Calculate the enthalpy of combustion of ethylene at 300K at constant pressure if its enthalpy of combustion at constant volume is `-"1406 kJ mol"^(-1)`.

Explanation

Let us interpret the information about enthalpy of formation by writing out the equations. It is important to note that the standard enthalpy of formation of pure elemental gases and elements is assumed to be zero under standard conditions. Thermochemical equation for the formation of methane from its constituent elements is,
C(graphite) + `2H_(2(g)) to CH_(4(g))`
`DeltaH_f^0= "X kJ mol"^(-1)` ....(i)
Thermo chemical equations for the combustion of given substances are ,
`H_(2(g)) +1//2 O_2 to H_2O_((l))`
`DeltaH^0=-285.8 "kJ mol"^(-1)` ....(ii)
C (graphite) + `O_2 to CO_2`
`DeltaH^0=-393.5 "kJ mol"^(-1)`....(iii)
`CH_(4(g)) +2O_2 to CO_(2(g)) +2H_2O_((l))`
`DeltaH^0=-890.4 "kJ mol"^(-1)`....(iv)
Since methane is in the product side of the required equation (i), we have to reverse the equation (iv)
`CO_(2(g)) +2H_2O_((l)) to CH_(4(g)) +2O_2`
`DeltaH^0=+890.4 "kJ mol"^(-1)`...(v)
In order to get equation (i) from the remaining ,(i) =[(ii) x 2 ] + (iii) +(v)
X=[(-285.8) x 2 ] + [ -393.5 ]+[+890.4]
=-74.7 kJ
Hence , the amount of energy required for the formation of 1 mole of methane is -74.7 kJ The heat of formation methane =-`74.7 "kJ mol"^(-1)`