S5 Chemistry – Unit 15: Entropy & Free Energy

15.1

Entropy

Entropy (S) Entropy is a measure of the disorder or randomness of a system — more precisely, it is a measure of the number of microstates (W) available to the system:

S = k_B ln W (Boltzmann equation; k_B = 1.381×10⁻²³ J/K)

Units of entropy: J/mol·K (or J/K for absolute entropy).
ΔS = S_final − S_initial (ΔS > 0: more disorder; ΔS < 0: less disorder)

Factors That Increase Entropy

Change of state: solid → liquid → gas. S(g) ≫ S(l) > S(s). Gases have enormous positional freedom.
Increase in number of moles of gas: more gas moles = more disorder.
Mixing: mixing of different substances always increases entropy.
Dissolution: dissolving a solid in water usually increases entropy (ions dispersed through solvent).
Increasing temperature: more thermal energy → more microstates accessible.
Increasing volume: more space → more positional microstates.
More complex molecules: more atoms → more vibrational/rotational modes → higher S.

Substance	S° (J/mol·K) at 298 K	Substance	S° (J/mol·K) at 298 K
H₂(g)	130.6	H₂O(l)	69.9
O₂(g)	205.1	H₂O(g)	188.7
N₂(g)	191.6	CO₂(g)	213.8
C(graphite)	5.7	CH₄(g)	186.3
Fe(s)	27.3	NaCl(s)	72.1
Cu(s)	33.2	NH₃(g)	192.5
Cl₂(g)	223.1	C₂H₅OH(l)	160.7
HCl(g)	186.9	C(diamond)	2.4

Example 1

Predicting the Sign of ΔS

Predict whether ΔS is positive or negative for each reaction. Justify.

CaCO₃(s) → CaO(s) + CO₂(g)
ΔS > 0: solid produces a gas — massive increase in disorder. n(gas) increases from 0 to 1.

N₂(g) + 3H₂(g) → 2NH₃(g)
ΔS < 0: 4 mol gas → 2 mol gas. n(gas) decreases — fewer microstates. More ordered product.

NaCl(s) → Na⁺(aq) + Cl⁻(aq)
ΔS > 0: ordered crystal lattice → dispersed ions in solution. Mixing and dissolution.

H₂O(g) → H₂O(l)
ΔS < 0: gas condenses to liquid. Huge decrease in positional freedom.

15.2

The Second Law of Thermodynamics

Second Law The total entropy of the universe always increases in any spontaneous process:

ΔS_universe = ΔS_system + ΔS_surroundings ≥ 0

For a spontaneous process: ΔS_univ > 0
For a reversible (equilibrium) process: ΔS_univ = 0
For a non-spontaneous (impossible) process: ΔS_univ < 0

Entropy Change of the Surroundings

When a reaction releases heat q to the surroundings at temperature T:

ΔS_surr = -ΔH_system / T (at constant T and P) - Exothermic reaction (ΔH < 0): ΔS_surr > 0 (surroundings gain entropy) - Endothermic reaction (ΔH > 0): ΔS_surr < 0 (surroundings lose entropy) The higher the temperature, the smaller the effect of heat transfer on ΔS_surr (at high T, the surroundings are already very disordered — adding more heat changes entropy less).

Example 2

Calculating ΔS°rxn from Absolute Entropies

Calculate ΔS°rxn for: N₂(g) + 3H₂(g) → 2NH₃(g)
S° (J/mol·K): N₂=191.6; H₂=130.6; NH₃=192.5

ΔS°_rxn = ΣS°(products) − ΣS°(reactants)

ΣS°(products) = 2×S°(NH₃) = 2×192.5 = 385.0 J/mol·K

ΣS°(reactants) = S°(N₂) + 3×S°(H₂) = 191.6 + 3×130.6 = 191.6 + 391.8 = 583.4 J/mol·K

ΔS°_rxn = 385.0 − 583.4 = −198.4 J/mol·K (negative — 4 mol gas → 2 mol gas)

15.3

The Third Law of Thermodynamics

Third Law The entropy of a perfect crystalline substance is zero at absolute zero (0 K).

S(perfect crystal, 0 K) = 0

This law establishes an absolute scale for entropy — unlike enthalpy (where only differences ΔH can be measured), absolute values of S can be determined. This is why standard entropies S° are absolute values (not relative to elements in standard states).

Absolute Entropy and Temperature

As temperature increases from 0 K, the entropy of any substance increases because more energy levels become thermally accessible. The entropy rises continuously, with sharp increases at phase transitions (melting and boiling).

At constant pressure: dS = Cp dT / T Integrating: S(T) = S(0) + ∫[0→T] Cp/T dT For a perfect crystal: S(0) = 0 (Third Law) At phase transitions: - Melting: ΔS_fus = ΔH_fus / T_m - Boiling: ΔS_vap = ΔH_vap / T_b (Trouton's rule: ≈ 88 J/mol·K for most liquids)

15.4

Gibbs Free Energy

Gibbs Free Energy (G) The Gibbs free energy G combines enthalpy and entropy into a single criterion for spontaneity at constant T and P:

G = H − TS

ΔG = ΔH − TΔS

A reaction is spontaneous if ΔG < 0 (system can do useful work).
At equilibrium: ΔG = 0.
Non-spontaneous: ΔG > 0.

Standard Gibbs Free Energy Change

ΔG° = ΔH° - TΔS° (at temperature T, using standard values) ΔG°rxn = Σ ΔG°f(products) - Σ ΔG°f(reactants) where ΔG°f = standard Gibbs free energy of formation (ΔG°f of any element in standard state = 0) Relationship to equilibrium constant K: ΔG° = -RT ln K (R = 8.314 J/mol·K) At equilibrium: ΔG = 0 (not ΔG°) When Q < K: ΔG < 0 (spontaneous forward) When Q > K: ΔG > 0 (spontaneous reverse)

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ΔG vs ΔG° ΔG° is the free energy change when all reactants and products are in standard states (1 mol/L, 298 K). It is a fixed number for a given reaction at a given temperature.
ΔG is the free energy change at any specified conditions. ΔG = ΔG° + RT ln Q.

15.5

Spontaneity Criteria

The Four ΔH / ΔS Combinations

Whether a reaction is spontaneous depends on the signs of ΔH and ΔS and the temperature T:

Case 1: ΔH < 0, ΔS > 0

ΔG = ΔH − TΔS < 0 at ALL temperatures.
Always spontaneous.
Example: combustion reactions (exothermic, produces gas).

Case 2: ΔH > 0, ΔS < 0

ΔG = ΔH − TΔS > 0 at ALL temperatures.
Never spontaneous.
Example: reverse Haber process (endothermic, fewer gas moles).

Case 3: ΔH < 0, ΔS < 0

ΔG < 0 only at LOW temperatures (enthalpy dominates).
Example: condensation of water vapour (exothermic but less disordered).

Case 4: ΔH > 0, ΔS > 0

ΔG < 0 only at HIGH temperatures (entropy dominates).
Example: decomposition of CaCO₃ (endothermic but more disordered).

Example 3

Calculating ΔG° and Checking Spontaneity

For: N₂(g) + 3H₂(g) → 2NH₃(g) at 298 K.
ΔH° = −92.4 kJ/mol; ΔS° = −198.4 J/mol·K.
(a) Calculate ΔG° at 298 K. (b) Is the reaction spontaneous at 298 K? (c) At what temperature does ΔG° = 0?

ΔG° = ΔH° − TΔS° = −92,400 − (298 × −198.4) = −92,400 + 59,123 = −33,277 J/mol = −33.3 kJ/mol

ΔG° < 0 → spontaneous at 298 K. The enthalpy term (−92.4 kJ) dominates over the unfavourable entropy term (+59.1 kJ).

At ΔG° = 0: T = ΔH°/ΔS° = −92,400/(−198.4) = 466 K. Above 466 K, ΔG° > 0 (non-spontaneous). This explains why the Haber process uses moderate temperatures.

15.6

Temperature Effects on ΔG

Crossover Temperature

The temperature at which a reaction changes from spontaneous to non-spontaneous is the crossover temperature:

At T_cross: ΔG = 0 Therefore: ΔH = T_cross × ΔS T_cross = ΔH / ΔS (units: K; use consistent units — both in J)

ΔG vs T for the four ΔH/ΔS combinations

Example 4

Crossover Temperature

For: CaCO₃(s) → CaO(s) + CO₂(g): ΔH° = +178 kJ/mol; ΔS° = +165 J/mol·K. Find the temperature above which the reaction becomes spontaneous.

Case 4 (ΔH > 0, ΔS > 0) — spontaneous only above T_cross.

T_cross = ΔH°/ΔS° = 178,000/165 = 1079 K ≈ 806°C

Above ~806°C: ΔG < 0 → CaCO₃ decomposes spontaneously. Industrial lime kilns operate at >900°C for this reason.

15.7

The van't Hoff Equation

Van't Hoff Equation Relates the equilibrium constant K to temperature T:

ln K = −ΔH°/(RT) + ΔS°/R

For change in K between two temperatures:
ln(K₂/K₁) = −(ΔH°/R)(1/T₂ − 1/T₁)

Plot of ln K vs 1/T: straight line, slope = −ΔH°/R, intercept = ΔS°/R.

Derivation

ΔG° = −RT ln K and ΔG° = ΔH° − TΔS° ‪∴ −RT ln K = ΔH° − TΔS° Divide by −RT: ln K = −ΔH°/(RT) + ΔS°/R Subtracting for two temperatures T₁ and T₂: ln(K₂/K₁) = −(ΔH°/R)(1/T₂ − 1/T₁) = (ΔH°/R)(1/T₁ − 1/T₂)

Exothermic (ΔH° < 0)

As T increases: K decreases. Equilibrium shifts left. Higher T disfavours exothermic direction — consistent with Le Chatelier.

Endothermic (ΔH° > 0)

As T increases: K increases. Equilibrium shifts right toward products. Higher T favours endothermic direction.

Example 5

Van't Hoff — K at New Temperature

N₂O₄(g) ⇌ 2NO₂(g): K = 0.113 at 298 K; ΔH° = +57.2 kJ/mol. Calculate K at 350 K.

ln(K₂/K₁) = −(ΔH°/R)(1/T₂ − 1/T₁) = −(57200/8.314)(1/350 − 1/298)

= −6879 × (0.002857 − 0.003356) = −6879 × (−4.99×10⁻⁴) = +3.43

K₂/K₁ = e³·⁴³ = 31.0; K₂ = 0.113 × 31.0 = 3.50 (endothermic → K increases with T)

Example 6

Calculating K from ΔG°

2SO₂(g) + O₂(g) → 2SO₃(g): ΔH° = −197.8 kJ/mol; ΔS° = −187.9 J/mol·K. Find ΔG° and K at 298 K.

ΔG° = −197,800 − (298)(−187.9) = −197,800 + 55,994 = −141,806 J/mol = −141.8 kJ/mol

ln K = 141,806/(8.314×298) = 57.23; K = e⁵⁷·²³ = 7.1×10²⁴ (extremely large — very favourable)

15.8

Coupled Reactions and Free Energy

Coupled Reactions A non-spontaneous reaction (ΔG > 0) can be driven by coupling it to a spontaneous reaction (ΔG < 0), provided the combined ΔG₁ + ΔG₂ < 0. This is fundamental to biochemistry.

Principle

Reaction A (non-spontaneous): ΔG_A > 0 Reaction B (spontaneous): ΔG_B < 0 Combined: ΔG_total = ΔG_A + ΔG_B Condition for spontaneous coupling: |ΔG_B| > |ΔG_A|

Biological Example: ATP

ATP + H₂O → ADP + Pᵢ ΔG° = −30.5 kJ/mol Synthesis of glutamine (non-spontaneous alone): Glutamate + NH₃ → Glutamine + H₂O ΔG° = +14.2 kJ/mol Coupled: Glutamate + NH₃ + ATP → Glutamine + ADP + Pᵢ ΔG°_total = +14.2 + (−30.5) = −16.3 kJ/mol ✓ Spontaneous!

Maximum Work and Electrochemistry

Maximum useful work = ΔG For electrochemical cells: ΔG° = −nFE°cell (n = moles electrons; F = 96,500 C/mol; E° = standard cell voltage) Combining ΔG° = −RT ln K and ΔG° = −nFE°: E°cell = (RT/nF) ln K = (0.0592/n) log K at 298 K

Example 7

ΔG° and E°cell

Zn(s)|Zn²⁺(aq)||Cu²⁺(aq)|Cu(s): E°cell = +1.10 V, n = 2. Calculate ΔG° and K at 298 K.

ΔG° = −nFE° = −2 × 96,500 × 1.10 = −212,300 J/mol = −212.3 kJ/mol

ln K = 212,300/(8.314×298) = 85.73; K = e⁸⁵·⁷³ = 1.8×10³⁷ (reaction goes essentially to completion)

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Exercises

Predict the sign of ΔS° and justify: (a) 2H₂O(l) → 2H₂(g) + O₂(g); (b) Ag⁺(aq) + Cl⁻(aq) → AgCl(s); (c) CO₂(g) → CO₂(s); (d) C₆H₁₂O₆(s) + 6O₂(g) → 6CO₂(g) + 6H₂O(l).
(a) ΔS > 0: liquid → gases. 3 mol gas produced from liquid. Huge increase in disorder.
(b) ΔS < 0: dispersed aqueous ions → ordered solid precipitate. Lattice formed, decrease in disorder.
(c) ΔS < 0: gas → solid. Very large decrease in disorder.
(d) ΔS slightly negative: reactants: 6 mol gas + 1 mol solid; products: 6 mol gas + 6 mol liquid. Gas moles unchanged; losing the ordered solid but gaining liquid water. Net slightly negative.
Calculate ΔS°rxn for: 2SO₂(g) + O₂(g) → 2SO₃(g).
S° (J/mol·K): SO₂(g)=248.1; O₂(g)=205.1; SO₃(g)=256.6.
ΔS° = 2S°(SO₃) − [2S°(SO₂) + S°(O₂)]
= 2×256.6 − [2×248.1 + 205.1]
= 513.2 − 701.3 = −188.1 J/mol·K
(Negative: 3 mol gas → 2 mol gas)
For: ΔH° = −241.8 kJ/mol; ΔS° = +44.4 J/mol·K. Calculate ΔG° at (a) 298 K and (b) 1000 K. Spontaneous at each temperature?
(a) ΔG°(298) = −241,800 − (298×44.4) = −241,800 − 13,231 = −255.0 kJ/mol → spontaneous
(b) ΔG°(1000) = −241,800 − (1000×44.4) = −286.2 kJ/mol → spontaneous
Case 1 (ΔH<0, ΔS>0): always spontaneous at all temperatures.
For N₂(g) + 3H₂(g) → 2NH₃(g): K = 977 at 298 K and ΔH° = −92.4 kJ/mol. Calculate ΔG°(298) and K at 500 K.
ΔG°(298) = −RT ln K = −8.314×298×ln(977) = −2478×6.884 = −17.1 kJ/mol

ln(K₅₀₀/K₂⁹⁸) = −(−92400/8.314)(1/500−1/298) = 11113×(−1.356×10⁻³) = −15.07
K₅₀₀ = 977 × e⁻¹⁵·ᾆ⁷ = 977 × 2.83×10⁻⁷ = 2.77×10⁻⁴
(K drops dramatically at 500 K — low yield at high temperature)
Calculate E°cell for the Daniel cell given ΔG° = −212.3 kJ/mol and n = 2. Then find K at 298 K.
E° = −ΔG°/(nF) = 212,300/(2×96,500) = +1.10 V
ln K = 212,300/(8.314×298) = 85.73; K = e⁸⁵·⁷³ = 1.8×10³⁷
Explain coupled reactions using the ATP example. State the condition for spontaneous coupling.
Coupled reactions: a non-spontaneous process (ΔG > 0) is driven by coupling with a spontaneous process (ΔG < 0). ΔG_total = ΔG₁ + ΔG₂ < 0 required.
Condition: |ΔG(driving)| > |ΔG(driven)|.
ATP example: ATP hydrolysis releases ΔG° = −30.5 kJ/mol. Couples to drive biosynthesis reactions up to +30.5 kJ/mol. E.g. glutamine synthesis (ΔG° = +14.2) coupled with ATP gives net −16.3 kJ/mol → spontaneous.

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Unit Test

ℹ️

Instructions Total: 50 marks | Time: 55 minutes | Show all working. R = 8.314 J/mol·K.

Section A — Short Answer

30 marks

Q1 [4 marks]

Calculate ΔS°rxn for: CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l)
S° (J/mol·K): CH₄=186.3; O₂=205.1; CO₂=213.8; H₂O(l)=69.9. Explain the sign in terms of states of matter.

ΔS° = [S°(CO₂) + 2S°(H₂O(l))] − [S°(CH₄) + 2S°(O₂)]
= [213.8 + 2×69.9] − [186.3 + 2×205.1]
= 353.6 − 596.5 = −242.9 J/mol·K
Negative: 3 mol gas → 1 mol gas + 2 mol liquid. Gas → liquid gives large decrease in entropy.

Q2 [5 marks]

2NO(g) + O₂(g) → 2NO₂(g): ΔH° = −114.6 kJ/mol; ΔS° = −146.5 J/mol·K.
(a) ΔG° at 298 K; (b) Spontaneous at 298 K?; (c) T_cross; (d) Spontaneous above or below T_cross?

(a) ΔG°(298) = −114,600 − 298×(−146.5) = −114,600 + 43,657 = −70.9 kJ/mol
(b) ΔG° < 0 → spontaneous at 298 K
(c) T_cross = 114,600/146.5 = 782 K (509°C)
(d) Case 3 (ΔH<0, ΔS<0): spontaneous below T_cross. Above 782 K, ΔG > 0.

Q3 [5 marks]

HF(g) ⇌ H(g) + F(g): K = 1.0×10⁻¹³ at 298 K and 2.0×10⁻⁷ at 1000 K.
(a) ΔG° at 298 K; (b) ΔH° via van't Hoff; (c) Exo- or endothermic? Consistent with Le Chatelier?

(a) ΔG° = −RT ln K = −8.314×298×ln(10⁻¹³) = −2478×(−29.93) = +74.2 kJ/mol
(b) ln(2×10⁻⁷/10⁻¹³) = ln(2×10⁶) = 14.51
14.51 = −(ΔH°/8.314)(1/1000 − 1/298) = −(ΔH°/8.314)(−2.356×10⁻³)
ΔH° = 14.51×8.314/2.356×10⁻³ = +51.2 kJ/mol
(c) Endothermic. Le Chatelier: increasing T increases K → more dissociation. Consistent.

Q4 [5 marks]

Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s): E°cell = +1.10 V, n = 2, ΔH° = −218.7 kJ/mol.
(a) ΔG°; (b) K at 298 K; (c) ΔS° from ΔG° = ΔH° − TΔS°; (d) Verify ΔG°.

(a) ΔG° = −nFE° = −2×96500×1.10 = −212.3 kJ/mol
(b) ln K = 212300/(8.314×298) = 85.73; K = 1.8×10³⁷
(c) ΔS° = (ΔH° − ΔG°)/T = (−218700 + 212300)/298 = −6400/298 = −21.5 J/mol·K
(d) ΔG° = −218700 − 298×(−21.5) = −218700 + 6407 = −212,293 J ≈ −212.3 kJ/mol ✓

Q5 [5 marks]

State and explain the Second Law. Use it to explain spontaneous: (a) heat flow hot → cold; (b) gas expansion; (c) crystal dissolving in water.

Second Law: In any spontaneous process, ΔS_universe = ΔS_system + ΔS_surroundings > 0.

(a) Heat q from hot (T_h) to cold (T_c): ΔS_total = q/T_c − q/T_h > 0 (since T_c < T_h). Universe entropy increases → spontaneous. Reverse would give ΔS < 0 → impossible.

(b) Gas expanding: more volume → more positional microstates → W increases → S increases → ΔS_universe > 0 → spontaneous.

(c) Crystal dissolving: ions from ordered lattice into disordered solution → enormous increase in microstates → ΔS_system ≫ 0. Even if endothermic (ΔS_surr < 0), net ΔS_universe > 0 → spontaneous.

Q6 [6 marks]

2H₂O₂(l) → 2H₂O(l) + O₂(g): ΔH° = −196.1 kJ/mol.
S° (J/mol·K): H₂O₂=109.6; H₂O(l)=69.9; O₂=205.1.
(a) ΔS°; (b) ΔG° at 298 K; (c) K at 298 K; (d) K at 310 K; (e) Why is catalase needed biologically?

(a) ΔS° = [2×69.9 + 205.1] − 2×109.6 = 344.9 − 219.2 = +125.7 J/mol·K
(b) ΔG° = −196,100 − 298×125.7 = −196,100 − 37,459 = −233.6 kJ/mol
(c) ln K = 233,600/(8.314×298) = 94.27; K ≈ 10⁴¹
(d) ln(K₃¹₀/K₂⁹⁸) = −(−196100/8.314)(1/310−1/298) = 23584×(−1.30×10⁻⁴) = −3.07; K₃¹₀ ≈ K₂⁹⁸ × e⁻³·ᾆ⁷ ≈ ~5×10³⁹
(e) K is enormous → thermodynamically very favourable. But decomposition is very slow without a catalyst (high activation energy). Catalase lowers activation energy to allow rapid decomposition of toxic H₂O₂ in cells. Thermodynamics determines feasibility; kinetics determines rate.

Section B — Extended Response

20 marks

Q7 [10 marks]

(a) State the Second and Third Laws of Thermodynamics. What does each tell us about entropy? [4]
(b) CO(g) + ½O₂(g) → CO₂(g): ΔH°=−283.0 kJ/mol; ΔS°=−86.8 J/mol·K.
(i) ΔG° at 298 K and 1500 K; (ii) K at both temperatures; (iii) Why is this reaction used in furnaces at high T despite decreasing K? [6]

(a) Second Law: ΔS_universe > 0 for any spontaneous process. Natural processes move toward maximum disorder; the "arrow of time" is defined by increasing universal entropy.
Third Law: S(perfect crystal, 0 K) = 0. Gives an absolute reference for entropy, allowing absolute S° values to be tabulated (unlike enthalpy, where only ΔH is measurable).

(b)(i)
ΔG°(298) = −283,000 − 298×(−86.8) = −283,000 + 25,866 = −257.1 kJ/mol
ΔG°(1500) = −283,000 − 1500×(−86.8) = −283,000 + 130,200 = −152.8 kJ/mol

(b)(ii)
K(298): ln K = 257,100/(8.314×298) = 103.7; K ≈ 10⁴⁵
K(1500): ln K = 152,800/(8.314×1500) = 12.25; K ≈ 2.1×10⁵

(b)(iii) K decreases dramatically (10⁴⁵ → 10⁵) but both are still very large → still spontaneous. Furnaces use high T for kinetic reasons: CO combustion has a high activation energy. At room T, thermodynamically favourable but too slow. High T provides the activation energy for a practical combustion rate. This illustrates that thermodynamics (ΔG) says whether a reaction CAN occur, but kinetics determines HOW FAST.

Q8 [10 marks]

(a) Derive the van't Hoff equation from ΔG° = ΔH° − TΔS° and ΔG° = −RT ln K. Explain how a ln K vs 1/T graph yields ΔH° and ΔS°. [5]
(b) Haber process: N₂ + 3H₂ ⇌ 2NH₃: ΔH° = −92.4 kJ/mol; ΔS° = −198.4 J/mol·K.
(i) Show spontaneous at 298 K but not at 800 K; (ii) K at 298 K; (iii) K at 700 K via van't Hoff; (iv) Explain the industrial conditions (450°C, 200 atm, Fe catalyst). [5]

(a) ΔG° = −RT ln K and ΔG° = ΔH° − TΔS°
∴ −RT ln K = ΔH° − TΔS° → divide by −RT:
ln K = −ΔH°/(RT) + ΔS°/R
y = mx + c form: y = ln K; x = 1/T; slope = −ΔH°/R; intercept = ΔS°/R
ΔH° = −slope × R; ΔS° = intercept × R
Negative slope → endothermic (K increases with T). Positive slope → exothermic.

(b)(i)
ΔG°(298) = −92,400 − 298×(−198.4) = −92,400 + 59,123 = −33.3 kJ/mol → spontaneous ✓
ΔG°(800) = −92,400 − 800×(−198.4) = −92,400 + 158,720 = +66.3 kJ/mol → non-spontaneous ✓
T_cross = 92,400/198.4 = 466 K.

(b)(ii) ln K = 33,277/(8.314×298) = 13.43; K = e¹³·⁴³ = 677

(b)(iii) ln(K₇₀₀/K₂⁹⁸) = −(−92400/8.314)(1/700−1/298) = 11113×(−1.927×10⁻³) = −21.42
K₇₀₀ = 677 × e⁻²¹·⁴² = 677 × 4.45×10⁻¹⁰ = 3.0×10⁻⁷ (very small — poor yield at 700 K)

(b)(iv) Industrial conditions:
• Low T favours thermodynamics (higher K, better yield) but too slow → compromise at ~450°C
• High P (200 atm): 4 mol gas → 2 mol gas → Le Chatelier: high P shifts right → more NH₃
• Fe catalyst: lowers activation energy → acceptable rate at 450°C without sacrificing K
• Without catalyst, even 450°C too slow. Thermodynamics + kinetics both must be optimised.

← Unit 14: Enthalpy Changes 🏠 Course Home

Entropy & Gibbs Free Energy

Entropy

Factors That Increase Entropy

Predicting the Sign of ΔS

The Second Law of Thermodynamics

Entropy Change of the Surroundings

Calculating ΔS°rxn from Absolute Entropies

The Third Law of Thermodynamics

Absolute Entropy and Temperature

Gibbs Free Energy

Standard Gibbs Free Energy Change

Spontaneity Criteria

The Four ΔH / ΔS Combinations

Case 1: ΔH < 0, ΔS > 0

Case 2: ΔH > 0, ΔS < 0

Case 3: ΔH < 0, ΔS < 0

Case 4: ΔH > 0, ΔS > 0

Calculating ΔG° and Checking Spontaneity

Temperature Effects on ΔG

Crossover Temperature

Crossover Temperature

The van't Hoff Equation

Derivation

Exothermic (ΔH° < 0)

Endothermic (ΔH° > 0)

Van't Hoff — K at New Temperature

Calculating K from ΔG°

Coupled Reactions and Free Energy

Principle

Biological Example: ATP

Maximum Work and Electrochemistry

ΔG° and E°cell

Exercises

Interactive Quiz

Unit 15 Quiz — Entropy & Free Energy

Unit Test

Section A — Short Answer

Section B — Extended Response

Course Complete! All 15 Units Done.

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