Question

The heat of atomisation of `PH_(3)(g)` and `P_(2)H_(4)(g)` are `954 kJ mol^(-1)` and `1485 kJ mol^(-1)` respectively. The `P-P` bond energy in `kJ mol^(-1)` is
A. `213`
B. `426`
C. `318`
D. `1272`

Answer 1

Correct Answer - A
Atomisation energy of `PH_(3)=3xxBE(P-H)`
`=954 kJ "mole"^(-1)`
`:. BE(P-H)=(954 kJ mol^(-1))/(3)=318 kJ mol^(-1)`
Atomisation energy of `P_(2)H_(4)`
`=BE(P-H)xx4+BE(P-P)`
`=1485 kJ mol^(-1)`
`:. BE (P-P)`
`= "Atomisation energy of" P_(2)H_(4)-4xxBE(P-H)`
`=1485-4xx318`
`=1485-1272=213 kJ mol^(-1)`