Gas Equilibrium from Partial Pressures

Introduction

Gas-equilibrium calculations often provide total pressure and mole amounts rather than each species' partial pressure. The bridge is the mole fraction. Because reactions can change total gas moles, calculate the final total before turning an ICE table into a Kp expression.

Core explanation

For an ideal gas mixture, y i = n i/Σn j and p i = y iPtotal. If the mixture has 1 mol A and 3 mol B at total pressure 200 kPa, y A = 0.25 and y B = 0.75, so p A = 50 kPa and p B = 150 kPa. These are the pressures used in the equilibrium quotient, not the raw mole amounts.

For N₂O₄(g) ⇌ 2NO₂(g), suppose initially n₀ N₂O₄ and degree of dissociation α. Equilibrium n(N₂O₄) = n₀(1 − α), n(NO₂) = 2n₀α, and total n = n₀(1 + α). Therefore y(N₂O₄) = (1 − α)/(1 + α) and y(NO₂) = 2α/(1 + α). Multiplying each by total pressure gives p N2O4 and p NO2. The unnormalized pressure quotient is p NO2²/p N2O4, with standard-pressure normalization needed for a rigorous dimensionless K.

Substituting the mole fractions gives a useful ideal expression: p NO2²/p N2O4 = [4α²Ptotal]/(1 − α²) in a consistent pressure-number convention. The total pressure appears because the reaction has one more gas mole on the product side. A higher total pressure at fixed temperature generally suppresses dissociation for this reaction, consistent with favoring the side with fewer gas moles.

One must not set each p i equal to n iPtotal without dividing by total moles. That mistake can produce partial pressures summing far above Ptotal. A basic check is Σp i = Ptotal in an ideal mixture. Another is that each nonnegative partial pressure must lie between zero and total pressure.

If product gas is initially present or another inert gas is present, include all gases in the mole-fraction denominator. The inert species need not appear in the K expression, but it can affect mole fractions and partial pressures under a specified total-pressure constraint. Keep the physical setup explicit.

Step-by-step reasoning

1. Use stoichiometry or α to find equilibrium moles of every gas. 2. Sum all gas moles, including inert gases in the mixture. 3. Calculate y i and p i = y iPtotal. 4. Substitute reactive partial pressures into Kp and check Σp i.

Visual explanation

Draw a gas-mixture pie chart with slices proportional to equilibrium mole amounts. Label each slice's fraction, then multiply every fraction by the common total pressure.

Real-world analogy

A group bill is divided among participants according to their shares. A person's share of the bill is their fraction of the group times the total, just as partial pressure is mole fraction times total pressure.

Real-world example

An equilibrium vessel can be analyzed by measuring total pressure and gas composition. Those readings yield partial pressures used to calculate a pressure-based equilibrium constant.

Why?

Why recalculate total moles after dissociation? The reaction may change the number of gas particles, so initial and final mole-fraction denominators differ substantially in some cases.

Common misconception

“The product's mole amount is its mole fraction.” Mole fraction divides that amount by all gas moles, including reactants and any inert species.

Worked example

At equilibrium, a mixture contains 1.00 mol A and 2.00 mol B at Ptotal = 150 kPa for A(g) ⇌ 2B(g). Total is 3.00 mol, so p A = 50 kPa and p B = 100 kPa. The unnormalized pressure quotient is p B²/p A = 100²/50 = 200 kPa. With standard pressure p° = 100 kPa, dimensionless K is (p B/p°)²/(p A/p°) = 1²/0.5 = 2.

Quick check

1. What must ideal-gas partial pressures sum to? Answer: The total pressure of the gas mixture.

Exam focus

Calculate total equilibrium gas moles first. Use standard-pressure normalization if a dimensionless K is requested.

Advanced insight

For real gases, replace ideal p i/p° activities with fugacity ratios. Mole fractions still describe composition, but fugacity coefficients correct their relationship to chemical potential at high pressure.

Summary

Mole fractions turn equilibrium gas amounts into partial pressures, which enter Kp with stoichiometric powers. Recompute total moles after reaction and keep pressure conventions consistent.

Practice questions

1. A mixture has 2 mol A and 3 mol B at 500 kPa. Find p A ideally. Answer: (2/5)(500) = 200 kPa. 2. Does inert gas enter the mole-fraction denominator? Answer: Yes, if it is present in the gas mixture; it may not enter the reaction's K expression. 3. Why is p i = n iPtotal generally wrong? Answer: n i is an amount, not a fraction; divide by total gas moles first.