In the reaction 2SO2(g) + O2(g) ⇌ 2SO3(g), what happens to the equilibrium when

In the reaction 2SO2(g) + O2(g) ⇌ 2SO3(g), what happens to the equilibrium when the volume of the container is decreased? (2023)

In the reaction 2SO2(g) + O2(g) ⇌ 2SO3(g), what happens to the equilibrium when the volume of the container is decreased? (2023)

Step 1: Understand the reaction: 2SO2(g) + O2(g) ⇌ 2SO3(g).
Step 2: Identify the 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.
Step 6: Apply Le Chatelier's principle: the system will shift to reduce the pressure.
Step 7: Determine which side has fewer moles of gas: the right side has 2 moles, while the left has 3 moles.
Step 8: Conclude that the equilibrium will shift to the right side (towards SO3) because it has fewer moles of gas.

No concepts available.