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Kinetics assumptions

  1. Jan 23, 2014 #1
    I would like to understand under what circumstances rate laws can leave out rate terms from certain reactions, in general, as a process of approximation. I am familiar with the basic method regarding the steady-state approximation already. Any books, websites or articles which are good or comprehensive on this would be appreciated as suggestions and/or links.

    An example of where I am stuck: in the following problem (see the question), why is it that when writing the differential rate laws d[I1]/dt and d[I2]/dt, the equilibrium between I1 and I2 is omitted?

    e.g. my original equation was

    d[I1]/dt = k1[A][M] -k-1[I1] -k2[I1] +k-3[I2] -k3[I1] ≈ 0

    but the solutions gives

    d[I1]/dt = k1[A][M] -k-1[I1] -k2[I1] ≈ 0

    omitting the terms concerning the equilibrium between the intermediates I1 and I2. How is this further approximation - that +k-3[I2] -k3[I1] is negligible - justified for the question given?

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  3. Jan 24, 2014 #2


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    For the first question, well you ask about laws that are written that I can't see written. But if you mean the figure you could complain that it is not written very clearly: it would have been clearer if the author had written a vertical pair of arrows between I1 and I2 and not duplicated the I1. But you may often find such things in books and other publications for typesetting reasons - the authors think what they mean is obvious so try to not make such heavy weather of them! :smile:

    In that spirit and for your further question I think it is often best to solve the questions in textbooks first, and having done that you can look back and hopefully understand what their explanations :devil: of how to do it meant! :biggrin: . I cannot read on my screen whether certain symbols are = or ≈ but the latter should apply to the concentration of any intermediate.

    The second equation I don't know for sure. It would make sense if the rate constants for decomposition to P of I1 and of I2 were the same. Question (a) posits I1 and I2 are equal, so half the intermediates are I1 and half are I2 but each decomposes at the same rate so the result would be the same as if all the intermediates were I1 (and it would be whatever fractions the intermediates were).

    I find it difficult to know what question (b) is after. P does not enter into the steady state.

    I suggest you just solve the general problem: get the steady-state equation for dP/DT (= - dA/dt) . And if you like solve it to give A or P as a function of t. They seem to be asking about subclasses of this which could give you differences in the equations, but the forms of the equations would not be changed and it does not look to me there would be any way you could distinguish the cases.
  4. Jan 25, 2014 #3
    Ah I see now. For (a), I misread the question to have it suggest that the rates of change of [I1] and [I2] with time are constant, not the concentrations themselves. If [I1]=[I2]=0 is assumed, then the question in my first post is answered: my general steady-state equation reduces directly to the one they've used.

    In this problem, they have given the two conditions they want you to use along with the steady-state approximation. With all this given, it's easy to get the problem itself right. What I would like to know is, in what cases (in terms of rate constants, initial concentrations, etc.) would these approximations - [I1]=[I2]=0 in (a), and the approximation from part (b) - be valid? i.e. assuming these approximations are justified, in each case, can we say what therefore must be true about the relative values of some of the rate constants, initial concentrations, etc.?
  5. Jan 25, 2014 #4


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    I do not like the way this has been formulated and do not understand what is in the author's mind.

    To say anything further give us the steady state formula for dP/dt you have obtained.

    Then you say it's easy to get the problem right, but as I don't understand what the problem is give us your solution and then maybe I will.

    For your first sentence the rates of change of concentrations are indeed zero, not just constant, in a steady state, by definition. On the other hand [I1]=[I2]=0 means dP/dt = 0 - there is no reaction.

    (Just like for half the questions I answer on here. :grumpy:)
  6. Jan 26, 2014 #5
    You're right, my step in the derivation now seems unjustified. From the steady-state formula in the OP:

    d[I1]/dt = k1[A][M] -k-1[I1] -k2[I1] +k-3[I2] -k3[I1] ≈ 0

    I can't see a good reason, for which there is evidence in the question, to neglect +k-3[I2] -k3[I1] as ≈ 0 yet -k-1[I1] -k2[I1] does not have to be approximately equal to 0.

    If the question is valid, we must justify neglecting the last two terms of my steady-state formula, since the form of the steady-state formula given in the solutions was:

    d[I1]/dt = k1[A][M] -k-1[I1] -k2[I1] ≈ 0

    Do you know how we could do this, or whether it is even a reasonable assumption to make for a situation like this?
  7. Jan 26, 2014 #6


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    If you will just get and give here, and it is a matter of a minute or two, the formula for dP/dt, assuming that I1 and I2 are in a steady state, then you will have got most of what is worth attention in this mechanism, and afterwards we can consider any matters arising.

    d[I2]/dt = 0 will give you a simple relation between [I1]ss and [I2]ss with which to reduce d[I1]/dt = 0 to something easy.
  8. Jan 26, 2014 #7
    Ok, I ultimately ended up with

    [tex]\frac{d(c_P)}{dt} = k_1 \cdot c_A \cdot c_M \cdot (\frac{k_2(k_{-3}+k_4)+k_3k_4}{(k_{-3}+k_4)(k_{-1}+k_2+k_3)-k_3k_{-3}})[/tex]

    where cP=[P], cA=[A] and cM=[M],

    but this hasn't answered any of my questions? The solutions manual follows a simpler approach by neglecting the rate terms coming from the equilibrium between I2 and I1. How is this justified, and under what conditions (in terms of rate constants) would this be valid?
  9. Jan 26, 2014 #8


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    I think your equation is OK, I got the same.

    I find it difficult to answer this

    as I don't have your solutions manual. Unless it is the page copied onto your first post, in which case I don't see where it says this.

    It might be considering a simpler mechanism in which there is no I2.
    It might be considering the case that I1 is almost in equilibrium with I2, what I call quasi-equilibrium. The condition for this is that k3 and k-3 are much larger than k2 and k4. In that case the equations and their derivation do simplify.

    You should note that making any of these assumptions or not, or adding any number of extra steps I2⇔ I3 → P , I4, I5,... does not change the form of the resulting experimentally verifiable equation for dP/dt which remains just k[A][M] where k is a measurable constant made up from one doesn't know what others.

    (The form does change however if there are alternate bimolecular paths leading to P bit I do not know of any non-theoretical example of that and would be glad to hear.)
    Last edited: Jan 26, 2014
  10. Jan 26, 2014 #9
    The problem comes from this PDF, which also contains the solution directly beneath it (see Page 3 of 10):

    If k3 and k-3 are much larger than k2 and k4, surely we could reduce d[I1]/dt to

    d[I1]/dt = k1[A][M] + k-3[I2] - k3[I1] ≈ 0

    by approximating all rate terms with k2 or k4 to 0 (negligible compared to the terms with k3, k-3 or k1).

    But as you can see from the equation for d[I1]/dt in the first line of the derivation in the solutions, this is not the line they have taken ...
  11. Jan 26, 2014 #10
    It looks to me like they just did the problem incorrectly. They definitely should not have neglected the terms which they did neglect. My answer agrees with yours and epenguin's.

  12. Jan 27, 2014 #11


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    As previously, start with d[I2]/dt = 0 rather and you will see. A more careful way of saying what I would have said - "Assume I1 and I2 are nearly in equilibrium, then [I2 ] = (k3/k-3)[I1 ]".

    There appears to be some nonsense around this point in the manuscript, so just work out what makes sense without having to conform to every single thing that is written. I may come back on the other points later.
  13. Jan 27, 2014 #12
    Thanks for confirming.

    I see. This would follow logically from the pre-equilibrium approximation, but since they don't say anything about k3, k-3 being much larger than the other rate constants, we probably shouldn't assume this.

    On a slightly different note, is it that a rate-determining step dominates the rate terms, rather than being negligible? For example, if I have a given set of steps in a mechanism, e.g.

    A+B → C (k1)
    C+D → E (k2)
    E → 2F (k3)

    and the step C+D→E is "rate-determining" (or we could say that this step is slow and the other two are fast), how do we simplify our rate laws to account for this approximation? e.g. instead of d[C]/dt=k1[A]-k2[C][D], what could we simplify this to? Instead of d[A]/dt=-k1[C], what could we simplify this to? (presumably until [A] is 0, if the first reaction is considered fast d[A]/dt will be considered ∞, i.e. all A and B are instantly reacted into C?)

    This follows on from what I was trying to get at before about what to do with the differential rate laws if, e.g. you know that k3,k-3>>k2,k4.
  14. Jan 27, 2014 #13


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    No should or shouldn't really, this is just exhibiting various models of reactions. In the general case of that model you have the steady state equation you got, in the case you have that gross >> inequality (the quasi-equilibrium case) the equation simplifies. But, to repeat, the only experimental handle we have on it that is specified till now is we can vary [A] and [M] and measure [P] against time. Two input variables and one measurable output variable for a mechanism with three or six rate constants. And with just those handles you can't distinguish steady state from quasiequilibrium and one intermediate (I1) or any number of other intermediates.

    You can't assume always there is a 'rate-determining step' but quite often one essential step on the pathway to product is indeed much slower than others and so is rate-determining. All the reactions preceding it come to an equilibrium so you can simplify. In the case you mention the equilibrium strongly favours C (you write it as irreversible) so everything has become C and you'd really be observing the reaction C + D → → F.

    Then things depend on times of reactions. The simplest oldest kind of kinetics, you mix A and B and measure say and measure the appearance of F over time. But if say your first reaction takes minutes while you observe F over hours then in the first minutes you will able to observe an acceleration in the first minutes of the rate of production of F which can thereafter become constant. However if that first reaction takes milliseconds the same thing can be observed with special apparatus. There is a different method for observing chemical reaction rates over about every timescale, down to the order of 10-12 sec, the lifetime of the protonated state of a typical weak acid.

    Sometimes as conditions (e.g. [H+]) are varied a rate step ceases to be rate-determining and you get a change in the kinetic law.
    Last edited: Jan 28, 2014
  15. Jan 28, 2014 #14


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    Now looked at pp. 34-36. Oof! :yuck: . Unless this is just supplementary equation illustrations additional to something explained better in a lecture. Two or three simple ideas seem to be concealed and jumbled together here. I don't want to undermine your teacher, but here we don't want to see students undermined which is worse.

    Firstly slightly irritating he has A and M combine to form something he calls I1. In most normal treatments this is an AM complex which would be called AM so people following the argument could easily see what it is.

    He introduces product P onto the RHS of the equations so he says now you can integrate it if [M] is constant (!not small) enough. Big deal. Or he could just have put the LHS equal to - d[A]/dt. Integrating - d[A]/dt = k[A] is not difficult.

    Then what the rest seems to be about is this. In many cases [AM] is small or very small compared to [A] and [M]. E.g. frequently in specific acid or base catalysis where one of the participants is H+ or OH- and AM would be MH+ or MOH-. Then the overall reaction is first order in M and first order in A. But in other cases there is an equilibrium between A, M and AM (or a steady state) where [AM] is a significant fraction of all the M. Suppose you are in a situation where nearly all the M is in the complex. Then further increasing [A] will not increase the reaction rate - it has become zero order not first in [A]. This is called saturation. It happens almost always in steady state enzyme kinetics, and frequently in heterogeneous catalysis but not only. You get this hyperbolic form of the p 36 eq. which is called the Michaelis equation. Normal texts would introduce it with a simple mechanism, he chooses to introduce it with a more complicated one.
    Last edited: Jan 28, 2014
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