Heat from enthalpy of formation of water at constant pressure

[Cogswell]

Homework Statement

I'm doing physics and I don't know if this counts as physics or Chemistry (sorry if this does not belong here)

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).

Homework Equations

$$\Delta H = \Delta U + P \Delta V$$

$$\Delta U = Q + W$$

The Attempt at a Solution

So firstly I started off with $\Delta H = \Delta U + P \Delta V$

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 $-286kJ$

$$\Delta H = \Delta U + P \Delta V$$

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

$$\Delta U = 101300 \cdot (0.0336) - 286 000$$

$$\Delta U = -282596.32$$

We also know that:

$$\Delta U = Q + W$$

But there is no external work done on it and so

$$\Delta U = Q$$

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

Does that seem right?

 

[DrClaude]

We also know that:

$$\Delta U = Q + W$$

But there is no external work done on it and so

$$\Delta U = Q$$

What about the $P \Delta V$ you just calculated? I think you are misinterpreting the meaning of "internal energy".

Otherwise, you seem to have the correct result for the amount of the heat released due to the change in internal energy, apart from the insane precision on the number you quote.