[EmperorTeepo]

Hi everyone. I recently took a test in my general chemistry 102 class regarding chemical kinetics. Although the professor did mention the existence of "a" in the following equation in his notes, I ignored it considering the fact that nearly every textbook and online source never noted this.

In the equation for finding the half life of a substance for a first-order equation, you have:

t_1/2= ln(2)/k.

The question reads:

The decomposition of N₂O₅ is 1st order. 2N₂O₅ ---> 4NO₂ + O₂ is 1st order. If the rate constant is
5.90 x 10^-4  s^-1, what is the half life of this reaction?

He used the equation t(1/2)= ln(2)/ak, where the "a" is the coefficient 2 of the reactant. If what he says is true, then pretty much all reaction rates in my book are wrong. He's very adamant that there is, in fact, an a there.

What is your say on this matter?


My answer: 1.17 x 10^3 s

The "correct" answer: 5.87 x 10^2 s.

[stewie griffin]

It's been quite a while since I've done anything with kinetics, but as far as I understand when you have first order kinetics the half life is simply ln(2)/k. What is your professor's reasons for including this "a" term? It seems he's using the "a" term to be the coefficient that shows up in the stoichiometry... but if that's indeed what he's doing then that doesn't seem to make sense. What if I write the reaction instead as just N₂O₅ -> 2NO₂ + .5 O₂? Now the "a" term is 1, not 2.

[EmperorTeepo]

That's exactly my argument. There are other ways of balancing it other than the way he has done so. If I balance it another way, according to him, won't that just throw off t(1/2) and/or other values?

I may have to see him in person and argue this point. I just need to make sure that this is 100% true. Has anyone else heard of this magical "a" coefficient?

[orgoclear]

The question reads:

The decomposition of N₂O₅ is 1st order. 2N₂O₅ ---> 4NO₂ + O₂ is 1st order.

Since the value of half life is given for this particular reaction (balanced in this particular manner). Then the factor of 'a' is correct. because the rate expression becomes -d[N2O5]/2dt = k[N2O5]

Then integrating, we do get that magical factor of 2.

This is what physics has to say.

I dont know much about the chem. part of it. i.e. whether half life is expressed in terms of the balanced chemical reaction or if it is a  constant. (i do believe that it is constant independent of the stoichiometry as ive never seen C having a half life of 2*5770 yrs due to a change in the representation of the equation)

if it is the latter then the factor of 2 does not arise

however, if it is the former case, then the factor of 2 comes

[stewie griffin]

Re: orgoclear
I'm confused by what you're saying. If you are trying to relate disappearance of starting material to appearance of product, then yes the coefficients in the balanced equation are certainly involved. But I'm not so sure about the coefficients having an effect on the half-life.
You say "Since the value of the half life is given for this particular reaction (balanced in this particular manner). Then the factor of "a" is correct". Well, the value of the half-life is NOT given... indeed that's what we are trying to find. Now perhaps you meant to say that the value of the rate constant k is given... however k is a CONSTANT and thus does not change just because one changes the coefficients in the balanced equation.
Reading straight out of Anslyn and Dougherty's "Modern Physical Organic Chemistry", it says that "For a first order reaction the half-life is t_1/2=ln(2)/k=0.693/k." I'm not sure where this "a" factor is coming from... and Anslyn and Dougherty are fairly smart dudes.

[renge ishyo]

-d[N2O5]/2dt = k[N2O5]

This is probably where the confusion is. Rate laws are not necessarily based on the overall stoichiometry of the reaction, but are based on the results of experiment. If the reaction rate was based on the overall stoichiometry, we would expect this reaction to be second order in N₂O₅:

2N₂O₅   4NO₂ + O₂

rate = - d[N₂O₅]/2dt = k[N₂O₅]²

Note that the order of this reaction would be second order if the balanced stoichiometry determined the rate. However, the question specifically tells us that the reaction is 1st order so we know that the stoichometry of the overall reaction DOES NOT determine the rate. In fact, a multi-step mechanism is proposed for this reaction (I "borrowed" this from: http://www.scribd.com/doc/11579757/Examvillecom-Molecularity-Mechanism-and-Rate-Law-of-a-Reaction-Review-Notes but you can get it in many introductory chem books that might not be as sloppy as this article...).

N₂O₅   NO₂ + NO₃        slow (rate determinining)
N₂O₅ + NO₃ 3NO₂ + O₂      fast
---------------------------------------------------------------------------------------------------------
2N₂O₅   4NO₂ + O₂

Now, the rate is ONLY dependent on the equation written for the slow step of this reaction scheme. Applying orgoclears method to the first step of the mechanism we get:

rate = - d[N₂O₅]/dt = k[N₂O₅]

The mysterious factor of two disappears. I am with the student on this one, but I would still find the reaction worked out in your textbook or another introductory textbook so you can show your teacher that the reaction should probably be this way (and be nice and respectful when you show this to him! Remember, he's doing his best and this was an honest misunderstanding I think).

[Dan]

- d[N₂O₅]/dt = k[N₂O₅]

It's been so long I had to grab a piece of paper and integrate between the limits [N₂O₅]=X, t=0 and [N₂O₅]=X/2, t=t_1/2. I get:

lnX - ln(X/2) = kt_1/2

therefore

ln2 = kt_1/2

I also agree with the student...

[orgoclear]

Attached are pages from our physical chem book..

I have attached some similar looking questions as the one given..

It has also introduced the factor of '2'

and the given rate law expression is -d[N2O5]/2dt = k[N2O5]

i would be very obliged if i have some clarifications..

[renge ishyo]

And here is a typical general chemistry book showing it done the other way as I showed before :

http://books.google.com/books?id=_vRm5tiUJcsC&pg=PA572&lpg=PA572&dq=rate+law+N2O5&source=bl&ots=baOIThXP-f&sig=qhhSqzuLSHnt3aP42JMwBHABNfw&hl=en&ei=BgKIS5D6OYSusgPQ2ZCGAw&sa=X&oi=book_result&ct=result&resnum=5&ved=0CB0Q6AEwBA#v=onepage&q=rate%20law%20N2O5&f=false

So which approach is correct? Having seen the analysis now (thanks orgo!) I think both *can* be correct, but the one that is in agreement with the way chemists do things is the approach taken by the student. Still, it depends on which value of the rate constant, k, you are giving in the problem statement.

Here is what I mean: in general chemistry we base the rate law on the disappearance of the reactant (which is monitored spectrophotometrically or whatever). This is why all the rate laws are relative to N₂O₅; it is the reactant. If you look at orgoclear's book, you will see that the you only have one rate listed that described the disappearance of the reactant (note that this is identical to the rate equation that I provided in my post above):

(i) - d[N₂O₅]/dt = k₁[N₂O₅]

The other two equations are *rate of formation* of products (note that these are positive rates in the differential form and also note that they are given with respect to the concentration of N₂O₅, the reactant):

(ii) d[NO₂]/dt = k₂[N₂O₅]
(iii) d[O₂]/dt = k₃[N₂O₅]

Now this other "k" the book lists is NOT the same k as the one for the disappearance of reactant that we get from our data (which is k₁)...this k is actually k₃, a rate of formation constant for oxygen! Look at the book solution for EX.17 and you will see this clearly displayed:

k = k₁/2 = k₂/4 = k₃ (or k = k₃)

From our experimental data, we will NOT get k₃ as our constant experimentally because we are measuring a decrease in the concentration of N₂O₅ to determine the rate! Sure, we *could* convert the data after the fact to give the rate of formation of O₂ with respect to the concentration of N₂O₅ (the book orgoclear provided shows how to do this), but why do more work? You do not need k₃ to use the kinetics as shown below.

The integrated rate law looks like this:

ln [A] = - k₁t + ln[A₀]

I have marked the k as k₁ to emphasize the point. If A₀ is your initial contration of N₂O₅ and [A] is your concentration of N₂O₅ at time t, then the rate constant will be the one associated with the rate of decrease in N₂O₅ concentration with time which is given as k₁. Since the half life equation is based on the concentration of N₂O₅ decreasing by half in time: t_1/2, then you can solve for t_1/2 from the integrate rate law as follows:

ln [1] = - k₁t_1/2 + ln[2]

or ln[2]/k₁ = t_1/2

as given in all the textbooks.

Now, is the teacher wrong? Strictly speaking no, but err...it depends on which k he is giving you. If he gives you the rate constant for the rate of formation of O₂ with respect to N₂O₅ (which we can get by converting to k₃ from the data obtained from k₁...why he would do this I have no idea), then the half life equation becomes (on substitution k₁=2k₃):

ln[2]/2(k₃) = t_1/2

This is simply not the way we do things in chemistry And if you took the experimentally determined rate constant and plugged it into the above equation expecting an accurate half life then your answer WOULD be wrong (experimentally, we get k₁, not k₃).

I hope this cleared everything up.

[orgoclear]

@renge ishyo: that post really clarified things up a lot..

But i still have one nagging doubt that is the solved ex. 20 of the book that i attached..

In that the 'k' given is k1 or k3 according to the notation that you have used?

If nothing is mentioned, then what do we take k as? (k1 isnt it?)

Then what the book has done .. will it be correct or wrong?

[renge ishyo]

orgoclear, the book is using k₃ for the "k" in example 20 as well. Here's my work for example 20 to demonstrate:

Start with the integrated rate law:

ln[A] = - k₁t + ln[A₀]

Substitute k₁ = 2(k₃)

ln[A] = - 2(k₃)t + ln [A₀]

Plug in the values and solve for [A]

ln [A] = -2(.0084)(60) + ln[(2.50/5.00)]

[A] = .182500932 mol/L

For the moles of A at t=60: (.182500932 mol/L)(5.00L) = .9125 mol (in agreement with the answer key)

Is this approach wrong? No, but it is not consistent with the way we typically do things. Since we have already gone down the rabbit hole a bit, I will try to show you a few more reasons (aside from getting k₁ from the experimental data) why most chemists use k₁ and not k₃ when they describe the kinetics for this reaction.

First, comes the problem of assumptions. If we did not have example #17 to tell us that k=k₃, and not that k=k₁ or even k=k₂, then how would we ever guess which constant they wanted us to use? Would it be wise to assume that they wanted us to use k₃ for the rate of formation of oxygen (?!?) instead of the more direct k₁ for the rate of decomposition of N₂O₅ when solving this problem? Probably not. For that matter, how would we know that they weren't giving us k₂, the rate constant for the formation of NO₂? You can also write a correct rate law for k₂ in which the half life equation would change:

- 2d[N₂O₅]/dt = k₂[N₂O₅]

t_1/2 = ln(2)/(1/2)(k₂) = ln(2)/(.5)k₂

This approach is "correct" as well, so long as I specify that I am giving you rate constant for the rate of formation of NO₂ with respect to the concentration of N₂O₅.

Using k=k₁ as the standard for this problem occurs for a number of reasons. First, if we agree to group all of the constants together before calling it our "k" prior to integration then ALL of the alternative rate equations reduce to the same equation:

When starting with k₃
-------------------------
-1/2d[N₂O₅]dt = k₃[N₂O₅]  
-d[N₂O₅]/dt = 2k₃[N₂O₅]  
-d[N₂O₅]/dt = k₁[N₂O₅] (since k₁ = 2k₃)

When starting with k₂
----------------------------
-2d[N₂O₅]/dt = k₂[N₂O₅]  
-d[N₂O₅]/dt = k₂/2[N₂O₅]  
-d[N₂O₅]/dt = k₁[N₂O₅] (since k₁ = k₂/2)

When starting with k₁
----------------------------
-d[N₂O₅]/dt = k₁[N₂O₅]  
no change, this equation already has all the constants grouped together in its "reduced" form.

The second reason to use k₁ is that so long as you agree to use this most "reduced" rate constant it gives only ONE half life equation for a first order reaction that is *independent* of any stoichiometric coefficients:

t_1/2 = ln(2)/k₁

On the other hand, the other half life equations need correction factors added to their k's for the answers to come out the same:

t_1/2 = ln(2)/2k₃ = ln(2)/.5k₂ = ln(2)/k₁

So the teacher and the book are not "wrong", but their approach is A) not the simplest approach to getting the correct answer and B) not in agreement with the rest of the world which would very much like to use the simplest approach to getting the correct answer. In fact, if that book overcomplicates such a simple idea as this, then I shudder to think of what the rest of it might be like...