Kp: equilibria in the gas phase

Partial pressure

Air at sea level presses at about 101 kPa. Seventy-eight molecules in every hundred are nitrogen, twenty-one are oxygen, and the last one is mostly argon. Each gas behaves as if the others were not there: the nitrogen alone accounts for 78% of the pressure, about 79 kPa, and the oxygen for about 21 kPa. On the summit of Everest the total is about 34 kPa, so a climber breathes oxygen at about 7 kPa, a third of what the lungs are used to.

The fraction of the molecules that belong to one gas is its mole fraction. Pressure comes from molecules hitting the walls, and at a given temperature each molecule contributes the same on average whatever it is, so the partial pressure of a gas is its mole fraction multiplied by the total pressure.

x(A) = amount of A ÷ total amount of gasMole fractions have no units and add up to 1.

p(A) = x(A) × total pressurePartial pressures add up to the total pressure.

A mixture of 3.00 mol of nitrogen and 1.00 mol of hydrogen at 200 kPa has x(N₂) = 3.00 ÷ 4.00 = 0.750, so p(N₂) = 0.750 × 200 = 150 kPa, and p(H₂) = 0.250 × 200 = 50 kPa. The two add up to 200 kPa, as they must.

Describe the drawing

Two stacked pressure bars on a kilopascal axis. The first, for 3.00 mol of nitrogen and 1.00 mol of hydrogen, has a dark cyan nitrogen band from 0 to 150 kPa labelled mole fraction 0.750 and an amber hydrogen band from 150 to 200 kPa labelled mole fraction 0.250, with a bracket beside it reading total pressure 200 kPa equals 150 plus 50. The second, for air at 101 kPa, stacks nitrogen at 79 kPa, oxygen at 21 kPa and a thin sliver of other gases at 1 kPa.

Partial pressure can stand in for concentration. From pV = nRT, the partial pressure of one gas is p = (n ÷ V)RT, so at a fixed temperature it is proportional to that gas's concentration. Pressures are what a chemical engineer measures, and they give an equilibrium constant suited to gases.

Mole fraction

The amount in moles of one gas in a mixture divided by the total amount in moles of all the gases.

Partial pressure

The pressure that one gas in a mixture would exert if it alone occupied the whole volume; its mole fraction multiplied by the total pressure.

Total pressure

The pressure of a gas mixture, equal to the sum of the partial pressures of all the gases in it.

Writing Kp

Kp is built from the equation exactly as Kc is, with the partial pressure of each gas at equilibrium in place of its concentration. For the Haber equilibrium:

Kp = p(NH₃)² ÷ (p(N₂) × p(H₂)³)For N₂(g) + 3H₂(g) ⇌ 2NH₃(g); each p is a partial pressure at equilibrium.

The units come from the expression, as they do for Kc. With pressures in kPa, the Haber Kp has units of kPa² ÷ kPa⁴ = kPa⁻². For the dissociation of N₂O₄ into NO₂ they are kPa² ÷ kPa = kPa, and for the formation of hydrogen iodide they cancel.

Questions give pressures in kPa or Pa, and the unit of Kp follows whichever is used: the Haber Kp with pressures in Pa has units of Pa⁻², and its value is a million times smaller than in kPa⁻², because each of the two surplus pressures on the bottom is a thousand times larger. Work the units out from the expression every time rather than carrying them over from Kc.

Every species in these equilibria is a gas. Kp is written only for the gases in an equilibrium, and the expression uses p( ) for a partial pressure, never square brackets, which mean a concentration.

Kp

The equilibrium constant for a reaction between gases, written with the partial pressure of each gas at equilibrium in place of its concentration.

Kp from amounts and total pressure

A typical question gives the amounts mixed, one amount at equilibrium and the total pressure. The route has five steps: an ICE table in moles, the total amount at equilibrium, each mole fraction, each partial pressure, and then Kp. The total amount usually changes as the reaction proceeds, because the number of moles of gas changes, so it is worked out from the equilibrium row, not the initial one.

Describe the drawing

An extended ICE table for dinitrogen tetroxide forming nitrogen dioxide at 298 K and 100 kPa, with columns for N₂O₄, NO₂ and the total. Initial moles 0.0500, 0 and 0.0500; change minus 0.0095 and plus 0.0190; equilibrium 0.0405, 0.0190 and a total of 0.0595, which is boxed with the note divide by this, not by 0.0500. Two extra shaded rows give mole fractions 0.6807 and 0.3193 and partial pressures 68.07 and 31.93 kPa, and beneath the table Kp is worked out as 31.93 squared over 68.07, which is 15.0 kPa.

Dividing by the initial amount, 0.0500 mol, instead of the equilibrium total gives mole fractions that add up to more than 1 and a Kp that is wrong. The two checks in the working, mole fractions adding up to 1 and partial pressures adding up to the total, catch it.

The same vessel at a different total pressure would settle with different mole fractions and give the same 15.0 kPa: Kp has not changed; the mole fractions have.

Set out in rows, part (a) is the mole-fraction route in full. The first three rows are the ICE table of Kc. The total column is what Kp adds: it grows by 0.0690 mol because one mole of gas becomes two, and every mole fraction is taken from its equilibrium value, 0.1690. The last two rows each carry a check: mole fractions add up to 1, and partial pressures to the total pressure.

The expression α²P ÷ (1 − α²) holds whenever one mole of gas splits into two different gases. When it splits into two molecules of the same gas, as N₂O₄ does, the amount of NO₂ is 2α, the top of the expression carries (2α)², and Kp = 4α²P ÷ (1 − α²). Neither is worth learning by heart: building it from the ICE table each time takes four lines and avoids mixing the two up.

Some questions skip the first steps and give the partial pressures at equilibrium directly, or give the mole fractions and the total pressure. The calculation then starts part-way down the same route. What does not change is that every number substituted into Kp is a partial pressure at equilibrium, never an amount in moles and never a pressure from before the reaction.

Degree of dissociation

The fraction of the molecules of a substance that have dissociated when equilibrium is reached, written α.

What changes Kp

Kp, like Kc, depends only on temperature. The dinitrogen tetroxide example shows what a change of pressure does instead. Double the total pressure to 200 kPa by halving the volume, and at that instant each partial pressure doubles: p(N₂O₄) = 136.1 kPa and p(NO₂) = 63.9 kPa. Put into the Kp expression, they give 63.9² ÷ 136.1 = 30.0 kPa, twice Kp.

That value is the reaction quotient in partial pressures, Qp. It is greater than Kp, so NO₂ molecules pair up faster than N₂O₄ molecules split, and the position shifts left until the ratio is back to 15.0 kPa. The shift removes molecules, so with the total pressure held at 200 kPa, by pushing the plunger a little further in, the new mixture has p(N₂O₄) = 152.2 kPa and p(NO₂) = 47.8 kPa, and 47.8² ÷ 152.2 = 15.0 kPa. Left at the halved volume, the total would settle near 191 kPa instead, with NO₂ still a smaller share.

Describe the drawing

Three stacked bars of partial pressure on a shared kilopascal axis. The first, equilibrium at 100 kPa, has 68.1 kPa of dinitrogen tetroxide under 31.9 kPa of amber nitrogen dioxide, Qp 15.0. The second, the instant the volume is halved, is twice as tall with 136.1 and 63.9 kPa and Qp 30.0, marked Qp greater than Kp. The third, the new equilibrium at 200 kPa after the shift towards dinitrogen tetroxide, has 152.2 and 47.8 kPa and Qp back at 15.0; nitrogen dioxide's share falls from 32 to 24 percent.

Mole fractions show the same thing algebraically. Each partial pressure is x × P, so Kp = (x(NO₂)P)² ÷ (x(N₂O₄)P) = (x(NO₂)² ÷ x(N₂O₄)) × P. At a fixed temperature Kp is fixed, so if P doubles, the ratio of mole fractions must halve, and that can only happen by NO₂ turning into N₂O₄. For the Haber equilibrium the same working gives the mole-fraction ratio multiplied by P⁻², so a higher pressure must raise the share of ammonia. The power of P is the moles of gas on the right minus those on the left.

The squared term is the reason. Doubling every pressure multiplies the top of this expression by four and the bottom by two. In any equilibrium with more moles of gas on one side than the other, the powers do not balance, so a change of total pressure pushes the ratio away from Kp and the position moves until the ratio is restored. With equal moles of gas on each side, the factors cancel and the position does not move. Le Chatelier's principle for pressure follows from this.

A catalyst speeds up both directions equally and changes neither Kp nor the position.

Temperature changes Kp itself. The dissociation of N₂O₄ is endothermic, ΔH = +57 kJ mol⁻¹, so Kp rises with temperature, to roughly thirty times its 298 K value by 350 K. A sealed tube of the gas turns a deeper brown as it is warmed, because the position moves to match the new, larger Kp. For an exothermic reaction such as the Haber process, Kp falls as the temperature rises.

Reaction quotient, Qp

The Kp expression evaluated with the partial pressures at any moment; it equals Kp only when the mixture is at equilibrium.

Other boards: solids in Kp, and pressures in atmospheres

OCR and Edexcel also set Kp for heterogeneous equilibria, where a solid takes part, and Edexcel's specification, 11.1, states Kp with partial pressures in atmospheres; Cambridge questions often use atmospheres too, and OCR questions sometimes do. One atmosphere is 101.325 kPa, taken as 101 kPa in most questions, and the method does not change.

A solid is left out of Kp. For limestone in a sealed container, Kp = p(CO₂), so at a given temperature the carbon dioxide always settles at the same pressure, whatever the amount of solid. When the gas side holds two gases, the solid is left out twice over: out of the Kp expression, and out of the total amount of gas used for the mole fractions. A solid exerts no pressure on the walls of the vessel, so it has no partial pressure. The blast furnace supplies the standard case, where carbon dioxide rising through hot coke is turned into the carbon monoxide that reduces the iron ore.

C(s) + CO₂(g) ⇌ 2CO(g)ΔH = +173 kJ mol⁻¹. Kp = p(CO)² ÷ p(CO₂); the carbon does not appear.

Raising the pressure shifts this equilibrium to the left, towards one mole of gas from two; the carbon counts for nothing in that comparison. The forward reaction is endothermic, so Kp rises steeply with temperature. In the hottest part of a blast furnace, near atmospheric pressure and far above 1025 K, almost all the carbon dioxide that meets the coke becomes carbon monoxide.

Converting Kp between units needs the power. Kp for the Haber equilibrium at 500 °C is 1.45 × 10⁻⁵ atm⁻². The expression has two more pressures on the bottom than on the top, so the conversion factor is squared: Kp = 1.45 × 10⁻⁵ ÷ 101² = 1.42 × 10⁻⁹ kPa⁻². Dividing by 101 once gives a value about a hundred times too large.

Atmosphere, atm

A unit of pressure equal to 101.325 kPa, the mean pressure of the air at sea level.

Heterogeneous equilibrium

An equilibrium with species in more than one phase; only the gases appear in Kp, and a solid is left out of both the expression and the total amount of gas.