Calculate the ethalpy of combustion of ethylene (g) to form `CO_(2)` (gas) and `H_(2)O` (gas) at 298 K and 1 atmospheric pressure. The enthalpies of formation of `CO_(2),H_(2)O` and `C_(2)H_(4)` are `-393.7, -241.8+52.3` kJ per mole respectively.

Explanation

We are given `:`
(i) `C(s) +O_(2)(g) rarr CO_(2)(g), DeltaH^(@) = - 393.5kJ mol^(-1)`
(ii) `H_(2)(g) + (1)/(2) O_(2)(g) rarr H_(2)O(g), Delta H^(@) = - 241.8 kJ mol^(-1)`
(iii) `2C(s) + 2H_(2)(g) rarr C_(2)H_(4)(g), Delta H^(@)= + 52.3 kJ mol^(-1)`
We aim at `:` `C_(2)H_(4)(g) + 3O_(2)(g) rarr 2CO_(2)(g) + 2H_(2)O(g) , Delta _(c ) H^(@) = ?`
` 2 xx` Eqn. `(i) + 2 xx ` Eqn. (ii) - Eqn. (iii) gives
`ul{(" "2C(s) ,+2O_(2)(g)rarr2CO_(2)(g)+O_(2)(g),+2H_(2)(g)+2H_(2)O(g)),(-2C(s),,-2H_(2)(g)-C_(2)H_(4)(g)):}`
`3O_(2)(g)rarr 2CO_(2) +2H_(2)(g)-C_(2)H_(4)(g)`
or `C_(2)H_(4)(g)+3O_(2)(g) rarr 2CO_(2)(g) + 2H_(2)O(g)`
`Delta_(c) H^(@) = 2 ( -393.5) + (-241.8) - (52.3) = -1322.9 kJ mol^(-1)`
Alternative Method ` :`
We aim at `:` `C_(2)H_(4)(g)+3O_(2)(g) rarr 2CO_(2)(g)+2H_(2)O(g)`
We are given `:` `Delta_(f) H_((CO_(2)))^(@) = - 393.5 kJ mol^(-1)`
`Delta_(f) H_((H_(2)O))^(@)= - 241.8 kJ mol^(-1)`
`Delta _(f) H_((C_(2)H_(4)))^(@)= + 52.3 kJ mol^(_1)`
`Delta_(r)H^(@) = ("Sum of " Delta_(f)H^(@) " values of Products")-("Sum of" Delta_(f)H^(@)" values of Reactants")`
`=[ 2 xx Delta_(f) H^(@) (CO_(2))+2xxDelta_(f)H^(@)(H_(2)O)]-[Delta_(f)H^(@)(C_(2)H_(4))+3xxDelta_(f)H^(@)(O_(2))]`
`=[ 2 xx ( -393.5) + 2 xx ( -241.8) ] - [ ( 52.3) + 0] ( :' Delta_(f) H^(@)` for elementary subtance `=0 )`
`=[ -787 .0 -483.6 ] - 52.3 = - 1322.9 kJ mol^(-1)`