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.

Combustion enthalpy of carbon, Hydrogen and Methane at 25°C temperatu

1.	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.
Answer» `890.4` Kj mol`""^(-1)` `-298.8` KJ mol`""^(-1)` `-74.7` KJ mol`""^(-1)` `-107.7` KJ mol`""^(-1)` Solution :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)`