Edexcel International A Level Chemistry: Entropy (YCH11)

This guide teaches Topic 12A: Entropy from the Pearson Edexcel International Advanced Subsidiary/Advanced Level in Chemistry specification (YCH11), Issue 1, September 2017. It covers learning outcomes 12.1 to 12.11. Pearson places Topic 12 in Unit 4: Rates, Equilibria and Further Organic Chemistry, which is examined at International A2 (IA2) level. Part 12B, lattice energy, is a separate part of the same topic and is taught on its own page.

Every Edexcel International A Level Chemistry topic is listed on the course hub, and you can print the Edexcel Chemistry checklist. After this guide, use the entropy revision notes to condense it and the entropy practice questions to test it. To find weak spots across the course, try a free diagnostic.

The outcomes in Topic 12A

Outcome	Skill
12.1	Explain why enthalpy change alone cannot decide whether a reaction happens
12.2	Describe entropy as disorder: random dispersal of molecules and of energy quanta between molecules
12.3	State how entropy depends on temperature and state; perfect crystals at 0 K have zero entropy
12.4	Treat the natural direction of change as the direction of increasing total entropy, including gas spreading through a room
12.5	Explain entropy changes in changes of state, dissolving an ionic lattice, and reactions that change the number of moles
12.6	Use ΔStotal = ΔSsystem + ΔSsurroundings
12.7	Calculate ΔSsystem from the entropies of reactants and products
12.8	Calculate ΔSsurroundings = −ΔH/T, and hence ΔStotal
12.9	Explain how feasibility depends on the balance of the two terms and on temperature; calculate the temperature at which a reaction becomes feasible
12.10	Explain that a reaction can occur if ΔStotal is positive even when one of the other terms is negative
12.11	Distinguish thermodynamic stability from kinetic stability

Entropy values are in J mol⁻¹ K⁻¹ and enthalpy changes are usually in kJ mol⁻¹, so you will often need to convert.

12.1 Why enthalpy is not enough

Most reactions that happen on their own give out heat, but “exothermic” does not mean “will happen”. Some endothermic changes happen by themselves at room temperature:

• Ammonium nitrate dissolves in water and the solution gets colder.
• Solid barium hydroxide octahydrate, Ba(OH)₂·8H₂O, mixed with solid ammonium chloride reacts on its own. The mixture gets cold enough to freeze water on the outside of the flask, and ammonia is given off.

If enthalpy decided everything, none of these could happen. Entropy must also count.

12.2 and 12.3 What entropy measures

Entropy, S, measures the disorder of a system. In this course you describe disorder in two linked ways:

1. Dispersal of particles. Molecules spread out at random. The more ways there are to arrange them, the higher the entropy.
2. Dispersal of energy. Energy is held in small packets, called quanta, shared between molecules. The more ways those quanta can be spread among the molecules, the higher the entropy.

Three facts follow.

• Entropy rises with temperature. Hotter particles move more and have more energy quanta to share, so there are more possible arrangements.
• Entropy rises from solid to liquid to gas. Particles in a solid only vibrate about fixed positions. In a liquid they move past each other. In a gas they move freely through the whole container. So S(solid) < S(liquid) < S(gas) for the same substance.
• A perfect crystal at zero kelvin has zero entropy. Every particle is in a fixed place and there is no thermal energy to spread, so there is only one possible arrangement. Above 0 K every substance, element or compound, has a positive standard entropy, S⦵.

12.4 The natural direction of change

Change happens naturally in the direction that increases the total entropy. Open a bottle of perfume in one corner of a room and the smell soon reaches the far side. Nothing pushes the molecules there. Random motion spreads them out, because there are vastly more arrangements with the molecules spread through the room than with them all in the bottle.

12.5 Predicting the sign of ΔSsystem

Changes of state

Melting, boiling and sublimation increase entropy; freezing and condensing decrease it. When a liquid boils at its boiling point, the liquid and its vapour sit at equilibrium, so ΔStotal = 0 and ΔSsystem = ΔvapH/T.

For benzene, ΔvapH = +30.8 kJ mol⁻¹ at its boiling point of 353 K (use these values):

ΔSsystem = 30 800 / 353 = +87.3 J mol⁻¹ K⁻¹

Notice the conversion from kJ to J before dividing.

Dissolving a solid ionic lattice

When an ionic solid dissolves, the ordered lattice breaks up and the ions spread through the water. That increases entropy. But each ion attracts a shell of water molecules, which become more ordered than they were in the bulk liquid. That decreases entropy. The overall ΔSsystem depends on the balance:

• For ions with a small charge and a large radius, the water shells are loosely held. Dispersal of the ions wins, and ΔSsystem is usually positive.
• For small, highly charged ions, many water molecules are held tightly. The ordering of water can win, and ΔSsystem can be negative.

Reactions that change the number of moles

The biggest effect comes from gases. A reaction that produces more moles of gas than it uses has a positive ΔSsystem. One that uses up gas has a negative ΔSsystem. Count moles of gas first; if that does not change, look at state changes.

Examples, with the kind of reason you should give:

• 2NaHCO₃(s) → Na₂CO₃(s) + CO₂(g) + H₂O(g): 0 mol gas → 2 mol gas, so ΔSsystem is positive.
• C₂H₄(g) + H₂O(g) → C₂H₅OH(g): 2 mol gas → 1 mol gas, so ΔSsystem is negative.

12.6 System and surroundings

The system is the reacting chemicals. The surroundings are everything else. The total entropy change is:

ΔStotal = ΔSsystem + ΔSsurroundings

A reaction is feasible (it can happen on its own) when ΔStotal is positive.

12.7 Calculating ΔSsystem

ΔSsystem = ΣS⦵(products) − ΣS⦵(reactants)

Multiply each standard entropy by its coefficient in the balanced equation.

Worked example 1. Calculate ΔSsystem for the complete combustion of ethanol, forming liquid water:

C₂H₅OH(l) + 3O₂(g) → 2CO₂(g) + 3H₂O(l)

Use these values of S⦵ / J mol⁻¹ K⁻¹: C₂H₅OH(l) 160.7; O₂(g) 205.0; CO₂(g) 213.6; H₂O(l) 69.9.

ΣS(products)  = 2(213.6) + 3(69.9)  = 636.9
ΣS(reactants) = 160.7 + 3(205.0)    = 775.7
ΔSsystem      = 636.9 − 775.7       = −138.8 J mol⁻¹ K⁻¹

The sign makes sense: 3 mol of gas become 2 mol, and the water ends up liquid.

12.8 Calculating ΔSsurroundings and ΔStotal

Heat given out by a reaction spreads into the surroundings and increases their entropy; heat taken in lowers it. The size of the effect depends on temperature:

ΔSsurroundings = −ΔH / T

• ΔH is the enthalpy change of the reaction (the system), converted to J mol⁻¹.
• T is in kelvin.
• The minus sign means an exothermic reaction (ΔH negative) gives a positive ΔSsurroundings.

Worked example 1, continued. ΔcH⦵ of ethanol = −1367 kJ mol⁻¹. Find ΔStotal at 298 K.

ΔSsurroundings = −(−1 367 000) / 298 = +4587 J mol⁻¹ K⁻¹
ΔStotal        = −138.8 + 4587      = +4448 J mol⁻¹ K⁻¹

ΔStotal is positive, so the combustion is feasible even though ΔSsystem is negative. This is outcome 12.10 in action.

Worked example 2: an endothermic change that happens. Ammonium nitrate dissolves: NH₄NO₃(s) → NH₄⁺(aq) + NO₃⁻(aq). Use these values: ΔsolH = +25.7 kJ mol⁻¹; S⦵ / J mol⁻¹ K⁻¹: NH₄NO₃(s) 151.1; NH₄⁺(aq) 113.4; NO₃⁻(aq) 146.4.

ΔSsystem       = (113.4 + 146.4) − 151.1 = +108.7 J mol⁻¹ K⁻¹
ΔSsurroundings = −25 700 / 298          = −86.2 J mol⁻¹ K⁻¹
ΔStotal        = 108.7 + (−86.2)        = +22.5 J mol⁻¹ K⁻¹

The surroundings lose entropy (they cool), but the system gains more. ΔStotal is positive, so the salt dissolves. This answers the puzzle in 12.1.

12.9 and 12.10 Feasibility and temperature

The balance of the two terms

ΔH	ΔSsystem	ΔSsurroundings	Feasible?
negative	positive	positive	at all temperatures
positive	negative	negative	at no temperature
negative	negative	positive	when the surroundings term is larger: at lower temperatures
positive	positive	negative	when the system term is larger: at higher temperatures

Why temperature matters

ΔSsystem changes only a little with temperature, so you can treat it as constant. ΔSsurroundings has T on the bottom. For ΔH = −100 kJ mol⁻¹:

T / K	ΔSsurroundings / J mol⁻¹ K⁻¹
200	+500
400	+250
800	+125

Higher temperatures shrink ΔSsurroundings, so its contribution to ΔStotal gets smaller: the same heat makes less difference to surroundings that are already hot and disordered. At high temperature the sign of ΔSsystem takes over.

Calculating the temperature at which a reaction becomes feasible

The changeover happens when ΔStotal = 0:

0 = ΔSsystem − ΔH/T    so    T = ΔH / ΔSsystem

with ΔH in J mol⁻¹.

Worked example 3. When heated, sodium hydrogencarbonate breaks down:

2NaHCO₃(s) → Na₂CO₃(s) + CO₂(g) + H₂O(g) ΔH = +136 kJ mol⁻¹

Use these values of S⦵ / J mol⁻¹ K⁻¹: NaHCO₃(s) 101.7; Na₂CO₃(s) 135.0; CO₂(g) 213.6; H₂O(g) 188.7.

Step 1, ΔSsystem:

ΣS(products)  = 135.0 + 213.6 + 188.7 = 537.3
ΣS(reactants) = 2(101.7)               = 203.4
ΔSsystem      = +333.9 J mol⁻¹ K⁻¹

Step 2, check 298 K:

ΔSsurroundings = −136 000 / 298 = −456.4
ΔStotal        = 333.9 − 456.4  = −122.5 J mol⁻¹ K⁻¹   → not feasible

Step 3, find the changeover temperature:

T = 136 000 / 333.9 = 407 K

Above about 407 K, ΔStotal is positive and the decomposition is feasible. This assumes ΔH and ΔSsystem stay roughly constant with temperature.

Using ΔG (optional)

The specification says you may use ΔG = ΔH − TΔSsystem in answers, but it is not a requirement. Since ΔG = −TΔStotal, “ΔStotal positive” and “ΔG negative” are the same test, and the changeover temperature is still T = ΔH/ΔSsystem.

12.11 Thermodynamic and kinetic stability

A positive ΔStotal tells you a reaction can happen. It says nothing about how fast.

• Thermodynamic stability: a substance is thermodynamically unstable relative to some products if ΔStotal for forming them is positive.
• Kinetic stability: a substance is kinetically stable if the activation energy for its reaction is so high that the rate at room temperature is negligible.

Worked example 4. For C(diamond) → C(graphite), ΔH = −1.9 kJ mol⁻¹. Use S⦵ values of 2.4 (diamond) and 5.7 (graphite) J mol⁻¹ K⁻¹. At 298 K:

ΔSsystem       = 5.7 − 2.4      = +3.3
ΔSsurroundings = +1900 / 298    = +6.4
ΔStotal        = +9.7 J mol⁻¹ K⁻¹

Diamond is thermodynamically unstable relative to graphite, yet it does not turn into graphite in any measurable time: every carbon atom is held by four strong covalent bonds, so the activation energy is very high. Diamond is kinetically stable. Paper in air is similar: thermodynamically unstable, but nothing happens until a flame supplies enough energy to overcome the activation energy.

Suggested practicals in the specification

The specification suggests investigating reactions in terms of disorder and enthalpy change, including dissolving ammonium nitrate in water, reacting ethanoic acid with ammonium carbonate, burning magnesium ribbon in air, and mixing Ba(OH)₂·8H₂O with ammonium chloride. For each, say whether the temperature rises or falls and give the sign of ΔSsystem with a reason. Ethanoic acid with ammonium carbonate releases carbon dioxide gas, so ΔSsystem is positive. Burning magnesium has a negative ΔSsystem (a gas is used up and a solid forms), but its large heat release makes ΔSsurroundings large and positive.

Common errors

• Dividing ΔH in kJ mol⁻¹ by T and adding it to ΔSsystem in J mol⁻¹ K⁻¹. Convert first.
• Losing the minus sign in ΔSsurroundings = −ΔH/T, so that an exothermic reaction appears to lower the entropy of the surroundings.
• Forgetting coefficients when summing entropies, or taking the entropy of an element as zero.
• Saying a reaction with negative ΔSsystem “cannot happen”. Only ΔStotal decides.
• Saying a feasible reaction “will happen quickly”. Feasibility and rate are separate ideas.
• Giving a changeover temperature but not stating which side of it (hotter or colder) the reaction goes.

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.