Question

What is the enthalpy of reaction for the decomposition of calcium carbonate?

CaCO3(s) Right arrow. CaO(s) + CO2(g)

kJ

Answer 1

Use standard enthalpies of formation (ΔHf° at 298 K):

ΔHf°(CaCO3, s) = −1206.9 kJ·mol−1
ΔHf°(CaO, s) = −635.5 kJ·mol−1
ΔHf°(CO2, g) = −393.5 kJ·mol−1

ΔH°rxn = [ΔHf°(CaO) + ΔHf°(CO2)] − ΔHf°(CaCO3)
= (−635.5 + −393.5) − (−1206.9)
= +177.9 kJ·mol−1 ≈ +178 kJ·mol−1

So the decomposition is endothermic, requiring about 178 kJ per mole of CaCO3 (at 298 K; values vary slightly with data source and temperature).