Suppose the formation of tert-butanol proceeds by the following mechanism: step elementary reaction rate constant 1 (CH₂), CBr(aq) (CH₂), C* (aq) + Br (aq) k₁ 2 (CH₂), C ¸C²(aq) + OH¯(aq) → (CH₂)₂COH(aq) k₂ Suppose also k₁k₂. That is, the first step is much faster than the second. Write the balanced chemical equation for the overall chemical reaction: Write the experimentally- observable rate law for the overall chemical reaction. Note: your answer should not contain the concentrations of any intermediates. Express the rate constant k for the overall chemical reaction in terms of K₁, K₂, and (if necessary) the rate constants k.₁ and k.2 for the reverse of the two elementary reactions in the mechanism. 0 rate = k = 0 0-0 4 0² 00 X
Suppose the formation of tert-butanol proceeds by the following mechanism: step elementary reaction rate constant 1 (CH₂), CBr(aq) (CH₂), C* (aq) + Br (aq) k₁ 2 (CH₂), C ¸C²(aq) + OH¯(aq) → (CH₂)₂COH(aq) k₂ Suppose also k₁k₂. That is, the first step is much faster than the second. Write the balanced chemical equation for the overall chemical reaction: Write the experimentally- observable rate law for the overall chemical reaction. Note: your answer should not contain the concentrations of any intermediates. Express the rate constant k for the overall chemical reaction in terms of K₁, K₂, and (if necessary) the rate constants k.₁ and k.2 for the reverse of the two elementary reactions in the mechanism. 0 rate = k = 0 0-0 4 0² 00 X
Chapter13: Kinetic Methods
Section: Chapter Questions
Problem 4P
Related questions
Question
Suppose the formation of tert-butanol proceeds by the following mechanism:
| step | elementary reaction | rate constant |
|---|---|---|
|
1
|
CH33CBr
→
CH33C+
+
Br−
|
k1
|
|
2
|
CH33C+
+
OH−
→
CH33COH
|
k2
|
Suppose also
k1
≫
k2
. That is, the first step is much faster than the second.
Transcribed Image Text:Suppose the formation of tert-butanol proceeds by the following mechanism:
step
elementary reaction
rate constant
1
(CH3), CBr(aq) → (CH3), C (aq) + Br (aq)
k₁
2 (CH3), C (aq) + OH (aq) → (CH₂), COH(aq)
k₂
Suppose also k₁k₂. That is, the first step is much faster than the second.
Write the balanced
chemical equation for the
overall chemical reaction:
Write the experimentally-
observable rate law for the
overall chemical reaction.
Note: your answer should
not contain the
concentrations of any
intermediates.
Express the rate constant
k for the overall chemical
reaction in terms of K₁, K₂,
and (if necessary) the rate
constants k.₁ and K-2 for
the reverse of the two
elementary reactions in
the mechanism.
0
rate = k
k = 0
ローロ
00
믐
X
8
olo
Ar
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