Lime Kiln Below the 'Decomposition Temperature'
A lime producer in the Penedès feeds 12.0 t h⁻¹ of limestone (98.0% CaCO3, Mr 100.09) into a rotary kiln at 1100 K, fired with natural gas. Part of the off-gas is recycled, so the CO2 partial pressure inside the kiln is held at 0.30 atm. Thermodynamic data for CaCO3(s) → CaO(s) + CO2(g) are ΔH° = +178.0 kJ mol⁻¹ and ΔS° = +160.5 J K⁻¹ mol⁻¹, taken as constant over 1000–1300 K. The plant engineer reads that decomposition 'requires 1109 K' and orders extra fuel to push the kiln hotter, while the production manager argues that recycling CO2 is saving energy. A winter shutdown is planned at 1050 K. R = 8.314 J K⁻¹ mol⁻¹. The kiln's residence time is about 40 min, and the lime is known to sinter and lose reactivity above about 1250 K.
1. Calculate the threshold temperature for decomposition at P(CO2) = 1 atm and explain the signs of ΔH° and ΔS°. What does this say about the engineer's 1109 K figure?
Read the worked answer
At the threshold, ΔG° = 0, so T = ΔH°/ΔS° = 178000 J mol⁻¹/160.5 J K⁻¹ mol⁻¹ = 1109 K.
Sign analysis: ΔH° > 0 because breaking the carbonate lattice to give a gas plus a solid costs energy; ΔS° > 0 because one mole of dense solid is replaced by one mole of gas (plus a solid), producing a large increase in disorder. Both terms are large, which is why the decomposition is only favourable at high temperature: the −TΔS term must outweigh the positive enthalpy.
At exactly 1109 K and P(CO2) = 1 atm the system is at equilibrium (K = 1). Below that temperature, with pure CO2 at 1 atm, ΔG° > 0 and limestone does not decompose; above it, decomposition is spontaneous under standard conditions. So the figure is a standard-state threshold, valid only when the CO2 partial pressure is 1 atm - it is not a universal 'operating temperature'. The engineer has taken a textbook constant as a plant specification, which is the first error to correct.
2. Calculate ΔG for the decomposition at 1100 K when P(CO2) = 0.30 atm. Is the kiln actually able to calcine limestone?
Read the worked answer
First get ΔG° at 1100 K (assuming ΔH° and ΔS° constant): ΔG° = ΔH° − TΔS° = 178.0 − 1100 × 0.1605 = 178.0 − 176.6 = +1.4 kJ mol⁻¹. So at 1 atm CO2 the reaction would be slightly non-spontaneous, consistent with being just below 1109 K.
Now correct for the actual CO2 pressure: ΔG = ΔG° + RT ln Q, with Q = P(CO2)/p° = 0.30 (solids have unit activity). RT ln 0.30 = 8.314 × 1100 × (−1.204) = −1.10 × 10⁴ J = −11.0 kJ mol⁻¹. ΔG = +1.4 + (−11.0) = −9.6 kJ mol⁻¹.
Negative, so the decomposition is spontaneous at 1100 K with a CO2 partial pressure of 0.30 atm. The engineer's claim that the kiln is 'too cold' is wrong for this atmosphere: the recycled off-gas lowers the CO2 partial pressure and shifts the equilibrium to the right, exactly as Le Chatelier predicts. The driving force is still modest (−9.6 kJ mol⁻¹), so the kiln should not be run far below about 1050 K without checking the equilibrium calculation again.
3. At what temperature does decomposition become favourable with P(CO2) = 0.30 atm? Use this to evaluate the manager's claim about CO2 recycling and the winter 1050 K shutdown.
Read the worked answer
Set ΔG = 0 with the pressure term included: 0 = ΔH° − TΔS° + RT ln Q 0 = 178000 − T(160.5) + T(8.314)(ln 0.30) 0 = 178000 − T(160.5) + T(−10.01) 0 = 178000 − T(170.5) T = 178000/170.5 = 1044 K.
So at 0.30 atm CO2 the threshold falls from 1109 K to about 1044 K, a drop of about 65 K. The manager is right in principle: removing CO2 (or sweeping it away with recycled or fresh gas) lowers the required temperature and saves fuel. But the margin is thin: at the planned 1050 K shutdown, ΔG is only just negative (about −1 kJ mol⁻¹), so any rise in CO2 partial pressure, or a colder spot in the bed, will stop calcination.
Judgement: recycling is beneficial but cannot be pushed indefinitely - CO2 recycling also reduces the O2 available for combustion and raises the CO2 content of the gas. Also, thermodynamics says nothing about speed: at 1050 K the 40 min residence time may be insufficient for coarse limestone, and raising the temperature above 1250 K to compensate would sinter the lime and destroy its reactivity. The correct decision is to stay near 1100–1150 K with a controlled CO2 partial pressure and adequate residence time, rather than either 'more fuel' or 'lower temperature' alone.