Worked example 1
Predict the sign of ΔS for CaCO₃(s) → CaO(s) + CO₂(g).
Try it first: Count the moles of gas on each side.
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Thermochemistry & ThermodynamicsEntropyContent level: Core 22 min
Entropy and the Second Law
What you'll be able to do: Predict the sign of an entropy change and calculate ΔS from standard molar entropies.
Best after: Energy, Heat and the First Law
Introduction
Enthalpy alone cannot explain why ice melts on a warm day or why gases mix without being pushed. The missing ingredient is entropy, a measure of how many ways the energy and particles can be arranged.
These are recommended, not required. You can start this lesson at any time.
Learning objectives
- Predict the sign of ΔS from a balanced equation
- Calculate ΔS from standard molar entropy values
- State the second and third laws of thermodynamics
- Distinguish system entropy from universe entropy
Lesson
What entropy counts
Entropy is a measure of how many microscopic arrangements of position and energy give the same overall state. More available arrangements means higher entropy. Describing it as disorder is a rough shorthand that fails for cases such as protein folding, so counting arrangements is the safer picture.
Predicting the sign
Entropy increases going solid to liquid to gas, on heating, on dissolving most ionic solids, and whenever the total moles of gas rise. When gas moles change, that term overwhelms everything else.
Count moles of gas on each side first. If they differ, you already have the sign of ΔS(system).
Standard molar entropies
Tabulated as S in J/mol K, these are absolute values rather than changes, because the third law sets the entropy of a perfect crystal at 0 K to exactly zero. Every substance above absolute zero therefore has a positive S, elements included.
ΔS(rxn) = S n S(products) - S n S(reactants)
Entropies are in J/mol K while enthalpies are in kJ/mol. Convert before combining them in a Gibbs calculation.
The second law
A process is spontaneous when the total entropy of the universe increases. The system alone is allowed to become more ordered, as when water freezes, provided the surroundings gain more entropy than the system loses.
ΔS(universe) = ΔS(system) + ΔS(surroundings) > 0 for a spontaneous change
Entropy of the surroundings
Heat released by an exothermic reaction spreads into the surroundings and raises their entropy, and it does so more effectively at low temperature. This is why exothermic reactions are so often spontaneous, and it is the idea the Gibbs equation packages up.
ΔS(surroundings) = -ΔH(system) / T
Key ideas
Definition
Entropy (S)
A measure of the number of microscopic arrangements available to a system, in J/mol K.
Rule
Gas moles rule
An increase in the total moles of gas gives a positive ΔS(system).
Rule
Third law
A perfect crystal at 0 K has zero entropy, which makes absolute entropies measurable.
Rule
Second law
For a spontaneous change the entropy of the universe increases.
Key concept
Elements are not zero
Unlike ΔHf, standard molar entropies of elements are positive.
Equation
Reaction entropy
ΔS(rxn) = S n S(products) - S n S(reactants)
- S = standard molar entropy in J/mol K
- n = stoichiometric coefficient
Equation
Entropy change of the surroundings
ΔS(surroundings) = -ΔH(system)/T
- T = absolute temperature in K
Worked examples
Worked example 2
Calculate ΔS for N₂(g) + 3 H₂(g) → 2 NH₃(g) given S: N₂ = 191.6, H₂ = 130.7, NH₃ = 192.8 J/mol K.
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Worked example 3
Water freezes at -10 C, yet the system becomes more ordered. Explain how this is consistent with the second law.
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Common mistakes
Assuming any process that orders the system violates the second law.
Why it's wrong: The law applies to the universe, not the system alone.
Check instead: Include the surroundings term before judging spontaneity.
Combining S in J/mol K with ΔH in kJ/mol without converting.
Why it's wrong: The entropy term ends up 1000 times too large.
Check instead: Convert entropy to kJ/mol K, or enthalpy to J/mol, before combining.
Giving elements a standard entropy of zero.
Why it's wrong: Only a perfect crystal at 0 K has zero entropy.
Check instead: Look up the tabulated positive value for the element.
Predicting ΔS from the total number of particles rather than gas particles.
Why it's wrong: Gases carry far more entropy per mole than solids or liquids.
Check instead: Count moles of gas first and let that decide the sign.
Practice this skill
No practice questions are available for this topic yet. You can still practice the whole unit.
What you should now know
Entropy measures the number of microscopic arrangements available to a system. It rises with temperature, with the number of gas particles, on melting and vaporising, and on dissolving most solids. Gases dominate every qualitative prediction: if the moles of gas increase, ΔS(system) is almost certainly positive. Standard molar entropies are absolute values in J/mol K and are never zero for a substance above 0 K, so unlike formation enthalpies, elements have non-zero entries. ΔS is still products minus reactants, and the second law requires the entropy of the universe to increase for any spontaneous change.
- Entropy counts available microscopic arrangements
- Rising moles of gas means a positive ΔS(system)
- Standard molar entropies are absolute and positive, elements included
- ΔS = products minus reactants, scaled by coefficients
- Spontaneity requires the entropy of the universe to increase
Sources and further reading
- Chemistry 2e, Section 16.2: EntropyPaul Flowers, Klaus Theopold, Richard Langley, William R. Robinson · OpenStax, Rice University · Chapter 16.2View source
Chemistry 2e, OpenStax, Rice University, licensed CC BY 4.0. License