Question 7 of 7: Reading the Ellingham Diagram — Al/Al₂O₃ Equilibria and the Ca–MgO Reduction
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Exams — December 2015 — 10-Met-A1 Metallurgical Thermodynamics. Three-hour, closed-book exam using an approved (Casio or Sharp) calculator; candidates were told to state any interpretive assumptions. Any five of the seven questions constitute a complete paper — all seven are solved below for completeness. All questions are of equal value (20 marks each out of 100).
Reference texts: Gaskell, D. R., Introduction to the Thermodynamics of Materials (2nd–5th ed.) — the exam's own Ellingham diagram (Fig. 9-3) is reproduced from this text, which also supplies the standard-state 298 K enthalpy/entropy data used in Question 7; supporting 298 K entropies for CO(g), CO₂(g), H₂(g) and H₂O(g) from the NIST–JANAF Thermochemical Tables.
Question 7: Reading the Ellingham Diagram — Al/Al₂O₃ Equilibria and the Ca–MgO Reduction (20 marks)
Given. The attached Ellingham diagram (Fig. 9-3, Gaskell) plots $\Delta G^\circ=RT\ln P_{O_2}$ for oxide-formation reactions against temperature, with nomographic CO/CO₂ and H₂/H₂O scales.
[Figure not reproduced: Ellingham diagram (Gaskell Fig. 9-3). See the official exam paper or the cited reference text.]
Figure 2 — The exam’s attached Ellingham diagram (Gaskell, Fig. 9-3), reproduced from the source paper.
Check
Parts (a)–(d) are computed from the same 298 K standard-state data the printed diagram is built from — $\Delta H^\circ_f$ and $S^\circ$ for Al₂O₃, MgO, CaO and the relevant elements/O₂ are Gaskell's own tabulated constants (Al₂O₃: $\Delta H^\circ_f=-1{,}675{,}700$ J/mol, $S^\circ=50.9$ J/mol·K; MgO: $-601{,}700$ J/mol, 26.9 J/mol·K; CaO: $-634{,}900$ J/mol, 38.1 J/mol·K). Each line is then the same straight-line approximation $\Delta G^\circ(T)=\Delta H^\circ-T\Delta S^\circ$ that makes the printed lines straight, used here in place of reading the printed nomographic scales by eye — a chart reading, worked algebraically, to the same accuracy the diagram itself carries. This does not correct for the small kinks the real diagram would show at a metal's own melting/boiling point (e.g. Mg boils near 1090°C); the single straight-line fit from 298 K data is the same simplification the printed chart itself uses away from those marked break points.
Find. (a) $P_{O_2}$ on the Al/Al₂O₃ line at 1600°C. (b) CO/CO₂ ratio matching the Al/Al₂O₃ oxygen potential at 1600°C. (c) H₂/H₂O ratio matching the Al/Al₂O₃ oxygen potential at 1600°C. (d) $\Delta G^\circ$ for Ca+MgO=CaO+Mg at 1300°C. (e)/(f) Why the C/CO₂ and C/CO lines have the slopes they do.
Approach. Build the straight-line $\Delta G^\circ(T)=\Delta H^\circ-T\Delta S^\circ$ for the Al/Al₂O₃ line (per mole O₂) from 298 K formation data; parts (a)–(c) read $P_{O_2}$ off that line directly or match it against a CO/CO₂ or H₂/H₂O gas-ratio line built the same way; part (d) takes the difference of the Ca/CaO and Mg/MgO lines; parts (e)/(f) use $\text{slope}=-\Delta S^\circ$ and the change in moles of gas across each reaction.
(a) Al/Al₂O₃ line, 1600°C ($T=1873.15$ K). For $\tfrac43\text{Al}+\text{O}_2=\tfrac23\text{Al}_2\text{O}_3$: $\Delta H^\circ=\tfrac23(-1{,}675{,}700)=-1{,}117{,}133$ J, $\Delta S^\circ=\tfrac23(50.9)-\tfrac43(28.3)-205.1=-208.9$ J/K, so $\Delta G^\circ(T)=-1{,}117{,}133+208.9T$. At 1873.15 K, $\Delta G^\circ=-725{,}832$ J, and since $\Delta G^\circ=RT\ln P_{O_2}$:
$$P_{O_2}=\exp\!\left(\frac{-725{,}832}{(8.314)(1873.15)}\right)=\boxed{5.7\times10^{-21}\ \text{atm}}.$$
Al₂O₃ is one of the most stable oxides on the entire chart, so this equilibrium oxygen pressure is extraordinarily low — aluminium metal is essentially incompatible with any measurable oxygen partial pressure at this temperature.
(b) CO/CO₂ ratio, 1600°C. Build the $2\text{CO}+\text{O}_2=2\text{CO}_2$ line the same way ($\Delta H^\circ=2(-393{,}500)-2(-110{,}500)=-566{,}000$ J, $\Delta S^\circ=2(213.7)-2(197.9)-205.1=-173.5$ J/K): $\Delta G^\circ_{2CO}(1873.15)=-566{,}000+173.5(1873.15)=-241{,}008$ J, with $K_{2CO}=P_{CO_2}^2/(P_{CO}^2P_{O_2})=\exp(-\Delta G^\circ_{2CO}/RT)=5.26\times10^{6}$. Matching the Al/Al₂O₃ oxygen potential from part (a):
$$\frac{P_{CO}}{P_{CO_2}}=\frac{1}{\sqrt{K_{2CO}\,P_{O_2,Al}}}=\boxed{5.8\times10^{6}}.$$
An overwhelmingly CO-rich atmosphere is required — consistent with Al₂O₃ being far more stable than CO₂, so almost no CO₂ can be tolerated before Al starts to oxidize.
(c) H₂/H₂O ratio, 1600°C. The $2\text{H}_2+\text{O}_2=2\text{H}_2\text{O}$ line ($\Delta H^\circ=2(-241{,}800)=-483{,}600$ J, $\Delta S^\circ=2(188.7)-2(130.6)-205.1=-88.9$ J/K) gives $\Delta G^\circ_{2H_2}(1873.15)=-483{,}600+88.9(1873.15)=-317{,}077$ J and $K_{2H_2}=6.96\times10^{8}$. By the same construction as part (b):
$$\frac{P_{H_2}}{P_{H_2O}}=\frac{1}{\sqrt{K_{2H_2}\,P_{O_2,Al}}}=\boxed{5.0\times10^{5}}.$$
(d) $\Delta G^\circ$ for Ca + MgO = CaO + Mg at 1300°C ($T=1573.15$ K). This reaction is (Ca+⅓O₂=CaO) minus (Mg+⅓O₂=MgO), so $\Delta G^\circ_{rxn}=\Delta G^\circ_{CaO}-\Delta G^\circ_{MgO}$. For CaO ($\Delta H^\circ_f=-634{,}900$ J/mol, $S^\circ(\text{CaO})=38.1$, $S^\circ(\text{Ca})=41.6$ J/mol·K): $\Delta S^\circ_{CaO}=38.1-41.6-102.55=-106.05$ J/K, so $\Delta G^\circ_{CaO}(1573.15)=-634{,}900+106.05(1573.15)=-468{,}067$ J. For MgO ($\Delta H^\circ_f=-601{,}700$ J/mol, $S^\circ(\text{MgO})=26.9$, $S^\circ(\text{Mg})=32.7$ J/mol·K): $\Delta S^\circ_{MgO}=26.9-32.7-102.55=-108.35$ J/K, so $\Delta G^\circ_{MgO}(1573.15)=-601{,}700+108.35(1573.15)=-431{,}249$ J. Then
$$\Delta G^\circ_{rxn}=-468{,}067-(-431{,}249)=\boxed{-36.8\ \text{kJ/mol}}.$$
Negative, so calcium spontaneously reduces MgO at 1300°C — consistent with the Ca/CaO line sitting below the Mg/MgO line on the diagram at this temperature (the thermodynamic basis of calciothermic reduction).
(e) Why C(s)+O₂(g)=CO₂(g) is nearly horizontal. The reaction consumes 1 mol of gas (O₂) and produces 1 mol of gas (CO₂) — the moles of gas are unchanged ($\Delta n_{gas}=0$). Since a mole of gas carries far more entropy ($\sim$200 J/mol·K) than the solid carbon it replaces contributes or removes, $\Delta n_{gas}=0$ leaves $\Delta S^\circ$ small. As the line's slope on the $\Delta G^\circ$ vs. $T$ plot is $-\Delta S^\circ$, a near-zero $\Delta S^\circ$ gives a near-zero slope — the line runs almost horizontally.
(f) Why 2C(s)+O₂(g)=2CO(g) runs downward. Here 1 mol of gas (O₂) becomes 2 mol of gas (CO) — a net increase in gas moles, $\Delta n_{gas}=+1$. This produces a large positive $\Delta S^\circ$ (more gas-phase disorder is created than is lost from the solid carbon consumed). Since slope $=-\Delta S^\circ$, a positive $\Delta S^\circ$ gives a negative slope: $\Delta G^\circ$ becomes more negative as $T$ increases, so the line runs downward, eventually crossing below every other oxide's line at sufficiently high $T$ — the thermodynamic basis of carbothermic reduction (e.g. the blast furnace).
