Chemical Thermodynamics and Thermochemistry

1The first law, and the sign of work

Chemistry writes the first law as ΔU = q + w. Both terms are counted from the system's point of view: q is positive when heat flows in, and w is positive when the surroundings do work on the system. Work against a constant external pressure is w = −P_ext ΔV, so a reaction that pushes the atmosphere back — any reaction whose gaseous moles increase, Δn_g > 0 — does work on its surroundings and its w is negative. Read that condition off Δn_g, not off whether a gas appears: ammonia synthesis makes two moles of gas and still has Δn_g = −2, so the atmosphere does work on it and its w is positive.

Note which pressure appears. It is P_ext, the pressure the system is pushing against, not the gas's own pressure. The two are equal only in the limiting reversible case, and the difference between them is the whole of the next concept.

Figure. Heat arriving and work leaving are counted on the same ledger, both from the system's side. The arrow marked P_ext points inward because that is the pressure resisting the piston — the gas's own pressure is not what appears in w = −P_ext ΔV.

Counting the energy

1. Heat in, positiveq > 0 when the system absorbs heat; an exothermic reaction has q < 0.
2. Expansion out, negativew = −P_ext ΔV. Growing volume against a resisting pressure spends energy, so w < 0.
3. The rest is ΔUWhatever the two add up to is the change in internal energy, on any path.

2State functions, path functions and reversible work

U and H are state functions: their changes depend only on where you started and where you finished. Neither q nor w is. Take the same gas between the same two states by two different routes and you get two different heats and two different works — only their sum is fixed, which is precisely what the first law asserts.

The extreme case is the reversible path, where P_ext is kept infinitesimally below the gas's own pressure the whole way, so w = −nRT ln(V₂/V₁) for an isothermal change. That is the largest amount of work an expansion can ever deliver. Let the gas slam out against a single low pressure instead and you get much less, and you cannot get it back.

Figure. Both routes run between the same two marked states. The solid curve is the gas's own pressure, tracked all the way down; the dashed line is the constant P_ext of the single-step expansion. The work is the area beneath each — 5.70 kJ under the curve, 2.23 kJ under the flat line. During the irreversible step the gas has no single pressure at all, which is why the dashed line is labelled P_ext and not P.

3Enthalpy versus internal energy

Reactions in a sealed bomb are held at constant volume, where no expansion work is possible and q = ΔU. Reactions in a flask are held at constant pressure, where the system pays for whatever room it takes up. Enthalpy H = U + PV is defined so that this payment is already inside it: at constant pressure, q = ΔH.

For ideal gases the difference is ΔH = ΔU + Δn_g RT, and Δn_g counts gaseous moles only — products minus reactants. Solids and liquids never enter it, because their volumes are negligible. If Δn_g = 0 the two are identical. The same statement per mole per kelvin is Cp − Cv = R: heating a gas at constant pressure costs an extra R because the gas expands while you heat it.

Figure. Three straight lines through a common origin, slope Δn_g R. The dashed Δn_g = 0 line is the true zero of this axis — below it ΔH sits under ΔU, above it over. The marked point is the −4955 J of the ammonia example. The bottom rule of the frame is only the frame.

4Hess's law and formation enthalpies

Because H is a state function, the enthalpy change of a reaction depends only on the reactants and the products, never on the route between them. That is Hess's law, and it means you may invent any convenient route — through the free elements, through an intermediate nobody isolates — and add the steps up. Reverse a step and its sign flips; double it and its magnitude doubles.

The standard route goes through the elements, which is what a standard enthalpy of formation ΔHf° measures: the enthalpy of making one mole of a compound from its elements in their standard states. Elements in their standard states therefore have ΔHf° = 0 by definition, and every reaction becomes ΔHrxn = Σ ΔHf°(products) − Σ ΔHf°(reactants).

Figure. Two routes from methane to carbon dioxide and water: straight across, or up through the free elements and down again. +74.8 − 965.1 = −890.3, so the cycle closes. The boxes are stacked in order of enthalpy — elements highest, products lowest — but the gaps are not to scale; only the arrow labels carry numbers.

Building the cycle

1. Pick a hubChoose the free elements as the state every route passes through.
2. Climb, then fallUnmake the reactants into elements, then make the products from them.
3. Add and checkThe two legs must sum to the direct arrow, or a sign has been dropped.

5Bond enthalpies, and why the subtraction reverses

The second route to a reaction enthalpy tears every molecule down to free gaseous atoms and rebuilds it. Breaking a bond always costs energy and making one always releases it, so ΔH = Σ(bonds broken) − Σ(bonds formed). Every student who has just learned the formation rule writes this backwards at least once.

The orders differ because the reference states differ. Formation enthalpies are measured from the elements, which sit between the reactants and the products, so you subtract where you came from: products − reactants. Bond enthalpies are measured from free atoms, which sit far above both, so you climb the full height of the reactant bonds and descend the full height of the product bonds: broken − formed. Two different hubs, two different subtractions.

Figure. Enthalpy increases upward, so the free atoms sit at the top and the products at the bottom. This ladder is drawn to scale: the products lie only 97 kJ below the reactants, 4.3% of the 2249 kJ climb to the atoms. That sliver is the entire heat of the reaction, which is why the method is only ever an estimate.

6Entropy and the second law

Entropy is defined through heat: for a reversible change, ΔS = q_rev/T. The same heat is worth more entropy at a low temperature than at a high one, which is why the T sits underneath. For an ideal gas the change comes out as ΔS = nR ln(V₂/V₁) + nCv ln(T₂/T₁) — a logarithm, so it is the ratio of the volumes that matters, never the difference.

The second law is a statement about the universe, not about the system: a spontaneous change has ΔS_univ = ΔS_sys + ΔS_surr > 0. The system's own entropy is free to fall, and often does. Water freezing, a crystal forming from solution and ammonia condensing out of the Haber reactor are all spontaneous and all lower the system's entropy; in each case the heat dumped into the surroundings raises theirs by more.

Figure. One mole, isothermal. The curve is a logarithm, so it never flattens to a ceiling and never turns over — but it does climb ever more slowly. The two marked points make the point exactly: doubling the volume is worth 5.76 J K⁻¹, and doubling it twice is worth exactly twice that, however large the volumes have become.

7Gibbs energy and spontaneity

Judging spontaneity by ΔS_univ means knowing what the surroundings did, which is awkward. Gibbs energy folds the surroundings back into the system's own quantities: ΔG = ΔH − TΔS, and at constant temperature and pressure the change is spontaneous when ΔG < 0, at equilibrium when ΔG = 0, and non-spontaneous when ΔG > 0.

Read as a function of temperature this is a straight line: the intercept is ΔH and the slope is −ΔS. Four sign combinations give four lines. Exothermic with an entropy rise is negative everywhere; endothermic with an entropy fall is positive everywhere. Only the two mixed cases cross ΔG = 0, and only for those is T = ΔH/ΔS an actual temperature — when ΔH and ΔS have opposite signs the ratio is negative, which is no temperature at all.

Figure. All four lines use |ΔH| = 30 kJ mol⁻¹ and |ΔS| = 100 J mol⁻¹ K⁻¹, so they differ only in their signs. ΔG increases upward, so a line running downward on the page is a reaction becoming more spontaneous as it is heated — that is every line with ΔS > 0. The dashed rule through the middle is ΔG = 0; the two lines that never reach it are the two whose sign pair makes T = ΔH/ΔS negative.

Reading a sign pair

1. Same signs, real crossoverΔH and ΔS both positive or both negative: T = ΔH/ΔS is positive and real.
2. Which side is spontaneousEndothermic pairs go spontaneous above it, exothermic pairs below it.
3. Opposite signs, no crossoverThe ratio is negative, so the verdict is the same at every temperature.

8ΔG° and the equilibrium constant

A reaction does not run until the reactants are gone; it runs until ΔG reaches zero, and where that happens is fixed by the standard Gibbs energy change: ΔG° = −RT ln K = −2.303RT log K. A negative ΔG° puts K above one and the equilibrium on the product side; a positive ΔG° puts K below one; ΔG° = 0 gives K = 1 exactly.

Keep ΔG° and ΔG apart. ΔG° is a single number for the reaction under standard conditions and it fixes K. ΔG is what the mixture in front of you currently has, and it changes as the reaction proceeds, reaching zero at equilibrium. A reaction with ΔG° = +20 kJ mol⁻¹ is not a reaction that refuses to happen — it is one whose equilibrium sits at K = 3 × 10⁻⁴.

Figure. A straight line of slope −2.303RT, which at 298 K is 5.71 kJ mol⁻¹ per power of ten. ΔG° increases upward, so the line running down the page is K increasing. The marked crossing is K = 1 — the whole of the reactant-favoured half of the graph lies above and to the left of it, and it is nowhere near the bottom of the frame.