Q72E - Chemistry (OpenStax)

Question

Question: Using the standard enthalpy of formation data in Appendix G, determine which bond is stronger: the \({\rm{P - Cl}}\) bond in \({\rm{PC}}{{\rm{l}}_{\rm{3}}}{\rm{(g)}}\) or in \({\rm{PC}}{{\rm{l}}_{\rm{5}}}{\rm{(g)}}\)?

Step-by-Step Solution

Answer

Using the standard enthalpy of formation data, the \({\rm{P - Cl}}\) bond is stronger in \({\rm{PC}}{{\rm{l}}_3}(g)\).

Step 1: Concept Introduction

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

Step 2: Reactions and their Enthalpy

The reactions with their enthalpy values are –

\(\begin{array}{*{20}{l}}{{\rm{PC}}{{\rm{l}}_{\rm{3}}}{\rm{(g)}} \to \frac{{\rm{1}}}{{\rm{4}}}{{\rm{P}}_{\rm{4}}}{\rm{(s) + }}\frac{{\rm{3}}}{{\rm{2}}}{\rm{C}}{{\rm{l}}_{\rm{2}}}{\rm{(g)}}}&{{\rm{\Delta H}}_{\rm{1}}^{\rm{^\circ }}{\rm{ =  - \Delta H}}_{{\rm{f}}\left( {{\rm{PC}}{{\rm{l}}_{\rm{3}}}{\rm{(g)}}} \right)}^{\rm{^\circ }}}\\{\frac{{\rm{1}}}{{\rm{4}}}{{\rm{P}}_{\rm{4}}}{\rm{(s)}} \to {\rm{P(g)}}}&{{\rm{\Delta H}}_{\rm{2}}^{\rm{^\circ }}{\rm{ = \Delta H}}_{{\rm{f(P)(g)]}}}^{\rm{^\circ }}}\\{\frac{{\rm{3}}}{{\rm{2}}}{\rm{C}}{{\rm{l}}_{\rm{2}}}{\rm{(g)}} \to {\rm{3Cl(g)}}}&{{\rm{3\Delta H}}_{\rm{3}}^{\rm{^\circ }}{\rm{ = 3\Delta H}}_{{\rm{f(Cl(g))}}}^{\rm{^\circ }}}\end{array}\)\({\rm{PC}}{{\rm{l}}_{\rm{3}}}{\rm{(g)}} \to {\rm{P(g) + 3Cl(g)}}\)

Step 3: Bond Energy Calculation

\({\rm{PC}}{{\rm{l}}_{\rm{3}}}{\rm{(g)}} \to {\rm{P(g) + 3Cl(g)}}\)

\({\rm{PC}}{{\rm{l}}_{\rm{3}}}{\rm{(g)}} \to {\rm{P(g) + 3Cl(g)}}\)

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

The bond energy is calculated as –

\(\begin{array}{c}{{\rm{D}}_{{\rm{PC}}{{\rm{l}}_{\rm{3}}}}}{\rm{ = 287}}{\rm{.0 + 314}}{\rm{.64 + 3(121}}{\rm{.3)}}\\{\rm{ = 965}}{\rm{.54\;kJ}}\\{{\rm{D}}_{{\rm{PC}}{{\rm{l}}_{\rm{3}}}}}{\rm{ = }}\frac{{{\rm{965}}{\rm{.54}}}}{{\rm{3}}}\\{\rm{ = 321}}{\rm{.8\;kJ}}\end{array}\)

Step 4: Stronger bond

If we proceed in the manner as described above, then \({\rm{ - \Delta Hf}}\left( {{\rm{PC}}{{\rm{l}}_5}{\rm{(\;g)}}} \right){\rm{ = 374}}{\rm{.9 kJ}}\). Now, since \({\rm{5\;Cl(g)}}\) and one \({\rm{P(g)}}\) contribute \({\rm{921}}{\rm{.14\;kJ}}\), then \({\rm{D}}{{\rm{F}}_{{\rm{PC}}{{\rm{l}}_5}}}{\rm{ = 1296}}{\rm{.04\;kJ}}\) and \({{\rm{D}}_{{\rm{P - Cl}}}}{\rm{ = }}\frac{{{\rm{1296}}{\rm{.04}}}}{{\rm{5}}}{\rm{ = 259}}{\rm{.2\;kJ}}\).

Therefore, based on the calculations it can be said that \({\rm{P - Cl}}\) bond is stronger.