In the reaction 2SO2(g) + O2(g) ⇌ 2SO3(g), what happens to the equilibrium if the volume of the container is decreased?
Practice Questions
1 question
Q1
In the reaction 2SO2(g) + O2(g) ⇌ 2SO3(g), what happens to the equilibrium if the volume of the container is decreased?
Shift to the right
Shift to the left
No change
Increase the rate of reaction
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
1 item
Q
Q: In the reaction 2SO2(g) + O2(g) ⇌ 2SO3(g), what happens to the equilibrium if the volume of the container is decreased?
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: 8
Step 1: Understand that the reaction involves gases: 2SO2, O2, and 2SO3.
Step 2: Identify the total number of gas moles on each side of the reaction.
Step 3: Count the moles on the left side: 2 moles of SO2 + 1 mole of O2 = 3 moles total.
Step 4: Count the moles on the right side: 2 moles of SO3 = 2 moles total.
Step 5: Recognize that decreasing the volume of the container increases the pressure of the gases inside.
Step 6: Apply Le Chatelier's principle, which states that if a system at equilibrium is disturbed, it will shift to counteract the disturbance.
Step 7: Since there are fewer moles of gas on the right side (2 moles of SO3) compared to the left side (3 moles), the equilibrium will shift to the right.
Step 8: Conclude that the equilibrium will shift towards the production of SO3 when the volume is decreased.
