[ausstu89]

Hi, I'm hoping someone might be able to help me understand this topic.

ΔU= q + w

I know that in relation to a gas at constant pressure, the formula becomes ΔU = ΔH + (ΔnRT), and to find ΔH: ΔH = ΔU - (ΔnRT). To cut a long story (textbook question!) short, the ΔU for this particular combustion reaction (at constant pressure and temperature) is 5,142,000 J. The ΔnRT is (-2mol · 8.314 J K mol · 298K). This gives a ΔH of 5,147,000 J.

Mathematically, it makes sense. But (if I understand correctly) ΔU = the heat added/taken from the system + the work done by/to the system. In this example, the ΔH (-) is greater than the ΔU (also -), meaning there must have been work done to the system to replace some of the ΔU. I'm unsure how work is being "done to" the system in this reaction.

Thanks in advance!

[mjc123]

Mathematically, it makes sense.

No it doesn't. There seems to be some confusion here.
You say ΔU and ΔH are negative, but you quote positive values, with ΔH being more positive, despite ΔnRT being negative. If you are just working with the magnitudes of ΔH and ΔU, ignoring their signs, this is sloppy practice that easily leads to errors.
Your more serious error is that actually ΔH = ΔU + PΔV = ΔU + ΔnRT. w is the work done on the system; PΔV is the work done by the system.

(if I understand correctly) ΔU = the heat added/taken from the system + the work done by/to the system.

ΔU = the heat added to the system + the work done on the system. Heat taken out is negative q; work done by the system is negative w.
If Δn is -2 mol, work is done on the system by the external pressure as the gas contracts at constant pressure. Hence ΔH is more negative than ΔU.
Can you quote the actual question?

[ausstu89]

Thanks for getting back to me.

Sorry- you are right, I was being sloppy with negatives. ΔU and ΔH were negative values.

I think you may have answered by question, however. I think I was getting confused between the effects of constant pressure and constant volume. The image in my head was of a fixed volume for some reason! It makes perfect sense that the external pressure does work on the gas.

Thanks again.