[ptryon]

The IB chemistry guide (2016) states "In particular cases, such as a linear chain of
elementary reactions, no equilibria and only one significant activation barrier, the rate equation is equivalent to the slowest step of the reaction."

I can see why this is a valid simplification (i.e. if the activation energy barrier of one of the steps is particularly high, then a change in concentration of any species in this step will lead to a far bigger relative change in the number of particles that have an energy ≥ E_a compared to changes in concentration in other slower steps). So what happens when the slow step is not the first step in a sequence? For example, if the reaction was:

Overall: AB + C  AC + B
Fast step: AB  A + B
(Very) slow step: A + C  AC

Would the rate equation be:

Rate = k[AB][C] or would it be Rate = k[C] 

At first I thought it was Rate = k[AB][C] because the rate equation for the second step is
Rate = k[A][C] and changes in [A] depends directly on changes in [AB]. Therefore [AB] can be substituted into the equation in the place of the intermediate [A]. However, this substituted equation now depends on species from the first step- which can be ignored (if I am interpreting the IB's statement correctly). Therefore I now think it is Rate = k[C] but I am not sure!

Can anyone with a good background in Kinetics help?
Thank you

[mjc123]

You are (in your second thoughts) interpreting the IB statement incorrectly. The controlling rate equation is that of the slowest step, but you can't ignore species from earlier steps, because the concentration of species involved in the slow step may depend on them. Thus in your example
Rate = k[A][C] (I don't like that expression, by the way, because what do you mean by "rate"? Let's be more specific and say d[AC]/dt = k[A][C].)
But you have to be careful about how [A] depends on [AB]. For example, if the decomposition of AB is an equilibrium AB  A + B, and there is no other source of A or B, then
K = [A]{B}/[AB] = [A]²/[AB] so [A] = sqrt(K[AB]) and rate = k[AB]^1/2[C] (I write {B} to prevent bolding everything.)
If AB  A + B fast and irreversibly, then essentially before the slow step begins, [AB] = 0 and [A] = [AB]₀. You can then solve the second-order rate equation for A and C with [A]₀ = [AB]₀. This rate equation will not depend on the instantaneous concentration of AB, which is zero, but will depend on the initial concentration.
In neither case is rate = k[AB][C].

[ptryon]

Hi mjc! Thank you for taking the time to reply.

Am I right thinking that if the first step is fast and irreversible the rate equation would be:

d[AC]/dt = [AB]_o[C]? And that since [AB]_o is the initial rate with respect to AB and therefore the value of [AB]_o doesn't change throughout the reaction?

If this is the case, is it really a second order reaction overall? Or would it be referred to a pseudo second order? Would a graph of [C] versus t have a constant half life? (if so this is what I would expect from a first order reaction).

Again, thank you for your help

[mjc123]

No, d[AC]/dt = k[A][C]. [A] and [C] both vary with time in the manner of a second-order reaction, unless one of them is in large excess, when you get a pseudo-first-order reaction. The initial value of [A] at the start of the slow step is equal to the initial value of [AB] at the start of the reaction, assuming the first step goes to completion.

(And there was a mistake in my expression for the equilibrium case. As A reacts and B doesn't, the concentrations of A and B are not equal for t > 0, so [A] ≠ sqrt(K[AB]). If it was A₂ 2A, then [A] = sqrt(K[A₂]). For AB  A + B, [A] = K[AB]/{[AB]₀ - [AB]}. You see how complicated it is!)

[ptryon]

Hi, thank you- that does make sense. I read in one set of lecture notes from Oxford (http://vallance.chem.ox.ac.uk/pdfs/KineticsLectureNotes.pdf) that intermediates should not be included in an overall rate equation. So, although I can see why it is d[AC]/dt = k[A][C], there is the problem of A being an intermediate. Am I being too pedantic here?
Your thoughts are greatly appreciated...

[mjc123]

In the case of our hypothesis that step 1 is much faster than step 2, and irreversible, A is not really a "reactive intermediate". Essentially it is a product of reaction 1 that is a reactant for reaction 2. The rate of reaction 2 depends on [A], which doesn't depend on [AB], which is essentially zero by the time reaction 2 kicks off. (Though the initial rate of reaction 2 will depend on [A]₀, which equals [AB]₀.)
If the rates of reactions 1 and 2 are similar, we can say that
d[A]/dt = k₁[AB] - k₂[A][C]
and if we make the steady state assumption, we may be able to solve for [A] in terms of [AB] and [C]. But this is usually when k₁ < k₂ and [A] is small. If k₁ >> k₂, we don't get a steady state, we get effectively two sequential reactions, where 1 is virtually complete before 2 gets started.

[ptryon]

MJC you are a true legend! Thank you. This makes sense now and also it is completely consistent with the IB statement (i.e. the rate equation is equivalent to the step with a significantly high E_A barrier). This is different to how kinetics is normally taught at this level (most UK exam boards simply require students to substitute earlier species into the rate equation).
I do understand it better now! Thank you again 

[mjc123]

Happy to be of help.