Given that (at `25^(@)C)` :
`(1)/(2)H_(2)(g)+(1)/(2)O_(2)(g) to OH(g), Delta H^(0)= 38.95 " kJ mol"^(-1)`
`H_(2)(g)+(1)/(2)O_(2)(g) to H_(2)O (g), Delta H^(0) = -241.8 " kJ mol"^(-1)`
`H_(2)(g) to 2H(g), Delta H^(0)= 436.0 " kJ mol"^(-1)`
`O_(2)(g) to 2O (g), Delta H^(0)= 498.3 " kJ mol"^(-1)`
Calculate `Delta H^(0)` for the following reaction -
`(a) OH(g) to H(g)+O(g)`
(b) `H_(2)O(g) to 2H(g)+O(g)`

Explanation

`(i) (1)/(2)xx eq.(3)+(1)/(2)xx eq. (4)- eq(1).` we have, `OH(g) to H(g)+O(g)`
`Delta H^(0)=[((1)/(2)xx436+(1)/(2)xx498.3)-1xx 38.95] kJ`
`= 428.2 kJ`
(ii) `1 xx eq. (3)+(1)/(2)xx eq. (4)-1 xx eq (2)`, we have `H_(2)O(g)to 2H(g)+O(g)`
`Delta H^(0)= [1xx (436.0)+(1)/(2)(498.3)-(-241.8)] kJ`
= 926.95 kJ