Question

The enthalpy of dissociation of PH3 is 954 kj/mol and that of P2H4 is 1.485 Mj/mol , what is bond enthalpy of P-p bond

Answer 1

Plz go through the following attatchment carefully:

Answer 2

Explanation:

As it is given that enthalpy of dissociation for $PH_{3}$ is 954 kJ/mol. There are 1000 joules present in 1 kilo joule.

Hence, 954 kJ/mol will be equal to $954 kJ/mol \times \frac{1000J}{1 kJ/mol}$ = 954000 J/mol.

Bond enthalpy for P-H bond is $\frac{954000 J}{3}$ = 318000 J.

Hence, bond enthalpy for 4 P-H bonds will be calculated as follows.

                   $318000 J \times 4$ = 1272000 J

As there are $10^{6}$ joules present in 1 mega joules. Hence, converting 1.485 Mj/mol into joules as follows.

                   $1.485 Mj/mol \times \frac{10^{6}j}{1 Mj}$

                 = $1485 \times 10^{3} J$

Thus, calculate the bond enthalpy of P-p bond as follows.

             $B.E of P_{2}H_{4} = B.E of P-H + B.E of P-P$

where,      B.E = bond enthalpy

             $1485 \times 10^{3} J = 1272000 J + B.E of P-P$

                         B.E of P-P = 213000 J

Hence, we can conclude that the bond enthalpy of P-P bond is 213000 J.