For the reaction 2SO2(g) + O2(g) ⇌ 2SO3(g), if the volume of the container is decreased, what will happen to the equilibrium? (2021)
Practice Questions
1 question
Q1
For the reaction 2SO2(g) + O2(g) ⇌ 2SO3(g), if the volume of the container is decreased, what will happen to the equilibrium? (2021)
Shift to the right
Shift to the left
No change
Depends on the temperature
Decreasing the volume increases the pressure, and 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), if the volume of the container is decreased, what will happen to the equilibrium? (2021)
Solution: Decreasing the volume increases the pressure, and the equilibrium will shift towards the side with fewer moles of gas, which is the right side (2 moles of SO3).
Steps: 7
Step 1: Understand that the reaction involves gases: 2SO2, O2, and SO3.
Step 2: Note that the left side of the reaction (2SO2 + O2) has 3 moles of gas (2 moles of SO2 + 1 mole of O2).
Step 3: The right side of the reaction (2SO3) has 2 moles of gas.
Step 4: When the volume of the container is decreased, the pressure inside the container increases.
Step 5: According to Le Chatelier's principle, the equilibrium will shift to reduce the pressure.
Step 6: The equilibrium will shift towards the side with fewer moles of gas to decrease the pressure.
Step 7: Since the right side has fewer moles (2 moles of SO3), the equilibrium will shift to the right.
