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Wilhelm Ostwald

Friedrich Wilhelm Ostwald

A course in two strands: the life of the Riga cooper’s son who made physical chemistry a discipline, with its own chair, journal and school, and the science he left in every laboratory: the dilution law, the modern idea of catalysis, the nitric acid process and the physics of metastable solids.

Born 2 September 1853, Riga Died 4 April 1932, Leipzig Nobel Prize in Chemistry 1909 Level advanced undergraduate
Ostwald’s dilution law for acetic acid, \(K=\alpha^2c/(1-\alpha)=1.75\times10^{-5}\). The weak acid is barely ionized at 1 mol/L and almost fully ionized as the solution approaches infinite dilution.
Lesson 00

How to use this course

Wilhelm Ostwald is the organizer of physical chemistry. With Svante Arrhenius and Jacobus van ’t Hoff he built the theory of solutions and ions in the 1880s, and from Leipzig he trained a generation of chemists who took the subject to Europe, America and Japan. He also proposed the definition of a catalyst still taught today, patented the industrial route to nitric acid, and spent his last decades on colour, philosophy and the organization of knowledge. He was the most prominent chemist to reject atoms, and he publicly changed his mind.

Prerequisites for Part II
  • General chemistry: stoichiometry, equilibrium constants, acids and bases.
  • Elementary calculus: first-order differential equations and logarithms.
  • For Lessons 07 and 09, basic thermodynamics: Gibbs energy and chemical potential.
  • For Lesson 10, the ideal gas law and an outline of kinetic theory.

Each technical lesson opens with a paragraph for the general reader, then gives precise statements, derivations and exercises with folded solutions. The lessons follow Ostwald’s own path: affinity (1877–1884), electrolytes (1884–1891), analysis and catalysis (1891–1901), industry and solids (1897–1906), and the debate over atoms (1887–1909).

Reference

Timeline

1853Born 2 September (21 August, Old Style) in Riga, Livonia, in the Russian Empire, son of the master cooper Gottfried Wilhelm Ostwald and Elisabeth Leuckel.
1872Enters the University of Dorpat (now Tartu) to study chemistry.
1875Candidate’s examination; assistant to the physicist Arthur von Oettingen, then to the chemist Carl Schmidt.
1877Master’s thesis, Volumchemische Studien über Affinität.
1878Doctorate with Volumchemische und optisch-chemische Studien.
1880Marries Helene von Reyher on 24 April.
1881Appointed professor of chemistry at the Riga Polytechnic.
1884Reads Arrhenius’s dissertation on conductivity and travels to Sweden to meet him.
1887Chair of physical chemistry at Leipzig. Founds the Zeitschrift für physikalische Chemie with van ’t Hoff. Completes the Lehrbuch der allgemeinen Chemie.
1888The dilution law.
1889Launches the series Ostwalds Klassiker der exakten Wissenschaften; publishes the Grundriss der allgemeinen Chemie.
1894Defines catalysis as the acceleration of a slow reaction by a foreign substance. Founds the German Electrochemical Society, later the Bunsen Society. Die wissenschaftlichen Grundlagen der analytischen Chemie.
1895Lecture against “scientific materialism” at the Lübeck meeting of German scientists; Ludwig Boltzmann leads the opposition.
1897The step rule for the formation of solid phases.
1898New physico-chemical institute opens in Leipzig.
1901–02With Eberhard Brauer, oxidizes ammonia to nitric acid over platinum; pilot plant at Niederlehme, patents in 1902.
1902Founds the Annalen der Naturphilosophie.
1905–06First German exchange professor at Harvard; also lectures in other American universities.
1906Retires from Leipzig to his country house “Energie” at Großbothen, Saxony.
1908Fourth edition of the Grundriss: accepts the experimental evidence for atoms.
1909Nobel Prize in Chemistry, for catalysis, equilibria and reaction rates.
1911President of the German Monist League (to 1915); co-founds the Munich institute Die Brücke.
1914–17Begins his colour system; Die Farbenfibel and the first colour atlas.
1932Dies 4 April in a Leipzig hospital, aged 78. His urn is placed at Großbothen.
Part I · Formation
Lesson 01

Riga and Dorpat, 1853–1881

Friedrich Wilhelm Ostwald was born in Riga on 2 September 1853, a son of Gottfried Wilhelm Ostwald, a master cooper, and Elisabeth Leuckel. Riga was then a Baltic port of the Russian Empire whose merchants, craftsmen and professionals were largely German-speaking. The family belonged to the craftsmen’s class, and his father hoped the boy would become an engineer. He attended the Realgymnasium in Riga, a school stronger in science than in classics, and filled his free time with home chemistry experiments, painting and music. The habits stayed with him. He painted throughout his life and later wrote that art had given him refreshment and new courage.

In 1872 he entered the University of Dorpat, the German-language university of the Baltic provinces, now Tartu in Estonia. After a slow start he passed the candidate’s examination in 1875 and became an assistant, first to the physicist Arthur von Oettingen and then to the chemist Carl Schmidt. Ostwald later named these two men as the teachers who formed him: Oettingen gave him physics and measurement, Schmidt gave him chemistry and a laboratory.

A chemist who measured

His first research asked a question that chemists had been asking since the eighteenth century: how strongly does an acid hold a base? Instead of weighing products, Ostwald measured physical properties of solutions, their volume and their refractive index, and used the changes to see how a base divides itself between two competing acids (Lesson 04). The master’s thesis of 1877, Volumchemische Studien über Affinität, and the doctoral thesis of 1878, Volumchemische und optisch-chemische Studien, both used this method. On 24 April 1880 he married Helene von Reyher. They had five children; their son Wolfgang Ostwald, born in 1883, became a leading colloid chemist.

Lesson 02

The Riga Polytechnic and the textbook, 1881–1887

In 1881 Ostwald was appointed professor of chemistry at the Riga Polytechnic, at the age of 28. He continued to measure chemical affinity by new methods. He followed the acid-catalysed hydrolysis of methyl acetate and the inversion of cane sugar, and found that the order of acid strengths obtained from reaction rates agreed with the order from his volume measurements. This was the first sign that a single property of an acid controls very different phenomena.

In 1884 he received the doctoral dissertation of a young Swede, Svante Arrhenius, on the electrical conductivity of dilute solutions. Arrhenius proposed that part of an electrolyte in solution is “active” and that conductivity measures this part. Ostwald saw at once that conductivity tracked his affinity scale, and he travelled to Sweden in 1884 to meet Arrhenius. The two began a lifelong friendship and a scientific alliance. Ostwald invited Arrhenius to work with him in Riga, and he started measuring the conductivities of hundreds of acids himself.

The Lehrbuch

In the same years he wrote the Lehrbuch der allgemeinen Chemie, completed in 1887 in two volumes. It gathered everything known about the physical laws of chemistry: gases, solutions, thermochemistry, electrochemistry, affinity and reaction rates. It was the first systematic textbook of the field, and it made Ostwald the natural editor of a new subject.

Lesson 03

Leipzig and the Ionists, 1887–1895

In 1887 the Saxon government called Ostwald to Leipzig as professor of physical chemistry. The chair was the first appointed specifically in that subject. In the same year he founded the Zeitschrift für physikalische Chemie, with van ’t Hoff as co-editor. Its first volume carried van ’t Hoff’s theory of osmotic pressure and Arrhenius’s paper on electrolytic dissociation. Ostwald edited it until 1922, about a hundred volumes. Critics called the three men the “Ionists”, and the name stuck.

Leipzig became the world centre of the subject. Walther Nernst came as Ostwald’s assistant in 1887 and qualified as a lecturer there in 1889 before moving to Göttingen. Students and visitors included Arthur Amos Noyes, Willis Rodney Whitney, Theodore William Richards, Georg Bredig, Paul Walden, Max Le Blanc, Max Bodenstein, Herbert Freundlich, Frederick Donnan, James Walker and Kikunae Ikeda. Through Noyes and Whitney the Leipzig style reached MIT and the General Electric laboratory; through Ikeda it reached Tokyo. A new institute building opened in 1898.

Ostwald was also a tireless organizer. In 1889 he launched Ostwalds Klassiker der exakten Wissenschaften, cheap reprints of classic papers, which ran to more than 250 volumes. In 1892 he translated Josiah Willard Gibbs’s papers on heterogeneous equilibrium into German, which made them widely read in Europe. In 1894 he founded the German Electrochemical Society, renamed the Deutsche Bunsen-Gesellschaft in 1902. He was made a Geheimrat by the King of Saxony in 1899.

Part II · The chemistry
Lesson 04

Measuring affinity

For the general reader. Mix a base with two acids and it will divide itself between them. The stronger acid takes the larger share. Ostwald measured that share by watching how the volume of the solution changed, then showed that the same ranking of acids appears in reaction speeds and electrical conductivity. Behind all three lies one number for each acid.

Goals
  • Understand Ostwald’s volume-chemical method for the partition of a base between two acids.
  • Show with mass action that partition, catalytic rate and conductivity all measure the square root of the dissociation constant.

The method

Neutralizing an acid with a base changes the volume of the solution by an amount characteristic of the acid. Julius Thomsen had used heats of neutralization in the same way. Suppose neutralizing acid HA by the base changes the volume by \(\Delta V_A\) per mole and acid HB by \(\Delta V_B\). Add one mole of base to a mixture of one mole of each acid. If a fraction \(x\) of the base goes to HA, the observed change is

\[ \Delta V = x\,\Delta V_A + (1-x)\,\Delta V_B, \qquad x=\frac{\Delta V-\Delta V_B}{\Delta V_A-\Delta V_B}. \]

Ostwald called the ratio \(x/(1-x)\) the relative affinity, or avidity, of the two acids. He then measured the same acids by two other routes: the rates of acid-catalysed reactions, and later their conductivities. In modern language all three are governed by the acid dissociation constant \(K_a\).

Proposition (the three measures agree)

For two weak acids at equal concentration \(c\) with dissociation constants \(K_A, K_B\ll c\):

  1. the partition of one equivalent of base between equimolar amounts of the acids satisfies \(x/(1-x)=\sqrt{K_A/K_B}\);
  2. the ratio of hydrogen-ion concentrations, hence of rates of acid-catalysed reactions, is \(\sqrt{K_A/K_B}\);
  3. the ratio of molar conductivities is approximately \(\sqrt{K_A/K_B}\), when the two anions have similar mobilities.

Derivation of (1). The base settles into the exchange equilibrium \(\mathrm{A^-}+\mathrm{HB}\rightleftharpoons\mathrm{HA}+\mathrm{B^-}\), whose constant is \(K_B/K_A\). With \(x\) mol of A\(^-\), \(1-x\) of HA, \(1-x\) of B\(^-\) and \(x\) of HB,

\[ \frac{K_B}{K_A}=\frac{(1-x)^2}{x^2}\quad\Longrightarrow\quad \frac{x}{1-x}=\sqrt{\frac{K_A}{K_B}}. \]

Derivation of (2) and (3). For a weak acid alone, \([\mathrm{H^+}]\approx\sqrt{K_ac}\). The rate of an acid-catalysed reaction is proportional to \([\mathrm{H^+}]\), and the conductivity is proportional to the ion concentration \([\mathrm{H^+}]=[\mathrm{A^-}]\). The square root appears every time, which is why Ostwald’s three scales agreed with one another. Strong acids such as hydrochloric and nitric acid sit at the top of all three scales together.

Exercise 4.1A base is shared between equal amounts of formic acid (\(K_a=1.8\times10^{-4}\)) and acetic acid (\(K_a=1.75\times10^{-5}\)). What fraction goes to formic acid?

Show solution
\(x/(1-x)=\sqrt{1.8\times10^{-4}/1.75\times10^{-5}}=\sqrt{10.3}=3.21\), so \(x=3.21/4.21=0.76\). About three quarters of the base goes to formic acid.

Exercise 4.2Ostwald compared the rates at which 0.1 mol/L hydrochloric acid and 0.1 mol/L acetic acid invert cane sugar. Estimate the ratio of rates, assuming the rate is proportional to \([\mathrm{H^+}]\) and hydrochloric acid is fully dissociated.

Show solution
For acetic acid, \(\alpha^2c/(1-\alpha)=K\) with \(c=0.1\) gives \(\alpha\approx0.0131\), so \([\mathrm{H^+}]\approx1.31\times10^{-3}\). Hydrochloric acid gives \(0.1\). The ratio is about 76: the strong acid inverts sugar roughly 76 times faster. Ostwald’s measured ratios were of this order.

Exercise 4.3Sugar inversion is first order in sucrose. If 0.5 mol/L hydrochloric acid at 25 °C gives a rate constant of about \(1\times10^{-4}\ \mathrm{s^{-1}}\), how long does it take to invert 90% of the sugar?

Show solution
\(t=\ln10/k=2.303/10^{-4}\ \mathrm{s}\approx2.3\times10^4\ \mathrm{s}\), about 6.4 hours. The slowness is what made the reaction a convenient clock for Ostwald and his predecessor Ludwig Wilhelmy, who measured it with a polarimeter in 1850.
Lesson 05

Conductivity and the dilution law

For the general reader. Salt water conducts electricity because the salt splits into charged ions. Arrhenius proposed that a weak acid splits only partly, and more completely the more it is diluted. In 1888 Ostwald applied the law of mass action to that splitting and got a formula that predicted, from a single constant, how the conductivity of any weak acid changes with dilution. It was the first quantitative triumph of the ionic theory.

Goals
  • Derive Ostwald’s dilution law from mass action.
  • Use Kohlrausch’s law to find the limiting conductivity of a weak acid.
  • Extract a dissociation constant from conductivity data and see where the law fails.

Molar conductivity

Friedrich Kohlrausch had shown how to measure the conductivity \(\kappa\) of solutions with alternating current, avoiding electrolysis at the electrodes. The molar conductivity is \(\Lambda=\kappa/c\). For salts and strong acids \(\Lambda\) changes little with dilution. For weak acids it rises steeply. Kohlrausch also found that at infinite dilution each ion contributes independently:

\[ \Lambda_0=\lambda_+ + \lambda_- . \]

Arrhenius interpreted \(\alpha=\Lambda/\Lambda_0\) as the fraction of the electrolyte dissociated into ions.

Ostwald’s dilution law (1888)

For a binary weak electrolyte \(\mathrm{HA}\rightleftharpoons\mathrm{H^+}+\mathrm{A^-}\) at total concentration \(c\), with degree of dissociation \(\alpha=\Lambda/\Lambda_0\),

\[ K=\frac{\alpha^2c}{1-\alpha}=\frac{\Lambda^2c}{\Lambda_0(\Lambda_0-\Lambda)} \]

is independent of \(c\). In particular \(\alpha\to1\) as \(c\to0\), and \(\alpha\approx\sqrt{K/c}\) when \(\alpha\ll1\).

Derivation

At equilibrium \([\mathrm{H^+}]=[\mathrm{A^-}]=\alpha c\) and \([\mathrm{HA}]=(1-\alpha)c\). Mass action for dilute solutions, which van ’t Hoff had just justified by treating solutes like gases, gives

\[ K=\frac{[\mathrm{H^+}][\mathrm{A^-}]}{[\mathrm{HA}]}=\frac{(\alpha c)^2}{(1-\alpha)c}=\frac{\alpha^2c}{1-\alpha}. \]

Solving the quadratic, \(\alpha=\big(-K+\sqrt{K^2+4Kc}\big)/(2c)\). This is the curve in the header. Ostwald tested the law on more than two hundred organic acids and found constant \(K\) over wide ranges of dilution. The constant became the standard measure of acid strength, and it showed systematic regularities: chlorine substitution in acetic acid raised \(K\) sharply, and the effect weakened with distance from the carboxyl group.

Where the law fails

For strong electrolytes such as KCl, \(K\) computed this way drifts strongly with concentration. Kohlrausch had found empirically that \(\Lambda=\Lambda_0-A\sqrt c\) for these salts. The explanation came only in 1923, when Peter Debye and Erich Hückel showed that strong electrolytes are fully dissociated and that the fall in \(\Lambda\) comes from interionic attraction. Ostwald’s law is exact in spirit for weak electrolytes once activities replace concentrations.

Exercise 5.1At 25 °C the limiting molar conductivities are \(\Lambda_0(\mathrm{HCl})=426.2\), \(\Lambda_0(\mathrm{CH_3COONa})=91.0\) and \(\Lambda_0(\mathrm{NaCl})=126.5\ \mathrm{S\,cm^2\,mol^{-1}}\). Find \(\Lambda_0\) for acetic acid, which cannot be measured by extrapolation.

Show solution
By Kohlrausch’s independent migration, \(\Lambda_0(\mathrm{CH_3COOH})=\Lambda_0(\mathrm{HCl})+\Lambda_0(\mathrm{CH_3COONa})-\Lambda_0(\mathrm{NaCl})=426.2+91.0-126.5=390.7\ \mathrm{S\,cm^2\,mol^{-1}}\). The Na\(^+\) and Cl\(^-\) contributions cancel.

Exercise 5.2Acetic acid at \(c=0.0100\) mol/L has \(\Lambda=16.3\ \mathrm{S\,cm^2\,mol^{-1}}\). Find \(\alpha\) and \(K\). Then predict \(\Lambda\) at \(c=0.001\) mol/L.

Show solution
\(\alpha=16.3/390.7=0.0417\), \(K=\alpha^2c/(1-\alpha)=(0.0417)^2(0.01)/0.958=1.82\times10^{-5}\). At \(c=0.001\), solve \(\alpha^2(0.001)/(1-\alpha)=1.82\times10^{-5}\): \(\alpha\approx0.126\), so \(\Lambda\approx0.126\times390.7\approx49\ \mathrm{S\,cm^2\,mol^{-1}}\). Tenfold dilution raises \(\Lambda\) by a factor of about three, close to \(\sqrt{10}\).

Exercise 5.3Show that the dilution law can be written \(\dfrac1\Lambda=\dfrac1{\Lambda_0}+\dfrac{c\Lambda}{K\Lambda_0^2}\). How does this give \(\Lambda_0\) and \(K\) from data alone?

Show solution
From \(K\Lambda_0(\Lambda_0-\Lambda)=\Lambda^2c\), divide by \(K\Lambda_0^2\Lambda\): \(\dfrac{1}{\Lambda}-\dfrac1{\Lambda_0}=\dfrac{\Lambda c}{K\Lambda_0^2}\). Plot \(1/\Lambda\) against \(c\Lambda\): the intercept is \(1/\Lambda_0\) and the slope is \(1/(K\Lambda_0^2)\). This linearization, associated with Charles Kraus and William Bray, avoids extrapolating a steep curve.
Lesson 06

Indicators and the theory of analysis

For the general reader. Analytical chemistry in the 1880s was a collection of recipes. In 1894 Ostwald’s small book Die wissenschaftlichen Grundlagen der analytischen Chemie explained the recipes with ions and equilibrium. Its best-known chapter explained why a colour indicator changes colour and why the right indicator depends on the acid being titrated.

Goals
  • State Ostwald’s theory of acid-base indicators.
  • Derive the colour-change interval and choose an indicator for a titration.
Ostwald’s indicator theory (1894)

An indicator is a weak acid HIn (or weak base) whose undissociated form and anion have different colours. In a solution of hydrogen-ion concentration \([\mathrm{H^+}]\),

\[ \frac{[\mathrm{In^-}]}{[\mathrm{HIn}]}=\frac{K_{\mathrm{In}}}{[\mathrm{H^+}]}. \]

The eye sees a change when this ratio passes from about \(1/10\) to about \(10\), so the colour changes over a hundredfold range of \([\mathrm{H^+}]\) centred on \(K_{\mathrm{In}}\).

In the logarithmic notation introduced by Søren Sørensen in 1909, the transition interval is \(\mathrm{pH}=\mathrm{p}K_{\mathrm{In}}\pm1\). Ostwald’s theory was later refined, since many indicators also change structure when they lose a proton, but the equilibrium argument and its practical conclusion stand: a titration must use an indicator whose interval contains the pH at the equivalence point.

Hydrolysis

At the equivalence point of a weak acid titrated by a strong base the solution contains only the salt, and the anion reacts with water: \(\mathrm{A^-}+\mathrm{H_2O}\rightleftharpoons\mathrm{HA}+\mathrm{OH^-}\), with constant \(K_b=K_w/K_a\). The solution is alkaline, which is why phenolphthalein works for acetic acid and methyl orange does not.

Exercise 6.10.10 mol/L acetic acid is titrated with 0.10 mol/L sodium hydroxide. Find the pH at the equivalence point (\(K_a=1.75\times10^{-5}\), \(K_w=1.0\times10^{-14}\)). Which indicator should be used: methyl orange (interval 3.1–4.4) or phenolphthalein (8.2–10)?

Show solution
At equivalence the volume has doubled, so \([\mathrm{A^-}]=0.050\). \(K_b=10^{-14}/1.75\times10^{-5}=5.7\times10^{-10}\), \([\mathrm{OH^-}]=\sqrt{K_b\cdot0.050}=5.3\times10^{-6}\), pOH 5.27, pH 8.73. Phenolphthalein. Methyl orange would change colour far too early, while most of the acid is still unneutralized.

Exercise 6.2An indicator has \(K_{\mathrm{In}}=1\times10^{-7}\). What fraction is in the anion form at pH 6, 7 and 8?

Show solution
The fraction is \(f=K_{\mathrm{In}}/(K_{\mathrm{In}}+[\mathrm{H^+}])\). At pH 6: \(10^{-7}/(1.1\times10^{-6})=0.09\). At pH 7: 0.5. At pH 8: \(10^{-7}/(1.1\times10^{-7})=0.91\). The colour goes from about 9% to 91% of the anion form across two pH units, as the theory predicts. Bromothymol blue behaves roughly like this.
Lesson 07

Catalysis

For the general reader. Some substances make reactions go faster without being used up. Jöns Jacob Berzelius had named the effect catalysis in 1835 and attributed it to a mysterious “catalytic force”. Ostwald removed the mystery. A catalyst, he said, only changes the speed of a reaction that could happen anyway. It cannot make an impossible reaction possible, and it cannot change where the reaction ends up.

Goals
  • State Ostwald’s kinetic definition of a catalyst and the principle that catalysts do not shift equilibrium.
  • Relate rate enhancement to activation energy.
  • Solve the rate equation of an autocatalytic reaction.
Definition (Ostwald, 1894; 1901)

Catalysis is the acceleration of a slowly occurring chemical process by the presence of a foreign substance. In his 1901 lecture Über Katalyse Ostwald gave the general form: a catalyst is any substance that alters the rate of a chemical reaction without appearing in its final products.

The 1894 statement appeared in a short review in his own journal. Its starting point was thermodynamic. A homogeneous system that can lower its free energy by reacting must react, however slowly; the energy laws require only that the rate not be exactly zero. A catalyst changes that rate. It supplies no energy, so it cannot drive a reaction uphill.

Principle (a catalyst does not shift equilibrium)

A catalyst present in small amount accelerates the forward and reverse reactions by the same factor, so the equilibrium constant \(K=k_{\mathrm f}/k_{\mathrm r}\) is unchanged.

Argument. Suppose a catalyst shifted the equilibrium of a gas reaction that changes the number of molecules. Introducing and withdrawing the catalyst repeatedly would make the pressure rise and fall, and a piston could extract work at constant temperature indefinitely. That is a perpetual motion machine of the second kind. In modern terms, \(K=e^{-\Delta G^\circ/RT}\) depends only on the initial and final states, and the catalyst is neither.

How much faster?

Arrhenius showed in 1889 that rate constants follow \(k=A\,e^{-E_a/RT}\). A catalyst opens a path with a lower activation energy. If it lowers \(E_a\) by \(\Delta E\) without changing \(A\), the rate rises by \(e^{\Delta E/RT}\) in both directions.

Autocatalysis

In some reactions a product is itself a catalyst. Ostwald studied the hydrolysis of esters, which produces an acid that speeds the hydrolysis. If the rate is proportional to both the remaining reactant and the catalytic product, with \(x\) the amount converted,

\[ \frac{dx}{dt}=k\,(x_0+x)(a-x), \]

where \(a\) is the initial reactant and \(x_0\) the initial catalyst. The reaction starts slowly, accelerates, then slows: the S-shaped curve later recognized in population growth and in clock reactions.

Exercise 7.1A catalyst lowers the activation energy of a reaction by 20 kJ/mol at 300 K. By what factor does it multiply the rate constant? What happens to the equilibrium constant?

Show solution
\(e^{20000/(8.314\times300)}=e^{8.02}\approx3.0\times10^3\). Both \(k_{\mathrm f}\) and \(k_{\mathrm r}\) rise by this factor, since the transition state is shared, so \(K\) is unchanged. Equilibrium is reached about 3000 times sooner.

Exercise 7.2Solve the autocatalytic rate law with \(x(0)=0\), and find the time of maximum rate.

Show solution
Partial fractions: \(\dfrac{dx}{(x_0+x)(a-x)}=\dfrac{1}{a+x_0}\Big(\dfrac{1}{x_0+x}+\dfrac{1}{a-x}\Big)dx\). Integrating, \(\ln\dfrac{a(x_0+x)}{x_0(a-x)}=(a+x_0)kt\), so \(x(t)=\dfrac{a x_0\,(e^{(a+x_0)kt}-1)}{a+x_0e^{(a+x_0)kt}}\). The rate \(k(x_0+x)(a-x)\) is a downward parabola in \(x\), maximal at \(x=(a-x_0)/2\), reached at \(t^*=\dfrac{1}{(a+x_0)k}\ln\dfrac{a}{x_0}\) (for \(x_0<a\)).

Exercise 7.3Platinum catalyses \(2\mathrm{SO_2}+\mathrm{O_2}\rightleftharpoons2\mathrm{SO_3}\). Explain from the principle above why a better catalyst does not raise the equilibrium yield at a given temperature, and why industry nonetheless values a catalyst that works at lower temperature.

Show solution
At fixed \(T\) and pressure the yield is set by \(K(T)\), which the catalyst does not change. The reaction is exothermic, so \(K\) falls as \(T\) rises (van ’t Hoff’s equation). A catalyst active at lower \(T\) lets the plant operate where \(K\) is larger while the rate stays acceptable. The catalyst changes which equilibrium is practically reachable, not the equilibrium itself.
Lesson 08

Nitric acid from ammonia

For the general reader. Nitric acid is needed for fertilizers and explosives. Around 1900 almost all of it came from Chilean saltpetre, a finite supply across an ocean. Ostwald and his assistant Eberhard Brauer showed that ammonia burned over hot platinum for a tiny fraction of a second gives nitric oxide instead of nitrogen. The Ostwald process, combined a decade later with ammonia from the Haber–Bosch process, still makes nearly all the world’s nitric acid.

Goals
  • Write the chemistry of the process and its overall balance.
  • Understand why selectivity, not equilibrium, is the problem the catalyst solves.

History

Frédéric Kuhlmann had observed in 1838 that ammonia and air over platinum give oxides of nitrogen, but ammonia was scarce and the idea lay dormant. From 1900 Ostwald and Brauer studied the reaction systematically. A pilot plant at Niederlehme, near Berlin, produced acid from February 1902, and Ostwald patented the process that year. Production moved to Gerthe near Bochum, using ammonia from coking plants, and reached about 300 kg of nitric acid a day by 1906. When Fritz Haber and Carl Bosch made ammonia from the air from 1913, the two processes together freed Germany from imported nitrates, a fact with grave consequences in the First World War.

The chemistry

\[ \begin{aligned} 4\,\mathrm{NH_3}+5\,\mathrm{O_2}&\longrightarrow4\,\mathrm{NO}+6\,\mathrm{H_2O} &\quad \Delta H&=-905\ \mathrm{kJ}\\ 2\,\mathrm{NO}+\mathrm{O_2}&\longrightarrow2\,\mathrm{NO_2} &\quad \Delta H&=-114\ \mathrm{kJ}\\ 3\,\mathrm{NO_2}+\mathrm{H_2O}&\longrightarrow2\,\mathrm{HNO_3}+\mathrm{NO} &\quad \Delta H&=-117\ \mathrm{kJ} \end{aligned} \]

The difficulty lies in the first step. The competing reaction \(4\,\mathrm{NH_3}+3\,\mathrm{O_2}\to2\,\mathrm{N_2}+6\,\mathrm{H_2O}\) is thermodynamically even more favourable, and nitric oxide itself is unstable with respect to its elements. Equilibrium would give almost no NO. The Ostwald–Brauer insight was kinetic: on platinum the oxidation to NO is extremely fast, so a contact time of a small fraction of a second converts the ammonia before the slower routes to nitrogen take over. The process is a pure application of Ostwald’s view of catalysis: the catalyst chooses among reactions by speed. Modern plants use platinum–rhodium gauzes at about 900 °C.

Overall balance

With the NO from the absorption step recycled, the process is \(\mathrm{NH_3}+2\,\mathrm{O_2}\longrightarrow\mathrm{HNO_3}+\mathrm{H_2O}\).

Exercise 8.1Find the maximum mass of nitric acid obtainable from one tonne of ammonia. If the plant converts 96% of the ammonia to NO and later steps lose 2% of the nitrogen, what is the actual yield?

Show solution
Molar masses 17.03 and 63.01, so the maximum is \(63.01/17.03=3.70\) t. Actual: \(3.70\times0.96\times0.98\approx3.48\) t.

Exercise 8.2Show that the overall equation follows from the three steps when the NO from absorption is recycled, and find the heat released per mole of ammonia in the first step.

Show solution
Take \(\tfrac14\) of the first step, \(\tfrac32\) of the second and \(\tfrac12\) of the third: \(\mathrm{NH_3}+\tfrac54\mathrm{O_2}\to\mathrm{NO}+\tfrac32\mathrm{H_2O}\); \(\tfrac32\mathrm{NO}+\tfrac34\mathrm{O_2}\to\tfrac32\mathrm{NO_2}\); \(\tfrac32\mathrm{NO_2}+\tfrac12\mathrm{H_2O}\to\mathrm{HNO_3}+\tfrac12\mathrm{NO}\). Adding, the half mole of NO from absorption cancels against the recycle, and the sum is \(\mathrm{NH_3}+2\mathrm{O_2}\to\mathrm{HNO_3}+\mathrm{H_2O}\). Check: N 1 = 1, H 3 = 1 + 2, O 4 = 3 + 1. Heat of the first step: \(905/4\approx226\) kJ per mole of NH\(_3\), enough to keep the gauze hot without external heating once lit.
Lesson 09

Metastability: the step rule and ripening

For the general reader. When a solid forms from a solution or a melt, it rarely appears at once in its most stable form. Ostwald noticed that nature tends to go downhill in steps, first producing a less stable form that later turns into a more stable one. He also saw that in a population of crystals the big ones grow while the small ones dissolve. Both effects matter today in pharmaceuticals, metallurgy and nanoparticle synthesis.

Goals
  • State the step rule and recognize its examples.
  • Derive the size dependence of solubility and use it to explain Ostwald ripening.
Ostwald’s step rule (1897)

When a system can pass into several more stable states, it usually goes first to the nearest one, the least stable of the forms accessible to it, rather than directly to the most stable state. This is an empirical tendency, not a theorem.

Examples: phosphorus vapour condenses as white phosphorus, which slowly turns red; calcium carbonate precipitates first as an amorphous phase and vaterite before calcite; titanium dioxide often forms anatase before rutile. The usual explanation is kinetic. A less stable phase is often structurally closer to the liquid and has a lower interfacial energy, so its nucleation barrier is lower and it forms faster.

Small particles are more soluble

A small crystal of radius \(r\) has a large surface relative to its volume. The surface energy \(\gamma\) raises the chemical potential of its molecules by \(2\gamma V_m/r\), where \(V_m\) is the molar volume. Ostwald published the consequence for solubility in 1900, and Herbert Freundlich refined it in 1909.

Ostwald–Freundlich equation \[ \ln\frac{c(r)}{c(\infty)}=\frac{2\gamma V_m}{rRT}, \]

where \(c(r)\) is the solubility of particles of radius \(r\) and \(c(\infty)\) that of a flat surface. It is the solution analogue of William Thomson’s (Lord Kelvin’s) equation for the vapour pressure of droplets.

Ripening. In a suspension of crystals of different sizes, the solution is supersaturated with respect to the large ones and undersaturated with respect to the small ones. Material diffuses from small to large: the small crystals dissolve and the large ones grow. Ostwald described this coarsening in 1896. Ilya Lifshitz, Vitaly Slyozov and Carl Wagner showed in 1961 that, when diffusion controls the rate, the mean radius obeys \(\langle r\rangle^3-\langle r\rangle_0^3\propto t\).

Exercise 9.1For a salt with \(\gamma=0.10\ \mathrm{J\,m^{-2}}\) and \(V_m=5.0\times10^{-5}\ \mathrm{m^3\,mol^{-1}}\) at 298 K, how much more soluble are particles of radius 10 nm than large crystals? And of radius 1 µm?

Show solution
\(\dfrac{2\gamma V_m}{rRT}=\dfrac{2(0.10)(5.0\times10^{-5})}{(10^{-8})(8.314)(298)}=0.404\), so \(c(r)/c(\infty)=e^{0.404}=1.50\): 50% more soluble. At 1 µm the exponent is 0.004 and the excess is 0.4%. The effect matters only below about 100 nm.

Exercise 9.2Under the Lifshitz–Slyozov–Wagner law, a precipitate’s mean radius grows from 0.1 µm to 1 µm in 10 hours. Neglecting the starting size, how long until it reaches 2 µm?

Show solution
\(\langle r\rangle^3\propto t\). Neglecting \(0.1^3\) against \(1^3\), reaching 1 µm takes 10 h, so \(k=0.1\ \mu\mathrm m^3/\mathrm h\). Reaching 2 µm needs \(\langle r\rangle^3=8\), i.e. 80 h in total, another 70 hours. Doubling the radius takes eight times as long: chemists use this when they “digest” a precipitate to make it filterable.
Lesson 10

Energetics and the atom

For the general reader. In the 1890s nobody had seen an atom. Ostwald argued that science should describe only what can be measured, and that energy, not matter in motion, is the fundamental reality. For fifteen years he was the most prominent opponent of atoms in chemistry. Then experiments on the jiggling of tiny particles in water persuaded him, and in 1908 he said so in print.

Goals
  • Understand the energetics programme and the 1895 Lübeck debate.
  • Compute a Brownian displacement from the Einstein–Stokes relation, the measurement that settled the question.

Energetics

Ostwald’s Leipzig inaugural lecture of 1887 already presented energy and its transformations as the unifying principle of science. From 1893 he argued that the atomic hypothesis was unnecessary: chemical laws could be stated as relations among forms of energy and their intensity and capacity factors. Georg Helm developed a similar programme. Ernst Mach shared the scepticism about atoms from a different, phenomenalist position.

At the meeting of German scientists and physicians in Lübeck in September 1895, Helm and Ostwald presented energetics, Ostwald under the title of overcoming scientific materialism. Ludwig Boltzmann, supported by Felix Klein, attacked the programme in the discussion. Boltzmann’s central objection was that energetics could not account for irreversibility and the second law, which kinetic theory explained statistically. Arnold Sommerfeld, then a young listener, later recalled that Boltzmann’s arguments won the day. Ostwald and Boltzmann remained on friendly terms, and Boltzmann later spent 1900–1902 as a professor at Leipzig, Ostwald’s university.

The reversal

Two lines of evidence changed Ostwald’s mind. J. J. Thomson had identified the electron in 1897 and measured its charge-to-mass ratio, and Jean Perrin’s measurements from 1908 confirmed Albert Einstein’s 1905 theory of Brownian motion in detail. In the preface to the fourth edition of his Grundriss der allgemeinen Chemie (1908), Ostwald accepted that the atomic hypothesis had become a well-founded scientific theory. He kept energetics as a philosophy of nature for the rest of his life.

Einstein–Stokes relation (1905)

A sphere of radius \(a\) in a fluid of viscosity \(\eta\) at temperature \(T\) diffuses with coefficient

\[ D=\frac{RT}{N_A}\,\frac{1}{6\pi\eta a}, \qquad \langle x^2\rangle=2Dt \ \text{ along any one axis}. \]

Measuring \(\langle x^2\rangle\), \(a\), \(\eta\) and \(T\) therefore gives Avogadro’s number \(N_A\).

The relation joins a visible quantity, the spread of a particle seen in a microscope, to the number of molecules in a mole, a word Ostwald himself introduced into chemistry. A world without molecules would have no such spread.

Exercise 10.1Estimate the root-mean-square displacement along one axis, in one minute, of a sphere of radius 0.5 µm in water (\(\eta=1.0\times10^{-3}\) Pa s) at 293 K.

Show solution
\(D=\dfrac{8.314\times293}{6.022\times10^{23}\times6\pi\times10^{-3}\times0.5\times10^{-6}}=4.3\times10^{-13}\ \mathrm{m^2/s}\). Then \(\sqrt{2Dt}=\sqrt{2\times4.3\times10^{-13}\times60}=7.2\times10^{-6}\) m, about 7 µm: easily seen in a microscope.

Exercise 10.2Perrin measured a displacement and solved for \(N_A\). If a particle of radius 0.5 µm showed \(\langle x^2\rangle=5.0\times10^{-11}\ \mathrm{m^2}\) in 60 s under the conditions of Exercise 10.1, what value of \(N_A\) follows?

Show solution
\(D=\langle x^2\rangle/2t=4.17\times10^{-13}\). \(N_A=RT/(6\pi\eta aD)=\dfrac{2436}{9.42\times10^{-9}\times4.17\times10^{-13}}=6.2\times10^{23}\ \mathrm{mol^{-1}}\). Perrin’s values from several methods clustered between about 6 and 7 \(\times10^{23}\).

DiscussionOstwald argued that hypotheses about unobservable entities should be avoided when the measurable relations suffice. Which parts of Lessons 04–09 could be stated without atoms, and which would be hard to formulate without them?

Part III · Life and legacy
Lesson 11

Großbothen, the prize and the world, 1906–1932

By 1905 Ostwald was tired of teaching and in conflict with the Leipzig philosophy faculty. In 1905–06 he went to the United States as the first German exchange professor at Harvard and lectured at other American universities. In 1906 he retired at 52 to his country house at Großbothen in Saxony, bought in 1901, which he named “Energie”. He worked there as a private scholar for the rest of his life.

In 1909 he received the Nobel Prize in Chemistry in recognition of his work on catalysis and his investigations of the fundamental principles governing chemical equilibria and rates of reaction. He had been nominated repeatedly since 1904, and his nominators praised the teacher and organizer as much as the experimenter. As a laureate he gained the right to nominate, and he proposed Albert Einstein for the physics prize. In 1901 the young Einstein had written to Ostwald asking for an assistant’s post and received no reply.

Philosopher and reformer

Ostwald turned his energetics into an ethics, summed up in an “energetic imperative”: do not waste energy, convert it to its most useful form. From 1911 to 1915 he presided over the German Monist League founded by Ernst Haeckel, writing popular “Sunday sermons” for a scientific worldview opposed to church influence. Monist circles, Ostwald among them, also promoted eugenic ideas that were common before the First World War and are rightly judged harshly today.

He campaigned for an international auxiliary language, first Esperanto and then Ido, to which he gave much of his prize money. In 1911 he co-founded Die Brücke in Munich, an institute for the organization of intellectual work. It proposed a “world format” for printed paper with sides in the ratio \(1:\sqrt2\), so that halving a sheet preserves its shape. The institute went bankrupt in 1914, but the idea, carried forward by his former assistant Walter Porstmann, became the German standard DIN 476 in 1922 and through it the A-series paper sizes used worldwide.

Colour

From 1914, at the request of the Deutscher Werkbund, Ostwald built a quantitative colour system. Each colour was described by its hue and its proportions of white and black, arranged in a double cone with white and black at the apexes. Die Farbenfibel and his first colour atlas appeared in 1916–17. The system was disputed by artists such as Adolf Hölzel, but it influenced De Stijl, the Bauhaus and later colour-order systems. He founded the journal Die Farbe in 1927.

Exercise 11.1Show that a rectangle with sides \(1:\rho\), \(\rho>1\), cut in half across its long side, gives two rectangles of the same shape if and only if \(\rho=\sqrt2\). If the largest sheet has area 1 m², find its sides.

Show solution
Halving the long side of a \(1\times\rho\) sheet gives two sheets of size \(1\times\tfrac\rho2\). Their long-to-short ratio is \(1/(\rho/2)=2/\rho\) (for \(\rho<2\)). Same shape means \(2/\rho=\rho\), so \(\rho^2=2\). With sides \(w\) and \(\sqrt2w\) and area \(\sqrt2w^2=1\), \(w=2^{-1/4}\) m \(\approx0.841\) m and the long side is \(2^{1/4}\approx1.189\) m: DIN A0, 841 × 1189 mm.
Lesson 12

Death and legacy

Ostwald remained active at Großbothen into his late seventies, writing on colour, philosophy and the history of science, and completing a three-volume autobiography, Lebenslinien (1926–27). He died in a Leipzig hospital on 4 April 1932, aged 78. His urn was placed in the old quarry on his estate. The house, laboratory and library are preserved, and the grounds have been the Wilhelm-Ostwald-Park since 2009.

Legacy

  • A discipline: with van ’t Hoff and Arrhenius he made physical chemistry a field with its own chair, journal, textbooks and society. Leipzig students founded or led physical chemistry in the United States, Britain, Russia and Japan.
  • Electrolytes: the dilution law and his hundreds of dissociation constants made the ionic theory quantitative, and his indicator theory gave analytical chemistry a rational basis.
  • Catalysis: his kinetic definition is still the textbook definition, and the Ostwald process still makes most of the world’s nitric acid.
  • Materials: the step rule, Ostwald ripening and the Ostwald–Freundlich equation are standard tools in crystallization and colloid science. The Ostwald viscometer is still in laboratory catalogues.
  • Atoms: his reversal of 1908 is a classic case of a leading scientist yielding to new experiments.
  • Beyond chemistry: the Klassiker series, the 1:√2 paper format and a colour system.
  • Honours: Faraday Lectureship of the Chemical Society (1904); foreign membership of the US National Academy of Sciences, the American Academy of Arts and Sciences and the Royal Netherlands Academy; the Wilhelm Exner Medal (1923); a crater on the far side of the Moon.

Final projectChoose one of Ostwald’s results from Part II. Write a two-page account for a strong undergraduate that states it, reproduces one of his numerical tests with modern data, and explains whether the result depends on the atomic hypothesis he opposed when he found it.

Reference

Sources and further reading

Biography

  1. Wilhelm Ostwald, Biographical, NobelPrize.org.
  2. Wilhelm Ostwald, Facts, NobelPrize.org (prize motivation).
  3. Wilhelm Ostwald, Wikipedia; German Wikipedia for the Leipzig years, the nitric acid plants and Die Brücke.
  4. W. Ostwald, Lebenslinien: Eine Selbstbiographie, 3 vols., Klasing, Berlin, 1926–27.
  5. Energeticism, Wikipedia, on the Lübeck debate and the 1908 reversal.
  6. R. S. Root-Bernstein, The Ionists: Founding Physical Chemistry, 1872–1890, PhD thesis, Princeton University, 1980.

Chemistry

  1. W. Ostwald, Zur Theorie der Elektrolyte, Z. phys. Chem. 2 (1888): the dilution law.
  2. W. Ostwald, Die wissenschaftlichen Grundlagen der analytischen Chemie, Engelmann, Leipzig, 1894.
  3. W. Ostwald, Studien über die Bildung und Umwandlung fester Körper, Z. phys. Chem. 22 (1897) 289–330: the step rule.
  4. W. Ostwald, Catalysis (1901 lecture, Z. Elektrochem. 7, 995–1003), English excerpt in the Le Moyne College classic chemistry collection.
  5. Ostwald process, Wikipedia (reactions and enthalpies).
  6. Ostwald’s rule and Ostwald ripening, Wikipedia.
  7. P. Atkins and J. de Paula, Physical Chemistry, Oxford University Press: conductivity, Kohlrausch’s law, catalysis.
  8. A. Einstein, Über die von der molekularkinetischen Theorie der Wärme geforderte Bewegung von in ruhenden Flüssigkeiten suspendierten Teilchen, Ann. Phys. 17 (1905) 549–560.