Question

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

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

kJ
A 2-column table with 6 rows. Column 1 is labeled Compound with entries upper C a upper O solid, Upper C a upper C upper O subscript 3 solid, Upper C subscript 6 upper H subscript 6 gas, Upper C upper O subscript 2 gas, Upper H subscript 2 upper O gas, Upper N upper O subscript 2 gas. Column 2 is labeled Delta H subscript f in kilojoules per molecule with entries negative 157.3, negative 1,207.1, 82.90, negative 393.5, negative 241.82, 33.84.

Answer 1

Use enthalpies of formation:

ΔH°rxn = ΣΔH°f(products) − ΣΔH°f(reactants)
= [ΔH°f(CaO) + ΔH°f(CO2)] − ΔH°f(CaCO3)
= [−157.3 + (−393.5)] − (−1207.1)
= −550.8 + 1207.1
= +656.3 kJ

So the decomposition is endothermic by 6.56×10^2 kJ per mole of CaCO3 (≈656 kJ·mol−1).