# Heat of combustion for octane

I'm a Mechanical Engineering student and I have been working on a school project but I want to go a bit beyond that and it has led me here. I'm designing an engine cylinder, and I figured I should be able to calculate the heat of combustion of the working fluid.

Please remember that I have no chemistry textbook and all I have been doing up to now is using the internet and some vague memory from high school.

So I wrote the balanced equation,

$$\ce{C8H18 + 12.5\left(O2 + 3.76 N2\right) -> 8 CO2 + 9 H2O + 23.5 N2}$$

using that I was able to compute the air-to-fuel ratio, individual masses for the $1.8\ \mathrm L$ volume, etc. I settled on a compression ratio of 8 which after some thermodynamic computation (if neccessary I'll present them) I got to the following state,

$$T_2 = 650.9 \; \text{K} \\ P_2 = 1779 \; \text{kPa}$$

basically those are the conditions prior to the spark. I'm trying to compute the heat of combustion using those properties. Using Mathematica I found out that for Octane the heat of combustion is somewhere along,

$$\Delta H=5474\;\frac{\text{kJ}}{\text{mol}}$$

but I assume this is for STP, my interest is to calculate for my conditions at state 2. After some bit of research I noticed one has to use the specific heats of the participants, but the website stated 'find these at the end of the book' – which I don't have one. To be honest, it will probably take me days to figure out this problem on my own using the internet. So does anyone know of a shortcut such as an online database or some public tables that display specific heats (maybe as a function of temp) for octane. Or how would you go about solving my problem?

Remember that I don't have much knowledge regarding advanced chemistry, and most of my interest lies in the field of engineering.

• First, nitrogen does not participate in the reaction, so remove it (BTW, it's not balanced). You would still need the same volume of air since oxygen is about 21% of air. Could you tell me how much octane is in the cylinder when it is ignited? Also, is there an excess of oxygen?
– LDC3
Sep 26, 2014 at 0:55
• balanced equation is 2C8H18+25O2----> 18H2O+16CO2 May 22, 2018 at 3:13
• Remarkably, also this exercise (like many textbooks and lectures) selects the combustion reaction of octane of all the possible combustion reactions of gasoline to show the principle of combustion, although gasoline actually does not contain a high concentration of octane.
– user7951
May 22, 2018 at 6:51

Enthalpy of a system is a state function, meaning, it does not depend upon the path you take to reach the end state. What that means in this case is that the change in the enthalpy of combustion will be the heat required to raise the temperature of the combustion material to the new temperature and to increase the pressure. Essentially, think of it as if the combustion is occurring at STP, and you must construct a path from the beginning state to the end state which will pass that combustion at STP. In this case, that would be:

1. Initial state of reactants -> STP reactants
2. Heat of combustion at STP
3. STP products -> final state of products

Your final state will be the same as your initial state, 650.9 K/1779 kPa. The important point is that the heat in the change of the reactants to STP will be different from the change of the products back to the initial state. This difference would be reflected if you found a heat of combustion at some condition other than STP.

Nitrogen is a spectator to the reaction, so you can completely ignore it for the sake of reaction. However, you do need to account for raising it to the final pressure/temperature if either is changing. I'll assume you've got that under control.

Now, the issue comes in at the fact these are somewhat extreme conditions, at least in pressure. To calculate this properly would take a great deal of time and effort. That said, there may be corners you are capable of cutting in the process. The general process you need to follow is:

1. Take octane and $\ce{O2}$ from 1779 kPa to 1 atm, 650.9 K to $T_{1,\mathrm i}$.
2. Take octane and $\ce{O2}$ from $T_\mathrm i$ to 25 °C at constant pressure (using $C_p$).
3. Use the heat of combustion.
4. Take $\ce{H2O}$ and $\ce{CO2}$ from 1 atm to 1779 kPa, 25 °C to $T_{2,\mathrm i}$.
5. Take $\ce{H2O}$ and $\ce{CO2}$ from $T_{2,\mathrm i}$ to 650.9 K.

The overall enthalpy change of this process will be the heat of combustion at your conditions. Now, you'll have to decide for yourself about steps 1 and 4, as you need to be considering intermolecular interactions, and thus need to pick an equation of state (not ideal gas). I feel like I am not necessarily knowledgeable enough in this specific field to offer suggestions in that regard. I personally would just pick something like VDW, but there might be a specific standard for combustion processes. Further, only one book that I have access to right now that has the proper temperature-dependent-equation form for $C_p$, and it doesn't have one for octane. These are the ones for the other gases involved:

$\ce{O2}$: $C_p = 25.46 + 1.519 10^{-2} T - 0.715 10^{-5} T^2 + 1.311 10^{-9} T^3$

$\ce{CO2}$: $C_p = 22.243 + 5.977 10^{-2} T - 3.499 10^{-5} T^2 + 7.464 10^{-9} T^3$

$\ce{H2O}$: $C_p = 32.218 + 0.192 10^{-2} T + 1.055 10^{-5} T^2 - 3.953 10^{-9} T^3$

From Chemical and Engineering Thermodynamics (Sandler).

I wish I could help more. Hopefully this explanation at least points you in the right direction.