Question 7 of 7: Reading the Ellingham Diagram — Cu/Cu₂O, Cr/Cr₂O₃ and Ti/TiO₂ Equilibria
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Exams — May 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 enthalpy/entropy data used below; supporting 298 K entropies for the elements, CO, CO₂, H₂, H₂O(g) and Cu₂O/Cr₂O₃/TiO₂/SiO₂ from the NIST–JANAF Thermochemical Tables.
Question 7: Reading the Ellingham Diagram — Cu/Cu₂O, Cr/Cr₂O₃ and Ti/TiO₂ Equilibria (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 Cu₂O, Cr₂O₃, TiO₂, SiO₂ and the relevant elements/gases are Gaskell's own tabulated constants (Cu₂O: $\Delta H^\circ_f=-168{,}600$ J/mol, $S^\circ=93.14$ J/mol·K; Cr₂O₃: $-1{,}139{,}700$ J/mol, 81.2 J/mol·K; TiO₂: $-944{,}000$ J/mol, 50.6 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.
Find. (a) $P_{O_2}$ on the Cu/Cu₂O line at 500°C. (b) CO/CO₂ ratio matching the Cr/Cr₂O₃ oxygen potential at 400°C. (c) H₂/H₂O ratio matching the Ti/TiO₂ oxygen potential at 600°C. (d) $\Delta G^\circ$ for Ti+SiO₂=TiO₂+Si at 1500°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 each metal/oxide line (per mole O₂) from 298 K formation data; parts (a)–(c) read $P_{O_2}$ off a line directly or match it against a CO/CO₂ or H₂/H₂O gas-ratio line; part (d) subtracts the Si/SiO₂ line from the Ti/TiO₂ line; parts (e)/(f) use $\text{slope}=-\Delta S^\circ$ and the change in moles of gas across each reaction.
(a) Cu/Cu₂O line, 500°C ($T=773.15$ K). For $4\text{Cu}+\text{O}_2=2\text{Cu}_2\text{O}$: $\Delta H^\circ=2(-168{,}600)=-337{,}200$ J, $\Delta S^\circ=2(93.14)-4(33.15)-205.1=-151.4$ J/K, so $\Delta G^\circ(T)=-337{,}200+151.4T$. At 773.15 K, $\Delta G^\circ=-220{,}130$ J, and since $\Delta G^\circ=RT\ln P_{O_2}$:
$$P_{O_2}=\exp\!\left(\frac{-220{,}130}{(8.314)(773.15)}\right)=\boxed{1.3\times10^{-15}\ \text{atm}}.$$
Cu₂O is one of the least stable oxides on the chart, consistent with $P_{O_2}\gg$ that of the more reactive metals lower on the diagram — still far below air's 0.21 atm, which is why copper does tarnish (slowly) in air.
(b) Cr/Cr₂O₃ line, 400°C ($T=673.15$ K), matched to a CO/CO₂ atmosphere. For $\tfrac43\text{Cr}+\text{O}_2=\tfrac23\text{Cr}_2\text{O}_3$: $\Delta H^\circ=\tfrac23(-1{,}139{,}700)=-759{,}800$ J, $\Delta S^\circ=\tfrac23(81.2)-\tfrac43(23.77)-205.1=-182.7$ J/K, giving $\Delta G^\circ_{Cr}(673.15)=-636{,}840$ J and $P_{O_2}=\exp(\Delta G^\circ_{Cr}/RT)=3.8\times10^{-50}$ atm. Build the $2\text{CO}+\text{O}_2=2\text{CO}_2$ line the same way ($\Delta H^\circ=-566{,}000$ J, $\Delta S^\circ=-173.1$ J/K): $\Delta G^\circ_{2CO}(673.15)=-449{,}480$ J, with $K_{2CO}=P_{CO_2}^2/(P_{CO}^2P_{O_2})=\exp(-\Delta G^\circ_{2CO}/RT)$. Matching the Cr/Cr₂O₃ oxygen potential:
$$\frac{P_{CO}}{P_{CO_2}}=\frac{1}{\sqrt{K_{2CO}\,P_{O_2,Cr}}}=\boxed{1.9\times10^{7}}.$$
An overwhelmingly CO-rich atmosphere is needed — Cr₂O₃ is far more stable than CO₂, so almost no CO₂ can be tolerated before Cr starts to oxidize.
(c) Ti/TiO₂ line, 600°C ($T=873.15$ K), matched to a H₂-H₂O atmosphere. For $\text{Ti}+\text{O}_2=\text{TiO}_2$: $\Delta H^\circ=-944{,}000$ J, $\Delta S^\circ=50.6-30.7-205.1=-185.2$ J/K, giving $\Delta G^\circ_{Ti}(873.15)=-782{,}290$ J and $P_{O_2}=1.6\times10^{-47}$ atm. The $2\text{H}_2+\text{O}_2=2\text{H}_2\text{O}$ line ($\Delta H^\circ=-483{,}600$ J, $\Delta S^\circ=-88.9$ J/K) gives $\Delta G^\circ_{2H_2}(873.15)=-405{,}980$ J. By the same construction as part (b):
$$\frac{P_{H_2}}{P_{H_2O}}=\frac{1}{\sqrt{K_{2H_2}\,P_{O_2,Ti}}}=\boxed{1.8\times10^{11}}.$$
(d) $\Delta G^\circ$ for Ti + SiO₂ = TiO₂ + Si at 1500°C ($T=1773.15$ K). This reaction is (Ti+O₂=TiO₂) minus (Si+O₂=SiO₂), so $\Delta G^\circ_{rxn}=\Delta G^\circ_{Ti}-\Delta G^\circ_{Si}$. $\Delta G^\circ_{Ti}(1773.15)=-944{,}000+185.2(1773.15)=-615{,}600$ J. For SiO₂ ($\Delta H^\circ_f=-910{,}900$ J, $S^\circ=41.5$ J/mol·K, $S^\circ(\text{Si})=18.8$ J/mol·K): $\Delta S^\circ_{Si}=41.5-18.8-205.1=-182.4$ J/K, so $\Delta G^\circ_{Si}(1773.15)=-910{,}900+182.4(1773.15)=-587{,}500$ J. Then
$$\Delta G^\circ_{rxn}=-615{,}600-(-587{,}500)=\boxed{-28.1\ \text{kJ/mol}}.$$
Negative, so titanium spontaneously reduces silica at 1500°C — consistent with the Ti line sitting (just) below the Si line on the diagram at this temperature.
(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 (∼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).
