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General Chemistry II · Chemical Thermodynamics

Temperature Dependence of Spontaneity

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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. Quick check
  8. Study tools
  9. Sources & references

In 30 seconds

Because ΔG = ΔH − TΔS, the signs of ΔH and ΔS — and the temperature — jointly decide spontaneity. There are four cases: (1) ΔH < 0, ΔS > 0 → always spontaneous; (2) ΔH > 0, ΔS < 0 → never spontaneous (reverse is always spontaneous); (3) ΔH < 0, ΔS < 0 → spontaneous only at low T; (4) ΔH > 0, ΔS > 0 → spontaneous only at high T. The last two have a crossover temperature where ΔG = 0: T = ΔH/ΔS.

Why this matters

Temperature control is how chemists and engineers make unfavorable reactions favorable: driving endothermic syntheses at high temperature (steam reforming, lime production) or stabilizing products by cooling (condensation, crystallization). It also explains everyday phase behavior — ice melts above 0 °C, water boils above 100 °C — as a single ΔH/ΔS crossover logic.

The college version

Core Concept

Because ΔG = ΔH − TΔS, the signs of ΔH and ΔS — and the temperature — jointly decide spontaneity. There are four cases: (1) ΔH < 0, ΔS > 0 → always spontaneous; (2) ΔH > 0, ΔS < 0 → never spontaneous (reverse is always spontaneous); (3) ΔH < 0, ΔS < 0 → spontaneous only at low T; (4) ΔH > 0, ΔS > 0 → spontaneous only at high T. The last two have a crossover temperature where ΔG = 0: T = ΔH/ΔS.

Key Ideas

  • Enthalpy- vs. entropy-driven. Cases 3 and 4 are a tug-of-war; temperature tips the balance.
  • Crossover temperature. Set ΔG = 0 ⇒ T = ΔH/ΔS (with ΔH and ΔS in consistent units).
  • Low-T spontaneity (ΔH<0, ΔS<0). Exothermic and ordering (e.g., freezing, condensation) wins when T is small because the TΔS penalty shrinks.
  • High-T spontaneity (ΔH>0, ΔS>0). Endothermic and disordering (melting, boiling, many decompositions) wins when T is large because the TΔS reward grows.
  • Signs stay fixed in sign tables. ΔH and ΔS are usually treated as roughly constant over modest temperature ranges, so the crossover logic holds.

Equations and Variables

SymbolMeaningCommon units
ΔGGibbs free-energy changekJ/mol
ΔHEnthalpy changekJ/mol
ΔSEntropy changeJ/(mol·K)
T_crossCrossover temperature (ΔG = 0)K

T_cross = ΔH / ΔS (consistent units)

How It Works

  1. Determine the signs of ΔH and ΔS from the process (phase change, gas-mole count, bond breaking).
  2. Assign the case from the four combinations.
  3. For the temperature-dependent cases, solve T_cross = ΔH/ΔS.
  4. Decide which side of T_cross is spontaneous by testing the sign of ΔG on either side.

Worked Example

Predict the temperature dependence of CaCO₃(s) → CaO(s) + CO₂(g), and find the crossover temperature.

This is the thermal decomposition of limestone. ΔH° = +178.1 kJ/mol (endothermic; bonds broken), and ΔS° = +160.5 J/(mol·K) = +0.1605 kJ/(mol·K) (a gas is produced, so entropy rises).

This is case 4 (ΔH > 0, ΔS > 0) → spontaneous only at high temperature.

  • T_cross = ΔH/ΔS = 178.1 kJ/mol ÷ 0.1605 kJ/(mol·K) = 1110 K (≈ 837 °C).

Below 1110 K, ΔG > 0 and limestone is stable; above 1110 K, ΔG < 0 and it decomposes to quicklime and CO₂ — the basis of lime kilns.

How it works

  1. Determine the signs of ΔH and ΔS from the process (phase change, gas-mole count, bond breaking).
  2. Assign the case from the four combinations.
  3. For the temperature-dependent cases, solve T_cross = ΔH/ΔS.
  4. Decide which side of T_cross is spontaneous by testing the sign of ΔG on either side.

Common confusions

  • "Exothermic reactions are spontaneous at all temperatures." — Wrong. If ΔS < 0 (e.g., freezing), they're spontaneous only below T_cross.
  • "Endothermic reactions can never be spontaneous." — Wrong. If ΔS > 0, they become spontaneous above T_cross (melting, boiling).
  • "Compute T_cross in °C." — Wrong. ΔH/ΔS gives Kelvin; convert afterward.
  • "Mixing kJ and J is fine." — Wrong. ΔH (kJ) and ΔS (J/K) must be converted to the same energy unit before dividing.
  • "The four cases are four different equations." — Wrong. They're all the same equation ΔG = ΔH − TΔS with different sign combinations.

Quick review

  • ΔG = ΔH − TΔS; temperature scales the entropy term.
  • (−,+) always spontaneous; (+,−) never; (−,−) low-T only; (+,+) high-T only.
  • Crossover: T = ΔH/ΔS where ΔG = 0.
  • Below T_cross, ΔH dominates the sign; above it, TΔS dominates.
  • Higher T always favors the higher-entropy state.
Eli, the EliExplains learning guide

Eli explains

The same idea, in plain words

Explain it like I’m 10

Think of a reaction as a debate between two voters: one votes "go" if heat is released (ΔH), the other votes "go" if things get more spread out (ΔS). Temperature is a microphone turned toward the "spread out" voter. When the room is cold, that microphone is quiet, so the heat voter decides. When the room gets hot, the spread-out voter shouts louder and louder until he overrules the heat voter. The temperature where the vote ties is the "crossover" — exactly where ice melts or limestone turns to lime. (The limit: this is a cartoon; real ΔH and ΔS drift slightly with temperature, but the "which voter wins when" logic is exactly right.)

Worked example

Worked Example

Predict the temperature dependence of CaCO₃(s) → CaO(s) + CO₂(g), and find the crossover temperature.

This is the thermal decomposition of limestone. ΔH° = +178.1 kJ/mol (endothermic; bonds broken), and ΔS° = +160.5 J/(mol·K) = +0.1605 kJ/(mol·K) (a gas is produced, so entropy rises).

This is case 4 (ΔH > 0, ΔS > 0) → spontaneous only at high temperature.

  • T_cross = ΔH/ΔS = 178.1 kJ/mol ÷ 0.1605 kJ/(mol·K) = 1110 K (≈ 837 °C).

Below 1110 K, ΔG > 0 and limestone is stable; above 1110 K, ΔG < 0 and it decomposes to quicklime and CO₂ — the basis of lime kilns.

Key takeaways

  • ### High-Yield Facts
  • Four cases: (−,+ ) always spontaneous; (+,−) never spontaneous; (−,−) low-T only; (+,+) high-T only.
  • Crossover temperature: T = ΔH/ΔS (units must match: both kJ or both J).
  • ΔH < 0, ΔS < 0 ⇒ spontaneous below T_cross (e.g., freezing, condensation).
  • ΔH > 0, ΔS > 0 ⇒ spontaneous above T_cross (e.g., melting, boiling, CaCO₃ decomposition).
  • At T_cross, ΔG = 0: the system is at equilibrium (e.g., melting/boiling point).
  • Raising T always favors the direction with more entropy (more gas, looser phase).

Quick check

1 question here. Answers stay hidden until you check.

Question 1 of 1

For a reaction with ΔH > 0 and ΔS > 0, increasing temperature:

Choose an answer, then check it.

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Study tools & related lessonsYou’ll learn to · Related

You’ll learn to

  • Enumerate the four sign combinations of ΔH and ΔS and predict the temperature dependence of spontaneity for each.
  • Calculate the crossover (transition) temperature at which a process becomes spontaneous.
  • Explain why some processes are always spontaneous, some never, and some only above or below a threshold.

Sources & references

  1. OpenStax, *Chemistry 2e*, Ch. 16.4 "Free Energy." https://openstax.org/books/chemistry-2e/pages/16-4-free-energy
  2. NIST Chemistry WebBook (thermochemical data). https://webbook.nist.gov/chemistry/
  3. PubChem, National Library of Medicine. https://pubchem.ncbi.nlm.nih.gov/

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

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