The heats of combustion of `C_(2)H_(4(g)), C_(2)H_(6(g))` and `H_(2(g))` are `-1405,-1558.3` and `-285.6 kJ` respectively. Calculate heat of hydrogenation of ethylene.

Explanation

We have to find
`C_(2)H_(4)(g) +H_(2)(g) rarr C_(2)H_(6)(g), DeltaH = ?`
Givne, `C_(2)H_(4)(g) +3O_(2)(g) rarr 2CO_(2)(g)+ 2H_(2)O(g), DeltaH =- 1409.5 kJ …(i)`
`C_(6)H_(6)(g) +7//2O_(2)(g) rarr 2CO_(2)(g) +3H_(2)O(g), DeltaH =- 1558.3 kJ....(ii)`
`H_(2) + 1//2O_(2)(g) rarr H_(2)O(g), DeltaH =- 285.6 kJ ...(iii)`
Adding equations (i) and (iii)
`C_(2)H_(4)(g) +7//2_(2)(g)+H_(2)(g) rarr 2CO_(2)(g)+3H_(2)O(g) DeltaH =- 1695.1 kJ ..(iv)`
Subtracting equation (ii) from equation (iv)
`{:(C_(2)H_(6)(g)+7//2O_(2)(g)rarr2CO_(2)(g)+3H_(2)O(g)DeltaH),(ul(" DeltaH =-1558.3 kJ..(iv)" )),(C_(2)H_(4)(g)+H_(2)(g)rarrC_(2)H_(6)(g),DeltaH =- 136.8 kJ):}`
Heat of hydrogenation of `C_(2)H_(4) =- 136.8 kJ`