Question

Question: Using the standard enthalpy of formation data in Appendix G, show how the standard enthalpy of formation of \({\rm{HCl(g)}}\) can be used to determine the bond energy.

Short answer

Using the standard enthalpy of formation of \({\rm{HCl(g)}}\), the bond energy is determined as \({\rm{431}}{\rm{.6\;kJ}}\).

Step-by-step solution

Concept Introduction

Bond energy, also known as mean bond enthalpy or average bond enthalpy in chemistry, gives the measure of the bond strength.

Reactions and their Enthalpy

The reactions with their enthalpy values are –

\(\begin{array}{l}{\rm{HCl(g)}} \to \frac{{\rm{1}}}{{\rm{2}}}{{\rm{H}}_{\rm{2}}}{\rm{(\;g) + }}\frac{{\rm{1}}}{{\rm{2}}}{\rm{C}}{{\rm{l}}_{\rm{2}}}{\rm{(\;g) \Delta H}}_{\rm{1}}^{\rm{^\circ }}{\rm{ = \Delta H}}_{{\rm{f(HCl(g))}}}^{\rm{^\circ }}\\\frac{{\rm{1}}}{{\rm{2}}}{{\rm{H}}_{\rm{2}}}{\rm{(g)}} \to {\rm{H(g) \Delta H}}_{\rm{2}}^{\rm{^\circ }}{\rm{ = \Delta H}}_{{\rm{f(H(g))}}}^{\rm{^\circ }}\\\frac{{\rm{1}}}{{\rm{2}}}{\rm{C}}{{\rm{l}}_{\rm{2}}}{\rm{(b)}} \to {\rm{Cl(g) \Delta H}}_{\rm{3}}^{\rm{^\circ }}{\rm{ = \Delta H}}_{{\rm{f(Cl(g))}}}^{\rm{^\circ }}\end{array}\)

Bond Energy Calculation

The net reaction is –

\({\rm{HCl(g)}} \to {\rm{H}}(g) + {\rm{Cl}}(g)\)

\({{\rm{D}}_{{\rm{HCl}}}}{\rm{ = \Delta }}{{\rm{H}}_{{\rm{29}}{{\rm{8}}^{\rm{0}}}}}{\rm{ = \Delta H}}_{\rm{1}}^{\rm{0}}{\rm{ + \Delta H}}_{\rm{2}}^{\rm{0}}{\rm{ + \Delta H}}_{\rm{3}}^{\rm{0}}\)

The bond energy is calculated as –

\(\begin{array}{c}{{\rm{D}}_{{\rm{HCl}}}}{\rm{ = - ( - 92}}{\rm{.3\;kJ) + 218\;kJ + 121}}{\rm{.3\;kJ}}{{\rm{D}}_{{\rm{HCl}}}}\\{\rm{ = 431}}{\rm{.6\;kJ}}\end{array}\)

Therefore, the value for bond energy is \({\rm{431}}{\rm{.6\;kJ}}\).