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.

Answer

416

Explanation

The given equations are :
(i) `C(s) + 2H_(2)(g) to CH_(4)(g) Delta_(r)H^(@) = -74.8 kJ`
(ii) `C(s) to C(g) Delta_(r)H^(@) = 719.6 kJ`
(iii) `H_(2)(g) to 2H(g) Delta_(r)H^(@) = +435.4 kJ`
The required equation is
`CH_(4)((g) to C(g) + 4H(g)`
Multiply eq.(iii) by 2 and add to eq(ii)
`(iii) xx 2 "2H_(2)(g) to 4H(g) Delta_(r)H^(@) = 870.8 kJ`
(ii) `C(s) to C(g) Delta_(r)H^(@) = 719.6 kJ`
Adding
(iv) `C(s) + 2H_(2)(g) to C(g) + 4H(g) Delta_(r)H^(@) = 1590.4 kJ`
Subtract eq. (i) from eq.(iv)
`C(s) + 2H_(2)(g) to CH_(4)(g) Delta_(r)H^(@) = -74.8 kJ`
Subtracting `CH_(4)(g) to C(g) + 4H(g) Delta_(r)H^(@) = 1665.2 kJ`
This gives the enthalpy of dissociation of four moles of C-H bonds. Hence, bond enthalpy of C-H bond is `Delta_(C-H)H^(@) = (1665.2)/(4) = 416.3 kJ mol^(-1)`.