Ellingham-Diagram Equilibria for Ti, Ca and Si Oxides at 1600°C: 21-Mat-A1 Thermodynamics, December 2017, Q7

Question 7: Ellingham-Diagram Equilibria for Ti, Ca and Si Oxides at 1600°C (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, and reusing the paper's own $\Delta H^\circ_f$ values for $CO_2$ and $H_2O(g)$ from Question 4 for internal consistency), so computing from that 298 K data reproduces the diagram's own readings to full precision rather than eyeballing a ruler position. Parts (e) and (f) are physical constants read directly off the metal/oxide-line kinks the diagram marks "m" and "b" and are not derived from the 298 K formation data.

Given. $T=1600\,{}^{\circ}\text{C}=1873$ K (also $T=673$ K for part (d)). 298 K formation data (Gaskell/NIST–JANAF): $Ti+O_2=TiO_2$: $\Delta H^\circ_f=-944.0$ kJ mol⁻¹, $S^\circ(TiO_2)=50.6$, $S^\circ(Ti)=30.7$ J K⁻¹mol⁻¹. $2Ca+O_2=2CaO$: $\Delta H^\circ_f(CaO)=-635.1$ kJ mol⁻¹, $S^\circ(CaO)=39.7$, $S^\circ(Ca)=41.6$ J K⁻¹mol⁻¹. $Si+O_2=SiO_2$: $\Delta H^\circ_f=-910.7$ kJ mol⁻¹, $S^\circ(SiO_2)=41.5$, $S^\circ(Si)=18.8$ J K⁻¹mol⁻¹. 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 kJ or J K⁻¹mol⁻¹ as shown. $S^\circ(O_2)=205.2$ J K⁻¹mol⁻¹ throughout.

Find. (a) $P_{O_2}$ for Ti/$TiO_2$ at 1600°C. (b) CO/$CO_2$ ratio for Ca/CaO at 1600°C. (c) $H_2$/$H_2O$ ratio for Si/$SiO_2$ at 1600°C. (d) $\Delta G^\circ$ at 400°C for $Ti+SiO_2=TiO_2+Si$. (e) Melting point of Mn. (f) Boiling point of Mg.

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

Figure 9-3 (Gaskell) — Ellingham diagram for the oxides referenced in this question, as attached to the exam paper.

Approach. Build each oxide's $\Delta G^\circ(T)=\Delta H^\circ-T\Delta S^\circ$ line from its 298 K data; read $P_{O_2}$ directly from the metal/oxide line via $\Delta G^\circ=RT\ln P_{O_2}$; couple to the CO/$CO_2$ or $H_2$/$H_2O$ gas-phase lines by equating the $P_{O_2}$ each produces; and take the difference of two oxide lines for the metallothermic reduction in (d).

Oxide-formation free-energy lines. $\Delta S^\circ_{Ti}=S^\circ(TiO_2)-S^\circ(Ti)-S^\circ(O_2)=50.6-30.7-205.2=-185.3$ J K⁻¹, so $\Delta G^\circ_{Ti}(T)=-944{,}000+185.3T$. Likewise for $2Ca+O_2=2CaO$: $\Delta S^\circ_{Ca}=2(39.7)-2(41.6)-205.2=-209.0$ J K⁻¹, $\Delta G^\circ_{Ca}(T)=-1{,}270{,}200+209.0T$ (per 2 mol CaO, i.e. per mol $O_2$). And for $Si+O_2=SiO_2$: $\Delta S^\circ_{Si}=41.5-18.8-205.2=-182.5$ J K⁻¹, $\Delta G^\circ_{Si}(T)=-910{,}700+182.5T$.
(a) Oxygen partial pressure for Ti/$TiO_2$ at 1873 K. $\Delta G^\circ_{Ti}(1873)=-944{,}000+185.3(1873)=\boxed{-596{,}933\ \text{J} \,(-596.9\ \text{kJ})}$. For $Ti(s)+O_2(g)=TiO_2(s)$ with both solids at unit activity, $K=1/P_{O_2}$, so $\Delta G^\circ=RT\ln P_{O_2}$: $$\ln P_{O_2}=\dfrac{-596{,}933}{(8.314)(1873)}=-38.33\ \Rightarrow\ \boxed{P_{O_2}=10^{-16.65}\ \text{atm}\approx2.2\times10^{-17}\ \text{atm}}.$$
(b) CO/$CO_2$ ratio for Ca/CaO at 1873 K. $\Delta G^\circ_{Ca}(1873)=-1{,}270{,}200+209.0(1873)=\boxed{-878{,}743\ \text{J}}$. For the coupling reaction $2CO+O_2=2CO_2$: $\Delta S^\circ_{CO}=2(213.8)-2(197.7)-205.2=-173.0$ J K⁻¹, $\Delta H^\circ=2(-393{,}500)-2(-110{,}500)=-566{,}000$ J, giving $\Delta G^\circ_{CO/CO_2}(1873)=-566{,}000+173.0(1873)=-241{,}971$ J. Subtracting from the Ca line ($Ca+2CO_2=CaO$-equivalent construction, $K\propto(P_{CO}/P_{CO_2})^2$): $$\ln\!\left(\dfrac{P_{CO_2}}{P_{CO}}\right)=\dfrac{\Delta G^\circ_{Ca}-\Delta G^\circ_{CO/CO_2}}{2RT}=\dfrac{-878{,}743-(-241{,}971)}{2(8.314)(1873)}=-20.44\ \Rightarrow\ \boxed{\dfrac{P_{CO}}{P_{CO_2}}\approx7.6\times10^{8}}.$$ An extremely CO-rich (essentially oxygen-free) atmosphere is required — Ca is the most reactive metal on the chart, so its line sits far below the CO/$CO_2$ coupling line even at 1600°C.
(c) $H_2$/$H_2O$ ratio for Si/$SiO_2$ at 1873 K. $\Delta G^\circ_{Si}(1873)=-910{,}700+182.5(1873)=\boxed{-568{,}878\ \text{J}}$. 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, $\Delta H^\circ=2(-241{,}800)=-483{,}600$ J, giving $\Delta G^\circ_{H_2/H_2O}(1873)=-483{,}600+89.0(1873)=-316{,}903$ J. By the same construction: $$\ln\!\left(\dfrac{P_{H_2O}}{P_{H_2}}\right)=\dfrac{\Delta G^\circ_{Si}-\Delta G^\circ_{H_2/H_2O}}{2RT}=\dfrac{-568{,}878-(-316{,}903)}{2(8.314)(1873)}=-8.09\ \Rightarrow\ \boxed{\dfrac{P_{H_2}}{P_{H_2O}}\approx3.3\times10^{3}}.$$
(d) $\Delta G^\circ$ at 400°C for $Ti+SiO_2=TiO_2+Si$. At $T=673$ K: $\Delta G^\circ_{Ti}(673)=-944{,}000+185.3(673)=-819{,}293$ J; $\Delta G^\circ_{Si}(673)=-910{,}700+182.5(673)=-787{,}878$ J. Subtracting the Si line from the Ti line ($Ti+O_2=TiO_2$ minus $Si+O_2=SiO_2$ gives exactly this reaction): $$\Delta G^\circ=\Delta G^\circ_{Ti}-\Delta G^\circ_{Si}=-819{,}293-(-787{,}878)=\boxed{-31.4\ \text{kJ}\,\text{mol}^{-1}}.$$ The Ti line lies below (more negative than) the Si line at 400°C, so the reaction is spontaneous — metallic titanium can reduce silica to silicon, consuming itself to form rutile.
(e) Melting point of Mn. The kink marked "m" on the $2Mn+O_2=2MnO$ line of the attached diagram falls at the standard melting point of manganese, $\boxed{1246\,{}^{\circ}\text{C}}$ (1519 K).
(f) Boiling point of Mg. The kink marked "b" on the $2Mg+O_2=2MgO$ line falls at the standard boiling point of magnesium, $\boxed{1090\,{}^{\circ}\text{C}}$ (1363 K) — well below most of the diagram's 1600°C ceiling, which is why the Mg line shows a distinct change of slope partway across the chart.

Quantity	Value at 1600°C (unless noted)
$\Delta G^\circ_{Ti}$, $\Delta G^\circ_{Ca}$, $\Delta G^\circ_{Si}$	−596.9, −878.7, −568.9 kJ/mol
(a) $P_{O_2}$ (Ti/$TiO_2$)	≈2.2×10⁻¹⁷ atm
(b) CO/$CO_2$ (Ca/CaO)	≈7.6×10⁸
(c) $H_2$/$H_2O$ (Si/$SiO_2$)	≈3.3×10³
(d) $\Delta G^\circ$ at 400°C ($Ti+SiO_2=TiO_2+Si$)	−31.4 kJ/mol
(e) Melting point of Mn	1246°C
(f) Boiling point of Mg	1090°C

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 $\ln(\text{ratio})=(\Delta G^\circ_{oxide}-\Delta G^\circ_{couple})/2RT$; difference of two oxide lines gives the metallothermic reduction $\Delta G^\circ$; a phase-transition kink is a slope change, never a jump, in $\Delta G^\circ(T)$.

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/$CO_2$ and $H_2$/$H_2O$ side scales are 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/$CO_2$ or $H_2$/$H_2O$ scale directly for a very reactive metal (Ca, Ti, Si, Al) without noticing the reading falls outside the printed scale range — the physically correct answer still exists and is obtained directly from the 298 K data, as done here, rather than by extrapolating a ruler.

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 Ti, Ca and Si Oxides at 1600°C

Question 7: Ellingham-Diagram Equilibria for Ti, Ca and Si Oxides at 1600°C (20 marks)