What is the effect of decreasing the volume of the container on the equilibrium of the reaction 2N2O(g) ⇌ 2N2(g) + O2(g)?
Practice Questions
1 question
Q1
What is the effect of decreasing the volume of the container on the equilibrium of the reaction 2N2O(g) ⇌ 2N2(g) + O2(g)?
Shift to the left
Shift to the right
No change
Increase the rate of reaction
Decreasing the volume increases the pressure, shifting the equilibrium towards the side with fewer moles of gas. The left side has 2 moles (N2O) compared to 3 moles (N2 + O2), so it shifts to the left.
Questions & Step-by-step Solutions
1 item
Q
Q: What is the effect of decreasing the volume of the container on the equilibrium of the reaction 2N2O(g) ⇌ 2N2(g) + O2(g)?
Solution: Decreasing the volume increases the pressure, shifting the equilibrium towards the side with fewer moles of gas. The left side has 2 moles (N2O) compared to 3 moles (N2 + O2), so it shifts to the left.
Steps: 6
Step 1: Understand that the reaction involves gases: 2N2O(g) ⇌ 2N2(g) + O2(g).
Step 2: Identify the number of gas moles on each side of the reaction.
Step 3: Count the moles: The left side (2N2O) has 2 moles, and the right side (2N2 + O2) has 3 moles.
Step 4: Know that decreasing the volume of the container increases the pressure of the gases inside.
Step 5: Remember that according to Le Chatelier's principle, the equilibrium will shift to reduce the pressure.
Step 6: Since the left side has fewer moles (2 moles) compared to the right side (3 moles), the equilibrium will shift to the left.
