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The enthalpy for the following reaction ...

The enthalpy for the following reaction (at `25^(@)C` are given below):
(i) `½H_(2(g)) + ½ O_(2(g)) rarr OH_((g))` = 10.06 Kcal
(ii) `H_(2(g)) rarr 2H_((g))` 104.18 Kcal
(iii) `O_(2(g)) rarr 2O_((g))` 118.32 Kcal
Calculate the O-H bond energy in the hydroxyl radical (give your answer in terms of `10^(5)` cal)

Text Solution

Verified by Experts

The correct Answer is:
1
`(1)/(2)H_(2) rarr (1)/(2) O_(2) rarr H, 10.06`
`(1)/(2) H_(2) rarr H, (104.18)/(2) = 52.09`
`(1)/(2) O_(2) rarr O, (118.32)/(2) = 59.16`
`O + H rarr OH, Delta H = 10.06 - 52.09 - 59.16`
`= -101.19` kcal
Bond energy `= 1.0119 xx 10^(5)` cal
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