Combustion enthalpy of carbon, Hydrogen and Methane at 25°C temperature are `395.5` KJ mol`""^(1-) , -284.8` kJ mol`""^(-1) and -890.4 "kJ mol"^(-1)` respectively. Calculate the standard enthalpy of formation of methane for the same temperature.

Options

A. `890.4` Kj mol`""^(-1)`
B. `-298.8` KJ mol`""^(-1)`
C. `-74.7` KJ mol`""^(-1)`
D. `-107.7` KJ mol`""^(-1)`

Answer

C

Explanation

Combusion reaction of Carbon :
`C_((s) ) + O_(2(g)) to CO_(2(g))`
`Delta H_(1) = -395.5 "KJ mol"^(-1)`
Combusion reaction of Hydrogen :
`H_(2(g)) + (1)/(2) O_(2(g)) to H_(2) O_((l))`
`Delta H_(2) = -284.8 "KJ mol"^(-1)`
Combusion of Methane :
`CH_(4(g)) + 2O_(2(g)) to CO_(2(g)) + 2H_(2) O_((l))`
`DeltaH_(3) = -890.4 "KJ mol"^(-1)`
Reaction of formation of Methane :
`C_((s)) + 2H_(2(g)) to CH_(4(g)) Delta H= (?)`
`=-395.5 -2 xx 284.8 - 890.4`
`=-74.7 "KJ mol"^(-1)`