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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

Chemistry: The Molecular Science
Chemistry: The Molecular Science
5th Edition
ISBN:9781285199047
Author:John W. Moore, Conrad L. Stanitski
Publisher:John W. Moore, Conrad L. Stanitski
Chapter11: Chemical Kinetics: Rates Of Reactions
Section: Chapter Questions
Problem 83QRT
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Suppose the formation of tert-butanol proceeds by the following mechanism:

step elementary reaction rate constant
1
CH33CBr
(aq) 
→
 
CH33C+
(aq) 
+
 
Br−
(aq)
k1
2
CH33C+
(aq) 
+
 
OH−
(aq) 
→
 
CH33COH
(aq)
k2

Suppose also 

k1

≫

k2

. That is, the first step is much faster than the second.

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
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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