According to several authentic sources (for example, ChemLibreTexts), for elementary reactions, the order is equal to molecularity according to the rate law.

Example: For an elementary reaction $$\ce{aA +bB -> cC +dD}$$ the rate of reaction is given by $$R = k [\ce{A}]^{a} [\ce{B}]^{b},$$ where $k$ is the rate constant and $[\cdot]$ represents concentration.

I am aware that, in real solutions, concentrations are replaced by activities; however, the structure of the real rate law is still same as that obtained from the ideal rate law.

I wanted to know if there are any exceptions to this. Is there an elementary reaction for which the rate law is not obeyed, and the molecularity is not equal to order?

  • $\begingroup$ I think it would hold perfectly, if not for things like fast two-step vs one step ambiguity. $\endgroup$
    – Mithoron
    May 4, 2023 at 14:08
  • 1
    $\begingroup$ Your first equation tells us nothing about the way the species react other than $a$ moles of $[A]$ plus $b$ of $[B]$ react and so on. The example by Poutnik below for $H_2/Br_2$ clearly shows this where the rate $R$ is not of the form you quote even when $a=b=1$. Order is an experimental quantity only found with great difficulty. If you like, the connection between $R$ and your chemical equation is fiction and based on assumptions about the mechanism as a way or trying to fit mechanism to experiment. $\endgroup$
    – porphyrin
    May 4, 2023 at 17:19
  • $\begingroup$ @porphyrin the rate law helps us in this regard; if it is known that a reaction is elementary, we can determine the order theoretically using just the stoichiometry of the reaction. $\endgroup$
    – ananta
    May 4, 2023 at 21:04
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    $\begingroup$ @anata , 'if is known that the reaction is elementary' this is just the point, you do not know this unless an experiment is done, even atom+atom. $\endgroup$
    – porphyrin
    May 5, 2023 at 7:18

1 Answer 1


It is true that, for elementary reactions, the reaction rate order is equal to molecularity.

The point is, for $a + b \gt 3$ ( and often for $a + b = 3$ or even $a + b = 2$ ), the reaction is not elementary.

Typical case is $\ce{Br2 + H2 -> 2 HBr}$. It looks like it may be an elementary reaction, but it is not. It has the loopbacked chain reaction schema. Its reaction rate is a rational function with some non-integer exponents:

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

( $k$ and $j$ are rate constants for particular reactions steps withing the reaction kinetic schema. )

Another exception are reactions of the zeroth order, with the rate not depending on the analyte concentration, typically photolytic reactions.

  • $\begingroup$ As far as I know, all zero order reactions are complex (and not elementary) reactions, these do not make an exception to the rate law. $\endgroup$
    – ananta
    May 4, 2023 at 13:21
  • $\begingroup$ Hmm, $\ce{Cl2 ->[h\nu] 2 Cl}$ does not seem complex. at least far from the steady state when recombination matches photolysis. $\endgroup$
    – Poutnik
    May 4, 2023 at 13:23
  • $\begingroup$ What is the rate expression for the reaction $\ce{Cl2 ->[h\nu] 2 Cl}$? Actually, even my first thought was photodissociation reactions. I wanted to confirm if this is the case. $\endgroup$
    – ananta
    May 4, 2023 at 13:24
  • $\begingroup$ R_forward = k. I[W/m2], net rate R = k.I - k2 . (p(Cl2))^2 $\endgroup$
    – Poutnik
    May 4, 2023 at 13:28
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    $\begingroup$ I have not said it is an elementary reaction, I was just hinting it may look like one. As, OTOH, $\ce{I2 + H2 -> 2 HI}$ is not a chain reaction, but elementary reaction, forming $\ce{I2 \bond{...}H2}$ activated complex. // j is rate constant for particular steps in the reaction schema. $\endgroup$
    – Poutnik
    May 4, 2023 at 14:02

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