For the reaction 2N2O(g) ⇌ O2(g) + 2N2(g), what happens to the equilibrium position if the volume decreases?
A. does nothing
B. shifts to the left
C. doubles
D. shifts to the right
I don't know how volume can decrease with pressure, temp constant. But assuming it can, then the reaction will "undo" the change, shifting to right.
When the volume of a reaction container decreases, the equilibrium position of a reaction may shift to restore equilibrium. According to Le Chatelier's principle, if the volume decreases, the system will try to decrease the total pressure by shifting in the direction that has fewer gas molecules.
In this reaction, there are a total of 3 gas molecules on the reactant side (2 molecules of N2O and 1 molecule of O2) and 2 gas molecules on the product side (2 molecules of N2). Therefore, the reaction has more gas molecules on the reactant side.
When the volume decreases, the system will shift to the side with fewer gas molecules in order to decrease the total pressure. Therefore, the equilibrium position will shift to the right, resulting in an increase in the concentration of products (O2 and N2) and a decrease in the concentration of reactants (N2O).
Based on this, the correct answer is D. The equilibrium position shifts to the right.
To determine what happens to the equilibrium position when the volume decreases, we need to consider Le Chatelier's principle. According to Le Chatelier's principle, when a system at equilibrium is subjected to a change, it will react in a way that minimizes the effect of the change and restore equilibrium.
In this case, when the volume is decreased, it means the reaction container becomes smaller. This change is a "concentration effect" since the reaction involves gases. When the volume decreases, the concentration of the gases will increase.
Now let's consider the balanced equation:
2N2O(g) ⇌ O2(g) + 2N2(g)
According to Le Chatelier's principle, the equilibrium position will shift in a direction that reduces the concentration effect. In other words, the system will react in a way that reduces the concentrations of the gases.
Since the volume decrease will increase the concentration of the gases, the system will react by shifting the equilibrium to the side with fewer moles of gas molecules. In this case, the right side of the equation shows fewer moles of gas since O2(g) and 2N2(g) are combined. Therefore, the equilibrium position will shift to the right.
Therefore, the correct answer is D. The equilibrium position shifts to the right.