The standard molar enthalpies of formation of H 2 O(l) and H 2 O 2 (l) are -285 and -187 kJ/mol, respectively. Molar enthalpies of vaporization of H 2 O and H 2 O 2 are 44 and 51 kJ, respectively. The

Question

The standard molar enthalpies of formation of H 2 O(l) and H 2 O 2 (l) are -285 and -187 kJ/mol, respectively. Molar enthalpies of vaporization of H 2 O and H 2 O 2 are 44 and 51 kJ, respectively. The bond dissociation enthalpy of O 2 (g) is 496 kJ/mol. Calculate the bond dissociation enthalpy (in kJ/mol) of O-O bond in H 2 O 2 , assuming that the O-H bond enthalpy is the same in both H 2 O and H 2 O 2 .

TestHub · Accepted Answer

Δ f H(H 2 O, g) = -285 + 44 = -241; Δ f H(H 2 O 2 , g) = -187 + 51 = -136 kJ/mol. For H 2 O 2 (g) → H 2 O(g) + 1/2O 2 (g): ΔH = -241 + 136 = -105 kJ. The O-H terms cancel: x - 496/2 = -105, so x = 248 - 105 = 143 kJ/mol. The answer is 0143.

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