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Lesson 5 established the theoretical framework: a process is thermodynamically feasible at constant temperature and pressure when ΔG = ΔH − TΔS < 0, with the four sign-combinations of ΔH and ΔS giving a qualitative decision rule and T = ΔH/ΔS marking the crossover. This lesson takes that framework into the wild — through industrial extraction of metals from their oxides (with Ellingham diagrams as graphical feasibility maps), thermal decomposition of Group 2 carbonates, coupled reactions in biochemistry where ATP hydrolysis drags otherwise unfavourable steps into spontaneity, and the quantitative bridges ΔG° = −RT ln K (equilibrium) and ΔG° = −nFE_cell° (electrochemistry). One caveat recurs: thermodynamics tells you whether a reaction can happen, never how fast.
Spec mapping (AQA 7405): This lesson maps to §3.1.8 (thermodynamics — Gibbs free energy and feasibility), building directly on lesson 5 (theoretical framework — entropy, ΔG = ΔH − TΔS, four sign-combinations). It cross-references §3.1.7 and §3.1.11 (oxidation, reduction and redox equilibria / electrode potentials and electrochemical cells — anchors the ΔG° = −nFE_cell° link), §3.1.6 and §3.1.10 (equilibria and Kp / acids and bases — anchors the ΔG° = −RT ln K link), and §3.2 (inorganic chemistry, Period 3 and Group 2 — provides metal-extraction and carbonate-decomposition case studies). Refer to the official AQA specification document for the exact wording of each section.
Assessment objectives: AO1 recalls the four sign-combinations of ΔH and ΔS and their feasibility implications, together with the qualitative shape of Ellingham diagrams and the statement ΔG° = −RT ln K. AO2 computes the crossover temperature for industrial decomposition reactions, chooses reducing agents from comparative ΔG data, and applies ΔG° = −RT ln K and ΔG° = −nFE° to numerical problems. AO3 rationalises industrial process choices in terms of thermodynamics, kinetics, economics and environmental impact, and synthesises the framework across feasibility, equilibrium and electrochemistry.
At constant temperature and pressure:
ΔG = ΔH − TΔS
A process is thermodynamically feasible (spontaneous) when ΔG < 0. The four sign combinations summarise the qualitative behaviour:
| ΔH | ΔS | ΔG behaviour | Comment |
|---|---|---|---|
| − | + | always negative | feasible at all T |
| + | − | always positive | never feasible |
| − | − | negative at low T | feasible below T_crossover |
| + | + | negative at high T | feasible above T_crossover |
The crossover temperature, found by setting ΔG = 0, is:
T_crossover = ΔH / ΔS
Critical reminder: Feasibility is a thermodynamic statement, not a kinetic one. ΔG < 0 tells you the reaction can go; it says nothing about the activation energy or the rate. Diamond is thermodynamically unstable with respect to graphite at 298 K — yet your engagement ring is in no immediate danger. The distinction between feasibility and rate is the single most-tested A-level subtlety in this topic.
Most metals occur in nature as oxides or as compounds easily converted to oxides on roasting. The central problem of extractive metallurgy is reduction:
MO(s) + reducing agent → M(s) + oxidised by-product
The most economically attractive reducing agent is carbon (as coke), because it is abundant, cheap, and the by-product (CO or CO₂) escapes as a gas, driving the reaction forward via Le Chatelier's principle. The reaction of interest is:
MO(s) + C(s) → M(s) + CO(g) [reaction 1]
Whether reaction 1 is feasible depends on ΔG, and crucially on T. Reaction 1 is the sum of two half-reactions:
Reaction 1 = reaction 3 − reaction 2. Reaction 3 has ΔS > 0 (net gain of ½ mol gas), so ΔG(3) becomes more negative as T rises. Reaction 2 has ΔS < 0 (gas consumed in forming a solid oxide), so ΔG(2) becomes less negative as T rises. Subtract: ΔG(1) becomes more negative as T rises. High temperature favours carbon reduction.
For lead oxide (PbO), ΔG° for carbon reduction becomes negative below 1000 K — easy and cheap. Lead has been smelted from galena (PbS, roasted to PbO, then reduced with carbon) for several thousand years.
For iron oxide (Fe₂O₃), the crossover is around 1100 K. The blast furnace operates at ~1900 K at the tuyères, well above the crossover, and the reduction proceeds rapidly with CO as the immediate reducing agent:
Fe₂O₃(s) + 3CO(g) → 2Fe(l) + 3CO₂(g)
For aluminium oxide (Al₂O₃), the corresponding crossover for carbon reduction is above 2300 K — higher than the melting point of most furnace linings, and uneconomic in equipment and energy. Worse, any aluminium formed reacts back with carbon to give Al₄C₃, contaminating the product. Aluminium is therefore extracted by electrolysis of a molten Al₂O₃/cryolite (Na₃AlF₆) mixture at ~1300 K (~950 °C — cryolite lowers the melting point of Al₂O₃ to a workable ~1300 K), driven by electrical work rather than by the favourability of carbon oxidation. The energetic cost (~14 kWh per kg) is why aluminium smelters sit next to hydroelectric power and why recycled aluminium is so much cheaper than primary metal.
The takeaway: smelting vs electrolysis is a feasibility-vs-cost decision mediated by ΔG(T). Where ΔG for carbon reduction is comfortably negative at a practical T, you smelt. Where it is not, you electrolyse — and pay the electricity bill.
The argument above can be made visual. An Ellingham diagram plots ΔG° per mole of O₂ consumed against temperature for a family of metal-oxidation reactions:
M(s) + ½O₂(g) → MO(s) (for divalent oxides; analogous for others)
Each line is essentially straight (ΔG° = ΔH° − TΔS°, treated as linear in T over the displayed range), sloping upward (positive gradient) because ΔS° is negative — gas being consumed — making ΔG° less negative as T rises. Different metals give parallel-ish lines at different vertical positions. The lower a line on the diagram, the more stable the corresponding oxide (more negative ΔG° of formation).
Now overlay the carbon-oxidation line:
C(s) + ½O₂(g) → CO(g)
This reaction gains gas (½O₂ → CO is a net gain of ½ mole of gas), so ΔS° is positive and the line slopes downward — ΔG° becomes more negative as T rises. Sooner or later the carbon line crosses each metal-oxide line. Below the crossing temperature, the metal-oxide line is lower (more negative ΔG°), so the metal oxide is more stable than CO — carbon cannot reduce the oxide. Above the crossing temperature, the carbon line is lower, so CO is more stable than the metal oxide — carbon reduces the oxide. The crossing point identifies the minimum temperature for carbon-reduction feasibility for that metal.
For lead this crossing is at relatively low T (~1000 K); for iron, around 1100 K; for chromium, around 1500 K; for aluminium, above 2300 K. Reading the diagram immediately tells the metallurgist which reducing agent works at what temperature. (AQA treats Ellingham diagrams qualitatively; you are not expected to memorise specific values, only the principle of the construction and the link between line slope and ΔS°.)
Synoptic point: The slope of an Ellingham line equals −ΔS°. Steep upward slopes mean a large negative ΔS° (lots of gas consumed). The carbon-to-CO line slopes downward because that reaction produces gas. The geometry of the diagram is entropy on display.
Heating a Group 2 metal carbonate decomposes it to the oxide and carbon dioxide:
MCO₃(s) → MO(s) + CO₂(g)
This reaction has:
By the four-sign rule, this is a (ΔH > 0, ΔS > 0) reaction — not feasible at low T, feasible above the crossover T = ΔH/ΔS.
Typical data (per mole of carbonate decomposed):
| Carbonate | ΔH° / kJ mol⁻¹ | ΔS° / J K⁻¹ mol⁻¹ | T_crossover / K | T_crossover / °C |
|---|---|---|---|---|
| MgCO₃ | +117 | +175 | 668 | 395 |
| CaCO₃ | +178 | +161 | 1106 | 833 |
| SrCO₃ | +234 | +171 | 1369 | 1096 |
| BaCO₃ | +269 | +172 | 1564 | 1291 |
The crossover temperatures rise steadily down the group. ΔS° is roughly constant (the same mole of CO₂ is liberated in each case), so the trend is driven mainly by ΔH°: decomposition becomes progressively more endothermic from MgCO₃ to BaCO₃.
The lattice enthalpy of the oxide (MO) is more negative than that of the carbonate (MCO₃) — the smaller, more charge-dense oxide ion packs more efficiently than the larger carbonate. As the cation grows down the group (Mg²⁺ to Ba²⁺), the lattice enthalpies of both MO and MCO₃ become less negative, but the oxide lattice enthalpy falls more steeply because the small O²⁻ is more sensitive to cation size than the diffuse CO₃²⁻. The result is that the energetic incentive to convert MCO₃ → MO (the difference in lattice enthalpy) shrinks down the group, so ΔH° for decomposition grows more positive, and T_crossover rises.
This is consistent with the polarising-power argument from inorganic chemistry: small, high-charge-density cations (Mg²⁺) polarise the carbonate anion strongly, destabilising MCO₃ relative to MO + CO₂; large cations (Ba²⁺) polarise weakly, so MCO₃ is relatively stable. The thermodynamic trend and the bonding rationalisation are two views of the same physics.
A reaction with ΔG > 0 is non-spontaneous on its own — but it can be made to go by coupling it to a strongly exergonic partner reaction. Provided the two share a common intermediate, the overall ΔG is the sum of the individual ΔG values (Hess's law in free-energy form, since G is a state function).
The cell's universal energy currency is adenosine triphosphate (ATP). Its hydrolysis to adenosine diphosphate (ADP) and inorganic phosphate (Pi) is strongly exergonic:
ATP + H₂O → ADP + Pi ΔG° ≈ −31 kJ mol⁻¹
(The biologically standard ΔG°' — at pH 7 — is around −30 kJ mol⁻¹; the value under cellular conditions, where ATP, ADP and Pi are not at standard concentrations, is closer to −50 kJ mol⁻¹. AQA accepts the standard figure.)
Consider the first step of glycolysis, the phosphorylation of glucose by hexokinase:
glucose + Pi → glucose-6-phosphate + H₂O ΔG° ≈ +14 kJ mol⁻¹ (uphill — would not run alone)
Couple this to ATP hydrolysis (sum the two steps; the Pi cancels):
glucose + ATP → glucose-6-phosphate + ADP ΔG° ≈ +14 + (−31) = −17 kJ mol⁻¹ (downhill — runs)
The coupling is enzymatic — hexokinase holds glucose and ATP in close enough proximity for the phosphate to transfer in a single concerted step, so no free Pi is liberated. The mechanism makes the two reactions effectively a single event; the thermodynamic accounting simply adds the ΔG values.
This same trick — couple to ATP hydrolysis — is used throughout metabolism for biosynthesis, active transport, muscle contraction and nerve-impulse propagation. The point for A-level chemistry is conceptual: thermodynamic feasibility is additive, and coupling unfavourable reactions to favourable ones is a universal strategy, in industry and in biology alike.
The bridge from thermodynamics to equilibrium is one equation:
ΔG° = −RT ln K
where K is the thermodynamic equilibrium constant (Kp for gases, Kc for solutions, etc.; R = 8.314 J K⁻¹ mol⁻¹; T in K; ΔG° in joules per mole, not kilojoules — unit conversion is the most common error in numerical exercises).
| ΔG° | K | Interpretation |
|---|---|---|
| Large negative | K ≫ 1 | products strongly favoured at equilibrium |
| Slightly negative | K > 1 | products modestly favoured |
| Zero | K = 1 | equal amounts at equilibrium |
| Slightly positive | K < 1 | reactants modestly favoured |
| Large positive | K ≪ 1 | reactants strongly favoured |
For N₂(g) + 3H₂(g) ⇌ 2NH₃(g), take ΔG° = −16.1 kJ mol⁻¹ = −16 100 J mol⁻¹ at 700 K.
ln K = −ΔG° / (RT) = −(−16 100) / (8.314 × 700) = 16 100 / 5820 = 2.77
K = e^2.77 = 16.0 (dimensionless thermodynamic K; numerically equal to Kp expressed in standard pressure units)
The K is modest but positive — at 700 K and standard pressures, products are favoured, but only by an order of magnitude. The industrial Haber process compensates by running at high pressure (Le Chatelier), increasing the effective driving force toward NH₃. Note that ΔG° depends on T (because TΔS° contributes), so K is also T-dependent — quantitatively expressed in the van't Hoff equation (see "Going Further").
ΔG° gives the equilibrium position under standard conditions. It says nothing about the rate of approach to equilibrium — that is the province of kinetics and the activation energy. A reaction with K = 10²⁰ can still take a million years (graphite → diamond at high pressure is one example; the gas-phase formation of NH₃ at room temperature, without a catalyst, is another). Industrial ammonia synthesis combines a strongly negative ΔG° (large K) with an iron catalyst (low activation energy) and elevated temperature (acceptable rate). The catalyst lowers Ea without altering ΔG° or K.
The link from thermodynamics to electrochemistry is:
ΔG° = −nFE_cell°
where n is the number of moles of electrons transferred per mole of reaction, F is the Faraday constant (96 485 C mol⁻¹), and E_cell° is the standard cell e.m.f. in volts. Multiplying out gives ΔG° in joules per mole. The equation says: the maximum non-expansion work obtainable from an electrochemical cell (which equals −ΔG°) is the charge transferred (nF) times the potential difference driving it (E_cell°).
For 2H₂(g) + O₂(g) → 2H₂O(l), ΔG° = −474 kJ mol⁻¹ (= 2 × −237 kJ mol⁻¹ per mole of water). The cell reaction transfers 4 electrons per mole of O₂:
E_cell° = −ΔG° / (nF) = 474 000 / (4 × 96 485) = 474 000 / 385 940 = 1.23 V
This is the maximum theoretical e.m.f. of a hydrogen–oxygen fuel cell at standard conditions. Real fuel cells deliver substantially less (~0.7 V under load) because of activation overpotentials, ohmic losses and concentration polarisation — kinetic, not thermodynamic, limitations.
The reverse calculation also matters: measuring E_cell° experimentally gives ΔG° for the cell reaction, and via ΔG° = −RT ln K, the equilibrium constant — the favoured route to K for redox systems where direct concentration measurements are difficult.
A chemical engineer considers two routes to product Z, starting from different feedstocks:
| Route | ΔH° / kJ mol⁻¹ | ΔS° / J K⁻¹ mol⁻¹ |
|---|---|---|
| A | −80 | −150 |
| B | +40 | +120 |
At what temperature ranges is each route feasible? At what temperature would the two routes have equal ΔG?
Route A: (ΔH < 0, ΔS < 0) — feasible at low T, infeasible at high T. Crossover T = ΔH/ΔS = (−80 000) / (−150) = 533 K. Feasible below 533 K.
Route B: (ΔH > 0, ΔS > 0) — feasible at high T, infeasible at low T. Crossover T = ΔH/ΔS = 40 000 / 120 = 333 K. Feasible above 333 K.
Both routes feasible in the window 333 K < T < 533 K (60 °C to 260 °C). In this window, the choice depends on which has the more negative ΔG. Setting ΔG_A = ΔG_B:
ΔH_A − TΔS_A = ΔH_B − TΔS_B −80 000 − T(−150) = 40 000 − T(120) −80 000 + 150T = 40 000 − 120T 270T = 120 000 T = 444 K
Below 444 K route A has the more negative ΔG (more feasible); above 444 K route B does. The thermodynamic optimum is route A near room temperature (high driving force) or route B at higher T (entropy-driven). Kinetics, capital cost, feedstock availability and product purity would then refine the decision. Thermodynamics frames the choice; engineering closes it.
Repeat after the textbook: ΔG < 0 means the reaction CAN happen; it does NOT mean the reaction WILL happen at observable rate.
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