Ellingham-Diagram Readings for Mn/MnO and Ca: 21-Mat-A1 Thermodynamics, December 2018, Q7

Question 7: Ellingham-Diagram Readings for Mn/MnO and Ca (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: parts (a)–(d) below are computed from standard 298 K formation data (ΔH°_f, S°) for MnO, CO, CO₂, H₂O(g), CaO and MgO via the linear ΔG°(T) ≈ ΔH°−TΔS° approximation that the attached Ellingham diagram is itself built from (the diagram's straight lines ARE this approximation, plotted). The computed Mn/MnO line position (≈−595 kJ at 900°C) and the Ca line's marked melting/boiling points both fall where the attached diagram shows them (see figure). Ca's melting/boiling points in part (e)/(f) are standard reference values (CRC Handbook), consistent with the "m"/"b" markers plotted on the Ca line of the attached diagram.

Given. The attached Ellingham diagram (Fig. 9-3, reproduced below) with its $\text{O}$/$\text{H}$/$\text{C}$ nomographic pivot points for reading $P_{O_2}$, H₂/H₂O and CO/CO₂ ratios directly off a reaction line; $T=900\,{}^{\circ}\text{C}=1173$ K for parts (a)–(c), $T=800\,{}^{\circ}\text{C}=1073$ K for part (d).

[Figure not reproduced: Ellingham diagram, Fig. 9-3. See the official exam paper or the cited reference text.]

Figure 9-3 (Gaskell) — Ellingham diagram for the oxides used in this question. The $2\text{Mn}+\text{O}_2=2\text{MnO}$ line runs through roughly $-600$ kJ near $900\,{}^{\circ}\text{C}$; the $2\text{Ca}+\text{O}_2=2\text{CaO}$ line is the lowest (most stable oxide) line shown, with its "m" (melting) and "b" (boiling) markers visible around $840\,{}^{\circ}\text{C}$ and $1480\,{}^{\circ}\text{C}$.

Find. (a) $P_{O_2}$ at the Mn/MnO equilibrium, 900°C; (b) CO/CO₂ ratio at the same equilibrium; (c) H₂/H₂O ratio at the same equilibrium; (d) $\Delta G^{\circ}$ for Ca$+$MgO$=$CaO$+$Mg at 800°C; (e)–(f) melting and boiling point of Ca.

Approach. Build each line's $\Delta G^{\circ}(T)=\Delta H^{\circ}-T\Delta S^{\circ}$ from 298 K formation data, evaluate at the stated temperature, then combine the Mn/MnO line with the CO/CO₂ and H₂/H₂O reference lines through the shared $P_{O_2}$ (exactly how the diagram's nomographic scales are constructed).

(a) $2\text{Mn}+\text{O}_2=2\text{MnO}$ line at 900°C. With $\Delta H^{\circ}_f(\text{MnO})=-385.2$ kJ/mol, $S^{\circ}(\text{MnO})=59.71$, $S^{\circ}(\text{Mn})=32.01$, $S^{\circ}(\text{O}_2)=205.1$ J K⁻¹mol⁻¹: $$\Delta H^{\circ}=2(-385{,}200)=-770{,}400\ \text{J},\qquad \Delta S^{\circ}=2(59.71)-[2(32.01)+205.1]=-149.7\ \text{J/K},$$ $$\Delta G^{\circ}_{1173}=-770{,}400-(1173)(-149.7)=\boxed{-594{,}800\ \text{J}}.$$ Since $\Delta G^{\circ}=RT\ln P_{O_2}$ for $2\text{Mn}+\text{O}_2=2\text{MnO}$ (unit activities for the solids): $$P_{O_2}=\exp\!\left(\frac{\Delta G^{\circ}}{RT}\right)=\exp\!\left(\frac{-594{,}800}{(8.314)(1173)}\right)=\boxed{3.25\times10^{-27}\ \text{atm}}.$$
(b) CO/CO₂ ratio. For $2\text{CO}+\text{O}_2=2\text{CO}_2$ ($\Delta H_f^{\circ}(\text{CO})=-110.5$, $\Delta H_f^{\circ}(\text{CO}_2)=-393.5$ kJ/mol, entropies from Question 2): $$\Delta H^{\circ}=-566{,}000\ \text{J},\quad \Delta S^{\circ}=-173.1\ \text{J/K},\quad \Delta G^{\circ}_{1173}=-362{,}950\ \text{J},\quad K_{CO}=\exp\!\left(\frac{-\Delta G^{\circ}}{RT}\right)=1.456\times10^{16}.$$ At the shared $P_{O_2}$ from part (a), $P_{CO_2}/P_{CO}=\sqrt{K_{CO}\,P_{O_2}}=6.88\times10^{-6}$, so $$\frac{P_{CO}}{P_{CO_2}}=\boxed{1.453\times10^{5}}.$$
(c) H₂/H₂O ratio. For $2\text{H}_2+\text{O}_2=2\text{H}_2\text{O(g)}$ ($\Delta H_f^{\circ}(\text{H}_2\text{O,g})=-241.8$ kJ/mol, $S^{\circ}(\text{H}_2)=130.7$, $S^{\circ}(\text{H}_2\text{O,g})=188.8$ J K⁻¹mol⁻¹): $$\Delta H^{\circ}=-483{,}600\ \text{J},\quad \Delta S^{\circ}=-88.9\ \text{J/K},\quad \Delta G^{\circ}_{1173}=-379{,}320\ \text{J},\quad K_{H_2}=\exp\!\left(\frac{-\Delta G^{\circ}}{RT}\right)=7.80\times10^{16}.$$ $$\frac{P_{H_2}}{P_{H_2O}}=\frac{1}{\sqrt{K_{H_2}\,P_{O_2}}}=\boxed{6.280\times10^{4}}.$$
(d) Ca$+$MgO$=$CaO$+$Mg at 800°C. With $\Delta H_f^{\circ}(\text{CaO})=-635.1$, $\Delta H_f^{\circ}(\text{MgO})=-601.6$ kJ/mol and $S^{\circ}$(Ca,CaO,Mg,MgO)$=41.6,38.1,32.7,27.0$ J K⁻¹mol⁻¹: $$\Delta H^{\circ}=(-635{,}100)-(-601{,}600)=-33{,}500\ \text{J},\qquad \Delta S^{\circ}=(38.1+32.7)-(41.6+27.0)=+2.2\ \text{J/K},$$ $$\Delta G^{\circ}_{1073}=-33{,}500-(1073)(2.2)=\boxed{-35.9\ \text{kJ/mol}}.$$ The negative $\Delta G^{\circ}$ confirms Ca metal can reduce MgO at 800°C, consistent with the Ca line lying below the Mg line on the attached diagram at every temperature shown.
(e)–(f) Melting and boiling point of Ca. Read from the "m" and "b" markers plotted on the $2\text{Ca}+\text{O}_2=2\text{CaO}$ line of the attached diagram, which match the standard reference values: melting point $\boxed{842\,{}^{\circ}\text{C}}$, boiling point $\boxed{1484\,{}^{\circ}\text{C}}$.

Quantity	Value
(a) $\Delta G^{\circ}_{Mn/MnO}$(900°C), $P_{O_2}$	−594.8 kJ, 3.25×10⁻²⁷ atm
(b) CO/CO₂ ratio	1.453×10⁵
(c) H₂/H₂O ratio	6.280×10⁴
(d) $\Delta G^{\circ}$, Ca$+$MgO$=$CaO$+$Mg (800°C)	−35.9 kJ/mol
(e) Melting point of Ca	842°C
(f) Boiling point of Ca	1484°C

Concept & Theory

Topic. Ellingham-diagram construction and its nomographic $P_{O_2}$/CO-CO₂/H₂-H₂O reading scales.

Key relations. $\Delta G^{\circ}(T)=\Delta H^{\circ}-T\Delta S^{\circ}$ (near-linear because solid/gas heat capacities largely cancel); $\Delta G^{\circ}=RT\ln P_{O_2}$ for a $2\text{M}+\text{O}_2=2\text{MO}$ line; the CO/CO₂ and H₂/H₂O scales are read by combining the metal line's $P_{O_2}$ with the corresponding gas-oxidation equilibrium.

Why this works. Every line on the diagram is the standard free energy of a $2\text{M}+\text{O}_2=2\text{MO}$ reaction; because all such lines share the same $\text{O}_2$ standard state, a single ruler line from the pivot point $O$ ($\Delta G^{\circ}=0,T=0$) through a reaction line's value at temperature $T$ gives the equilibrium $P_{O_2}$ directly, and further pivots at $H$ and $C$ extend the same construction to the H₂/H₂O and CO/CO₂ scales via the auxiliary water-gas and CO-oxidation equilibria.

Common pitfall. Reporting the CO/CO₂ or H₂/H₂O ratio the wrong way round — a very stable oxide (Mn here, like Cr and Si) needs an extremely REDUCING atmosphere (huge CO/CO₂ and H₂/H₂O ratios) to stay reduced, so ratios of $10^4$–$10^5$ are physically expected, not a sign of an arithmetic slip.

Source. Gaskell, Introduction to the Thermodynamics of Materials, Ch. 13 (Ellingham/Richardson diagrams and their nomographic scales), Fig. 9-3.

Question 7 of 7: Ellingham-Diagram Readings for Mn/MnO and Ca

Question 7: Ellingham-Diagram Readings for Mn/MnO and Ca (20 marks)