I am new to kinetics so please explain it from basic, If there is a reaction -

$$\ce{A + B + C -> D}$$

In first order reaction rate is given by-

$\mathrm R = k[\mathrm A]$ or $\mathrm R = k[\mathrm B]$ or $\mathrm R = k[\mathrm C]$

my question is why cant $\mathrm R = k[\mathrm A]^{1/3}[\mathrm B]^{1/3}[\mathrm C]^{1/3}$ even though order of reaction is still $1$,----- is it by definition that 1st order reaction always depend on concentration of single reactant, or is there any experimental proof for this?.

  • 3
    $\begingroup$ because the definition of first order reaction is that it depends on concentration of single reactant $\endgroup$ Commented Jun 2, 2020 at 12:55
  • $\begingroup$ @chemstackisunhelpful I am sorry but your answer is unhelpful $\endgroup$
    – maverick
    Commented Jun 2, 2020 at 13:06
  • $\begingroup$ pl explain it briefly $\endgroup$
    – maverick
    Commented Jun 2, 2020 at 13:07
  • $\begingroup$ $\ce{A + B + C -> D}$ is not a first order reaction. $\endgroup$ Commented Jun 2, 2020 at 13:16
  • 1
    $\begingroup$ No. If $\ce{A + B + C -> D}$, then the rate law would be $r = k[\ce{A}][\ce{B}][\ce{C}]$, if the rate law is determined to be $r = k[\ce{A}]$, then the reaction cannot be elementary. $\endgroup$ Commented Jun 2, 2020 at 17:15

1 Answer 1


As a purely mathematical definition, a first order reaction is one in which the rate is directly proportional to concentration of reactants. The thing you should look for is physical examples of this. This only happens if you have reactions where one compound is decreasing and the other one is abundance (pseudo - first order) , or , radioactive decay where the decay rate is proportional to number of atoms at a point.

The reaction you have shown with three reactants would most probably me not first order unless two of the three are abundant. So, the solution to the confusion is to think of the physically possible scenarios which the equation models instead of creating hypothetical situations and force the math to model that.

  • $\begingroup$ Yeah thanks for answer , after so long, I had already finished chemical kinetics !! $\endgroup$
    – maverick
    Commented Jul 9, 2020 at 13:47
  • $\begingroup$ Did my answer help? ^^ $\endgroup$
    – Babu
    Commented Jul 9, 2020 at 13:48
  • $\begingroup$ Yeah you have written almost same, about what I thought after getting no help, But thank you for the answer it really helps a lot, to clear concepts $\endgroup$
    – maverick
    Commented Jul 9, 2020 at 13:51

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