Consider the following system at equilibrium where AH° = -198 kJ, and Kc = 34.5, at 1150 K: 2SO2(g) + O₂(g) ⇒ 2SO3(g) When 0.24 moles of O₂ (g) are added to the equilibrium system at constant temperature: The value of Ke O increases O decreases O remains the same The value of Qc O is greater than Kc O is equal to Kc O is less than Ke The reaction must O run in the forward direction to reestablish equilibrium Orun in the reverse direction to reestablish equilibrium O remain in the current position, since it is already at equilibrium The concentration of SO2 will O increase O decrease O remain the same
Consider the following system at equilibrium where AH° = -198 kJ, and Kc = 34.5, at 1150 K: 2SO2(g) + O₂(g) ⇒ 2SO3(g) When 0.24 moles of O₂ (g) are added to the equilibrium system at constant temperature: The value of Ke O increases O decreases O remains the same The value of Qc O is greater than Kc O is equal to Kc O is less than Ke The reaction must O run in the forward direction to reestablish equilibrium Orun in the reverse direction to reestablish equilibrium O remain in the current position, since it is already at equilibrium The concentration of SO2 will O increase O decrease O remain the same
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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![Consider the following system at equilibrium where \(\Delta H^\circ = -198 \text{ kJ}\), and \(K_c = 34.5\) at 1150 K:
\[2\text{SO}_2(g) + \text{O}_2(g) \rightleftharpoons 2\text{SO}_3(g)\]
When 0.24 moles of \(\text{O}_2(g)\) are added to the equilibrium system at constant temperature:
1. **The value of \(K_c\)**
- increases
- decreases
- remains the same
2. **The value of \(Q_c\)**
- is greater than \(K_c\)
- is equal to \(K_c\)
- is less than \(K_c\)
3. **The reaction must**
- run in the forward direction to reestablish equilibrium
- run in the reverse direction to reestablish equilibrium
- remain in the current position, since it is already at equilibrium
4. **The concentration of \(\text{SO}_2\) will**
- increase
- decrease
- remain the same
This analysis involves Le Chatelier's principle, which predicts how the system will adjust to a change, such as the addition of reactants.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F6e7279b6-864f-464c-a490-1592f397f56d%2Fa0461afb-5f22-4e98-a33d-bbc419d1f762%2Fqnvkf3u_processed.png&w=3840&q=75)
Transcribed Image Text:Consider the following system at equilibrium where \(\Delta H^\circ = -198 \text{ kJ}\), and \(K_c = 34.5\) at 1150 K:
\[2\text{SO}_2(g) + \text{O}_2(g) \rightleftharpoons 2\text{SO}_3(g)\]
When 0.24 moles of \(\text{O}_2(g)\) are added to the equilibrium system at constant temperature:
1. **The value of \(K_c\)**
- increases
- decreases
- remains the same
2. **The value of \(Q_c\)**
- is greater than \(K_c\)
- is equal to \(K_c\)
- is less than \(K_c\)
3. **The reaction must**
- run in the forward direction to reestablish equilibrium
- run in the reverse direction to reestablish equilibrium
- remain in the current position, since it is already at equilibrium
4. **The concentration of \(\text{SO}_2\) will**
- increase
- decrease
- remain the same
This analysis involves Le Chatelier's principle, which predicts how the system will adjust to a change, such as the addition of reactants.
Expert Solution
Step 1: Relation between reaction quotient and equilibrium constant!
Answer:
- When QC=KC, system is at equilibrium.
- When QC<KC, system is not a equilibrium and in order to reestablish the equilibrium, reaction will move in forward direction, so that value of QC can increase and become equal to KC.
- When QC>KC, system is not a equilibrium and in order to reestablish the equilibrium, reaction will move in reverse direction, so that value of QC can decrease and become equal to KC.
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