4.2Unit 4. Rates, Equilibria and Further Organic Chemistry
Entropy — IAL Edexcel Linear (Chemistry) Revision Notes & Topical Questions
5 min read 13 parts Lesson 2 of 56 1 quiz
Lesson summary
4.2 Entropy — disorder, entropy changes, total entropy and feasibility — IAL EDEXCEL LINEAR (CHEMISTRY) notes: Explain entropy in terms of arrangements of particles and energy, calculate system, surroundings and total entropy changes, and use them to judge feasibility and kinetic…
- Entropy, S
- Melting increases entropy
- Boiling increases entropy greatly
- Condensing and freezing decrease entropy
- Second law
- Entropy (S)
Part 1 of 11Why entropy is needed
Enthalpy change alone cannot explain why endothermic reactions occur, because the products end in a higher-energy state.
Entropy, S
- Entropy, S a measure of the number of possible arrangements of the particles and their energy in a system;
- in effect, how disordered the system is.
An increase in entropy makes a system energetically more favourable, because its energy is more spread out.
Units
- J K−1 mol−1.
Part 2 of 11Arrangements of particles and energy
1
Distribution of molecules
- Five molecules that can be in either of two jars have W = 25 = 32 arrangements; 100 molecules have 2100.
- The number of arrangements rises rapidly with the number of particles.
- S = k ln W, where k is the Boltzmann constant.
2
Distribution of energy
- Energy exists in packets called quanta, shared among particles.
- More quanta or more particles give more ways of arranging the energy, so entropy increases with temperature.
3
Diffusion and spreading of heat both increase entropy.
4
A perfect crystal at 0 K has S = 0; elements have non-zero standard entropies, unlike ΔfH.
Part 3 of 11Entropy and changes of state
1Melting increases entropy
- Melting increases entropy particles in a regular lattice become a more random arrangement and can slide past one another.
2Boiling increases entropy greatly
- Boiling increases entropy greatly gas particles move freely and are far apart.
3Condensing and freezing decrease entropy
- Condensing and freezing decrease entropy particles become more ordered, so there are fewer ways to arrange the energy.
4
Order of entropy: solid < liquid < gas.
5
Entropy rises when temperature is raised because particles become more disordered.
6
Complex molecules have higher entropy than simple ones.
Part 4 of 11Reactions and dissolving
1
Producing a gas increases entropy.
- CaCO3(s) → CaO(s) + CO2(g) forms a gas from a solid.
- (NH4)2CO3 reacting with ethanoic acid forms a solution and CO2, so entropy rises although the reaction is endothermic.
2
Dissolving ammonium nitrate is endothermic (ΔsolH = +25.7 kJ mol−1) but the ions are free to move, so entropy increases.
- For ionic solids, breaking the lattice raises disorder while solvent bonding lowers it, so the sign of ΔH is hard to predict.
3
Two solids reacting, e.g. Ba(OH)2·8H2O with NH4Cl, form NH3 gas and a liquid, so entropy rises even though the reaction is endothermic.
4
When two solids react, the entropy change depends on the physical states of the products.
Part 5 of 11Entropy change of the system
1
ΔSsystem = ΣS⦵(products) − ΣS⦵(reactants).
2
Multiply each S⦵ by its balancing number and use data-book values, which are never zero for elements.
3
Entropy changes are much smaller than enthalpy changes, so they are quoted in J K−1 mol−1.
4
An increase in the number of moles of gas almost always gives a positive ΔSsystem.
| Change | Sign of ΔS system | Reason |
|---|---|---|
| Solid to liquid to gas | Positive | Particles more free to move |
| More moles of gas formed | Positive | More possible arrangements |
| Dissolving a solid | Usually positive | Ions spread through solvent |
| Gas to solid | Negative | Particles become ordered |
| Increase in temperature | Positive | More energy quanta to distribute |
Part 6 of 11Entropy change of the surroundings
Entropy change of the surroundings
ΔSsurroundings = −ΔH ÷ T, with T in kelvin and ΔH converted from kJ mol−1 to J mol−1 (× 1000).
An exothermic reaction passes energy to the surroundings, so ΔSsurroundings is positive.
The same amount of energy transferred to cooler surroundings gives a larger entropy change than to hotter ones.
Part 7 of 11Total entropy and feasibility
ΔStotal = ΔSsystem + ΔSsurroundings.
Second law
- Second law in any spontaneous change the total entropy increases, so a reaction is feasible if ΔStotal > 0.
ΔStotal is positive when:
- both ΔSsystem and ΔSsurroundings are positive,
- ΔSsurroundings is positive and ΔSsystem is negative, but ΔSsurroundings is larger,
- ΔSsurroundings is negative and ΔSsystem is positive, but ΔSsystem is larger.
Na + Cl2 → NaCl has negative ΔSsystem but a large exothermic ΔSsurroundings, so it is feasible.
Ethanoic acid with ammonium carbonate is endothermic, but a gas forms, so ΔSsystem outweighs ΔSsurroundings.
| ΔH | ΔS system | Feasible? |
|---|---|---|
| Negative | Positive | At all temperatures |
| Positive | Negative | Never |
| Negative | Negative | Only at low temperature |
| Positive | Positive | Only at high temperature |
Part 8 of 11Gibbs energy and the temperature of feasibility
1
ΔG = ΔH − TΔSsystem, with ΔS in kJ K−1 mol−1; a reaction is feasible when ΔG ≤ 0.
2
Exothermic with positive ΔS is feasible at all temperatures; endothermic with negative ΔS is never feasible.
3
Endothermic with positive ΔS becomes feasible at high temperature, e.g. metal extraction in the blast furnace at about 1500 °C.
- The minimum temperature is where ΔG = 0: T = ΔH ÷ ΔSsystem.
4
ΔStotal = R ln K links entropy to the equilibrium constant: a large positive value gives a large K.
Part 9 of 11Thermodynamic and kinetic stability
Thermodynamic and kinetic stability
Thermodynamic feasibility says whether a reaction can happen, not how fast.
A reaction with a high activation energy is kinetically stable and may not occur at room temperature.
- Methane and oxygen do not react until ignited.
- H2O2 → H2O + ½O2 is feasible but needs MnO2, which gives a lower Ea route.
- Diamond to graphite is feasible but extremely slow.
ΔH and ΔS vary slightly with temperature, and non-standard conditions change the values.
Part 10 of 11More from the specification
- Substance
- A material with only one type of particle.
Part 11 of 11More on this topic
More on this topic
Temperature has the largest effect.
- A much larger fraction of molecules exceeds the activation energy, shown as the area beyond Ea on a Maxwell-Boltzmann curve.
Higher pressure favours the side with fewer gas molecules.
Higher temperature favours the endothermic direction.
Key terms
| Term | Meaning |
|---|---|
| Entropy (S) | A measure of the number of ways particles and their energy can be arranged, in J K−1 mol−1. |
| Standard molar entropy | The entropy of one mole of a substance under standard conditions. |
| Entropy change of surroundings | ΔS surroundings = −ΔH/T. |
| Total entropy change | ΔS total = ΔS system + ΔS surroundings. |
| Feasible reaction | A reaction with a positive total entropy change (ΔG ≤ 0). |
| Gibbs energy change | ΔG = ΔH − TΔS system; feasible when ΔG ≤ 0. |
| Kinetic stability | A feasible reaction that does not occur at a noticeable rate because of a high activation energy. |
Avoid these mistakes
1
Giving ΔS in kJ — use J K−1 mol−1 and convert ΔH to J mol−1 before ΔS surroundings = −ΔH/T.
2
Using the wrong sign in ΔS surroundings — it is −ΔH/T.
3
Assuming elements have zero standard entropy — only ΔfH of elements is zero.
4
Assuming a feasible reaction is fast — a high activation energy can make it very slow.
5
Saying a decrease in system entropy means no reaction — the surroundings may compensate.
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