[PFScience]

I'm doing some kinetics revision and I have come across quite a notable difference of opinion regarding the integrated rate equation for 2nd order reactions.

Most sources quote for 2nd order reactions:

Differential rate law: - d[A]/dt = k[A]²

Integrated rate law: 1/[A]_t - 1/[A]_o = kt

However, I have seen a few chemical kinetic 'purists' quote the following:

Differential rate law: - d[A]/dt = 2k[A]²

Integrated rate law: 1/[A]_t - 1/[A]_o = 2kt


The crux of the argument is that in the first version they shouldnt be using a unimolecular reaction as the basis for a second order reaction. They should instead be using the form A + A = products as second order reactions are bimolecular.

This then creates the differential rate equation:

-1/2 d[A]/dt = k[A]²

And so thus: - d[A]/dt = 2k[A]²

What are everybody's thoughts? Most sources seem to prefer the first version  but some kinetic purists are adamant that the second version is correct. I'm in a dilemma because I can see how both could be viewed - but ultimately only one can win!!

Will be interesting to get some thoughts! 

[Irlanur]

it is a matter of convention and depends on the stoichiometric equation one is using. obviously the k is different in the two cases. In general the rate constants only have a meaning together with the stochiometric equaiton.  the rest is a matter of taste.

[Irlanur]

they shouldnt be using a unimolecular reaction as the basis for a second order reaction.

I think this doesnt make too much sense as an argument...

[PFScience]

Much appreciated.... its very frustrating to have these anomalies crop up.

The difference between k[A]² and 2k[A]² is significant!

[PFScience]

Still struggling a bit with this one, it seems to be a real puzzle, and I can't seem to find any clear answers!

In terms of second order (elementary) reactions, there are essentially 3 possibilities:

1) - d[A]/dt = 2k[A]²

I understand how this could occur as you would have...

A + A  products
2A  products

-1/2 d[A]/dt = k[A]²

2) -d[A]/dt = k[A] {B}

I understand how this could occur as each reactant is present in the same concentration and is disappearing at the same rate.

3) -d[A]/dt = k[A]²

This is the one that has me stumped and it is the one that is often quoted in books. Because its an elementary reaction, the stoichiometry of the reactants is equal to their reaction order, So, therefore, in order for the reaction to be second order with respect to A, you would have to have 2 moles of A present initially - but this cant be the case because it would then take the form of the rate law in example 1.

The only way I can seem to make this work is:

rate=-d[A]/dt = -d[A]/dt =d[prod]/dt

rate=k[A]²

Therefore..   -d[A]/dt = k[A]²

In essence this is like example 1 but I have treated each molecule of reactant A as a separate entity and not 2A.


Any ideas...??

[Yggdrasil]

k is a constant.  2*k is also a constant.  Would it make more sense if equation 3 was -d[A]/dt = k'[A]^2?

[mjc123]

There is no such thing as THE rate constant for a reaction. It depends on the rate expression you use. This should always be stated when a rate constant is quoted, but unfortunately it often isn't. Thus, for example, if you write
1) 2A  B + C
d[A]/dt = -2k[A]²
2) A  1/2 B + 1/2 C
d[A]/dt = -k'[A]²; k' = 2k
3) 478A  239B + 239C
d[A]/dt = -478k''[A]²; k'' = k/239
None of these is "right" or "wrong". 3 is rather inconvenient; and if 1 is an elementary reaction it makes sense to use that form; but they are all equivalent.

[jasongnome]

Much appreciated.... its very frustrating to have these anomalies crop up.

The difference between k[A]² and 2k[A]² is significant!

Maybe but there is no difference at all between k[A]² and 2k'[A]² where k=2k'.

[PFScience]

Cheers all for the replies - I will have a quick think and reply to each of you.

[PFScience]

k is a constant.  2*k is also a constant.  Would it make more sense if equation 3 was -d[A]/dt = k'[A]^2?

Well using k' notation indicates the use of the isolation technique, so in your example:

k' = effective rate constant for reaction of A and where k'=k{B}₀

So am I right in thinking that the only way that -d[A]/dt = k[A]^2 can work out is that if its a complex reaction and not an elementary one? As its a complex reaction, the partial orders of reaction have nothing to do with the reaction stoichiometry?


Or alternatively, viewed still as elementary, if we say:

A + B = products

Where:

{A}₀={B}₀

Then:

-d{A}/dt = -d{B}/dt = k{A}{B} (2nd order overall)

Or simplified:

-d{A}/dt = k{A}^2 (2nd order overall and since {A}₀={B}₀)


Is either of these approaches on the right track?! 

[mjc123]

Well using k' notation indicates the use of the isolation technique

No it does not. It merely serves to distinguish one rate constant from another. Thus in my example above, k' is the rate constant in the expression d[A]/dt = -k'[A]². It is NOT equal to k[B ]₀ (wherever [B ]₀ might have come from in this instance).
There are just not enough letters in the alphabet for all the things we want to talk about in science, so many of them have to be used for a number of different things. You may have read in a book about the isolation technique, where the authors used k' for k[B ]₀, but you must not assume this is a unique usage and k' can never mean anything else.

So am I right in thinking that the only way that -d[A]/dt = k[A]^2 can work out is that if its a complex reaction and not an elementary one?

No. Generally -d[A]/dt = k[A]^2 is an empirical rate equation, indicating that the rate depends on the square of [A]. It doesn't necessarily imply anything about the mechanism in terms of elementary reactions. If you go the other way around, and say the elementary reaction is 2K  products, then it is reasonable to postulate that for that elementary reaction rate is proportional to [A]². Whether you write that as k[A]² or 2k[A]² or 478k[A]² is immaterial, but the value of k in each expression will be different, which is why I used k, k' and k'' above.

[PFScience]

Thanks for the reply, sorry for the tardy reply - had a massively manic week with work + studies + life!

Right OK....

In regards to k' - yes I see what you mean, the confusion was just down to different notation use. When talking about different rate constants I was always taught to use k₁, k₂, k₃... etc

So looking at the 2 differing view points:

r = 2k[A]² and r = k[A]²

What in essence you are saying is, the key part is the k[A]² part.

The multiplier before the rate constant, k, will be purely down to the stoichiometric coefficient for that particular elementary step, and as such shouldn't really be the focus as it will vary depending on the equation.

Hence the use of r=k'[A]² - ie) the rate is equal to the square of the concentration of reactant A multiplied by the product of the rate constant and whatever the relevant stoichiometric coefficient may be? 

Better ?