Question

Given: 2H2O2 → 2H2O + O2 structure of H2O2: H–O–O–H Bond Bond Energy (kJ/mol) O–H 459 O=O 494 O–O 142 Based on the given bond energies, what is the enthalpy change for the chemical reaction? A. -352 kJ B. -210 kJ C. -176 kJ D. -105 kJ

Answer 1

Answer:

B. - 210 kJ

Explanation:

∵ ΔHrxn = ∑(bond energies)products - ∑(bond energies)reactants.

• The bond formation in the products releases energy (exothermic).

• The bond breaking in the reactants requires energy (endothermic).

The products:

• H₂O contains 2 O-H (- 459 kJ/mol) bonds.

• O₂ contain 1 O=O (- 494 kJ/mol) bond.

The reactants:

• H₂O₂ contain 2 O–H (459 kJ/mol) bonds and 1 O–O (142 kJ/mol) bond.

∵ ΔHrxn = ∑(bond energies)products - ∑(bond energies)reactants.

∴ ΔHrxn = [2 (2 x (O–H bond energy) + (1 x (O=O bond energy)] - 2 [(2 x (O–H bond energy) + (1 x (O–O bond energy)] = [2 (2 x - 459 kJ/mol) + (1 x - 494 kJ/mol)] - 2 [(2 x 459 kJ/mol) + (1 x 142 kJ/mol)] = (- 2330 kJ) + (2120 kJ) = - 210 kJ.