Ellingham-Diagram Equilibria for Si/SiO₂ and the Ti–SiO₂ Reduction: 21-Mat-A1 Thermodynamics, May 2017, Q7

Question 7: Ellingham-Diagram Equilibria for Si/SiO₂ and the Ti–SiO₂ Reduction (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Check: this question supplies the attached Gaskell Ellingham diagram (Fig. 9-3, reproduced below) for a graphical straightedge reading. That graphical method is reproduced here exactly by direct calculation — the diagram's oxide lines are themselves plots of the same $\Delta G^\circ=\Delta H^\circ-T\Delta S^\circ$ straight-line approximation built from each oxide's 298 K formation enthalpy and entropy (Gaskell Ch. 9 data, cross-checked against NIST–JANAF), so computing from that 298 K data reproduces the diagram's own readings to full precision rather than eyeballing a ruler position. For the very stable Si and Ti oxide lines, the CO/CO₂ and H₂/H₂O readings in (b)–(c) fall above the printed right-hand nomographic scale range and would need graphical extrapolation on the physical chart; the calculation below gives the exact value directly.

Given. $T=1100\,{}^{\circ}\text{C}=1373.15$ K. 298 K formation data (Gaskell/NIST–JANAF): $Si+O_2=SiO_2$: $\Delta H^\circ_f=-910.7$ kJ mol⁻¹, $S^\circ(SiO_2)=41.5$, $S^\circ(Si)=18.8$ J K⁻¹mol⁻¹. $Ti+O_2=TiO_2$: $\Delta H^\circ_f=-944.0$ kJ mol⁻¹, $S^\circ(TiO_2)=50.6$, $S^\circ(Ti)=30.7$ J K⁻¹mol⁻¹. Coupling reactions: $2CO+O_2=2CO_2$ ($\Delta H^\circ_f(CO)=-110.5$, $S^\circ(CO)=197.7$, $\Delta H^\circ_f(CO_2)=-393.5$, $S^\circ(CO_2)=213.8$) and $2H_2+O_2=2H_2O(g)$ ($S^\circ(H_2)=130.7$, $\Delta H^\circ_f(H_2O,g)=-241.8$, $S^\circ(H_2O,g)=188.8$), all J K⁻¹mol⁻¹. $S^\circ(O_2)=205.2$ J K⁻¹mol⁻¹ throughout.

Find. $P_{O_2}$, the CO/CO₂ ratio and the H₂/H₂O ratio in equilibrium with Si+SiO₂ at 1100 °C; $\Delta G^\circ$ at 1100 °C for $Ti+SiO_2=TiO_2+Si$; and the reason Ellingham lines have no vertical jump at a metal's melting/boiling point.

[Figure not reproduced: Ellingham diagram (Gaskell Fig. 9-3) for oxide formation free energies vs. temperature, with CO/CO2 and H2/H2O nomographic scales. See the official exam paper or the cited reference text.]

Figure 2 — Gaskell Fig. 9-3, the Ellingham diagram attached to the exam. The Si+O₂=SiO₂ and Ti+O₂=TiO₂ lines sit near the bottom of the chart, well below the base-metal oxide lines — both are computed directly below from the same 298 K data the diagram is constructed from.

Approach. Build $\Delta G^\circ(T)=\Delta H^\circ-T\Delta S^\circ$ for $Si+O_2=SiO_2$ and $Ti+O_2=TiO_2$ from 298 K data, evaluate at $T=1373.15$ K, then read $P_{O_2}$ directly from $\Delta G^\circ_{Si}=RT\ln P_{O_2}$; couple to the $2CO+O_2=2CO_2$ and $2H_2+O_2=2H_2O$ lines (same construction) to get the CO/CO₂ and H₂/H₂O ratios; and take the difference of the two oxide lines for part (d).

Oxide-formation free energies at 1373.15 K. $\Delta S^\circ_{Si}=S^\circ(SiO_2)-S^\circ(Si)-S^\circ(O_2)=41.5-18.8-205.2=-182.5$ J K⁻¹, so $$\Delta G^\circ_{Si}=-910{,}700-(1373.15)(-182.5)=\boxed{-660.1\ \text{kJ}\,\text{mol}^{-1}}.$$ Likewise $\Delta S^\circ_{Ti}=50.6-30.7-205.2=-185.3$ J K⁻¹, giving $$\Delta G^\circ_{Ti}=-944{,}000-(1373.15)(-185.3)=\boxed{-689.6\ \text{kJ}\,\text{mol}^{-1}}.$$
(a) Oxygen partial pressure. For $Si(s)+O_2(g)=SiO_2(s)$ with both solids at unit activity, $K=1/P_{O_2}$, so $\Delta G^\circ_{Si}=-RT\ln(1/P_{O_2})=RT\ln P_{O_2}$: $$\ln P_{O_2}=\dfrac{\Delta G^\circ_{Si}}{RT}=\dfrac{-660{,}100}{(8.314)(1373.15)}=-57.82\ \Rightarrow\ \boxed{P_{O_2}=10^{-25.1}\ \text{atm}\approx7.7\times10^{-26}\ \text{atm}}.$$
(b) CO/CO₂ ratio. For $2CO+O_2=2CO_2$: $\Delta S^\circ_{CO}=2(213.8)-2(197.7)-205.2=-173.0$ J K⁻¹, $\Delta G^\circ_{CO}(1373.15\ \text{K})=-566{,}000-(1373.15)(-173.0)=-328.4$ kJ. Subtracting from the Si line ($Si+2CO_2=SiO_2+2CO$, $K=(P_{CO}/P_{CO_2})^2$): $$\ln\!\left(\dfrac{P_{CO}}{P_{CO_2}}\right)=\dfrac{\Delta G^\circ_{CO}-\Delta G^\circ_{Si}}{2RT}=\dfrac{-328{,}400-(-660{,}100)}{2(8.314)(1373.15)}=14.53\ \Rightarrow\ \boxed{\dfrac{P_{CO}}{P_{CO_2}}\approx2.0\times10^{6}}.$$ An extremely CO-rich (strongly reducing) atmosphere is required — consistent with SiO₂ being far more stable than the base-metal oxides plotted above it on the chart.
(c) H₂/H₂O ratio. For $2H_2+O_2=2H_2O(g)$: $\Delta S^\circ_{H_2}=2(188.8)-2(130.7)-205.2=-89.0$ J K⁻¹, $\Delta G^\circ_{H_2}(1373.15\ \text{K})=-483{,}600-(1373.15)(-89.0)=-361.4$ kJ. By the same construction: $$\ln\!\left(\dfrac{P_{H_2}}{P_{H_2O}}\right)=\dfrac{\Delta G^\circ_{H_2}-\Delta G^\circ_{Si}}{2RT}=\dfrac{-361{,}400-(-660{,}100)}{2(8.314)(1373.15)}=13.08\ \Rightarrow\ \boxed{\dfrac{P_{H_2}}{P_{H_2O}}\approx4.8\times10^{5}}.$$
(d) $\Delta G^\circ$ for $Ti+SiO_2=TiO_2+Si$. Subtracting the Si line from the Ti line ($Ti+O_2=TiO_2$ minus $Si+O_2=SiO_2$ gives exactly this reaction): $$\Delta G^\circ=\Delta G^\circ_{Ti}-\Delta G^\circ_{Si}=-689.6-(-660.1)=\boxed{-29.5\ \text{kJ}\,\text{mol}^{-1}}.$$ The Ti line lies below (more negative than) the Si line at 1100 °C, so the reaction is spontaneous — metallic titanium can reduce silica to silicon, consuming itself to form rutile.
(e) No discontinuity at a phase transformation. At a metal's melting or boiling point, its formation reaction acquires an extra term for the metal's enthalpy of fusion/vaporisation, and simultaneously its entropy of fusion/vaporisation, related by $\Delta S_{trans}=\Delta H_{trans}/T_{trans}$ exactly at that temperature. Because $\Delta G^\circ=\Delta H^\circ-T\Delta S^\circ$ is continuous in $T$ even though its slope ($-\Delta S^\circ$) changes abruptly, the line on the Ellingham diagram shows a sharp change in gradient (a kink, marked "m" or "b" on the chart) at the transition temperature but never a vertical jump in $\Delta G^\circ$ itself — the free energy is a continuous function of state even where its derivative is not.

Quantity	Value at 1100 °C
$\Delta G^\circ$ (Si+O₂=SiO₂)	−660.1 kJ/mol
$\Delta G^\circ$ (Ti+O₂=TiO₂)	−689.6 kJ/mol
(a) $P_{O_2}$	≈7.7×10⁻²⁶ atm
(b) CO/CO₂	≈2.0×10⁶
(c) H₂/H₂O	≈4.8×10⁵
(d) $\Delta G^\circ$ (Ti+SiO₂=TiO₂+Si)	−29.5 kJ/mol

Concept & Theory

Topic. Reading and reconstructing an Ellingham diagram: oxide stability, coupled gas equilibria, and the physical meaning of the diagram's Richardson nomographic scales.

Key relations. $\Delta G^\circ(T)=\Delta H^\circ-T\Delta S^\circ$ (each oxide line); $\Delta G^\circ=RT\ln P_{O_2}$ at the metal/oxide equilibrium; coupling to $2CO+O_2=2CO_2$ or $2H_2+O_2=2H_2O$ via $\ln(\text{ratio})=(\Delta G^\circ_{couple}-\Delta G^\circ_{oxide})/2RT$; difference of two oxide lines gives the metallothermic reduction $\Delta G^\circ$.

Why this works. Every line on an Ellingham diagram is literally a plot of $\Delta G^\circ=\Delta H^\circ-T\Delta S^\circ$ versus $T$ for one metal/oxide formation reaction; the CO/CO₂ and H₂/H₂O side scales are just a graphical shortcut for reading off the equilibrium gas ratio that would produce the same $P_{O_2}$, obtained by coupling the oxide line to the corresponding gas-phase oxidation line through the common $O_2$ term. A lower (more negative) line always wins a metallothermic reduction against a higher line, because the reaction $\Delta G^\circ$ is simply the vertical separation between the two lines at that temperature.

Common pitfall. Reading the CO/CO₂ or H₂/H₂O scale directly for a very stable oxide (Si, Ti, Al) without noticing the reading falls outside the scale printed on the chart — the physically correct answer still exists, but requires either extrapolating the straightedge construction or, as here, computing it directly from the 298 K data.

Source. Gaskell, Introduction to the Thermodynamics of Materials, Ch. 9 (Ellingham/Richardson diagrams, oxide stability, metallothermic reduction).

Question 7 of 7: Ellingham-Diagram Equilibria for Si/SiO₂ and the Ti–SiO₂ Reduction

Question 7: Ellingham-Diagram Equilibria for Si/SiO₂ and the Ti–SiO₂ Reduction (20 marks)