For the reaction 2SO2(g) + O2(g) ⇌ 2SO3(g), what happens to the equilibrium if the volume of the container is decreased? (2022)
Practice Questions
1 question
Q1
For the reaction 2SO2(g) + O2(g) ⇌ 2SO3(g), what happens to the equilibrium if the volume of the container is decreased? (2022)
Equilibrium shifts to the left
Equilibrium shifts to the right
No effect on equilibrium
Equilibrium constant changes
Decreasing the volume increases the pressure, and according to Le Chatelier's principle, the equilibrium will shift towards the side with fewer moles of gas, which is the right side (2 moles of SO3).
Questions & Step-by-step Solutions
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Q
Q: For the reaction 2SO2(g) + O2(g) ⇌ 2SO3(g), what happens to the equilibrium if the volume of the container is decreased? (2022)
Solution: Decreasing the volume increases the pressure, and according to Le Chatelier's principle, the equilibrium will shift towards the side with fewer moles of gas, which is the right side (2 moles of SO3).
Steps: 6
Step 1: Understand that the reaction involves gases: 2SO2, O2, and SO3.
Step 2: Recognize that decreasing the volume of the container increases the pressure of the gases inside.
Step 3: Recall Le Chatelier's principle, which states that if a system at equilibrium is disturbed, it will shift to counteract the disturbance.
Step 4: Identify the number of moles of gas on each side of the reaction: the left side has 3 moles (2SO2 + O2) and the right side has 2 moles (2SO3).
Step 5: Since the pressure increases when the volume decreases, the equilibrium will shift towards the side with fewer moles of gas to reduce the pressure.
Step 6: Conclude that the equilibrium will shift to the right side, producing more SO3.
