What is the chemistry behind why reactions of order >1 exist? And what role does the number of activated complexes formed have in determining reaction rate?

  • $\begingroup$ I believe you should breakdown your question into different questions. All of them are good questions, but together, the compound question becomes too broad to answer. $\endgroup$ Mar 6 '14 at 8:06
  • $\begingroup$ I agree with @SatwikPasani, breaking these down will help the answers stay organized. $\endgroup$
    – jonsca
    Mar 6 '14 at 9:22

The following answers might be vague or overtly simplified or basic, but the questions by nature are fairly open-ended, aren't they? feel free to inquire further if something isn’t clear enough.

  1. Take $$\ce{2NO2 \to 2NO + O2}$$ You can (extremely basically) think of this reaction as two $\ce{NO2}$ particles colliding, (forming a complex), then coming out as $3$ molecules, $2$ being $\ce{NO}$ and $1$ being $\ce{O2}$. But you needed two $\ce{NO2}$ molecules to collide. Collisions are unlikely, first the two molecules have to find each other, then they have to have enough energy, and be in the right orientation. I’m not sure about multiple activated complexes, but I would only imagine that the reaction becomes even less likely if more than one activated complex is necessary for it; that basically means you need a second reaction from the first activated complex, which means a second collision. (likely slower rates? Someone else feel free to correct if I’m wrong) Three molecules colliding at once ($3$ rd order reaction): way less likely, and that’s why it’s not studied so much, it’s much harder getting $3$ molecules to collide (especially in the gas phase without) in the right orientation with enough energy.

  2. By general rate law do you mean finding the constant $k$? Or deriving its form entirely? Finding the orders for each reactant and the constant can only be done experimentally by using different concentrations and seeing what comes out. I think there is more than one way to derive the general rate law’s form. One way is to take the steady state approximation. It has a useful problem that might explain more clearly. The steady state approximation is a commonly made assumption in various situations. Essentially, it assumes that the intermediate is consumed as soon as it’s produced (so there’s no net accumulation of intermediate – which makes sense for many situations). Using the info gained from this approximation you can use the intermediates and the relations that result to form the general rate law. Of course, it turns out the $k$ is composed of multiple constants.

  3. It’s true because they told you it was. This gives you details that you wouldn’t have necessarily been able to deduce just by having the reaction equation there. Importantly, it’s there to tell you straight up that the equation you have can’t be broken down to further reactions... Again you can think of it like this: $\ce{a A}$ molecules have to collide with $\ce{b B}$ molecules to produce a single activated complex. That’s one step. If they didn’t specify that, maybe there could have been intermediate steps that could change the rate law.

  4. Maybe it'll be easier to see if you consider extremes. Say Reaction $R$ can be divided into $4$ reactions, $r1, r2, r3, r4$. Now, say $r1, r2$, and $r4$ happen in microseconds, while $r3$ takes $100$ seconds. Already you're at the point where you can basically ignore the times of $r1, r2$, and $r4,$ because they're so small compared to the time $r3$ took (comparing $1\mathrm E-6$ to $1\mathrm E+2$ here). So as you can see in this example, $r3$ was literally the rate determining step.


  • $\begingroup$ I have closed this as too broad, but when the OP breaks down the questions, you can rearrange your responses for each one and paste them as answers to the new questions. $\endgroup$
    – jonsca
    Mar 6 '14 at 9:23
  • $\begingroup$ Yes, for question 2 I do mean entirely. Why is each component of the rate law correct and significant in determining rate? I get the idea that higher concentration means higher rate, but why do we even have a constant? $\endgroup$
    – user11629
    Mar 6 '14 at 19:01
  • $\begingroup$ Also, with regard to question 3, the answer was not helpful because it doesn't address why mole ratios in a single-step reaction equate to corresponding reaction orders in A and B. Please elaborate. $\endgroup$
    – user11629
    Mar 6 '14 at 19:04
  • $\begingroup$ Hopefully someone else can answer #2 in depth. For 3: The problem is specifying the conditions for you. You need experimental data to determine rate law for a chemical reaction. Just looking at the mole ratios isn't a valid way of determining the rate law. Hope that clarifies both a bit of 2 and 3. $\endgroup$
    – rch
    Mar 7 '14 at 7:40

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