[neil]

Does anyone know what the enthalpy change of reaction for the following reaction is?

4C₃H₅N₃O₉ (l)    -->    6N₂ (g) + O₂ (g) + 12CO₂ (g) + 10H₂O (g)

 I got -1814kJmol^-1 but i think its wrong.

[Mitch]

show your work.

[neil]

I got:
5 x (H-C) + 2 x (C-C) + 3 x (C-O) + 6 x (O-N) + 3 x (O=N) =
5 X 413    + 2 x 347   + 3 x 358    + 6 x 214   + 3 x 587     = +6878kJmol^-1

1.5 x (N=N) + 0.25 x (O=O) + 6 x (C=O) + 5 x (H-O) =
1.5 x 924 + 0.25 x 498 + 6 x 805 + 5 x 464 = -8660.5kJmol^-1

+6878 -8660.5 = -1782.5kJmol^-1

Thats just more confusing because i got a different answer!

[GCT]

I got:
5 x (H-C) + 2 x (C-C) + 3 x (C-O) + 6 x (O-N) + 3 x (O=N) =
5 X 413    + 2 x 347   + 3 x 358    + 6 x 214   + 3 x 587     = +6878kJmol^-1

1.5 x (N=N) + 0.25 x (O=O) + 6 x (C=O) + 5 x (H-O) =
1.5 x 924 + 0.25 x 498 + 6 x 805 + 5 x 464 = -8660.5kJmol^-1

+6878 -8660.5 = -1782.5kJmol^-1

Thats just more confusing because i got a different answer!

 a better way to approach this is to find the net standard enthalpy from the standard enthalpy of each compound.  Find the standard enthalpies from a list in your text or somewhere else, I believe that most, if not all, will be listed.