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In Class 12 you meet Nernst through the equation that gives a cell's EMF at any concentration. In Class 11 thermodynamics, you meet the third law, often without his name attached. The Nobel Prize went for the second idea. This post explains what Nernst's heat theorem says, how it connects to ΔG = ΔH – TΔS, how his laboratory tested it, and what it means for the problems you solve.
Who was Walther Nernst?
The Nobel Prize biography says Walther Hermann Nernst was born on 25 June 1864 in Briesen, West Prussia. He studied physics and mathematics at Zurich, Berlin and Graz, where Ludwig Boltzmann taught, and graduated at Würzburg in 1887. He then joined Wilhelm Ostwald at Leipzig, working alongside van 't Hoff and Arrhenius. In 1894 he founded the Institute for Physical Chemistry and Electrochemistry at Göttingen, and in 1905 moved to Berlin. He also invented the Nernst lamp, an electric light with a ceramic body. Both his sons were killed in the First World War. He died on 18 November 1941.
The Nobel facts page gives the prize motivation as "in recognition of his work in thermochemistry", and notes that he received the 1920 prize one year later, in 1921.
Where did the Nernst equation come from?
According to the biography, in 1889 Nernst explained how galvanic cells work by assuming an "electrolytic pressure of dissolution" that pushes ions from an electrode into solution, opposed by the osmotic pressure of the ions already dissolved. In his Nobel lecture he called his equation for the electrode potential one "which has been used successfully for a long time, and one might say daily, by numerous investigators". For worked EMF problems, see our guide to electrochemistry and the Nernst equation.
What does the heat theorem say?
The older idea, Berthelot's principle, said the heat a reaction gives out measures its chemical affinity, its driving force. Nernst found that this works in many cases but fails badly in others, such as melting. While studying gas equilibria at high temperatures, he noticed regular relationships between equilibrium and heat. After what he called "very lengthy, tedious calculations", he published Über die Berechnung chemischer Gleichgewichte aus thermischen Messungen (The calculation of chemical equilibria from thermal measurements) in 1906.
He stated the principle in his lecture like this: "in all cases chemical affinity and evolution of heat become identical at low temperatures". Not just at absolute zero, he stressed, but "the two curves become mutually tangential in the vicinity of absolute zero". The biography says this heat theorem is "known as the Third Law of Thermodynamics".
How does this connect to ΔG = ΔH – TΔS?
Here is an illustrative calculation. The numbers are chosen to show the idea, not taken from a real reaction.
Suppose a reaction has ΔH = –10.0 kJ/mol and ΔS = –20.0 J/K/mol.
At 298 K: TΔS = 298 × (–20.0) = –5960 J/mol = –5.96 kJ/mol. So ΔG = ΔH – TΔS = –10.0 – (–5.96) = –4.04 kJ/mol. Berthelot's principle, which equates affinity with heat, would be off by nearly 6 kJ/mol.
At 50 K, if ΔS stayed the same: TΔS = 50 × (–20.0) = –1000 J/mol = –1.00 kJ/mol, so ΔG = –9.0 kJ/mol. The gap has shrunk.
Near 0 K: T goes to zero, and Nernst's theorem says ΔS goes to zero as well. The TΔS term vanishes, so ΔG and ΔH meet, and they meet smoothly rather than at an angle.
The NCERT Class 11 thermodynamics chapter states the modern form: "The entropy of any pure crystalline substance approaches zero as the temperature approaches absolute zero." It adds why this matters: the third law "permits the calculation of absolute values of entropy of pure substance from thermal data alone". That is exactly what Nernst wanted, equilibrium from heat measurements.
Common mistakes in these problems:
Mixing J and kJ. ΔS is usually in J/K/mol and ΔH in kJ/mol. Convert before subtracting.
Sign errors. ΔG = ΔH – TΔS. A negative ΔS makes –TΔS positive. Write the formula out before substituting.
Applying the third law to everything. NCERT notes it is confined to pure crystalline solids; solutions and supercooled liquids do not have zero entropy at 0 K.
How did Nernst test his theorem?
By measuring specific heats down to very low temperatures, which took him years. In his lecture he said his institute carried out "more than 100 experimental studies" that confirmed the theorem. They found that specific heats of solids become very small at low temperatures. Einstein had begun to explain this with quantum theory, and Peter Debye then showed that at very low temperatures they rise with the cube of absolute temperature, which Nernst's laboratory confirmed for many substances. The biography adds that the theorem was soon applied to industrial problems, including calculations for ammonia synthesis. Nernst also mentioned that it had helped Meghnad Saha and others decide astrophysical questions.
What can a student take from Nernst's way of working?
Compare like with like. He first saw his pattern only when he compared reactions in which the number of molecules changed by the same amount. Group similar problems before looking for a rule.
Do the tedious part. His theorem came after long calculations, and its proof after a hundred experiments. Neat ideas usually sit on dull work.
Test consequences, not just claims. He checked his law by predicting specific heats and measuring them.
If ΔG, ΔH and ΔS problems keep going wrong at the same step, the Fermi Problem Library has a step-by-step solver that explains the reasoning rather than printing an answer.
Tonight, redo the illustrative example at 100 K and 10 K, then write one sentence explaining why the third law lets you look up absolute entropies in a table.
