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Q.

The standard molar enthalpies of formation of HO (l) and H2O2(l) are -286 and -188 /mol} respectively. Molar enthalpies of vaporization Of H2O (l) and H2O2(l) are 44 and 53 kJ respectively. The bond dissociation enthalpy ofO2 (.g) is 498 kJ/mol. Calculate the bond dissociation enthalpy (in kJ/mol) of 0-0 bond in H2O2 , assuming that the bond dissociation enthalpy of O-H bond is same in both H2O and H2O2 .

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answer is 142.

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Detailed Solution

H2O(l)+12O2(g)H2O2(l)

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H2O(g)+12O2(g)H2O2(g)        ΔH=12×498x

44+249x53=188+286

x= 142

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