The standard enthalpy of atomization of \(PCl_3\) (g) is 195 kcal/mol. What will be 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)?

318.5 kcal/mol

The correct answer is option 4. 318.5 kcal/mol.

Given,

Standard enthalpy of atomization of \(PCl_3\) \(=\) \(195\, \ kcal/mol\)

Change in bond dissociation energies for \(PCl_5\):

(i) Axial \(P-Cl\) bonds are \(10\%\) lower

(ii) Equitorial \(P-Cl\) bonds are \(10\%\) higher.

Bond enthalpy of \(P-Cl\) in \(PCl_3\) is

\(\frac{195}{3}\, \ kcal/mol\)

or, \(65\, \ kcal/mol\)

Thus, bond dissociation energy of axial \(P-Cl\) in \(PCl_5\) is

\(65 - 10\% \, \ of\, \ 65\)

\(= 65 - \frac{10}{100} \times 65\)

\(= 65 - 6.5 \)

\(= 58.5\, \ kcal/mol\)

Also, bond dissociation energy of equitorial \(P-Cl\) in \(PCl_5\) is

\(65 + 10\% \, \ of\, \ 65\)

\(= 65 + \frac{10}{100} \times 65\)

\(= 65 + 6.5 \)

\(= 71.5\, \ kcal/mol\)

Thus, enthalpy of atomization of \(PCl_5\)

\(= (3 \times 58.5) + (2 \times 71.5)\)

\(= 175.5 + 143\)

\(=318.5\, \ kcal/mol\)