Gc

299678

The use the bond energies in the table to determine delta H for the formation of hydrazine, N2H4 from nitrogen and hydrogen according to the equation.
N2 (g) + 2H2(g) -> N2H4 (g)

Bond energies KJ/mol
N-N 159
N=N 201
N triple bond N 941
H-H 436
H-N 389

loveoforganic

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10+ Year Member
(941 + 436*2) - (159 + 389*4)

?

akeurogh

7+ Year Member
Hey so the answer is going be delta H=+458

The formula you use here is (bonds broken)-(bonds formed)

You have to break 2 bonds b/w the nitrogen. One with a bond energy of 941 KJ/mol and one with 201 KJ/mol.

Do you need me to go into more detail?

loveoforganic

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10+ Year Member
Given those numbers akeurogh, it seems like the value given is for the entire triple bond, isn't it? A single pi bond shouldn't be stronger than a sigma bond.

akeurogh

7+ Year Member
oh yeah you're right! nice catch!

sfoksn

10+ Year Member
Hm... I have gotten +109 as my final answer..

Cuz.. you have to break the triple bond of N2, which is +941, and break two H2 bonds which are +436 each.. So that comes out to be +1812..

Then you have to make single N2 bond, which is -159, and make 4 N-H bond, which is -386 each..

So I think total is +109.

Maybe I am completely wrong

Thank you.

+ 98.

10+ Year Member
5+ Year Member

sfoksn

10+ Year Member
Hmm I'm got +98 too, when I did it with a calculator haha.

10+ Year Member
Hm... I have gotten +109 as my final answer..

Cuz.. you have to break the triple bond of N2, which is +941, and break two H2 bonds which are +436 each.. So that comes out to be +1812..

Then you have to make single N2 bond, which is -159, and make 4 N-H bond, which is -386 each..

So I think total is +109.

Maybe I am completely wrong