Appearance
10.4 — Reactions, the Mole and Equilibrium
Mix hydrogen and iodine in a sealed flask at 450 °C and hydrogen iodide forms. Wait long enough and the composition stops changing — but not because the reaction stopped. Both directions are still running, at exactly equal rates.
That distinction is the whole of this chapter, and getting it wrong is the most common error in chemistry.
The mole
Chemistry counts particles and weighs grams, and the mole is the conversion.
One mole is 6.02214076\times10^{23} entities, exact by definition since the 2019 SI redefinition (Chapter 1.1). Before that it was defined as the number of atoms in 12 g of carbon-12.
Why that number? So that one mole of a substance weighs its atomic or molecular mass in grams. Carbon's atomic mass is 12, so one mole of carbon weighs 12 g and contains 6.022\times10^{23} atoms.
Avogadro's number is not a physical constant in the way c or h are. It is a counting convention, chosen to make laboratory bookkeeping convenient.
How big is it? A mole of grains of sand would cover the entire surface of the Earth to a depth of about 5 metres. A mole of seconds is 1.9\times10^{16} years, over a million times the age of the universe.
And there are about 5000 moles of water molecules in your body, which is 3\times10^{27} molecules.
Working with moles
n = \frac{m}{M} = \frac{N}{N_A} = \frac{PV}{RT}\ \text{(for a gas)}
Worked example: how many atoms in a gold ring? A 5.0 g ring, M_{\text{Au}} = 197 g/mol.
n = \frac{5.0}{197} = 0.0254\ \text{mol}
N = 0.0254\times6.022\times10^{23} = 1.53\times10^{22}\ \text{atoms}
Fifteen thousand billion billion atoms in a wedding ring.
Worked example: molar volume of a gas. At standard temperature and pressure, 273.15 K and 100 kPa:
V = \frac{nRT}{P} = \frac{(1)(8.314)(273.15)}{100000} = 0.02271\ \text{m}^3 = 22.71\ \text{L}
Any gas, 22.7 litres per mole — a direct consequence of the ideal gas law of Chapter 3.2 and of Avogadro's insight that equal volumes contain equal numbers.
Stoichiometry
A balanced equation is a statement about moles, not about masses.
2\text{H}_2+\text{O}_2 \to 2\text{H}_2\text{O}
Two moles of hydrogen react with one of oxygen, not two grams with one gram.
Worked example: the limiting reagent. 10.0 g of hydrogen and 100.0 g of oxygen are ignited. How much water forms and what is left over?
n_{\text{H}_2} = \frac{10.0}{2.016} = 4.96\ \text{mol}, \qquad n_{\text{O}_2} = \frac{100.0}{32.00} = 3.125\ \text{mol}
Required ratio is 2:1. With 4.96 mol of H₂ you need 2.48 mol of O₂ and you have 3.125. So hydrogen runs out first — it is the limiting reagent.
n_{\text{H}_2\text{O}} = 4.96\ \text{mol} \quad\Longrightarrow\quad m = 4.96\times18.02 = 89.4\ \text{g}
Oxygen left over:
3.125-2.48 = 0.645\ \text{mol} = 20.6\ \text{g}
Check by mass conservation: 10.0+100.0 = 110.0 g in, and 89.4+20.6 = 110.0 g out. ✔
Percent yield compares what you actually get with the theoretical maximum:
\%\ \text{yield} = \frac{\text{actual}}{\text{theoretical}}\times100
Real reactions rarely reach 100 %, because of side reactions, losses in handling, and — most fundamentally — because many reactions reach equilibrium before completion.
Equilibrium
A reversible reaction runs both ways. Once the forward and reverse rates are equal, the concentrations stop changing.
\text{H}_2+\text{I}_2 \rightleftharpoons 2\text{HI}
This is dynamic, not static. Both reactions continue at full speed. Proof: run the reaction with deuterium and ordinary hydrogen and you find HD forming, which is only possible if bonds are being broken and remade continuously.
Two things are constantly confused and should not be:
Equilibrium is about the destination. How far the reaction goes.
Kinetics is about the journey. How fast it gets there.
A reaction can be enormously favourable and infinitely slow. Diamond turning to graphite is thermodynamically downhill and takes longer than the age of the universe. A mixture of hydrogen and oxygen sits stably in a flask for years and explodes instantly with a spark. Both are kinetics, not thermodynamics.
The equilibrium constant
For a general reaction:
a\text{A}+b\text{B} \rightleftharpoons c\text{C}+d\text{D}
\boxed{K = \frac{[\text{C}]^c[\text{D}]^d}{[\text{A}]^a[\text{B}]^b}}
Products over reactants, each raised to its coefficient.
Reading K:
| K | Meaning |
|---|---|
| >10^{3} | Essentially complete |
| 10^{-3} to 10^{3} | Substantial amounts of both |
| <10^{-3} | Barely proceeds |
Conventions that matter: pure solids and pure liquids are omitted, because their "concentration" does not change. Gases may be expressed as partial pressures, giving K_p instead of K_c, and the two are related by:
K_p = K_c(RT)^{\Delta n}
where \Delta n is the change in the number of gas moles.
Where K comes from
It is not an empirical fit. It follows from thermodynamics, and Chapter 10.6 derives the connection:
\boxed{\Delta G^\circ = -RT\ln K}
Read it: the equilibrium constant is fixed by the standard free energy change, which is fixed by the enthalpy and entropy changes. K is a thermodynamic quantity, not a kinetic one, which is why catalysts cannot change it.
Worked example. A reaction has \Delta G^\circ = -20.0 kJ/mol at 298 K.
\ln K = -\frac{\Delta G^\circ}{RT} = \frac{20000}{(8.314)(298)} = \frac{20000}{2477.6} = 8.072
K = e^{8.072} = 3.2\times10^{3}
Strongly product-favoured. And note the sensitivity: at \Delta G^\circ = -40 kJ/mol, K would be 10^{7}. A factor of two in energy gives four orders of magnitude in K, because the relationship is exponential.
The reaction quotient
Q has the same form as K but uses the concentrations at any moment, not just at equilibrium.
| Comparison | Direction |
|---|---|
| Q < K | Forward — too few products |
| Q = K | At equilibrium |
| Q > K | Reverse — too many products |
This is the practical tool. Measure what you have, compute Q, compare with K, and you know which way the system will move.
Le Chatelier's principle
A system at equilibrium, when disturbed, shifts to partially oppose the disturbance.
Note "partially". The system never fully undoes the change; it moves to a new equilibrium somewhere in between.
Concentration
Add a reactant and the reaction shifts forward. In Q-and-K terms: adding to the denominator makes Q < K, so the system moves forward until they match.
Remove a product and it also shifts forward. This is exploited constantly — distilling off a product as it forms drives a reaction to completion that would otherwise stop halfway.
Pressure
Only for gases, and only when the number of gas moles changes.
\text{N}_2(g)+3\text{H}_2(g) \rightleftharpoons 2\text{NH}_3(g)
Four moles of gas become two. Increasing the pressure shifts it towards ammonia, because fewer molecules occupy less volume and relieve the pressure.
A reaction with equal moles on both sides is unaffected by pressure.
And adding an inert gas at constant volume changes nothing, because it does not alter the partial pressures of the participants. This trips people up regularly.
Temperature
Temperature is different from the others, and this is the point most often missed: it is the only disturbance that changes K itself.
Treat heat as a reactant or product:
Exothermic (\Delta H < 0): heat is a product. Raising the temperature shifts it backwards and K falls.
Endothermic (\Delta H > 0): heat is a reactant. Raising the temperature shifts it forwards and K rises.
The van 't Hoff equation makes it quantitative:
\ln\frac{K_2}{K_1} = -\frac{\Delta H^\circ}{R}\left(\frac{1}{T_2}-\frac{1}{T_1}\right)
Worked example. A reaction with \Delta H^\circ = -92 kJ/mol has K = 6.0\times10^{5} at 298 K. What is K at 700 K?
\ln\frac{K_2}{K_1} = -\frac{-92000}{8.314}\left(\frac{1}{700}-\frac{1}{298}\right) = 11066\times(0.001429-0.003356)
= 11066\times(-1.927\times10^{-3}) = -21.32
\frac{K_2}{K_1} = e^{-21.32} = 5.5\times10^{-10}
K_2 = (6.0\times10^{5})(5.5\times10^{-10}) = 3.3\times10^{-4}
Nine orders of magnitude smaller. Raising the temperature from 25 °C to 427 °C destroyed the yield entirely.
That reaction is ammonia synthesis, and this calculation is the central problem of the Haber process.
Catalysts
A catalyst does not change K and does not shift the equilibrium.
It speeds up the forward and reverse reactions by exactly the same factor, because it lowers the activation barrier for both directions equally (Chapter 10.6). It gets you to the same destination faster.
Claiming a catalyst improves yield at equilibrium is a thermodynamic impossibility — if it did, you could couple the catalysed and uncatalysed reactions and extract work from nothing, violating the second law (Chapter 3.4).
The Haber process, in full
This is the case study that ties everything together, and Chapter 9.6 established the stakes: about half the nitrogen atoms in your body passed through this reaction.
\text{N}_2(g)+3\text{H}_2(g) \rightleftharpoons 2\text{NH}_3(g), \qquad \Delta H = -92\ \text{kJ/mol}
Three requirements, and two of them conflict.
Requirement 1: high yield. The reaction is exothermic, so low temperature favours ammonia.
Requirement 2: high yield. Four gas moles become two, so high pressure favours ammonia.
Requirement 3: acceptable rate. The N≡N triple bond is 945 kJ/mol (Chapter 9.6), so the activation energy is enormous and low temperature means an impossibly slow reaction.
Requirements 1 and 3 are in direct opposition.
The equilibrium yields:
| Temperature | 100 atm | 200 atm | 400 atm |
|---|---|---|---|
| 200 °C | 81 % | 86 % | 94 % |
| 400 °C | 25 % | 36 % | 47 % |
| 600 °C | 5 % | 8 % | 13 % |
At 200 °C the yield is excellent and the rate is negligible. At 600 °C the rate is fine and the yield is 5 %.
The compromise, and it is genuinely a compromise:
450 °C — hot enough for an acceptable rate, giving about 30 % at pressure rather than the 90 % that thermodynamics would allow.
200 atmospheres — as high as the plant can be built for. Higher pressure always helps and the vessels become prohibitively expensive.
An iron catalyst with potassium and aluminium oxide promoters. This is the piece that makes it work, because it allows a usable rate at 450 °C rather than the 700 °C otherwise needed. It does not improve the equilibrium yield at all — it lets you operate at a temperature where the equilibrium yield is tolerable.
Continuous removal. The ammonia is condensed out (it liquefies at -33 °C while N₂ and H₂ do not) and the unreacted gases are recycled. Removing the product keeps Q below K permanently, so the reaction never actually reaches equilibrium and the overall conversion approaches 98 %.
The scale: about 150 million tonnes of ammonia a year, consuming roughly 1–2 % of world energy and about 3–5 % of natural gas production. The hydrogen comes from methane, which is why fertiliser prices track gas prices and why ammonia is a major target for decarbonisation.
Carl Bosch's contribution was the engineering, and it was as hard as the chemistry. Hydrogen at 200 atm and 450 °C attacks steel, diffusing in and reacting with the carbon to form methane, which embrittles the vessel. Bosch's solution was a double-walled reactor: a soft iron liner that hydrogen passes through harmlessly, inside a steel pressure vessel drilled with holes to let the hydrogen escape rather than build up. He received the Nobel Prize in 1931.
Solubility equilibrium
A saturated solution is at equilibrium between dissolved and undissolved solid.
\text{AgCl}(s) \rightleftharpoons \text{Ag}^+(aq)+\text{Cl}^-(aq)
The solid is omitted, so:
K_{sp} = [\text{Ag}^+][\text{Cl}^-] = 1.8\times10^{-10}
Worked example: silver chloride's solubility. Let s be the molar solubility:
s^2 = 1.8\times10^{-10} \quad\Longrightarrow\quad s = 1.34\times10^{-5}\ \text{mol/L}
In grams, with M = 143.3:
1.34\times10^{-5}\times143.3 = 1.9\times10^{-3}\ \text{g/L}
Under two milligrams per litre.
The common ion effect. Now dissolve AgCl in 0.10 M NaCl instead of pure water. The chloride is already 0.10 M:
s\times0.10 = 1.8\times10^{-10} \quad\Longrightarrow\quad s = 1.8\times10^{-9}\ \text{mol/L}
Seven thousand times less soluble. This is Le Chatelier applied to solubility, and it is the standard laboratory method for recovering a precipitate quantitatively — wash it with a solution containing one of its own ions rather than with water.
And it is why hard water leaves scale. Calcium carbonate's K_{sp} is 3.3\times10^{-9}, and heating drives off dissolved CO₂, which shifts the carbonate equilibrium and precipitates CaCO₃ inside kettles and pipes.
Where this shows up in your life
Fertiliser — the Haber process, and roughly half the world's food.
Blood pH. The bicarbonate buffer holds your blood at 7.4, and Chapter 10.5 works it through. Hyperventilating removes CO₂, shifts the equilibrium, and raises blood pH — which is why breathing into a bag helps.
Carbonated drinks. CO₂ is dissolved under pressure; opening the bottle drops the pressure and Le Chatelier drives it out of solution.
Scuba diving. Nitrogen dissolves in blood at depth under pressure, and ascending too fast drops the pressure faster than it can be exhaled, so bubbles form in tissue. The bends is Le Chatelier's principle in a diver.
Haemoglobin. Oxygen binding is an equilibrium, and it shifts with pH and CO₂ concentration — the Bohr effect — so oxygen is released preferentially in active tissue where CO₂ is high.
Ocean acidification. Roughly a quarter of emitted CO₂ dissolves in seawater, shifting the carbonate equilibrium and lowering the carbonate ion concentration that shell-forming organisms need. Ocean pH has fallen from 8.2 to 8.1 since 1750, which sounds tiny and is a 30 % increase in hydrogen ion concentration, because pH is logarithmic.
What the next chapter fixes
One family of equilibria is important enough to deserve its own treatment. Acids and bases run every biological process, and the same electron-transfer logic that governs them also governs batteries, corrosion and metabolism. Chapter 10.5 derives pH from first principles, explains how buffers hold a pH steady against added acid, works through redox balancing properly, and derives the Nernst equation that gives every battery its voltage.