The decomposition of calcium carbonate, CaCO3(s) CaO(s) + CO2(g), has the following values for free energy and enthalpy at 25.0°C. G = 130.5 kJ/molH = 178.3 kJ/molWhat is the entropy of the reaction? Use G = H – TS.

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160.3 J/(molK)

Unit Test

The decomposition of calcium carbonate, CaCO3(s) CaO(s) + CO2(g), has the following values for free energy and enthalpy at 25.0°C. G = 130.5 kJ/molH = 178.3 kJ/molWhat is the entropy of the reaction? Use G = H – TS.