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Thermodynamics of materials

For the decomposition of calcium carbonate, CaCO₃(s) → CaO(s) + CO₂(g), take the standard reaction enthalpy as ΔH° = +178 kJ/mol and the standard reaction entropy as ΔS° = +161 J/(mol·K), both independent of temperature. Which value is closest to the temperature at which decomposition starts to proceed spontaneously in the standard state with a CO₂ partial pressure of 1 bar (the temperature at which ΔG° = 0)?

  1. AAbout 833 K
  2. BAbout 1106 K
  3. CAbout 1379 K
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Answer: B. About 1106 K

Setting ΔG° = ΔH° − TΔS° = 0 gives T = ΔH°/ΔS° = 178×10³ J/mol ÷ 161 J/(mol·K) ≈ 1106 K, which is about 833℃ on the Celsius scale. Above this temperature ΔG° < 0, and decomposition proceeds even with a CO₂ partial pressure of 1 bar. About 833 K takes the Celsius value, obtained by subtracting 273 from the absolute temperature, and attaches the unit K to it. About 1379 K is the value obtained by adding 273 instead. Dividing ΔH° in kJ by ΔS° in J gives about 1.1 K, so it is also important to make the units consistent. In the actual calcining of limestone, lowering the CO₂ partial pressure lets decomposition proceed at lower temperatures.

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