In simple terms
A friendly intro before the formal notes — no formulas yet.
Two drives, one decision
A reaction happens on its own when it lowers the usable energy of the system. Two things pull on that: enthalpy (reactions like to release heat) and entropy (the universe likes energy and matter spread out). Gibbs free energy is the single number that weighs both and tells you which way the reaction really goes.
Imagine tidying a room versus letting it drift into mess. Left alone, things spread out and get disordered — that is entropy rising, and it happens for free. Bringing order back (a highly structured product) costs something. A reaction proceeds spontaneously when the books balance in favour of going: either it releases enough heat, or it spreads energy and matter out enough, or both. Temperature sets how much weight the 'spreading out' side of the scale carries.
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Read the process and predict the sign of — is energy and matter becoming more spread out (gas made, solid dissolving) or more confined?
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Find (given or calculated) and , and make their units match: in kJ, so convert from J K⁻¹ to kJ K⁻¹ by dividing by 1000.
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Substitute into with in kelvin.
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Judge the sign: spontaneous, non-spontaneous, at equilibrium — and if the sign can flip, find the crossover temperature .
Explore the concept
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Key formulas
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Full topic notes
Formal explanation with the rigour you need for the exam.
Entropy: the dispersal of energy and matter
Entropy, symbol , is a measure of how spread out the energy and matter of a system are. The more ways the particles and their quanta of energy can be arranged — the number of microstates, — the higher the entropy. A system left to itself moves towards the macrostate that can be realised in the greatest number of ways, which is why energy and matter tend to disperse rather than concentrate. 'Disorder' is a useful shorthand, but the precise idea is dispersal: a gas fills its container, a hot object shares its energy with a cold one, and a drop of dye spreads through water, all because the dispersed arrangement is overwhelmingly more probable. Standard molar entropies, , in J K⁻¹ mol⁻¹, are tabulated in the data booklet and are positive for every substance.
State of matter: entropy rises solid → liquid → gas. , because gas particles access far more positions and energy arrangements.
Moles of gas: for a reaction, the sign of is usually decided by whether the number of moles of GAS goes up (positive ) or down (negative ).
Dissolving: a solid dissolving generally increases entropy as the ordered lattice breaks up and the ions or molecules disperse through the solvent.
Temperature: heating a substance raises its entropy, as more energy levels become populated and the energy is more widely distributed.
Predicting the sign of ΔS
Many exam marks come simply from predicting whether is positive or negative and justifying it in one line. The reliable rule is to count the moles of gas on each side: gas contributes far more entropy than liquids or solids, so a change that makes more gas has , and a change that consumes gas has . When the moles of gas are unchanged, look to dissolving, state changes, or an increase in the total number of particles.
Calculating the standard entropy change of a reaction
Because standard molar entropies are tabulated for individual substances, the entropy change of a reaction is a 'products minus reactants' sum, exactly like a Hess's-law enthalpy calculation. Each substance's is multiplied by its coefficient in the balanced equation.
Gibbs free energy and the condition for spontaneity
Enthalpy and entropy pull in their own directions, and Gibbs free energy combines them into one deciding quantity. At constant temperature and pressure, the change in Gibbs free energy is . Its sign is the verdict on spontaneity, because — a negative is exactly a positive total entropy change of the universe, which the second law requires for a spontaneous change.
— the reaction is spontaneous (thermodynamically feasible) in the forward direction.
— the reaction is non-spontaneous forward (the reverse direction is spontaneous).
— the system is at equilibrium, with no net drive either way.
Spontaneous is not the same as fast: says whether a reaction can go, never how quickly.
Two traps decide most of the marks on Gibbs calculations. First, UNITS: is in kJ mol⁻¹ but is in J K⁻¹ mol⁻¹ — divide by 1000 before you substitute, or the term comes out a thousand times too large. Second, TEMPERATURE must be in kelvin (). Get these two right and the arithmetic is easy.
How ΔH, ΔS and temperature decide spontaneity
Because the entropy term carries a factor of , temperature can change the sign of — but only in two of the four sign combinations. When and agree (both favour going, or both oppose), temperature cannot overturn the verdict. When they disagree, temperature is the tie-breaker: raising it increases the weight of the term until it dominates.
In the two temperature-dependent cases there is a crossover temperature at which the sign flips. It is found by setting , the point of equilibrium, and rearranging.
Both quantities must be in matching energy units before dividing (both in J, or both in kJ). For an endothermic, entropy-increasing reaction (, ) this is the temperature ABOVE which the reaction becomes spontaneous; for the enthalpy-driven case (, ) it is the temperature BELOW which it stays spontaneous.
, : both terms make negative — spontaneous at all temperatures.
, : both terms make positive — never spontaneous.
, : enthalpy-driven — spontaneous only at low temperature, where is small.
, : entropy-driven — spontaneous only at high temperature, where overcomes .
Model answer — marked the way our engine marks it
This is the showcase for a calculation topic. In Paper 2 the marks are analytic: each is tied to a specific line of working. A method mark (M) is earned for a correct process — the right substitution, the right rearrangement — even if a number is wrong; an accuracy mark (A) is earned for the correct value and DEPENDS on the method being present. Error-carried-forward (ECF) means a wrong number early on is only penalised once: later marks are awarded on your figures provided the method is sound. None of that protection exists unless the method is written down. Study how each of the five marks below attaches to one line.
Common mistakes examiners penalise
Mixing J and kJ in — is in kJ mol⁻¹ but is in J K⁻¹ mol⁻¹. Divide by 1000 first, or the term is a thousandfold too large and the answer is nonsense.
Using temperature in °C — must be in kelvin. Add 273 to a Celsius value before it enters the equation.
Getting the spontaneity condition backwards — spontaneous means (negative), NOT positive. A positive is non-spontaneous.
Confusing spontaneous with fast — tells you a reaction is feasible, never how quickly it goes. Rate is a separate, kinetic question.
Sign errors from the double negative — when is negative, becomes a POSITIVE contribution to . Track the signs carefully.
Wrong crossover units — in both quantities must be in the same energy unit (both J or both kJ). Dividing kJ by J K⁻¹ gives a temperature 1000× too small.
Predicting from the total number of species instead of moles of gas — the sign is dominated by the change in moles of GAS; a change from solid+aqueous to solid can still be entropy-decreasing.
Quoting a sign for with no reason — exam marks for predicting require the justification (moles of gas / dispersal), not just the sign.
Where this leads
Gibbs free energy is the hinge between the thermodynamics you have here and equilibrium later in the course: is linked to the equilibrium constant, so a more negative means an equilibrium lying further towards products. The same explains why raising temperature can drive an endothermic reaction and shift its position of equilibrium, and it underpins electrochemistry, where cell potentials are Gibbs energies in disguise. Master the habit — predict the sign of , match your units, substitute, judge the sign of — and the rest of physical chemistry becomes variations on a method you already own.
Worked examples
See the formulas applied — reveal one step at a time, like the exam.
Predict, with a reason, the sign of for each change: (a) ; (b) ; (c) . [3]
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(a) Count moles of gas: 3 mol gas (2 SO₂ + 1 O₂) → 2 mol gas (2 SO₃). Gas moles DECREASE, so energy and matter become less dispersed: (negative). [1]
Calculate the standard entropy change for the decomposition of calcium carbonate: / J K⁻¹ mol⁻¹: CaCO₃(s) = 92.9, CaO(s) = 39.7, CO₂(g) = 213.8. Comment on the sign. [3]
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Step 1 — sum the products. J K⁻¹ mol⁻¹. [M1: products sum]
For a reaction, kJ mol⁻¹ and J K⁻¹ mol⁻¹. Calculate at 298 K and determine whether the reaction is spontaneous, then find the temperature above which it becomes spontaneous. [5]
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Model answer — full working.
How it all connects
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Glossary
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Quick check
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Revision flashcards
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Entropy ()
A measure of how dispersed the energy and matter of a system are — the number of ways () the particles and their energy can be arranged (microstates). More ways of arranging means higher entropy. Units: J K⁻¹ mol⁻¹.
Key takeaways
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State of matter: entropy rises solid → liquid → gas. , because gas particles access far more positions and energy arrangements.
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Moles of gas: for a reaction, the sign of is usually decided by whether the number of moles of GAS goes up (positive ) or down (negative ).
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Dissolving: a solid dissolving generally increases entropy as the ordered lattice breaks up and the ions or molecules disperse through the solvent.
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Temperature: heating a substance raises its entropy, as more energy levels become populated and the energy is more widely distributed.
Practice — then mark it
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Get a Paper 2 thermodynamics question marked: calculate ΔG, decide spontaneity, and find the crossover temperature with full working
Get a Paper 2 thermodynamics question marked: calculate ΔG, decide spontaneity, and find the crossover temperature with full working
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