What is the enthalpy of reaction for the
decomposition of calcium carbonate?
caco3(s) → cao(s) + co2(g)
kj

The enthalpy of decomposition is  656.3  KJ/mol.

Given that the enthalpy of decomposition is obtained from;

ΔHrxn = ∑ΔHf(products) -  ∑ΔHf(reactants)

Now, the species involved in the reaction are;

ΔHf(CaO)(s)) = -157.3 KJ/mol

ΔHf((CO2)(g)) = -393.5  KJ/mol

ΔHf(CaCO3)(s)) = -1207.1 KJ/mol

Substituting values;

ΔHrxn = [(-393.5) + ( -157.3)] - ( -1207.1) KJ/mol

ΔHrxn =(( -550.8) + 1207.1)  KJ/mol

ΔHrxn = 656.3  KJ/mol

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Explanation:

The formula for calculating the enthalpy change of a reaction by using the enthalpies of formation of reactants and products is:

                         CaCO₃(s) ⟶ CaO(s) + CO₂(g)

ΔH°f/kJ·mol⁻¹:    -1207.1          -157.3    -393.5

fe 55.85 g/mol and h2o 18.02 g/mol.