Lesson 1 · 30 min

Entropy: counting the ways

Why does heat always flow from hot to cold? Because that way of sharing energy has unimaginably more arrangements. Start from the microstates of Einstein solids, understand S = k ln W, then compute entropy changes of phase changes, mixing and reactions.

In this lesson

  • Distinguish macrostates from microstates and understand entropy as S = k ln W
  • Explain why heat flows from hot to cold as the macrostate with the most microstates
  • Calculate entropy changes for phase changes (ΔS = ΔH/T), ideal mixing and gas expansion
  • Compute reaction entropies ΔrS° from standard molar entropies and predict their sign from the change in moles of gas

A cup of hot tea on the table only ever cools; it never draws heat from the air to get hotter — although that would conserve energy just as well. The ammonium nitrate in a cold pack absorbs heat as it dissolves, yet it dissolves by itself; open a partition and a gas rushes into a vacuum without any change in its energy. The first law only checks that energy balances; it says nothing about direction. The quantity that decides direction is entropy, and at heart it is surprisingly simple: counting.

Predict first

Two “solids” touch: A has 60 oscillators, B has 40, and all 60 energy quanta start in B (B hot, A cold). After the quanta have hopped randomly between oscillators for a long time, how many quanta will A most probably hold?

Pick an answer and lock it in before you look. A wrong guess teaches more than no guess.

Entropy counts the ways

A system can be described at two levels: a macrostate says only “this much energy in A, that much in B”; a microstate specifies how many quanta each individual oscillator holds. One macrostate usually corresponds to an astronomically large number of microstates, WW. The basic assumption of statistical mechanics is that in an isolated system every accessible microstate is equally likely. So the macrostate with the largest WW is almost certain to be what we see.

Boltzmann defined entropy as S=kln⁡WS = k\ln W. The logarithm turns huge products into sums: for two solids together W=WAWBW = W_\mathrm{A}W_\mathrm{B}, so S=SA+SBS = S_\mathrm{A} + S_\mathrm{B}. The second law of thermodynamics can then be put like this: an isolated system moves spontaneously toward macrostates with more microstates, and its total entropy never decreases. Heat does not flow from hot to cold because something pushes it; it flows because there are vastly more ways to share the energy that way.

S=kln⁡WΔSuniverse≥0S = k\ln W\qquad \Delta S_\text{universe} \ge 0
Boltzmann’s entropy and the second law. For a mole of substance, k becomes R = N_A k = 8.314 J/(K·mol).
kk
Boltzmann constant, 1.380649 × 10⁻²³ J/K
WW
number of microstates consistent with the macrostate

On the bench, a system of 60 + 40 oscillators still visibly jitters; scale it up “×100” and the peak becomes tall and narrow, with q_A spending more than 99.9 % of the time within ±5 % of its most probable value (for 60 + 40 oscillators even ±20 % covers only about 88 %). A real solid has 10²³ atoms, and the chance of energy flowing back is too small to be seen even once in the age of the universe. The second law is a statistical law, but for everyday objects it is iron-clad.

Measuring entropy with heat

Clausius gave entropy a thermodynamic definition long before its statistical meaning: for heat absorbed reversibly, ΔS=qrev/T\Delta S = q_\text{rev}/T. The same heat raises the entropy of a cold object more — a cold object holds little energy, so a little extra opens up relatively many new ways to share it.

  • A phase change at its normal transition temperature is reversible, so ΔtrsS=ΔtrsH/Ttrs\Delta_\text{trs}S = \Delta_\text{trs}H/T_\text{trs}. Many liquids have entropies of vaporization close to 85 J/(K·mol) (Trouton’s rule), because turning a liquid into gas at 1 bar multiplies the volume each molecule can roam by roughly the same factor; water, alcohols and ammonia have hydrogen-bonded, more ordered liquids and clearly larger values (109 for water).
  • Isothermal expansion of an ideal gas: ΔS=nRln⁡(V2/V1)\Delta S = nR\ln(V_2/V_1); doubling the volume adds Rln⁡2=5.76R\ln 2 = 5.76 J/K per mole.
  • Ideal mixing: ΔmixS=−nR∑xiln⁡xi\Delta_\text{mix}S = -nR\sum x_i\ln x_i; mixing equal amounts of two gases again gives 5.76 J/K per mole — the enthalpy does not change, the whole drive is entropy.
ΔS=qrevTΔtrsS=ΔtrsHTtrs\Delta S = \frac{q_\text{rev}}{T}\qquad \Delta_\text{trs}S = \frac{\Delta_\text{trs}H}{T_\text{trs}}
qrevq_\text{rev}
heat absorbed by the system along a reversible path
TtrsT_\text{trs}
normal melting or boiling point / K
Phase changes · mixing · reactions

Where entropy changes come from

Trouton’s rule ΔvapS ≈ 84 J/(K·mol) HeliumNeonNitrogenArgonOxygenMethaneKryptonXenonHydrogen sulfideChlorineAcetoneBromineHexaneTetrachloromethaneBenzeneMercury AmmoniaMethanolEthanolWater H₂O0100200300400500600Normal boiling point T_b / K0204060Enthalpy of vaporization ΔvapH / (kJ/mol)
  • ordinary liquids
  • hydrogen-bonded liquids
  • Trouton’s rule ΔvapS ≈ 84 J/(K·mol)

ΔvapS=40650 J/mol373.1 K=108.9 J/(K⋅mol)\Delta_\mathrm{vap}S = \frac{40650\ \text{J/mol}}{373.1\ \text{K}} = 108.9\ \text{J/(K·mol)}

The same substance’s entropy of fusion: 22.0 J/(K·mol) — far smaller than for vaporization, because a liquid is almost as crowded as the solid while the gas occupies about a thousand times the volume.

Computed: ΔS = ΔH/T from CRC normal boiling/melting points and transition enthalpies; ideal mixing entropy ΔmixS = −R Σ x ln x; standard molar entropies S° and ΔrS° from the 298.15 K data table (CODATA/NBS/CRC). The Trouton constant is the mean over non-associated liquids boiling above 100 K.
Most liquids sit near Trouton’s line (about 85 J/(K·mol)); the hollow hydrogen-bonded liquids all lie above it. Pick water to see that its entropy of fusion is only 22 J/(K·mol): solid and liquid differ far less than liquid and gas. Then switch to the “Mixing” tab and drag the mole fraction.

Standard molar entropies and reaction entropies

The third law of thermodynamics: a perfect crystal at 0 K has only one arrangement, W=1W = 1, so S=0S = 0. Heating from 0 K bit by bit and adding up qrev/Tq_\text{rev}/T measures a substance’s absolute entropy at 298 K, the standard molar entropy S∘S^\circ (unlike enthalpy, entropy has a true zero). Some patterns:

  • gases ≫ liquids > solids: HX2O(g)\ce{H2O(g)} 188.8, HX2O(l)\ce{H2O(l)} 70.0 J/(K·mol);
  • heavier atoms and more complex molecules have larger S°: He 126.2, Xe 169.7; CHX4\ce{CH4} 186.3, CX3HX8\ce{C3H8} 270.3;
  • hard, orderly crystals have very small S°: diamond 2.4, graphite 5.7.

A reaction’s entropy change is the difference between the S° of products and reactants. Rule of thumb: if the moles of gas increase, ΔrS° is usually positive; if they decrease, negative.

ΔrS∘=∑ν S∘(products)−∑ν S∘(reactants)\Delta_\mathrm{r}S^\circ = \sum \nu\,S^\circ(\text{products}) - \sum \nu\,S^\circ(\text{reactants})
S∘S^\circ
standard molar entropy / (J K⁻¹ mol⁻¹), 298.15 K, 1 bar
Phase changes · mixing · reactions

Where entropy changes come from

NX2(g)+3 HX2(g)⟶2 NHX3(g)\ce{N2(g)} + 3\,\ce{H2(g)} \longrightarrow 2\,\ce{NH3(g)}

  • NX2(g)\ce{N2(g)} − 1 × 191.6
  • HX2(g)\ce{H2(g)} − 3 × 130.7
  • NHX3(g)\ce{NH3(g)} + 2 × 192.8
ΔrS° = Σν·S°(products) − Σν·S°(reactants)
−198.1 J/K
Δn(gas)
−2

Rule of thumb: more moles of gas usually means ΔS > 0; fewer means ΔS < 0. This reaction follows the rule of thumb.

Computed: ΔS = ΔH/T from CRC normal boiling/melting points and transition enthalpies; ideal mixing entropy ΔmixS = −R Σ x ln x; standard molar entropies S° and ΔrS° from the 298.15 K data table (CODATA/NBS/CRC). The Trouton constant is the mean over non-associated liquids boiling above 100 K.
Ammonia synthesis turns 4 mol of gas into 2, so ΔrS° is strongly negative; limestone releases CO₂ and ΔrS° is positive. Look at ammonium nitrate dissolving: no gas at all, yet ΔrS° = +109 J/K — the ions freed from the lattice gain many more arrangements.

Worked example

The entropy change of ammonia synthesis

From S° (J K⁻¹ mol⁻¹): NX2\ce{N2} 191.61, HX2\ce{H2} 130.68, NHX3\ce{NH3} 192.77, calculate ΔrS° for NX2(g)+3 HX2(g)⟶2 NHX3(g)\ce{N2(g) + 3H2(g)} \longrightarrow \ce{2NH3(g)}.

Concept check

Which process has a negative entropy change?

Try the calculation

Benzene boils at 353.24 K with an enthalpy of vaporization of 30.72 kJ/mol. Find its entropy of vaporization (J K⁻¹ mol⁻¹).

J/(K·mol)

Key takeaways

  • Entropy counts microstates: S = k ln W; an isolated system moves spontaneously toward macrostates with larger W.
  • Heat flows from hot to cold because that way of sharing energy has vastly more arrangements — the second law, ΔS_universe ≥ 0.
  • ΔS = q_rev/T; phase changes ΔS = ΔH/T (Trouton ≈ 85 J/(K·mol)); an equimolar ideal mix gives R ln 2 per mole.
  • The third law gives absolute entropies S°; ΔrS° = ΣνS°(products) − ΣνS°(reactants), usually positive when the gas increases.
Progress is stored only in this browser and never uploaded.

Data and references

  1. D. V. Schroeder, An Introduction to Thermal Physics, Addison-Wesley (2000), Ch. 2–3 (Einstein solids, S = k ln W, temperature).
  2. P. Atkins, J. de Paula, J. Keeler, Atkins’ Physical Chemistry, 11th ed. (2018), Topics 3A–3C (entropy, third law).
  3. W. M. Haynes (ed.), CRC Handbook of Chemistry and Physics, 97th ed. (2016), “Enthalpy of vaporization” and “Enthalpy of fusion”.
  4. J. D. Cox, D. D. Wagman, V. A. Medvedev, CODATA Key Values for Thermodynamics (1989) — standard molar entropies.
  5. F. L. Lambert, “Disorder — a cracked crutch for supporting entropy discussions”, J. Chem. Educ. 79, 187 (2002).