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Consider this gas phase reaction,
${Cl}_{2} (g) + {CHCl}_{3} (g) ⟶ HCl (g) + {CCl}_{4} (g)$
The reaction is found experimentally to follow this rate law.
$Rate = k [{CHCl}_{3}] {[{Cl}_{2}]}^{1 / 2}$
Based on this information, what conclusions can be drawn about this proposed mechanism?
$Step 1 : {Cl}_{2} (g) ⟶ 2 Cl (g)$
$Step 2: Cl (g) + {CHCl}_{3} (g) ⟶ HCl (g) + {CCl}_{3} (g)$
$Step 3: Cl (g) + {CCl}_{3} (g) ⟶ {CCl}_{4} (g)$

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Step 1 is the rate-determining step.

Step 3 is the rate-determining step.

Step 2 is the rate-determining step.

The rate-determining step cannot be identified.

answer is B.

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Detailed Solution

If RDS were, Step 2,

$R = k [Cl] [{CHCl}_{3}]$ ------equation 1

Starting from the reversible step 1,

$K_{c} = \frac{[{Cl}^{2}}{[{Cl}_{2}]}$

$K_{c} [{Cl}_{2}] = [Cl]^{2}$

$[Cl] = {K_{c}}^{1 / 2} {[{Cl}_{2}]}^{1 / 2}$

Putting the [Cl] value in equation 1,

$R = {kK}_{c}^{1 / 2} {[{Cl}_{2}]}^{1 / 2} [{CHCl}_{3}]$

$R = k^{'} [{CHCl}_{3}] {[{Cl}_{2}]}^{1 / 2}$

$∴ Step 2 is RDS$

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