Molecules and Reactions

Why a reaction goes, how fast, and how far: kinetics and equilibrium, acids and bases, and the electron transfers behind every battery.

Counting a reaction

A chemical equation is not a sentence about substances but an accounting identity, and almost every quantitative question in chemistry is answered by taking that identity seriously.

The previous course ended with the bond: two atoms approach, the energy falls into a minimum, and a molecule exists. This one starts one level up, where molecules meet and come apart again. Before any of the interesting questions can be asked, whether a reaction goes, how fast, and how far, we need to be able to count what goes into it and what comes out, and to do that with masses measured on a real balance.

What a balanced equation claims

Write the combustion of methane as $C + 2 \to C + 2 O$ and you have asserted exactly two things. Every atom present on the left is present on the right, one carbon, four hydrogens and four oxygens on each side, and the total charge is the same on both sides. That is all. The arrow makes no claim about speed, about how far the reaction proceeds, or about whether it happens at all. Methane and oxygen sit together indefinitely at room temperature without reacting, and the equation is still true.

Balancing is therefore a problem in linear constraints rather than an art. Put unknown coefficients on every species, write one equation per element, and solve. For a hydrocarbon $C_{x} H_{y}$ burning to carbon dioxide and water, carbon forces $x$ molecules of $C$ , hydrogen forces $y / 2$ of water, and the oxygen count on the right is then $2 x + y / 2$ , so the oxygen requirement is fixed at

C_{x} H_{y} + (x + \frac{y}{4}) \to x C + \frac{y}{2} O

Octane, , needs $8 + 18 / 4 = 12.5$ molecules of oxygen per molecule of fuel, which is why the equation is usually written doubled to clear the fraction. Nothing was guessed.

The constraints do not always have a unique answer up to scale. Write the reaction of carbon with oxygen allowing both $CO$ and $C$ as products and two independent balanced equations exist, so any mixture of them balances too. That is not a defect in the arithmetic: it is the arithmetic reporting, correctly, that the products depend on conditions and are not determined by conservation alone. Whenever a balancing problem has more than one independent solution, a chemical decision has been left unmade.

State symbols carry real information and are worth writing. The combustion of methane releases $890$ kJ per mole with liquid water as the product and $88$ kJ less with water vapour, so $O(l)$ and $O(g)$ are not interchangeable labels.

The mole is a count, and since 2019 an exact one

A balanced equation counts molecules; a balance in a laboratory weighs grams. The mole is the bridge. Since 20 May 2019 it is defined by fixing the Avogadro constant at exactly $N_{A} = 6.02214076 \times 10^{23}$ per mole, so a mole is a number in the same sense that a dozen is, and the mole of a substance is that many of its formula units.

The redefinition matters more than it looks. Before it, the mole was tied to $12$ grams of carbon-12, and the Avogadro constant was a measured quantity with an uncertainty. Now the count is exact and the molar mass of carbon-12 is what carries the uncertainty, at $11.9999999958$ g mol⁻¹ with an uncertainty in the last digit. For any chemical purpose the shift is invisible, and molar masses in grams per mole remain numerically equal to relative atomic masses. What it buys is a definition that does not depend on a particular substance.

Molar mass is then a sum over the formula. Water is $2 (1.008) + 15.999 = 18.015$ g mol⁻¹, ammonia $14.007 + 3 (1.008) = 17.031$ , iron(III) oxide $2 (55.845) + 3 (15.999) = 159.687$ . These values are averages over natural isotopic abundance, which is exactly what a bulk sample contains, and they are the reason chemistry can ignore isotopes almost everywhere and cannot ignore them in a mass spectrum.

The numbers involved are worth feeling once. A drop of water of $0.05$ grams is $0.05 / 18.015 = 2.78 \times 10^{- 3}$ mol, and therefore $1.67 \times 10^{21}$ molecules. A single carbon-12 atom weighs $12 / N_{A} = 1.99 \times 10^{- 23}$ grams. No experiment on a bench ever handles fewer than about $10^{15}$ molecules, which is why the statistical statements later in this course, about average rates and equilibrium positions, are so extraordinarily reliable.

Finding a formula by burning it

Before a reaction can be balanced, the formulas in it have to be known, and the classical way to get one is to destroy the compound in a controlled way. In combustion analysis a weighed sample is burned in excess oxygen, and the carbon dioxide and water produced are absorbed and weighed separately. Every carbon in the sample ends as one $C$ and every two hydrogens as one $O$ , so the masses of the products give the moles of carbon and hydrogen directly. Anything left over in the original mass, when the compound contains only carbon, hydrogen and oxygen, is oxygen.

Example. Burning $6.51$ mg of vitamin C gives $9.76$ mg of $C$ and $2.66$ mg of $O$ . Its molar mass is known from mass spectrometry to be $176$ g mol⁻¹. What is its molecular formula?

Carbon: $9.76 / 44.009 = 0.2218$ mmol, weighing $0.2218 \times 12.011 = 2.664$ mg. Hydrogen: each water carries two, so $2 \times 2.66 / 18.015 = 0.2953$ mmol, weighing $0.298$ mg. The oxygen is the remainder, $6.51 - 2.664 - 0.298 = 3.548$ mg, which is $3.548 / 15.999 = 0.2218$ mmol. The three amounts are in the ratio $0.2218 : 0.2953 : 0.2218$ , or $1 : 1.331 : 1$ , and multiplying by three clears it to $3 : 4 : 3$ . The empirical formula is , of mass $88.06$ g mol⁻¹. The measured molar mass is twice that, so the molecular formula is .

Now you. Burning $5.000$ mg of a compound of carbon, hydrogen and oxygen gives $7.33$ mg of $C$ and $3.00$ mg of $O$ . Its molar mass is $90.08$ g mol⁻¹. Find the molecular formula.

Answer

Carbon: $7.33 / 44.009 = 0.1666$ mmol, or $2.001$ mg. Hydrogen: $2 \times 3.00 / 18.015 = 0.3331$ mmol, or $0.336$ mg. Oxygen by difference: $5.000 - 2.001 - 0.336 = 2.663$ mg, which is $0.1665$ mmol. The ratio is $1 : 2 : 1$ , so the empirical formula is $CO$ at $30.03$ g mol⁻¹. Since $90.08 / 30.03 = 3.00$ , the molecular formula is , lactic acid.

Note what the method cannot do. It gives the ratio of atoms and, with a molar mass, the formula, but nothing about how those atoms are joined. Lactic acid and glyceraldehyde are both and burn identically. Structure comes from spectroscopy, not from a balance.

The limiting reagent

Mix reactants in a ratio other than the one the equation demands and one of them runs out first. Everything after that moment is idle, so the yield is set by whichever reactant is exhausted soonest, the limiting reagent. The test is not which reactant there is less of, but which has the smallest amount when divided by its coefficient.

Example. An ammonia plant feeds $28.0$ kg of nitrogen and $6.00$ kg of hydrogen into $+ 3 \to 2 N$ . Which limits, and what mass of ammonia can be made?

The molar masses are $28.014$ and $2.016$ g mol⁻¹, giving $28000 / 28.014 = 999.5$ mol of nitrogen and $6000 / 2.016 = 2976$ mol of hydrogen. Consuming all the nitrogen would need $3 \times 999.5 = 2999$ mol of hydrogen, which is more than there is, so hydrogen limits. It makes $\frac{2}{3} \times 2976 = 1984$ mol of ammonia, which at $17.031$ g mol⁻¹ is $33.8$ kg. The nitrogen consumed is $2976 / 3 = 992$ mol, leaving $7.4$ mol, or about $0.21$ kg, unreacted.

Now you. A blast furnace charge is $500$ kg of iron(III) oxide, $F$ at $159.687$ g mol⁻¹, and $250$ kg of carbon monoxide at $28.01$ g mol⁻¹, reacting as $F + 3 CO \to 2 Fe + 3 C$ . Which limits, and what mass of iron is produced?

Answer

The oxide gives $500000 / 159.687 = 3131$ mol and the monoxide $250000 / 28.01 = 8925$ mol. Full reduction of the oxide would need $3 \times 3131 = 9393$ mol of carbon monoxide, so the monoxide limits. It reduces $8925 / 3 = 2975$ mol of oxide, giving $5950$ mol of iron, which at $55.845$ g mol⁻¹ is $332$ kg.

Industrially the limiting reagent is a choice rather than an accident. The expensive reactant is made limiting so that none of it is wasted, and the cheap one is fed in excess, which is also why an excess of oxygen is used in combustion analysis.

Yield, and where the rest of it went

The mass calculated above is the theoretical yield, what the equation permits. The actual yield is what the flask contains after the reaction, the workup and the purification, and the ratio of the two, as a percentage, is the percentage yield. If the ammonia plant above produced $28.7$ kg rather than $33.8$ , the yield is $28.7 / 33.8 = 84.9$ per cent.

A yield below one hundred per cent has three quite different causes, and telling them apart is the whole of process chemistry. The reaction may have reached equilibrium before consuming the limiting reagent, which is a thermodynamic limit and the subject of the fourth and fifth lessons. It may still be running when the experimenter stopped, a kinetic limit, which is the sixth and seventh. Or some of the reagent may have gone into a different reaction entirely, a side product, which is a matter of mechanism. Only the third is a loss of material; the first two are situations where the missing reagent is still sitting in the flask.

Atom economy asks a different question, and one that a percentage yield hides. It is the mass of the desired product divided by the total mass of all products, taken from the balanced equation at one hundred per cent conversion. A synthesis with a ninety per cent yield that discards two thirds of its atoms as by-product is worse, in waste terms, than one with a sixty per cent yield and no by-product at all. The distinction has driven a good deal of industrial redesign since Barry Trost named it in 1991.

The extent of reaction

Tracking every substance separately is redundant, because the coefficients tie them together. Assign each species a stoichiometric number $ν_{i}$ , negative for reactants and positive for products, so that for $+ 3 \to 2 N$ we have $ν = - 1$ , $- 3$ and $+ 2$ . Then a single variable $ξ$ , the extent of reaction, measured in moles, fixes every amount at once:

n_{i} = n_{i, 0} + ν_{i} ξ

At $ξ = 0$ nothing has happened, and $ξ$ increases as the reaction runs forward. Its maximum is set by the limiting reagent, at the smallest value of $n_{i, 0} / | ν_{i} |$ over the reactants. One number now describes the whole mixture, which is the reason this variable is worth the notation: the rate of reaction in the sixth lesson is $d ξ / d t$ divided by volume, and the equilibrium of the fourth lesson is the particular $ξ$ at which the free energy stops falling.

Example. A vessel is charged with $5.00$ mol of $S$ and $3.00$ mol of , which react as $2 S + \to 2 S$ . At one moment $ξ = 1.80$ mol. Give the composition, and the largest $ξ$ the charge allows.

With $ν = - 2$ , $- 1$ and $+ 2$ : sulfur dioxide is $5.00 - 2 (1.80) = 1.40$ mol, oxygen is $3.00 - 1.80 = 1.20$ mol, sulfur trioxide is $2 (1.80) = 3.60$ mol. The total is $6.20$ mol, down from $8.00$ , because the reaction consumes three molecules of gas for every two it makes, and the mole fraction of the product is $3.60 / 6.20 = 0.581$ . The limit is the smaller of $5.00 / 2 = 2.50$ and $3.00 / 1 = 3.00$ , so $ξ_{\max} = 2.50$ mol, set by the sulfur dioxide.

Now you. A vessel holds $4.00$ mol of and $9.00$ mol of , reacting as $+ 3 \to 2 N$ . Give the composition at $ξ = 1.20$ mol, the total amount of gas, and $ξ_{\max}$ .

Answer

Nitrogen $4.00 - 1.20 = 2.80$ mol, hydrogen $9.00 - 3 (1.20) = 5.40$ mol, ammonia $2 (1.20) = 2.40$ mol, totalling $10.60$ mol against $13.00$ at the start. The limit is the smaller of $4.00 / 1$ and $9.00 / 3 = 3.00$ , so $ξ_{\max} = 3.00$ mol, set by the hydrogen.

What the arithmetic cannot say

Everything in this lesson follows from conservation, and conservation is silent about direction. The equation for the synthesis of ammonia is exactly as well balanced as the equation for its decomposition, and the arithmetic that predicts $33.8$ kg of product would predict the reverse yield with equal confidence. Yet a sealed flask of nitrogen and hydrogen at room temperature contains essentially no ammonia, and a flask of ammonia left alone does not fall apart.

Something other than counting decides which way a mixture moves, and the first candidate is energy. Reactions that go tend to release heat: methane burns and warms the room, and nobody has to be persuaded that the reverse will not happen spontaneously. Making that intuition quantitative means being able to compute the heat of a reaction from tabulated data, for reactions nobody has run, which is the next lesson. It also means finding out, at the end of it, that the intuition is wrong.

The heat of a reaction

The heat given out by a reaction can be computed from tables for reactions that have never been carried out, and the reason that works is that heat measured in the right way is a property of the change rather than of the apparatus.

The previous lesson counted what a reaction consumes and produces, and ended with the observation that conservation says nothing about direction. Energy is the obvious candidate for what does. This lesson makes the energy of a reaction computable; the next one shows that computing it is not enough.

Heat measured in an open flask is a property of the reaction

Heat is not a property of a system. How much of an energy change arrives as heat rather than as work depends on the path taken, so in general asking for the heat of a process has no answer until the constraints are stated. Chemistry gets lucky, because almost every reaction of interest happens in a vessel open to the atmosphere, which fixes the pressure.

At constant pressure the only work a reaction usually does is pushing the atmosphere back, $p Δ V$ . Conservation of energy then gives $Δ U = Q_{p} - p Δ V$ , and rearranging,

Q_{p} = Δ U + p Δ V = (U_{2} + p V_{2}) - (U_{1} + p V_{1})

The heat is the change in the combination $U + p V$ , evaluated at the start and the end and nowhere in between. That combination is the enthalpy $H$ , and since $U$ , $p$ and $V$ are all properties of the state, so is $H$ . The result is $Δ H = Q_{p}$ : at constant pressure, with only expansion work, the heat absorbed is the change in a state function.

This is why every table in chemistry is a table of enthalpies. A reaction run in a flask, in a bucket or in a chemical plant gives the same $Δ H$ , because the $p Δ V$ term the atmosphere absorbs has already been folded into the bookkeeping. Nothing is stored in the $p V$ term in any physical sense; it is the correct accounting for the commonest constraint. A reaction with $Δ H < 0$ releases heat and is exothermic, one with $Δ H > 0$ absorbs it and is endothermic.

Formation enthalpies, and Hess's law as a consequence

Enthalpy has no absolute zero, so only differences are tabulated, and the convention picks one reference. The standard enthalpy of formation $Δ_{f} H^{ominus}$ is the enthalpy change on forming one mole of a substance from its elements in their standard states at $10^{5}$ Pa, and an element in its standard state is assigned zero. Graphite is zero and diamond is $+ 1.9$ kJ mol⁻¹, because graphite is the stable form at ordinary conditions.

Because $H$ is a state function, the enthalpy change around any closed path is zero, so a reaction enthalpy does not depend on the route. That is Hess's law, published by Germain Hess in 1840, before the First Law of thermodynamics was settled. It is not an extra assumption but a restatement of what a state function is, and it licenses routing every reaction through the elements:

Δ_{r} H^{ominus} = \sum ν_{i} Δ_{f} H^{ominus} (i)

with the stoichiometric numbers of the previous lesson, negative for reactants and positive for products. The practical payoff is that a table of a few thousand formation enthalpies gives the enthalpy of any reaction that can be written between them, including reactions too slow, too violent or too impure to measure.

Example. The thermite reaction is $F(s) + 2 Al(s) \to A(s) + 2 Fe(s)$ . Given $Δ_{f} H^{ominus} = - 824.2$ kJ mol⁻¹ for $F$ and $- 1675.7$ for $A$ , find $Δ_{r} H^{ominus}$ .

Aluminium and iron are elements in their standard states, so both are zero. The products sum to $- 1675.7$ and the reactants to $- 824.2$ , giving $Δ_{r} H^{ominus} = - 1675.7 - (- 824.2) = - 851.5$ kJ mol⁻¹. Released into a small mass with no cooling, that is enough to raise the iron produced past its melting point of $1538$ °C, which is exactly what the reaction is used for when welding rail.

Now you. Find $Δ_{r} H^{ominus}$ for $2 S(g) + 3 (g) \to 2 S(g) + 2 O(l)$ , given $Δ_{f} H^{ominus} = - 20.6$ kJ mol⁻¹ for $S$ , $- 296.8$ for $S$ and $- 285.8$ for liquid water.

Answer

Products: $2 (- 296.8) + 2 (- 285.8) = - 1165.2$ kJ mol⁻¹. Reactants: $2 (- 20.6) = - 41.2$ , oxygen being zero. So $Δ_{r} H^{ominus} = - 1165.2 + 41.2 = - 1124.0$ kJ mol⁻¹.

The state symbols are load-bearing. Taking the water as vapour instead would change the answer by $2 \times 44 = 88$ kJ, the enthalpy of vaporisation of two moles, and that difference is precisely why natural gas has two quoted heating values that differ by about ten per cent.

Bond enthalpies: a cruder estimate that says why

Formation enthalpies give the number without explaining it. Bonds do the reverse. A reaction breaks some bonds and makes others, breaking costs energy and making releases it, so to a first approximation

Δ_{r} H \approx \sum (bonds broken) - \sum (bonds formed)

The values used are mean bond enthalpies, averaged over many molecules, since the strength of a C to H bond is not quite the same in methane as in ethanol. That averaging is the source of the method's error and also of its reach: one table of about thirty numbers covers organic chemistry.

Example. Estimate the enthalpy of combustion of methane to $C$ and water vapour, using mean bond enthalpies of $413$ kJ mol⁻¹ for C to H, $498$ for O to O double, $799$ for C to O double in carbon dioxide and $463$ for O to H.

Breaking four C to H bonds and two oxygen double bonds costs $4 (413) + 2 (498) = 2648$ kJ. Forming two carbon to oxygen double bonds and four O to H bonds releases $2 (799) + 4 (463) = 3450$ kJ. The estimate is $2648 - 3450 = - 802$ kJ mol⁻¹, against $- 802.5$ from formation enthalpies.

Now you. Estimate $Δ_{r} H$ for the hydrogenation of ethene, $+ \to$ , given C to C double $614$ , H to H $436$ , C to C single $347$ and C to H $413$ kJ mol⁻¹. Compare with the value from $Δ_{f} H^{ominus} = + 52.4$ kJ mol⁻¹ for ethene and $- 84.0$ for ethane.

Answer

Broken: the carbon double bond and the hydrogen bond, $614 + 436 = 1050$ kJ. Formed: one carbon single bond and two new C to H bonds, $347 + 2 (413) = 1173$ kJ. The estimate is $- 123$ kJ mol⁻¹; the formation data give $- 84.0 - 52.4 = - 136.4$ . The estimate is out by $13$ kJ mol⁻¹, about ten per cent.

The agreement in the methane example is better than the method deserves, and treating it as typical would be a mistake. Errors of $10$ to $30$ kJ mol⁻¹ are normal, and they become much worse wherever a real molecule is not a collection of independent bonds. Benzene is the standard casualty: bond enthalpies predict its hydrogenation enthalpy about $150$ kJ mol⁻¹ too negative, because the delocalised ring is more stable than three isolated double bonds. Bond enthalpies also apply only to gases, since they say nothing about the energy of holding a liquid or a lattice together. Use them to see why a reaction is exothermic, and formation enthalpies when the number has to be right.

Measuring it: two calorimeters

Both tables above are ultimately built from measurements, and the measurement means choosing a constraint. A coffee cup calorimeter, an insulated vessel open to the air, holds the pressure fixed and so measures $Δ H$ directly. It suits reactions in solution: neutralisations, dissolutions, precipitations. The heat released warms the solution, and if the solution is dilute its specific heat capacity is close to that of water, $4.18$ J g⁻¹ K⁻¹.

Example. Mixing $50.0$ mL of $1.00$ mol dm⁻³ hydrochloric acid with $50.0$ mL of $1.00$ mol dm⁻³ sodium hydroxide, both at $21.0$ °C, raises the temperature to $27.8$ °C. Taking the mixture as $100.0$ g of water, find the enthalpy of neutralisation per mole of water formed.

The heat gained by the solution is $q = m c Δ T = 100.0 \times 4.18 \times 6.8 = 2842$ J. The acid supplies $0.0500$ mol of protons and the base $0.0500$ mol of hydroxide, so $0.0500$ mol of water forms. Since the solution gained that heat, the reaction lost it: $Δ H = - 2842 / 0.0500 = - 56.8$ kJ mol⁻¹, against a tabulated $- 57.3$ . The shortfall is heat that went into the polystyrene cup and the thermometer rather than the liquid.

Now you. The same experiment is done with $50.0$ mL of each solution at $2.00$ mol dm⁻³, again treating the mixture as $100.0$ g of water with $c = 4.18$ J g⁻¹ K⁻¹, and the true value of $- 57.3$ kJ mol⁻¹. What temperature rise should a perfect calorimeter show?

Answer

Now $0.100$ mol of water forms, releasing $0.100 \times 57300 = 5730$ J. So $Δ T = 5730 / (100.0 \times 4.18) = 13.7$ K. Doubling the concentration doubles the rise, because both the heat and the mass of solution scale, but only the heat depends on concentration.

A bomb calorimeter seals the sample in a rigid steel vessel under excess oxygen, so the volume rather than the pressure is fixed. No expansion work is possible, and the heat measured is $Δ U$ , not $Δ H$ . The two are related by $Δ H = Δ U + Δ n_{gas} R T$ , where only the change in moles of gas matters, since solids and liquids occupy a negligible volume by comparison. For methane, three moles of gas become one, so $Δ n_{gas} = - 2$ and the correction is $2 \times 2.48 = 5.0$ kJ mol⁻¹, half a per cent and far larger than the instrument's error.

The same reaction at a different temperature

Tables are at $298.15$ K and reactions are run wherever they are run. The correction follows from Hess's law again: instead of reacting at $T_{2}$ , cool the reactants to $298$ K, react there, and warm the products back. The two temperature changes cost the heat capacities of the reactants and products respectively, so

Δ_{r} H (T_{2}) = Δ_{r} H (T_{1}) + Δ_{r} C_{p} (T_{2} - T_{1})

with $Δ_{r} C_{p}$ the same stoichiometric sum over molar heat capacities. This is Kirchhoff's law, and treating $Δ_{r} C_{p}$ as constant is an approximation good over a hundred kelvin or so.

Apply it to ammonia synthesis, $+ 3 \to 2 N$ , with $Δ_{r} H^{ominus} (298) = - 91.8$ kJ mol⁻¹ and molar heat capacities of $29.12$ , $28.82$ and $35.06$ J mol⁻¹ K⁻¹. Then $Δ_{r} C_{p} = 2 (35.06) - 29.12 - 3 (28.82) = - 45.5$ J K⁻¹, and at $500$ K, a typical plant temperature, $Δ_{r} H = - 91.8 - 0.0455 (202) = - 101.0$ kJ mol⁻¹. The reaction is nine per cent more exothermic hot than the table suggests, which matters when the reactor has to be cooled continuously.

The sign of $Δ_{r} C_{p}$ has a plain reading. It is negative here because four moles of gas become two, and fewer molecules store less thermal energy, so the products carry less of the temperature change than the reactants did.

Dissolving: two large numbers that nearly cancel

Dissolving a salt is a good test of the machinery, because it splits cleanly into two steps that can be looked at separately. Pull the lattice apart into gaseous ions, which costs the lattice enthalpy, then let water surround each ion, which releases the hydration enthalpy. For sodium chloride the lattice enthalpy is $+ 787$ kJ mol⁻¹ and the hydration enthalpies of the two ions sum to about $- 783$ , so the enthalpy of solution is the difference of two numbers near $800$ : about $+ 4$ kJ mol⁻¹, against a measured $+ 3.9$ .

That near-cancellation is the whole story of solubility enthalpies. Both terms scale with charge and with the inverse of ionic radius, so they track each other, and what survives is a small residue whose sign is genuinely hard to predict. Sodium chloride absorbs a little heat as it dissolves, lithium chloride releases $37$ kJ mol⁻¹, and ammonium nitrate absorbs $25.7$ . That last one is sold in instant cold packs, and it is about to cause trouble.

Exothermic is not the same as spontaneous

The intuition that reactions happen because they release energy is old, respectable and false. Marcellin Berthelot stated it as a principle in 1867: every spontaneous change is the one that releases the most heat. It accounts for a great many reactions, which is why it survived as long as it did.

The counterexamples are on any bench. Ammonium nitrate dissolves eagerly in water while chilling the beaker, with $Δ_{sol} H^{ominus} = + 25.7$ kJ mol⁻¹. Ice in a room at $10$ °C melts, absorbing $6.01$ kJ mol⁻¹. Mix solid barium hydroxide octahydrate with solid ammonium thiocyanate and the two powders liquefy while the beaker drops to around $- 20$ °C, cold enough to freeze it to a wooden block, absorbing some $80$ kJ per mole. All three go uphill in enthalpy and all three go anyway.

Notice what the three have in common: in every case the products are more spread out, more mobile, more numerous than the reactants. A crystal becomes free ions in solution; a solid becomes a liquid; two solids become a liquid and a gas. Something is being gained that is not energy, and gained enough to pay an energy penalty of tens of kilojoules per mole.

Naming that quantity, tabulating it as reliably as formation enthalpies are tabulated, and combining it with $Δ H$ into a single criterion that decides the direction of any reaction, is the next lesson.

Entropy and free energy

An instant cold pack gets colder while it dissolves, which means something other than energy is deciding what happens, and that something can be measured and tabulated as precisely as an enthalpy.

The previous lesson made reaction enthalpies computable and then produced three reactions that absorb heat and happen anyway. What they had in common was that the products are more spread out than the reactants. This lesson turns that observation into a number.

Counting the arrangements

Take a crystal of a salt and a beaker of water. Before dissolving, every ion sits at a fixed lattice site; afterwards each ion can be anywhere in the liquid, and the water molecules around it have been reorganised too. The number of microscopic arrangements consistent with the second situation is unimaginably larger than the number consistent with the first.

Ludwig Boltzmann's proposal, engraved on his tombstone, is that this count is a thermodynamic quantity:

S = k_{B} \ln Ω

where $Ω$ is the number of microscopic states compatible with the macroscopic condition and $k_{B} = 1.381 \times 10^{- 23}$ J K⁻¹. The logarithm is not decoration. Put two independent systems side by side and their state counts multiply, while any sensible extensive property must add, and the logarithm is what turns one into the other.

Entropy also has a purely thermodynamic definition that never mentions atoms: for a reversible transfer of heat, $d S = δ Q_{rev} / T$ . The two definitions agree, which is one of the deeper results in physics, and for chemistry the second is what makes entropy measurable. Melting ice at $273.15$ K absorbs $6.01$ kJ mol⁻¹ reversibly, so the entropy of fusion is $6010 / 273.15 = 22.0$ J mol⁻¹ K⁻¹, a number obtained with a thermometer and a heater.

The Second Law is then the statement that the entropy of an isolated system never falls. Applied to the universe as a whole, that is the criterion we were missing: a change happens if the total entropy of system plus surroundings increases.

The Third Law makes entropy absolute

Enthalpy has no zero, but entropy does. The Third Law says that the entropy of a perfect crystalline substance approaches zero as the temperature approaches absolute zero: at $T = 0$ a perfect crystal has one arrangement, $Ω = 1$ , so $S = 0$ . That fixes the origin, and it means entropies can be tabulated as absolute values rather than as changes.

Measuring one means integrating the heat capacity from as near zero kelvin as an experiment can reach, adding the entropy of each phase change on the way up. The result is the standard molar entropy $S^{ominus}$ , in J mol⁻¹ K⁻¹, and unlike a formation enthalpy it is never zero for an element: graphite is $5.74$ , hydrogen gas $130.68$ , oxygen $205.14$ .

The values behave as the counting argument predicts. Gases greatly exceed liquids, which exceed solids: water is $69.91$ as a liquid and $188.83$ as a vapour, and that difference of $119$ J mol⁻¹ K⁻¹ is the biggest single lever in most reactions involving water. Bigger and floppier molecules have more entropy, since there are more ways to distribute energy among their vibrations: methane is $186.26$ and octane vapour $467$ . Harder substances have less, because stiff bonds mean widely spaced vibrational levels and fewer accessible states, which is why diamond sits at $2.38$ .

The Third Law also has known exceptions, and they are informative rather than embarrassing. Carbon monoxide freezes into a lattice where each molecule can point either way, and the mismatch is too small for the crystal to sort itself out before it stops moving. That leaves $2^{N}$ arrangements at absolute zero, predicting a residual entropy of $R \ln 2 = 5.76$ J mol⁻¹ K⁻¹, against a measured $4.6$ . Ice has a similar disorder in its hydrogen bonds, predicted at about $3.37$ and measured at $3.4$ . The law is a statement about perfect crystals, and a frozen-in disorder is a failure to be perfect.

The entropy of a reaction

Because $S^{ominus}$ values are absolute, a reaction entropy is a stoichiometric sum in exactly the way a reaction enthalpy is, with no reference state to cancel:

Δ_{r} S^{ominus} = \sum ν_{i} S^{ominus} (i)

The sign is usually predictable before any arithmetic, from the change in the number of moles of gas. Gases dominate the sum, so a reaction that consumes gas has a negative $Δ_{r} S$ and one that produces gas a positive one.

Example. Find $Δ_{r} S^{ominus}$ for the combustion of methane, $C(g) + 2 (g) \to C(g) + 2 O(l)$ , using $S^{ominus} = 186.26$ , $205.14$ , $213.74$ and $69.91$ J mol⁻¹ K⁻¹.

Products: $213.74 + 2 (69.91) = 353.56$ . Reactants: $186.26 + 2 (205.14) = 596.54$ . So $Δ_{r} S^{ominus} = 353.56 - 596.54 = - 242.98$ J mol⁻¹ K⁻¹. Three moles of gas become one, and two of the products are liquid, so the mixture ends far more ordered than it began. A reaction everyone agrees is spontaneous has a strongly negative entropy change, which already shows that system entropy alone is not the criterion either.

Now you. Find $Δ_{r} S^{ominus}$ for $2 (g) + (g) \to 2 O(l)$ , using $S^{ominus} = 130.68$ for hydrogen, $205.14$ for oxygen and $69.91$ for liquid water.

Answer

Products: $2 (69.91) = 139.82$ . Reactants: $2 (130.68) + 205.14 = 466.50$ . So $Δ_{r} S^{ominus} = - 326.68$ J mol⁻¹ K⁻¹. Three moles of gas become two moles of liquid, which is about as large a decrease as ordinary chemistry offers.

The surroundings are part of the account

Methane burns, and its entropy change is $- 243$ J K⁻¹ per mole. The Second Law is not violated, because the system is not isolated: the reaction dumps $890$ kJ into the room.

Heat arriving in the surroundings raises their entropy. If the surroundings are large enough to stay at constant temperature and pressure, the heat they receive is $- Δ H$ and they receive it reversibly, so

Δ S_{surr} = - \frac{Δ H}{T}

The division by $T$ carries the physical content: the same joule of heat buys more entropy in a cold place than a hot one, because it makes a larger relative difference to a system with little thermal energy already. For burning methane at $298$ K, $Δ S_{surr} = 890500 / 298.15 = 2987$ J K⁻¹, which dwarfs the system's $- 243$ , and the total is strongly positive.

The same accounting settles the melting of ice, which the previous lesson left as a puzzle. At $283$ K the ice gains $6010 / 273.15 = 22.0$ J K⁻¹ per mole as it melts while the room loses $6010 / 283.15 = 21.2$ , so the total is $+ 0.78$ J mol⁻¹ K⁻¹ and the ice melts. At $263$ K the surroundings would lose $22.9$ , the total would be negative, and it does not. The melting point is where the two terms cancel exactly.

Free energy: the criterion in system quantities alone

Requiring $Δ S_{total} > 0$ is correct but inconvenient, because it asks about the universe when the experimenter only has a flask. Substituting the expression for the surroundings fixes that:

Δ S_{total} = Δ S - \frac{Δ H}{T}

Multiply by $- T$ , which is negative and so reverses the inequality, and define the result as the Gibbs free energy change:

Δ G = Δ H - T Δ S = - T Δ S_{total}

A process at constant temperature and pressure happens spontaneously when $Δ G < 0$ . Every quantity on the right belongs to the system, which is why this combination and not the entropy is the working criterion of chemistry. Josiah Willard Gibbs published it in 1876 in a journal so obscure that Maxwell had to draw European attention to it.

$Δ G$ has a second reading worth carrying: it is the maximum non-expansion work a process can deliver. A reaction with $Δ G = - 474$ kJ mol⁻¹ can in principle supply that much electrical work in a fuel cell, and this is exactly the quantity a battery converts, as the last lesson of this course will use.

Example. For $2 (g) + (g) \to 2 O(l)$ , with $Δ_{r} H^{ominus} = - 571.6$ kJ mol⁻¹ and $Δ_{r} S^{ominus} = - 326.68$ J mol⁻¹ K⁻¹, find $Δ_{r} G^{ominus}$ at $298.15$ K.

Convert the entropy to kJ before combining, a step worth checking every time: $T Δ S = 298.15 \times (- 0.32668) = - 97.4$ kJ mol⁻¹. So $Δ_{r} G^{ominus} = - 571.6 - (- 97.4) = - 474.2$ kJ mol⁻¹, which agrees with twice the tabulated formation free energy of liquid water, $- 237.1$ . The reaction is strongly favoured despite its large entropy penalty, because the enthalpy term is larger still.

Now you. For $(g) + 3 (g) \to 2 N(g)$ , $Δ_{r} H^{ominus} = - 91.8$ kJ mol⁻¹ and $Δ_{r} S^{ominus} = - 198.75$ J mol⁻¹ K⁻¹. Find $Δ_{r} G^{ominus}$ at $298.15$ K.

Answer

$T Δ S = 298.15 \times (- 0.19875) = - 59.3$ kJ mol⁻¹, so $Δ_{r} G^{ominus} = - 91.8 + 59.3 = - 32.5$ kJ mol⁻¹, matching twice the tabulated $- 16.4$ for ammonia. Negative, so at room temperature and standard conditions the synthesis is favoured.

Temperature picks the winner

The two terms of $Δ G = Δ H - T Δ S$ can agree or fight, which gives four cases. If $Δ H < 0$ and $Δ S > 0$ the reaction goes at every temperature; if the signs are reversed it goes at none. The interesting cases are the two where they conflict, because there the temperature decides, and the crossover is where $Δ G = 0$ :

T_{cross} = \frac{Δ H}{Δ S}

An endothermic reaction with a positive entropy change is impossible when cold and spontaneous when hot. That is every decomposition that releases a gas, every evaporation, and much of extractive metallurgy.

Example. Limestone decomposes as $CaC(s) \to CaO(s) + C(g)$ . Using $Δ_{f} H^{ominus} = - 1206.9$ , $- 635.1$ and $- 393.5$ kJ mol⁻¹, and $S^{ominus} = 92.9$ , $39.75$ and $213.74$ J mol⁻¹ K⁻¹, find $Δ_{r} G^{ominus}$ at $298$ K and the temperature above which the reaction becomes favourable.

$Δ_{r} H^{ominus} = - 635.1 - 393.5 + 1206.9 = + 178.3$ kJ mol⁻¹ and $Δ_{r} S^{ominus} = 39.75 + 213.74 - 92.9 = + 160.59$ J mol⁻¹ K⁻¹. At $298.15$ K, $Δ_{r} G^{ominus} = 178.3 - 298.15 (0.16059) = + 130.4$ kJ mol⁻¹, hopelessly unfavourable, which is why buildings made of limestone stay up. The crossover is $T = 178300 / 160.59 = 1110$ K, or $837$ °C. Real lime kilns run at $900$ to $1000$ °C, and the measured temperature at which carbon dioxide reaches one bar over the solid is about $1170$ K. The estimate is out by five per cent because it holds $Δ H$ and $Δ S$ fixed over eight hundred kelvin, which Kirchhoff's law says they are not.

Now you. Magnesium carbonate decomposes as $MgC(s) \to MgO(s) + C(g)$ , with $Δ_{r} H^{ominus} = + 100.7$ kJ mol⁻¹ and $Δ_{r} S^{ominus} = + 174.98$ J mol⁻¹ K⁻¹. Estimate the crossover temperature, and say whether magnesium carbonate or calcium carbonate is easier to decompose.

Answer

$T = 100700 / 174.98 = 575$ K, or $302$ °C. The entropy changes are similar, since both release one mole of gas, so the crossover is set by the enthalpy, and magnesium carbonate needs far less: its smaller cation binds the carbonate ion more strongly and polarises it, weakening it. Magnesium carbonate is much the easier to decompose.

What the criterion still does not deliver

Two things are missing, and both are the business of later lessons.

The first is time. $Δ_{r} G^{ominus}$ for the combustion of methane is $- 818$ kJ mol⁻¹, about as favourable as chemistry gets, and a mixture of methane and air in a sealed bottle will sit unchanged for a century. Free energy says which way the mixture would go if it went. It says nothing whatever about how long that takes, and nothing in this lesson can be repaired to make it say so. That is the sixth lesson.

The second is subtler and immediate. Take the criterion literally and a reaction with $Δ G < 0$ should run until a reactant is exhausted, since the sign does not change on the way. Reactions do not do this. Heat calcium carbonate at $1100$ K in a closed vessel and it stops with solid and gas both present; dissolve acetic acid in water and most of it stays intact. Something makes $Δ G$ climb to zero at a composition short of completion, and stop there.

The missing piece is that $Δ_{r} G^{ominus}$ is a standard value, referring to a specific composition, and the actual $Δ_{r} G$ depends on how much of each substance is present. Working out that dependence gives the equilibrium constant, and it is the next lesson.

The equilibrium constant

A reaction with a negative free energy change should run until a reactant is gone, and almost none of them do, which means the criterion of the previous lesson is incomplete rather than wrong.

The previous lesson established that a change at constant temperature and pressure goes when $Δ G < 0$ , where $Δ G = Δ H - T Δ S$ . Take that literally and nothing should ever stop halfway. Yet acetic acid in water is barely ionised, calcium carbonate heated in a sealed vessel reaches a fixed carbon dioxide pressure and stays there, and a reactor full of nitrogen and hydrogen makes some ammonia and then makes no more. This lesson repairs the criterion, and the repair turns out to be the most useful single equation in chemistry.

Why the free energy stops falling

The quantity that must decrease is the free energy of the whole mixture, and the fault in the naive picture is treating that as a straight line between pure reactants and pure products. It is not, because a mixture has an entropy that neither pure state has.

Follow the extent of reaction $ξ$ from the first lesson. At $ξ = 0$ the vessel holds pure reactants; at $ξ_{\max}$ pure products. If mixing contributed nothing, $G$ would run linearly between the two ends and its minimum would be at whichever end is lower. But at every intermediate composition both reactants and products are present, and the entropy of mixing them is positive, which pulls $G$ down in the middle and away from both ends. The mixing term is largest at intermediate compositions and falls to zero at each pure limit, with an infinite slope there, and that infinite slope is decisive: however unfavourable a reaction, the first trace of product always lowers $G$ .

So $G$ plotted against $ξ$ is a curve with a minimum strictly inside the range, and the system slides downhill to that minimum and stops. Equilibrium is the bottom of that curve, where

{(\frac{\partial G}{\partial ξ})}_{T, p} = 0

The derivative on the left is what $Δ_{r} G$ actually means: the slope of the free energy with respect to extent, in joules per mole of reaction, at the composition the mixture currently has. A reaction with $Δ_{r} G^{ominus} = - 32$ kJ mol⁻¹ does not have that slope throughout; it has it at one particular reference composition, and the slope climbs toward zero as products build up.

This also explains why no reaction ever goes fully to completion. Complete conversion is a pure state, where the mixing term has infinite slope in the reverse direction, so the last trace of reactant never disappears. Sometimes what remains is one molecule in $10^{60}$ , which is completion for any practical purpose, but the position of equilibrium is always strictly inside.

Activity, and the composition dependence

To make this quantitative we need how the free energy of one substance depends on how much of it is there. For an ideal gas at constant temperature the answer follows from the work of isothermal compression, and comes out as a logarithm. Writing the molar free energy of a species as its chemical potential $μ$ ,

μ_{i} = μ_{i}^{ominus} + R T \ln a_{i}

where $a_{i}$ is the activity, a dimensionless measure of how much of the substance is present relative to its standard state. For a gas it is the partial pressure divided by $10^{5}$ Pa; for a dissolved solute it is the concentration divided by $1$ mol dm⁻³; and for a pure solid or pure liquid it is $1$ , because a pure substance is already in its standard state. That last rule is why solids never appear in an equilibrium expression: adding more solid does not change its activity, and a lump of calcium carbonate has the same chemical potential whether there is a gram of it or a tonne.

These are the ideal forms. Real gases at high pressure and real ions in solution deviate, and the honest statement is $a_{i} = γ_{i} c_{i} / c^{ominus}$ with an activity coefficient $γ_{i}$ that approaches $1$ at infinite dilution. For a $0.1$ mol dm⁻³ solution of a simple salt, $γ$ is already around $0.78$ , so calculations that ignore it are good to a few per cent at best in ionic systems, and better than that in dilute or gaseous ones. Everything below uses $γ = 1$ , and it is worth remembering that this is where the error lives.

The reaction quotient and the exact relation

Now assemble the mixture. The slope of $G$ with respect to $ξ$ is the stoichiometric sum of the chemical potentials, $Δ_{r} G = \sum ν_{i} μ_{i}$ , so substituting the expression above and collecting the logarithms gives

Δ_{r} G = Δ_{r} G^{ominus} + R T \ln Q

where the reaction quotient $Q$ is the product of the activities each raised to its stoichiometric number, products on top and reactants underneath. For $+ 3 \to 2 N$ it is $Q = a (N)^{2} / [a () a ()^{3}]$ .

Two things follow at once. First, $Δ_{r} G^{ominus}$ is simply $Δ_{r} G$ when every activity is $1$ , which is what "standard conditions" means and why a standard free energy change never on its own tells you what a real flask will do. Second, at equilibrium the slope is zero, and the value $Q$ has taken there is by definition the equilibrium constant $K$ :

Δ_{r} G^{ominus} = - R T \ln K

This is the bridge the whole course has been building toward. On the left are tabulated enthalpies and entropies measured in a calorimeter; on the right is the composition a reactor will actually reach. $K$ is dimensionless, because every activity is, and it depends on temperature alone: not on pressure, not on the starting amounts, not on whether a catalyst is present.

The exponential makes it a violent amplifier. At $298$ K, $R T = 2.479$ kJ mol⁻¹, so every $5.7$ kJ mol⁻¹ of free energy is a factor of ten in $K$ .

Example. Ammonia synthesis has $Δ_{r} G^{ominus} = - 32.5$ kJ mol⁻¹ at $298.15$ K. Find $K$ . Then do the same for the combustion of methane, $Δ_{r} G^{ominus} = - 818.0$ kJ mol⁻¹.

For ammonia, $\ln K = 32500 / (8.314 \times 298.15) = 13.11$ , so $K = 5.0 \times 10^{5}$ . Equilibrium lies well over toward ammonia, which is a fact about the room temperature reaction that no ammonia plant has ever been able to use. For methane, $\ln K = 818000 / 2479 = 330$ , so $K = 2 \times 10^{143}$ . There is no meaningful sense in which any methane remains, and equally no sense in which a bottle of methane and air is in danger of reacting by itself.

Now you. Dinitrogen tetroxide dissociates as $(g) \to 2 N(g)$ , with $Δ_{f} G^{ominus} = 97.9$ kJ mol⁻¹ for and $51.3$ for $N$ . Find $Δ_{r} G^{ominus}$ and $K$ at $298.15$ K.

Answer

$Δ_{r} G^{ominus} = 2 (51.3) - 97.9 = + 4.7$ kJ mol⁻¹. Then $\ln K = - 4700 / 2479 = - 1.896$ , so $K = 0.150$ . Positive but small, so at standard pressure the mixture is mostly the dimer with an appreciable amount of the brown monomer, which is what the tube looks like.

Computing an equilibrium composition

With $K$ in hand the composition follows from a bookkeeping table: initial amounts, change in terms of one unknown, equilibrium amounts. The unknown is the extent of reaction in whatever units the constant uses, and the equation to solve is $K$ set equal to the quotient of those equilibrium values.

Example. $1.00$ mol of is placed in a $10.0$ dm³ vessel at $298.15$ K, where $K = 0.150$ . Find the equilibrium partial pressures and the degree of dissociation.

The initial pressure is $n R T / V = 1.00 \times 0.083145 \times 298.15 / 10.0 = 2.479$ bar, using $R$ in bar dm³ mol⁻¹ K⁻¹. Let $x$ bar of dissociate. Then at equilibrium $p () = 2.479 - x$ and $p (N) = 2 x$ , and since the standard pressure is $1$ bar the activities are these numbers. So

K = \frac{(2 x)^{2}}{2.479 - x} = 0.150

which rearranges to $4 x^{2} + 0.150 x - 0.3719 = 0$ and gives $x = 0.287$ bar. The equilibrium pressures are $p (N) = 0.573$ bar and $p () = 2.192$ bar, and the degree of dissociation is $0.287 / 2.479 = 11.6$ per cent. Checking, ${0.573}^{2} / 2.192 = 0.150$ .

Now you. Repeat with the same $1.00$ mol in a $40.0$ dm³ vessel, so the initial pressure is $0.620$ bar. Find the degree of dissociation and compare.

Answer

Now $4 x^{2} + 0.150 x - 0.0930 = 0$ , giving $x = 0.1348$ bar, so $p (N) = 0.270$ and $p () = 0.485$ bar. The degree of dissociation is $0.1348 / 0.620 = 21.8$ per cent, nearly double. $K$ has not changed at all; the composition has, because diluting a reaction that makes more molecules pushes it forward.

That last observation is the whole of the next lesson in miniature. The constant stayed constant and the answer moved.

Where the constant is easier than it looks

Two shortcuts save a great deal of algebra. When $K$ is very small, the change $x$ is small compared with the initial amounts, so the denominators can be left at their starting values, giving an explicit answer rather than a quadratic. The rule of thumb is that this is safe when $x$ comes out below five per cent of the smallest initial value, and the discipline is to compute $x$ and then check that, because when the approximation fails it fails silently. Weak acids in the ninth lesson are where this matters most.

The second is that a constant can be assembled from others. Reverse a reaction and $K$ becomes $1 / K$ ; double the coefficients and it becomes $K^{2}$ ; add two reactions and their constants multiply. All three follow from $Δ_{r} G^{ominus} = - R T \ln K$ and the fact that free energies add while logarithms turn addition into multiplication.

One conversion is worth stating explicitly. If a gas phase constant is written in concentrations rather than pressures, the two differ by the ideal gas relation $p = c R T$ applied to each species, so $K_{p} = K_{c} (R T / p^{ominus})^{Δ n}$ where $Δ n$ is the change in moles of gas. When $Δ n = 0$ , as in the hydrogen and iodine reaction below, the two are numerically the same and the distinction can be ignored.

Q against K says which way

The reaction quotient is defined for any composition, not just the equilibrium one, and comparing it with $K$ answers the practical question directly. From $Δ_{r} G = R T \ln (Q / K)$ , obtained by substituting the definition of $K$ into the earlier expression, the sign of $Δ_{r} G$ is the sign of $\ln (Q / K)$ . If $Q < K$ there is too little product and the reaction runs forward; if $Q > K$ it runs backward; if $Q = K$ nothing happens on average, though both directions continue at the molecular level.

Example. Hydrogen and iodine react as $+ \to 2 HI$ with $K = 54.3$ at $698$ K. A vessel holds all three at $0.0100$ , $0.0100$ and $0.0200$ mol dm⁻³. Which way does it go, and where does it end if it starts instead from $1.00$ mol dm⁻³ each of hydrogen and iodine?

$Q = {0.0200}^{2} / (0.0100 \times 0.0100) = 4.0$ , well below $54.3$ , so the mixture makes more hydrogen iodide. Starting from $1.00$ each, let $x$ react: at equilibrium the reactants are $1.00 - x$ each and the product is $2 x$ , so $K = 4 x^{2} / (1 - x)^{2}$ . Since both sides are perfect squares, take the square root: $2 x / (1 - x) = \sqrt{54.3} = 7.369$ , giving $x = 0.787$ . The equilibrium concentrations are $0.213$ mol dm⁻³ for each reactant and $1.573$ for hydrogen iodide.

Now you. For the same reaction and constant, a vessel holds at $0.0500$ , at $0.0200$ and $HI$ at $0.400$ mol dm⁻³. Which way does the reaction run?

Answer

$Q = {0.400}^{2} / (0.0500 \times 0.0200) = 0.160 / 0.00100 = 160$ , which exceeds $54.3$ , so $Δ_{r} G > 0$ in the forward direction and hydrogen iodide decomposes until $Q$ falls to $54.3$ .

What moves it, and what does not

Everything in this lesson holds the temperature fixed, because $K$ is a function of temperature and of nothing else. Change the amounts and $Q$ moves while $K$ stays put, and the system responds by returning $Q$ to $K$ . Change the temperature and $K$ itself moves, which is a different kind of change and needs a different equation.

That distinction, together with the case of compressing a gas mixture, where the amounts do not change but the activities do, is the content of the next lesson. Applied to the synthesis of ammonia the three answers pull against each other, and the resolution is one of the more consequential compromises in industrial history.

Moving an equilibrium

A chemical plant is a machine for moving an equilibrium, and the three levers available to it push in directions that do not agree.

The previous lesson fixed the position of equilibrium with $Δ_{r} G^{ominus} = - R T \ln K$ , and ended with a hint: diluting dinitrogen tetroxide nearly doubled its dissociation while $K$ did not move at all. That is the pattern to generalise. Some changes move the composition and leave the constant alone; only one moves the constant.

Le Chatelier's principle, and why it needs care

Henri Le Chatelier stated the rule in 1884: a system at equilibrium, when disturbed, shifts so as to partially oppose the disturbance. Add a reactant and it is consumed; heat an exothermic reaction and it runs backward; compress a mixture and it moves toward fewer molecules.

The principle is genuinely useful and genuinely unreliable, and it is worth knowing which parts of it are which. It gives a direction and never a magnitude, so it cannot tell you whether a shift is worth engineering. It is stated in terms of "the disturbance", which is not well defined when several things change at once. And it has honest counterexamples: add nitrogen to an ammonia equilibrium that is already very rich in nitrogen and the mole fraction of ammonia goes down, not up, because the dilution of the hydrogen outweighs the mass action of the added nitrogen.

Everything the principle gets right follows from two facts already established, so the safe procedure is to use those instead. First, $K$ depends on temperature and on nothing else. Second, the system moves so as to return $Q$ to $K$ . Every case below is one of those two.

Adding a reagent: $Q$ moves and $K$ does not

Pour more of a reactant into a mixture at equilibrium and its activity rises, so $Q$ falls below $K$ , so $Δ_{r} G$ becomes negative and the reaction runs forward until $Q$ has climbed back. The new composition is not the old one plus the addition; it is a fresh equilibrium calculation with the disturbed mixture as the starting point.

Example. The mixture $+ \to 2 HI$ at $698$ K, with $K = 54.3$ , sits at $[] = [] = 0.213$ and $[HI] = 1.573$ mol dm⁻³. Now add $1.000$ mol dm⁻³ of hydrogen. Find the new equilibrium.

Immediately after the addition, $Q = {1.573}^{2} / (1.213 \times 0.213) = 9.58$ , well below $54.3$ , so the reaction goes forward. Let $x$ be the amount of iodine consumed:

\frac{(1.573 + 2 x)^{2}}{(1.213 - x) (0.213 - x)} = 54.3

Solving gives $x = 0.152$ mol dm⁻³, so the new concentrations are $[] = 1.061$ , $[] = 0.061$ and $[HI] = 1.877$ mol dm⁻³. The added hydrogen converted most of the remaining iodine, taking it from $0.213$ to $0.061$ : this is the standard industrial trick of driving a reaction to completion in the expensive reagent by flooding it with the cheap one.

Now you. Without solving anything, say what happens to that same equilibrium if hydrogen iodide is continuously removed as it forms, and why a plant might do that.

Answer

Removing the product lowers $Q$ below $K$ continuously, so the reaction never reaches equilibrium and keeps running forward. In principle it can be driven to any conversion, limited only by the rate. This is why ammonia is condensed out of the recycle loop and why esterifications are run with the water distilled off.

Squeezing a gas mixture

Compression is the interesting case, because the amounts of substance do not change at all. What changes is the activity of each gas, which is its partial pressure, and the activities appear in $Q$ with different powers on the two sides.

Write the constant in terms of mole fractions and total pressure. Since $p_{i} = x_{i} P$ , each activity carries a factor of $P$ , and collecting them gives

K = K_{x} {(\frac{P}{p^{ominus}})}^{Δ n}

where $K_{x}$ is the same quotient in mole fractions and $Δ n$ is the change in moles of gas. So if $Δ n$ is positive, raising $P$ must lower $K_{x}$ , and the mixture shifts toward the reactants; if $Δ n$ is zero, pressure does nothing whatever. That is Le Chatelier's pressure rule, derived, with the exception it always forgets attached.

Example. For $\to 2 N$ at $298$ K, $K = 0.150$ . Find the degree of dissociation $α$ at total pressures of $1.00$ bar and $10.0$ bar.

Starting from one mole of the dimer, $α$ moles dissociate to give $2 α$ of the monomer and $1 - α$ of the dimer, totalling $1 + α$ . The mole fractions are $2 α / (1 + α)$ and $(1 - α) / (1 + α)$ , so

K = \frac{4 α^{2}}{1 - α^{2}} \cdot \frac{P}{p^{ominus}}

using the difference of two squares to tidy the denominator. At $P = 1.00$ bar, $4 α^{2} / (1 - α^{2}) = 0.150$ , giving $α^{2} = 0.150 / 4.150$ and $α = 0.190$ . At $P = 10.0$ bar the left side must equal $0.0150$ , giving $α = 0.061$ . Compressing tenfold cuts the dissociation from $19$ per cent to $6$ .

Now you. Find $α$ for the same equilibrium at $P = 0.100$ bar.

Answer

Now $4 α^{2} / (1 - α^{2}) = 0.150 / 0.100 = 1.50$ , so $α^{2} = 1.50 / 5.50 = 0.273$ and $α = 0.522$ . At a tenth of a bar the gas is more than half dissociated, and the tube looks distinctly brown.

One trap is worth stating plainly. Pumping an inert gas such as argon into a vessel of fixed volume raises the total pressure and shifts nothing at all, because the partial pressures of the reacting species are unchanged and $Q$ is built from those. Adding argon while holding the total pressure fixed, so the vessel expands, is a genuine dilution and shifts the equilibrium exactly as expansion does. The total pressure is not what the system responds to.

Temperature is the only lever that moves $K$

Combine the two expressions for the standard free energy change, $Δ_{r} G^{ominus} = Δ_{r} H^{ominus} - T Δ_{r} S^{ominus}$ and $Δ_{r} G^{ominus} = - R T \ln K$ , and divide through by $- R T$ :

\ln K = - \frac{Δ_{r} H^{ominus}}{R T} + \frac{Δ_{r} S^{ominus}}{R}

Everything about the temperature dependence is in that one line. Treating $Δ_{r} H^{ominus}$ and $Δ_{r} S^{ominus}$ as constant over the range of interest and writing it at two temperatures, the entropy term cancels and leaves the van 't Hoff equation:

\ln \frac{K_{2}}{K_{1}} = - \frac{Δ_{r} H^{ominus}}{R} (\frac{1}{T_{2}} - \frac{1}{T_{1}})

Read the sign. For an endothermic reaction $Δ H$ is positive, so raising $T$ makes $1 / T$ smaller and the bracket negative, and $K$ increases. For an exothermic one $K$ falls. Le Chatelier's temperature rule, with the additional information that the size of the shift is set by $Δ H$ and by nothing else: a reaction with a small enthalpy change is nearly indifferent to temperature however favourable it is.

Example. $\to 2 N$ has $K = 0.150$ at $298.15$ K and $Δ_{r} H^{ominus} = 2 (33.2) - 9.16 = + 57.2$ kJ mol⁻¹. Find $K$ at $350$ K.

\ln \frac{K_{2}}{K_{1}} = - \frac{57240}{8.314} (\frac{1}{350} - \frac{1}{298.15}) = - 6886 \times (- 4.969 \times 10^{- 4}) = 3.42

So $K_{2} = 0.150 \times e^{3.42} = 4.59$ . A rise of $52$ K multiplies the constant by thirty, which is why the sealed tube of nitrogen dioxide used in lecture demonstrations goes from pale to deep brown when dropped in hot water.

Now you. Ammonia synthesis has $K = 5.0 \times 10^{5}$ at $298.15$ K and $Δ_{r} H^{ominus} = - 91.8$ kJ mol⁻¹. Estimate $K$ at $700$ K.

Answer

$\ln (K_{2} / K_{1}) = (91800 / 8.314) (1 / 700 - 1 / 298.15) = 11042 \times (- 1.925 \times 10^{- 3}) = - 21.26$ , so $K_{2} = 5.0 \times 10^{5} \times e^{- 21.26} = 2.9 \times 10^{- 4}$ . Nine orders of magnitude lost, which is the central problem of the ammonia industry.

A thermometer that measures an enthalpy

The relation runs backward as well, and this is how a great many reaction enthalpies are actually obtained. Measure $K$ at several temperatures, plot $\ln K$ against $1 / T$ , and the result is a straight line of slope $- Δ_{r} H^{ominus} / R$ and intercept $Δ_{r} S^{ominus} / R$ . Both quantities come out of composition measurements, with no calorimeter anywhere in the experiment.

Two points suffice in principle. Using the pair above, $\ln (4.59 / 0.150) = 3.42$ over an interval of $1 / 350 - 1 / 298.15 = - 4.969 \times 10^{- 4}$ K⁻¹ gives a slope of $- 6886$ K, so $Δ_{r} H^{ominus} = 6886 \times 8.314 = 57.2$ kJ mol⁻¹, recovering what we put in.

In practice a real plot uses many temperatures and its curvature is informative. A van 't Hoff plot is straight only while $Δ_{r} H^{ominus}$ is constant, and Kirchhoff's law from the second lesson says it is not, so any measurement over a few hundred kelvin bends slightly. The curvature is a measurement of $Δ_{r} C_{p}$ , which is otherwise awkward to get. This is the routine way biochemists obtain binding enthalpies for a protein and a ligand, where the reaction is too dilute to warm anything measurably.

Ammonia: three answers that disagree

Now put the three levers on one reaction. Nitrogen and hydrogen make ammonia with $Δ_{r} H^{ominus} = - 91.8$ kJ mol⁻¹ and $Δ n = - 2$ . So the equilibrium wants a low temperature, since the reaction is exothermic, and a high pressure, since it consumes gas. Taking $\ln K = 11042 / T - 23.91$ from the standard values and solving for the equilibrium mole fraction of ammonia from a stoichiometric feed gives:

Temperature	at 10 bar	at 200 bar
$473$ K (200 °C)	$53$ %	$87$ %
$573$ K (300 °C)	$20$ %	$68$ %
$673$ K (400 °C)	$7$ %	$46$ %
$773$ K (500 °C)	$2$ %	$28$ %

These are computed from thermodynamic data alone and agree with the measured equilibrium yields to within two or three percentage points across the whole table, which is a reasonable check on everything in the last three lessons.

Read the first column and the answer looks obvious: run it cold. At $473$ K and $10$ bar you keep half the feed as ammonia with a cheap vessel. Fritz Haber's problem, from 1905 onward, was that at $473$ K the reaction does not happen. Nitrogen's triple bond is $945$ kJ mol⁻¹ and nothing at $200$ °C breaks it at a usable rate. The equilibrium is inviting and inaccessible.

The industrial solution is a compromise on all three axes at once. Run hot enough for the iron catalyst to work, $400$ to $450$ °C, accept that this costs most of the equilibrium yield, and buy some of it back with pressure, $150$ to $250$ bar, which is expensive in steel and in compression work. Then, because even the equilibrium yield is not reached in one pass through the converter, condense the ammonia out of the product stream and recycle the unreacted gas, which is the trick from the second section of this lesson used to push the overall conversion above $95$ per cent. Carl Bosch's engineering of that loop between 1909 and 1913, and Alwin Mittasch's search across some twenty thousand catalyst samples, are what turned Haber's bench result into a process that now fixes more nitrogen than the entire biosphere.

What a yield chart cannot tell you

Every number in that table is a statement about where the mixture ends up, and not one of them is a statement about when. The $473$ K column is the proof: thermodynamics says a mixture of nitrogen and hydrogen at $200$ °C and $10$ bar should be half ammonia, and the experiment says that after a week it is not measurably anything.

Nothing in the last three lessons can be repaired to fix this, because free energy is a function of the end states and knows nothing about the route between them. The rate depends on the route: on which bonds have to break first, on how much energy a collision needs, on whether a surface is available to help. It is a separate theory with separate measurements, and it starts in the next lesson with the plainest possible question, how the composition of a reacting mixture changes with time.

Rate laws

Thermodynamics says a mixture of hydrogen and oxygen should be water, and the mixture will sit in a flask for a thousand years without becoming any, so the question of how fast is a separate science with its own measurements.

The previous five lessons answered where a reaction ends up. None of them can be adapted to say when, because free energy depends only on the initial and final states and a rate depends entirely on the route between them. This lesson starts that second theory from the plainest possible measurement: the composition of a mixture as a function of time.

A rate everyone can agree on

Take $2 \to 4 N +$ . Nitrogen dioxide appears four times as fast as oxygen does, so "the rate" is ambiguous until we say the rate of what. The fix is the extent of reaction from the first lesson. Every amount obeys $n_{i} = n_{i, 0} + ν_{i} ξ$ , so dividing each rate of change by the stoichiometric number gives one quantity that all of them share:

v = \frac{1}{V} \frac{d ξ}{d t} = \frac{1}{ν_{i}} \frac{d [i]}{d t}

For the reaction above, $v = - \frac{1}{2} d [] / d t = \frac{1}{4} d [N] / d t = d [] / d t$ . The sign convention is built in, since $ν$ is negative for reactants, so $v$ comes out positive for a reaction running forward. Its units are mol dm⁻³ s⁻¹.

The division by volume matters when the volume changes, and is a nuisance rather than a principle. Everything below is at constant volume, where concentrations behave.

Measuring one

A rate is a slope, so measuring one means following a concentration in time without disturbing the mixture. The methods divide by how fast the reaction is.

Anything with a coloured species is followed by absorbance, since the Beer-Lambert law makes absorbance proportional to concentration, and a spectrophotometer samples continuously without removing anything. A reaction that changes the number of moles of gas is followed by pressure at constant volume, which is how the decomposition of dinitrogen pentoxide was originally studied. Reactions that change the ion count are followed by conductivity, and those involving a chiral substance by optical rotation.

Slower reactions can be sampled instead: withdraw an aliquot, quench it by chilling or by destroying the catalyst, and analyse at leisure. Faster ones need the stopped-flow method, which mixes two streams in about a millisecond and watches the mixing chamber, or, below that, flash photolysis, which starts the reaction with a light pulse. Manfred Eigen, Ronald Norrish and George Porter shared the 1967 Nobel Prize for pushing the accessible range down to microseconds; laser techniques have since reached femtoseconds, fast enough to watch a bond break.

For the arithmetic, two experimental designs recur. In the initial rates method the reaction is run several times from different starting concentrations and only the first few per cent is used, so the concentrations are still known and no product has accumulated. In the integrated method one run is followed to substantial conversion and the whole curve is fitted. The next two sections take them in that order.

The rate law is an experimental fact

For many reactions the rate turns out to depend on concentrations as a product of powers:

v = k [A]^{a} [B]^{b}

Here $k$ is the rate constant, which depends on temperature but not on concentration, and the exponents are the orders: order $a$ with respect to A, and $a + b$ overall. The units of $k$ are whatever makes the equation dimensionally consistent, so they change with the overall order, which is a useful check on an answer.

The essential point, and the one most often got wrong, is that the orders are measured and not derived. They are not the stoichiometric coefficients. The decomposition of dinitrogen pentoxide has a coefficient of $2$ and is first order. The gas phase reaction of hydrogen with iodine has coefficients of $1$ and $1$ and happens to be first order in each, which for a century was taken as evidence that it was a single collision, wrongly. And $2 \to 3$ has the rate law $v = k []^{2} []^{- 1}$ , in which a product appears with a negative order, meaning oxygen inhibits its own formation. No reading of the balanced equation predicts that.

Some rate laws are not of this form at all. The formation of hydrogen bromide from its elements obeys

v = \frac{k [] [B]^{1 / 2}}{1 + k' [HBr] / [B]}

measured by Max Bodenstein in 1906, which has no single order and changes shape as the reaction proceeds. A rate law like that is not a nuisance: it is a fingerprint of the mechanism, and the eighth lesson shows how to read it.

Orders by initial rates

The method is to change one concentration at a time and see what the rate does. Doubling a concentration doubles a first order rate, quadruples a second order one, and leaves a zero order one alone.

Example. The oxidation of iodide by peroxodisulfate, $+ 2 \to 2 S +$ , gives these initial rates at $298$ K.

$[]$ / mol dm⁻³	$[]$ / mol dm⁻³	rate / mol dm⁻³ s⁻¹
$0.038$	$0.060$	$1.4 \times 10^{- 5}$
$0.076$	$0.060$	$2.8 \times 10^{- 5}$
$0.076$	$0.030$	$1.4 \times 10^{- 5}$

Find the rate law and $k$ .

Between the first two rows the iodide is fixed and the persulfate doubles, and the rate doubles: first order in persulfate. Between the second and third the persulfate is fixed and the iodide halves, and the rate halves: first order in iodide. So $v = k [] []$ , second order overall, and $k = 1.4 \times 10^{- 5} / (0.038 \times 0.060) = 6.1 \times 10^{- 3}$ dm³ mol⁻¹ s⁻¹. Note that the order in iodide is one although the equation needs two of them.

Now you. For $2 NO + \to 2 N$ at $298$ K, doubling $[NO]$ from $0.0100$ to $0.0200$ mol dm⁻³ at fixed $[] = 0.0100$ raises the rate from $7.1 \times 10^{- 3}$ to $2.84 \times 10^{- 2}$ mol dm⁻³ s⁻¹, and doubling $[]$ instead raises it to $1.42 \times 10^{- 2}$ . Give the rate law, the overall order and $k$ with its units.

Answer

Doubling nitric oxide multiplies the rate by four, so the order in it is two; doubling oxygen doubles the rate, so that order is one. Thus $v = k [NO]^{2} []$ , third order overall, and $k = 7.1 \times 10^{- 3} / ({0.0100}^{2} \times 0.0100) = 7.1 \times 10^{3}$ dm⁶ mol⁻² s⁻¹.

Integrating the first order law

The other approach follows one run to completion, which needs the rate law solved as a differential equation. First order is the case worth doing in full, because it recurs everywhere.

Write $- d [A] / d t = k [A]$ , separate the variables and integrate from the start to time $t$ :

\ln \frac{[A]}{[A]_{0}} = - k t so [A] = [A]_{0} e^{- k t}

Two consequences are worth naming. A plot of $\ln [A]$ against $t$ is a straight line of slope $- k$ , which is the test for first order behaviour. And the half-life, the time to fall to half of any starting value, is

t_{1 / 2} = \frac{\ln 2}{k}

independent of concentration. That independence is the signature: a first order reaction takes as long to go from $0.01$ to $0.005$ mol dm⁻³ as from $1.0$ to $0.5$ . It is why radioactive decay, which is first order by nature, has a half-life worth tabulating, and why a drug cleared by a first order process has a dose-independent elimination time.

Example. Dinitrogen pentoxide decomposes with $k = 4.87 \times 10^{- 3}$ s⁻¹ at $65$ °C. Starting from $0.1000$ mol dm⁻³, find the half-life and the concentration after $300$ s.

The half-life is $\ln 2 / k = 0.6931 / 4.87 \times 10^{- 3} = 142$ s. After $300$ s, $[A] = 0.1000 e^{- 4.87 \times 10^{- 3} \times 300} = 0.1000 e^{- 1.461} = 0.0232$ mol dm⁻³. As a check, $300$ s is a little over two half-lives, and two half-lives would leave $0.025$ .

Now you. For the same reaction and rate constant, how long does it take for $90$ per cent of the pentoxide to decompose, and does the answer depend on the starting concentration?

Answer

Ninety per cent gone leaves a tenth, so $k t = \ln 10 = 2.303$ and $t = 2.303 / 4.87 \times 10^{- 3} = 473$ s. It does not depend on the starting concentration, because only the ratio appears in the integrated law.

Zero and second order, and how to tell them apart

The same integration for other orders gives different straight lines, and comparing which plot is straight is how an order is assigned from a single run.

Order	Integrated law	Straight line	Half-life
$0$	$[A] = [A]_{0} - k t$	$[A]$ against $t$	$[A]_{0} / 2 k$
$1$	$\ln [A] = \ln [A]_{0} - k t$	$\ln [A]$ against $t$	$\ln 2 / k$
$2$	$1 / [A] = 1 / [A]_{0} + k t$	$1 / [A]$ against $t$	$1 / (k [A]_{0})$

The half-life column is the quickest diagnostic. A zero order reaction's successive half-lives get shorter, each half of the one before; a first order reaction's are all equal; a second order reaction's double each time. Watching a decay curve halve three times and noting whether the intervals grow, stay or shrink identifies the order before any plotting.

Zero order sounds strange but is common wherever something other than the reactant concentration is the bottleneck: a saturated enzyme, a saturated catalyst surface, a photochemical reaction limited by the light supply. Ethanol is cleared from human blood at close to zero order above about $0.02$ per cent by volume, because the alcohol dehydrogenase is saturated, which is why blood alcohol falls linearly rather than exponentially.

Example. These concentrations were recorded for a decomposition at constant temperature.

$t$ / s	$0$	$100$	$200$	$300$
$[A]$ / mol dm⁻³	$0.01000$	$0.00648$	$0.00479$	$0.00380$

Find the order and the rate constant.

Test first order: the ratio over the first interval is $0.648$ and over the second $0.739$ , which are not equal, so the concentration is not falling exponentially. Test second order by taking reciprocals: $100.0$ , $154.3$ , $208.6$ , $262.9$ dm³ mol⁻¹. Those rise by exactly $54.3$ every $100$ s, so $1 / [A]$ is linear in time and the reaction is second order with $k = 54.3 / 100 = 0.543$ dm³ mol⁻¹ s⁻¹.

Now you. Another run gives $0.1000$ , $0.0614$ , $0.0378$ , $0.0232$ and $0.0143$ mol dm⁻³ at $0$ , $100$ , $200$ , $300$ and $400$ s. Find the order and the rate constant.

Answer

Each interval multiplies the concentration by the same factor, $0.614$ , so the decay is exponential and the reaction is first order. Then $k = - \ln (0.614) / 100 = 4.88 \times 10^{- 3}$ s⁻¹, and the half-life is $\ln 2 / k = 142$ s, consistent with the concentration falling from $0.1000$ to about $0.05$ somewhere between $100$ and $200$ s.

What the orders are hiding

Two honest limits close the lesson.

The first is that every integrated law above ignores the reverse reaction, so it describes only the early part of a run, before enough product has built up for the mixture to notice equilibrium. Near equilibrium the net rate falls to zero while both directions continue at full speed, and the correct treatment subtracts the reverse rate. This is why kinetic runs are usually analysed over the first fraction of the reaction, and why a reaction with a small $K$ cannot be studied this way at all without special handling.

The second is more interesting. Nothing so far explains why simple integer orders should turn up at all, why the order in iodide should be one when the equation needs two, or where a $- 1$ or a $1 / 2$ comes from. A rate law is a summary of a mechanism, and the orders are evidence about which molecules meet in the slowest step. Extracting the mechanism from the rate law is the eighth lesson.

Before that, there is a variable the rate constant is far more sensitive to than any concentration. Raising the temperature of a mixture by ten degrees typically doubles its rate, which is a much steeper response than any concentration effect, and it is the subject of the next lesson.

Temperature and activation energy

Warming a mixture by ten kelvin, a change of about three per cent in absolute temperature, commonly doubles the rate of a reaction, and no amount of pushing on concentrations produces a response that steep.

The previous lesson defined the rate constant $k$ and treated it as a fixed number for a given reaction. It is fixed only at a fixed temperature. This lesson is about how it varies, what that variation measures, and what it reveals about how a reaction happens at all.

The steepest dependence in chemistry

Doubling a concentration doubles a first order rate: an effect proportional to the change. Warming a reaction from $300$ K to $310$ K, which raises the average molecular kinetic energy by about three per cent, can double it too. Something is amplifying a small change enormously, and an exponential is the only function that behaves like that.

The mechanism of the amplification is that a reaction does not use the average molecule. It uses the rare ones in the high energy tail of the distribution, and the population of that tail is exponentially sensitive to temperature. The fraction of molecules with energy above $100$ kJ mol⁻¹ at $298$ K, using the Boltzmann factor $e^{- E / R T}$ , is $3.0 \times 10^{- 18}$ . At $308$ K it is $1.1 \times 10^{- 17}$ . Nothing about the bulk has changed appreciably, and the population that can react has nearly quadrupled.

The Arrhenius equation

Svante Arrhenius put this together in 1889, generalising an empirical fit of Jacobus van 't Hoff's. He proposed that the rate constant has the form

k = A e^{- E_{a} / R T}

with two parameters. $E_{a}$ is the activation energy, an energy barrier that reacting molecules must carry into the collision, and $A$ is the pre-exponential factor, which has the same units as $k$ and represents how often the encounter happens at all. The exponential is a Boltzmann factor: it is the fraction of encounters energetic enough to matter.

The immediate consequence is that taking logarithms gives a straight line,

\ln k = \ln A - \frac{E_{a}}{R} \cdot \frac{1}{T}

so plotting $\ln k$ against $1 / T$ gives a slope of $- E_{a} / R$ and an intercept of $\ln A$ . This is the Arrhenius plot, and it is how essentially every activation energy in the literature was obtained. Note its resemblance to the van 't Hoff plot of the fifth lesson, which is a real family likeness and not a coincidence: both are the logarithm of a Boltzmann factor against reciprocal temperature. They are not the same quantity, though. The van 't Hoff slope gives $Δ H$ , a difference between two states; the Arrhenius slope gives $E_{a}$ , the height of a barrier between them.

The familiar rule that ten degrees doubles a rate is a numerical accident, not a law. At $300$ K a barrier of $50$ kJ mol⁻¹ gives a factor of $1.9$ for a ten kelvin rise, which is where the rule comes from; a barrier of $100$ kJ mol⁻¹ gives $3.6$ , and a very low barrier gives almost nothing. The rule holds because activation energies around $50$ kJ mol⁻¹ are common in the reactions people happen to run near room temperature.

Measuring the barrier

Two rate constants at two temperatures are enough, by writing the equation twice and subtracting:

\ln \frac{k_{2}}{k_{1}} = \frac{E_{a}}{R} (\frac{1}{T_{1}} - \frac{1}{T_{2}})

Example. Dinitrogen pentoxide decomposes with $k = 3.38 \times 10^{- 5}$ s⁻¹ at $25$ °C and $4.87 \times 10^{- 3}$ s⁻¹ at $65$ °C. Find $E_{a}$ and $A$ .

The ratio of the constants is $144$ , so $\ln (k_{2} / k_{1}) = 4.97$ . The reciprocal temperatures are $1 / 298.15 = 3.3540 \times 10^{- 3}$ and $1 / 338.15 = 2.9572 \times 10^{- 3}$ K⁻¹, differing by $3.968 \times 10^{- 4}$ . Hence $E_{a} = 8.314 \times 4.97 / 3.968 \times 10^{- 4} = 1.042 \times 10^{5}$ J mol⁻¹, or $104$ kJ mol⁻¹. For $A$ , substitute back at either temperature: $A = k e^{E_{a} / R T} = 3.38 \times 10^{- 5} \times e^{42.02} = 6.0 \times 10^{13}$ s⁻¹.

That value of $A$ is worth a moment. For a unimolecular decomposition, $10^{13}$ s⁻¹ is roughly a molecular vibration frequency, which is exactly what it should be: the molecule tries to fall apart once per vibration and succeeds with the Boltzmann probability.

Now you. A reaction has $k = 1.20 \times 10^{- 4}$ s⁻¹ at $300$ K and $2.40 \times 10^{- 4}$ s⁻¹ at $310$ K. Find $E_{a}$ .

Answer

$\ln (k_{2} / k_{1}) = \ln 2 = 0.693$ , and $1 / 300 - 1 / 310 = 1.0753 \times 10^{- 4}$ K⁻¹. So $E_{a} = 8.314 \times 0.693 / 1.0753 \times 10^{- 4} = 5.36 \times 10^{4}$ J mol⁻¹, about $53.6$ kJ mol⁻¹. This is the barrier the ten degree rule secretly assumes.

Collision theory: the exponential is right and the prefactor is not

For a gas phase bimolecular reaction, $A$ can be computed rather than fitted, which is a genuine test of the picture. Kinetic theory gives the rate at which two species collide: the collision cross section $σ$ times the mean relative speed times the number densities. Per mole,

A_{coll} = σ ⟨ v_{rel} ⟩ N_{A} with ⟨ v_{rel} ⟩ = \sqrt{\frac{8 R T}{π μ}}

where $μ$ is the reduced molar mass. Every quantity is measurable independently, from viscosity data for $σ$ and from the formula for $μ$ .

Example. Estimate $A$ for $+ \to$ at $628$ K, with $σ = 0.46$ nm², and compare with the measured $A = 1.24 \times 10^{6}$ dm³ mol⁻¹ s⁻¹.

The reduced molar mass is $(2.016 \times 28.05) / (2.016 + 28.05) = 1.881$ g mol⁻¹, so $⟨ v_{rel} ⟩ = \sqrt{8 \times 8.314 \times 628 / (π \times 1.881 \times 10^{- 3})} = 2659$ m s⁻¹. Then $A_{coll} = 4.6 \times 10^{- 19} \times 2659 \times 6.022 \times 10^{23} = 7.4 \times 10^{8}$ m³ mol⁻¹ s⁻¹, which is $7.4 \times 10^{11}$ dm³ mol⁻¹ s⁻¹. The measured value is smaller by a factor of $1.7 \times 10^{- 6}$ .

Now you. The reaction $K + B \to KBr + Br$ has a measured $A$ about $4.8$ times its calculated collision value. What does a steric factor greater than one imply about the encounter?

Answer

The reaction happens more often than the two species collide, so the reactants must be interacting before they touch. This is the harpoon mechanism: an electron jumps from potassium to bromine at long range and the resulting ions are pulled together, giving an effective cross section much larger than the physical one.

That ratio of observed to calculated prefactor is the steric factor $P$ , and its size is the honest verdict on collision theory. The theory gets the temperature dependence right and the absolute rate wrong by up to six orders of magnitude, because it models molecules as featureless spheres for which any collision counts. Real molecules have to meet in a particular orientation, and the more complicated they are, the smaller the fraction of encounters that qualifies. $P$ is a fudge factor with a physical interpretation, which is better than a fudge factor without one, but it cannot be predicted from within the theory.

The transition state

The better picture follows the reacting pair along a reaction coordinate, the path from reactants to products through the configuration of highest energy. That maximum is the transition state, an arrangement in which old bonds are partly broken and new ones partly formed. It is not an intermediate: it sits at a maximum, not a minimum, and has no lifetime beyond a single vibration.

Drawing the energy against the reaction coordinate makes the relations plain. The barrier from the reactant side is the forward activation energy, the barrier from the product side the reverse one, and the difference between the two ends is the enthalpy change of the reaction. Hence, for an elementary step,

E_{a, forward} - E_{a, reverse} = Δ H

An exothermic step has the lower barrier in the forward direction, which is the grain of truth in the old idea that exothermic reactions are fast. It is only a grain: the barrier can be large in both directions, as it is for hydrogen and oxygen, where $Δ H$ is $- 572$ kJ mol⁻¹ and the mixture is stable indefinitely.

Henry Eyring and Michael Polanyi's transition state theory of 1935 treats the activated complex as being in equilibrium with the reactants and derives the rate constant from the free energy of activation, $Δ G^{‡}$ . Splitting that into $Δ H^{‡}$ and $Δ S^{‡}$ gives the prefactor a meaning: $A$ is controlled by the entropy of activation, which is strongly negative when two floppy molecules have to freeze into one rigid arrangement. That is the steric factor, arrived at from a theory instead of from a ratio.

Catalysis is a different route, not a push

A catalyst provides an alternative path with a lower activation energy, participating in the mechanism and being regenerated by it. Two consequences follow immediately from the equations above, and both are worth stating carefully.

First, the effect on the rate is exponential in the barrier reduction, so modest reductions produce enormous accelerations. Second, a catalyst cannot change the equilibrium position at all. It lowers the forward and reverse barriers by exactly the same amount, because both routes pass over the same new transition state, so $k_{f}$ and $k_{r}$ change by the same factor and their ratio, which is $K$ , does not move. A catalyst gets you to the same place sooner. Any claim that one improves a yield beyond equilibrium is a claim that energy is being created.

Example. Hydrogen peroxide decomposes with $E_{a} = 76$ kJ mol⁻¹ uncatalysed, $57$ kJ mol⁻¹ with iodide ion, and about $8$ kJ mol⁻¹ with the enzyme catalase. By what factor does each catalyst multiply the rate at $298$ K, if $A$ is unchanged?

For iodide the barrier falls by $19$ kJ mol⁻¹, giving $e^{19000 / 2479} = e^{7.67} = 2.1 \times 10^{3}$ . For catalase it falls by $68$ kJ mol⁻¹, giving $e^{27.4} = 8.2 \times 10^{11}$ . A reaction that would take a thousand years takes a hundredth of a second, which is why a drop of blood on a peroxide-soaked cut foams.

Now you. A catalyst lowers the activation energy of a reaction by $27$ kJ mol⁻¹. By what factor is the rate multiplied at $298$ K, assuming $A$ is unchanged?

Answer

$e^{27000 / (8.314 \times 298.15)} = e^{10.89} = 5.4 \times 10^{4}$ , a factor of about fifty thousand.

The assumption that $A$ is unchanged is a simplification, and often a bad one: a heterogeneous catalyst that binds a reactant to a surface changes the entropy of activation substantially. The barrier is the dominant term, but a full account has to include both.

What the barrier does not tell you

An activation energy is a property of a step, not of an equation. Measuring $E_{a} = 104$ kJ mol⁻¹ for the decomposition of dinitrogen pentoxide is a statement about whatever the slowest step of that decomposition is, and the balanced equation, with its two molecules of reactant, does not say what that step is.

This is the same gap the previous lesson found in the orders. Both the rate law and the activation energy are experimental summaries of something happening underneath, and both become interpretable only once the sequence of elementary steps is proposed. That sequence is a mechanism, and constructing one, deriving the rate law it predicts, and comparing it with the measurement, is the next lesson.

Reaction mechanisms

A balanced equation is a summary of a journey, and the rate law measured in the laboratory is evidence about the individual steps that journey is made of.

The sixth lesson found orders that bear no relation to the coefficients, and the seventh found an activation energy that belongs to no particular part of the equation. Both are symptoms of the same thing: reactions almost never happen the way they are written. This lesson builds the underlying sequence and shows how to test it.

Elementary steps and molecularity

An elementary step is a reaction that happens exactly as written, in a single encounter, with no intermediate. For an elementary step, and only for an elementary step, the rate law can be written down from the equation. A unimolecular step, one molecule falling apart or rearranging, is first order. A bimolecular step is first order in each of the two species, second order overall. That is not an empirical finding; it follows from what a collision is, since the chance of two particular molecules meeting is proportional to each concentration.

Termolecular steps, requiring three molecules at one point at one time, are possible but rare, because a triple encounter in a gas is orders of magnitude less likely than a double one. Anything requiring four is not seriously proposed. This is the strongest constraint on mechanism building: a balanced equation with large coefficients cannot possibly be elementary, so $2 \to 4 N +$ needs at least three separate steps whatever else is true.

A mechanism is a proposed set of elementary steps that add up to the overall equation. Species that are produced in one step and consumed in another are intermediates: real molecules with real, if short, lifetimes, unlike a transition state, which is a maximum on a path and has none.

The rate determining step

If one step in a sequence is much slower than the rest, the overall rate is that step's rate, and everything else is fast enough to be invisible. This is the rate determining step, and it is the simplest way a mechanism produces a rate law.

Example. Above $500$ K, $N + CO \to NO + C$ is observed to follow $v = k [N]^{2}$ , with carbon monoxide absent from the rate law entirely. Show that the mechanism

N + N \to N + NO (slow)

N + CO \to N + C (fast)

accounts for it.

First check the sum. Adding the two steps and cancelling one $N$ and the $N$ , which appear on both sides, gives $N + CO \to NO + C$ , the observed equation. Then the rate: the first step is bimolecular in nitrogen dioxide, so its rate is $k_{1} [N]^{2}$ , and since it is rate determining the overall rate is that. Carbon monoxide is consumed only in the fast step, which processes nitrate radical as soon as it appears, so its concentration cannot affect anything. The prediction matches, including the surprising absence.

Now you. The observed rate law for $2 N + \to 2 NF$ is $v = k [N] []$ . Is the single-step termolecular mechanism consistent with it? Propose a two step mechanism that is.

Answer

A single termolecular step would give $v = k [N]^{2} []$ , second order in nitrogen dioxide, which is not observed. A consistent mechanism is $N + \to NF + F$ (slow), followed by $N + F \to NF$ (fast). The steps sum correctly and the slow step gives exactly the observed rate law.

Notice what has and has not been shown. The mechanism is consistent with the data, which is the most any mechanism ever is. A rate law can rule mechanisms out; it can never prove one, since another sequence may predict the same law. Mechanisms are supported by detecting the intermediate, by isotopic labelling, by stereochemistry, and by the failure of every proposed alternative.

The steady state approximation

When no step is clearly slowest, the algebra needs a different tool. The steady state approximation assumes that a reactive intermediate, being consumed almost as fast as it is made, sits at a low and nearly constant concentration, so its net rate of change can be set to zero. Not because nothing is happening to it, but because formation and destruction have come into balance.

The classic application answers a question the sixth lesson left hanging: why is a unimolecular decomposition first order at all? A molecule cannot decompose unless something gives it the energy, and the only source is a collision, which is bimolecular. Frederick Lindemann's 1922 answer was that the two events are separate:

A + M rightleftharpoons A^{*} + M (k_{1}, k_{- 1}) A^{*} \to P (k_{2})

where $M$ is any collision partner and $A^{*}$ is an energised molecule. Apply the steady state to $A^{*}$ : it is formed at $k_{1} [A] [M]$ and destroyed at $k_{- 1} [A^{*}] [M] + k_{2} [A^{*}]$ . Setting the two equal and solving,

[A^{*}] = \frac{k_{1} [A] [M]}{k_{- 1} [M] + k_{2}} so v = k_{2} [A^{*}] = \frac{k_{1} k_{2} [A] [M]}{k_{- 1} [M] + k_{2}}

Example. Take the two limits of that expression and say what each predicts.

At high pressure $[M]$ is large, so $k_{- 1} [M]$ dominates $k_{2}$ in the denominator and $[M]$ cancels: $v = (k_{1} k_{2} / k_{- 1}) [A]$ , cleanly first order. Energisation is so frequent that it is effectively an equilibrium, and the decomposition of the energised molecule is what limits the rate. At low pressure $k_{2}$ dominates instead, leaving $v = k_{1} [A] [M]$ , second order: now collisions are rare, and every molecule that gets energised decomposes before it can be deactivated.

Now you. The observed first order rate constant is $k_{obs} = v / [A]$ . Show that $1 / k_{obs}$ is linear in $1 / [M]$ , and say why that is a useful test.

Answer

From the expression above, $k_{obs} = k_{1} k_{2} [M] / (k_{- 1} [M] + k_{2})$ , so inverting gives $1 / k_{obs} = k_{- 1} / (k_{1} k_{2}) + 1 / (k_{1} [M])$ . Plotting $1 / k_{obs}$ against $1 / [M]$ should give a straight line whose intercept and slope give the ratios of the rate constants. Real gas phase decompositions do fall off from first order at low pressure and the plot is roughly straight, which is why the mechanism is accepted; it curves at the low pressure end, which is why the modern treatment refines it.

A related shortcut is the pre-equilibrium, where a fast reversible step precedes a slow one. Then the first step stays at equilibrium and the intermediate's concentration is $K_{1}$ times the reactants'. The third order oxidation $2 NO + \to 2 N$ is the standard case: two nitric oxide molecules dimerise reversibly to , which then reacts with oxygen, giving $v = k_{2} K_{1} [NO]^{2} []$ without ever needing a triple collision. It also explains an oddity: this reaction gets slower as the temperature rises, because $K_{1}$ is for an exothermic dimerisation and falls with temperature faster than $k_{2}$ climbs.

Chain reactions

Some mechanisms regenerate their own intermediates, so a single initiation event drives many turnovers. The steps are classified as initiation, which creates the carriers, propagation, which consumes one and makes another, inhibition, which undoes propagation, and termination, which destroys carriers.

The formation of hydrogen bromide is the case that made the method's reputation. Bodenstein's 1906 rate law was

v = \frac{k [] [B]^{1 / 2}}{1 + k' [HBr] / [B]}

which resisted explanation for thirteen years. In 1919 Christiansen, Herzfeld and Polanyi independently proposed a chain: bromine dissociates to atoms, $Br + \to HBr + H$ , then $H + B \to HBr + Br$ regenerating the carrier, with $H + HBr \to + Br$ as inhibition and atom recombination as termination. Applying the steady state to both $H$ and $Br$ reproduces that entire expression, including the half power, which comes from the equilibrium concentration of bromine atoms varying as the square root of $[B]$ , and the inhibition term in the denominator, which comes from the product competing with bromine for hydrogen atoms.

Where termination is inefficient and one propagation step produces two carriers, the chain branches and the rate grows without limit. That is an explosion in the chemical sense, distinct from the merely thermal kind: the hydrogen and oxygen reaction has branching steps and shows explosion limits that depend on pressure in a way no thermal theory predicts. Combustion and polymerisation are both chain chemistry, which is why both are so sensitive to traces of inhibitor.

Enzymes and the saturating rate law

Biological catalysts bind their substrate first and react afterwards, which gives a rate law of a shape that turns up wherever a catalyst can be saturated. Write

E + S rightleftharpoons ES \to E + P

and apply the steady state to the complex $ES$ , using the fact that the total enzyme is conserved. The result, from Leonor Michaelis and Maud Menten in 1913 in the form Briggs and Haldane later justified, is

v = \frac{v_{\max} [S]}{K_{M} + [S]}

with $v_{\max} = k_{2} [E]_{0}$ and $K_{M} = (k_{- 1} + k_{2}) / k_{1}$ . The shape has two limits and a useful midpoint. When $[S] ≪ K_{M}$ the rate is first order in substrate; when $[S] ≫ K_{M}$ the enzyme is saturated and the rate is $v_{\max}$ , independent of substrate, which is the zero order behaviour of the sixth lesson. And when $[S] = K_{M}$ the rate is exactly half of $v_{\max}$ , which is how $K_{M}$ is measured.

Example. An enzyme has $K_{M} = 5.0 \times 10^{- 5}$ mol dm⁻³ and $v_{\max} = 2.0 \times 10^{- 3}$ mol dm⁻³ s⁻¹. Find the rate at $[S] = 1.0 \times 10^{- 5}$ and at $1.0 \times 10^{- 3}$ mol dm⁻³.

At the low concentration, $v = 2.0 \times 10^{- 3} \times 1.0 \times 10^{- 5} / (6.0 \times 10^{- 5}) = 3.3 \times 10^{- 4}$ mol dm⁻³ s⁻¹, which is $17$ per cent of the maximum. At the high one, $v = 2.0 \times 10^{- 3} \times 1.0 \times 10^{- 3} / (1.05 \times 10^{- 3}) = 1.9 \times 10^{- 3}$ , which is $95$ per cent. Raising the substrate concentration a hundredfold has raised the rate by less than a factor of six, because the enzyme was already running out of capacity.

Now you. For the same enzyme, at what substrate concentration is the rate three quarters of $v_{\max}$ ?

Answer

Set $[S] / (K_{M} + [S]) = 0.75$ , giving $[S] = 3 K_{M} = 1.5 \times 10^{- 4}$ mol dm⁻³. The general result is that reaching a fraction $f$ of the maximum needs $[S] = K_{M} f / (1 - f)$ , which diverges as $f$ approaches one: saturation is approached and never reached.

The turnover numbers involved are extraordinary. Catalase processes about $4 \times 10^{7}$ molecules of hydrogen peroxide per enzyme molecule per second, and carbonic anhydrase around $10^{6}$ , close to the limit set by how fast substrate can diffuse to the active site.

Kinetics and thermodynamics must agree

The two halves of this course have run in parallel, and one relation ties them together. Consider an elementary step at equilibrium. Equilibrium is dynamic: forward and reverse continue at equal rates, so $k_{f} [A] = k_{r} [B]$ , and therefore

K = \frac{[B]}{[A]} = \frac{k_{f}}{k_{r}}

The equilibrium constant of an elementary step is the ratio of its rate constants. This is not an extra assumption but a requirement, the principle of detailed balance: every individual step must be separately balanced at equilibrium, not merely the overall process. If a step measured $k_{f} = 1.6 \times 10^{- 2}$ and its reverse $2.95 \times 10^{- 4}$ in the same units, thermodynamics is obliged to report $K = 54.2$ for it, and if it reports something else, one of the measurements is wrong.

The same requirement forbids a cycle of three reactions all of which run preferentially in the same direction, which would be a chemical perpetual motion machine. And it connects to the previous lesson: since $K$ depends on temperature through $Δ H$ and each rate constant through its $E_{a}$ , taking logarithms of $K = k_{f} / k_{r}$ and differentiating recovers $E_{a, f} - E_{a, r} = Δ H$ for the step, which was drawn from the energy profile there and is now derived.

The general machinery is complete. The remaining lessons apply it to the two families of reaction that account for most of the chemistry anyone actually does, starting with the transfer of a proton.

Acids and bases

Water is very slightly ionised, and every acid and base calculation is a statement about which way that ionisation has been pushed.

The equilibrium machinery of the fourth lesson applies unchanged to proton transfer, and this is where it earns its keep, because the numbers involved span sixteen orders of magnitude and are therefore always handled as logarithms. Nothing new is assumed here beyond $K$ and the reaction quotient.

Three definitions, each for a different job

Arrhenius, in 1884, defined an acid as a substance that releases hydrogen ions in water and a base as one that releases hydroxide. It is correct as far as it goes and it excludes ammonia, which is plainly a base and contains no hydroxide.

The definition used throughout this lesson is Brønsted and Lowry's, from 1923: an acid is a proton donor and a base is a proton acceptor. The gain is that acidity becomes a relationship rather than a property, since a donor needs an acceptor. Every acid has a conjugate base, what remains when the proton has gone, and the reaction is always a competition between two bases for one proton:

CCOOH + O rightleftharpoons CCO +

Here water is the base. In the presence of hydrogen chloride water is also the base; in the presence of ammonia it is the acid. A substance that can do either is amphiprotic, and water's being so is why it is the reference solvent.

Lewis's definition, from the same year, is broader still: an acid is an electron pair acceptor. That covers boron trifluoride and metal ions, which have no proton to donate, and it is the definition organic chemistry works in. It is not needed below, where every reaction involves a proton moving.

The free proton itself does not exist in solution. It is , or more accurately a shifting cluster of several water molecules, and writing is a convenient abbreviation rather than a species.

Water ionises, slightly

Water transfers a proton to itself, $2 O rightleftharpoons + O$ , and since the solvent is a pure liquid with activity $1$ , the equilibrium constant contains only the two ions:

K_{w} = [] [O] = 1.0 \times 10^{- 14} at 25 °C

In pure water the two concentrations are equal, so each is $1.0 \times 10^{- 7}$ mol dm⁻³. Out of about $55$ moles of water in a cubic decimetre, roughly one molecule in $5 \times 10^{8}$ is ionised at any moment.

The product is the point. $K_{w}$ holds in every aqueous solution, so fixing one ion fixes the other, and an acidic solution still contains hydroxide, just very little of it.

$K_{w}$ is an equilibrium constant, so it depends on temperature, and the ionisation is endothermic ( $Δ H = + 55.8$ kJ mol⁻¹), so van 't Hoff says it rises with temperature. At $60$ °C, $K_{w} = 9.6 \times 10^{- 14}$ , so neutral water has $[] = 3.1 \times 10^{- 7}$ and a pH of $6.51$ . That water is not acidic: it has equal concentrations of both ions, which is what neutral means. Neutrality is pH $7$ only at $25$ °C, and blood at $37$ °C is neutral at about $6.8$ .

The pH scale

Søren Sørensen introduced the logarithmic measure in 1909 while working on beer at the Carlsberg laboratory:

pH = - \log_{10} a () \approx - \log_{10} []

with pOH defined the same way, and taking logarithms of $K_{w}$ gives $pH + pOH = 14.00$ at $25$ °C. The same operator applied to an equilibrium constant gives $p K = - \log_{10} K$ , which is used constantly below.

A strong acid is one that transfers its proton completely, so its concentration is the hydronium concentration and the pH follows immediately.

Example. Find the pH of $0.025$ mol dm⁻³ nitric acid, and of $0.0050$ mol dm⁻³ barium hydroxide.

Nitric acid is strong, so $[] = 0.025$ and $pH = - \log_{10} (0.025) = 1.60$ . Barium hydroxide is strong and supplies two hydroxides per formula unit, so $[O] = 0.010$ , giving $pOH = 2.00$ and $pH = 12.00$ .

Now you. Find the pH of $1.0 \times 10^{- 8}$ mol dm⁻³ hydrochloric acid. Be careful.

Answer

Not $8$ , which would make an acid alkaline. At this dilution the water's own ionisation dominates. Charge balance requires $[] = [C] + [O] = 10^{- 8} + K_{w} / []$ , a quadratic whose root is $[] = 1.05 \times 10^{- 7}$ , giving $pH = 6.98$ . Slightly acidic, as it must be, and the shortcut of reading the concentration off the label fails below about $10^{- 6}$ mol dm⁻³.

$K_{a}$ , $p K_{a}$ , and conjugate pairs

A weak acid is one whose proton transfer to water is an equilibrium with a constant well below one:

K_{a} = \frac{[] []}{[HA]}

Water is omitted, being the solvent. Acetic acid has $K_{a} = 1.75 \times 10^{- 5}$ , so $p K_{a} = 4.76$ ; chloroacetic acid $1.36 \times 10^{- 3}$ , so $2.87$ ; hydrofluoric acid $3.17$ ; ammonium ion $9.25$ . A smaller $p K_{a}$ means a stronger acid, and each unit is a factor of ten.

For the conjugate base, $K_{b}$ describes its reaction with water, and multiplying the two expressions together makes every term cancel except the ions of water:

K_{a} K_{b} = K_{w} or p K_{a} + p K_{b} = 14.00

So a strong acid has a negligibly weak conjugate base, which is why chloride does nothing in solution, and a weak acid has a conjugate base that matters: acetate has $K_{b} = 5.7 \times 10^{- 10}$ , small but not zero, which is the subject of the last section. One table of $K_{a}$ values therefore covers bases too.

Water also imposes a ceiling. Any acid stronger than hands its proton to the solvent completely, so hydrochloric, nitric and perchloric acids are indistinguishable in water at equal concentration. That is the levelling effect, and comparing them requires a less basic solvent such as acetic acid, which is how their true relative strengths are known.

The pH of a weak acid

The calculation is the equilibrium table of the fourth lesson. Let $x$ be the hydronium concentration produced; then the acid is depleted to $C - x$ and the conjugate base is $x$ , so $K_{a} = x^{2} / (C - x)$ . That is a quadratic, and the universal shortcut is to assume $x ≪ C$ and drop it from the denominator, giving $x = \sqrt{K_{a} C}$ .

The shortcut has to be checked rather than trusted. The convention is that it is acceptable when $x$ is under five per cent of $C$ , and the discipline is to compute $x$ and then look, because when it fails it fails quietly.

Example. Find the pH of $0.100$ mol dm⁻³ acetic acid, $K_{a} = 1.75 \times 10^{- 5}$ , and the fraction ionised.

The approximation gives $x = \sqrt{1.75 \times 10^{- 5} \times 0.100} = 1.32 \times 10^{- 3}$ mol dm⁻³, which is $1.3$ per cent of $0.100$ , comfortably within the limit. So $pH = - \log_{10} (1.32 \times 10^{- 3}) = 2.88$ . Solving the quadratic exactly gives $1.314 \times 10^{- 3}$ and the same pH to two decimals. Only about one molecule in eighty has given up its proton, which is what "weak" means quantitatively.

Now you. Find the pH of $0.0010$ mol dm⁻³ acetic acid, using the approximation first and then checking it.

Answer

The approximation gives $x = \sqrt{1.75 \times 10^{- 8}} = 1.32 \times 10^{- 4}$ , which is $13$ per cent of $0.0010$ and so not acceptable. Solving $x^{2} + 1.75 \times 10^{- 5} x - 1.75 \times 10^{- 8} = 0$ gives $x = 1.24 \times 10^{- 4}$ and $pH = 3.91$ , against $3.88$ from the shortcut. Note also that the fraction ionised has risen from $1.3$ to $12$ per cent on dilution, which is the equilibrium shifting toward more particles exactly as the fifth lesson said it would.

What makes one acid stronger than another

Three effects account for most of the variation, and all three are statements about the stability of the conjugate base, since that is what the acid becomes.

Down a group, bond strength dominates. Hydrofluoric acid is weak with $p K_{a} = 3.17$ while hydrochloric, hydrobromic and hydroiodic acids are all strong and increasingly so, because the hydrogen to halogen bond gets longer and weaker down the group. Electronegativity would predict the opposite order, and it loses.

Across a period, and for oxoacids, charge stabilisation dominates. Adding an oxygen to a chlorine oxoacid pulls electron density away from the resulting anion and spreads its negative charge over more atoms, and each addition drops $p K_{a}$ by roughly five units: hypochlorous $7.5$ , chlorous $1.9$ , chloric about $- 1$ , perchloric about $- 8$ . Sixteen orders of magnitude, from three oxygen atoms.

The same logic works at a distance, which is the inductive effect. Replacing one hydrogen of acetic acid with chlorine drops $p K_{a}$ from $4.76$ to $2.87$ , a factor of nearly eighty, although the chlorine is two atoms from the acidic proton. Adding two more chlorines gives trichloroacetic acid at $0.66$ , comparable to a mineral acid. The electronegative atom pulls electron density along the bonds and stabilises the anion.

Salts are not neutral

Dissolve sodium acetate in water and the solution is alkaline, because acetate is a base and takes a proton from water. Dissolve ammonium chloride and the solution is acidic. This is not a new phenomenon: it is the $K_{a} K_{b} = K_{w}$ relation being used from the other end.

Example. Find the pH of $0.100$ mol dm⁻³ ammonium chloride, given $K_{b} = 1.8 \times 10^{- 5}$ for ammonia.

Chloride is the conjugate base of a strong acid and does nothing. The ammonium ion is a weak acid with $K_{a} = K_{w} / K_{b} = 1.0 \times 10^{- 14} / 1.8 \times 10^{- 5} = 5.6 \times 10^{- 10}$ . Then $x = \sqrt{5.6 \times 10^{- 10} \times 0.100} = 7.5 \times 10^{- 6}$ , which is a tiny fraction of $0.100$ , so $pH = 5.13$ .

Now you. Is a solution of sodium acetate acidic, alkaline or neutral, and what is $K_{b}$ for acetate given $K_{a} = 1.75 \times 10^{- 5}$ for acetic acid?

Answer

Alkaline. Sodium does nothing, but acetate accepts a proton from water and releases hydroxide, with $K_{b} = K_{w} / K_{a} = 1.0 \times 10^{- 14} / 1.75 \times 10^{- 5} = 5.7 \times 10^{- 10}$ . For a $0.100$ mol dm⁻³ solution that gives $[O] = \sqrt{5.7 \times 10^{- 11}} = 7.6 \times 10^{- 6}$ , so $pOH = 5.12$ and $pH = 8.88$ .

Polyprotic acids ionise in stages, each with its own constant, and the constants fall steeply because pulling a second proton off an already negative ion is much harder. Carbonic acid has $p K_{a 1} = 6.35$ and $p K_{a 2} = 10.33$ , phosphoric acid $2.15$ , $7.20$ and $12.35$ . Separations of four or five units mean the stages barely overlap, so each can usually be treated on its own.

The mixture that resists

One case has been left out and it is the most useful of all. Every calculation above involved a weak acid alone or its conjugate base alone. Put both in the same solution, in comparable amounts, and something new happens: the pH becomes insensitive to added acid or base, and to dilution.

That is a buffer, and since blood, seawater and every enzyme assay depend on one, it deserves its own treatment. It also turns out to be the key to reading a titration curve, which is the next lesson.

Buffers and titrations

Adding a hundredth of a mole of strong acid to a litre of water takes it from pH 7 to pH 2, and adding the same amount to a litre of acetate buffer moves it by less than a tenth of a unit.

The previous lesson computed the pH of a weak acid on its own and of its conjugate base on its own. This lesson puts both in the same flask, which is the situation in blood, in seawater, in every enzyme assay and at every point of a titration except two.

What a buffer is

A buffer is a solution containing appreciable amounts of a weak acid and its conjugate base. It resists pH change because it has a reservoir of each: added hydroxide is consumed by the acid, added protons are consumed by the base, and neither reservoir is depleted quickly by a small addition.

The working equation comes straight from the definition of $K_{a}$ . Rearranged for the hydronium concentration,

[] = K_{a} \frac{[HA]}{[]}

and taking negative logarithms of both sides gives the Henderson-Hasselbalch equation, published by Lawrence Henderson in 1908 and put into logarithmic form by Karl Hasselbalch in 1917:

pH = p K_{a} + \log_{10} \frac{[]}{[HA]}

Two features do the work. The pH depends on the ratio of the two species, not on their absolute concentrations, so diluting a buffer tenfold changes its pH hardly at all. And a logarithm of a ratio near one is small and flat, so a substantial change in the ratio moves the pH very little: going from $1 : 1$ to $2 : 1$ costs only $0.30$ units.

The equation is an approximation, and it is worth knowing where it fails. It assumes that the amounts of $HA$ and actually present are the amounts mixed in, that is, that the acid's own ionisation and the base's own hydrolysis are negligible compared with them. That holds when both are present at concentrations well above $10^{- 3}$ mol dm⁻³ and the ratio is within about a factor of ten of one. It fails near the ends of a titration, where one component nearly vanishes, and it takes no account of activity coefficients, which in a real buffer of ionic strength $0.1$ shift the answer by around $0.1$ units.

How much it resists

Example. A litre of buffer contains $0.100$ mol of acetic acid ( $p K_{a} = 4.76$ ) and $0.100$ mol of sodium acetate. Find its pH, then the pH after adding $0.010$ mol of hydrogen chloride, ignoring the volume change. Compare with adding the same to a litre of pure water.

Initially the ratio is $1$ and the logarithm is zero, so $pH = 4.76$ . The added protons convert acetate to acetic acid mole for mole, giving $0.110$ mol of acid and $0.090$ mol of base. Then

pH = 4.76 + \log_{10} \frac{0.090}{0.110} = 4.76 - 0.09 = 4.67

a shift of $0.09$ units. The same addition to pure water gives $[] = 0.010$ and $pH = 2.00$ , a shift of five units. The buffer has absorbed the same chemistry with a fiftyfold smaller effect on the logarithm, which is roughly a factor of $10^{5}$ in hydronium concentration.

Now you. Take the same buffer and add $0.010$ mol of sodium hydroxide instead. Find the new pH.

Answer

The hydroxide converts acetic acid to acetate, giving $0.090$ mol of acid and $0.110$ of base, so $pH = 4.76 + \log_{10} (0.110 / 0.090) = 4.76 + 0.09 = 4.85$ . Symmetric with the acid case, because the ratio has been inverted.

Capacity, and how to choose a buffer

Resistance is finite. The buffer capacity $β$ is how many moles of strong acid or base per litre are needed to move the pH by one unit, and it depends on two things.

It is proportional to the total concentration of the pair, which is the size of the reservoirs. And it is greatest when the two are equal, which is when the ratio can be disturbed proportionally least, giving $β_{\max} = 0.576 C_{total}$ : for the $0.200$ mol dm⁻³ total above, about $0.115$ mol per pH unit. Beyond a ratio of about $10 : 1$ in either direction the capacity has fallen so far that the mixture is no longer usefully a buffer.

So choosing a buffer is choosing an acid whose $p K_{a}$ is close to the pH wanted, ideally within one unit, and then setting the ratio to fine-tune.

Example. Prepare a buffer at pH $7.40$ , the pH of blood, from dihydrogenphosphate and hydrogenphosphate, whose pair has $p K_{a} = 7.20$ . What ratio of the two is needed?

Rearranging the Henderson-Hasselbalch equation, $\log_{10} ([] / [HA]) = pH - p K_{a} = 0.20$ , so the ratio is $10^{0.20} = 1.58$ . Dissolving $1.58$ mol of the hydrogenphosphate salt for every $1.00$ mol of the dihydrogenphosphate gives the target, and any total concentration will do, with larger totals giving more capacity.

Now you. What ratio of acetate to acetic acid ( $p K_{a} = 4.76$ ) gives a buffer at pH $5.00$ ?

Answer

$\log_{10} ([] / [HA]) = 5.00 - 4.76 = 0.24$ , so the ratio is $10^{0.24} = 1.74$ . Close enough to one that the capacity is near its maximum, which is why acetate buffers are used across roughly pH $3.8$ to $5.8$ and not beyond.

Phosphate ( $p K_{a 2} = 7.20$ ) is the standard choice near neutrality and is why phosphate-buffered saline is ubiquitous in biology. Carbonate ( $p K_{a 1} = 6.35$ ) buffers blood at $7.4$ , helped by the fact that the carbon dioxide reservoir is open to the lungs, so one component is regulated by breathing rate rather than being fixed. Blood held outside $7.35$ to $7.45$ for long is fatal, and that stability is a buffer doing its job.

The four regions of a titration curve

A titration adds a strong base of known concentration to an acid of unknown concentration and follows the pH. For a weak acid the curve has four regions, each needing a different calculation, and the whole of this course so far is used across them.

Example. Titrate $25.00$ mL of $0.1000$ mol dm⁻³ acetic acid with $0.1000$ mol dm⁻³ sodium hydroxide. Find the pH at $0$ , $10.00$ , $12.50$ , $25.00$ and $30.00$ mL.

The flask starts with $2.500$ mmol of acid.

At $0$ mL there is a weak acid alone, so $[] = \sqrt{K_{a} C} = \sqrt{1.75 \times 10^{- 5} \times 0.1000} = 1.32 \times 10^{- 3}$ and $pH = 2.88$ .

At $10.00$ mL, $1.000$ mmol of base has converted $1.000$ mmol of acid to acetate, leaving $1.500$ mmol of acid. This is a buffer, and since both species sit in the same volume the ratio of amounts serves: $pH = 4.76 + \log_{10} (1.000 / 1.500) = 4.58$ .

At $12.50$ mL exactly half the acid has been converted, the ratio is one, and $pH = p K_{a} = 4.76$ . This is the half equivalence point, and it is the standard way of measuring a $p K_{a}$ : read the pH off the curve halfway to the end point.

At $25.00$ mL all the acid has become acetate. The flask holds $2.500$ mmol of acetate in $50.00$ mL, so $0.0500$ mol dm⁻³ of a weak base with $K_{b} = K_{w} / K_{a} = 5.71 \times 10^{- 10}$ . Then $[O] = \sqrt{5.71 \times 10^{- 10} \times 0.0500} = 5.35 \times 10^{- 6}$ , giving $pOH = 5.27$ and $pH = 8.73$ .

At $30.00$ mL there is $0.500$ mmol of excess hydroxide in $55.00$ mL, which is $9.09 \times 10^{- 3}$ mol dm⁻³, so $pOH = 2.04$ and $pH = 11.96$ . Past equivalence the weak acid is irrelevant and only the excess strong base counts.

Now you. For the same titration, find the pH after $20.00$ mL of base.

Answer

$2.000$ mmol of base leaves $0.500$ mmol of acid and makes $2.000$ mmol of acetate, so $pH = 4.76 + \log_{10} (2.000 / 0.500) = 4.76 + 0.60 = 5.36$ . Still in the buffer region, but at a ratio of $4 : 1$ the capacity is falling and the curve is starting to steepen.

The Henderson-Hasselbalch treatment of the buffer region degrades as equivalence is approached. At $24.90$ mL it returns $7.15$ , which is wrong, because with only $0.010$ mmol of acid left the assumption that the mixed amounts survive intact has collapsed. The last fraction of a millilitre needs the full equilibrium treatment or a numerical solution, which is exactly why the practical method is to locate the steepest point rather than to compute the region.

Why the equivalence point is not pH 7

The most common error in titration work is expecting neutrality at equivalence. Equivalence means the stoichiometric amounts have been matched, not that the solution is neutral, and what remains in the flask decides the pH.

Titrate a strong acid with a strong base and the product is sodium chloride, which does nothing, so equivalence is at pH $7.00$ . Titrate a weak acid with a strong base and the product is the conjugate base, which is alkaline: $8.73$ in the example above. Titrate a weak base with a strong acid and the product is acidic, typically around pH $5$ . Only the first case gives seven.

The shape of the curve differs as much as the end point. A strong acid with a strong base jumps from pH $3.70$ at $24.90$ mL to $10.30$ at $25.10$ mL: over six units across two drops. The weak acid curve above jumps from about $7$ to about $10$ over the same interval, a smaller and shallower step, because the buffer region leading into it has flattened everything. Titrating a very weak acid, below $K_{a}$ of about $10^{- 8}$ , gives no usable step at all.

Choosing an indicator

An indicator is itself a weak acid whose two forms differ in colour, so the same equilibrium applies: it appears in its acid colour when the pH is well below its own $p K_{a}$ and in its base colour well above, changing over a range of roughly $p K_{a} \pm 1$ .

The rule for choosing one follows from the previous section. The indicator's range must lie inside the vertical jump of that particular curve. Phenolphthalein changes between $8.3$ and $10.0$ , which sits neatly inside the jump for a weak acid titrated with a strong base and is therefore correct for the acetic acid case. Methyl orange changes between $3.1$ and $4.4$ , which is inside the jump for a strong acid with a weak base but far outside the acetic acid jump: use it there and the colour changes around $12$ mL, halfway through the buffer region, giving an answer wrong by half.

A pH meter avoids the choice altogether and is what any modern laboratory uses, but the indicator argument is worth keeping, because it is the same reasoning applied to a reagent instead of a sample.

Polyprotic curves

An acid with several protons gives a curve with several steps, provided the constants are far enough apart. Phosphoric acid, with $p K_{a}$ values of $2.15$ , $7.20$ and $12.35$ , shows two clear jumps, at one and at two equivalents of base; the third is lost because $12.35$ is too close to the solvent's own range for a step to develop.

Between two equivalence points the solution is a buffer, and at the first equivalence point of a diprotic acid the species present is the amphiprotic intermediate, which both donates and accepts. Its pH turns out to be close to the mean of the two constants that flank it, $pH \approx (p K_{a 1} + p K_{a 2}) / 2$ , independent of concentration. For carbonic acid that gives $(6.35 + 10.33) / 2 = 8.34$ , which is why a solution of sodium hydrogencarbonate sits near pH $8.3$ whatever its strength.

The same constant, applied to a solid

Every calculation in the last two lessons has been an equilibrium constant used on a proton transfer, with the shape of the arithmetic set by the constant and the shape of the answer by which species dominate.

Nothing in that machinery is specific to protons. Write the same expression for a salt sitting in contact with its own saturated solution and it gives the solubility, in grams per litre, of a substance that a table lists only as "slightly soluble". It also predicts something the tables do not: which insoluble salts dissolve in acid and which ignore it entirely, a question the last two lessons have already supplied the answer to. That is the next lesson.

Solubility and precipitation

An insoluble salt in water is not inert: it is dissolving and crystallising at equal rates, and the balance point between them is an equilibrium constant like any other.

The previous two lessons applied the constant of the fourth lesson to proton transfer. This one applies it to a solid in contact with its saturated solution, which needs no new theory at all, only the rule that a pure solid has activity $1$ .

Saturation is an equilibrium

Drop silver chloride into water and a little dissolves, $AgCl(s) rightleftharpoons A(aq) + C(aq)$ . The solid is pure, so it does not appear in the constant, and what remains is the solubility product:

K_{s p} = [A] [C] = 1.77 \times 10^{- 10} at 25 °C

The equilibrium is dynamic, which can be shown directly. Add silver chloride labelled with radioactive silver-110 to a saturated solution of ordinary silver chloride and the radioactivity appears in the solution within minutes, although the amount of solid never changes. Nothing is at rest; the two rates are equal.

Two consequences follow from the form of the constant. The product of the ion concentrations is fixed whatever their individual values, so raising one must lower the other. And $K_{s p}$ depends on temperature alone, in the way the fifth lesson established, with $Δ_{sol} H$ setting the direction: most salts dissolve endothermically and so get more soluble when heated, while calcium sulfate and the other salts that scale a kettle are the exceptions, dissolving exothermically and precipitating as the water warms.

Molar solubility from the constant

The molar solubility $s$ is how many moles of the salt dissolve per cubic decimetre of saturated solution. Getting it from $K_{s p}$ means writing each ion concentration in terms of $s$ , remembering that a salt of formula $A_{x} B_{y}$ releases $x$ and $y$ ions per formula unit.

Example. Find the molar solubility of silver chloride, $K_{s p} = 1.77 \times 10^{- 10}$ , and of calcium fluoride, $K_{s p} = 3.9 \times 10^{- 11}$ , and express each in grams per litre. The molar masses are $143.32$ and $78.07$ g mol⁻¹.

For silver chloride, dissolving $s$ moles gives $[A] = [C] = s$ , so $K_{s p} = s^{2}$ and $s = \sqrt{1.77 \times 10^{- 10}} = 1.33 \times 10^{- 5}$ mol dm⁻³, or $1.9 \times 10^{- 3}$ g dm⁻³.

For calcium fluoride, dissolving $s$ moles gives $[C] = s$ and $[] = 2 s$ , so $K_{s p} = s (2 s)^{2} = 4 s^{3}$ . Then $s = (3.9 \times 10^{- 11} / 4)^{1 / 3} = 2.14 \times 10^{- 4}$ mol dm⁻³, or $0.0167$ g dm⁻³. Note the factor of two entering both as a coefficient and as a power; forgetting the power is the standard mistake here.

Now you. Silver chromate, $ACr$ , has $K_{s p} = 1.12 \times 10^{- 12}$ and a molar mass of $331.73$ g mol⁻¹. Find its molar solubility, and compare it with silver chloride's.

Answer

Two silver ions per formula unit, so $[A] = 2 s$ and $[Cr] = s$ , giving $K_{s p} = 4 s^{3}$ and $s = (1.12 \times 10^{- 12} / 4)^{1 / 3} = 6.5 \times 10^{- 5}$ mol dm⁻³, or $0.022$ g dm⁻³. Silver chromate has a solubility product a hundred and sixty times smaller than silver chloride's and is nearly five times more soluble.

That comparison is the point of the exercise. Solubility products can be compared directly only between salts of the same stoichiometry, because the exponent relating $s$ to $K_{s p}$ differs otherwise. Ranking a $1 : 1$ salt against a $2 : 1$ salt by their constants gives the wrong order, as it does here.

The common ion effect

Since $K_{s p}$ is a product, supplying one of the ions from elsewhere forces the other down, and the solid that has to precipitate to make that happen is the salt itself. This is the common ion effect, and it is Le Chatelier's principle in a form that can be computed exactly.

Example. Find the solubility of silver chloride in $0.010$ mol dm⁻³ sodium chloride, and compare with its solubility in pure water.

The dissolved silver chloride contributes a negligible amount of chloride compared with $0.010$ , so $[C] = 0.010$ and $[A] = K_{s p} / 0.010 = 1.77 \times 10^{- 8}$ mol dm⁻³. Since every dissolved silver ion means one dissolved formula unit, that is the solubility: $750$ times lower than the $1.33 \times 10^{- 5}$ in pure water. The approximation is safe because $1.77 \times 10^{- 8}$ is indeed tiny beside $0.010$ .

Now you. Find the solubility of calcium fluoride in $0.010$ mol dm⁻³ sodium fluoride, given $K_{s p} = 3.9 \times 10^{- 11}$ .

Answer

With $[] = 0.010$ fixed by the added salt, $[C] = K_{s p} / (0.010)^{2} = 3.9 \times 10^{- 11} / 1.0 \times 10^{- 4} = 3.9 \times 10^{- 7}$ mol dm⁻³, which is the solubility. That is $550$ times lower than in pure water. The suppression is stronger than for silver chloride at the same added concentration, because the fluoride concentration enters squared.

This is why a precipitate is washed with a dilute solution of one of its own ions rather than with pure water, and why gravimetric analysis adds an excess of the precipitating reagent: a hundredfold suppression of solubility is the difference between losing one per cent of the sample down the sink and losing none of it.

There is an opposite effect that the calculation above ignores. Adding an inert salt with no ion in common, such as potassium nitrate, actually raises the solubility slightly, because the extra ionic atmosphere lowers the activity coefficients of the dissolving ions, so a larger concentration is needed to reach the same activity product. It is a small correction in dilute solution and a real one in seawater.

Acid dissolves some salts and not others

The most useful prediction in this lesson comes from combining $K_{s p}$ with the previous two lessons. If the anion of a salt is a base, adding acid removes it from solution by protonating it, the ion product falls below $K_{s p}$ , and more solid dissolves. If the anion is not a base, acid does nothing at all.

Chloride is the conjugate base of a strong acid, so it is not a base in any useful sense, and silver chloride is as insoluble in dilute nitric acid as in water. Carbonate is the conjugate base of a weak acid with $p K_{a 2} = 10.33$ , so it is a strong base, and calcium carbonate dissolves readily in acid. The reaction runs to completion because the product escapes:

CaC(s) + 2 \to C + O + C(g)

Rainwater in equilibrium with atmospheric carbon dioxide is a weak acid at about pH $5.6$ , which is enough. Limestone caves, karst landscapes and the pitting of marble statues in polluted air are all this equation running for a long time, and the reverse of it, carbon dioxide escaping from groundwater as it drips into an air-filled cavity, is what builds a stalactite. Fluoride is a weaker base ( $p K_{a}$ of $HF$ is $3.17$ ), so calcium fluoride is intermediate: appreciably more soluble in strong acid, unaffected by weak.

Metal hydroxides are the extreme case, because their anion is hydroxide itself and the pH is the direct control. Magnesium hydroxide has $K_{s p} = 5.6 \times 10^{- 12}$ , so its saturated solution has $s = (5.6 \times 10^{- 12} / 4)^{1 / 3} = 1.12 \times 10^{- 4}$ mol dm⁻³ and $[O] = 2.24 \times 10^{- 4}$ , giving a pH of $10.35$ . That is milk of magnesia: alkaline enough to neutralise stomach acid, and self-limiting, because as acid is consumed the pH cannot rise past the saturated value.

Will it precipitate?

For a mixture that is not yet at equilibrium, the test is the one from the fourth lesson: compute the ion product $Q$ and compare it with $K_{s p}$ . If $Q > K_{s p}$ the solution is supersaturated and solid should form; if $Q < K_{s p}$ any solid present dissolves.

Where two salts could form, whichever needs the lower concentration of the added reagent precipitates first, and if the gap is wide enough the two can be separated almost completely. That is selective precipitation, the basis of classical qualitative analysis.

Example. A solution is $0.010$ mol dm⁻³ in both chloride and chromate. Silver nitrate is added slowly. Which precipitates first, and what fraction of the first ion remains when the second begins to come down? Use $K_{s p} = 1.77 \times 10^{- 10}$ for $AgCl$ and $1.12 \times 10^{- 12}$ for $ACr$ .

Silver chloride starts when $[A] = K_{s p} / [C] = 1.77 \times 10^{- 10} / 0.010 = 1.77 \times 10^{- 8}$ mol dm⁻³. Silver chromate starts when $[A] = \sqrt{K_{s p} / [Cr]} = \sqrt{1.12 \times 10^{- 10}} = 1.06 \times 10^{- 5}$ mol dm⁻³, six hundred times higher. So the chloride comes down first. At the moment the chromate begins, the remaining chloride is $1.77 \times 10^{- 10} / 1.06 \times 10^{- 5} = 1.7 \times 10^{- 5}$ mol dm⁻³, so $99.83$ per cent of the chloride has already precipitated. This is exactly the Mohr method for chloride, in which chromate is the indicator and the first permanent red tinge of silver chromate marks the end point.

Now you. Equal volumes of $0.0020$ mol dm⁻³ silver nitrate and $0.0020$ mol dm⁻³ sodium chloride are mixed. Does a precipitate form?

Answer

Mixing equal volumes halves both concentrations, so $[A] = [C] = 0.0010$ mol dm⁻³ and $Q = 1.0 \times 10^{- 6}$ . That exceeds $K_{s p} = 1.77 \times 10^{- 10}$ by nearly four orders of magnitude, so silver chloride precipitates until the product falls to $K_{s p}$ .

Whether it precipitates promptly is a separate question, and one for the kinetics half of this course. Solutions can stay supersaturated for a long time when there is no surface for a crystal to start on, which is why sodium acetate hand warmers hold litres of liquid far past saturation until a metal disc is clicked.

Complexing dissolves the insoluble

The other way to remove an ion from solution is to tie it up in a complex ion, which works on the cation as acid works on the anion. Silver ion binds two ammonia molecules with a formation constant $K_{f} = 1.7 \times 10^{7}$ , so the overall dissolution of silver chloride in ammonia is the sum of two equilibria and its constant is the product:

K = K_{s p} K_{f} = 1.77 \times 10^{- 10} \times 1.7 \times 10^{7} = 3.0 \times 10^{- 3}

In $1.0$ mol dm⁻³ ammonia that gives a solubility of about $0.049$ mol dm⁻³, nearly four thousand times the value in water, which is why ammonia clears a silver chloride precipitate and why photographic fixer, which uses thiosulfate with a much larger formation constant still, removes unexposed silver halide from film.

Some hydroxides do this to themselves. Aluminium hydroxide dissolves in acid as a base and in excess alkali as an acid, forming $[Al(OH]^{-}$ , so its solubility curve has a minimum near pH $6$ and rises in both directions. Water treatment exploits exactly that minimum.

A closing caution on all of these numbers. Solubility products are the equilibrium constants most affected by the activity coefficients set aside in the fourth lesson, since they involve multiply charged ions at appreciable ionic strength, and published values for the same salt can differ by a factor of two or three between sources. Treat a calculated solubility as good to an order of magnitude unless the ionic strength has been dealt with properly.

Moving the electron itself

Both families of reaction covered so far move a proton: from acid to base, or from acid to the anion of a salt. In every case the electrons stay where they were.

The last two lessons of this course are about reactions in which the electrons themselves change owner. They need their own bookkeeping, because an electron transferred between two atoms in the same molecule leaves no visible trace in the formula, and the accounting device that makes them visible is the next lesson.

Oxidation and reduction

When magnesium burns in carbon dioxide there is no oxygen anywhere in the reactants, and the magnesium is oxidised anyway, which is a hint that the useful definition is not about oxygen.

The previous lessons moved protons between molecules. This one and the next move electrons, which is harder to see, because an electron shifted along a bond leaves the formula unchanged. The first job is a device that makes the shift visible.

What actually changes hands

The oldest definition is Lavoisier's: oxidation is combination with oxygen. It works for rusting and burning, and it fails for $2Mg + C \to 2MgO + C$ , where magnesium takes oxygen away from carbon, and it says nothing at all about $Mg + C \to MgC$ , which is chemically the same event.

The modern definition is about electrons. Oxidation is loss of electrons; reduction is gain. The two always occur together, since electrons do not accumulate anywhere, so the whole class is called redox. The species that takes electrons is the oxidising agent, and it is itself reduced; the species that supplies them is the reducing agent, and it is oxidised. Both halves of that sentence catch people out, and the reliable way to keep them straight is to track the electrons rather than the words.

For an ionic reaction the transfer is literal: magnesium atoms become $M$ and chlorine molecules become $C$ , two electrons each. For a covalent one, such as methane burning, nothing is transferred completely. Electron density shifts along polar bonds, which is a matter of degree. That is what the next section exists to handle.

Oxidation numbers: a fiction that works

The oxidation number of an atom is the charge it would have if every bond it took part in were fully ionic, with each shared pair assigned entirely to the more electronegative atom. It is an accounting convention, not a measurement. The carbon in methane has an oxidation number of $- 4$ and a real partial charge of a few tenths of an electron. Nothing is being claimed about physical charge.

The rules follow from that definition, and are worth deriving rather than memorising. An element on its own has zero, since the sharing is symmetric. A monatomic ion has its charge. Fluorine, the most electronegative element, always takes $- 1$ . Oxygen takes $- 2$ except when bonded to fluorine or to itself. Hydrogen takes $+ 1$ except when bonded to a metal, where it is the more electronegative partner and takes $- 1$ . And the numbers must sum to the charge on the species, which is the constraint that determines everything else.

Example. Assign oxidation numbers to sulfur in $S$ and in , to iron in $F$ , and to oxygen in and in $O$ .

In sulfuric acid, two hydrogens at $+ 1$ and four oxygens at $- 2$ leave $x + 2 - 8 = 0$ , so sulfur is $+ 6$ . In thiosulfate, $2 x - 6 = - 2$ gives $x = + 2$ . In magnetite, $3 x - 8 = 0$ gives $x = + 8 / 3$ , which is not a possible charge on any atom. In hydrogen peroxide the oxygens are bonded to each other, so that bond is shared equally and each oxygen is $- 1$ . In oxygen difluoride, fluorine outranks oxygen, so the fluorines are $- 1$ and oxygen is $+ 2$ .

Now you. Assign oxidation numbers to chromium in $C$ , to manganese in $Mn$ , to nitrogen in $N$ , and to carbon in glucose, .

Answer

Dichromate: $2 x - 14 = - 2$ , so chromium is $+ 6$ . Permanganate: $x - 8 = - 1$ , so manganese is $+ 7$ . Ammonium: $x + 4 = + 1$ , so nitrogen is $- 3$ . Glucose: $6 x + 12 - 12 = 0$ , so carbon averages $0$ , though the individual carbons in the real molecule range from $- 1$ to $+ 1$ .

Two of those answers show the limits honestly. The $+ 8 / 3$ in magnetite is an average over two iron(III) and one iron(II) in the real lattice. The $+ 2$ in thiosulfate is an average over one sulfur at about $+ 5$ and one at about $- 1$ , which is why isotopic labelling shows the two sulfurs behaving quite differently in reactions. Oxidation numbers are a device for counting electrons transferred, and they do that job perfectly; they are not a description of a molecule.

With them in hand, redox becomes visible: any reaction in which an oxidation number changes is a redox reaction. Carbon goes from $- 4$ in methane to $+ 4$ in carbon dioxide, so combustion is redox. In an acid and base neutralisation nothing changes at all, which is why proton transfer is a separate family.

Half reactions in acid

Balancing a redox equation by inspection is painful and unnecessary. Splitting it into two half reactions, balancing each, and combining them so the electrons cancel is mechanical.

The procedure in acidic solution is: balance the atom being oxidised or reduced, then balance oxygen by adding water, then hydrogen by adding , then charge by adding electrons. The electrons should come out on the left for a reduction and on the right for an oxidation, which is a check on the work.

Example. Balance the reaction of permanganate with oxalate in acid, $Mn + \to M + C$ .

Reduction half: manganese goes from $+ 7$ to $+ 2$ . Balancing oxygen with water and hydrogen with protons gives $Mn + 8 \to M + 4 O$ , whose charges are $+ 7$ on the left and $+ 2$ on the right, so five electrons are added to the left.

Oxidation half: $\to 2 C$ is already balanced in atoms, with charges $- 2$ and $0$ , so two electrons go on the right.

To cancel electrons, multiply the first by two and the second by five, giving ten each:

2 Mn + 5 + 16 \to 2 M + 10 C + 8 O

Check the charge: $- 2 - 10 + 16 = + 4$ on the left, $+ 4$ on the right. Check the atoms: manganese two each side, carbon ten, oxygen $8 + 20 = 28$ on the left and $20 + 8 = 28$ on the right.

Now you. Balance the oxidation of iron(II) by permanganate in acid, $Mn + F \to M + F$ .

Answer

The manganese half is as before, needing five electrons. The iron half is $F \to F + e^{-}$ , one electron, so it is multiplied by five:

Mn + 5 F + 8 \to M + 5 F + 4 O

Charges: $- 1 + 10 + 8 = + 17$ on the left, $+ 2 + 15 = + 17$ on the right.

Half reactions in base

In alkaline solution is not available at any useful concentration, so the balanced equation must not contain it. The quickest route is to balance in acid as above, then add enough $O$ to both sides to convert every into water, and cancel whatever water appears on both sides.

Take the oxidation of iodide by permanganate in base, which stops at manganese dioxide rather than going to $M$ . The half reactions are $Mn + 2 O + 3 e^{-} \to Mn + 4 O$ and $+ 6 O \to I + 3 O + 6 e^{-}$ . Doubling the first to match six electrons and combining gives

2 Mn + + O \to 2 Mn + I + 2 O

with charge $- 3$ on each side and nine oxygens on each side.

That the product is $Mn$ in base and $M$ in acid is not a detail of the method: it is chemistry, and it says that the oxidising power of permanganate depends on pH. The next lesson makes that dependence quantitative.

Ranking the agents

Some species take electrons more eagerly than others, and the ordering can be established by experiment before any theory. Put a strip of zinc into copper sulfate solution and it darkens with a deposit of copper while the blue fades, so zinc gives electrons to copper ions. Put copper into zinc sulfate and nothing happens. Repeating this over many pairs gives the activity series, in which each metal displaces every metal below it.

The series explains a great deal at a glance: why potassium and sodium have to be kept out of water, why iron rusts and gold does not, why aluminium is protected by an oxide film rather than by being unreactive, and why the metals known to antiquity are gold, silver and copper, the three that occur native because they are the hardest to oxidise. It also explains extraction: metals near the top are won by electrolysis, ones in the middle by reduction with carbon, and the ones at the bottom are simply dug up.

Among non-metals the halogens rank the same way. Chlorine displaces bromide from solution and bromine displaces iodide, so oxidising power falls down the group, which is the ordering used in the next lesson to build a numerical scale.

Redox titration

Because the electron count is exact, a redox reaction of known stoichiometry measures concentration as precisely as an acid and base titration does, and often more conveniently.

Permanganate is the classic titrant, for a reason worth noticing: it is intensely purple and its product $M$ is almost colourless, so the first drop in excess turns the flask permanently pink and no indicator is needed. It is not a primary standard, since solid potassium permanganate always carries some manganese dioxide, so it is standardised against pure sodium oxalate using the equation balanced above.

Example. $25.00$ mL of a solution of iron(II) in dilute sulfuric acid requires $22.45$ mL of $0.02000$ mol dm⁻³ potassium permanganate to reach the first permanent pink. Find the concentration of the iron(II).

The permanganate supplies $0.02245 \times 0.02000 = 4.490 \times 10^{- 4}$ mol. The balanced equation has five iron per permanganate, so the iron is $5 \times 4.490 \times 10^{- 4} = 2.245 \times 10^{- 3}$ mol, in $25.00$ mL. That gives $2.245 \times 10^{- 3} / 0.02500 = 0.08980$ mol dm⁻³.

Now you. A $0.5000$ g sample of iron ore is dissolved and all the iron reduced to iron(II). Titration needs $18.30$ mL of $0.02000$ mol dm⁻³ permanganate. Find the percentage of iron by mass, taking the molar mass of iron as $55.845$ g mol⁻¹.

Answer

Permanganate: $0.01830 \times 0.02000 = 3.660 \times 10^{- 4}$ mol, so iron is $1.830 \times 10^{- 3}$ mol, weighing $1.830 \times 10^{- 3} \times 55.845 = 0.1022$ g. As a percentage of the sample, $0.1022 / 0.5000 = 20.44$ per cent.

Dichromate is the other common titrant, less powerful but a genuine primary standard, stable in solution and usable in hydrochloric acid, which permanganate is not because it oxidises chloride. It needs a separate indicator, since chromium(III) is green rather than colourless. Iodine titrations form a third family, in which iodine liberated by an oxidising agent is titrated with thiosulfate to a starch end point, and this is how dissolved oxygen and residual chlorine in water are routinely measured.

When a substance reacts with itself

An element in an intermediate oxidation state can sometimes be oxidised and reduced at once, which is disproportionation. Copper(I) does it in water, $2 C \to Cu + C$ , which is why copper(I) salts are stable only as insoluble solids or complexes. Chlorine does it in cold alkali, $C + 2 O \to C + Cl + O$ , going to $- 1$ and $+ 1$ from $0$ , which is the industrial route to bleach. Warm the same mixture and hypochlorite disproportionates further to chloride and chlorate.

The reverse, comproportionation, brings two different states together to a middle one, as when iodate and iodide give iodine in acid.

Whether either happens is a thermodynamic question with a definite answer, and so far this lesson has provided none: nothing here says why copper(I) disproportionates and iron(II) does not, or why permanganate is a stronger oxidising agent in acid. Those need a number attached to each half reaction rather than a rank order.

Separating the two halves into different beakers, so the electrons have to travel through a wire, supplies exactly that number, and turns it into a voltage a meter can read. That is the last lesson.

Electrochemical cells

Put the two halves of a redox reaction in separate beakers and the electrons have to travel through a wire to get from one to the other, where they can be made to do work.

The previous lesson balanced redox equations and ranked the agents by displacement experiments, and ended with three questions it could not answer: why copper(I) disproportionates, why permanganate is weaker in neutral solution, and how far any of these reactions go. All three need a number attached to each half reaction. This lesson gets that number from a voltmeter.

Separating the halves

Drop zinc into copper sulfate and the reaction $Zn + C \to Z + Cu$ happens on the metal surface, releasing $219$ kJ mol⁻¹ as heat and nothing else. The electrons go straight from zinc atom to copper ion across a distance of a few tenths of a nanometre.

John Daniell's arrangement of 1836 separates them. Zinc sits in zinc sulfate, copper in copper sulfate, and the only electrical path between the two solutions is a salt bridge, a tube of electrolyte that carries ions but not electrons. The electrons must now go the long way, through an external wire, and on that journey they can turn a motor.

The vocabulary is fixed by function, not by sign. Oxidation happens at the anode, reduction at the cathode, always. In this cell the zinc is the anode, dissolving as $Z$ and leaving electrons behind, so it is the negative terminal. The salt bridge exists because without it the zinc solution would build up positive charge and the copper solution negative, and after a few nanocoulombs the resulting electric field would stop the reaction dead. Sulfate migrating one way and potassium the other keeps both solutions neutral.

The conventional shorthand is $Zn(s) | Z(aq) ‖ C(aq) | Cu(s)$ , with single bars for phase boundaries, the double bar for the salt bridge, and the anode written on the left.

A scale needs a zero

A voltmeter measures a difference, so no single electrode has a potential that can be measured on its own. The convention assigns zero to one of them: the standard hydrogen electrode, hydrogen gas at $10^{5}$ Pa bubbling over platinised platinum in a solution of unit hydrogen ion activity, is defined as $0.000$ V at all temperatures.

Every other half reaction is then measured against it and tabulated as a standard reduction potential $E^{ominus}$ , always written as a reduction. Zinc measures $- 0.76$ V, copper $+ 0.34$ V, silver $+ 0.80$ V, and the permanganate couple in acid $+ 1.51$ V. A positive value means the couple takes electrons more readily than hydrogen ions do, so a strong oxidising agent sits at the top of the table and a strong reducing agent at the bottom.

This is the activity series of the previous lesson made numerical, and the extra content is that the numbers can be subtracted. For any pair,

E_{cell}^{ominus} = E_{cathode}^{ominus} - E_{anode}^{ominus}

with both taken from the table as reductions. A positive cell potential means the reaction as written is spontaneous under standard conditions.

One point causes endless confusion and is worth stating flatly: potentials are not multiplied when a half reaction is scaled. Doubling $A + e^{-} \to Ag$ leaves $E^{ominus}$ at $0.80$ V. Potential is energy per unit charge, an intensive quantity, and doubling the reaction doubles both the energy and the charge.

Example. Find the standard cell potential of the Daniell cell, and of a cell combining copper with silver, given $E^{ominus} = - 0.76$ V for $Z / Zn$ , $+ 0.34$ for $C / Cu$ and $+ 0.80$ for $A / Ag$ .

For the Daniell cell, copper is reduced and zinc oxidised, so $E_{cell}^{ominus} = 0.34 - (- 0.76) = 1.10$ V. For the copper and silver cell, silver is the stronger oxidising agent, so silver is reduced and copper oxidised: $E_{cell}^{ominus} = 0.80 - 0.34 = 0.46$ V. The overall reaction is $Cu + 2 A \to C + 2 Ag$ , and the silver potential is not doubled to match the two electrons.

Now you. Using $E^{ominus} = - 0.44$ V for $F / Fe$ and $+ 0.15$ V for $S / S$ , find the cell potential and say which way the reaction runs.

Answer

The tin couple has the higher potential, so it is reduced and iron oxidised: $E_{cell}^{ominus} = 0.15 - (- 0.44) = 0.59$ V, positive, so $Fe + S \to F + S$ is spontaneous under standard conditions.

Potential is free energy per coulomb

The third lesson noted that $Δ G$ is the maximum non-expansion work a process can deliver. An electrochemical cell is a machine for collecting exactly that work, so the two quantities must be connected.

Moving $n$ moles of electrons through a potential difference $E$ transfers charge $n F$ , where $F = 96485$ C mol⁻¹ is the Faraday constant, the charge on a mole of electrons. The electrical work done is charge times potential, and when the cell is drawn on reversibly, meaning infinitely slowly against an almost equal opposing voltage, that work equals $Δ G$ :

Δ_{r} G = - n F E and at standard conditions Δ_{r} G^{ominus} = - n F E^{ominus}

The minus sign puts a positive potential with a negative free energy change, so a cell that reads positive is a cell whose reaction goes. Combining this with $Δ_{r} G^{ominus} = - R T \ln K$ from the fourth lesson gives a startling shortcut:

\ln K = \frac{n F E^{ominus}}{R T}

A voltmeter reads an equilibrium constant. This is the most sensitive method available for constants far from one, because the logarithm compresses them: at $298$ K, $0.0592$ V of cell potential per electron is a factor of ten in $K$ .

Example. For the Daniell cell, $E^{ominus} = 1.10$ V with $n = 2$ . Find $Δ_{r} G^{ominus}$ and $K$ .

$Δ_{r} G^{ominus} = - 2 \times 96485 \times 1.10 = - 2.12 \times 10^{5}$ J mol⁻¹, or $- 212$ kJ mol⁻¹. Then $\ln K = 212267 / (8.314 \times 298.15) = 85.6$ , so $K = 1.5 \times 10^{37}$ . The reaction goes essentially to completion, which is what a strip of zinc in copper sulfate looks like.

Now you. For the copper and silver cell, $E^{ominus} = 0.46$ V with $n = 2$ . Find $Δ_{r} G^{ominus}$ and $K$ .

Answer

$Δ_{r} G^{ominus} = - 2 \times 96485 \times 0.46 = - 8.88 \times 10^{4}$ J mol⁻¹, or $- 88.8$ kJ mol⁻¹. Then $\ln K = 88766 / 2478.8 = 35.8$ , so $K = 3.6 \times 10^{15}$ . Less extreme than the Daniell cell, and still complete for any practical purpose.

The Nernst equation

Standard conditions mean unit activities, which no working battery has. Take $Δ_{r} G = Δ_{r} G^{ominus} + R T \ln Q$ from the fourth lesson and divide throughout by $- n F$ :

E = E^{ominus} - \frac{R T}{n F} \ln Q

This is the Nernst equation, from Walther Nernst in 1889. At $298$ K, converting to base ten logarithms gives the form used in practice, since $R T \ln 10 / F = 0.0592$ V:

E = E^{ominus} - \frac{0.0592}{n} \log_{10} Q

Everything about a cell under real conditions is in that line. As a battery discharges, products accumulate, $Q$ rises and $E$ falls, reaching zero exactly when $Q = K$ : a flat battery is a cell at equilibrium, and the reason it is flat is that there is no free energy left to extract, not that anything has been used up in the ordinary sense.

Example. A Daniell cell has $[Z] = 1.0$ mol dm⁻³ and $[C] = 0.010$ mol dm⁻³. Find its potential.

The reaction is $Zn + C \to Z + Cu$ , and the solids have activity $1$ , so $Q = [Z] / [C] = 100$ . With $n = 2$ , $E = 1.10 - (0.0592 / 2) \log_{10} (100) = 1.10 - 0.059 = 1.04$ V. A hundredfold change in concentration has cost less than six per cent of the voltage, which is the logarithm at work and the reason cells hold their voltage well until they are nearly exhausted.

Now you. A concentration cell has copper electrodes in $0.0010$ and $1.0$ mol dm⁻³ copper sulfate, joined by a salt bridge. Its $E^{ominus}$ is zero, since both halves are the same couple. What is its potential?

Answer

The cell runs to equalise the two concentrations, so copper dissolves in the dilute half and deposits in the concentrated one, and $Q = 0.0010 / 1.0 = 10^{- 3}$ . Then $E = 0 - (0.0592 / 2) \log_{10} (10^{- 3}) = 0.089$ V. A cell driven entirely by a concentration difference, which is also how a pH meter works.

The Nernst equation also settles the pH question from the previous lesson. Permanganate is reduced as $Mn + 8 + 5 e^{-} \to M + 4 O$ , so $Q$ contains $[]^{- 8}$ and the potential depends on pH as $E = 1.51 - (0.0592 \times 8 / 5) pH$ . At pH $7$ that is $0.85$ V, a loss of two thirds of a volt, which is why permanganate stops at manganese dioxide in neutral solution instead of going to $M$ .

Real cells

A lead-acid cell delivers $2.05$ V from $Pb + Pb + 2 S \to 2 PbS + 2 O$ , and six in series give the familiar $12.3$ V. It is a rare case where the state of charge can be read directly, since the reaction consumes sulfuric acid and the electrolyte density falls from about $1.28$ to $1.10$ g cm⁻³ as it discharges. Its virtue is the ability to deliver hundreds of amperes briefly; its vice is that lead is heavy, at around $40$ W h kg⁻¹.

A lithium-ion cell delivers $3.6$ V and $250$ W h kg⁻¹ or more, and works quite differently: nothing dissolves, and lithium ions shuttle between two host lattices, graphite and a metal oxide, in which they sit between layers. Because the electrodes are not consumed, the cell survives thousands of cycles.

A hydrogen fuel cell is the cleanest illustration of the theory. Its reaction is $+ \frac{1}{2} \to O(l)$ , with $Δ_{r} G^{ominus} = - 237.1$ kJ mol⁻¹ from the third lesson and $n = 2$ , so $E^{ominus} = 237100 / (2 \times 96485) = 1.23$ V, which is what the cell measures. And since a fuel cell is limited by $Δ G$ rather than by the Carnot efficiency of a heat engine, its ceiling is $Δ G / Δ H = 237.1 / 285.8 = 83$ per cent, far above what any combustion engine can reach.

Corrosion and electrolysis

Rusting is a cell that nobody wanted. Iron oxidises at $- 0.44$ V where the metal is stressed or the oxygen supply is poor, the electrons travel through the metal itself, and oxygen is reduced elsewhere on the surface at $+ 0.40$ V in neutral water. The iron(II) formed is then oxidised further to hydrated iron(III) oxide, which flakes off rather than protecting what is beneath, unlike aluminium's oxide.

Knowing it is a cell tells you how to stop it. Attach a metal with a more negative potential, such as zinc at $- 0.76$ V or magnesium at $- 2.37$ V, and that metal becomes the anode and corrodes instead. This is sacrificial protection: the zinc blocks bolted to a ship's hull, and the galvanised coating on a steel bucket, which protects the steel even where the coating is scratched through, because protection is electrical rather than physical.

Run a cell backwards by applying a voltage greater than its own and the reaction is forced uphill. That is electrolysis, and the stoichiometry is exact, since electrons are counted in coulombs: the amount of substance transformed is $Q / n F$ , where $Q = I t$ . Michael Faraday established this in 1834, before anyone knew what an electron was, and it is one of the strongest early pieces of evidence that charge is carried in units.

The industrial scale of that arithmetic is worth seeing. Aluminium is produced as $A + 3 e^{-} \to Al$ , so one kilogram, which is $1000 / 26.982 = 37.06$ mol, needs $3 \times 37.06 \times 96485 = 1.07 \times 10^{7}$ C. A Hall-Héroult cell runs at about $4.5$ V, so the energy is $4.5 \times 1.07 \times 10^{7} = 4.8 \times 10^{7}$ J, or $13.4$ kW h per kilogram. Real smelters use $13$ to $15$ , the excess being resistive heating, and this is why aluminium plants are built next to hydroelectric dams and why recycling aluminium, which needs only melting, saves about $95$ per cent of the energy.

Where this leaves you

The course set out to answer three questions about a reaction: whether it goes, how far, and how fast. All three now have machinery behind them. Free energy decides direction, and $Δ_{r} G^{ominus} = - R T \ln K$ converts a table of enthalpies and entropies into a predicted yield. A rate law, measured rather than assumed, gives the time, and its temperature dependence gives a barrier height that belongs to a step rather than to an equation. Proton transfer, dissolution and electron transfer are the same equilibrium arithmetic applied three times.

What has been set aside throughout is worth naming. Activity coefficients have been taken as one, which is good to a few per cent in dilute solution and poor in seawater or in a battery. Rate constants have been treated as numbers to measure rather than to calculate, and calculating them from the potential energy surface is the subject of chemical dynamics. And nothing here says how to make a particular molecule rather than a mixture, which is synthesis, a discipline built on top of everything in this course rather than contained in it.

Molecules and Reactions, from libre.university