Chemistry: Atoms First 2e · Thermodynamics

The Second and Third Laws of Thermodynamics

10 min read
Thermodynamic values cited (ΔH°f NH₃ = −45.9 kJ/mol; ΔS°rxn = −197.7 J/K; standard molar entropies) are standard reference values; calculations and reasoning are original worked examples.
Want it in plain words first? Jump to Eli explains — the same idea, no jargon.
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

The first law of thermodynamics conserves energy, and the previous topics added entropy as the measure of dispersal. But a complete theory of direction needs two more laws. The states the master criterion for spontaneity: the entropy of the universe increases in every spontaneous process:

ΔSuniv = ΔSsys + ΔSsurr ≥ 0

A process is spontaneous only if the total entropy change of the universe is positive; at equilibrium the total is zero; a negative total means the reverse direction is spontaneous.

The completes the picture by giving entropy an absolute zero: the entropy of a at absolute zero is zero. This makes absolute (standard) molar entropies possible — the tabulated S° values used to calculate reaction entropies. This topic shows how to combine system and surroundings entropy changes to decide spontaneity quantitatively, and how the third law grounds entropy in a real zero point.

Why this matters

  • The universal spontaneity test: ΔSuniv > 0 is the criterion that works for every process — no exceptions, no special cases. Free energy (Topic 4) is just this test repackaged for constant T and P.
  • Understanding "the arrow of time": The second law explains why heat flows hot → cold, why gases mix, and why time has a direction — the universe's entropy is always increasing.
  • Predicting reactions without experiments: Combining ΔS°rxn with ΔH°rxn lets chemists predict whether a reaction can run at a given temperature.
  • Energy limits: The second law sets the maximum efficiency of heat engines and why perpetual motion machines are impossible.
  • Absolute entropies: The third law's zero point is why S° values are real, measurable quantities used in every ΔS°rxn calculation.
  • Exam value: Computing ΔSuniv to judge spontaneity and applying the third law's trends are classic assessment items.

The college version

Core Concepts

The second law: entropy of the universe never decreases

The second law is usually stated as: the entropy of the universe increases in a spontaneous process and remains constant for a reversible (equilibrium) process. Equivalently:

ΔSuniv = ΔSsys + ΔSsurr ≥ 0

The sign of ΔSuniv decides everything:

  • ΔSuniv > 0: process is spontaneous in the forward direction.
  • ΔSuniv = 0: process is at equilibrium (reversible; no net direction favored).
  • ΔSuniv < 0: process is nonspontaneous forward; the reverse is spontaneous.

Notice what the second law does not say: a system's entropy can decrease (water freezes, gas compresses) — the law only demands that the surroundings gain at least as much entropy as the system loses. Local order is always paid for by greater disorder elsewhere.

The surroundings' entropy change

At constant temperature and pressure, when the system exchanges heat with its surroundings, the surroundings' entropy changes by:

ΔSsurr = -ΔHsysT

  • Exothermic reaction (ΔHsys < 0): surroundings gain heat → ΔSsurr > 0.
  • Endothermic reaction (ΔHsys > 0): surroundings lose heat → ΔSsurr < 0.

The minus sign is the crux: heat leaving the system is heat entering the surroundings, so the two entropy changes always oppose each other. Spontaneity is decided by which tendency wins, with temperature weighting the surroundings' term (dividing by T makes the surroundings' entropy gain smaller at high temperature).

The combined calculation

Putting the pieces together, the spontaneity test for a process at constant T and P becomes:

ΔSuniv = ΔSsys - ΔHsysT

where ΔSsys comes from standard molar entropies (ΔS°rxn = ∑n S°(products) - ∑m S°(reactants)) and ΔHsys from heats of formation or calorimetry. Multiplying through by T gives TΔSuniv = TΔSsys - ΔHsys, which is exactly the negative of the Gibbs free energy change (ΔG = ΔH - TΔS) that Topic 4 develops — the same criterion in a more convenient form.

The third law: absolute zero of entropy

The third law of thermodynamics states that the entropy of a perfect crystal at absolute zero (0 K) is zero. At 0 K a perfect crystal has exactly one arrangement — every particle locked in its lattice site, all energy in its lowest state — so S = kB ln1 = 0.

Two consequences matter for calculations:

  1. Absolute entropies exist. Because entropy has a real zero, the entropy of any substance at any temperature can be measured (by heating it stepwise and integrating qrev/T). These are the standard molar entropies S° in tables — real absolute values, unlike enthalpies which are relative to an arbitrary zero.
  2. Entropy trends. S° increases with temperature, with molecular size and complexity, and with freedom of motion: S°(solid) < S°(liquid) ≪ S°(gas); larger molecules have more internal motions (vibrations, rotations) and higher entropy; allotropes differ (diamond vs. graphite).

Why entropy is "absolute" but enthalpy is not

Enthalpy is measured relative to a chosen reference (elements in their standard states = 0). Entropy is not: the third law provides a genuine zero at 0 K for perfect crystals. This is why S° values are all positive and why they can be tabulated for individual substances and combined by stoichiometry — no formation-reference bookkeeping needed.

Common Confusions

Do Not ConfuseWithDifference
System entropy decreaseViolation of the second lawThe second law constrains the universe (ΔSuniv ≥ 0); the system can lose entropy if the surroundings gain more.
ΔSsurr signΔHsys signThey are opposite: ΔSsurr = -ΔHsys/T. Exothermic (negative ΔH) gives positive surroundings entropy.
Spontaneous at one TSpontaneous at all TSpontaneity is temperature-dependent; exothermic reactions can become nonspontaneous at high T (NH₃ example).
ΔSuniv = 0"No process occurs"Equilibrium is dynamic — forward and reverse proceed at equal rates; the net change is zero.
Absolute entropyEnthalpy of formationS° has a real zero (third law); ΔH°f is relative to elements = 0. Both are tabulated, but only entropies are absolute.
Third law (0 K, perfect crystal)"Entropy is zero at absolute zero" in generalOnly for a perfect crystal; glasses, solutions, and defective crystals retain residual entropy even at 0 K.
Perfect crystalAny solidReal solids have defects; the perfect crystal is the idealization that makes S(0 K) = 0 exact.
Eli, the EliExplains learning guide

Eli explains

The same idea, in plain words

Explain it like I’m 10

If you drop a whole box of puzzle pieces, the pieces scatter across the floor — that's the universe getting more "spread out" (higher entropy), and it happens all by itself. But if you build the puzzle, you're making things more ordered in one spot — that's fine, as long as you (or something else) use up even more energy and create even more mess somewhere else. The second law says the total mess in the universe can never get smaller; the third law says if you cooled a perfect crystal to absolute zero, it would have exactly zero mess — the cleanest thing possible.

Worked example

Worked example 1 — is ammonia synthesis spontaneous at 298 K? For N2(g) + 3H2(g) → 2NH3(g):

  • ΔS°rxn = -197.7 J/K (from Topic 2, using S° values).
  • ΔH°rxn = -92.2 kJ/mol (standard heat of formation of NH₃ is -45.9 kJ/mol, doubled for 2 mol).

Step 1 — surroundings entropy at 298 K:

ΔSsurr = -ΔHsysT

ΔSsurr = -(-92.2 × 103 J/mol)298 K = +309 J/(mol·K)

(Convert kJ → J with  × 103, a dimensional-analysis step students often skip.)

Step 2 — total entropy change:

ΔSuniv = ΔSsys + ΔSsurr

ΔSuniv = (-197.7) + (+309) = +111 J/(mol·K)

Step 3 — verdict: ΔSuniv > 0, so the reaction is spontaneous at 298 K despite the system's entropy drop: the exothermic release of heat spreads enough energy into the surroundings to pay for the loss of gas-phase disorder. This is why the "system entropy decreased" fact alone never settles a question — the surroundings always get a vote.

Worked example 2 — same reaction at very high temperature. Suppose the reaction ran at 1500 K (ignoring that ΔH and ΔS change somewhat with T — a stated assumption). Recompute the surroundings term:

ΔSsurr = -(-92.2 × 103 J/mol)1500 K = +61.5 J/(mol·K)

ΔSuniv = (-197.7) + (61.5) = -136 J/(mol·K)

Now ΔSuniv < 0: the reaction is nonspontaneous at high temperature. The exothermic advantage shrinks (dividing by larger T), while the system's entropy deficit stays roughly constant — so high temperature flips the direction. This is the general rule: exothermic reactions become less favored at high T, and endothermic ones become more favored — the temperature dependence previewed in Topic 1, now computed exactly.

Worked example 3 — third law reasoning. Two solids at 298 K: ice (crystalline H₂O) and a 1.0 g sample of glassy (amorphous) silica. Which has higher entropy, and why?

  • Crystalline ice has regular, ordered packing — fewer microstates, lower entropy per gram.
  • Amorphous (glassy) silica has no long-range order — many more arrangements of the same atoms — so higher entropy.

The third law sharpens the contrast: cool a perfect crystal to 0 K and S → 0, but a glass frozen at 0 K would retain residual disorder (a "frozen-in" entropy) because it never reaches the perfect-crystal limit. Real materials always have some defects, which is why measured entropies approach but never quite hit zero — the law describes the ideal limit, and deviations are informative.

Key takeaways

  • Second law: ΔSuniv = ΔSsys + ΔSsurr ≥ 0. Spontaneous: > 0; equilibrium: = 0; nonspontaneous forward: < 0.
  • Surroundings term: ΔSsurr = -ΔHsys/T — exothermic reactions boost surroundings entropy; endothermic ones lower it.
  • The system's entropy may decrease; only the universe's total must not.
  • Combined test at constant T, P: ΔSuniv = ΔSsys - ΔHsys/T — the seed of the free-energy equation.
  • Third law: entropy of a perfect crystal at 0 K is zero (S = kB ln1 = 0).
  • S° values are absolute (real zero), all positive, and increase with T, molecular size/complexity, and phase freedom.
  • Entropy trends: solid < liquid ≪ gas; larger/complex molecules > small ones; diamond vs. graphite differ.
  • A process that is endothermic and entropy-decreasing (like ammonia synthesis) is nonspontaneous at low T — pressure and catalysts can't change thermodynamics, only kinetics and equilibrium position.

Check yourself

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

  1. State the second law in terms of ΔSuniv, and give the three possible outcomes and their meanings.

    Show answer

    ΔSuniv = ΔSsys + ΔSsurr ≥ 0. Positive → spontaneous forward; zero → equilibrium (reversible); negative → nonspontaneous forward (reverse is spontaneous).

  2. A reaction has ΔH°= -57.2 kJ/mol at 298 K. Calculate ΔSsurr.

    Show answer

    ΔSsurr = -ΔH/T = -(-57.2 × 103 J/mol)/298 K = +192 J/(mol·K).

  3. For N2(g) + 3H2(g) → 2NH3(g), ΔS°rxn = -197.7 J/K and ΔH°rxn = -92.2 kJ/mol. Is it spontaneous at 298 K? Show the calculation.

    Show answer

    ΔSsurr = +309 J/(mol·K); ΔSuniv = -197.7 + 309 = +111 J/(mol·K) > 0 → spontaneous at 298 K.

  4. Why does an exothermic reaction become less favorable as temperature increases?

    Show answer

    The surroundings' entropy gain is -ΔH/T; as T rises, the same heat release buys less surroundings entropy, so an exothermic reaction's advantage shrinks and it can flip to nonspontaneous.

  5. State the third law and explain why it makes standard molar entropies "absolute" while enthalpies of formation are relative.

    Show answer

    The third law sets S = 0 for a perfect crystal at 0 K — a genuine zero — so entropies are measured from absolute zero. Enthalpies have no such natural zero; they are referenced to elements in standard states.

  6. True or false: "The second law forbids any local decrease in entropy." Explain.

    Show answer

    False. The second law constrains the universe's total entropy; a system can decrease its entropy (freezing, compressing) as long as the surroundings' entropy increases by at least as much.

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Study tools & related lessonsKey vocabulary · Related

Key vocabulary

Second law of thermodynamics
The entropy of the universe increases in every spontaneous process.
Δ Suniv
Total entropy change of system plus surroundings.
Δ Ssurr
Entropy change of everything outside the system.
Third law of thermodynamics
Entropy of a perfect crystal at 0 K is zero.
Perfect crystal
A crystal with every particle in its ideal lattice position, no defects.
Absolute entropy
Entropy measured from the true zero at 0 K.
Standard molar entropy (S°)
Entropy of one mole in standard state at 298 K (J/(mol·K)).
Equilibrium (thermodynamic)
State where ΔSuniv = 0; forward and reverse balance.

Sources & references

  1. openstax.org — Chemistry Atoms First 2e

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

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