Differential equation for the rate of unimolecular reaction

Question

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 ?

Answer 1

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 $(2)$:

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