Question

A proposed mechanism for the reaction Cl2(g) + CHCl3(g)  HCl(g) + CCl4(g) in the atmosphere is Step 1: Cl2(g) 2 Cl(g) (very fast, reversible) Step 2: Cl(g) + CHCl3(g)  HCl(g) + CCl3(g) (slow) Step 3: Cl(g) + CCl3(g)  CCl4(g) (fast) What is the rate law for the overall reaction?

Answer 1

Answer: The rate law for the overall reaction is $\text{Rate}=k[Cl_2]^{1/2}[CCl_4][CHCl_3][CCl_3]^{-1}$

Explanation:

In a mechanism of the reaction, the slow step in the mechanism determines the rate of the reaction.

For the given chemical reaction:

$Cl_2(g)+CHCl_3(g)\rightarrow HCl(g)+CCl_4(g)$

The intermediate reaction of the mechanism follows:

Step 1:  $Cl_2(g)\rightleftharpoons 2Cl(g);\text{ (very fast)}$

Step 2:  $Cl(g)+CHCl_3(g)\rightarrow HCl(g)+CCl_3(g);\text{ (slow)}$

Step 3:  $Cl(g)+CCl_3(g)\rightarrow CCl_4(g);\text{ (fast)}$

As, step 2 is the slow step. It is the rate determining step.

Rate law for the reaction follows:

$\text{Rate}=K_2[Cl][CHCl_3]$           ......(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:

$K_1=\frac{[Cl]^2}{[Cl_2]}$

$[Cl]=\sqrt{K_1[Cl_2]}$

$K_3=\frac{[CCl_4]}{[Cl][CCl_3]}$

$[Cl]=\frac{[CCl_4]}{K_3[CCl_3]}$  

Putting the value of [Cl] in equation 1, we get:  

$\text{Rate}=K_2\times \sqrt{K_1[Cl_2]}\times \frac{[CCl_4]}{K_3[CCl_3]}\times [CHCl_3]\\\\\text{Rate}=k[Cl_2]^{1/2}[CCl_4][CHCl_3][CCl_3]^{-1}$

Hence, the rate law for the overall reaction is $\text{Rate}=k[Cl_2]^{1/2}[CCl_4][CHCl_3][CCl_3]^{-1}$