Entropy
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Entropy
Summary — What This Chapter Covers
- Entropy (S) measures disorder / the number of ways particles and energy can be arranged. More disorder = higher entropy.
- Entropy increases going solid → liquid → gas, and increases whenever a gas is produced, a solid dissolves, or particles are freed up to move.
- The driving force of a reaction isn't just enthalpy (ΔH) — it's the total entropy change, which includes the surroundings too.
- ΔS°total = ΔS°system + ΔS°surroundings — a reaction is spontaneous (feasible) when this is positive.
- ΔS°system is calculated from standard entropies: ΔS°system = ΣS°products − ΣS°reactants
- ΔS°surroundings is calculated from the enthalpy change: ΔS°surr = −ΔH / T
- Gibbs free energy, ΔG = ΔH − TΔS, combines both enthalpy and entropy into one number. A reaction is feasible when ΔG ≤ 0.
- Temperature controls whether some reactions are feasible — this is why iron extraction needs a blast furnace at ~1500°C.
- Feasible (thermodynamically favourable) ≠ fast. That's the difference between thermodynamic stability and kinetic stability.
1. Entropy — Introduction
1.1 Why enthalpy alone isn't enough
Here's the puzzle this chapter solves: most reactions we see every day are exothermic — energy flows out, products end up more stable, lower in energy. That makes intuitive sense; things "want" to roll downhill to a lower-energy state, like a ball rolling down a slope.
But then you meet endothermic reactions — the products end up in a higher energy state than the reactants. If the driving force were purely "systems want to lose energy," these reactions shouldn't happen at all. And yet they do — ice melts at room temperature, ammonium nitrate dissolves in water and the water goes cold, thermal decompositions happen. So enthalpy (ΔH) clearly isn't the whole story.
The missing piece is entropy — and once you add it in, both exothermic and endothermic reactions make sense under one single rule.
1.2 What entropy actually is
Think of entropy (symbol S) as a measure of how many different ways you could arrange the particles and their energy in a system and have it look basically the same. The more possible arrangements, the higher the entropy.
Imagine your bedroom. There is exactly one way for it to be perfectly tidy — every item precisely in its spot. But there are thousands of ways for it to be messy — a sock here, a book there, endless combinations. Messiness (disorder) has vastly more possible arrangements than order. That's entropy in a nutshell: disordered states have more ways to exist, so they have higher entropy.
Classic example: melting ice.
1.3 Distribution of molecules — the gas jar model
Picture two gas jars side by side, separated by a glass cover slip. One jar holds 5 molecules of orange bromine gas; the other is empty. The moment you remove the slip, the bromine doesn't stay put — it spreads out evenly across both jars. Why? Not because of some mysterious "force" pushing it apart, but purely because there are vastly more ways for the molecules to be spread out than to stay bunched in one jar.
Each molecule independently could be in the left jar or the right jar — 2 possible locations per molecule. For 5 molecules, the total number of arrangements, W, is:
Entropy is linked to W (the number of possible arrangements) by the Boltzmann equation:
1.4 Distribution of energy — quanta
Entropy isn't just about where particles are in space — it's also about how energy is shared among them. Energy comes in discrete "packets" called quanta — only whole numbers of quanta are possible (you can't have half a packet).
The more energy quanta available, and the more particles there are to share them between, the more ways there are to distribute that energy — and so the higher the entropy.
This is exactly why heating something increases its entropy — it's not just "more vibration," it's genuinely more ways for that thermal energy to be spread among the particles.
Explain, in terms of particle arrangement, why the entropy of water increases when it evaporates to steam.
If 3 molecules of gas could each be in one of two connected flasks, calculate the total number of possible arrangements, W.
2. Entropy Changes
2.1 Changes of state
Entropy always increases in this order as you heat a substance through its states:
- Melting (solid → liquid): the rigid lattice breaks down, particles can now rotate and slide past each other → entropy increases.
- Boiling (liquid → gas): particles become free to move anywhere, far apart from each other → entropy increases significantly (a much bigger jump than melting).
- Condensing (gas → liquid) or freezing (liquid → solid): particles become more ordered, fewer ways to arrange them → entropy decreases.
Notice the graph doesn't rise smoothly — it jumps sharply at the melting point and boiling point (these are the state changes), and rises only gradually within a single state (just from increasing particle vibration/movement as temperature increases).
2.2 Production of a gas
Any reaction that produces a gas from solids or liquids will show a big increase in entropy, because gas molecules are so much more disordered than condensed phases.
CaCO₃ (s) → CaO (s) + CO₂ (g)
A solid decomposes into another solid plus a gas. The CO₂ molecules are constantly moving and colliding — far more disordered than the solid reactant. So the system's entropy increases.
(NH₄)₂CO₃ (s) + 2CH₃COOH (aq) → 2CH₃COO⁻ (aq) + 2NH₄⁺ (aq) + CO₂ (g) + H₂O (l)
This reaction is actually endothermic (temperature drops slightly — energy is absorbed from the surroundings), yet it still happens. Why? Because a solid + aqueous solution turns into ions in solution plus a gas — a big jump in disorder. The system's entropy increase is large enough to drive the reaction even though enthalpy is working against it.
2.3 Dissolving a solid
NH₄NO₃ (s) + aq → NH₄⁺ (aq) + NO₃⁻ (aq) ΔH = +25.7 kJ mol⁻¹
Temperature drops when this dissolves (endothermic), but it still happens spontaneously — because the ions go from being locked in a rigid lattice to being free to move around in solution. That's a big increase in disorder (entropy).
2.4 Endothermic reactions between two solids
2NH₄Cl (s) + Ba(OH)₂·8H₂O (s) → BaCl₂·2H₂O (s) + 2NH₃ (g) + H₂O (l)
Two solids are mixed and the temperature drops sharply — a classic endothermic demonstration (you can even freeze a wet block of wood to a beaker with this reaction!). It happens because a gas (ammonia — you'd smell it, and it turns damp red litmus blue) and a liquid are produced from two ordered solids. Even though energy is absorbed, the huge jump in disorder makes the reaction spontaneous.
General rule: when mixing two solids, entropy change depends on the physical states of the products, not just the energy change. Two solids reacting to form more solid = low entropy change (very ordered on both sides). If a liquid or especially a gas appears among the products, expect entropy to increase substantially.
Predict whether ΔSsystem is positive or negative for the reaction: 2Na (s) + Cl₂ (g) → 2NaCl (s). Explain your reasoning.
3. Total Entropy Calculations
3.1 Why we need the surroundings too
Here's a genuine puzzle: the reaction between sodium and chlorine is extremely exothermic and very obviously happens — yet we just showed its entropy decreases (a gas becomes a solid, more order, not less). If disorder is the driving force, how can this reaction possibly be spontaneous?
The resolution: entropy isn't just about the reacting particles themselves (the "system") — it's also about the surroundings. When a highly exothermic reaction releases a huge burst of energy into the surroundings, that energy has to spread out among a colossal number of surrounding particles — creating an equally colossal increase in the number of ways that energy could be arranged. In other words, a massive entropy increase in the surroundings.
For sodium + chlorine: the energy released is so large that ΔS°surr is hugely positive — positive enough to completely outweigh the negative ΔS°sys. The total entropy change ends up positive, so the reaction is spontaneous.
Calculate the total entropy change in the formation of 1 mole of sodium chloride from its elements in their standard states, given:
ΔS°sys = −90.1 J K⁻¹ mol⁻¹ ΔS°surr = +1379 J K⁻¹ mol⁻¹
Answer:
ΔS°total = ΔS°sys + ΔS°surr
ΔS°total = −90.1 + 1379 = +1289 J K⁻¹ mol⁻¹
Since ΔS°total is positive, the reaction is spontaneous — exactly what we observe.
3.2 Calculating ΔS°system
Every substance — even pure elements — has a "standard entropy" value, S° (unlike enthalpy of formation, where elements in their standard state are defined as zero). You'll find these values in a data booklet.
2Mg (s) + O₂ (g) → 2MgO (s)
S°[Mg(s)] = 32.60 J K⁻¹ mol⁻¹ S°[O₂(g)] = 205.0 J K⁻¹ mol⁻¹ S°[MgO(s)] = 38.20 J K⁻¹ mol⁻¹
Answer:
ΔS°system = ΣS°products − ΣS°reactants
ΔS°system = (2 × 38.20) − (2 × 32.60 + 205.0)
ΔS°system = 76.4 − 270.2 = −193.8 J K⁻¹ mol⁻¹
Makes sense: a gas (O₂) is consumed to form only solids — big decrease in disorder.
N₂ (g) + 3H₂ (g) ⇌ 2NH₃ (g)
S°[N₂(g)] = 191.6 S°[H₂(g)] = 131 S°[NH₃(g)] = 192.3 (all J K⁻¹ mol⁻¹)
Answer:
ΔS°system = [2 × 192.3] − [191.6 + (3 × 131)]
ΔS°system = 384.6 − 584.6 = −200 J K⁻¹ mol⁻¹
4 moles of gas (1 N₂ + 3 H₂) become 2 moles of gas (2 NH₃) — fewer gas particles, fewer ways to arrange them, so entropy decreases, even though everything stays gaseous.
3.3 Calculating ΔS°surroundings
To find how much the surroundings' entropy changes, we use the enthalpy change of the reaction — because ΔH tells us exactly how much energy was transferred into (or out of) the surroundings.
Notice also that ΔS°surr depends on temperature: transferring the same amount of energy at a low temperature produces a bigger entropy change than transferring it at high temperature. Think of it like this — dropping a bucket of hot water into a cold lake changes the lake's "disorder" more dramatically than dropping the same bucket into an already-boiling pool.
Al₂O₃ (s) + 3C (s) → 2Al (s) + 3CO (g) ΔH° = +1336 kJ mol⁻¹, T = 298 K
Answer:
ΔS°surr = −ΔH / T (convert ΔH to J mol⁻¹ first: ×1000)
ΔS°surr = −(1336000) / 298
ΔS°surr = −4483 J K⁻¹ mol⁻¹
Makes sense: this reaction is endothermic (absorbs heat from the surroundings), so the surroundings lose thermal energy and become more ordered — entropy of surroundings decreases.
A reaction has ΔH° = −242 kJ mol⁻¹ at 298 K. Calculate ΔS°surroundings.
For the reaction CaCO₃ (s) → CaO (s) + CO₂ (g), use S°[CaCO₃(s)] = 92.9, S°[CaO(s)] = 39.7, S°[CO₂(g)] = 213.6 (all J K⁻¹ mol⁻¹) to calculate ΔS°system.
4. Reaction Feasibility
4.1 The feasibility rule
For a reaction to be spontaneous (chemists call this "feasible"), ΔStotal must be positive. There are three ways this can happen:
| ΔSsystem | ΔSsurroundings | Result |
|---|---|---|
| Positive | Positive | Both help → definitely feasible |
| Negative | Positive, and bigger | Surroundings "win" → feasible |
| Positive, and bigger | Negative | System "wins" → feasible |
Mg (s) + O₂ (g) → MgO (s)
A solid forms from a solid + a gas → ΔSsystem is negative. But this reaction is very exothermic, so ΔSsurroundings is huge and positive — big enough to overcome the negative system term. Total entropy is positive, so the reaction proceeds (spectacularly, if you've seen magnesium burn!).
2CH₃COOH (aq) + (NH₄)₂CO₃ (s) → 2CH₃COONH₄ (aq) + H₂O (l) + CO₂ (g)
This reaction is endothermic, so ΔSsurroundings is negative. But a gas is produced from a solid and liquid, so ΔSsystem is strongly positive — big enough to overcome the negative surroundings term. Total entropy is positive → spontaneous.
4.2 Gibbs Free Energy
Rather than calculating ΔSsys and ΔSsurr separately and adding them, chemists usually use a single combined quantity called Gibbs free energy (G), which bundles enthalpy and entropy together directly.
4.3 Temperature & feasibility
Because ΔG = ΔH − TΔS has a "−TΔS" term, temperature can flip a reaction from not-feasible to feasible (or vice versa), depending on the signs of ΔH and ΔS. There are four possible combinations:
Exothermic reactions (ΔH negative):
- If ΔSsystem is positive: both terms in ΔG are negative → ΔG is always negative → always feasible, at any temperature.
- If ΔSsystem is negative: the −TΔS term is positive. At low T, it's small and ΔH dominates → ΔG negative → feasible. At very high T, −TΔS grows large enough to overcome ΔH → ΔG positive → not feasible. (In practice, since entropy terms are usually much smaller than enthalpy terms, these reactions are usually feasible under normal conditions.)
Endothermic reactions (ΔH positive):
- If ΔSsystem is negative: both terms in ΔG are positive → ΔG is always positive → never feasible, at any temperature.
- If ΔSsystem is positive: at low T, −TΔS is small and can't overcome the positive ΔH → not feasible. At high T, −TΔS becomes large and negative enough to overcome ΔH → ΔG negative → feasible only at high temperature.
| ΔH | ΔS | ΔG | Spontaneous? | Why |
|---|---|---|---|---|
| Negative (exothermic) | Positive (more disorder) | Always negative | Always | Forward reaction spontaneous at any T |
| Positive (endothermic) | Negative (more order) | Always positive | Never | Reverse reaction spontaneous at any T |
| Negative | Negative | Negative at low T, positive at high T | Depends on T | Spontaneous only at low T (TΔS < ΔH) |
| Positive | Positive | Negative at high T, positive at low T | Depends on T | Spontaneous only at high T (TΔS > ΔH) |
The reaction between aluminium oxide and carbon is not feasible at room temperature:
Al₂O₃ + 3C (s) → 2Al (s) + 3CO₂ (g)
Given ΔH = +1336 kJ mol⁻¹ and ΔS = +581 J K⁻¹ mol⁻¹, find the temperature at which the reaction becomes feasible.
Answer:
Both ΔH and ΔS are positive, so ΔG becomes negative once TΔS > ΔH.
The threshold temperature is where ΔG = 0:
0 = ΔH − TΔS → T = ΔH / ΔS
Convert ΔS to kJ: 0.581 kJ K⁻¹ mol⁻¹
T = 1336 / 0.581 = 2299 K
Above 2299 K, this reaction becomes thermodynamically feasible — this is exactly the kind of reasoning behind why metal extraction reactions need extremely high furnace temperatures.
A reaction has ΔH = +58 kJ mol⁻¹ and ΔS = +176 J K⁻¹ mol⁻¹. Calculate the minimum temperature at which the reaction becomes feasible.
A reaction has ΔH = −120 kJ mol⁻¹ and ΔS = +40 J K⁻¹ mol⁻¹. Is this reaction feasible at all temperatures, no temperatures, or only some temperatures? Explain.
5. Thermodynamic vs. Kinetic Stability
This is one of the most important — and most commonly confused — ideas in this chapter: feasible does not mean fast.
- Thermodynamic stability/feasibility — whether a reaction can happen at all under given conditions. Governed entirely by ΔG (negative = feasible).
- Kinetic stability — how fast a feasible reaction actually proceeds. Governed by the activation energy (Ea). A reaction can be thermodynamically screaming "yes, go!" and still not visibly happen because the activation energy barrier is too high to overcome at room temperature.
CH₄ (g) + 2O₂ (g) → CO₂ (g) + 2H₂O (l)
This is hugely thermodynamically feasible (very negative ΔG) — yet a gas mixture of methane and oxygen sits around safely at room temperature indefinitely. Why? It's kinetically stable — the activation energy is high, so you need a spark or flame to give the molecules enough energy to actually react. Without that "kick," nothing happens, even though thermodynamics says it should.
H₂O₂ (l) → H₂O (l) + ½O₂ (g)
Thermodynamically feasible at 298 K, but the activation energy is high, so it decomposes only very slowly on its own. Adding a catalyst (MnO₂) provides an alternative reaction pathway with lower Ea — it doesn't change whether the reaction is feasible (that's fixed by ΔG), it just makes the already-feasible reaction happen much faster.
Diamond converting to graphite has a negative ΔG at room temperature, yet diamonds do not visibly turn into graphite on any human timescale. Explain this using the terms "thermodynamic" and "kinetic."
What to Memorise
| Term / Formula | Meaning |
|---|---|
| Entropy, S | Measure of the number of possible arrangements of particles and energy — i.e. how disordered a system is. Units: J K⁻¹ mol⁻¹ |
| S = k ln W | Boltzmann equation. k = 1.38 × 10⁻²³ J K⁻¹ mol⁻¹; W = number of possible arrangements |
| Entropy order | Solid < Liquid < Gas (increasing disorder) |
| ΔS°total = ΔS°sys + ΔS°surr | A reaction is feasible/spontaneous when ΔS°total is positive |
| ΔS°system = ΣS°products − ΣS°reactants | Calculated using standard entropy values (elements are NOT zero, unlike ΔHf) |
| ΔS°surr = −ΔH / T | Entropy change of surroundings, from the enthalpy change and temperature (K) |
| ΔG° = ΔH°reaction − TΔS°system | Gibbs free energy. Reaction feasible when ΔG° ≤ 0. Units: kJ mol⁻¹ (careful with unit conversions for ΔS!) |
| Thermodynamic feasibility | Whether ΔG is negative — whether a reaction can happen |
| Kinetic stability | Whether the reaction is fast or slow — governed by activation energy (Ea), completely separate from ΔG |
Concepts Checklist
Exam Tips & Common Mistakes
- Exam Tips & Common Mistakes
- 4.3 Temperature & feasibility
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