Which of the following shows little or no net reaction when the volume of the...
Which of the following shows little or no net reaction when the volume of the system is decreased?
- A)\(2O_{3(g)} \Leftrightarrow 3O_{2(g)}\)
- B)\(2NO_{2(g)} \Leftrightarrow N_2O_{4(g)}\)
- C)\(H_2 + I_{2(g)} \Leftrightarrow 2HI_{(g)}\)
- D)\(PCl_{5(g)} \Leftrightarrow PCl_{3(g)} + Cl_{2(g)}\)
Correct Answer: C)\(H_2 + I_{2(g)} \Leftrightarrow 2HI_{(g)}\)
Explanation
When the volume of a system is decreased, the pressure increases. According to Le Chatelier's principle, the system will respond by shifting the equilibrium in the direction that reduces the pressure. This usually means the reaction will shift towards the side with fewer moles of gas.
Let's examine each given reaction:
Option A: \(2O_{3(g)} \Leftrightarrow 3O_{2(g)}\)
In this reaction, there are 2 moles of gas on the left side and 3 moles of gas on the right side. If the volume is decreased, the reaction will shift to the left to reduce the pressure by having fewer gas moles.
Option B: \(2NO_{2(g)} \Leftrightarrow N_2O_{4(g)}\)
In this reaction, there are 2 moles of gas on the left side and 1 mole of gas on the right side. If the volume is decreased, the reaction will shift to the right to reduce the pressure by having fewer gas moles.
Option C: \(H_2 + I_{2(g)} \Leftrightarrow 2HI_{(g)}\)
In this reaction, there are 2 moles of gas on both sides. If the volume is decreased, the pressure increases, but there is no change in the number of moles of gas, so the equilibrium will not shift significantly. This is the correct answer.
Option D: \(PCl_{5(g)} \Leftrightarrow PCl_{3(g)} + Cl_{2(g)}\)
In this reaction, there is 1 mole of gas on the left side and 2 moles of gas on the right side. If the volume is decreased, the reaction will shift to the left to reduce the pressure by having fewer gas moles.
Therefore, Option C is the correct answer, as it shows little or no net reaction when the volume of the system is decreased.