Keq (PKapr CH3-C-OH H2N-C2H5 + CH3-C-o H3N-C2H5 + pka = 5 pKa = 11 log Keg = 11 - 5 = 6 or Keq = 106 ; Keq > 1 (favors in the forward reaction) NH4* + H20 pKa = 9 NH3 + H3O* pKa = -2 log Keq = -2 - 9 = -11 Keg = 10-11 Keg < 1, reaction favored in the reverse direction, %3D Try Write the product of the following acid base reactions. Also calculate the Keq for each reaction T. Ö + HCI: CH;CH2ÖH + CH3LI OH t CHq Li Ž NõLi t CH Chapter 5- Chemical Reactions and Mechanisms (307 Fall 2020) KSethi
Keq (PKapr CH3-C-OH H2N-C2H5 + CH3-C-o H3N-C2H5 + pka = 5 pKa = 11 log Keg = 11 - 5 = 6 or Keq = 106 ; Keq > 1 (favors in the forward reaction) NH4* + H20 pKa = 9 NH3 + H3O* pKa = -2 log Keq = -2 - 9 = -11 Keg = 10-11 Keg < 1, reaction favored in the reverse direction, %3D Try Write the product of the following acid base reactions. Also calculate the Keq for each reaction T. Ö + HCI: CH;CH2ÖH + CH3LI OH t CHq Li Ž NõLi t CH Chapter 5- Chemical Reactions and Mechanisms (307 Fall 2020) KSethi
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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![### Chapter 5: Chemical Reactions and Mechanisms
#### Problem Statement
**Task:** Write the product of the following acid-base reactions. Also, calculate the \( K_{eq} \) for each reaction.
#### Reactions and Calculations
1. **Reaction 1:**
\[
\text{CH}_3\text{C(=O)OH} + \text{H}_2\text{N-C}_2\text{H}_5} \rightleftharpoons \text{CH}_3\text{C(=O)}^- + \text{H}_3\text{N-C}_2\text{H}_5}^+
\]
- \( pK_a(\text{CH}_3\text{C(=O)OH}) = 5 \)
- \( pK_a(\text{H}_3\text{N-C}_2\text{H}_5}^+) = 11 \)
**Equilibrium Constant Calculation:**
\[
\text{log } K_{eq} = 11 - 5 = 6 \quad \text{or } K_{eq} = 10^6 \quad (\text{Favors the forward reaction})
\]
2. **Reaction 2:**
\[
\text{NH}_4^+ + \text{H}_2\text{O} \rightleftharpoons \text{NH}_3 + \text{H}_3\text{O}^+
\]
- \( pK_a(\text{NH}_4^+) = 9 \)
- \( pK_a(\text{H}_3\text{O}^+) = -2 \)
**Equilibrium Constant Calculation:**
\[
\text{log } K_{eq} = 2 - 9 = -11 \quad \text{or } K_{eq} = 10^{-11} \quad (\text{Reaction favored in the reverse direction})
\]
#### Additional Notes
- **Graph/Diagram Analysis:**
The diagrams illustrate the acid-base reactions with equilibrium arrows indicating the directionality. Protonation and deprotonation processes are highlighted.
- **Handwritten Calculations:**
- For Reaction 1, \( K_{eq} = 10^{\Delta pK](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F8aba95b7-533e-4661-834a-aa38ddc6056b%2Ffc793cf1-07e0-403a-afa4-837ab0b41644%2Fiy92s6_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Chapter 5: Chemical Reactions and Mechanisms
#### Problem Statement
**Task:** Write the product of the following acid-base reactions. Also, calculate the \( K_{eq} \) for each reaction.
#### Reactions and Calculations
1. **Reaction 1:**
\[
\text{CH}_3\text{C(=O)OH} + \text{H}_2\text{N-C}_2\text{H}_5} \rightleftharpoons \text{CH}_3\text{C(=O)}^- + \text{H}_3\text{N-C}_2\text{H}_5}^+
\]
- \( pK_a(\text{CH}_3\text{C(=O)OH}) = 5 \)
- \( pK_a(\text{H}_3\text{N-C}_2\text{H}_5}^+) = 11 \)
**Equilibrium Constant Calculation:**
\[
\text{log } K_{eq} = 11 - 5 = 6 \quad \text{or } K_{eq} = 10^6 \quad (\text{Favors the forward reaction})
\]
2. **Reaction 2:**
\[
\text{NH}_4^+ + \text{H}_2\text{O} \rightleftharpoons \text{NH}_3 + \text{H}_3\text{O}^+
\]
- \( pK_a(\text{NH}_4^+) = 9 \)
- \( pK_a(\text{H}_3\text{O}^+) = -2 \)
**Equilibrium Constant Calculation:**
\[
\text{log } K_{eq} = 2 - 9 = -11 \quad \text{or } K_{eq} = 10^{-11} \quad (\text{Reaction favored in the reverse direction})
\]
#### Additional Notes
- **Graph/Diagram Analysis:**
The diagrams illustrate the acid-base reactions with equilibrium arrows indicating the directionality. Protonation and deprotonation processes are highlighted.
- **Handwritten Calculations:**
- For Reaction 1, \( K_{eq} = 10^{\Delta pK
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