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bixtya [17]
3 years ago
6

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 f

ast, 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?
Chemistry
1 answer:
Gnesinka [82]3 years ago
8 0

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

<u>Explanation:</u>

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:

<u>Step 1:</u>  Cl_2(g)\rightleftharpoons 2Cl(g);\text{ (very fast)}

<u>Step 2:</u>  Cl(g)+CHCl_3(g)\rightarrow HCl(g)+CCl_3(g);\text{ (slow)}

<u>Step 3:</u>  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}

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