Consider the following three step mechanism for a reaction
Cl2 > 2Cl Fast
Cl+CHCl3 > HCl+CCl3 Slow
Cl+CCl3 > CCl4 Fast
What is the predicted rate law? How do you do it? The answer is rate=K[Cl2]^.5 [CHCl3]
The elemntary reaction steps and their rates law are:
1) Cl₂ ↔ 2Cl
=> r₁ = k₁₊∙[Cl₂] - k₁₋∙[Cl]² (fast) [ reversible
reaction]
2) Cl + CHCl₃ → HCl + CCl₃
=> r₂ = k₂∙[Cl]∙[CHCl₃] (slow)
3) CCl₃ + Cl → CCl₄
=> r₃ = k₃∙[Cl]∙[CCl₃] (fast)
Overall reaction is given by
CHCl₃ + Cl₂ → CCl₄ + HCl
The overall rate is determined by the slowest step, reaction
2):
r_overall = k₂∙[Cl]∙[CHCl₃]
This rate law contains concentration of intermediate Cl, which need
to be eliminated.
Since reaction1 is a fast reversible reaction, you can consider to
be permanently at equilibrium. That means
r₁ = k₁₊∙[Cl₂] - k₁₋∙[Cl]² = 0
<=>
k₁₊∙[Cl₂] = k₁₋∙[Cl]²
Use the last relation to express [Cl] in terms of reactant
[Cl₂]
[Cl] = √( (k₁₊/k₁₋)∙[Cl₂] )
substitute to overall rate law:
r_overall = k₂∙√( (k₁₊/k₁₋)∙[Cl₂] )∙[CHCl₃]
= k ∙√[Cl₂]∙[CHCl₃]
= k [Cl2]^0.5 [ CHCl3]
Consider the following three step mechanism for a reaction Cl2 > 2Cl Fast Cl+CHCl3 > HCl+CCl3...
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