What is the bond enthalpy of C-H bond of `CH_(4)`?

Explanation

The bond dissociation enthalpy for step wise dissociation of `CH_4` molecule are
`CH_(4(g)) to CH_(3(g)) +H_(g) , triangleH^0 =427.0kJ`
`CH_(3(g)) to CH_(2(g)) +H_(g) , triangleH^0 =418.4kJ`
`CH_(2(g)) to CH_(g) +H_(g) , triangleH^0 =460.2kJ`
`CH_(g) to CH_(g) +H_(g) , triangleH^0 =343.1kJ`
The enthalpy for each step is different due to the reason that in each step of dissociation different fragments of `CH_4` are involved.
But the average bond dissociation enthalpy is 1/4 of the total energy required to break one mole of `CH_(4(g))" into "C_(g) +4H_(g)`
C-H bond dissociation Enthalpy `=(427.0 +418.4 +460.2 +343.1)/(4)`
`=(1648.7)/(4)=412.2kJ`