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Chemistry 2e · Thermodynamics

Entropy

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Standard molar entropies are reference values at 25 °C (CaCO₃ 92.9, CaO 39.8, CO₂ 213.8, N₂ 191.6, H₂ 130.7, NH₃ 192.8, O₂ 205.2, H₂O(l) 69.9 J mol⁻¹ K⁻¹); ΔHvap(H2O) = 40.7 kJ mol-1 at 373.15 K. Small variations exist among reference tables.
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On this page 9 sections
  1. In 30 seconds
  2. Why this matters
  3. The college version
  4. Eli explains
  5. Worked example
  6. Key takeaway
  7. Check yourself
  8. Study tools
  9. Sources & references

In 30 seconds

Entropy, S, is the thermodynamic quantity that measures how broadly a system's energy is dispersed among available states. A gas released from a small flask spreads into an entire room; a hot metal block cools toward the temperature of its surroundings; two gases in a container mix on their own. In every case energy is being shared more widely, and entropy increases. The more ways a system's energy can be arranged (its microstates), the higher its entropy.

Entropy is a : its value depends only on the current state of the system, not on how that state was reached. It has absolute values (unlike enthalpy, which is measured relative to formation conventions) with units of energy per temperature per amount — joules per kelvin per mole, J mol⁻¹ K⁻¹. Because tabulated standard molar entropies S° are absolute, computing an entropy change is a simple products-minus-reactants sum:

ΔS°= ∑n S°(products) - ∑n S°(reactants)

Why this matters

Entropy explains the direction of change — the thermodynamic arrow of time. It tells you why heat never spontaneously flows from cold to hot, why a dropped egg never reassembles, and why spontaneous reactions are product-favored even when they absorb heat. Engineers design heat engines and refrigerators around entropy accounting; biologists study how living systems export entropy to their surroundings to maintain order locally; pharmaceutical scientists care that many drugs dissolve with an entropy gain. Entropy also predicts simple, testable chemistry: reactions that produce gas are favored entropically, which is why effervescence and explosions proceed so vigorously, and why some solids dissolve even when dissolution is endothermic.

The college version

Core Concepts

Entropy is energy dispersal, not "messiness"

The everyday gloss "entropy is disorder" is a useful memory hook but can mislead. The precise idea is microstates: the number of distinct arrangements of particles and energy that look identical at the macroscopic level. A gas in one corner of a flask has few microstates; spread through the whole flask, the same gas has enormously more. Because each is equally likely, the system overwhelmingly evolves toward the macrostate with more microstates — higher entropy. The Boltzmann formula expresses this:

S = kB lnW

where kB = 1.38 × 10-23 J K-1 is Boltzmann's constant and W counts microstates. More microstates → larger S.

Standard molar entropies are absolute

Standard molar entropy S° is the entropy of one mole of substance in its standard state at the reference temperature (commonly 25 °C). Two contrasts with enthalpy are essential:

  • No formation convention. There is no "entropy of formation"; S° is absolute, and elements in standard states have nonzero values (e.g., S°(O2) = 205.2 J mol-1K-1).
  • Values are small. Entropies are typically tens to hundreds of J mol⁻¹ K⁻¹, so entropy terms contribute kJ-scale free energies only when multiplied by T.

Predicting entropy changes qualitatively

Three rules of thumb cover most predictions:

  1. Phase: S(gas) ≫ S(liquid) > S(solid). Melting and vaporization increase entropy.
  2. Gas moles: reactions producing more gas molecules have ΔS > 0; consuming gas gives ΔS < 0. Solute particles from dissolving a solid also raise entropy.
  3. Temperature and complexity: entropy increases with temperature (more energy states populated) and with molecular complexity (more atoms, more ways to vibrate and rotate — e.g., S°(CO2) = 213.8 vs S°(O2) = 205.2 J mol-1K-1).

Computing ΔS°

Using tabulated S° values, apply

ΔS°= ∑n S°(products) - ∑n S°(reactants)

where n are the stoichiometric coefficients. The result carries units of J mol⁻¹ K⁻¹ per mole of reaction as written.

Entropy changes at phase transitions

At a transition temperature, the two phases are in equilibrium and the entropy change is exactly the heat absorbed divided by the absolute temperature:

ΔStrans = ΔHtransTtrans

For vaporization of water at its boiling point, this gives about 109 J mol⁻¹ K⁻¹ — a large entropy gain, consistent with liquid water's ordered hydrogen-bond network becoming free gas.

How It Works / Step-by-Step Process

  1. Write the balanced reaction; confirm physical states, since S° differs for gas/liquid/solid.
  2. Decide qualitatively first: does gas-mole count rise or fall? Does a phase change occur?
  3. Look up S° for every species (including elements — they are not zero).
  4. Apply ΔS°= ∑nS°(products) - ∑nS°(reactants), keeping J mol⁻¹ K⁻¹.
  5. Check the sign against your qualitative prediction; investigate discrepancies.
  6. For phase transitions, use ΔS = ΔH/T with T in kelvin and energy converted to joules.

Common Confusions

Common ConfusionCorrect Understanding
"Entropy is just messiness or disorder."The precise meaning is dispersal of energy among microstates; "disorder" is a rough proxy that fails for some systems (e.g., ordered ice has low entropy because few energy arrangements exist).
"Entropy always increases for a reaction's system."Only the universe's entropy must increase for a spontaneous process. A reaction's system can have ΔS < 0 (ammonia synthesis) if the surroundings gain entropy.
"Elements have S°= 0 like ΔH°f."No: entropies are absolute. S°(O2) = 205.2 J mol-1K-1 at 25 °C. Zero entropy occurs only at absolute zero (third law, next topic).
"Dissolving a solid always decreases entropy."Usually it increases it: the solid's ordered lattice breaks into freely moving ions — that is why many endothermic dissolutions are spontaneous.
"ΔS° changes with temperature the way ΔG does."ΔS° is tabulated at the reference temperature; the T dependence lives in the -TΔS term of ΔG.
"More gas moles always means the reaction is spontaneous."Entropy is only half the story; ΔH still matters through ΔG = ΔH - TΔS.
Eli, the EliExplains learning guide

Eli explains

The same idea, in plain words

Explain it like I’m 10

Think of ten coins: heads is "ordered," tails is "spread out." There is only one way to have all heads, but hundreds of ways to have a mix — so a mix is what you almost always see. Entropy counts the ways. Nature always drifts toward the arrangement that can happen in the most ways.

Worked examples

CaCO3(s) → CaO(s) + CO2(g). Tabulated S°: CaCO₃ 92.9, CaO 39.8, CO₂ 213.8 J mol⁻¹ K⁻¹.

ΔS°= [S°(CaO) + S°(CO2)] - [S°(CaCO3)]

ΔS°= (39.8 + 213.8) - (92.9) J mol-1K-1 = 253.6 - 92.9 = +160.7 J mol-1K-1

The positive sign matches the qualitative prediction: one mole of solid becomes one mole of solid plus one mole of gas, so dispersal increases. Note the units: subtracting J mol⁻¹ K⁻¹ values gives J mol⁻¹ K⁻¹ — no unit juggling needed because every term is an entropy.

N2(g) + 3H2(g) → 2NH3(g). With S°(N2) = 191.6, S°(H2) = 130.7, S°(NH3) = 192.8 J mol-1K-1:

ΔS°= 2(192.8) - [191.6 + 3(130.7)] J mol-1K-1

ΔS°= 385.6 - 583.7 = -198.1 J mol-1K-1

Four moles of gas become two, so entropy drops — the system becomes more ordered. This is exactly why the Haber process needs low temperatures thermodynamically (see Spontaneity), even though kinetics forces high ones.

At its normal boiling point, 373.15 K, water's enthalpy of vaporization is ΔHvap = 40.7 kJ mol-1. The entropy change is

ΔSvap = ΔHvapTb = 40.7 kJ mol-1373.15 K

Convert to joules so the result is in J mol⁻¹ K⁻¹:

ΔSvap = 40,700 J mol-1373.15 K = 109 J mol-1K-1

Dimensional analysis: J mol⁻¹ ÷ K = J mol⁻¹ K⁻¹. The measured value is about 109 J mol⁻¹ K⁻¹, confirming that a mole of water gains roughly 109 J of entropy per kelvin when it boils — a large gain that drives steam formation despite the energy cost.

Key takeaways

  • Entropy S = number of microstates / breadth of energy dispersal; state function; units J mol⁻¹ K⁻¹.
  • S(gas) ≫ S(liquid) > S(solid); melting, vaporization, and dissolution of solids raise entropy.
  • Reactions that increase gas moles have ΔS > 0; decreasing gas moles gives ΔS < 0.
  • Standard molar entropies S° are absolute — elements are not zero — so use the products-minus-reactants sum directly.
  • ΔS°= ∑nS°(products) - ∑nS°(reactants).
  • At a phase transition, ΔS = ΔH/T (use kelvin and matching energy units).
  • A negative ΔS for the system is fine: the second law requires the universe's entropy to increase (next topic).
  • Bigger, more complex molecules have higher S°; temperature raises entropy too.

Check yourself

6 review questions from the chapter. Try each one, then open the answer.

  1. State the two phase rules and the gas-mole rule for predicting the sign of ΔS.

    Show answer

    Gas ≫ liquid > solid; melting/vaporization increase entropy; reactions that produce more gas moles have ΔS > 0, and consuming gas gives ΔS < 0.

  2. Why are standard molar entropies of elements nonzero, unlike standard enthalpies of formation?

    Show answer

    Because entropy is an absolute property of the state (microstate count), with no formation reference; a perfect crystal at 0 K would be the only zero-entropy case.

  3. Compute ΔS° for 2H2(g) + O2(g) → 2H2O(l) given S°(H2) = 130.7, S°(O2) = 205.2, S°(H2O, l) = 69.9 J mol-1K-1.

    Show answer

    ΔS°= 2(69.9) - [2(130.7) + 205.2] = 139.8 - 466.6 = -326.8 J mol-1K-1 — three moles of gas become liquid, so entropy falls.

  4. Ice melts with ΔHfus = 6.01 kJ mol-1 at 273.15 K. What is ΔSfus?

    Show answer

    ΔSfus = 6010 J mol-1 / 273.15 K = 22.0 J mol-1K-1.

  5. A reaction's system has ΔS < 0 yet is spontaneous. How is that possible?

    Show answer

    The surroundings' entropy gain (from heat released) must outweigh the system's loss, so the universe's entropy still increases — the second law is satisfied.

  6. Does boiling water or freezing water have the larger entropy change per mole, and why?

    Show answer

    Vaporization: about 109 J mol⁻¹ K⁻¹ vs about 22 J mol⁻¹ K⁻¹ for melting, because gas has vastly more microstates than liquid, which has only somewhat more than solid.

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Study tools & related lessonsKey vocabulary · Related

Key vocabulary

entropy (S)
Measure of how broadly a system's energy is dispersed among microstates
microstate
One specific arrangement of particles and energy consistent with the macroscopic state
state function
Property depending only on current state, not the path taken
standard molar entropy (S°)
Absolute entropy of one mole in its standard state at the reference temperature
Boltzmann constant (kB)
1.38 × 10-23 J K-1; links microstates to entropy
Δ Strans = Δ Htrans/T
Entropy change when phases are at equilibrium during a transition
standard molar entropy, S°
Entropy of one mole of a substance in its standard state at a stated temperature, in J/(mol·K).

Sources & references

  1. openstax.org — Chemistry 2e

This lesson was adapted from the open educational references above; their licenses and attributions are preserved. See Copyright & Licensing.

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