1
$\begingroup$

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?.

$\endgroup$
  • 3
    $\begingroup$ because the definition of first order reaction is that it depends on concentration of single reactant $\endgroup$ – chemstackisunhelpful Jun 2 at 12:55
  • $\begingroup$ @chemstackisunhelpful I am sorry but your answer is unhelpful $\endgroup$ – maverick Jun 2 at 13:06
  • $\begingroup$ pl explain it briefly $\endgroup$ – maverick Jun 2 at 13:07
  • $\begingroup$ $\ce{A + B + C -> D}$ is not a first order reaction. $\endgroup$ – Martin - マーチン Jun 2 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. $\endgroup$ – Martin - マーチン Jun 2 at 17:15
1
$\begingroup$

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.

| improve this answer | |
$\endgroup$
  • $\begingroup$ Yeah thanks for answer , after so long, I had already finished chemical kinetics !! $\endgroup$ – maverick Jul 9 at 13:47
  • $\begingroup$ Did my answer help? ^^ $\endgroup$ – DDD4C4U Jul 9 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 Jul 9 at 13:51

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.