my Kaplan class/materials/anywhere...

EK just seems to kinda have that extra info.

 

[EECStoMed]

For some reason... I REALLY don't think we need to know that.
It's not covered in my Kaplan class/materials/anywhere...

EK just seems to kinda have that extra info.

hahs, it seems like you're the only one answering my question. NOW, refer to the other post because I just posted a reply to you!

 

[Dr. Pepper]

So according to EK, when the slow step is not the 1st step in an overall reaction, we must take in account the intermediates that are consumed in the slow step of the reaction. On the MCAT do we need to know this? Here's the example:

1. NO2 + NO2 -> NO3 + NO slow step
2. NO3 + CO -> NO2 + CO2 fast step

rate law: k2[NO2]^2

but if the speeds are reversed, then it's this:

1. NO2 + NO2 -> NO3 + NO fast step
2. NO3 + CO -> NO2 + CO2 slow step

rate law NOT: k2[NO3][CO] but IS: k2k1/k-1 [NO3][CO] k-1 = reverse reaction of step 1

The rate law is determined by the rate-determining step and all steps preceding the rate determining step. This is done to remove intermediates from the rate law.
-Dr. P.

 

[mrmilad]

but if the speeds are reversed, then it's this:

1. NO2 + NO2 -> NO3 + NO fast step
2. NO3 + CO -> NO2 + CO2 slow step

rate law NOT: k2[NO3][CO] but IS: k2k1/k-1 [NO3][CO] k-1 = reverse reaction of step 1


Ok as previously state the RATE law may NOT contain any intermediate molecules.

So the wrong answer would be: Rate = k2 [NO3][CO]

To solve this problem we need to find an equivalent expression that is equal to [NO3]. We can find this by using the fact that the rates forward and reverse reactions are equal. So taking the first step

1. NO2 + NO2 -> NO3 + NO (assuming this is an elementry step)

we have
Rate forward = Rate reverse ---> k1[NO2]^2=k-1[NO3][NO]
Solving for [NO3] = ( k1[NO2]^2 ) /( k-1[NO] )

Now we can substitute this expression inplace of [NO3] in our rate law.

Rate = k2 ( k1[NO2]^2 ) /( k-1[NO] ) * [CO] ---->
Rate = k2k1/k-1 * [NO2]^2[CO]/[NO]

This should be he official rate law which does not include the intermediates.

Thanks for the correction Acurax, yea I messed up the simplification.

 

[acurax]

I agree with mrmilad's method, but shouldn't the [NO] in the final rate law be in the denominator?

Something like...
Rate = (k2k1/k-1)*([NO2]^2/[NO])*[CO]

or maybe my algebra just needs work! 🙂

 

[MDwannabe7]

The rate law is determined by the rate-determining step and all steps preceding the rate determining step. This is done to remove intermediates from the rate law.
-Dr. P.

Thanks that answers a separate question that I had in regards to figuring out the rate determining step when given the rate law and a mechanism.

 

[madame99]

Funny you mention this cus I came across a practice problem almost like it today which was something like:
fast A2 + BC -->A2C + B
slow B + BC --> B2C

My first reaction was oh easy the rate of the reaction is k[BC] but thats not true cus is an intermediate and we have to take into account the preceding rxn so the actual overall rate is k[A2][BC]^2

 

[Mcatguy]

Funny you mention this cus I came across a practice problem almost like it today which was something like:
fast A2 + BC -->A2C + B
slow B + BC --> B2C

My first reaction was oh easy the rate of the reaction is k[BC] but thats not true cus is an intermediate and we have to take into account the preceding rxn so the actual overall rate is k[A2][BC]^2

I'm pretty sleepy, but that doesn't look right, it think it should be something like this, but someone can check I'm doing this in my head.

(K2k1/k-1)([A2][BC]^2)/[A2C]

 

[DrDre2001]

If you're in a Kaplan class, there's a good example of this on page 118 of the High Yield Problem solving guide. And it goes through the solution step by step and helped me understand what was going on 👍