The standard enthalpy of atomization of PCl 3 ( g ) is 195 kcal/mol . What will be the standard enthalpy of atomization of PCl 5 ( g ) , if the bond dissociation energies of axial P − Cl bonds in

Question

The standard enthalpy of atomization of PCl 3 ​ ( g ) is 195 kcal/mol . What will be the standard enthalpy of atomization of PCl 5 ​ ( g ) , if the bond dissociation energies of axial P − Cl bonds in PCl 5 ​ ( g ) are 10% lesser, and the bond dissociation energies of equatorial P − Cl bonds in PCl 5 ​ ( g ) are 10% higher than the bond dissociation energies of P − Cl bonds in PCl 3 ​ ( g ) ?

TestHub · Accepted Answer

The standard enthalpy of atomization of PCl₃(g) is the energy required to break all three P-Cl bonds.

Let $D$ be the average P-Cl bond dissociation energy in PCl₃(g).

Given: Enthalpy of atomization of PCl₃ = $3 D = 195 kcal/mol$ .

So, $D = \frac{195}{3} = 65 kcal/mol$ .

In PCl₅(g), there are 3 equatorial and 2 axial P-Cl bonds.

Equatorial bond energy = $1.10 \times D = 1.10 \times 65 = 71.5 kcal/mol$ .

Axial bond energy = $0.90 \times D = 0.90 \times 65 = 58.5 kcal/mol$ .

Enthalpy of atomization of PCl₅ = $(3 \times e q u a t or ia l b o n d e n er g y) + (2 \times a x ia l b o n d e n er g y)$

Enthalpy of atomization of PCl₅ = $(3 \times 71.5) + (2 \times 58.5) = 214.5 + 117 = 331.5 kcal/mol$ .

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