Question: Using the standard enthalpy of formation data in Appendix G, calculate the bond energy of the carbon-sulphur double bond in \({\rm{C}}{{\rm{S}}_{\rm{2}}}\).

Bond energy, also known as mean bond enthalpy or average bond enthalpy in chemistry, is a measure of the bond strength.It is the energy needed to dissociate the bonds into atoms.

The reactions with their enthalpy values are –

\(\begin{array}{*{20}{l}}{{\rm{C}}{{\rm{S}}_{\rm{2}}}{\rm{(g)}} \to {\rm{C( graphite ) + 2\;S(s)}}}&{{\rm{\Delta H}}_{\rm{1}}^{\rm{^\circ }}{\rm{ = \Delta H}}_{{\rm{f}}\left( {{\rm{C}}{{\rm{S}}_{\rm{2}}}{\rm{(g)}}} \right)}^{\rm{^\circ }}}\\{{\rm{C(graphite)}} \to {\rm{C(g)}}}&{{\rm{\Delta H}}_{\rm{2}}^{\rm{^\circ }}{\rm{ = \Delta H}}_{{\rm{f(Cl)(g))}}}^{\rm{^\circ }}}\\{{\rm{2\;S(s)}} \to {\rm{2\;S(9)}}}&{{\rm{2\Delta H}}_{\rm{3}}^{\rm{^\circ }}{\rm{ = 2\Delta H}}_{{\rm{f(S(g))}}}^{\rm{^\circ }}}\end{array}\)

The net reaction is –

\({\rm{C}}{{\rm{S}}_{\rm{2}}}{\rm{(g)}} \to {\rm{C(graphite) + 2\;S(g)}}\;\)

\(\begin{array}{c}{\rm{\Delta H}}_{{\rm{298}}}^{\rm{^\circ }}{\rm{ = \Delta }}{{\rm{H}}_{\rm{1}}}^{\rm{^\circ }}{\rm{ + \Delta H}}_{\rm{2}}^{\rm{^\circ }}{\rm{ + 2\Delta }}{{\rm{H}}_{\rm{3}}}^{\rm{^\circ }}\\{{\rm{D}}_{{\rm{C}}{{\rm{S}}_{\rm{2}}}}}{\rm{ = \Delta }}{{\rm{H}}^{\rm{^\circ }}}_{{\rm{298}}}{\rm{ = - \Delta }}{{\rm{H}}^{\rm{^\circ }}}_{{\rm{f}}\left( {{\rm{C}}{{\rm{S}}_{\rm{2}}}{\rm{(g)}}} \right)}{\rm{ + \Delta H}}_{{\rm{f(C(g))}}}^{\rm{^\circ }}{\rm{ + 2\Delta }}{{\rm{H}}^{\rm{^\circ }}}_{{\rm{f(S(g))}}}\end{array}\)

The bond energy is calculated as –

\(\begin{array}{c}{{\rm{D}}_{{\rm{C}}{{\rm{S}}_{\rm{2}}}}}{\rm{ = - 116}}{\rm{.9 + 716}}{\rm{.681 + 2(278}}{\rm{.81)}}\\{\rm{ = 1157}}{\rm{.4\;kJ}}{{\rm{D}}_{{\rm{C = S}}}}\\{\rm{ = }}\frac{{{\rm{1157}}{\rm{.4}}}}{{\rm{2}}}\\{\rm{ = 578}}{\rm{.7\;kJ}}\end{array}\)

Therefore, the value for bond energy is \({\rm{578}}{\rm{.7\;kJ}}\).