Chemistry 2e · Thermodynamics
The Second and Third Laws of Thermodynamics
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In 30 seconds
The first law of thermodynamics states that energy is conserved — but it says nothing about direction. A cup of hot coffee cools spontaneously, yet the reverse process (the room warming the coffee further) never happens even though both directions conserve energy. The second law supplies the missing direction: in any spontaneous process A process that proceeds without outside intervention once started. Full entry →, the total entropy of the universe increases. The third law gives entropy an absolute zero point, letting chemists tabulate actual (not relative) entropy values and calculate reaction entropy changes.
Together these laws turn "spontaneous" into something you can compute. A process is spontaneous if the entropy of the universe increases, ΔSuniv > 0. Because the universe is the system plus its surroundings, ΔSuniv = ΔSsys + ΔSsurr > 0. This topic shows how to evaluate both terms — the system's entropy change from tabulated standard entropies, and the surroundings' entropy change from the heat the process releases or absorbs.
Why this matters
- Predicting spontaneity without running the reaction: If you can calculate ΔSuniv, you can predict whether a reaction will proceed on its own — a huge shortcut.
- Every real engine and refrigerator obeys it: No engine converts heat completely into work (the second law in its Kelvin–Planck form), setting real efficiency limits on power plants and car engines.
- Phase changes and biological processes: Melting, boiling, dissolving, and even protein folding are governed by the balance between system and surroundings entropy.
- Foundation for free energy: The next topic defines Gibbs free energy as a convenient repackaging of ΔSuniv — master this and the next topic is mostly algebra.
- Exams: Expect to calculate ΔS°rxn from standard molar entropies, compute ΔSsurr = -ΔHsys/T, and justify spontaneity with signs.
The college version
Core Concepts
The second law: entropy of the universe increases
For any spontaneous process,
ΔSuniv = ΔSsys + ΔSsurr > 0
At equilibrium the process is reversible and ΔSuniv = 0. Two classic statements capture the same idea: heat flows spontaneously from hot to cold bodies (Clausius), and it is impossible to convert heat completely into work in a cyclic process (Kelvin–Planck). Note that ΔSsys alone can be negative — the system can become more ordered — as long as the surroundings gain enough entropy to keep the total positive.
The entropy change of the surroundings
At constant temperature and pressure, the surroundings exchange heat equal in magnitude but opposite in sign to the system's enthalpy change:
ΔSsurr = -ΔHsysT
An exothermic reaction (ΔHsys < 0) dumps heat into the surroundings, increasing their entropy; the minus sign makes ΔSsurr positive. An endothermic reaction cools its surroundings, so ΔSsurr < 0. This formula is why enthalpy data feed into spontaneity predictions.
Entropy changes at phase transitions
At the transition temperature, the system and surroundings are in equilibrium, so
ΔStrans = ΔHtransTtrans
Melting and boiling both increase entropy (disorder grows), so both have positive ΔS at the transition temperature. Boiling produces a much larger entropy increase than melting because a gas has vastly more accessible microstates than a liquid.
The third law and standard molar entropies
The third law states that the entropy of a perfectly ordered crystalline substance approaches zero as the temperature approaches absolute zero:
limT → 0 S = 0
This gives entropy an absolute scale, unlike enthalpy, which is only known relative to formation from elements. Standard molar entropy S° is the entropy of one mole of a pure substance in its standard state at a specified temperature (usually 298 K), with units J/(mol·K). Elements in their standard states have nonzero standard entropies — a key difference from ΔH°f and ΔG°f, which are zero for elements.
Calculating ΔS°rxn
For a reaction aA + bB → cC + dD:
ΔS°rxn = ∑n S°(products) - ∑m S°(reactants)
Useful trends for sign checks: forming gases increases entropy; reactions that reduce the number of gas moles typically have negative ΔS°; larger, more complex molecules have higher standard entropies.
How It Works / Step-by-Step Process
- Write the balanced chemical equation and note the physical states.
- Decide which term you need: ΔSsys from tabulated S° values, ΔSsurr from ΔHsys and T, or both for ΔSuniv.
- For ΔS°rxn, multiply each substance's S° by its stoichiometric coefficient, sum products, subtract sum of reactants.
- Keep units consistent: S° is in J/(mol·K), so ΔH must be converted to joules before using ΔSsurr = -ΔH/T.
- Check the sign against a physical trend (gas moles, phase changes) before trusting the arithmetic.
Common Confusions
| Do Not Confuse | With | Difference |
|---|---|---|
| ΔSsys < 0 | "The process is nonspontaneous." | The system can become more ordered as long as ΔSsurr is positive enough that ΔSuniv > 0. |
| Standard molar entropy S° | Standard enthalpy of formation ΔH°f | S° values are absolute (elements have nonzero values); ΔH°f = 0 for elements in standard states. |
| Melting point entropy | Boiling point entropy | Vaporization gives a much larger ΔS (gas microstates) than fusion. |
| "Spontaneous means fast." | "Spontaneous means thermodynamically favored." | Spontaneity says nothing about rate; thermodynamics and kinetics are separate. |
| Using ΔSsurr = -ΔH/T with kJ | Mixing units | ΔH must be in joules to match J/(mol·K) entropies. |
Eli explains
The same idea, in plain words
Explain it like I’m 10
The second law says that messiness (entropy) always increases in the universe when something happens by itself. Your messy room gets messier unless you clean it — cleaning takes energy. The third law says that if you cool a perfect crystal all the way to absolute zero, it has zero messiness, which gives scientists a starting point to measure how messy every substance is.
Worked example
Example 1: ΔS°rxn for ammonia synthesis
Synthesizing ammonia, N2(g) + 3H2(g) → 2NH3(g), is exothermic and industrially vital. Use standard molar entropies S°(N2) = 191.6 J/(mol·K), S°(H2) = 130.7 J/(mol·K), S°(NH3) = 192.8 J/(mol·K).
Formula:
ΔS°rxn = ∑n S°(products) - ∑m S°(reactants)
Substitution:
ΔS°rxn = 2(192.8) - [1(191.6) + 3(130.7)] J/(mol·K)
ΔS°rxn = 385.6 - 583.7 = -198.1 J/(mol·K)
The negative value makes sense: four moles of gas become two moles of gas, so the system becomes more ordered. Whether the reaction is spontaneous depends on whether the surroundings' entropy gain from the released heat outweighs this drop — exactly the calculation previewed above and formalized in the free-energy topic.
Example 2: Melting ice at 0 °C
Ice melts at 273.15 K with ΔHfus = 6.01 kJ/mol. Find the entropy change of the system at the transition.
Formula:
ΔSfus = ΔHfusTfus
Substitution (converting kJ to J for unit consistency):
ΔSfus = 6.01 × 103 J/mol273.15 K = 22.0 J/(mol·K)
Compare with vaporizing water at 373.15 K: ΔSvap = 40.7 × 103 / 373.15 = 109.1 J/(mol·K) — nearly five times more entropy than melting, the hallmark of a gas's enormous number of accessible microstates. At the melting point the process is at equilibrium (ΔSuniv = 0); slightly above it, melting is spontaneous; slightly below, freezing is.
Key takeaways
- Spontaneity requires ΔSuniv = ΔSsys + ΔSsurr > 0; equilibrium means ΔSuniv = 0.
- ΔSsurr = -ΔHsys/T: exothermic reactions increase surroundings entropy; endothermic reactions decrease it.
- At a phase transition, ΔStrans = ΔHtrans/Ttrans; melting and boiling both give positive ΔS.
- The third law assigns S = 0 to a perfect crystal at 0 K, making standard molar entropies absolute values in J/(mol·K).
- ΔS°rxn = ΣnS°(products) - ΣmS°(reactants); elements do not have zero S°.
- Gases have much larger entropies than liquids or solids; a decrease in gas moles usually means ΔS°< 0.
Check yourself
5 review questions from the chapter. Try each one, then open the answer.
State the second law in terms of the entropy of the universe, and give the equality condition.
Show answer
ΔSuniv > 0 for any spontaneous process; ΔSuniv = 0 at equilibrium (reversible process).
A reaction is exothermic with ΔHsys = -50 kJ at 298 K. What is ΔSsurr?
Show answer
ΔSsurr = -(-50,000 J)/298 K = +168 J/K. The released heat increases the surroundings' entropy.
Why is ΔS°rxn for N2(g) + 3H2(g) → 2NH3(g) negative, and how is that possible for a spontaneous process at 298 K?
Show answer
Four moles of gas become two (ΔS°rxn = -198.1 J/(mol·K)), so the system's entropy falls — but the exothermic heat release raises ΔSsurr enough that ΔSuniv > 0 at moderate temperatures.
What does the third law provide that enthalpy data cannot?
Show answer
An absolute entropy scale: S° values are real, measurable quantities, not differences relative to elements.
Which has the larger entropy change at its transition temperature: melting or boiling water? Why?
Show answer
Boiling: ΔSvap ≈ 109 J/(mol·K) versus ΔSfus ≈ 22 J/(mol·K), because gases have far more accessible microstates than liquids.
Study tools & related lessonsKey vocabulary · Related
Key vocabulary
- second law of thermodynamics
- In any spontaneous process, the entropy of the universe increases.
- standard molar entropy, S°
- Entropy of one mole of a substance in its standard state at a stated temperature, in J/(mol·K).
- surroundings entropy change, Δ Ssurr
- Entropy change of everything outside the system, set by heat exchange: -ΔHsys/T.
- spontaneous process
- A process that proceeds without outside intervention once started.
Sources & references
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