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the heats of atomization of PH3(g) and P2H4(g) are 954kj/mole and 1485kj/mole respectively .the P-P bond energy in kj/mole a) 213 b)426 c) 318 d) 1272


Suman kumar 27th Nov, 2019
Answer (1)
ADITYA KUMAR Student Expert 27th Nov, 2019
Hello there!

Greetings!

We have
Atomisation energy of PH3=3*BE(P-H)
=954


Then, BE(P-H)=954kJ/mol3=318 kJ/mol

Atomisation energy of P2H4=BE(P-H)*4+BE(P-P)

=1485 kJ/mol


Then, Atomisation energy of P2H4= 4*BE(P-H) +BE(P-P)
=1485

BE(P-P)= Atomisation energy of P2H4 - 4*BE(P-H)
=1485- 4*318=1485-1272
=213kJ/mol , so the correct option is A

Thankyou

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