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

The complete ethylene combustion reaction can be written as,
`C_(2)H_(4(g))+3O_(2(g))rarr 2CO_(2(g))+2H_(2)O_((l))`
`DeltaH=DeltaE+RT Deltan_((g))`,
where `Deltan_((g))=n_(p(g))-nP_(r(g))`.
`therefore" "Deltan_((g))=2-(3+1)=-2`.
Enthalpy of combustion at constant volume `=DeltaE=-"1406 kJ mol"^(-1)`
`therefore` Overall enthalpy of combustion
`=DeltaH_(c)=-1406+(-2xx8.134xx10^(-3)xx300)`
`=-1406-4.9884`
`DeltaH_(c )=-1410.9kJ mol^(-1)`.