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

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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
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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