# Heat of combustion

1. Oct 8, 2014

### Maylis

Hello,

Can the heat of combustion only be found experimentally? I am looking on this page at the heat of combustion of ethanol

http://www.ausetute.com.au/heatcomb.html

and I see that $\Delta H^{o}_c = 1368 \hspace{0.05 in} kJ/mol$. But if you calculate from its constituents

$C_{2}H_{6}O (l) + 3O_{2} (g) \rightarrow 2CO_{2} (g) + 3H_{2}O (l)$

The enthalpy of formation (calculated the same way as enthalpy of combustion??) is using the equation

$$\Delta H^{o}_{rxn} = \sum_{products} \nu_{i} \Delta H_{i}^{o} - \sum_{reactants} \nu_{i} \Delta H_{i}^{o}$$
where $\nu_{i}$ is the stoichiometric coefficient of a chemical species $i$

$[\Delta H_{f,ethanol}^{o} + 3(0 \hspace{0.05 in} kJ/mol)] - [(2)(-393.5 \hspace{0.05 in} kJ/mol) + 3(-285.8 \hspace{0.05 in} kJ/mol)] = 1368 \hspace{0.05 in} kJ/mol$

$\Delta H_{f,ethanol}^{o} = -276.4 \hspace{0.05 in} kJ/mol$

But to calculate the heat of formation of ethanol assumes a priori that you know the heat of combustion of ethanol. So is there a mathematical or theoretical way to calculate the heat of combustion, or is it purely from experiment?

Last edited: Oct 8, 2014
2. Oct 8, 2014

### Staff: Mentor

In theory one should be able to find bonding energy of all molecules involved using QM approach. If memory serves me well this never works perfectly (which I understand as "we are not there yet"). So we stick with methods that use some kind of experimental data.

But I am not confident in what I know, I can be missing some new developments.

3. Oct 8, 2014

### DrDu

This is also a question on the number of significant figures wanted and the availability of experimental data.
Quantum chemical calculations are used mainly to predict the heat of combustion of substances which haven't been synthesized yet. Google will spit up a lot of literature.

4. Oct 8, 2014

### Maylis

I think for my purposes it will just be experimental. I had always been confused between heats of combustion and heat of reaction. It seemed like they were identical, with the exception that a special name was given to reactions that involve combustion with oxygen. Maybe I was doing some mix up with heat of formation and heat of reaction.

5. Oct 8, 2014

### Staff: Mentor

If you have the heats of formation of the reactants and products of a reaction, then you can determine the heat of reaction. That is the power of tabulating heats of formation of species. Sometimes the heat of formation of a species (that goes into a heat of formation table) can be backed out of the heat of combustion, or from the heat of some other reaction that the species participates in. So some experiments are required to get heats of formation. But the advantage is that, by tabulating heats of formation, you don't have to measure the heat of every possible reaction. You only need to do enough experiments to measure the heats of formation. You can then use this data in predicting heats for reaction for reactions that have never been experimentally measured. This is a tremendously powerful tool for accurately predicting heat effects in chemical process calculations.

Chet

6. Oct 8, 2014

### Maylis

See, if you think of heat of reaction and heat of combustion as being the same thing, then there is a paradox.

To calculate the heat of reaction, you need the heat of formation. But to calculate the heat of formation, you need the heat of combustion.

7. Oct 8, 2014

### Staff: Mentor

Apparently, I didn't explain very well in my previous post. Measuring heat of combustion is not the only way to get the heat of formation of a substance. The idea is to get the heats of formation of substances by doing a minimum number of experiments. To see a better explanation than mine, see Smith and van ness.

8. Oct 9, 2014

### Staff: Mentor

Which is why we choose an arbitrary reference of zero for the elements in their most stable form.

9. Oct 9, 2014

### Maylis

I was reading smith and van ness section on Heat of Combustion (chapter 4, section 5 in the 7th edition) before I created this thread, which is what started this whole thing. The issue is how the number for the combustion of n-butane just comes out of nowhere. I know how they got the heats of reaction for the first two reactions. But 2,877,396 kJ/mol just seems to be placed in their with no justification.

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10. Oct 9, 2014

### Staff: Mentor

Looks like the combustion of butane, just reversed (flip the sign).

11. Oct 9, 2014

### Maylis

Okay forget about it, this is just going around in circles. I'm just going to accept that the enthalpy of combustion of a species is determined experimentally and should be looked up. Thank you!

12. Oct 9, 2014

### Yanick

Yes, heat of combustion is usually determined experimentally. It is a pretty trivial measurement by today's standards and very accessible. I think you may have it a little backwards.

To determine the heat of formation of butane, you need to have graphite and hydrogen gas in their standard states and react to form butane. Well good luck getting that done experimentally, it is not trivial. What is trivial is combusting butane (thereby getting the heat of combustion) and calculating the heat of formation from the chemical equation and the equation for hear of reaction (of which combustion is a special case). You will have 4 species in your chemical equation, oxygen (zero Hf), water (tabulated), carbon dioxide (tabulated), and butane. You can then solve for the heat of formation of butane by doing a relatively simple measurement, and looking up well known values, instead of having to figure out how to make butane from all of its elements in their standard state.

13. Oct 10, 2014

### Staff: Mentor

Yanick is right on target. I think the question is "where did the number for the heat of formation for butane that is listed in the table come from?" That answer was that, if you know the heat of formation of water and carbon dioxide, and you measure the heat of combustion of butane, you can back out the heat of formation of butane.

Chet

Last edited: Oct 10, 2014