For the reaction, `2H_(2)+O_(2) to 2H_(2)O,DeltaH=571.` bond energy of H-H=435, O=O=498, then calculate the average bond energy of O-H bond using the above data
A. 484
B. `-484`
C. `271`
D. `-271`

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1 Answers

Correct Answer - A
`DeltaH_(("reaction"))=sum` bond energies of reactants `-sum`bond energies of produts.
Given, `2H_(2)+O_(2)to2H_(2)O,DeltaH=-571`
`H-H=435,O=O` is 498, O-H is?
`DeltaH=(2xx"B E of "H_(2)+" B E of "O_(2))-2("B E of "H_(2)O)`
or `-571=(2xx435+498)-2xx`B E of `H_(2)O`
or `-571=870+498-2xx`Be of `H_(2)O`
or `2BE` of `H_(2)O =870+498+571=1939`
`therefore`BE of 1 `H_(2)O` molecule`=(1939)/(2)=969.5`
`therefore `BE of 1 O-H bond `=(969.5)/(2)`
(`because` `1H_(2)O` molecule has 2O-H bonds)=484.75

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