Deriving a rate equation

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1. Apr 9, 2015

annnoyyying

1. The problem statement, all variables and given/known data
Overall reaction: A + F = I + J (just take = as the equilibrium sign)
It is first order in both A and B.
The following mechanism has been proposed:
1. A + B = C + D
2. C + D = E + B (B is the catalyst)
3. E + F = G + H
4. G + H = I + J
These are all elementary steps and are all reversible.
Derive the mechanistic rate equation for the reaction.

2. Relevant equations
none, a matter of playing around with the equations above and making assumptions......

3. The attempt at a solution
The rate law (not mechanistic) should be r = k[A][F], according to the overall chemical equation.
C, D, E, G, H are all intermediates and their concentrations cannot be measured, so should not appear in the final mechanistic rate law. Steady state approximation can be applied to eliminate these when deriving the rate law. The concentration of B should not really change because it is consumed in 1 and regenerated in 2. 1 and 2 are in fast equilibrium.
The other thing I know is that F affects the rate of the reaction.
Is it possible to assume that the 3rd reaction as a rate determining, slow, forward step?

2. Apr 9, 2015

epenguin

There is no reason for such an assumption since there is no specific chemistry. Start by writing out the rate equation for each step. Afterwards can think about steady state assumptions.

3. Apr 11, 2015

annnoyyying

hmmm I've been trying for a few days and I still cannot get to the answer.
There must be some key piece of the puzzle I've missed.
Any tips?

4. Apr 11, 2015

epenguin

We haven't seen you start. Write a rate equation for the first step. If that is not obvious then follow the model you must have in your book

5. Apr 11, 2015

annnoyyying

redone below

Last edited: Apr 11, 2015
6. Apr 11, 2015

epenguin

Something like but it is not correct. I suggest you write instead of r, d[A]/dt etc. so everyone will know what you mean. For most of the steps you don't have to write a separate equation for the two components, e.g. you can write d[ I]/dt = d[J]/dt = ... , I hope it is obvious why. B is an exception because it is regenerated and not equal to d[A]/dt. So five equations in total.

Then, simply, all the reactions are elementary bimolecular yet you give a mixture of second order and first order terms.

7. Apr 11, 2015

annnoyyying

there was something wrong with the system when i was editing my reply which made it go all funny.
using b's make everything into bold, using i's make everything italic.....
let me do it again:
-d[A]/dt = k1[A][ b] - k-1[C][D]
-d[ b]/dt = k1[A][ b] - k-1[C][D] -k2[C][D] +K-2[E][ b]
-d[C]/dt = -d[D]/dt = k2[C][D] - k-3[E][ b] -k1[A][ b] +K-1[C][D]
-d[E]/dt = -k2[C][D] + k-2[E][ b] +K3 [E][F] -K-3[G][H]
-d[F]/dt = K3 [E][F] -K-3[G][H]
-d[G]/dt = -d[H]/dt = -k3[E][F] + k-3[G][H] +K4 [G][H] -K-4[J][i ]
d[ i]/dt = d[J]/dt = K4 [G][H] -K-4[J] [i ]

if i apply steady state approx then -d[C]/dt=-d[D]/dt=-d[E]/dt=-d[G]/dt=-d[H]/dt = 0 giving me 5 equations overall??

Last edited: Apr 11, 2015
8. Apr 11, 2015

epenguin

So far as I can see that is right and complete now. I presume the reaction is started with only A and B, or maybe only A, B, I, J. Some of the other concentrations are not independent so not only are some d/dt equal but some of the concentrations are equal, so the number of variables can be reduced. Then yes, you have to solve the steady state equations. By stages this is not quite as difficult as it may appear at first.
But if you find it too difficult make some further simplifying assumptions like I = J = 0 and you could get at least the intital steady-state rate and go more general afterwards.

9. Apr 11, 2015

annnoyyying

The reaction started with A, B and F. F is the reactant in the overall equation.
Can you be a bit more specific? Like in "Some of the other concentrations are not independent so not only are some d/dt equal but some of the concentrations are equal, so the number of variables can be reduced."
Does that mean say [C] = [D] so i can write anything thats[C][D] into [C]^2(thereby eliminating D), [G] = [H], eliminates H, [ i]= [J], eliminating J....?
But if you find it too difficult make some further simplifying assumptions like I = J = 0 and you could get at least the intital steady-state rate and go more general afterwards.
What does this mean?
I think I have to derive something that looks like d[J]/dt = k[A][F].
Thanks for the help so far.

Last edited: Apr 11, 2015
10. Apr 11, 2015

epenguin

That's it, [C] = [D] and [G] = [H]. Also if you start with [ i]= [J] = 0 these two will remain equal also when they are not 0. (And even if you don't start with that they are not independent and will always be related by a simple expression. Maybe not essential to incorporate this last at this stage.)

11. Apr 11, 2015

annnoyyying

ok,
so far i've got:
-d[A]/dt = k1[A][ B] - k-1[C]^2

As B is simutaneously consumed and generated,
-d[ B]/dt = k1[A][ B] - K-1[C]^2 - k-2[C]^2 + k-2[E][ B] = 0

As E is simutaneously consumed and generated,
-d[E]/dt = k3[E][F] - k-3[G]^2 + k-2[E][ B]- k2[C]^2 = 0

d[J]/dt = k4[G]^2 - k-2[J]^2
As i'm assuming [ i] = [J] = 0,
d[J]/dt = k4[G]^2 ????

I have no idea where this is going.
I've been stuck on this for the past week and a half and its driving me bonkers. Please give me a heads up.
(And I promise this is not some "no effort shown" act, see the pile of scribbled on paper in the recycling)

Also, if the steps 1, 2 and 4 are "fast" equilibrium steps (i.e. protonation and deprotonation), while step 3 is slower (a halonium ion formation step), will it make a difference? I have seen something called a "slow" equilibrium step in my lecture notes.......

12. Apr 11, 2015

epenguin

You have omitted dC/dt which you previously had from this last list - not sure it helps much.
From the first two equations you can eliminate C and hopefully will be able to keep on eliminating other things.
It does simplify if you have a fast equilibrium, as you can write the equilibrium equation which will enable easy elimination of a variable.
(For this kind of mechanism the steady state result works out as formally the same as with equilibria but with 'lumped constants' but we'll see that later.)
The variables you don't want to eliminate are A,F, I, J concentrations.

13. Apr 11, 2015

annnoyyying

I think the main problem I have with this is getting my head around the simutaneous equations.......
it would be much appreciated if you can give actual specific examples...... I'm begging you.
I keep on going around in circles.

so so far i've eliminated D and H by taking C = D and G = H
using the 1st 2 equations to eliminate C:
-dA/dt = k1AB -K-1 C^2
-dB/dt = k1[A][ B] - K-1[C]^2 - k-2[C]^2 + k-2[E][ B] = 0,
giving me C^2 = (k1AB + k-2EB)/(K-1 + k-2)??

If I= J then why can't I eliminate one of these?

If i apply the fast equilibrium to steps 1, 2 and 4, writing the equilibrium equations gives me:
kc1 = C^2/AB so C^2 = kc1AB
kc2 = EB/C^2 so C^2 = EB/kc2
so EB/kc2 = kc1AB, so EB = kc1kc2AB, E = kc1kc2A (and eliminating B)

kc4 = IJ/G^2 so G^2 = IJ/kc4
if I use -dF/dt = k3EF - k-3G^2
substituting the equations for E and G^2:
-dF/dt = k3kc1kc2AF - (k-3 IJ)/kc4????

Last edited: Apr 11, 2015
14. Apr 11, 2015

epenguin

OK if you substitute the C2 from the last line back into the other two you have eliminated it. You could also have eliminated it in the ususal simultaneous equation way. After that go about trying to eliminate E with the later equations involving it.

Sorry. Yes If they are 0 at start of reaction you can say I = J and get rid of one of them. As well as A you need keep B, Altogether keep the "inputs" A, B, F and output I or J.

If you make those 'quasi-equilibrium' assumptions it is simpler. Your result seems to be OK. (The differential equation can be solved but this was not asked).
B does not appear in this result because it is a catalyst, however by your assumptions the equilibrium it catalysed has at all times been reached, its concentration no longer matters. If instead you don't make this assumption you will find also a dependency on B concentration, as you were working towards in the first para above.

Last edited: Apr 12, 2015
15. Apr 12, 2015

annnoyyying

I'm going to go with the quasi equilibrium route for the moment (i.e. -dF/dt = k3kc1kc2AF - (k-3 IJ)/kc4 ) for various reasons (there are other questions prior to this). The I and J in this answer cannot be eliminated right?
I'll keep working on the non quasi equilibrium approach and i'll update when I get the answer.

Thank you so much for the effort, hope my stupidity haven't caused you much trouble.

Last edited: Apr 12, 2015
16. Apr 12, 2015

epenguin

You're welcome. Don't worry, we are getting more fed up with the opposite - threads that are left incompleted - so I hope to see your further considerations when ready.

I think you have correctly identified your main problem and it is the most common problem that a lot of students have with kinetics:
I.e. simple algebra, and it mostly is. very simple algebra. Like years ago they became quite proficient if they had two equations in, say x, y, z,...,, in getting an equation in which y, say, had been eliminated. But when instead of it being y in excercise 6 Chapter 3 of a book called 'Algebra' it's C2 and it's called Chemistry they can't do it any more.

That Is the most important take-home lesson from this, and I would say the second most important is don't content yourself with just formulae, but also use and develop kinetic intuition. Try and imagine from a physicochemkcal viewpoint what the form of equation is going to look like, and when you have a result try and think why you should have expected it beforehand. Then the intuition informs the math (and helps detect errors) and the math results instruct the intuition, and the whole thing means more so you can for example read a mechanism out of a rate equation or an experimental result.