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
Answer (1)
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
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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