Calculate the bond energy of `C - H` bond, given that the heat of formation of `CH_4`, heat of sublimation of carbon and heat of dissociation of `H_2` are `-74.8 + 719.6 and 435 kJ mol^(-1)` respectively.

Explanation

Here, we are given
`C(s) +2H_(2)(g)rarr CH_(4) (g), Delta_(f)H^(@)= - 74.8 kJ` ....(i)
`C(s)rarr C(g), Delta_(r) H^(@) = +719.6kJ ` ....(ii)
`H_(2)(g) rarr 2H(g), Delta_(r) H^(@) = + 435 .4 kJ` ...(iii)
We aim at `:`
Eqn. (ii) `+ 2 xx`Eqn. (iii) - Eqn (i) gives
`C(s) +2H_(2)(g) rarr C(g) + 4H(g)`
`- C(s) -2H_(2)(g) rarr - CH_(4) (g) `
`0 =C(g)+4H(g)-CH_(4)(g), Delta_(r) H^(@)= 719.6+2 (435.4)- (- 74.8)`
or ` CH_(4)(g) rarr C(g) + 4H(g), Delta H = + 1665.2 kJ`
This gives the enthalpy of dissociation of four moles of C-H bonds ( called enthalpy of atomization) .
Hence, bond energy for C-H bond (average value ) i.e.,
`Delta_(C-H) H^(@) = (1665.2)/(4)=416.3 kJ mol^(_1)`