Question

1. consider the following three step mechanism:

cl₂(g) ⇌ 2 cl(g) fast
cl(g) + chcl₃(g) → hcl(g) + ccl₃(g) slow
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: Combine all steps

Add the three elementary steps together:
$$\ce{Cl2(g) <=> 2 Cl(g)}$$
$$\ce{Cl(g) + CHCl3(g) -> HCl(g) + CCl3(g)}$$
$$\ce{CCl3(g) + Cl(g) -> CCl4(g)}$$

Step2: Cancel intermediates

Intermediates (species that appear as product and reactant) are $\ce{Cl(g)}$ (2 moles from first step, 1 + 1 moles consumed in second and third) and $\ce{CCl3(g)}$ (produced in second, consumed in third). Cancel them:
Reactants: $\ce{Cl2(g) + CHCl3(g)}$
Products: $\ce{HCl(g) + CCl4(g)}$

Intermediates are species produced in one step and consumed in another (not in overall reaction). From the steps:

• $\ce{Cl(g)}$: Produced in first step, consumed in second and third.
• $\ce{CCl3(g)}$: Produced in second step, consumed in third.

Step1: Determine rate-determining step

The slow step is $\ce{Cl(g) + CHCl3(g) -> HCl(g) + CCl3(g)}$. Rate law for slow step:
$$\text{Rate} = k_2 [\ce{Cl}] [\ce{CHCl3}]$$

Step2: Relate intermediate $[\ce{Cl}]$ to reactants

From fast equilibrium: $\ce{Cl2(g) <=> 2 Cl(g)}$, equilibrium constant $K = \frac{[\ce{Cl}]^2}{[\ce{Cl2}]}$, so $[\ce{Cl}] = \sqrt{K [\ce{Cl2}]}$.

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

Substitute $[\ce{Cl}]$ into the slow step’s rate law:
$$\text{Rate} = k_2 \cdot \sqrt{K [\ce{Cl2}]} \cdot [\ce{CHCl3}]$$
Let $k = k_2 \sqrt{K}$, so:
$$\text{Rate} = k [\ce{Cl2}]^{1/2} [\ce{CHCl3}]$$

Answer:

Overall reaction: $\boldsymbol{\ce{Cl2(g) + CHCl3(g) -> HCl(g) + CCl4(g)}}$

Part 2: Identify Intermediates