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

Explanation

The overall thermochemical equation for its atomisation reaction is
`CH_(4(g))toC_((g))+4H_(( g)),Delta_(a)H^(theta)_=1665kJ mol^(-1)`
In methane all the four C - H bonds are similar in bond length and energy. However the energies required to break the individual C -H bonds in each successive step differ.
`CH_(4(g))toCH_(3(g))+H_((g)),Delta_("bond")H^(theta)=+427kJmol^(-1)`
`CH_(3(g))toCH_(2(g))+H_((g)),Delta_("bond")H^(theta)=+439kJmol^(-1)`
`CH_(2(g))toCH_((g))+H_((g)),Delta_("bond")H^(theta)=+452kJmol^(-1)`
`CH_((g))toC_((g))+H_((g)),Delta_("bond")h^(theta)=+347kJmol^(-1)`
therefore
`CH_(4(g))toC_((g))+4H_((g)),Delta_(a)H^(theta)=1665 kJmol^(-1)`
In such cases we use mean bond enthalpy of C-H bond.
So in `CH_(4)Delta_(C-H)H^(theta)` is 1665 kJ`mol^(-1)//4=416kJ mol^(-1)`