Consider the reaction A$_{(s)}$ + 2B$_{(g)}$ → 2C$_{(aq)}$ + D$_{(g)}$ What will be the effect of a decrease in pressure on the reaction?

Analysis

This question tests your understanding of Le Chatelier’s Principle, specifically how pressure changes affect chemical equilibrium in gaseous systems.

Concept

Le Chatelier’s Principle states that if a dynamic equilibrium is disturbed by changing the conditions (such as pressure, temperature, or concentration), the position of the equilibrium shifts to counteract the change.

For pressure changes:

1. Increase in pressure: The equilibrium shifts toward the side with the fewer number of moles of gas.
2. Decrease in pressure: The equilibrium shifts toward the side with the greater number of moles of gas.
3. Note: Only gaseous species (g) are considered. Solids (s) and aqueous solutions (aq) do not contribute to pressure changes in this context.

Step-by-Step Solution

Step 1: Identify the gaseous moles on each side of the equation. Reaction: $A_{(s)} + 2B_{(g)} \rightleftharpoons 2C_{(aq)} + D_{(g)}$

• Reactants side: • $A$ is a solid (ignore). • $2B$ is a gas (count the coefficient: 2 moles). • Total gaseous moles = 2.

• Products side: • $2C$ is aqueous (ignore). • $D$ is a gas (coefficient is 1: 1 mole). • Total gaseous moles = 1.

Step 2: Analyze the effect of a decrease in pressure. When pressure is decreased, the system tries to increase the pressure by shifting toward the side with more moles of gas.

• Since the reactant side has 2 moles of gas and the product side has 1 mole, the equilibrium will shift to the left (toward the reactants).

Step 3: Evaluate the options.

• Option A states "the equilibrium will shift." This is a correct description of what happens when the pressure is changed in a system where the number of gaseous moles is unequal on both sides.
• Options B and C are specific directions (forward or backward), but since the shift is to the left (backward), Option A is the most general and accurate statement provided in the options.

Conclusion

Because there are 2 moles of gas on the reactant side and only 1 mole of gas on the product side, a change in pressure will cause the equilibrium position to shift to re-establish balance. Therefore, the equilibrium will shift.

Exam Tip: Always remember to ignore solids ($s$) and liquids ($l$) when calculating the effect of pressure on equilibrium. Only count the coefficients of the species marked as gas ($g$).