Fe,O3(s) + 3H,(g) 2 2Fe(s) + 3H,O(g) Choose the changes that will shift the equilibrium position to the left. (Select all that apply.) Decrease temperature O Add H2(g) OIncrease temperature O Add a catalyst Remove Fe,O3(s) ORemove H(g) O Decrease volume
Fe,O3(s) + 3H,(g) 2 2Fe(s) + 3H,O(g) Choose the changes that will shift the equilibrium position to the left. (Select all that apply.) Decrease temperature O Add H2(g) OIncrease temperature O Add a catalyst Remove Fe,O3(s) ORemove H(g) O Decrease volume
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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![### Understanding Shifts in Equilibrium for Endothermic Reactions
#### For the following endothermic reaction system at equilibrium:
\[ \text{Fe}_2\text{O}_3 (s) + 3\text{H}_2 (g) \rightleftharpoons 2\text{Fe} (s) + 3\text{H}_2\text{O} (g) \]
#### Choose the changes that will shift the equilibrium position to the left.
(Select all that apply.)
- [ ] Decrease temperature
- [ ] Add \(\text{H}_2 (g)\)
- [ ] Increase temperature
- [ ] Add a catalyst
- [ ] Remove \(\text{Fe}_2\text{O}_3 (s)\)
- [ ] Remove \(\text{H}_2 (g)\)
- [ ] Decrease volume
#### Explanation of the Impact of Changes on Equilibrium:
1. **Decrease Temperature**: In an endothermic reaction, decreasing the temperature shifts the equilibrium to the left as the system will try to produce more heat.
2. **Add \(\text{H}_2 (g)\)**: Adding more of the reactant \(\text{H}_2\) would shift the equilibrium to the right to consume the extra reactant, not to the left.
3. **Increase Temperature**: Increasing the temperature shifts the equilibrium to the right in endothermic reactions, as the system absorbs the additional heat.
4. **Add a Catalyst**: Adding a catalyst does not shift the position of equilibrium; it only speeds up the rate at which equilibrium is reached.
5. **Remove \(\text{Fe}_2\text{O}_3 (s)\)**: Removing a reactant shifts the equilibrium to the left to produce more of the reactant.
6. **Remove \(\text{H}_2 (g)\)**: Removing \(\text{H}_2\) would shift the equilibrium position to the left to create more \(\text{H}_2\).
7. **Decrease Volume**: Decreasing the volume increases the pressure. Since there are equal numbers of moles of gas on both sides (3 moles \(\text{H}_2\) on the reactant side and 3 moles \(\text{H}_2\text{O}\) on the product side), this change will](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F5fe141b1-9817-4d6d-9b09-948fe13c457c%2Fa312c832-38b6-4151-bbd5-8e7af5e8a17c%2Fj40n6_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Understanding Shifts in Equilibrium for Endothermic Reactions
#### For the following endothermic reaction system at equilibrium:
\[ \text{Fe}_2\text{O}_3 (s) + 3\text{H}_2 (g) \rightleftharpoons 2\text{Fe} (s) + 3\text{H}_2\text{O} (g) \]
#### Choose the changes that will shift the equilibrium position to the left.
(Select all that apply.)
- [ ] Decrease temperature
- [ ] Add \(\text{H}_2 (g)\)
- [ ] Increase temperature
- [ ] Add a catalyst
- [ ] Remove \(\text{Fe}_2\text{O}_3 (s)\)
- [ ] Remove \(\text{H}_2 (g)\)
- [ ] Decrease volume
#### Explanation of the Impact of Changes on Equilibrium:
1. **Decrease Temperature**: In an endothermic reaction, decreasing the temperature shifts the equilibrium to the left as the system will try to produce more heat.
2. **Add \(\text{H}_2 (g)\)**: Adding more of the reactant \(\text{H}_2\) would shift the equilibrium to the right to consume the extra reactant, not to the left.
3. **Increase Temperature**: Increasing the temperature shifts the equilibrium to the right in endothermic reactions, as the system absorbs the additional heat.
4. **Add a Catalyst**: Adding a catalyst does not shift the position of equilibrium; it only speeds up the rate at which equilibrium is reached.
5. **Remove \(\text{Fe}_2\text{O}_3 (s)\)**: Removing a reactant shifts the equilibrium to the left to produce more of the reactant.
6. **Remove \(\text{H}_2 (g)\)**: Removing \(\text{H}_2\) would shift the equilibrium position to the left to create more \(\text{H}_2\).
7. **Decrease Volume**: Decreasing the volume increases the pressure. Since there are equal numbers of moles of gas on both sides (3 moles \(\text{H}_2\) on the reactant side and 3 moles \(\text{H}_2\text{O}\) on the product side), this change will
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