Given in my book is the following reaction mechanism for the unimolecular $\ce{A -> B + C}$,

$$\ce{A + M ->[k_1] A^* + M} \\ \ce{A^* + M ->[k_2] A + M} \\ \ce{A^* ->[k_3] B + C}$$

Here $A^*$ is an $A$ molecule with enough vibrational energy to isomerize or decompose. In other words, part of kinetic energy of a bimolecular collision (between $A$ and $M$) has been used to raise an $A$ molecule to higher vibrational energy.

Since $A^*$ is never present at very high concentration we can use steady state appoximination to obtain an expression for rate of reaction in terms of $[A]$ and $[M]$.

$$\dfrac{d[A^*]}{dt} = k_1[A][M] - [A^*](k_2[M] + k_3) = 0\tag{1}$$ $$-\dfrac{d[A]}{dt} = k_3[A^*]\tag{2}$$

I don't understand how the $(2)$ equation is obtained, in my opinion it should be $$-\dfrac{d[A]}{dt} = k_1 [A][M]- k_2[A^*][M].$$

I did not understand how we get $k_3$ instead of $k_1$ and why not $[M]$ in the rate law ?

  • $\begingroup$ Your equation for the rate of change of $[A]$ is incorrect. It should be $$\frac{d[A]}{dt}=-k_{1}[A][M]+k_{2}[A^*][M].$$ $\endgroup$
    – Argon
    Jul 22, 2017 at 16:41

1 Answer 1


To start with, the equation you wrote for the rate of change of $[A]$ is not correct. It should instead be


Note that $[A]$ is consumed in the first reaction step and produced in the second. Now, rearrange equation $(1)$ to get


Plug in the above equation into the actual expression for $-d[A]/dt$ to get


This, of course, can be simplified to yield the desired equation $(2)$:


  • $\begingroup$ Sorry, that was a big typo on my part :). $\endgroup$ Jul 22, 2017 at 17:02
  • $\begingroup$ I see you edited your original post. It's still not entirely accurate. Species $A$ is consuned when it reacts with $M$. As such, it should be $k_{1}[A][M]$ for that step if you're describing $-d[A]/dt$. Also, it is not usually suggested to change the content of your post in a way that changes the original question. If you want to update an equation you suggested, you can clearly note the change without overwriting the original. Just for future reference! $\endgroup$
    – Argon
    Jul 22, 2017 at 17:04
  • $\begingroup$ I will take care from next time. Since I think it does not affect the original question in any way I will leave as edited. $\endgroup$ Jul 22, 2017 at 17:13

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