Answer
496 kJ/mol
Explanation
∆Hrxn = ∆HB (reactants) - ∆HB(products)
∆Hrxn = ( 2 × ∆HH-H + 1 × ∆HO=O) - ( 2 × ∆HH-H + 1 × ∆HO=O)
- 484kJ/mol = (2 × 436kJ/mol + 1 × ∆HO=O) - ( 4× 463kJ/mol)
- 484kJ/mol =( 872kJ/mol + ∆HO=O) - 1852 kJ/mol
∆HO=O = 496 kJ/mol
The enthalpy change for the following reaction is -484 kJ. Using bond energies, estimate the O-O...
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