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

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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.

### 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.
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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