Determine the enthalpy change of the following reaction: 2kclo3 -> 2kcl + 3o2 given enthalpies: kclo3: -391.4 kj/mol o2: 0 kj/mol kcl: -436.7 kj/mol

- 90.6 kJ/mol.

Explanation:

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

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

∴ ΔHrxn = [(2*ΔHf(KCl)) + (3*ΔHf(O₂))] - [(2*ΔHf(KClO₃))] = [(2*(- 436.7 kJ/mol)) + (3*(0)] - [(2*(- 391.4 kJ/mol)] = [- 873.4 kJ/mol] - [- 782.8 kJ/mol] = - 90.6 kJ/mol.