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)) to CO_(2(g)) + 2H_(2) O _((l)), DeltaH= -890.3` kJ/mol
`C_((s)) + O_(2(g)) to CO_(2(g)), DeltaH = -393.5` kJ/mol
`H_(2(g)) + (1)/(2) O_((g)) to H_(2) O_((l)) , DeltaH = -285.8` kJ/mol
`DeltaH= [2DeltaH_(H_(2) O)^(@) + DeltaH_(CO_(2) ) ^(@) ]-[DeltaH_(CH_(4) )^(@) ]`
`-890.3 = [2(-285.8)+ (-393.5)]-[DeltaH_(CH_(4) ) ]`
`DeltaH_(CH_(4) ) = -74.8` kJ/mol