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I need help with my physical chemistry lab report, attached is the question

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reaction of M-1. However, this mechanism is incomplete. The equilibrium represented by M-1
and M-2 has a mechanism of its own, as does reaction M-5.
For the case of reaction M-5, the corresponding mechanism is:
HOBr(aq)+I (aq)+H*(aq) >IBr(aq)+ H,O(I)
[M-6]
IBr(aq) +H,O(1) ->HOI(aq) + Br (aq) + H*(aq)
[M-7]
HOI(aq) + Br-(aq)+ H*(aq) >IBr(aq)+ H,O(1)
[M-8]
HOI(aq) +I (aq) + H*(aq) >I,(aq)+ H,O(1)
[M-3]
On the basis of known rate constants for these reactions and analogous reactions, the
argument is that the equilibrium represented by M-1 and M-2 is actually the rate-
determining step. For this equilibrium, the corresponding mechanism is proposed:
Bro; (aq) + H*(aq) > HBro; (aq)
[M-A]
HBro,(aq) >Bro; (aq) + H*(aq)
[M-B]
HBro,(aq)+I (aq) ->HBro; I (aq)
[M-C]
HBro, I (aq) - > HBro, (aq) +I (aq)
[M-D]
HBro, I (aq) +H*(aq) ->HBroz(aq)+ HOI(aq)
[M-E]
Of these reactions, M-A through M-E, M-E is the rate determining step if
KD >> KE[H*] leading to the rate law of:
Rate = [HOI k kcke Bro, III IIHP
dt
ksky
[Rate-M]
References
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14. Finally, using the reaction mechanisms [M-I] through [M-E] below, discuss whether or not
your data is consistent with the mechanism provided? To answer this question, you will need
to derive the overall rate law by showing the differential rate laws and assumptions for the
proposed mechanism. Does it matter that the rate law below is not expressed in terms of
[Bro; ]? Why? (This analysis is worth 10 pts on your lab report grade.)
Mechanism
In the presence of excess I, Eq. 14 would become:
Bro; (aq)+91 (aq)+6H*(aq)->31, (aq)+3H,O(1)+Br (aq)
[20]
In the presence of excess Bros . we have Eq. 14:
Bro, (aq)+61 (aq)+6H*(aq)->31, (aq)+3H,O(1)+Br (aq)
[14]
We have purposefully kept the concentrations in this experiment such that Bros is in excess. A
proposed mechanism for Eq. 14 is:
Bro, (aq)+I (aq)+2H*(aq)->HBrO,(aq)+HOI(aq)
[M-1]
HBrO, (aq)+HOI(aq)>Bro; (aq)+I(aq)+2H*(aq)
[M-2]
HOl(aq)+I (aq)+H*(aq)_>I,(aq)+H,O(1)
[M-3]
HBrO,(aq)+I(aq)+H*(aq)->HOBr(aq)+ HOI(aq)
[M-4]
HOBY(aq)+21 (aq)+H*(aq)_>1,(aq)+Br (aq)+H,OXI)
[M-5]
At this stage, the combination of M-I with two M-3, M-4 and M-5 satisfies the stoichiometry of
Eq. 14. The two reactions, M-1 and M-2, represent an equilibrium, with M-2 being the reverse
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