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Calculate the ethalpy of combustion of e...

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.

Text Solution

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Give `Delta_(f)H_((Co_(2)))^(0)=-393.5KJmol^(-1)`
`Delta_(f)H_((H_(2)O))^(0)=-241.8KJmol^(-1)`
`Delta_(f)H_((C_(2)H_(4)))^(0)=+52.3KJmol^(-1)` we need,
`C_(2)H_(4(g))+3O_(2(g))rarr2CO_(2(g))+2H_(2)O_((g))Deltaunderset(r)(H^(0))=?`
`Deltaunderset(r)(H^(0))=sumDelta_(f)H^(0)` of products - `sumDelta_(f)H^(0)` reactants
`=[2xxDelta_(f)H^(0)(CO_(2))+2xxDelta_(f)H^(0)(H_(2)O)]`
`[Delta_(f)H^(0)(C_(2)H_(4))+3xxDelta_(f)H^(0)(O_(2))]`
`=[2xx(-393.5)+2xx(-241.8)]-[52.3+0]`
`because Delta_(f)H^(0)` Element = 0
`therefore` Enthalpy of combustion of `C_(2)H_(4)` is `Deltaunderset(c)(H)=-1322.9KJmol^(-1)`
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