With the help of a thermochemical equation, calculate `Delta_(f)H^(@)` at 298 K for the following reactions :
`"C(graphite)" + O_(2)(g) rarr CO_(2)(g) , Delta H = -393.5 kJ//mol`
`H_(2)(g) + (1)/(2)O_(2)(g) rarr H_(2)O(l) , Delta_(f)H^(@) = -285.8 kJ//mol`
`CO_(2)(g) + 2H_(2)O(l) rarr CH_(4)(g) + 2O_(2)(g) , Delta_(f)H^(@) = +890.3 kJ//mol`

Explanation

`"C(graphite)" + O_(2)(g) rarr CO_(2)(g), Delta_(f)H^(@) = -393.5 kJ//mol` ….(1)
`H_(2)(g) + (1)/(2)O_(2)(g) rarr H_(2)O(l), Delta_(f)H^(@) = -285.8 kJ//mol` …(2)
`CO_(2)(g) + 2H_(2)O(l) rarr CH_(4)(g) + 2O_(2)(g) Delta_(f)H^(@) = +890.3 kJ//mol` ...(3)
Here as we want one mole of C(graphite) as reactant, we write it down as it is.
`"C(graphite)" + O_(2)(g) rarr CO_(2)(g) , Delta_(f)H^(@) = 393.5 kJ//mol` ...(1)
`2H_(2)(g)+O_(2)(g)rarr 2H_(2)O(l) , Delta_(f)H^(@) = 2(-285.8 kJ//mol) = -571.6 kJ//mol` ...(2)
`CO_(2)(g) + 2H_(2)O(l) rarr CH_(4)(g)+2O_(2)(g), Delta_(f)H^(@) = +890.3 kJ//mol` ...(3)
Adding 1,2 and 3 we obtain :
`"C(graphite)" + 2H_(2)O(l) rarr CH_(4)(g) , Delta_(f)H^(@) = -74.8 kJ//mol`