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Homework Help: Reaction Mechanisms

  1. Dec 4, 2013 #1


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    1. The problem statement, all variables and given/known data

    Reaction: 2A --> C
    Step 1: A <-> M (fast)
    Step 2: M + A -> C (slow)

    What is the observed rate law?

    2. Relevant equations

    Rate = k[A][...]

    3. The attempt at a solution

    The observed rate law seems to be k[A]^2.

    However, why is it so? Why does the rate depend on the concentration of A squared? I know that the rate of the reaction is determined by the slowest step, and that the steps in the reaction must add up to the overall reaction (it does in this case).

    What's the "for dummies" explanation for reaction mechanisms?
  2. jcsd
  3. Dec 4, 2013 #2


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    I think I have a grasp on what's going on. Somewhat. A few questions linger regarding the specifics and I think I'm still missing the forest for the trees. I know how to solve these types of problems now for a multiple choice test, yes, but I'm still looking for that deeper understanding. This is what I did:



    1) I thought we weren't supposed to mix stoichiometry with the rate law. As in, the exponents are equal to the stoichiometric coefficients of the rate-determining step.

    2) M is the intermediary molecule. I guess solving for the rate at which the intermediary molecule is produced and plugging it into the rate of the slow, rate-determining step helps us find the overall rate of the reaction, right?

    3) What is a mechanism? I'm assuming it's the intermediary steps, right?

    4) Why is the reactant rate proportional to the concentration of A squared? Why isn't there a linear relationship? Is it because there are two steps in the reaction?
    Last edited: Dec 4, 2013
  4. Dec 4, 2013 #3


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    Also I'm gonna work through this one and try to explain my grasp of the concept. K is an arbitrary constant.

    http://i.minus.com/j7kDVWw5XAiK1.png [Broken]

    (1) is a reversible step, so I'll start with that. The rate is: [Z] = k[X][Y]. The concentration of Z changes proportionally with the concentrations of X and Y. That's not too bad to visualize.

    (2) - from 2 we get:

    = k[Z][A] = k[X][Y][A].

    And since from 3, we have B --> C, therefore is proportional to [C].

    Therefore [C] = Rate of overall reaction = k[X][Y][A]

    Also by inspection three of the answer choices can be thrown out (these are 2, 3, and 5) immediately because all of them contain the intermediary Z, while the question specifically wants the reaction's OVERALL rate law.

    Still not completely understanding the underlying concept; just the mechanics! (No pun!)
    Last edited by a moderator: May 6, 2017
  5. Dec 5, 2013 #4


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    4) It is proportional to [A]2 you could say because your argument in the image is correct!

    But if you still need a 'for dummies' explanation you could say there is an equilibrium such that the concentration of molecules in the form that can react, M, is proportional to [A]; then the rate at which these M molecules react to form product is proportional to their frequency of collisions with A molecules which is again proportional to [A] so the result is proportionality to [A]2. This is saying in words what you said with formulae.

    It is not because there are two steps in the reaction. You could have a mechanism where there was nothing but collisions between A molecules a fraction of which collisions gave rise to the chemical reaction. That too would have a rate proportional to [A]2. You could have a mechanism in which M is formed as in our example, but then it undergoes a slow rate-limiting transformation into another form, M*, say, and this reacts with another a molecule, but so fast anyway that varying their concentration makes no difference to the overall rate which is just equal to the M→M* rate. Overall reaction rate would be proportional just to [A] in that case.

    So the fact there are always 2 molecules of A involved in a reaction with that stoichiometry doesn't mean always reaction rate proportionality to [A]2.
    Last edited: Dec 5, 2013
  6. Dec 5, 2013 #5
    To be more precise, you could write the equations as:


    And, you could solve the equations this way. However, if the equilibrium reaction occurs very rapidly, then [itex]-k_1[A]+k_{-1}[M]≈0[/itex], and you can approximate [M] by [itex][M]≈\frac{k_1[A]}{k_{-1}}[/itex] and thereby eliminate [M] as a parameter in the solution. You are then left with:

    The advantage of being able to make this approximation is that you get to eliminate [M] as a parameter in the analysis (with little loss of accuracy).
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