Pressure Equilibrium Constant Kp

Introduction

Gas equilibria are often measured through pressures rather than concentrations. Kp uses the partial pressures of gaseous species in the balanced reaction. Pure solids and liquids do not enter its simple expression, and total pressure must be separated into species-specific partial pressures before substitution.

Core explanation

For aA(g) + bB(g) ⇌ cC(g), a common ideal-gas expression is Kp = p C^c/(p A^a p B^b) when numerical partial pressures are used in a stated standard-pressure convention. The rigorous thermodynamic form uses dimensionless pressure activities p i/p°. The powers again follow balanced coefficients. A reaction's Kp value depends on temperature and on how the equation is written.

An ideal gas mixture has p i = y iPtotal, where y i is the species' gas mole fraction. If a sealed mixture contains 2 mol A and 1 mol B at total pressure 300 kPa, y A = 2/3 and y B = 1/3, giving p A = 200 kPa and p B = 100 kPa. These partial pressures, not 300 kPa for each gas, belong in a Kp expression.

For N₂ + 3H₂ ⇌ 2NH₃, Kp has p NH3² in the numerator and p N2 p H2³ in the denominator. Because the gas mole counts differ across sides, changing total pressure can affect composition at fixed temperature. Yet Kp itself remains constant at that temperature for the given standard-state convention; pressure changes Qp and the mixture shifts until Qp again equals Kp.

Units and reference pressure require care. Using kPa numbers rather than bar numbers in an unnormalized expression can change a numerical value if the net pressure exponent is nonzero. Activity factors p i/p° avoid this ambiguity and make the thermodynamic K dimensionless. In introductory work, follow the supplied convention and do not combine constants built from incompatible pressure units.

Kp does not include a non-gaseous pure solid even if the solid is essential to the reaction. For CaCO₃(s) ⇌ CaO(s) + CO₂(g), the simple equilibrium expression is the relative CO₂ pressure, while the two pure solid activities are one while both phases are present. A change in the quantity of solid alone does not alter equilibrium pressure under those assumptions.

Step-by-step reasoning

1. Balance the reaction and mark gas, solid and liquid phases. 2. Write partial-pressure factors for gases with coefficient powers. 3. Convert gas mole fractions to partial pressures if needed. 4. Use one consistent standard-pressure or textbook numerical convention.

Visual explanation

Draw a total-pressure bar divided according to gas mole fractions. Move each segment's pressure into the appropriate place in a Kp fraction.

Real-world analogy

A team's total budget is divided among departments; one department's share is not the entire budget. Partial pressures likewise divide total ideal-gas pressure among species.

Real-world example

Gas-phase reactor measurements often include total pressure and composition. The species mole fractions convert those readings into partial pressures for a Kp calculation at the measured temperature.

Why?

Why use partial pressure rather than total pressure for each gas? Each species contributes only its own fraction of ideal-gas pressure, and equilibrium depends on each species' chemical potential.

Common misconception

“All gases in a mixture have the same pressure equal to Ptotal.” They share the same container, but each has a partial pressure y iPtotal under the ideal-mixture model.

Worked example

For A(g) ⇌ 2B(g), let equilibrium y A = 0.40 and y B = 0.60 at Ptotal = 100 kPa. Then p A = 40 kPa and p B = 60 kPa. The unnormalized pressure quotient is p B²/p A = 60²/40 = 90 kPa. A dimensionless thermodynamic expression instead uses each p divided by a chosen p°, so state the pressure convention when reporting Kp numerically.

Quick check

1. What is p X for y X = 0.25 at total ideal-gas pressure 200 kPa? Answer: 0.25(200) = 50 kPa.

Exam focus

Use partial pressures and balanced exponents. Keep pressure units or standard-state normalization consistent and omit pure solids from simple Kp expressions.

Advanced insight

At high pressure, real-gas fugacities replace ideal partial-pressure activities. The thermodynamic equilibrium constant still uses activity-like dimensionless factors, while fugacity coefficients account for nonideal gas interactions.

Summary

Kp expresses gas equilibrium through species partial pressures raised to stoichiometric powers. Mole fractions give those pressures in an ideal mixture, and standard-state conventions keep the rigorous constant dimensionless.

Practice questions

1. Write Kp for H₂(g) + I₂(g) ⇌ 2HI(g). Answer: p HI²/(p H2 p I2) in a consistent pressure convention. 2. Does pure CaCO₃(s) appear in Kp for its decomposition? Answer: No. Its pure-solid activity is treated as one while present. 3. If y A = 0.2 at total 500 kPa, what is p A? Answer: 100 kPa under ideal-mixture behavior.