With the help of thermochemical equations given below, determine `Delta_(r )H^(Θ)` at `298 K` for the following reaction:
`C("graphite")+2H_(2)(g) rarr CH_(4)(g),Delta_(r )H^(Θ) = ?`
`C("graphite")+O_(2)(g) rarr CH_(2)(g),
Delta_(r )H^(Θ) = -393.5 kJ mol^(-1)` ...(1)
`H_(2)(g) +1//2O_(2)(g) rarr H_(2)O(l)`,
`Delta_(r )H^(Θ) = -285.8 kJ mol^(-1)` ...(2)
`CO_2(2)(g)+2H_(2)O(l) rarr CH_(4)(g)+2O_(2)(g)`,
`Delta_(r )H^(Θ) = +890.3 kJ mol^(-1)` ...(3)

Explanation

Here we want one mole of `C_(("graphite"))` as reactant, so we write down eauation (i), we want two mole of `H_(2)(g)` as reactant, so we multiply equation (ii) by `2`, we want one mole of `CH_(4)(g)` as product, so we write down equation (iii) as such,
`C_(("graphite"))+O_(2)(g) rarr CO_(2)(g), Delta_(r )H^(Θ)=-393.5 kJ mol^(-1)` ...(i)
`2H_(2)(g)+O_(2) rarr 2H_(2)O(l), Delta_(r)H^(Θ)=2(-285.8 kJ mol^(-1))` ...(ii)
`CO_(2)(g)+2H_(2)O(l) rarr CH_(4)(g)+2O_(2)(g), Delta_(r)H^(Θ)=+890.3 kJ mol^(-1)` ...(iii)
Adding we obtain
`C_(("graphite"))+2H_(2)(g) rarr CH_(4)(g), Delta_(r)H^(Θ)=-74.8 kJ mol^(-1)`