Differential equation for the rate of unimolecular reaction

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



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> $$\ce{A + M ->[k\_1] A^\* + M} \\ \ce{A^\* + M ->[k\_2] A + M} \\ \ce{A^\* ->[k\_3] B + C}$$
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> 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.
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> 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 $[M]$.
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> $$\dfrac{d[A^\*]}{dt} = k\_1[A][M] - [A^\*](k\_2[M] + k\_3) = 0\tag{1}$$
> $$-\dfrac{d[A]}{dt} = k\_3[A^\*]\tag{2}$$
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> 


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 ?




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 $(2)$:


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


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