Consider the following system at equilibrium at 25°C. PCI3(9) + Cl,(9) = PCI_(9) AG° = –92.50 k) What will happen to the ratio of partial pressure of PCI, to partial pressure of PCI, if the temperature is raised? decrease increase no change Explain completely. (Select all that apply.) O The reaction is exothermic. The equilibrium will shift to the left when the temperature is raised. The reaction is endothermic. The equilibrium will shift to the right when the temperature is raised. AS is negative. Not enough information to determine if the reaction is endothermic or exothermic. Not enough information to determine the sign of AH. O AH is negative. Not enough information to determine the sign of AS. AS is positive. AH is positive.
Consider the following system at equilibrium at 25°C. PCI3(9) + Cl,(9) = PCI_(9) AG° = –92.50 k) What will happen to the ratio of partial pressure of PCI, to partial pressure of PCI, if the temperature is raised? decrease increase no change Explain completely. (Select all that apply.) O The reaction is exothermic. The equilibrium will shift to the left when the temperature is raised. The reaction is endothermic. The equilibrium will shift to the right when the temperature is raised. AS is negative. Not enough information to determine if the reaction is endothermic or exothermic. Not enough information to determine the sign of AH. O AH is negative. Not enough information to determine the sign of AS. AS is positive. AH is positive.
Just having trouble with the last part of this question, thanks!
Transcribed Image Text:### Equilibrium Reaction Analysis
**Chemical Reaction at Equilibrium:**
Consider the following system at equilibrium at 25°C:
\[ \text{PCl}_3(g) + \text{Cl}_2(g) \rightleftharpoons \text{PCl}_5(g) \quad \Delta G^\circ = -92.50 \, \text{kJ} \]
**Question:**
What will happen to the ratio of partial pressure of \(\text{PCl}_5\) to partial pressure of \(\text{PCl}_3\) if the temperature is raised?
- **Decrease** (Selected and Correct)
- Increase
- No Change
**Explanation:**
Select all that apply:
- [ ] The reaction is exothermic.
- [ ] The equilibrium will shift to the left when the temperature is raised.
- [ ] The reaction is endothermic.
- [ ] The equilibrium will shift to the right when the temperature is raised.
- [ ] \(\Delta S\) is negative.
- [ ] Not enough information to determine if the reaction is endothermic or exothermic.
- [ ] Not enough information to determine the sign of \(\Delta H\).
- [ ] \(\Delta H\) is negative.
- [ ] Not enough information to determine the sign of \(\Delta S\).
- [ ] \(\Delta S\) is positive.
- [ ] \(\Delta H\) is positive.
### Analysis:
1. **Reaction Direction:**
- Given \(\Delta G^\circ = -92.50 \, \text{kJ}\), the reaction is spontaneous under standard conditions.
- Raising the temperature typically affects the position of equilibrium according to Le Chatelier's principle.
2. **Thermodynamic Considerations:**
- Since \(\Delta G^\circ\) is negative, the reaction is favorable in the forward direction.
- \(\Delta H\) is generally negative for exothermic reactions, suggesting heat is released.
3. **Equilibrium Shift:**
- For an exothermic reaction such as this, increasing the temperature would shift the equilibrium to the left, towards reactants (\(\text{PCl}_3\) and \(\text{Cl}_2\)), decreasing the ratio of \(\text{PCl}_5\).
This comprehensive approach aids in understanding equilibrium shifts in response to temperature changes and the
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![### Equilibrium Reaction Analysis
**Chemical Reaction at Equilibrium:**
Consider the following system at equilibrium at 25°C:
\[ \text{PCl}_3(g) + \text{Cl}_2(g) \rightleftharpoons \text{PCl}_5(g) \quad \Delta G^\circ = -92.50 \, \text{kJ} \]
**Question:**
What will happen to the ratio of partial pressure of \(\text{PCl}_5\) to partial pressure of \(\text{PCl}_3\) if the temperature is raised?
- **Decrease** (Selected and Correct)
- Increase
- No Change
**Explanation:**
Select all that apply:
- [ ] The reaction is exothermic.
- [ ] The equilibrium will shift to the left when the temperature is raised.
- [ ] The reaction is endothermic.
- [ ] The equilibrium will shift to the right when the temperature is raised.
- [ ] \(\Delta S\) is negative.
- [ ] Not enough information to determine if the reaction is endothermic or exothermic.
- [ ] Not enough information to determine the sign of \(\Delta H\).
- [ ] \(\Delta H\) is negative.
- [ ] Not enough information to determine the sign of \(\Delta S\).
- [ ] \(\Delta S\) is positive.
- [ ] \(\Delta H\) is positive.
### Analysis:
1. **Reaction Direction:**
- Given \(\Delta G^\circ = -92.50 \, \text{kJ}\), the reaction is spontaneous under standard conditions.
- Raising the temperature typically affects the position of equilibrium according to Le Chatelier's principle.
2. **Thermodynamic Considerations:**
- Since \(\Delta G^\circ\) is negative, the reaction is favorable in the forward direction.
- \(\Delta H\) is generally negative for exothermic reactions, suggesting heat is released.
3. **Equilibrium Shift:**
- For an exothermic reaction such as this, increasing the temperature would shift the equilibrium to the left, towards reactants (\(\text{PCl}_3\) and \(\text{Cl}_2\)), decreasing the ratio of \(\text{PCl}_5\).
This comprehensive approach aids in understanding equilibrium shifts in response to temperature changes and the](https://content.bartleby.com/qna-images/question/a50ae452-740d-4509-9969-24b1353a5a48/d49d1782-ac21-4730-9de5-bb17fb44494c/16883v8_processed.png)
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