The enthalpy of combustion of methane, graphite and dihydrogen at 298K are -890.3 kJ `mol^(-1)` ,- 393.5 kJ `mol^(-1)` and - 285.8 kJ `"mol"^(-1)` respectively. Enthalpy of formation of `CH_(4(g))` will be ____

Explanation

`CH_4 + 3O_2 to CO_2 +2H_2O " " DeltaH_1=-890.3 "kJ mol"^(-1)` …(1)
`C+O_2 to CO_2 " " DeltaH_2=-"393.5 kJ mol"^(-1)` ….(2)
`H_2 + 1//2 O_2 to H_2O " " DeltaH_3=-"285.8 kJ mol"^(-1)`…(3)
Equation (1) is reversed
Equation (3) x 2
`cancel(CO_2) + cancel(2H_2O) to CH_4 +cancel(2O_2) " " DeltaH_1=+890.3 "kJ mol"^(-1)`
`C+cancel(O_2) to cancel(CO_2) " " DeltaH_2 =-393.5 "kJ mol"^(-1)`[Eq.(2) as such]
`2H_2+cancel(O_2)to cancel(2H_2O) " " DeltaH_3=-571.6 "kJ mol"^(-1)` [Eq.(3) x 2 ]
`C+2H_2 to CH_4 " " DeltaH_f ` of methane =`DeltaH_1+DeltaH_2+DeltaH_3`
`DeltaH_f=+890.3-965.1=-74.8 "kJ mol"^(-1)`