CuO$_s$   +  H$_2$($_g$)  ⇌ Cu$_s$   +  H$_2$O($_g$) In the equation above, the effect of increased pressure on the equilibrium position is that

Analysis

The question asks how an increase in pressure affects the position of the chemical equilibrium for the reaction: CuO(s) + H₂(g) ⇌ Cu(s) + H₂O(g)

Concept: Le Chatelier’s Principle

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 equilibrium shifts to counteract the change.

When considering the effect of pressure on a gaseous equilibrium:

1. We only count the moles of gaseous substances. Solids (s) and liquids (l) are ignored because their volumes are negligible compared to gases.
2. If the number of moles of gas is the same on both sides of the equation, a change in pressure will have no effect on the equilibrium position.
3. If there are more moles of gas on one side, increasing the pressure will shift the equilibrium toward the side with fewer moles of gas to reduce the pressure.

Step-by-Step Solution

1. Identify the gaseous species: • Reactants: H₂(g) — there is 1 mole of gas. • Products: H₂O(g) — there is 1 mole of gas. • (Note: CuO and Cu are solids and are excluded from the pressure calculation).

2. Compare the moles of gas: • Total moles of gas on the left = 1 • Total moles of gas on the right = 1

3. Determine the effect: Since the number of moles of gas is equal on both sides (1 mole = 1 mole), changing the pressure does not favor either the forward or the reverse reaction. Therefore, the equilibrium position remains unchanged.

Conclusion

Exam Tip: Always look at the state symbols (s, l, g, aq) carefully! Many students make the mistake of counting the solid CuO or Cu in their pressure calculations. Remember: Pressure only affects systems where there is a change in the number of gaseous moles.