Pressure and Volume Changes

Introduction

Compressing a gaseous equilibrium mixture changes the partial pressures of all gas species before reaction readjustment. A common rule says compression favors the side with fewer gas moles, but that rule has conditions. The reaction quotient provides the precise reason and reveals when no shift occurs.

Core explanation

At fixed temperature in an ideal gas mixture, reducing volume by a factor of two initially doubles every species' concentration and partial pressure before reaction proceeds. If Qp contains product pressure powers totaling νproducts and reactant powers totaling νreactants, uniform scaling by factor f changes Qp by f^Δn, with Δn = νproducts − νreactants. Kp remains fixed because temperature has not changed.

For N₂ + 3H₂ ⇌ 2NH₃, Δn = 2 − 4 = −2. Compression f > 1 lowers Qp relative to Kp, so net forward reaction favors the side with fewer gas moles, ammonia. Expansion f < 1 raises Qp and favors reactants. The shift does not make the final pressure necessarily lower than its precompression value; it partly counters the imposed pressure increase.

For H₂ + I₂ ⇌ 2HI, Δn = 2 − 2 = 0. Uniform compression multiplies numerator and denominator pressure products by the same factor squared, so Qp is unchanged and ideal equilibrium composition does not shift. The pressure rises, but reaction direction is not driven by that uniform scaling.

Only gaseous coefficients count in this rule. Pure solids and liquids are omitted from Qp, and aqueous systems require their own activity analysis. Increasing pressure by adding an inert gas at fixed volume does not change reactive gas partial pressures in the ideal model, while adding inert gas at fixed total pressure expands the vessel and can act like dilution. The mechanical constraint must be specified.

Temperature should remain constant for the simple compression rule. A rapid compression may also heat a gas transiently; if temperature changes, K can change, requiring a separate analysis. Similarly nonideal gases at high pressure may not follow p i = n iRT/V exactly.

Step-by-step reasoning

1. Count gas coefficients on both sides to find Δn. 2. Specify constant temperature and how volume or pressure changes. 3. Determine whether reactive partial pressures scale uniformly. 4. Use Qnew/Qold = f^Δn to predict net direction.

Visual explanation

Draw a piston compressing a gas mixture. Put four reactant gas tokens on one side of the equation and two product gas tokens on the other, then mark the net shift toward fewer gas tokens.

Real-world analogy

Crowding a room may favor a process that reduces the number of separate occupied spaces, though the molecular explanation depends on pressure factors rather than conscious avoidance of crowding.

Real-world example

High pressure is used in ammonia synthesis partly because the balanced gas reaction has fewer product-side gas moles. Engineers also consider energy and equipment costs.

Why?

Why is there no ideal shift when Δn = 0? Uniform pressure scaling contributes equal total powers above and below Qp, so their factors cancel exactly in the quotient.

Common misconception

“Any pressure increase shifts every gas equilibrium.” A uniform pressure change leaves Q unchanged when gaseous mole counts are equal, and inert-gas effects depend on constraints.

Worked example

For 2SO₂(g) + O₂(g) ⇌ 2SO₃(g), Δn = 2 − 3 = −1. If a fixed-temperature compression doubles every reactive partial pressure instantly, Qp changes by 2⁻¹ = 1/2. A mixture previously at equilibrium now has Qp < Kp and moves net forward until a new equilibrium forms.

Quick check

1. Does compressing H₂ + I₂ ⇌ 2HI shift the ideal equilibrium composition at fixed temperature? Answer: No. Δn gas = 0, so uniform pressure scaling leaves Qp unchanged.

Exam focus

Count gas moles, not all formula coefficients. Distinguish compression from inert-gas addition and state constant-temperature assumptions.

Advanced insight

For real gases, fugacity coefficients may change with pressure, so even a Δn = 0 reaction can have nonideal composition effects. The simple no-shift result belongs to the ideal-gas approximation.

Summary

Uniform ideal-gas compression changes Q by a factor f^Δn while K stays fixed at constant temperature. It favors fewer gas moles when Δn is negative and gives no shift for Δn zero.

Practice questions

1. What side is favored by compression in N₂ + 3H₂ ⇌ 2NH₃? Answer: The ammonia product side, with fewer gas moles. 2. How does expansion affect a reaction with more gaseous products than reactants? Answer: It generally favors the product side under ideal, constant-temperature conditions. 3. Does total pressure alone specify the inert-gas effect? Answer: No. State whether volume or total pressure is held fixed.