[sundrops]

Use the given average bond energy values to estimate H for the following reaction in the gas phase.
N2 + 3H2  2NH3


DNN = 941 kJ/mol
DH-H = 432 kJ/mol
DH-N = 391 kJ/mol

I have no clue how to do this! I've gotten it wrong 4 times now, only a couple more tries.

Here's what I've tried:

deltaH = [sum of bonds formed] - [sum of bonds broken]
delta H = [(941) + 3(432)] - [2(391)]
delta H = 1455

and when that didn't work I put a negative sign in fron and it still din't work. :S

What am I doing wrong?

[sundrops]

DNN is supposed to read DN(triple bond)N

[Donaldson Tan]

bond forming is exothermic, so u have to include a negative sign to your bond-forming terms. bond-breaking is endothermic, so you should include a positive sign to your bond-breaking terms. Moreover, your workings for "[2(391)]" suggest that 2 N-H bonds are formed. However, there are 6 N-H bonds in 2 NH3 molecules. Please rectify your mistake.

[sundrops]

N2 + 3H2 ---> 2NH3

so I would go:

deltaH = [(-941) + 3(-432)] - [6(391)]
deltaH = -4583 kJ

but that didn't work - is there somethin g else that I am missing?

[sundrops]

ok, i got it!  

Escept here's what I did:

delta H = [(941) + 3(432)] - 6(391)
delta H = 2237-2346
delta H = -109 kJ

which was the answer.

but I'm a little confused to why the bond forming sign needed to be positive when as you said geodome it is an exothermic reaction, and why the bond breaking side was also positive.
geodome - please explain, I'm really confuzzled

thanks for your help so fa geodome! its been awesome!

[Mitch]

deltaH = [sum of bonds formed] - [sum of bonds broken]

[Donaldson Tan]

i was thinking along this line:

N2 + 3H2 -> 2NH3
dH = break(1 N=N and 3 H-H) + form (6 N-H)
     = (+ 941 + 3(432)) + (6(-391))

ignore this if u find this irrelevant. it's just an alternative mean to interpret the same equation.

[sundrops]

kay, thanks!