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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![**Equilibrium Constant Expression for the Given Reaction**
**Reaction:**
\[ C_6H_{10}O_4 (aq) + 2 \text{OH}^- (aq) \rightarrow 2 \text{CH}_3 \text{COO}^- (aq) + C_2H_6O_2 (aq) \]
To write the equilibrium constant expression (K_eq) for this reaction, follow these steps:
1. **Identify the reactants and products**:
- Reactants: \( C_6H_{10}O_4 (aq) \) and \( 2 \text{OH}^- (aq) \)
- Products: \( 2 \text{CH}_3 \text{COO}^- (aq) \) and \( C_2H_6O_2 (aq) \)
2. **Set up the general form of the equilibrium constant expression**:
\[ K_{eq} = \frac{[\text{products}]}{[\text{reactants}]} \]
3. **Apply the stoichiometric coefficients as exponents in the expression**:
\[ K_{eq} = \frac{[\text{CH}_3 \text{COO}^-]^2 [\text{C}_2 \text{H}_6 \text{O}_2]}{ [C_6H_{10}O_4] [\text{OH}^-]^2 } \]
Replace the placeholders with their respective components from the reaction:
\[ K_{eq} = \frac{[ \text{CH}_3 \text{COO}^-]^2 [C_2H_6O_2] } { [C_6H_{10}O_4] [\text{OH}^-]^2 } \]
**Visual Explanation:**
In this task's image, there is a rectangular box on the left side, which is presumably for inputting the equilibrium expression. On the right side, there is a series of icons, including a check mark (✓) likely for submission, and undo or redo arrows for editing.
Therefore, when writing the equilibrium constant expression (K_eq) for the reaction provided, it should be stated as:
\[ K_{eq} = \frac{[2 \text{CH}_3 \text{COO}^-]^2 [C_2H_6O](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F3991c7bd-acfb-40ed-bb56-29190279c7cb%2F5b3af2fe-8f30-458f-8d6d-3fdb6d8cfd70%2Frytl5x_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Equilibrium Constant Expression for the Given Reaction**
**Reaction:**
\[ C_6H_{10}O_4 (aq) + 2 \text{OH}^- (aq) \rightarrow 2 \text{CH}_3 \text{COO}^- (aq) + C_2H_6O_2 (aq) \]
To write the equilibrium constant expression (K_eq) for this reaction, follow these steps:
1. **Identify the reactants and products**:
- Reactants: \( C_6H_{10}O_4 (aq) \) and \( 2 \text{OH}^- (aq) \)
- Products: \( 2 \text{CH}_3 \text{COO}^- (aq) \) and \( C_2H_6O_2 (aq) \)
2. **Set up the general form of the equilibrium constant expression**:
\[ K_{eq} = \frac{[\text{products}]}{[\text{reactants}]} \]
3. **Apply the stoichiometric coefficients as exponents in the expression**:
\[ K_{eq} = \frac{[\text{CH}_3 \text{COO}^-]^2 [\text{C}_2 \text{H}_6 \text{O}_2]}{ [C_6H_{10}O_4] [\text{OH}^-]^2 } \]
Replace the placeholders with their respective components from the reaction:
\[ K_{eq} = \frac{[ \text{CH}_3 \text{COO}^-]^2 [C_2H_6O_2] } { [C_6H_{10}O_4] [\text{OH}^-]^2 } \]
**Visual Explanation:**
In this task's image, there is a rectangular box on the left side, which is presumably for inputting the equilibrium expression. On the right side, there is a series of icons, including a check mark (✓) likely for submission, and undo or redo arrows for editing.
Therefore, when writing the equilibrium constant expression (K_eq) for the reaction provided, it should be stated as:
\[ K_{eq} = \frac{[2 \text{CH}_3 \text{COO}^-]^2 [C_2H_6O
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