From the data given below at 298 K for the reaction :
`CH_(4)(g) + 2O_(2)(g) rarr CO_(2)(g) + 2H_(2)O(l)`
Calculate the enthalpy of formation of `CH_(4)(g)` at 298 K.
Enthalpy of reaction is = -893.5 kJ
Enthalpy of formation of `CO_(2)(g) = 393. kJ mol^(-1)`
Enthalpy of formation of `H_(2)O(l) = 286.0 kJ mol^(-1)`.

Explanation

`Delta H = Delta H_(f)CO_(2)(g) + 2Delta H_(f)H_(2)O(l) - Delta H_(f)CH_(4)(g) - Delta H_(f)O_(2)(g)`
`-890.5 kJ = -393.5 kJ + 2 xx -286 kJ - Delta H_(f)CH_(4)(g) - 0`
`Delta H_(f)CH_(4) = -75.0 kJ`.