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Enthalpy of combustion of H2 to form H20 at 298K is -298KJ/mole. Bond enthalpies of H-H and O=O are 433KJ/mole and 492KJ/mole respectively. Find bond enthalpy of O-H ?

Enthalpy of combustion of H2 to form H20 at 298K is -298KJ/mole. Bond enthalpies of H-H and O=O are 433KJ/mole and 492KJ/mole  respectively. Find bond enthalpy of O-H ?

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Grade:12th pass

4 Answers

Naveen Kumar
askIITians Faculty 60 Points
9 years ago
The reaction can be given as:
H2....+......(1/2)O2..................................>......H2O.......+......298kJ/mol
In this reaction, for combustion of 1 mole of H2, one mole of H-H bond and ½ mole of O-O bond will be broken and 2 mole of O-H bond will be formed.The braeking of bond is endothermic process and formation of bond is exothermic process.
applying energy conservation,
433KJ....+....(1/2)*492kJ= 298KJ+2*EO-H
EO-H=(433+246-298)*1/2 kJ
Damini Parihar
10 Points
9 years ago
answer is not matching , answr given is B 
Damini Parihar
10 Points
9 years ago
answer is not matching , answr given is B 
Naveen Kumar
askIITians Faculty 60 Points
9 years ago
Dear student just replace the sign of the energy released because it is exothermically released. what you will get is also bond energy released in the formation of O-H bond.
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