# Stuck on chem hw HELP

1. Nov 15, 2009

### yaganon

First of all, how do you know if a combustion process produce h2o(g) or h2o(l)??? It depends on the temperature, but how do you figure that out?

I found the heat released during the combustion processes, which should equate the work done by the system (theoretically). But according to w = -nRT[ln(V2/V1)], I need the initial volume and final volume. The pressure remains constant (I believe), and V can be figured by PV=nRT with P=1 atm throughout. And with that, I need the temperature.

to figure out the temperature, I need the specific heat. But what specific heat do I need? those of the reactants or the products??? my brain iis hurrrts badd

btw, here are the reactants: C8H18 (Octane), C2H5OH (ethanol), methane ch4, h2, and nh4.

Last edited: Nov 15, 2009
2. Nov 15, 2009

### pzona

This should probably be in the homework section, but I'll try to answer as much as I can. Combustions usually produce more than 373K of heat, so most of the time it's safe to assume it's gaseous water that is produced, at least initially. I'm pretty sure you need the specific heat of the environment in which the reaction is taking place, not the reactants or products. I could be wrong, but that's what I think I remember from high school chem. Also, do you have to use these specific equations? Work can also be expressed by P(delta V), or it can be figured out by E=q+w. Do either of these help?

3. Nov 16, 2009

### yaganon

mmm not really

I was thinking that if you're operating under standard conditions, water would be liquid since it would be something like

2H2(g) + O2(g) --> 2H2O(l) + ...KJ, in which case the heat would evaporate the liquid water.

4. Nov 16, 2009

### Staff: Mentor

Combustion always produce gaseous water (after all, products of combustion are hot), but standard enthalpy - as reported in tables - is for products at STP. That means liquid.

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5. Nov 16, 2009

### NotJohnson

Thinking about it logically, for one reaction of the type (g) + (g) -> products, any water produced would have to be gaseous since there would only be a few molecules of it, not enough to condense into liquid. If you're assuming an isolated system I would say it is safe to assume gaseous water since any liquid water would be condensed and removed from the system.

If you want to do it using numbers, either a specific heat for the environment, or you're going to need the specific heat for all reactants and products since they will all be present in the system, absorbing energy and raising temperature. Unless you're assuming the reaction has gone to completion, in that case, just products.

6. Nov 17, 2009

### Staff: Mentor

I can't see any logic here. Number of molecules is irrelevant. Besides, nothing stops you from burning tonnes of gas.

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methods

7. Nov 17, 2009

### NotJohnson

and nothing stops tonnes of gas from increasing the volume to become just as sparse. I'm saying that in a gaseous system the spread of molecules (by definition of gas) makes it incredibly unlikely that water will be formed in a liquid phase, because by definition liquid requires densely (relative to gas) packed molecules. I'm not saying a liquid phase won't form over time, it just won't be instantaneous.

8. Nov 17, 2009

### Staff: Mentor

Sorry but I still have problems understanding what you mean. Basically you write "it won't, but it will". Either either. Condensation - no matter how unlikely you think it is - occurs all the time around us, and by definition it requires gaseous phase.

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