Table 3. Thermodynamic Cycles Used to Calculate Formation Enthalpies (ΔH°f.ox) of MLi2Ti6O14 at 25 °C.
| reaction | ΔH (kJ/mol) |
|---|---|
| MLi2Ti6O14(S,25°C) → MO(soln,25°C) + Li2O(soln,25°C) + 6TiO2(soln,25°C) | ΔH1 = ΔHds (MLi2Ti6O14) |
| Li2O(S,25°C) → Li2O(soln,25°C) | ΔH2 = −77.21 ± 2.4423 |
| Na2O(S,25°C) → Na2O(soln,25°C) | ΔH4 = −195.90 ± 4.2327 |
| TiO2(S,25°C) → TiO2(soln,25°C) | ΔH3 = 73.70 ± 0.3928 |
| SrO(s,25°C) → SrO(soln,800°C) | ΔH5 = −127.48 ± 1.84a |
| BaO(s,25°C) → BaO(soln,800°C) | ΔH6 = −181.22 ± 2.32a |
| PbO(s,25°C) → PbO(soln,25°C) | ΔH7 = −14.34 ± 0.38a |
| MO(s,25°C) + Li2O(s,25°C) + 6TiO2(s,25°C) → MLi2Ti6O14(s,25°C) | ΔH8 = ΔH°f,ox (MLi2Ti6O14) |
| ΔH8 = ΔH°f(Na2Li2Ti6O14) = −ΔH1 + ΔH2 + 6ΔH3 + ΔH4 | |
| ΔH8 = ΔH°f(SrLi2Ti6O14) = −ΔH1 + ΔH2 + 6ΔH3 + ΔH5 | |
| ΔH8 = ΔH°f(BaLi2Ti6O14) = −ΔH1 + ΔH2 + 6ΔH3 + ΔH6 | |
| ΔH8 = ΔH°f(PbLi2Ti6O14) = −ΔH1 + ΔH2 + 6ΔH3 + ΔH7 | |
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