Which of the following reactions would shift to the right to favor the product side upon increasing the pressure (through a reduction in the volume of the reaction vessel) ?
I. H₂ (g) + C₂ N₂(g) 2 HCN (g)
II. CO (g) + Br₂(g) CO Br₂(g)
III. 6 CO₂ (g) + 6 H₂ O ( ) C₆ H₁₂ O₆ (s) + 6 O₂(g)

Question

Which of the following reactions would shift to the right to favor the product side upon increasing the pressure (through a reduction in the volume of the reaction vessel) ?
I. H₂ (g) + C₂ N₂(g)

2 HCN (g)
II. CO (g) + Br₂(g) Which of the following reactions would shift to the right to favor the product side upon increasing the pressure (through a reduction in the volume of the reaction vessel) ? I. H₂ (g) + C₂ N₂(g) 2 HCN (g) II. CO (g) + Br₂(g) CO Br₂(g) III. 6 CO₂ (g) + 6 H₂ O ( ) C6 H12 O6 (s) + 6 O₂(g) A) I only B) II only C) III only D) I and III E) II and III

CO Br₂(g)
III. 6 CO₂ (g) + 6 H₂ O ( Which of the following reactions would shift to the right to favor the product side upon increasing the pressure (through a reduction in the volume of the reaction vessel) ? I. H₂ (g) + C₂ N₂(g) 2 HCN (g) II. CO (g) + Br₂(g) CO Br₂(g) III. 6 CO₂ (g) + 6 H₂ O ( ) C6 H12 O6 (s) + 6 O₂(g) A) I only B) II only C) III only D) I and III E) II and III

)

C₆ H₁₂ O₆ (s) + 6 O₂(g)

Quizplus · Accepted Answer

According to Le Chatelier's principle, when the pressure is increased, a reaction will shift towards the side with fewer gas molecules to decrease the pressure. In reaction II, there are only two gas molecules on the reactant side and one gas molecule on the product side. Therefore, the reaction will shift to the right to produce more product and decrease the pressure. In reactions I and III, the number of gas molecules is the same on both sides, so they will not be affected by the change in pressure.

Which of the Following Reactions Would Shift to the Right