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# Help with bond enthalpy? ?

What would be the reaction enthalpy of the following reaction? The units of the bond energies are in joules (see chart). N2 + H3O+ --> NO2- + NH3

O-H 463

N≡O 742

N-H 391

N≡N 941

According to the answer key, the answer is 395 J, however, after I balanced the equation (3,4,2,4 from left to right) the answer I get is 719 J. Can someone help explain how the answer is what it is?

I got 719 J by subtracting the bonds breaking (reactants; 8379 J) by the bonds forming (products; 7660 J)

### 2 Answers

- ChemTeamLv 71 month ago
First, this is the equation from Roger's answer:

N2 + 2 H3O{+} → NO2{-} + NH3 + 3 H{+}

The key is to realize that H^+ and H3O^+ are the same thing. so cancel two H^+ to get:

N2 + 2 H2O → NO2{-} + NH3 + H{+}

HNO2 is a weak acid, so write it in an unionized form:

N2 + 2H2O → HNO2 + NH3

Do the bond enthalpy calculation on that equation and you get 395. However, you won't need N≡O, but you will need N=O. Look up the Lewis structure to confirm what bonds to use with regard to HNO2.

- Roger the MoleLv 71 month ago
Your fundamental trouble is that you did not balance the equation. Your number of atoms is OK, but you've got positive charges on the left and negative charges on the right.

The only way I see to fix it is to add hydrogen ions, as is often done with redox reactions:

N2 + 2 H3O{+} → NO2{-} + NH3 + 3 H{+}

When I try working from that, I get 790 J, which is exactly double your given answer, but I don't see where the extra factor of 2 comes from, unless the question was asking about the given equation in units of J per mole of H3O{+}, which is a strange thing to do, especially without telling you.

- ...Show all comments
Wasn't something we learned in class or even close to regarding the charges and a student instructor made the question so I'm just going to assume that it was a mistake. Thanks for the help

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Doesn't the fact that this reaction is not in the gas phase negate the use of bond energies?