Temperature of thermochemical reaction between propane and nitrous oxide

I'm trying to calculate the properties of the combustion process using propane and nitrous oxide. When I tried to nail down the combustion temperature, the result looks just off to me. I went through several times with the process but since the reaction should produce less enthalpy change than using pure oxygen gas and my results is above the standard combustion temperature with pure oxygen, it is definitely wrong somewhere.

The process, how I got my solution is the following:

$$C_3H_8 + 10 N_2O => 3 CO_2 + 4 H_2O + 10 N_2 + \Delta_rH$$

$$\Delta_rH = \Delta_fH_p - \Delta_fH_r$$

Where the r subscript means reactants and p is for products.

$$\Delta_fH(C_3H_8) = -104.7 kJ/mol$$ $$\Delta_fH(CO_2) = -393.5 kJ/mol$$ $$\Delta_fH(H_2O) = -241.8 kJ/mol$$ $$\Delta_fH(N_2O) = +82.05 kJ/mol$$ $$\Delta_fH(N_2) = 0 kJ/mol$$

Which amounts to the following conclusion:

$$\Delta_rH = (3 * -393.5 kJ + 4 * -241.8 kJ) - (1 * -104.7 kJ + 10 * 82.05 kJ) = -2863.5 kJ$$ $$\Delta_rH = -2863.5 kJ / 0.484 kg = -5916.3223 kJ/kg$$

This value is for 10 mols of Nitrous Oxide and 1 mol Propane. Since both molecules have the molar mass around 44 g / mol, it's essentially the stoichometric ratio of 1 to the 10.

So far, so good (I think). The result seems reasonable to me, as I went through the same process using only O2, which ended up considerable more as expected (it only takes 5 part of O2 so the reaction produces more enthalpy with the same mass, plus O2 molar weight is 32.)

In the next step I used the following formula:

$$\Delta_rH = \sum_{i=1}^m n_i \int_{T_0}^{T_r}C_pdT$$

I assumed here that the: $$\int_{T_0}^{T_r}C_pdT$$ part can be simplified with the assumption that the Cp won't change with the temperature. I'm well aware that this isn't the exact case, and I've found that for example water vapour has a particular big swing over the temperature in terms of heat capacity on a constant pressure. However, I tried to work with the values I could find around the internet, and in the case of water vapour I worked with the temperature that looked like a quite good mean value. So I continued like this:

$$\int_{T_0}^{T_r}C_pdT = \Delta T * \sum_{i=1}^m n_iC_p$$ $$T * \sum_{i=1}^m n_iC_p = \Delta_rH$$ $$T = \frac{\Delta_rH} {\sum_{i=1}^m n_iC_p}$$ $$T = 2863500 J / (3 * 37.135 J/K + 4 * 53.1 J/K + 10 * 29.12 J/K) = 4656.06 K$$

This value is obviously not good, given that the pure oxygen reaction combustion temperature with propane is 3093 K Ref. Now, is there some deeply flawed assumption in my process? Can it be that the simplification of that integral expression led me down on a wrong calculation in the first place? If so, how can I access to the exact functions of these heat capacities for the gases I'm using?

• Mols and enthalpies... this might get a better response at chemistry.stackexchange.com – user10851 Oct 4 '13 at 17:08
• Thermodynamics and statistical mechanics/physical chemistry are one of the blurry places where physics and chemistry come together we just have to share the sandbox. I don't think that this is quite off-topic here, but Chris is right that you might get more and/or better answers on Chem.SE. If you want it migrated just flag for moderator attention and ask. – dmckee --- ex-moderator kitten Oct 4 '13 at 18:40
• As I understand, the problem here is more on the thermodynamics side of things. Perhaps it is misleading that I started out from the chemical reaction, but I just wanted to show the entire problem to place it in to proper context. – progician Oct 5 '13 at 15:06
• Note that ThermoD is on topic for Chem too. I don't mind migrating this (I mod Chem too, it seems OK for both sites), but if you want it to stay here I can go with that :) – Manishearth Oct 6 '13 at 8:05

So, I think I found the answer to my problem. I was right when I suspected the integral expression to be overly simplified. Eventually I found the necessary data in my book, the Cengel, Boles: Thermodynamics, An Engineering Approach.

Basically, the heat capacity of common gases is a function of temperature. Taken these gases as ideal gases, the heat capacity function can be approximated with the following polynomial:

$$C_p = a + b \Delta T + c \Delta T^2 + d \Delta T^3$$

So the integral expression of the enthalpy change becomes:

$$\int C_pdT = adT + \frac{bdT^2}{2} + \frac{cdT^3}{3} + \frac{bdT^4}{4}$$

In the book's Appendix A, the molar heat capacity function's coefficients for common gases can be found in the table A-2C. Unfortunately however, these heat capacity coefficients are listed within the temperature range of 273K to 1800K which is unfortunately slightly too low for determining the adiabatic flame temperature. I indeed fell in to this trap, because when I tried to narrow it down, I got a temperature of >7500K which is obviously wrong. Never the less, using the linear coefficient only, the temperature falls in to the range of the possible solution cca 3000-3200K.

Also, since the integral value is a quartic function, the solution gets a bit tricky, and therefore it is easier to narrowing down to the right temperature.

EDIT:

Could anybody point me to some more accurate heat capacity functions/data at high temperatures for the common gases of N2, CO2, H2O (>1800K)? It's quite annoying that because of the lack of data I can't put a more precise number on this.

if the reaction is not problem, the answer is proper. because nitrous oxide decomposition release energy. then the temperature is high than react with oxygen. In fact the temperature is high than reality. because the reaction is not correct. the reaction production include many other species.