Respuesta :
Answer: The rate law for the overall reaction is [tex]\text{Rate}=k[Cl_2]^{1/2}[CCl_4][CHCl_3][CCl_3]^{-1}[/tex]
Explanation:
In a mechanism of the reaction, the slow step in the mechanism determines the rate of the reaction.
For the given chemical reaction:
[tex]Cl_2(g)+CHCl_3(g)\rightarrow HCl(g)+CCl_4(g)[/tex]
The intermediate reaction of the mechanism follows:
Step 1: [tex]Cl_2(g)\rightleftharpoons 2Cl(g);\text{ (very fast)}[/tex]
Step 2: [tex]Cl(g)+CHCl_3(g)\rightarrow HCl(g)+CCl_3(g);\text{ (slow)}[/tex]
Step 3: [tex]Cl(g)+CCl_3(g)\rightarrow CCl_4(g);\text{ (fast)}[/tex]
As, step 2 is the slow step. It is the rate determining step.
Rate law for the reaction follows:
[tex]\text{Rate}=K_2[Cl][CHCl_3][/tex] ......(1)
As, [Cl] is not appearing as a reactant in the overall reaction. So, we apply steady state approximation in it.
Applying steady state approximation for Cl from step 1 and step 3, we get:
[tex]K_1=\frac{[Cl]^2}{[Cl_2]}[/tex]
[tex][Cl]=\sqrt{K_1[Cl_2]}[/tex]
[tex]K_3=\frac{[CCl_4]}{[Cl][CCl_3]}[/tex]
[tex][Cl]=\frac{[CCl_4]}{K_3[CCl_3]}[/tex]
Putting the value of [Cl] in equation 1, we get:
[tex]\text{Rate}=K_2\times \sqrt{K_1[Cl_2]}\times \frac{[CCl_4]}{K_3[CCl_3]}\times [CHCl_3]\\\\\text{Rate}=k[Cl_2]^{1/2}[CCl_4][CHCl_3][CCl_3]^{-1}[/tex]
Hence, the rate law for the overall reaction is [tex]\text{Rate}=k[Cl_2]^{1/2}[CCl_4][CHCl_3][CCl_3]^{-1}[/tex]
