S5 Chemistry Practice Book · Unit 15: Entropy and free energy

Unit 15 Entropy and free energy

Key unit competence: Predict the feasibility of chemical reactions using entropy, enthalpy and Gibbs free energy.

Key ideas

Entropy (S) measures the disorder, or the number of ways energy and particles can be arranged. S(gas) > S(liquid) > S(solid). Unit: J K⁻¹ mol⁻¹. A perfect crystal at 0 K has zero entropy.
Entropy increases (ΔS positive) when: a solid melts or a liquid boils, a solid dissolves, the number of moles of gas increases, temperature rises, gases mix. ΔS°_r = ΣS°(products) − ΣS°(reactants).
Second law of thermodynamics: in a spontaneous process the total entropy of the universe increases: ΔS_total = ΔS_system + ΔS_surroundings > 0, with ΔS_surroundings = −ΔH ÷ T.
Gibbs free energy: ΔG = ΔH − TΔS (ΔS in kJ K⁻¹ mol⁻¹ when ΔH is in kJ). A reaction is feasible (spontaneous) when ΔG < 0, at equilibrium when ΔG = 0, not feasible when ΔG > 0.
Four cases: ΔH < 0, ΔS > 0: feasible at all T. ΔH > 0, ΔS < 0: never feasible. ΔH < 0, ΔS < 0: feasible at LOW T. ΔH > 0, ΔS > 0: feasible at HIGH T. The temperature at which feasibility changes is T = ΔH ÷ ΔS.
'Feasible' does not mean fast: a reaction with negative ΔG may be very slow because of a high activation energy (e.g. diamond → graphite, petrol in air). ΔG°_r = ΣΔG°_f(products) − ΣΔG°_f(reactants).

Common misconceptions

"All exothermic reactions are spontaneous." No: Feasibility depends on ΔG = ΔH − TΔS; an exothermic reaction with a large entropy decrease may not be feasible at high T.
"A feasible reaction happens quickly." No: ΔG says nothing about rate; a high activation energy can make it extremely slow.
"Entropy and enthalpy have the same units." No: S is in J K⁻¹ mol⁻¹ and ΔH in kJ mol⁻¹: convert before using ΔG = ΔH − TΔS.

Worked examples

Example 1. Explain, using ΔG = ΔH − TΔS, why water freezes below 0 °C and ice melts above 0 °C.

Solution

For melting: ΔH = +6.01 kJ mol⁻¹ (endothermic), ΔS = +22.0 J K⁻¹ mol⁻¹ (more disorder).

At T > 273 K, TΔS > ΔH, so ΔG < 0: melting is feasible.

At T < 273 K, TΔS < ΔH, so ΔG > 0 for melting; the reverse (freezing) has ΔG < 0.

At 273 K, ΔG = 0: ice and water are at equilibrium.

Example 2. Predict the sign of ΔS for: (a) 2Mg(s) + O₂(g) → 2MgO(s); (b) NH₄Cl(s) → NH₃(g) + HCl(g); (c) H₂O(l) → H₂O(g).

Solution

(a) Negative: 1 mol of gas is used up and solid is formed.

(b) Positive: a solid gives 2 mol of gas.

Rule: count moles of gas first; then consider changes of state and dissolving.

Practice questions

Define entropy.
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A measure of the disorder of a system, i.e. the number of ways its particles and energy can be arranged.
Common mistake: Saying entropy is the heat content.
State the second law of thermodynamics.
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The total entropy of an isolated system (the universe) increases in a spontaneous process.
Common mistake: Saying energy can be destroyed.
Give the condition for feasibility in terms of ΔG.
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ΔG must be negative (ΔG < 0).
Common mistake: Saying ΔH must be negative.
Convert ΔS = −120 J K⁻¹ mol⁻¹ to kJ K⁻¹ mol⁻¹.
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−0.120 kJ K⁻¹ mol⁻¹.
Common mistake: Using −120 directly with ΔH in kJ.
Why does a reaction with ΔH > 0 and ΔS > 0 become feasible on heating?
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TΔS grows with temperature until it is larger than ΔH, making ΔG negative.
Common mistake: Saying heat makes ΔH negative.
Predict the sign of ΔS for dissolving sugar in water.
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Positive: the ordered crystal breaks up and molecules spread through the solution.
Common mistake: Saying negative because a liquid forms.
What is the third law of thermodynamics?
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The entropy of a perfect crystal at 0 K is zero.
Common mistake: Saying entropy is zero at 0 °C.
Calculate ΔS_surroundings at 300 K for ΔH = −60 kJ mol⁻¹.
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ΔS_surr = −ΔH ÷ T = 60 000 ÷ 300 = +200 J K⁻¹ mol⁻¹.
Common mistake: Forgetting the minus sign.
Why may a feasible reaction not take place?
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Its activation energy is too high, so the rate is negligibly slow.
Common mistake: Saying ΔG must be positive.
Which of ΔH or ΔS dominates ΔG at very low temperatures?
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ΔH, because TΔS becomes very small.
Common mistake: Saying ΔS dominates.

Matching and drag-and-drop

Matching

Match each item in column A with the correct answer in column B. Write the letter next to each number.

Column A	Column B
1. The entropy change for CaCO₃(s) → CaO(s) + CO₂(g) is:	a. ΔG = ΔH − TΔS
2. The Gibbs free energy equation is:	b. the entropy of the universe increases
3. A reaction is feasible when ΔG is:	c. positive
4. The second law of thermodynamics states that in a spontaneous process:	d. negative

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1 → c (positive) · 2 → a (ΔG = ΔH − TΔS) · 3 → d (negative) · 4 → b (the entropy of the universe increases)

A gas is formed from a solid.
G combines enthalpy and entropy.
ΔG < 0.
ΔS_total > 0.

Drag and drop
Choose the correct word or phrase from the box for each gap. One word in the box is not needed.
T = ΔH ÷ ΔS · zero · −ΔH ÷ T · T = ΔG ÷ ΔH
1. At equilibrium, ΔG is .
2. The temperature at which a reaction just becomes feasible is .
3. ΔS of the surroundings is equal to .
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1: zero · 2: T = ΔH ÷ ΔS · 3: −ΔH ÷ T
- No net tendency to change.
- Set ΔG = 0.
- Heat given out raises the entropy of the surroundings.

Matching

Match each item in column A with the correct answer in column B. Write the letter next to each number.

Column A	Column B
1. Which state of a substance has the highest entropy?	a. Water evaporating
2. A reaction that is endothermic with a decrease in entropy is:	b. never feasible
3. Which process has a positive entropy change?	c. Gas
4. The unit of standard entropy is:	d. J K⁻¹ mol⁻¹

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1 → c (Gas) · 2 → b (never feasible) · 3 → a (Water evaporating) · 4 → d (J K⁻¹ mol⁻¹)

Gas particles are the most disordered.
Both terms make ΔG positive.
Liquid to gas increases disorder.
Entropy per kelvin per mole.

Drag and drop
Choose the correct word or phrase from the box for each gap. One word in the box is not needed.
increases entropy · the entropy increase makes ΔG negative · has no effect on entropy · feasible at all temperatures
1. Mixing two gases at the same temperature .
2. Dissolving ammonium nitrate in water is endothermic yet spontaneous because .
3. For ΔH < 0 and ΔS > 0, the reaction is .
Show answer
1: increases entropy · 2: the entropy increase makes ΔG negative · 3: feasible at all temperatures
- Particles spread into more space.
- Ions spread out in solution.
- Both terms make ΔG negative.

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