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Cl2(g) ↔ 2Cl(g)

Cl(g) + CHCl3 → HCl(g) + CCl3 (slow)

CCl3 + Cl(g) → CCl4(g) (fast)


a) The overall rate law is equal to the rate law of the second step
as it appears above


b) k1[Cl2] = k-1[Cl]2


c) Cl(g) does not appear in the overall reaction


d) If this is a correct mechanism the observed law would be
Rate = k[Cl2]1/2[CHCl3]


thank you in advance!

2007-02-06 04:50:07 · 3 answers · asked by Anonymous in Science & Mathematics Chemistry

There could be more than 1 correct choice. i need everything that is true?

2007-02-06 05:15:59 · update #1

3 answers

option a) The overall rate law is equal to the rate law of the second step as it appears above.

That is because the slowest step in a reaction is the one that rules the overall reaction's rate.

2007-02-06 05:07:04 · answer #1 · answered by CHESSLARUS 7 · 0 0

WAIT WAIT
I am sorry but as per my knowledge this is an example for free radical substitution and a photchemical reaction where rate of the reaction does not depend on the concentration of the reactants[zero order reaction]. In this case the rate law is
rate =k

2007-02-06 07:04:02 · answer #2 · answered by manidhar 2 · 0 0

Aww chemistry i'd positioned the first 2 statements down. imagine of a field with liquid that is going to evaporate. the floor section if small makes the technique really person-friendly, i'm not able to keep in recommendations although precisely. means should be intense for molecules that get away the floor section.

2016-11-25 20:22:46 · answer #3 · answered by rensing 4 · 0 0

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