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.

Question

Revolt · Accepted Answer

160.3 J/(molK)

Unit Test — 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.