[salocime]

I want to make sure I'm doing the right thing.

I'm doing questions that are asking me to calculate the enthalpy change of a reaction based on 2 or 3 thermochemical equations. In this one question, two of the equations have ΔH˚ with kJ, but then one ΔH˚  with kJ/mol. They aren't the same units. Can I still add all the numbers?

The full question in case it's helpful:

2N₂(g) + 5O₂(g) --> 2N₂O₅(g)

(1) 2H₂(g) + O₂(g) --> 2H₂O(l)   ΔH˚ =-572 kJ
(2) N₂O₅(g) + H₂O(l) --> 2HNO₃(l)  ΔH˚ =-77 kJ
(3) ½N₂(g) + 3/2 O₂(g) + 1/2 H₂(g) --> HNO₃(l)   ΔH˚ =-174 kJ/mol


My solution:
(1) 2H₂O(l) --> 2H₂(g) + O₂(g) ΔH˚ =572 kJ
2x(2) 4HNO₃(l) --> 2N₂O₅(g) + 2H₂O(l)  ΔH˚ = 154 kJ
4x(3) 2N₂(g) + 6O₂(g) + 2H₂(g) --> 4HNO₃(l)  ΔH˚ = -696 kJ/mol

------------------------------------------------
2N₂(g) + 5O₂(g) --> 2N₂O₅(g)    ΔH˚=

[cliverlong]

Interesting question.

My books are tantalisingly contradictory on this.

I have always worked on the basis that standard enthalpy changes are for the formation of one mole of product and the units are kJ mol^-1. The number of moles of the reactants are adjusted to ensure one mole of product is formed. e.g. H₂ + 1/2 O₂ -> H₂O . How one handles multiple products I have never got my head around

However, the big Housecroft textbook says if one produces 2 moles of product then the standard enthalpy change (delta H) is double that of the one mole case yet the units are still kJ mol^-1

Looking at your example I would go for the kJ mol^-1 unit indicating one mole of product and the kJ unit indicating the moles as in the version of the equation given  (i.e. scaled to avoid fractions).

But I may be wrong and will definitely interested if anyone can come up with a robust reference on how to handle this.

Clive