[Spraygun]

Hi,

Im writing a coursework for chemistry this year and have to calculate the temperature that will develop at theburning process of Pentane+Oxygen.

First of all I figured the Enthalpy, which is 3.276kJ/mole.

In the experiment we did at school wasnt enough space for such a mole configuration, so I figured that  a Volume of 0.808l is enough space for 3.305*10^-2 mole.
21% of that are Oxygen, which will be needed for the pentane to burn.

21%*3.305*10^-2 mole = 6.940*10^-3mole O2

1mole Pentane needs 8 mole Oxygen to burn, so 1:8 for Pentane

6.940*10^-3mole O2 / 8 = 8,676*10^-4mole Pentane

I used the same method to figure the CO2 5:8 and H2O 6:8
CO2 = 4,338*10^-3 mole
H2O = 5.205*10^-3 mole

Now I multiplied the actaul used amount of Pentane with the Enthalpy, to find out how much energy will be freed.

8,676*10^-4mole * 3.276kJ/mole = 2.842kJ

Now comes the tricky part.
Given the product gases H2O and CO2, as well as the N2(78% of the mole at the beginnig of the reaction) are ideal gases with two degrees of freedom will make their heat capacity at constant volume:
H20 = 5.205*10^-3 mole * 4 * R     =  0.173 J/K
CO2 = 4,338*10^-3 mole *3.5 * R   =  0.126 J/K
N2 = 2.78*10^-2 mole *3.5*R         =  0.750 J/K

Cv = 1.050 J/K

Now I  thought it would be possible to calculate the temperature through this formula:
Q=C*dT
With Q being dH=2.842kJ.
dT=dH/C
dT=2842 J /1.050 J/K
dT=3005K

I have my doubts that such a small amount of pentane will create a heat of 3005K. But I dont have anything to compare it with, due to my lack of expierience on account of me being a student in year 12
Could anyone verify this calculation ?

[Valdorod]

What are you using as your decimal mark a comma or a period?

6.940*10^-3mole O2 / 8 = 8,676*10^-4mole Pentane

It seems that you are mixing your marks.  Using a period as the decimarl mark you should get

8.676E-4 mole of pentane

8,676*10^-4mole * 3.276kJ/mole = 2.842kJ

Which would make the above calculation = 2.842E-3 KJ or 0.284 J


By the way typical combustion temperatures for heating oil and natural gas are around 2000 °C.

Valdo

[Spraygun]

Hi,

yeah its a Problem to do with the different software I use.

So decimal mark is a ,

Every . is supposed to be a decimal mark in this calculation.

So it is indeed 2,842kJ of energy, so 2842 J

Sorry, I didnt mind this problem when writing it.