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stareleven

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Q: The mechanism for the decomposition of NO2Cl is

N02Cl ---> N02 + Cl
N02Cl + Cl ---> N02 + Cl2

What would the predicted rate law be if the second step in the mechanism were the rate determining step?

It would be great if anyone can help me on this! Thanks in advance!
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stareleven said:
Q: The mechanism for the decomposition of NO2Cl is

N02Cl ---> N02 + Cl
N02Cl + Cl ---> N02 + Cl2

What would the predicted rate law be if the second step in the mechanism were the rate determining step?

It would be great if anyone can help me on this! Thanks in advance!
Click to expand...

It's not MCAT realistic without four choices. So, let's spice it up a bit and see who responds.

A) rate = k[NO2Cl]
B) rate = k[NO2Cl][Cl]
C) rate = k[NO2][Cl]
D) rate = k[NO2Cl]exp2
 
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jgberken said:
really? i thought it would be B...because the rate limiting step determines the reaction rate, not the overall reaction...
whats the ans berkteach?
Click to expand...
Yeah, you're right... my bad. Wasn't paying attention.
 
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You're almost there guys.

Keep in mind that you cannot measure the concentrations on intermediates...ie the concentration of Cl...
 
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bluemonkey said:
You're almost there guys.

Keep in mind that you cannot measure the concentrations on intermediates...ie the concentration of Cl...
Click to expand...

When did they pass that law?

I think you may be mistaking an intermediate for a transition state. Physically, you can measure the concentration of an intermediate (using UV absorption, etc...). Intermeidates have a finite, measurable lifetime, which is what makes it different than a transition state, so you can measure their concentration and detect their presence.

Think about the steady-state approximation on which Michaelis-Menten kinetic is built. It applies when the enzyme is saturated and thus the enzyme·substrate complex (an intermeidate) is in constant concentration.

jgberken is right on the mark that the rate law is based on the slowest (rate-determining) step. Now the mathematics might get cumbersome, where we determine the [Cl] to be equal to some other value based on our steady-state equilibrium, but the basic rate law is simply rate = k[Reactant in the rds].
 
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BRteach...you are awesome!!!
I just spent the last hour trying to find out whether intermediates are included in rate laws...and I couldn't find anything saying that they weren't.
So, rate laws are determined by only reactants in the rate-determining step. wonderful! thank you :D
 
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BerkReviewTeach said:
When did they pass that law?

I think you may be mistaking an intermediate for a transition state. Physically, you can measure the concentration of an intermediate (using UV absorption, etc...). Intermeidates have a finite, measurable lifetime, which is what makes it different than a transition state, so you can measure their concentration and detect their presence.

Think about the steady-state approximation on which Michaelis-Menten kinetic is built. It applies when the enzyme is saturated and thus the enzyme·substrate complex (an intermeidate) is in constant concentration.

jgberken is right on the mark that the rate law is based on the slowest (rate-determining) step. Now the mathematics might get cumbersome, where we determine the [Cl] to be equal to some other value based on our steady-state equilibrium, but the basic rate law is simply rate = k[Reactant in the rds].
Click to expand...

still confused, so A or B?:(
 
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hmm...are you sure it's B or for that matter any of those choices?
MY own thought is that if the rate determining step(slow) is 2nd and you have intermediates, then you take the rate law of the slow step, rate=k[NO2Cl][Cl] and solve for Cl from the fast reaction to get rid of the unstable Cl which will probably have a low unknown concentration as well.
So I'd say make the fast reaction into an equilibrium reaction and say k1[NO2Cl]=k-1[NO2][Cl] and solving for Cl you get [Cl]=k1/k-1 *[NO2Cl]/[Cl]
plug this Cl into the slow step to get rate=k [NO2Cl]*[NO2Cl]/[Cl]
 
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I agree with matt46. Intermediates generally are NOT measured in reactions. The correct form includes initial reactants and/or products, but not intermediates. While the overall rate law is indeed determined by the rate determining step, many intermediates are extremely short lived or difficult to measure.

To Berkeley Teach, I am not confusing an intermediate with a transition state, but thank you for your suggestion.
 
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bluemonkey said:
I agree with matt46. Intermediates generally are NOT measured in reactions. The correct form includes initial reactants. While the overall rate law is indeed determined by the rate determining step, many intermediates are extremely short lived or difficult to measure.

To Berkeley Teach, I am not confusing an intermediate with a transition state, but thank you for your suggestion.
Click to expand...

That's how I was taught it. I can vaguely remember the reasonign, but basically that I was told not to put the intermediates in there.
 
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bluemonkey said:
You're almost there guys.

Keep in mind that you cannot measure the concentrations on intermediates...ie the concentration of Cl...
Click to expand...

Monkey is right. You do not include intermediates in rate laws. The steps here represent a mechanism for an overall equation, so you can't have reactants in the rate law that do not take part in the initial overall reaction.

http://bouman.chem.georgetown.edu/S02/lect5/lect5.htm

"A rate law should only consist of concentrations of reactants and/or products, no intermediates."

The first step in the mechanism is fast and in equilibrium:

NO2Cl <=> NO2 + Cl

The second step (slow) is the rate limiting step, so you derive the rate law from the 2nd step in the mechanism:

NO2Cl + Cl -> NO2 + Cl2

Now the rate law would be:

rate law = k2[NO2Cl][Cl]

But here, the two mechanisms are for an overall equation and if we look at the overall equation:

 2 NO2Cl <=> 2 NO2 + Cl2

Cl is neither a product or a reactant, and the rate law has to be expressed in terms of the reactants/products of the overall equation. So we have to get the rate law for this reaction by solving for the Cl using the rates from the first mechanism. We can only do this because the first step is fast and in equilibrium (I looked it up).

So the rates for the first step would be:

rate = k1[NO2Cl]

And the reverse rate would be:

reverse rate = k-1[NO2][Cl]

Since the first reaction is fast and in equilibrium, we can say that:

k1[NO2Cl] = k-1[NO2][Cl]

Now all we have to do is solve for [Cl] and substitute it into our rate law.


rate law = k1k2[NO2Cl]^2
.....................k-1[NO2]
 
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fileserver said:
I was thinking that and it seem to be difficult because you would get the product in the rate law. :p
Click to expand...

Yeah it's tricky. If you had the extra information, the [NO2] concentration would be grouped together with the other k's in the equation. This would make the equation a lot simpler to look at, and combine the k's into one SUPER k! Kinda like Voltron!
 
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SketchLazy said:
Yeah it's tricky. If you had the extra information, the [NO2] concentration would be grouped together with the other k's in the equation. This would make the equation a lot simpler to look at, and combine the k's into one SUPER k! Kinda like Voltron!
Click to expand...


Thanks. It's important to know this because rates/rate laws are definately the love-child of the mcat. That and Voltron.
 
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For the MCAT, they probably wouldn't give you:

k2[NO2Cl][Cl]

As an answer to the question 'What is the rate law?'
Although it's not common to leave the rate law with an intermediate in it, it is still *correct*. The other answer which does not include Cl, and only includes the products and reactants is also correct and would probably be the answer expected for the MCAT.
 
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Okay, so I just did a problem like the one posted above. Here it is. According to the book it is correct:

The slow step determines the rate of the reaction! However, if the fast step comes before the slow step, the fast step will contribute to the overall rate law.

1) NO(g) + Br&#8322;(g) &#8594; NOBr&#8322;(g) FAST
2) NOBr&#8322; + NO(g) &#8594; 2NOBr(g) SLOW – rate determining

Thus, the rate = k[NOBr2][NO]; however, NOBr&#8322; is dependent on the fast step! So, assume that the fast step is at equilibrium, so the forward rate = reverse rate. I.e. rateforward=kf[NO][Br2] = ratereverse=kr[NOBr2]. Rearrange:

[NOBr2]=(rateforward/ratereverse) = (kf/kr)[NO][Br2]; substitute
rate = k(kf/kr)[NO][NO][Br&#8322;]
rate = k(kf/kr)[NO]2[Br&#8322;]; sometimes kkf/kr will be replace by a single constant ko.

QUESTION: Now, is it simply a coincidence that if I were to add up the two steps, I get the right answer?

Would we see something like this on the MCAT? :eek:
 
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tncekm said:
Okay, so I just did a problem like the one posted above. Here it is. According to the book it is correct:

The slow step determines the rate of the reaction! However, if the fast step comes before the slow step, the fast step will contribute to the overall rate law.

1) NO(g) + Br&#8322;(g) &#8594; NOBr&#8322;(g) FAST
2) NOBr&#8322; + NO(g) &#8594; 2NOBr(g) SLOW – rate determining

Thus, the rate = k[NOBr2][NO]; however, NOBr&#8322; is dependent on the fast step! So, assume that the fast step is at equilibrium, so the forward rate = reverse rate. I.e. rateforward=kf[NO][Br2] = ratereverse=kr[NOBr2]. Rearrange:

[NOBr2]=(rateforward/ratereverse) = (kf/kr)[NO][Br2]; substitute
rate = k(kf/kr)[NO][NO][Br&#8322;]
rate = k(kf/kr)[NO]2[Br&#8322;]; sometimes kkf/kr will be replace by a single constant ko.

QUESTION: Now, is it simply a coincidence that if I were to add up the two steps, I get the right answer?

Would we see something like this on the MCAT? :eek:
Click to expand...

What do you mean 'add up the two steps'?
The way you've outlined solving the problem already is the correct (and only?) way to do it.

And yes, you could see something like this on the MCAT.
 
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I took G-Chem long ago, but I remember something to the effect of being able to "add" the sides up to get the overall reaction, and cancel out the stuff that is on both sides.

Step 1 + Step 2 = Overall Reaction
NO + NO + BR&#8322; + NOBR&#8322; --> NOBr&#8322; + 2NOBr&#8322;

Overall Reaction
2NO + Br&#8322; --> 2NOBr&#8322;
rate=k&#8320;[NO]&#178;[Br] = the same as I determined above.

So, the rate law I determined algebraically is the rate law of the overall reaction.

EDIT: This was nothing more than a coincidence, see below for details!
 
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tncekm said:
I took G-Chem long ago, but I remember something to the effect of being able to "add" the sides up to get the overall reaction, and cancel out the stuff that is on both sides.

Step 1 + Step 2 = Overall Reaction
NO + NO + BR&#8322; + NOBR&#8322; --> NOBr&#8322; + 2NOBr&#8322;

Overall Reaction
2NO + Br&#8322; --> 2NOBr&#8322;
rate=k&#8320;[NO]&#178;[Br] = the same as I determined above.

So, the rate law I determined algebraically is the rate law of the overall reaction.
Click to expand...

Yeah I think that's valid. The only trouble you'll run into is figuring out what Ko is in terms of K, Kf and Kr. Disclaimer: I learned genchem for the MCAT (it was by far my worst subject to start out with).
 
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Okay, well, I'll keep the disclaimer in mind and continue to do things the slow way for now :)

Thanks, though! :thumbup:
 
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No, it looks like the answer was "neither D nor B" (more like, none of the above although B is "more correct"). It was the answer SketchLazy put up. I reviewed that problem again, and considered my question, and it was just a coincidence.

SketchLazy's work was perfect, and the rate law it gave him was not equal to the overall reaction like mine was; in my case, it appears to have been a coincidence!

Having said that, I think it may be "correct" to include an intermediate to show what the rate law is, but its not conventional to leave the intermediate in there since convention requires that the overall rate law only include products and reactants, not intermediates. So, I guess "B" is most correct, but "not correct".
 
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tncekm said:
I took G-Chem long ago, but I remember something to the effect of being able to "add" the sides up to get the overall reaction, and cancel out the stuff that is on both sides.

Step 1 + Step 2 = Overall Reaction
NO + NO + BR&#8322; + NOBR&#8322; --> NOBr&#8322; + 2NOBr&#8322;

Overall Reaction
2NO + Br&#8322; --> 2NOBr&#8322;
rate=k&#8320;[NO]²[Br] = the same as I determined above.

So, the rate law I determined algebraically is the rate law of the overall reaction.
Click to expand...

Hey tncekm,
I'm not so sure about this anymore...I think it works for this problem, but I'm not certain it works for all problems. Check out sketchlazy's post and try your method vs. his. Does it work?
 
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Just beat you to it :) I determined it was a coincidence after reviewing SketchLazy's work--his work was correct, and it didn't reflect the "overall reaction".
 
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tncekm said:
No, it looks like the answer was "neither D nor B" (more like, none of the above although B is "more correct"). It was the answer SketchLazy put up. I reviewed that problem again, and considered my question, and it was just a coincidence.

SketchLazy's work was perfect, and the rate law it gave him was not equal to the overall reaction like mine was; in my case, it appears to have been a coincidence!

Having said that, I think it may be "correct" to include an intermediate to show what the rate law is, but its not conventional to leave the intermediate in there since convention requires that the overall rate law only include products and reactants, not intermediates. So, I guess "B" is most correct, but "not correct".
Click to expand...
Well...isn't B just as correct as Sketchlazy's answer? If Sketchlazy's ans was up, I would have chose that over B...but B seems like a fine ans, assuming that the Cl (intermediate) conc can be determined.
 
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No, SketchLazy is right, you can't include intermediates in the rate law. I guess if the choices above were the only ones give on the MCAT then B would the answer on the MCAT that would be "most correct".

http://courses.chem.psu.edu/Chem112/Spring/MJBnotes/Mechanisms_with_fast_initial_step.pdf

Note: keep in mind that when they use Keq, they're just saying that the reaction is at equilibrium (rate forward = rate reverse) and Keq = k_forward/k_reverse.
 
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tncekm said:
No, SketchLazy is right, you can't include intermediates in the rate law. I guess if the choices above were the only ones give on the MCAT then B would the answer on the MCAT that would be "most correct".

http://courses.chem.psu.edu/Chem112/Spring/MJBnotes/Mechanisms_with_fast_initial_step.pdf

Note: keep in mind that when they use Keq, they're just saying that the reaction is at equilibrium (rate forward = rate reverse) and Keq = k_forward/k_reverse.
Click to expand...


the MCAT is a *****. It will use that card when it has the chance. Know and understand now that intermediates are NOT in the rate law as ^ and Sketchlazy have said. I missed numerous questions on practice mcats because of this. I know from past experience that the mcat has put that answer AS WELL as sketchlazy's answer just to mind**** you.:smuggrin:

And I figured your reasoning was coincidental, but I don't think the answer would include the reverse rate constant "k" in there, would it? Woudl we have to further derive another answer from sketch's?:confused:
 
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The premise to solving this problem is that the fast step, which precedes the slow step, is at equilibrium. That is, the rate forward = the rate backward. So:

rate_forward = kf[reactants] = kr[products] = rate_reverse

so, rate_foraward/rate_reverse = kf[reactants]/kr[products]

from above, you will solve for the intermediate and replace the intermediate in the rate law. So yeah, the reverse k is part of the equation to the best of my knowledge.

Sketch "does" include kr, he just calls it k-1 and he calls kf, k1.
 
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tncekm said:
The premise to solving this problem is that the fast step, which precedes the slow step, is at equilibrium. That is, the rate forward = the rate backward. So:

rate_forward = kf[reactants] = kr[products] = rate_reverse

so, rate_foraward/rate_reverse = kf[reactants]/kr[products]

from above, you will solve for the intermediate and replace the intermediate in the rate law. So yeah, the reverse k is part of the equation to the best of my knowledge.

Sketch "does" include kr, he just calls it k-1 and he calls kf, k1.
Click to expand...
Have the problems you've seen have the answer with a single, new rate const? or do we actually have to keep track of the rate constants?
 
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I've never seen a problem like this other than when I brought it up here, and it was more a description of how to deal with this kind of problem rather than an actual problem.

Having said that, it was made clear that both k_observer and the kk/k version were correct, and acceptable. You may want t keep track of the rate constants just for kicks, especially since its not that hard to do and since k_observed is the product of those three.
 
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tncekm said:
I've never seen a problem like this other than when I brought it up here, and it was more a description of how to deal with this kind of problem rather than an actual problem.

Having said that, it was made clear that both k_observer and the kk/k version were correct, and acceptable. You may want t keep track of the rate constants just for kicks, especially since its not that hard to do and since k_observed is the product of those three.
Click to expand...
I haven't seen a problem like this either...but tpr never addresses this, so that could be the reason. in the tpr workbook, the right answer is always the one with the intermediate.
 
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Weird... dunno what to tell ya. But, I'm going with what I just learned as a precautionary measure :)
 
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stareleven said:
Q: The mechanism for the decomposition of NO2Cl is

N02Cl ---> N02 + Cl
N02Cl + Cl ---> N02 + Cl2

What would the predicted rate law be if the second step in the mechanism were the rate determining step?

It would be great if anyone can help me on this! Thanks in advance!
Click to expand...

The answer is
rate = K [No2Cl]
taking a look at the attached picture. this is the reason why.

You have to wright the rate law for both chemical equations. since we are looking at No2Cl we wright it as d[No2Cl]/ dt = rate law 1 + rate law 2.
if the chemical is created then the rate law is positive. if it is being consumed then it is negative.
 

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This thread is so confusing. :eyebrow:

How I learned it was the rate is based on the slow step, so rate = k[NO2Cl][Cl].

Since you cannot (or are not supposed to, semantics) keep intermediates in the overall rate law, you have to replace the [Cl]. The source of [Cl] comes from the previous elementary step NO2Cl ---> N02 + Cl, which shows that [Cl] is dependent on [NO2Cl], so you replace [Cl] in the rate law equation with [NO2Cl].

This gives you rate = k[NO2Cl]^2, or D.

:shrug:
 
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Formation said:
This gives you rate = k[NO2Cl]^2, or D.
Click to expand...

It should be proportional to [NO2Cl]^2 for the logic you mention above but not exactly equivalent. If you do a steady state approximation with [Cl], you'll get something that is second order with respect to [NO2Cl] as expected, with a messy denominator. It also depends on NO2 and there must be additional approximations to get rid of that as well but that's way beyond the scope of the MCAT.
 
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