Energies requried to break the individual C-H bonds in each successive step differ
`CH_(4)(g)toCH_(3)(g)+H(g),Delta_("bond")H^(ө)=439" kJ "mol^(-1)`
`CH_(2)(g)toCH(g)+H(g),Delta_("bond")H^(ө)=439kJmol^(-1)`
`CH(g)toC(g)+H(g),Delta_(bond)H^(ө)=347kJmol^(-1)`.
In such cases we use

Options

A. bond association enthalpy of C-H bond, `Delta_(C-H)H^(ө)=416kJmol^(-1)`
B. Bond association enthalpy of C-H bond, `Delta_(C-H)H^(ө)=1665kJmol^(-1)`
C. mean bond enthalpy of C-H bond, `Delta_(C-H)H^(ө)=416kJmol^(-1)`
D. none of these

Answer

C

Explanation

The overall, thermochemical equation for its atomisation reaction is given below
`CH_(4)(g) to C(g)+4H(g),Delta_(a)H^(ө)=1665" kJ "mol^(-1)`
In methane, all the four C-H bonds are identical in bond length and energy. However, the energies required to break the individual C-H bonds in each successive step differ. in such case we use mean bond enthalpy of C-H bond, e.g. In `CH_(4),Delta_(C-H)H^(ө)` is calculated as
`Delta_(C-H)H^(ө)=(1)/(4)(Delta_(a)H^(ө))=(1)/(4)(1665" kJ "mol^(-1))=416kJ" "mol^(-1)`.