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Chemistry · Glossary

What is Inversion temperature?

Definition 2.19 University Chemistry — Year 2 · Chapter 2 — Entropy and Free Energy of Reaction

When ΔrH∘\Delta_r H^\circ and ΔrS∘\Delta_r S^\circ have the same sign, the inversion temperature of the reaction is the temperature where ΔrG∘\Delta_r G^\circ changes sign; in the Ellingham approximation, Ti=ΔrH∘/ΔrS∘T_i = \Delta_r H^\circ/\Delta_r S^\circ.

The enthalpy and entropy terms of CaCO3(s) -> CaO(s) + CO2(g) in the Ellingham approximation. Below the inversion temperature the enthalpy cost wins and limestone is stable under 1\, bar of carbon dioxide; above it the entropy term wins and limestone decomposes.
The enthalpy and entropy terms of CaCOX3(s)→CaO(s)+COX2(g)\ce{CaCO3(s) -> CaO(s) + CO2(g)} in the Ellingham approximation. Below the inversion temperature the enthalpy cost wins and limestone is stable under 1 bar1\,\mathrm{bar} of carbon dioxide; above it the entropy term wins and limestone decomposes.
The four cases of _r G = _r H - T _r S (Ellingham approximation). Combustions are in the upper left box; the decomposition of limestone and the cold pack in the upper right; the synthesis of ammonia in the lower left.
The four cases of ΔrG∘=ΔrH∘−TΔrS∘\Delta_r G^\circ = \Delta_r H^\circ - T\Delta_r S^\circ (Ellingham approximation). Combustions are in the upper left box; the decomposition of limestone and the cold pack in the upper right; the synthesis of ammonia in the lower left.

Examples

Example 2.21 (Two tests of the approximation)

For limestone, ΔrCp∘=42.80+37.13−81.88=−2.0 J/(K mol)\Delta_r C^\circ_p = 42.80 + 37.13 - 81.88 = -2.0\,\mathrm{J}/(\mathrm{K}\,\mathrm{mol}): the approximation is excellent, and ΔrG∘=178.3−T×0.1606\Delta_r G^\circ = 178.3 - T \times 0.1606 gives Ti=1110 KT_i = 1110\,\mathrm{K}. For ammonia, ΔrG∘(700)≈−91.9+700×0.1981=+46.8 kJ/mol\Delta_r G^\circ(700) \approx -91.9 + 700 \times 0.1981 = +46.8\,\mathrm{kJ}/\mathrm{mol}, while the tables give 2ΔfG∘(NHX3,700)=+54.4 kJ/mol2\Delta_f G^\circ(\ce{NH3}, 700) = +54.4\,\mathrm{kJ}/\mathrm{mol}: here ΔrCp∘=−44 J/(K mol)\Delta_r C^\circ_p = -44\,\mathrm{J}/(\mathrm{K}\,\mathrm{mol}) and the error is 8 kJ/mol8\,\mathrm{kJ}/\mathrm{mol}, a factor of 4 on the equilibrium constant at that temperature.

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