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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What does it mean forward or backward and does the k=5.0×10^23 even play a role in this
![**Equilibrium Calculation: Aluminum Nitrate and Sodium Fluoride Reaction**
When an aluminum nitrate solution is added to a sodium fluoride solution, the following equilibrium is established:
\[
\text{Al}^{3+} (aq) + 6 \text{F}^{-} (aq) \rightleftharpoons \text{AlF}_6^{3-} (aq)
\]
For this equilibrium, \( K = 5.0 \times 10^{23} \).
**Task:**
Choose the correct equilibrium constant expression and the direction favored at equilibrium.
**Options:**
1. \( K = \frac{[\text{AlF}_6^{3-}]}{[\text{Al}^{3+}][\text{F}^{-}]^6} \) (Reverse)
2. \( K = \frac{[\text{AlF}_6^{3-}]}{[\text{Al}^{3+}][\text{F}^{-}]^6} \) (Forward)
3. \( K = \frac{[\text{Al}^{3+}][\text{F}^{-}]^6}{[\text{AlF}_6^{3-}]} \) (Reverse)
4. \( K = \frac{[\text{AlF}_6^{3-}]}{[\text{Al}^{3+}][\text{F}^{-}]^6} \) (Forward)
5. \( K = \frac{1}{[\text{Al}^{3+}][\text{F}^{-}]^6} \) (Forward)
**Explanation:**
- The equilibrium constant \( K \) is given as a very large value (\( 5.0 \times 10^{23} \)), suggesting that the forward reaction is favored, leading to the formation of \(\text{AlF}_6^{3-}\) from \(\text{Al}^{3+}\) and \(\text{F}^{-}\) ions.
- The correct equilibrium constant expression is based on the concentrations of the products over reactants:
\[
K = \frac{[\text{AlF}_6^{3-}]}{[\text{Al}^{3+}][\text{F}^{-}]^6}
\]
- Therefore, the choice that correctly describes this scenario is **Option 2**, indicating the reaction proceeds](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fb0ffec19-af41-4c50-af1b-66e883fc2bb8%2F95aa2e94-e724-457b-abf8-0aba9a89910e%2Ft5rsovn_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Equilibrium Calculation: Aluminum Nitrate and Sodium Fluoride Reaction**
When an aluminum nitrate solution is added to a sodium fluoride solution, the following equilibrium is established:
\[
\text{Al}^{3+} (aq) + 6 \text{F}^{-} (aq) \rightleftharpoons \text{AlF}_6^{3-} (aq)
\]
For this equilibrium, \( K = 5.0 \times 10^{23} \).
**Task:**
Choose the correct equilibrium constant expression and the direction favored at equilibrium.
**Options:**
1. \( K = \frac{[\text{AlF}_6^{3-}]}{[\text{Al}^{3+}][\text{F}^{-}]^6} \) (Reverse)
2. \( K = \frac{[\text{AlF}_6^{3-}]}{[\text{Al}^{3+}][\text{F}^{-}]^6} \) (Forward)
3. \( K = \frac{[\text{Al}^{3+}][\text{F}^{-}]^6}{[\text{AlF}_6^{3-}]} \) (Reverse)
4. \( K = \frac{[\text{AlF}_6^{3-}]}{[\text{Al}^{3+}][\text{F}^{-}]^6} \) (Forward)
5. \( K = \frac{1}{[\text{Al}^{3+}][\text{F}^{-}]^6} \) (Forward)
**Explanation:**
- The equilibrium constant \( K \) is given as a very large value (\( 5.0 \times 10^{23} \)), suggesting that the forward reaction is favored, leading to the formation of \(\text{AlF}_6^{3-}\) from \(\text{Al}^{3+}\) and \(\text{F}^{-}\) ions.
- The correct equilibrium constant expression is based on the concentrations of the products over reactants:
\[
K = \frac{[\text{AlF}_6^{3-}]}{[\text{Al}^{3+}][\text{F}^{-}]^6}
\]
- Therefore, the choice that correctly describes this scenario is **Option 2**, indicating the reaction proceeds
Expert Solution
Step 1
Chemical kinetics is a branch of chemistry in which we deal with speed at which rate of reaction occurs. In the given question, it has been asked to choose the correct option for equilibrium constant and direction of reaction.
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