1.

For the reaction, 2H_(2)O(g) rarr 2H_(2)(g) + O_(2)(g), Delta H = 571.6 kJ Delta_(f) H^(theta) of water is:

Answer»

285.8kJ
`-285.8kJ`
1143.2kJ
`-1143.2kJ`

SOLUTION :`Delta_(f) H (H_(2)O)` MAY be written as
`H_(2)(G) + (1)/(2) O_(2)(g) rarr H_(2)O(g)`
`Delta_(f)H (H_(2)O) = - 571.6/2 = - 285.8 KJ`


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