Calculate the bond energy of C-=C in `C_(2)H_(2)` from the following data :
(i) `C_(2)H_(2)(g)+2(1)/(2)O_(2)(g) to 2CO_(2)(g)+H_(2)O,DeltaH=-310"kcal"`
(ii) `C(s)+O_(2)(g) to CO_(2)(g),DeltaH=-94"kcal"`
(iii) `H_(2)(g)+(1)/(2)O_(2)(g) to H_(2)O(g),DeltaH=-68 "kcal"`
Bond energy of C-H bonds =99 kcal
Heat of atomisation of C=171 kcal
Heat of atomisation of H=52 kcal

Explanation

Let us first calculate heat of formation of `C_(2)H_(6)` from which we can calculate bond energy of `C-=C` bond.
`2C(s)+H_(2) to C_(2)H_(2)(g)` , `DeltaH=?`
Applying the inspection method , i.e.,
`[2xx` Eqn. (ii) `+` Eqn. (iii) - Eqn. (i)] we get
`2C(s)+2O_(2)(g)+H_(2)(g)+(1)/(2)O_(2)(g)-C_(2)H_(2)(g)-2(1)/(2)O_(2) to 2CO_(2)(g)+H_(2)O(g)-2CO_(2)(g)-H_(2)O(g), DeltaH=2xx(-94)+(-68)-(-310)kcal`
or `2C(s)+H_(2)(g) to `C_(2)H_(2)(g) `, `DeltaH=54` kcal
Now heat changes for reactants
Heat of atomisation of 2 moles of `C=2xx171` kcal
Heat of atomisation of 2 moles of `H=2xx52` kcal
And heat change for the product `(H-C-=C-H)`
Heat of formation of 2 moles of C-H bonds `=-2xx99` kcal
Heat of formation of 1 mole of `C-=C` bonds `=x` (say)
Summing up, we get heat of formation of `C_(2)H_(2)`,
`2xx171+2xx52-2xx99+x=54`
`x=-194` kcal
Hence bond energy of `C-=C` bond in `C_(2)H_(2)` is `+194` kcal