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

Options

A. `-74.8 kJ mol^(-1)`
B. `-52.27 kJ mol^(-1)`
C. `+74.8 kJ mol^(-1)`
D. `+52.26 kJ mol^(-1)`

Explanation

(i) `CH_(4(g))+2O_(2(g))rarrCO_(2(g)+2H_(2)O_((g))`
`" "DeltaH=-890.3 kJ"mol^(-1)`
(ii) `C_((s))+O_(2(g))rarrCO_(2(g))`
`" "DeltaH=-393.5kJ "mol"^(-1)`
(iii) `2H_(2(g))_O_(2(g))rarr2H_(2)O_((g))`
`" " DeltaH=-285.8 kJ "mol"^(-1)`
Thus, the desired equation is the one that represents the formation of `CH_(4(g))` i.e.,
`C_((s))+2H_(2(g))rarrCH_(4(g))`
`Delta_(f)H_(CH_(4))=Delta_(c)H_(c)+2Delta_(c)H_(H_(2))-Delta_(c)H_(cO_(2))`
`=[-393.5+(-285.8)-(-890.3)]kJ "mol"^(-1)`
`=-7.48 kJ"mol^(-1)`
`:.` Enthalpy of formation of `CH_(4(g))=-74.8kJ"mol"^(-1)`
Hence,alternative (i) is correct.