The enthalpies of combustion of `C_(2)H_(4)(g), C_(2)H_(6)(g)` and `H_(2)` are - 1409.5 kJ, -1558.3 kJ and -285.6 kJ, respectively.Caalculate the enthalpy of hydrogenation of ethylene.

Answer

`-136.8 kJ "mol"^(-1)`

Explanation

(i)`C_(2)H_(4)(g) + 3O_(2)(g) to 2CO_(2)(g) + 2H_(2)O(l)`
`DeltaH = -1409.5 kJ`
(ii) `C_(2)H_(6)(g) + 7/2O_(2)(g) to 2CO_(2)(g) + 3H_(2)O(l) `
`DeltaH = -1558.3 kJ`
(iii)` H_(2)(g) + 1/2O_(2)(g) to H_(2)O(l) DeltaH = -285.6 kJ`
The required equation is
`C_(2)H_(4)(g) + H_(2)(g) to C_(2)H_(6)(g) DeltaH = ?`
Add eq. (i) and (iii)
(iv) `C_(2)H_(4)(g) + H_(2)(g)+ 7/2O_(2)(g) to 2CO_(2)(g) + 3H_(2)O(l) DeltaH = -1695.1 kJ`
Subtract eq.(ii) from eq.(iv)
(ii) `C_(2)H_(6)(g) + 7/2O_(2)(g) to 2CO_(2)(g) + 3H_(2)O(l)`
`DeltaH = -1558.3 kJ`

Subtracting `C_(2)H_(4)(g) + H_(2)(g) to C_(2)H_(6)(g)`
`DeltaH = -136.8 kJ "mol"^(-1)`.