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Enthalpy of formation of water

I'm doing physics and I don't know if this counts as physics or Chemistry (sorry if this is in the wrong section)

Consider the combustion of H2 with 0.5 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 collapsing the atmosphere? (Treat the volume of the liquid water as negligible).


So firstly I started off with ΔH=ΔU+PΔV\Delta H = \Delta U + P \Delta V

The volume of one mole of any ideal gas is 22.4L and I have 1.5 moles of gas to start off with so the initial volume is 0.0336m^3
Since we neglect the volume of the water (final volume), then delta V equals -0.0336m^3

We know the enthalpy of formation of water is -286kJ

ΔH=ΔU+PΔV\Delta H = \Delta U + P \Delta V

−286000=ΔU+101300⋅(−0.0336)⏟heat energy from collapsing the atmosphere-286 000 = \Delta U + \underbrace{101300 \cdot (-0.0336)}_{\text{heat energy from collapsing the atmosphere}}

ΔU=101300⋅(0.0336)−286000 \Delta U = 101300 \cdot (0.0336) - 286 000

ΔU=−282596.32 \Delta U = -282596.32

We also know that:

ΔU=Q+W \Delta U = Q + W

But there is no external work done on it and so

ΔU=Q \Delta U = Q

And so the heat formed from a decrease in the internal energy of the system is -282,596.32 Joules?

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