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
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
Section: Chapter Questions
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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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.
![### Formation of tert-Butanol Mechanism
#### Reaction Mechanism:
The formation of tert-butanol proceeds by the following mechanism:
| Step | Elementary Reaction | Rate Constant |
|------|----------------------|---------------|
| 1 | \( (CH_3)_3CBr(aq) \rightarrow (CH_3)_3C^+(aq) + Br^-(aq) \) | \( k_1 \) |
| 2 | \( (CH_3)_3C^+(aq) + OH^-(aq) \rightarrow (CH_3)_3COH(aq) \) | \( k_2 \) |
Suppose also \( k_1 \gg k_2 \). That is, the first step is much faster than the second.
### Tasks
1. **Write the balanced chemical equation for the overall chemical reaction:**
\[
\boxed{}
\]
2. **Write the experimentally-observable rate law for the overall chemical reaction.**
**Note:** Your answer should **not** contain the concentrations of any intermediates.
\[
\text{rate} = k \boxed{}
\]
3. **Express the rate constant \( k \) for the overall chemical reaction in terms of \( k_1 \), \( k_2 \), and (if necessary) the rate constants \( k_{-1} \) and \( k_{-2} \) for the reverse of the two elementary reactions in the mechanism.**
\[
k = \boxed{}
\]
### Interpretation of Graphs/Diagrams
- **Graph Explanation:**
No graphs or diagrams are present in the prompt.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F9757754f-3312-4a82-9461-9956fc98753f%2Fbfe1477c-dad9-492a-ba7f-6cc6188d5a85%2Fenan4au_processed.png&w=3840&q=75)
Transcribed Image Text:### Formation of tert-Butanol Mechanism
#### Reaction Mechanism:
The formation of tert-butanol proceeds by the following mechanism:
| Step | Elementary Reaction | Rate Constant |
|------|----------------------|---------------|
| 1 | \( (CH_3)_3CBr(aq) \rightarrow (CH_3)_3C^+(aq) + Br^-(aq) \) | \( k_1 \) |
| 2 | \( (CH_3)_3C^+(aq) + OH^-(aq) \rightarrow (CH_3)_3COH(aq) \) | \( k_2 \) |
Suppose also \( k_1 \gg k_2 \). That is, the first step is much faster than the second.
### Tasks
1. **Write the balanced chemical equation for the overall chemical reaction:**
\[
\boxed{}
\]
2. **Write the experimentally-observable rate law for the overall chemical reaction.**
**Note:** Your answer should **not** contain the concentrations of any intermediates.
\[
\text{rate} = k \boxed{}
\]
3. **Express the rate constant \( k \) for the overall chemical reaction in terms of \( k_1 \), \( k_2 \), and (if necessary) the rate constants \( k_{-1} \) and \( k_{-2} \) for the reverse of the two elementary reactions in the mechanism.**
\[
k = \boxed{}
\]
### Interpretation of Graphs/Diagrams
- **Graph Explanation:**
No graphs or diagrams are present in the prompt.
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