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)?
Answer & explanation
Correct answer: option 4
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 \(PCl_5\)
\(= (3 \times 58.5) + (2 \times 71.5)\)
\(= 175.5 + 143\)
\(=318.5\, \ kcal/mol\)