Edexcel International A Level Chemistry: Entropy (YCH11) – Revision Notes

For full explanations and worked examples, read the entropy study guide first. These notes condense Topic 12A: Entropy of the Pearson Edexcel International Advanced Subsidiary/Advanced Level in Chemistry specification (YCH11), Issue 1, September 2017, outcomes 12.1 to 12.11. Topic 12 is examined in Unit 4: Rates, Equilibria and Further Organic Chemistry (IA2).

Once these notes feel secure, work through the entropy practice questions. Tick off each outcome on the printable Edexcel Chemistry checklist, find the rest of the course on the Edexcel International A Level Chemistry hub, and use a free diagnostic to spot gaps.

Definitions to know word for word

Term	Meaning
Entropy, S	A measure of the disorder of a system: the random dispersal of molecules, and of energy quanta between molecules
Standard entropy, S⦵	The entropy of one mole of a substance in its standard state; units J mol⁻¹ K⁻¹; always positive above 0 K
System	The reacting chemicals
Surroundings	Everything outside the system (solvent, container, air)
Feasible	Able to happen on its own; requires ΔStotal > 0
Thermodynamically stable	ΔStotal for the change is negative, so the change is not feasible
Kinetically stable	The change is feasible, but the activation energy is so high that the rate is negligible

The three equations

Equation	Use	Units and traps
ΔSsystem = ΣS⦵(products) − ΣS⦵(reactants)	Outcome 12.7	Multiply each S⦵ by its coefficient; elements are not zero
ΔSsurroundings = −ΔH / T	Outcome 12.8	ΔH in J mol⁻¹, T in K; keep the minus sign
ΔStotal = ΔSsystem + ΔSsurroundings	Outcomes 12.6, 12.8	Both terms in J mol⁻¹ K⁻¹ before adding

Changeover temperature (when ΔStotal = 0):

T = ΔH / ΔSsystem      (ΔH in J mol⁻¹)

Optional, allowed by the specification but not required: ΔG = ΔH − TΔSsystem, and ΔG = −TΔStotal. Feasible when ΔG is negative.

Outcome 12.1: why enthalpy is not enough

Endothermic changes can happen on their own at room temperature. Learn two:

• Dissolving ammonium nitrate in water: the solution cools.
• Mixing solid Ba(OH)₂·8H₂O with solid ammonium chloride: the mixture becomes very cold and ammonia is given off.

So ΔH alone does not decide whether a reaction occurs.

Outcomes 12.2 and 12.3: entropy and its trends

• More ways to arrange the particles, or to share energy quanta between them, means higher entropy.
• Entropy increases with temperature.
• For one substance: S(solid) < S(liquid) < S(gas).
• A perfect crystal at 0 K has zero entropy: one arrangement, no thermal energy to spread.

Outcome 12.4: direction of change

The natural direction of change is the direction in which total entropy increases. A gas released in one part of a room spreads through all of it by random motion, because a spread-out arrangement is far more probable.

Outcome 12.5: sign of ΔSsystem in three situations

Situation	Usual sign	Reason
Melting, boiling, sublimation	+	Particles gain freedom of movement
Freezing, condensing	−	Particles lose freedom of movement
Dissolving an ionic solid	Usually +, can be −	Ions spread out (+), but water molecules become ordered round the ions (−); small, highly charged ions order the most water
More moles of gas formed	+	Gases have by far the highest entropies
Fewer moles of gas formed	−	Gas is used up

At a boiling point, ΔStotal = 0, so ΔSsystem(vaporisation) = ΔvapH/T.

Small worked reminder. Take ΔvapH of methanol as +35.3 kJ mol⁻¹ at its boiling point of 338 K. ΔSsystem = 35 300/338 = +104 J mol⁻¹ K⁻¹. Remember to convert kJ to J before dividing.

Method in steps: is a reaction feasible at temperature T?

1. Write out the balanced equation with state symbols.
2. ΔSsystem = ΣS(products) − ΣS(reactants), using coefficients.
3. Convert ΔH from kJ mol⁻¹ to J mol⁻¹ (multiply by 1000).
4. Convert T to kelvin if needed (add 273).
5. ΔSsurroundings = −ΔH/T.
6. ΔStotal = ΔSsystem + ΔSsurroundings, with sign and units.
7. Conclude: positive means feasible at T; negative means not feasible at T.

Method in steps: at what temperature does it become feasible?

1. Find ΔSsystem and ΔH (in J mol⁻¹).
2. Check the signs. If ΔH and ΔSsystem have the same sign, a changeover temperature exists.
3. T = ΔH / ΔSsystem.
4. State the direction: ΔH and ΔSsystem both positive → feasible above T. Both negative → feasible below T.

Outcomes 12.9 and 12.10: the four cases

ΔH	ΔSsystem	Result
−	+	Feasible at every temperature
+	−	Never feasible
−	−	Feasible below T = ΔH/ΔSsystem
+	+	Feasible above T = ΔH/ΔSsystem

Why temperature matters: T is on the bottom of −ΔH/T, so raising T shrinks the size of ΔSsurroundings. At high temperature, ΔSsystem dominates. A reaction can go even though one entropy term is negative, as long as ΔStotal is positive.

Small worked reminder. ΔH = +52 kJ mol⁻¹ and ΔSsystem = +130 J mol⁻¹ K⁻¹. Changeover T = 52 000/130 = 400 K. Feasible above 400 K.

Outcome 12.11: two kinds of stability

• Thermodynamic stability depends on ΔStotal (whether the change can happen).
• Kinetic stability depends on activation energy (how fast it happens).
• Example: petrol in air is thermodynamically unstable with respect to carbon dioxide and water, but kinetically stable at room temperature. It needs a spark.

Must-know distinctions

• Feasible vs fast. A positive ΔStotal says nothing about rate.
• ΔSsystem vs ΔStotal. Only ΔStotal decides feasibility.
• S⦵ of an element vs ΔfH⦵ of an element. S⦵ is positive; ΔfH⦵ is zero.
• J vs kJ. Entropy in J mol⁻¹ K⁻¹; enthalpy in kJ mol⁻¹.
• Above vs below. Endothermic with positive ΔSsystem: feasible above the changeover temperature. Exothermic with negative ΔSsystem: feasible below it.

Quick self-test

1. Give the sign of ΔSsystem for CO₂(s) → CO₂(g), with a reason.
2. Give the sign of ΔSsystem for 2H₂S(g) + 3O₂(g) → 2SO₂(g) + 2H₂O(l), with a reason.
3. Which has the higher entropy at 300 K, liquid bromine or gaseous bromine? Explain.
4. Calculate ΔSsurroundings at 310 K for a reaction with ΔH = −58.0 kJ mol⁻¹.
5. CaO(s) + H₂O(l) → Ca(OH)₂(s), ΔH = −65 kJ mol⁻¹. Use these S⦵ values / J mol⁻¹ K⁻¹: CaO(s) 39.8; H₂O(l) 69.9; Ca(OH)₂(s) 83.4. Calculate ΔSsystem and ΔStotal at 298 K.
6. A reaction has ΔH = +88 kJ mol⁻¹ and ΔSsystem = +160 J mol⁻¹ K⁻¹. Find the temperature above which it is feasible.
7. A reaction has ΔH = −120 kJ mol⁻¹ and ΔSsystem = −200 J mol⁻¹ K⁻¹. Is it feasible at 700 K? Show how you decide.
8. For ΔH = −50 kJ mol⁻¹, calculate ΔSsurroundings at 250 K and at 500 K. What does the comparison show?
9. Why does a perfect crystal at 0 K have zero entropy?
10. A mixture of hydrogen and oxygen can be kept for years at room temperature. Use the ideas of stability to explain this.
11. Suggest why ΔSsystem for dissolving a salt containing a small 3+ cation can be negative.

Answers

1. Positive. Solid becomes gas, so the particles gain freedom of movement.
2. Negative. 5 mol of gas become 2 mol of gas, and the water forms as a liquid.
3. Gaseous bromine. Gas particles move freely through the whole container, so there are far more ways to arrange them and their energy.
4. ΔSsurroundings = −(−58 000)/310 = +187 J mol⁻¹ K⁻¹.
5. ΔSsystem = 83.4 − (39.8 + 69.9) = −26.3 J mol⁻¹ K⁻¹. ΔSsurroundings = +65 000/298 = +218.1. ΔStotal = +191.8 J mol⁻¹ K⁻¹, so feasible.
6. T = 88 000/160 = 550 K.
7. Changeover T = 120 000/200 = 600 K; both terms negative, so feasible only below 600 K. At 700 K: ΔSsurroundings = +171.4, ΔStotal = −200 + 171.4 = −28.6 J mol⁻¹ K⁻¹. Not feasible.
8. +200 and +100 J mol⁻¹ K⁻¹. Doubling T halves ΔSsurroundings: heat makes less difference to hotter surroundings.
9. Every particle is fixed in one position and there is no thermal energy to spread, so there is only one arrangement.
10. ΔStotal for forming water is positive, so the mixture is thermodynamically unstable. The activation energy is high, so the rate at room temperature is negligible: the mixture is kinetically stable until a spark or flame starts it.
11. The small, highly charged cation holds many water molecules tightly in an ordered shell. The loss of entropy from ordering the water can be larger than the gain from spreading the ions out.

Where marks are usually lost

• Adding a ΔH/T value in kJ mol⁻¹ K⁻¹ to a ΔSsystem in J mol⁻¹ K⁻¹, giving an answer out by a factor of 1000.
• Dropping the minus sign in −ΔH/T, so an exothermic reaction seems to lower the entropy of the surroundings.
• Forgetting coefficients, for example using S(H₂) once when the equation has 3H₂.
• Treating an element’s standard entropy as zero.
• Predicting the sign of ΔSsystem from the number of all moles when the gas moles tell a different story.
• Giving a changeover temperature with no “above” or “below”.
• Concluding “not feasible” from a negative ΔSsystem alone, without working out ΔStotal.
• Explaining kinetic stability with ΔH or entropy instead of activation energy.
• Leaving out units or the sign on a final ΔS value.

Official syllabus

Pearson Edexcel International Advanced Subsidiary/Advanced Level in Chemistry (XCH11/YCH11) specification, Issue 1, September 2017, published by Pearson Education Limited. Topic 12: Entropy and Energetics, part 12A: Entropy, outcomes 12.1 to 12.11.