For the reaction:
`2H_(2)(g) +O_(2)(g) rarr 2H_(2)O(g), DeltaH =- 571 kJ`
bond enegry of `(H-H) = 435 kJ` and of `(O=O) = 498 kJ`. Then, calculate the average bond enegry of `(O-H)` bond using the above data.

Options

A. 484
B. `-484`
C. 271
D. `-271`

Answer

A

Explanation

`2H_(2) + O_(2) rarr 2H_(2)O, Delta H= -571`
`Delta H_("reaction") = 2 xx B.E (H_(2)) + B.E (O_(2))- 2 xx B.E (H_(2)O)`
`therefore -571= (2 xx 435) + 498 - 2 xx B.E (H_(2)O) or 2 xx B.E (H_(2)O)= 870+ 498+ 571`
`B.E (H_(2)O)= (1939)/(2)= 969.5`
`B.E(O- H)= (969.5)/(2)= 484.75`