Started reading Levine's Physical Chemistry [1] yesterday, and I hit upon (another) wall.

On page 517, Chapter 15, she includes a reaction resulting in the formation of $\ce{HBr}$ from $\ce{H2}$ and $\ce{Br2}$ (it's a homogeneous, gas-phase reaction) along with the corresponding rate equation:

$$\ce{H2 + Br2 -> 2HBr} \qquad r= \frac{k[\ce{H2}][\ce{Br2}]^{1/2}}{1 + j[\ce{HBr}]/[\ce{Br2}]},$$

where $k$ is the rate constant and $j$ is...um...just a(nother) constant O:)

Levine goes on to mention:

[For this reaction] the concept of order does not apply.

Now I'm still new to the whole Chemical Kinetics thing... But what's this?!

How come the "concept of order does not apply" to this reaction? I've never come across this situation in any book so far. Well, the reaction seems trivial (though the rate equation says otherwise...); I fail to understand why an order can't be assigned to this (overall) reaction.

(Note: Until I know otherwise, I'm assuming "concept of order doesn't apply", is the same as saying "this reaction has no order". Do correct me if I'm wrong in doing so...I tend to miss out on subtleties...)

Even having an order of zero still counts as having an "order". But what could Levine possibly be talking about when she says that the idea of an "order" doesn't apply to this reaction? What are the (physical) implications of a reaction lacking an "order"?

My questions, explicitly put:

  1. Why can't an order be assigned to the above-mentioned reaction? Or can it?
  2. What are the physical implications of a reaction having "no order" (as compared to reactions with an order)?


  1. Levine, I. N. Physical chemistry, 6th ed.; McGraw-Hill: Boston, 2009. ISBN 978-0-07-253862-5.
  • 1
    $\begingroup$ there is some more information in replies to this question chemistry.stackexchange.com/questions/79080/… $\endgroup$
    – porphyrin
    Aug 9, 2017 at 12:19
  • $\begingroup$ @porphyrin Gee thanks, that helped! Your answer there would work here...maybe you could slightly alter it and re-post it here? It'd get my upvote. Your call though O:) $\endgroup$ Aug 9, 2017 at 15:53
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    $\begingroup$ @paracetamol No prob, great question! An edit is the least I can do. $\endgroup$
    – andselisk
    Aug 9, 2017 at 15:58

1 Answer 1


We don't assign orders. An order either simply is there, or it isn't. In particular, this expression doesn't look like $\ce{\underbrace{(something)}_{\substack{\text{may include }\ce{[H2]},\\ \text{but not }\ce{[Br2]}}}\cdot[Br2]^a}$ for any $a$, hence there is no order.

Alternatively, if we look at the reaction at its initial moments when $\ce{[HBr]}\approx0$, the denominator simplifies to just $1$, and we may say that the reaction has order $1\over2$ with respect to bromine.

Having or not having an order doesn't have any particularly far-reaching physical consequences. Moreover, it is not even an intrinsic property of a reaction; it may depend on the approximation used, as you saw above.

The scary rate equation comes from the so-called steady state approximation and the detailed reaction mechanism with intermediates and stuff, which is about two times longer than my answer, so I'd better refrain from including it. Well, here is the first line, just to give you the taste of the whole thing: $$\ce{Br2}\underset{k_{-1}}{\stackrel{k_1}{\longleftrightarrows}}\ce{2Br^.}$$

To tell the truth, most real reactions are like that. The well-known $r=k\ce{[A]^a[B]^b}$ expression with stoichiometric coefficients for exponents is but a great oversimplification. It applies to reactions with no intermediates, which are quite rare.


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