3
$\begingroup$

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 ?

$\endgroup$
  • $\begingroup$ 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 '17 at 16:41
4
$\begingroup$

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

$$-\frac{d[A]}{dt}=k_{1}[A][M]-k_{2}[A^*][M].$$

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

$$k_{1}[A][M]=[A^*](k_{2}[M]+k_{3}).$$

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

$$-\frac{d[A]}{dt}=[A^*](k_{2}[M]+k_{3})-k_{2}[A^*][M].$$

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

$$-\frac{d[A]}{dt}=k_{3}[A^*].$$

$\endgroup$
  • $\begingroup$ Sorry, that was a big typo on my part :). $\endgroup$ – user8277998 Jul 22 '17 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 '17 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. $\endgroup$ – user8277998 Jul 22 '17 at 17:13

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.