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The heat of atomisation of PH(3)(g) and ...

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`

Text Solution

Verified by Experts

The correct Answer is:
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)`
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