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

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Chemistry

24 September, 01:24

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

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Tripp David · Answer 1

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.