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

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.

What is the question mate