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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

Respuesta :

superman1987

Answer:

- 90.6 kJ/mol.

Explanation:

  • The enthalpy change of the reaction (ΔHrxn) is the difference between the sum of heat of formation of products and reactants.

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

  • For the reaction: 2KClO₃ → 2KCl + 3O₂,

∴ 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.

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