Equilibrium · Lesson 4 of 14

Gas Pressures and Heterogeneous Equilibria

“Choose the correct pressure or concentration expression and understand why pure phases are omitted.”

Learning Objectives

• Derive the relation between concentration and pressure forms for ideal gases. • Calculate the gas-mole change from a balanced equation. • Construct heterogeneous-equilibrium expressions using pure-phase activities. • Use consistent pressure units and distinguish a numerical convention from a dimensionless constant.

Partial pressure describes the contribution of one gas

A gas mixture exerts a total pressure, but each species contributes a partial pressure. For an ideal mixture, each contribution is the pressure that the species would exert alone in the same volume at the same temperature. This lets us describe the composition by pressures instead of concentrations, which is convenient when gases are measured with pressure instruments.

Pressure and concentrationLaTeX

p_{i} = \frac{n _{i}}{V} R T = c_{i} R T

Here p_i is the partial pressure of species i, n_i its amount, V the common gas volume, c_i its molar concentration, T the absolute temperature and R the gas constant.

For pressures in bar and concentrations in moles per litre, use $R = 0.0831 L bar mo l^{- 1} K^{- 1}$ at the precision used here. One bar is $1 0^{5}$ pascals. Do not combine a pressure in pascals with an R value written for bar and litre. In a mixture, use the individual partial pressures in Kp, not the total pressure repeated for every species.

Gas-pressure expressionLaTeX

K_{p} = \frac{p _{C}^{c} p _{D}^{d}}{p _{A}^{a} p _{B}^{b}}

This conventional notation suppresses the division of each gas pressure by its standard pressure. The phase and standard-state details are explained below.

Changing from concentrations to pressures

Replace every partial pressure in the expression by its concentration multiplied by RT. Each product term contributes one factor of RT for every unit of its coefficient; each reactant term contributes the corresponding denominator factor. The remaining overall exponent counts gaseous product moles minus gaseous reactant moles.

Conversion in the textbook numerical conventionLaTeX

K_{p} = K_{c} (R T)^{Δ n_{g}}, Δ n_{g} = \sum ν_{gas, products} - \sum ν_{gas, reactants}

Count only gaseous coefficients. This form uses consistent chosen concentration and pressure units; all species must behave approximately ideally.

For hydrogen plus iodine vapour forming two HI molecules, $Δ n_{g} = 2 - 2 = 0$ , so the numerical pressure and concentration forms coincide. For ammonia synthesis, $Δ n_{g} = 2 - 4 = - 2$ , so $K_{p} = K_{c} (R T)^{- 2}$ . Solids and liquids do not contribute to this gas-mole change.

Example — A concentration constant for phosphorus pentachloride

Problem
At 500 K, PCl5, PCl3 and Cl2 have equilibrium concentrations 1.41 M, 1.59 M and 1.59 M. Calculate Kc for $PC l_{5} (g) ⇌ PC l_{3} (g) + C l_{2} (g)$ .

1.Write $K_{c} = [PC l_{3}] [C l_{2}] / [PC l_{5}]$ .
2.Substitute $(1.59) (1.59) /1.41$ .
3.The numerical value is $1.79$ . For this reaction, the gas-mole change is $2 - 1 = 1$ , so Kp would equal this value multiplied by RT in the matching numerical convention.

Example — Converting a constant for NOCl

Problem
For $2NOCl (g) ⇌ 2NO (g) + C l_{2} (g)$ , $K_{c} = 3.75 \times 1 0^{- 6}$ at 1069 K. Find the numerical Kp with bar pressures.

1.The gas-mole change is $(2 + 1) - 2 = 1$ .
2.Use $K_{p} = K_{c} R T = (3.75 \times 1 0^{- 6}) (0.0831) (1069)$ .
3.This gives $K_{p} = 3.33 \times 1 0^{- 4}$ . Multiplying by RT must not turn this into 0.033; that would be a factor-of-100 arithmetic error.

Example — A water–gas shift mixture

Problem
At 800 K, Kc = 4.24 for CO(g) + H2O(g) forming CO2(g) + H2(g). Initially CO and steam each have concentration 0.100 M and the products are absent. Find the equilibrium concentrations.

1.Let x M of each product form. The two reactant concentrations become 0.100 − x, while each product concentration is x.
2.The equation is x squared divided by (0.100 − x) squared = 4.24. Both concentrations are nonnegative, so x / (0.100 − x) = the positive square root of 4.24, or 2.059.
3.Solving gives x = 0.06731 M. Therefore CO2 and H2 are each 0.06731 M; CO and H2O are each 0.03269 M. Their ratio checks Kc. A quadratic method gives the same acceptable root and a second root beyond the initial reactant concentration, which must be rejected.

Pure solids and liquids have fixed activities

A heterogeneous equilibrium involves more than one phase. For calcium carbonate decomposing into calcium oxide and carbon dioxide, two solid phases coexist with a gas phase. Adding more of a pure solid changes its amount but not its composition or activity. Its thermodynamic activity is taken as one in its standard pure phase, so it does not appear as a variable in the equilibrium expression.

Calcium carbonate decompositionLaTeX

CaC O_{3} (s) ⇌ CaO (s) + C O_{2} (g), K_{p} = \frac{p _{C O_{2}}}{p ^{\circ}}

Both solid phases must remain available for this equilibrium to fix the CO2 pressure. The standard pressure p° is 1 bar.

If the CO2 equilibrium pressure is 2.0 bar at a specified temperature, the dimensionless Kp is $2.0 bar /1 bar = 2.0$ . More solid does not increase this equilibrium pressure while both pure solid phases are present. If one phase is exhausted, the old coexistence condition cannot simply be imposed.

A heterogeneous equilibrium— Different amounts of the same pure solids leave the equilibrium CO2 pressure unchanged at fixed temperature, provided both solids remain.

For $Ni (s) + 4CO (g) ⇌ Ni (CO)_{4} (g)$ , nickel is omitted but the fourth power of the CO pressure remains. For water evaporating into a gas space, pure liquid water is omitted but its vapour is included. In a dilute aqueous ester-hydrolysis expression, water as the solvent can be treated as having nearly constant activity. In a concentrated liquid reaction mixture where water is a varying component, it cannot automatically be removed merely because its phase label is liquid.

Standard states make the thermodynamic constant dimensionless

A logarithm requires a dimensionless argument. The thermodynamic equilibrium constant is built from activities, each measured relative to a specified standard state. For an ideal gas, activity is its partial pressure divided by the standard pressure. For a dilute solute, concentration divided by a standard concentration is an approximation to activity. Concentrated ionic solutions require activity corrections.

Standard-state version of the conversionLaTeX

K_{p} = K_{c} (\frac{R T c ^{\circ}}{p ^{\circ}})^{Δ n_{g}}

Here c° = 1 mol per litre and p° = 1 bar. RTc°/p° is dimensionless. The familiar numerical conversion results when their stated units are used consistently.

The chapter’s temperature table illustrates that each reaction has its own strong temperature dependence. For example, the supplied ammonia-synthesis values decrease from $6.8 \times 1 0^{5}$ at 298 K to 41 at 400 K and $3.6 \times 1 0^{- 2}$ at 500 K. For the reverse NO2 dimerisation direction, dissociation of N2O4 becomes more favourable as temperature rises. Comparing temperatures is meaningful only when the written reaction and standard-state convention are kept the same.

Example — A gas–solid equilibrium calculation

Problem
At 1000 K, the numerical Kp is 3.0 for $C O_{2} (g) + C (s) ⇌ 2CO (g)$ . Initially CO2 has partial pressure 0.48 bar and CO is absent. Solid carbon remains present. Find equilibrium pressures at fixed volume.

1.Let the CO2 partial pressure decrease by x bar. Stoichiometry makes the CO partial pressure rise by 2x bar.
2.The equation is $(2 x)^{2} / (0.48 - x) = 3.0$ , giving $4 x^{2} + 3 x - 1.44 = 0$ .
3.The physically acceptable root is $x = (- 3 + 32.04) /8 = 0.33255$ . Thus $p_{CO} = 0.6651 bar$ and $p_{C O_{2}} = 0.14745 bar$ . The negative root is impossible.

Common mistake

Omitting a pure phase from K does not mean removing it from the vessel, and “all substances labelled liquid are omitted” is not a valid rule. Also specify the units behind a quoted numerical Kp: changing pressure units without standardising can change an unnormalised numerical ratio.

Quick check

For N2 + 3H2 forming 2NH3, what is the gas-mole change?

Quick check

Which species is omitted from K for CaCO3(s) forming CaO(s) + CO2(g)?

Quick check

When do numerical Kp and Kc coincide in matching conventional units?

Quick check

What must be substituted into a gas-pressure equilibrium expression?

Quick check

The thermodynamic equilibrium constant is dimensionless because

Quick check

At fixed temperature, adding more pure CaCO3 to a system already containing CaCO3, CaO and equilibrium CO2 generally

Practice Problems

Find the gas-mole change for PCl5(g) forming PCl3(g) + Cl2(g), and for 2SO2(g) + O2(g) forming 2SO3(g).
Convert Kc = 0.020 to Kp at 500 K for a reaction with gas-mole change +1, using R = 0.0831 in bar–litre units.
Write Kp for Ni(s) + 4CO(g) forming Ni(CO)4(g).
Explain why adding a pure solid does not necessarily change the pressure above a heterogeneous equilibrium.
A pure-liquid component and a component of a concentrated liquid mixture both carry the label (l). Explain why their activities need not be treated in the same way.
Write dimensionless Kp for a reaction with equilibrium ratio pB squared / pA, making the standard-pressure factors explicit.

Key Takeaways

Use partial pressures for gases and count only gas coefficients in the gas-mole change. Pure-phase activities are constant, but the phases must still be available. Consistent units and standard states connect numerical textbook expressions to dimensionless thermodynamic constants.

Previous · Lesson 3

The Equilibrium Law and Concentration Constants

Next · Lesson 5

Using K, Q, and Gibbs Energy

Partial pressure describes the contribution of one gas

Changing from concentrations to pressures

Pure solids and liquids have fixed activities

Standard states make the thermodynamic constant dimensionless

Quiz

Practice Problems

Key Takeaways