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. Enthalpy 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)`

Answer

A

Explanation

`CH_(4)(g)+2O_(2)(g) rarr CO_(2)(g)+2H_(2)O(l) Delta H_(1)^(0)`
`{:(C(s)+O_(2)(g) rarr CO_(2)(g),DeltaH_(2)^(0)),(H_(2)(g)+1/2O_(2)(g) rarr H_(2)O(l),Delta H_(5)^(0)):}/((-Delta H_(1)^(0))+(Delta H_(2)^(0))+2(Delta H_(3)^(0)))`
`(890.3) + (-3.93.5) + 2 (-258.8) = -74.8 kJ//mol`.