At `25^(@)C` and 1 atm, heat of formation of `C_(2)H_(4) (g), CO_(2)(g) and H_(2)O(l)` are `52 " kJ mol"^(-1), -394 " kJ mol"^(-1) and -286 " kJ mol"^(-1)` respectively. Calculate the heat of combustion of `C_(2)H_(4)(g)`.

Explanation

`2C("graphite,s")+2H_(2)(g) to C_(2)H_(4)(g), Delta H^(0)=52 kJ " mol"^(-1)`
`C("graphite,s")+O_(2)(g)to CO_(2)(g), Delta H^(0)=- 394 " kJ mol"^(-1)`
`H_(2)(g)+(1)/(2)O_(2)(g) to H_(2)O(l), Delta H^(0)= - 286 " kJ mol"^(-1)`
`2 xx "equation" [2]+2xx "equation"[3] - "equation"[1]`, we have `C_(2)H_(4)(g)+3O_(2)(g) to 2CO_(2)(g)+2H_(2)O(l)`
`Delta H^(0)+ [2(-394)+2 (-286)-52]kJ= -1412 kJ`