Entropy

Spec Mapping — OCR H432 Module 5.2.2 — Entropy and free energy, covering entropy as a measure of the dispersal of energy among accessible microstates, the units of standard molar entropy ($\text{J K}^{-1}\text{ mol}^{-1}$J K−1 mol−1), qualitative prediction of the sign of $\Delta_r S^\circ$Δr​S∘ from changes in state, moles of gas and molecular complexity, the Third Law of Thermodynamics ($S = 0$S=0 for a perfect crystal at 0 K), and quantitative calculation of $\Delta_r S^\circ$Δr​S∘ from tabulated absolute entropies via $\Delta_r S^\circ = \sum S^\circ_{\text{products}} - \sum S^\circ_{\text{reactants}}$Δr​S∘=∑Sproducts∘​−∑Sreactants∘​ (refer to the official OCR H432 specification document for exact wording).

Enthalpy alone is not enough to predict whether a reaction will happen. A piece of ammonium nitrate stirred into water absorbs heat — the beaker chills noticeably — yet the dissolving proceeds enthusiastically without any external nudge. Ice melts spontaneously at 1 °C even though melting is endothermic. To explain these endothermic-but-spontaneous processes (and, more importantly, to predict which way unfamiliar reactions will go) you need a second thermodynamic state function alongside $H$H: the entropy, $S$S. Entropy quantifies the spread of energy and particles over the microstates available to a system; a system shifts spontaneously towards configurations with more microstates because such configurations are simply more numerous. This lesson defines entropy at the particulate level, walks through five sign-prediction examples (including the deceptive Haber case), four quantitative $\Delta_r S^\circ$Δr​S∘ calculations using tabulated $S^\circ$S∘ data, and the Third-Law foundation that anchors absolute entropies to a zero-point. Lesson 7 will combine $\Delta_r H^\circ$Δr​H∘ and $\Delta_r S^\circ$Δr​S∘ into the Gibbs free energy $\Delta G^\circ = \Delta H^\circ - T\Delta S^\circ$ΔG∘=ΔH∘−TΔS∘ to make feasibility quantitative.

Key Equation: the standard entropy change of a reaction is $\Delta_r S^\circ = \sum S^\circ_{\text{products}} - \sum S^\circ_{\text{reactants}}$Δr​S∘=∑Sproducts∘​−∑Sreactants∘​ where each $S^\circ$S∘ is the standard molar entropy in $\text{J K}^{-1}\text{ mol}^{-1}$J K−1 mol−1 (units: joules, NOT kilojoules) at 298 K and 100 kPa, and each term is multiplied by its stoichiometric coefficient from the balanced equation.

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What is entropy?

Entropy ($S$S) is a state function whose magnitude reflects how many ways the energy of a system can be distributed over its accessible microstates — the specific arrangements of particles and quantised energies that share the same macroscopic state (temperature, pressure, composition). Boltzmann's statistical definition makes this concrete:

$S = k_B \ln W$S=kB​lnW

where $k_B = 1.38 \times 10^{-23}$kB​=1.38×10−23 J K$^{-1}$−1 is the Boltzmann constant and $W$W is the number of microstates corresponding to the macroscopic state. A larger $W$W → larger $S$S. The Boltzmann formula is not examined at A-Level, but the underlying idea — entropy counts microstates — is the conceptual foundation OCR expects you to articulate.

A more familiar picture: a deck of 52 cards has exactly one fully-sorted arrangement but $52! \approx 8 \times 10^{67}$52!≈8×1067 shuffled arrangements. Shuffled decks have vastly higher entropy than sorted decks; they are also vastly more probable, which is why shuffling is irreversible in practice.

At the particulate level:

• A crystalline solid has particles locked at lattice sites with only small vibrational motion — very few microstates — low $S$S.
• A liquid has particles in close contact but free to move past one another — more microstates — intermediate $S$S.
• A gas has particles moving randomly through a large volume with a wide spread of kinetic energies — vastly more microstates — high $S$S.

Key Definitions:

• Entropy ($S$S) — a measure of the dispersal of energy and matter over accessible microstates.
• Standard molar entropy ($S^\circ$S∘) — the absolute entropy of one mole of a pure substance at 298 K and 100 kPa, in $\text{J K}^{-1}\text{ mol}^{-1}$J K−1 mol−1. Always positive (Third Law).
• Entropy change ($\Delta S^\circ$ΔS∘) — products minus reactants, in $\text{J K}^{-1}\text{ mol}^{-1}$J K−1 mol−1. Can be positive or negative.
• Microstate — a specific arrangement of particles and energy quanta consistent with the macroscopic state.

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States of matter and entropy

The clearest qualitative guide to entropy is physical state. The transition $\text{solid} \to \text{liquid} \to \text{gas}$solid→liquid→gas corresponds to ever-larger increases in entropy, and the largest single jump is at vaporisation.

State	Particle arrangement	Typical $S^\circ$S∘ / $\text{J K}^{-1}\text{ mol}^{-1}$J K−1 mol−1
Solid	regular, fixed lattice sites	30 – 100
Liquid	disordered, free to move past each other	70 – 200
Gas (small molecule)	random, large volume, high KE	150 – 250
Gas (complex molecule)	many rotational/vibrational modes	200 – 400

graph LR
  A["Solid<br/>low S"] -->|"melting"| B["Liquid<br/>medium S"]
  B -->|"vaporising large jump"| C["Gas<br/>high S"]
  C -->|"dissolving in liquid"| D["Gas in solution<br/>lower than free gas"]

Vaporisation gives a particularly large entropy jump because the volume per particle expands from a few cubic nanometres (liquid) to a few cubic centimetres (gas at 1 bar, 298 K) — a roughly thousand-fold increase in accessible volume. Trouton's rule (an empirical observation) gives $\Delta_{\text{vap}} S^\circ \approx +85$Δvap​S∘≈+85 J K$^{-1}$−1 mol$^{-1}$−1 for many simple liquids — a useful order-of-magnitude check.

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Predicting the sign of $\Delta_r S^\circ$Δr​S∘

The dominant rule of thumb is count the moles of gas. Gases contribute so much more $S^\circ$S∘ than condensed phases (factor ~3–10) that a change in $n_{\text{gas}}$ngas​ almost always dictates the sign.

1. $\Delta n_{\text{gas}} > 0$Δngas​>0 (more gas formed) → $\Delta S^\circ$ΔS∘ positive, usually large (+100 to +200 J K$^{-1}$−1 mol$^{-1}$−1).
2. $\Delta n_{\text{gas}} < 0$Δngas​<0 (gas consumed) → $\Delta S^\circ$ΔS∘ negative, usually large.
3. $\Delta n_{\text{gas}} = 0$Δngas​=0 → sign depends on subtler effects (solid/liquid transitions, complexity changes); $\Delta S^\circ$ΔS∘ is usually small.
4. Solid dissolving in water → usually $\Delta S^\circ > 0$ΔS∘>0 (ordered crystal becomes mobile aqueous ions). Exception: very highly charged ions order the water shell so strongly that the total entropy of mixing falls (e.g. $\text{Al}^{3+}$Al3+, $\text{Mg}^{2+}$Mg2+).
5. Mixing two miscible substances → $\Delta S^\circ > 0$ΔS∘>0.
6. Increase in molecular complexity at constant $n_{\text{gas}}$ngas​ → small positive $\Delta S^\circ$ΔS∘ (more vibrational/rotational modes per molecule).

Worked example 1 — combustion of hydrogen

$2\text{H}_2\text{(g)} + \text{O}_2\text{(g)} \rightarrow 2\text{H}_2\text{O(l)}$2H2​(g)+O2​(g)→2H2​O(l)

Moles of gas: $3 \to 0$3→0. Entropy plummets. $\Delta_r S^\circ \approx -327$Δr​S∘≈−327 J K$^{-1}$−1 mol$^{-1}$−1 — strongly negative, dominated by the loss of three moles of gas to a condensed phase.

Worked example 2 — thermal decomposition of limestone

$\text{CaCO}_3\text{(s)} \rightarrow \text{CaO(s)} + \text{CO}_2\text{(g)}$CaCO3​(s)→CaO(s)+CO2​(g)

Moles of gas: $0 \to 1$0→1. Entropy rises sharply: $\Delta_r S^\circ = +161$Δr​S∘=+161 J K$^{-1}$−1 mol$^{-1}$−1. This positive $\Delta S^\circ$ΔS∘ is what makes high-temperature kiln operation work — see Lesson 7 for the $T\Delta S$TΔS term that overtakes the endothermic $\Delta H$ΔH.

Worked example 3 — dissolving ammonium nitrate (the cold-pack)

$\text{NH}_4\text{NO}_3\text{(s)} \rightarrow \text{NH}_4^+\text{(aq)} + \text{NO}_3^-\text{(aq)}$NH4​NO3​(s)→NH4+​(aq)+NO3−​(aq)

Two ions liberated from a single solid formula unit, with full freedom in solution. $\Delta_r S^\circ \approx +108$Δr​S∘≈+108 J K$^{-1}$−1 mol$^{-1}$−1. The positive $\Delta S^\circ$ΔS∘ is large enough to overpower the endothermic $\Delta_{\text{sol}} H^\circ \approx +26$Δsol​H∘≈+26 kJ mol$^{-1}$−1, making $\Delta G^\circ < 0$ΔG∘<0 at room temperature (Lesson 7). This is precisely the chemistry behind disposable cold packs — the entropy term wins.

Worked example 4 — the Haber synthesis

$\text{N}_2\text{(g)} + 3\text{H}_2\text{(g)} \rightarrow 2\text{NH}_3\text{(g)}$N2​(g)+3H2​(g)→2NH3​(g)

Moles of gas: $4 \to 2$4→2, so $\Delta n_{\text{gas}} = -2$Δngas​=−2. Entropy falls: $\Delta_r S^\circ = -198$Δr​S∘=−198 J K$^{-1}$−1 mol$^{-1}$−1. The reaction would not happen at all on entropy grounds; what saves it is the strongly exothermic $\Delta H^\circ = -92$ΔH∘=−92 kJ mol$^{-1}$−1, which heats the surroundings and creates a much larger positive $\Delta S^\circ_{\text{surr}}$ΔSsurr∘​. At 298 K $\Delta G^\circ < 0$ΔG∘<0, but only by a slim margin — at high $T$T the reaction loses feasibility (Lesson 7).

Worked example 5 — gas-into-solution

$\text{CO}_2\text{(g)} \rightarrow \text{CO}_2\text{(aq)}$CO2​(g)→CO2​(aq)

Loss of a mole of gas to solvation. $\Delta_r S^\circ < 0$Δr​S∘<0 (typically about $-100$−100 J K$^{-1}$−1 mol$^{-1}$−1). This is why fizzy drinks decarbonate on warming — the entropy term $-T\Delta S$−TΔS becomes more punishing at higher $T$T, pushing CO$_2$2​ out of solution.

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Calculating $\Delta_r S^\circ$Δr​S∘ from tabulated $S^\circ$S∘

Standard molar entropies are absolute (not relative) quantities, anchored to $S = 0$S=0 at $T = 0$T=0 K (Third Law). Tabulated $S^\circ$S∘ values therefore include all the entropy a substance has accumulated from absolute zero up to 298 K. To find the entropy change of a reaction, apply:

$\Delta_r S^\circ = \sum \nu_i S^\circ_{\text{products}} - \sum \nu_j S^\circ_{\text{reactants}}$Δr​S∘=∑νi​Sproducts∘​−∑νj​Sreactants∘​

where $\nu_i$νi​ is the stoichiometric coefficient. Units: $\text{J K}^{-1}\text{ mol}^{-1}$J K−1 mol−1 throughout. Critically — and the single most common mark-scheme deduction in this topic — entropy is in joules per kelvin per mole, not kilojoules. Substituting J for kJ in $\Delta G = \Delta H - T\Delta S$ΔG=ΔH−TΔS at Lesson 7 will give an answer 1000× too large.

Worked example 6 — $\Delta_r S^\circ$Δr​S∘ for limestone decomposition

$\text{CaCO}_3\text{(s)} \rightarrow \text{CaO(s)} + \text{CO}_2\text{(g)}$CaCO3​(s)→CaO(s)+CO2​(g)

Substance	$S^\circ$S∘ / J K$^{-1}$−1 mol$^{-1}$−1
CaCO$_3$3​(s)	93
CaO(s)	40
CO$_2$2​(g)	214

$\Delta_r S^\circ = [S^\circ(\text{CaO}) + S^\circ(\text{CO}_2)] - S^\circ(\text{CaCO}_3) = (40 + 214) - 93 = +161\;\text{J K}^{-1}\text{ mol}^{-1}$Δr​S∘=[S∘(CaO)+S∘(CO2​)]−S∘(CaCO3​)=(40+214)−93=+161J K−1 mol−1

Positive — consistent with $\Delta n_{\text{gas}} = +1$Δngas​=+1.

Worked example 7 — $\Delta_r S^\circ$Δr​S∘ for ammonia synthesis

$\text{N}_2\text{(g)} + 3\text{H}_2\text{(g)} \rightarrow 2\text{NH}_3\text{(g)}$N2​(g)+3H2​(g)→2NH3​(g)

Substance	$S^\circ$S∘ / J K$^{-1}$−1 mol$^{-1}$−1
N$_2$2​(g)	192
H$_2$2​(g)	131
NH$_3$3​(g)	192

$\Delta_r S^\circ = [2 \times 192] - [192 + 3 \times 131] = 384 - 585 = -201\;\text{J K}^{-1}\text{ mol}^{-1}$Δr​S∘=[2×192]−[192+3×131]=384−585=−201J K−1 mol−1

Negative — consistent with $\Delta n_{\text{gas}} = -2$Δngas​=−2. Note the 2× coefficient for NH$_3$3​ and the 3× for H$_2$2​ — failing to apply the balanced-equation coefficients is one of the two commonest arithmetic errors in this topic.

Worked example 8 — combustion of ethanol

$\text{C}_2\text{H}_5\text{OH(l)} + 3\text{O}_2\text{(g)} \rightarrow 2\text{CO}_2\text{(g)} + 3\text{H}_2\text{O(l)}$C2​H5​OH(l)+3O2​(g)→2CO2​(g)+3H2​O(l)

Substance	$S^\circ$S∘ / J K$^{-1}$−1 mol$^{-1}$−1
C$_2$2​H$_5$5​OH(l)	161
O$_2$2​(g)	205
CO$_2$2​(g)	214
H$_2$2​O(l)	70

$\Delta_r S^\circ = [2(214) + 3(70)] - [161 + 3(205)] = (428 + 210) - (161 + 615) = 638 - 776 = -138\;\text{J K}^{-1}\text{ mol}^{-1}$Δr​S∘=[2(214)+3(70)]−[161+3(205)]=(428+210)−(161+615)=638−776=−138J K−1 mol−1

Negative because three moles of gaseous O$_2$2​ disappear into two moles of gaseous CO$_2$2​ plus three moles of liquid water — net $\Delta n_{\text{gas}} = -1$Δngas​=−1, plus the loss of three highly-mobile water gas-phase contributions.

Worked example 9 — formation of water vapour vs liquid water

$\text{H}_2\text{(g)} + \tfrac{1}{2}\text{O}_2\text{(g)} \rightarrow \text{H}_2\text{O(l)}$H2​(g)+21​O2​(g)→H2​O(l)

Substance	$S^\circ$S∘ / J K$^{-1}$−1 mol$^{-1}$−1
H$_2$2​(g)	131
O$_2$2​(g)	205
H$_2$2​O(l)	70
H$_2$2​O(g)	189

To liquid water: $\Delta_r S^\circ = 70 - [131 + \tfrac{1}{2}(205)] = 70 - 233.5 = -163.5$Δr​S∘=70−[131+21​(205)]=70−233.5=−163.5 J K$^{-1}$−1 mol$^{-1}$−1.

To water vapour: $\Delta_r S^\circ = 189 - 233.5 = -44.5$Δr​S∘=189−233.5=−44.5 J K$^{-1}$−1 mol$^{-1}$−1.

The difference of $189 - 70 = +119$189−70=+119 J K$^{-1}$−1 mol$^{-1}$−1 is the vaporisation entropy of water — close to Trouton's rule prediction.

Worked example 10 — phase change AND mole-of-gas change combined (decomposition of dinitrogen tetroxide)

The cleanest test of qualitative sign prediction is a single reaction in which both the physical state of one species changes and the number of moles of gas changes. Consider the decomposition of liquid N$_2$2​O$_4$4​ into gaseous NO$_2$2​:

$\text{N}_2\text{O}_4\text{(l)} \rightarrow 2\text{NO}_2\text{(g)}$N2​O4​(l)→2NO2​(g)

Two effects pile up additively in the same direction: