Question

The enthalpy of dissociation of $P{{H}_{3}}$ is 954 KJ/mol and that of ${{P}_{2}}{{H}_{4}}$ is 1.485 MJ/mol. What is the bond enthalpy of P – P bond?
(A) 213 KJ/mol
(B) 413 KJ/mol
(C) 200 KJ/mol
(D) Given data is incorrect

Answer 1

Here the bond dissociation enthalpy of $P{{H}_{3}}$ and ${{P}_{2}}{{H}_{4}}$ is given to us. Use the bond dissociation energy of $P{{H}_{3}}$ to find the bond enthalpy of one P – H bond. Use it in the bond enthalpy of ${{P}_{2}}{{H}_{4}}$ to find the required enthalpy of P – P.

Complete step by step solution:
In a crystalline solid, when the ions are combined, energy is released and this energy is known as lattice energy and the change in energy in this process is basically the lattice enthalpy.
We use the term lattice enthalpy for both bond formation as well as dissociation. However, to be more appropriate about the energy required for breaking of the crystal lattice and the formation of the crystal lattice, we use the terms bond dissociation enthalpy and bond formation enthalpy respectively.
Bond dissociation enthalpy is basically the energy required for the breaking a solid crystal into its corresponding ions.
Here, the bond dissociation enthalpy of $P{{H}_{3}}$ is given to us as 954 KJ/mol. Therefore, we can write that-
$P{{H}_{3}}\to P+3H\text{ ; }\Delta \text{H = 954 KJ/mol}$
Therefore, bond dissociation enthalpy of a single P – H bond will be $\dfrac{954}{3}KJ/mol$ i.e. 318 KJ/mol.
Also, the bond dissociation energy of ${{P}_{2}}{{H}_{4}}$ is given to us as 1.485 MJ/mol i.e. 1485 KJ/mol.
Therefore, we can write that-
${{P}_{2}}{{H}_{4}}\to 2P+4H\text{ ; }\Delta \text{H = 1485 KJ/mol}$
Here, $2\Delta {{H}_{P-P}}+4\Delta {{H}_{P-H}}=1485KJ/mol$
Or, putting the value of bond dissociation enthalpy of P – H bond in the above equation we will get-
$\begin{aligned} & \Delta {{H}_{P-P}}+4\times 318=1485KJ/mol \\\ & or,\Delta {{H}_{P-P}}=1485KJ/mol-1272KJ/mol=213KJ/mol \\\ \end{aligned}$
We can see from the above calculation that the bond enthalpy of P – P bond is 213 KJ/mol.

Therefore, the correct answer is option [A] 213 KJ/mol.

Note: There are two factors that affect the bond dissociation enthalpy of any substance.
-Firstly it is the charge on the ions on the ions. Higher is the charge on each ion, higher will be the lattice energy and the second factor is the radius. Lower the internuclear distances among the ions in an ionic crystal, higher will be their lattice energy i.e. closer the two ions are, higher is the force of attraction between them.