Question

1. consider the following three step mechanism:

$ce{cl2(g) <=> 2 cl(g)}$ fast
$ce{cl(g) + chcl3(g) -> hcl(g) + ccl3(g)}$ slow
$ce{ccl3(g) + cl(g) -> ccl4(g)}$ fast
what is the overall reaction?
identify any intermediates.
derive a rate law for this reaction.

Explanation:

Part 1: Overall Reaction

Step1: Write all reactions

Reaction 1: $\ce{Cl_{2}(g)
ightleftharpoons 2Cl(g)}$ (fast)
Reaction 2: $\ce{Cl(g) + CHCl_{3}(g)
ightarrow HCl(g) + CCl_{3}(g)}$ (slow)
Reaction 3: $\ce{CCl_{3}(g) + Cl(g)
ightarrow CCl_{4}(g)}$ (fast)

Step2: Sum all reactions

Add reactants on left and products on right:
$\ce{Cl_{2}(g) + 2Cl(g) + CHCl_{3}(g) + CCl_{3}(g)
ightarrow 2Cl(g) + HCl(g) + CCl_{3}(g) + CCl_{4}(g)}$

Step3: Cancel intermediates

Cancel $\ce{2Cl(g)}$ and $\ce{CCl_{3}(g)}$ (appear on both sides):
$\ce{Cl_{2}(g) + CHCl_{3}(g)
ightarrow HCl(g) + CCl_{4}(g)}$

Part 2: Identify Intermediates

Intermediates are species formed and consumed in the mechanism.

• $\ce{Cl(g)}$: Formed in Reaction 1, consumed in Reactions 2 and 3.
• $\ce{CCl_{3}(g)}$: Formed in Reaction 2, consumed in Reaction 3.

Part 3: Derive Rate Law

Step1: Rate-determining step (RDS)

The slow step (Reaction 2) determines the rate:
Rate = $k_2 [\ce{Cl}] [\ce{CHCl_{3}}]$

Step2: Equilibrium for fast step

Reaction 1 is fast equilibrium: $K_1 = \frac{[\ce{Cl}]^2}{[\ce{Cl_{2}}]} \implies [\ce{Cl}] = \sqrt{K_1 [\ce{Cl_{2}}]}$

Step3: Substitute $[\ce{Cl}]$ into rate law

Rate = $k_2 \sqrt{K_1 [\ce{Cl_{2}}]} [\ce{CHCl_{3}}]$
Let $k = k_2 \sqrt{K_1}$, so Rate = $k [\ce{Cl_{2}}]^{1/2} [\ce{CHCl_{3}}]$

Answer:

s:

• Overall Reaction: $\boldsymbol{\ce{Cl_{2}(g) + CHCl_{3}(g)

ightarrow HCl(g) + CCl_{4}(g)}}$

• Intermediates: $\boldsymbol{\ce{Cl(g)}}$ and $\boldsymbol{\ce{CCl_{3}(g)}}$
• Rate Law: $\boldsymbol{\text{Rate} = k [\ce{Cl_{2}}]^{1/2} [\ce{CHCl_{3}}]}$ (where $k = k_2 \sqrt{K_1}$)