Us the bond energies in the table to determine delta H for the formation of hydrazine, N2H4,?

Question

from nitrogen and hydrogen accordint to this equation

N2 9g) + 2H2 (g) -> N2H4

N-N 159 bond energies Kj/mol

N-N double bond 201

N-N triple 941

H-H = 436

H-N 389

I also need to know lewis structure of N2H4

Answer 1

∆H°f = ∑∆H(bonds broken) -∑∆H(bonds formed)

∆H°f = [(N≡N) + 2 * (H-H)] - [4 * (N-H) + (N-N)]

∆H°f = [(941) + 2 * (436)] - [4 * (389) + (159)]

∆H°f = +98 kJ/mol.

The structure is H2NNH2, all bonds are single covalent bonds, though I'm sure you figured that from the 2nd half of the calculations above. Both nitrogens have a lone electron pair.