Chemical equilibrium · Lesson 4

The one change that moves K

Heating speeds up every reaction, yet for many equilibria it lowers the yield. Temperature is the only disturbance that changes K itself, and it forces industry into a compromise between rate and yield.

In 1909 Fritz Haber showed that nitrogen from the air could be combined with hydrogen to make ammonia, and within a few years Carl Bosch had scaled the process up to industrial size. Ammonia became fertiliser, and fertiliser fed a world whose population has since quadrupled. It has been estimated that about half of the nitrogen atoms in your body have passed through a Haber–Bosch reactor.

The chemistry is simple: N2(g) + 3H2(g) ⇌ 2NH3(g), ΔH = −92 kJ mol−1. The engineering is a puzzle. The reaction is exothermic, so the cooler the reactor, the more ammonia there is at equilibrium. But nitrogen’s triple bond is very strong, and at low temperatures the reaction is so slow that equilibrium would take years to reach. Heat the reactor and the reaction speeds up, but the yield falls.

Industry settles on a compromise of about 400–450 °C with an iron catalyst. This lesson explains why temperature is the one change that alters the equilibrium constant itself, and why it pulls rate and yield in opposite directions for an exothermic reaction.

Temperature is different

Changing a concentration, a volume or adding a catalyst never changes the value of K. Changing the temperature does. The rule depends on the sign of ΔH:

  • If the forward reaction is endothermic (ΔH positive), raising the temperature increases K: the equilibrium shifts to the right.
  • If the forward reaction is exothermic (ΔH negative), raising the temperature decreases K: the equilibrium shifts to the left.

Le Chatelier’s principle gives the same prediction if you treat heat as a reactant or product. Adding heat is partially counteracted by the direction that absorbs heat, the endothermic direction. Cooling favours the exothermic direction, which releases heat.

EquilibriumΔH of forward reactionWhen heated
N2O4(g) ⇌ 2NO2(g)+57 kJ mol−1K increases; darker brown
Co(H2O)62+(aq) + 4Cl−(aq) ⇌ CoCl42−(aq) + 6H2O(l)positiveK increases; bluer
Fe3+(aq) + SCN−(aq) ⇌ FeSCN2+(aq)negativeK decreases; paler red
N2(g) + 3H2(g) ⇌ 2NH3(g)−92 kJ mol−1K decreases; less ammonia

Why the endothermic direction wins when you heat

Look at any energy profile. The endothermic direction always has the larger activation energy, by exactly the size of ΔH, because it starts from the lower-energy side of the same peak.

In Lesson 1 you saw that the higher the activation energy, the more sensitive a rate is to temperature. So when an equilibrium mixture is heated, both rates increase, because a greater fraction of collisions have enough energy in both directions. But the endothermic rate increases by a greater factor. For a while the endothermic direction is faster, so there is a net reaction in that direction until the rates are equal again.

Heating shifts the energy distribution to the right. The fraction of collisions beyond the higher barrier (endothermic direction) increases by a much larger factor than the fraction beyond the lower barrier.
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Energy landscape

Each ball is a particle in the reactant or product well. Raise the temperature and compare how the forward and reverse hop counts change. Then switch the forward reaction to exothermic.

What the graphs look like

A temperature change is the only disturbance in which both rates jump by different factors while no concentration jumps. The concentrations simply start to drift towards their new equilibrium values.

Heating N2O4(g) ⇌ 2NO2(g): both rates jump, the endothermic forward rate by more. They converge at a new, higher rate.
The concentrations do not jump. NO2 rises and N2O4 falls gradually to the new equilibrium.
Heating the exothermic Haber equilibrium: this time the reverse (endothermic) rate jumps further, so the system shifts left.
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Heating and cooling an equilibrium

Heat or cool the N2O4/NO2 mixture and watch both rates on the chart. Predict first, then check.

Rate versus yield: the industrial compromise

For the Haber process the conditions that give a high yield and those that give a high rate are not the same:

ConditionEffect on yieldEffect on rateIndustrial choice
Temperaturelow temperature favours NH3 (exothermic)high temperature is fasterabout 400–450 °C, a compromise
Pressurehigh pressure favours NH3 (4 mol gas → 2 mol)high pressure is fasterabout 200 atm, limited by the cost and safety of equipment
Catalystno effectmuch fasteriron catalyst, which makes a lower temperature workable
Removing NH3increases conversion–ammonia is liquefied and removed; unreacted gas is recycled

The catalyst does not improve the equilibrium yield at all, but it matters enormously: because it makes the reaction fast at a lower temperature, the plant can run cooler than it otherwise could, and a cooler reactor means a higher yield.

Every disturbance in one table

ChangeRates at that instantShiftK
Add a reactantforward jumpsforwardunchanged
Add a productreverse jumpsreverseunchanged
Decrease the volume (gases)both jump; the side with more gas particles jumps moretowards fewer gas particlesunchanged
Dilute a solutionboth fall; the side with more dissolved particles falls moretowards more dissolved particlesunchanged
Catalystboth jump equallynoneunchanged
Inert gas at constant volumeno changenoneunchanged
Raise the temperatureboth jump; the endothermic direction jumps moreendothermic directionchanges

A full-mark temperature explanation

  1. Increasing the temperature increases the average kinetic energy, so a greater fraction of collisions have at least Ea: both the forward and reverse rates increase.
  2. The endothermic direction has the larger activation energy, so its rate increases by a greater factor.
  3. So there is a net reaction in the endothermic direction until the rates are equal again. State the effect on the amounts (and colour) and on K.
Now put it to workFive worked examples, twenty practice questions and ten harder problems on this lesson.
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