Chapter 18, Problem 68A

Interpretation Introduction

Interpretation: The reaction in which a change in pressure affects the equilibrium position is to be identified. The equilibrium position change for that reaction is to be explained.

Concept Introduction: When the pressure decreases, the equilibrium will move toward the side of the reaction that produces more moles of gas as a product. When the pressure increases, the equilibrium will move toward the side of the reaction that produces fewer moles of gas as a product.

Answer to Problem 68A

The change in pressure affects the equilibrium position of the reaction (c). The equilibrium position moves toward the left side when the pressure increases and the left side when the pressure decreases because it has fewer moles of gas on the left side.

Explanation of Solution

The given reactions are as follows:

1. ${\text{H}}_2\left(g\right)+{\text{F}}_2\left(g\right)\to 2\text{HF}\left(g\right)$
2. ${\text{SO}}_2\left(g\right)+{\text{NO}}_2\left(g\right)\to \text{NO}\left(g\right)+{\text{SO}}_3\left(g\right)$
3. ${\text{2H}}_2\text{O}\left(g\right)\to {\text{2H}}_2\left(g\right)+{\text{O}}_2\left(g\right)$

In reaction (a) $1$ mole of ${\text{H}}_2$ gas reacts with $1$ mole of ${\text{F}}_2$ gas will give $2$ moles of $\text{HF}$ gas. In this reaction, the equilibrium position does not affect by a change in pressure because the number of moles on both sides of the reaction is the same.

In reaction (b) $1$ mole of ${\text{SO}}_2$ gas reacts with $1$ mole of ${\text{NO}}_2$ gas will give $1$ mole of $\text{NO}$ gas and 1 mole of ${\text{SO}}_3$ gas. In this reaction, the equilibrium position does not affect by a change in pressure because the number of moles on both sides of the reaction is the same.

In reaction (c) $2$ moles of ${\text{H}}_2\text{O}$ gas will give $2$ mole of ${\text{H}}_2$ gas and $1$ mole of ${\text{O}}_2$ gas. In this reaction, the equilibrium position affects by a change in pressure because it has $2$ moles of gas on the left side and $3$ moles of gas on the right side. When the pressure increases, the equilibrium shifts toward the left side of the reaction. When the pressure decreases, the equilibrium shifts toward the right side of the reaction.

Conclusion

The change in pressure affects the equilibrium position of the reaction (c). The equilibrium position moves toward the left side when the pressure increases and the left side when the pressure decreases because it has fewer moles of gas on the left side.