How Is Heat Energy Distributed in H2 Combustion?

[quasar_4]

Homework Statement

Consider the combustion of one mole of H2 with 1/2 mole of O2 under standard conditions. How much of the heat energy produced comes from a decrease in the internal energy of the system and how much comes from work done by the collapsing atmosphere? Treat the volume of the liquid water as negligible.

Homework Equations

We've got enthalpy, which is given by H = U + PV. Also, we know from a statement in the book that the enthalpy of formation for the chemical reaction is -286 kJ.

The Attempt at a Solution

If I understand the concept of enthalpy correctly, the idea is that enthalpy is the total energy "cost" for either creating or annihilating something. This consists of the "cost" of energy both from the object itself, and also from the need to make "room" or "collapse" the air around it. So this problem wants me to take the total cost (the enthalpy) and break it down into these constituent parts.

But I don't see at all how to do this. I know intuitively that most of the "cost" comes from the change in internal energy, not the collapse of the atmosphere. But, I don't know how to quantify this.

I am thinking maybe I can do the following, but am hoping for some feedback on it:

I can take the ideal gas law, PV=nRT, and use it to define the volume for 1 mole of water. Assuming the pressure is constant, then I could calculate the work done to "make room" for the liquid water. And then since I know the final enthalpy, I could subtract to get the other part (the change in internal energy).

But is that right? Or do I need to find the volume for 1 mole of H2 and the 1/2 mole of O2, and add them together instead (and would that even be the right way to find the whole volume)? I guess it's not clear to me which I should use.

 

[quasar_4]

never mind, it wasn't bad. I figured it out.