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### AuthorTopic: Finding Pressures and Temperatures of a Combustion Reaction  (Read 216 times) !function(d,s,id){var js,fjs=d.getElementsByTagName(s)[0];if(!d.getElementById(id)){js=d.createElement(s);js.id=id;js.src="https://platform.twitter.com/widgets.js";fjs.parentNode.insertBefore(js,fjs);}}(document,"script","twitter-wjs"); (function() {var po = document.createElement("script"); po.type = "text/javascript"; po.async = true;po.src = "https://apis.google.com/js/plusone.js";var s = document.getElementsByTagName("script")[0]; s.parentNode.insertBefore(po, s);})();

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#### SpagBol

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##### Finding Pressures and Temperatures of a Combustion Reaction
« on: February 08, 2019, 03:21:42 AM »

Equation:

5C12H22O11(s) + 48KNO3(s)  24K2CO3(s) +36CO2(g) + 55H2O(g) + 24N2(g)

I know the mass'/moles of the products and reactants. I think I've correctly worked out that ΔH = -20224kJ.

What i would like to know is, how to work out the final pressure once the reaction is complete, in a closed container of a known volume. And also a rough estimate of the temperature by the end of the reaction.

I thought you would use real gas laws...? Like 𝑝=𝑅𝑇/(𝑉𝑚−𝑏)−𝑎/(𝑉𝑚^2 ) for each of the gases produced, but unsure on how the change in temperature would effect pressures etc. Or even if any of this is possible...

Any help pointing me in the right direction would be great!
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#### chenbeier

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##### Re: Finding Pressures and Temperatures of a Combustion Reaction
« Reply #1 on: February 08, 2019, 08:45:40 AM »

You get 115 mol gaseous products together. If you know the volume of the container then you can calculate the pressure by a given temperature.
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#### Enthalpy

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##### Re: Finding Pressures and Temperatures of a Combustion Reaction
« Reply #2 on: February 10, 2019, 02:27:20 PM »

Both attempts are wrong.

This reaction is far too hot to be complete. It will not produce just CO2 and H2O like combustions in abundant air do, but lots of CO, plus H2, KOH, and many more.

The proportion of CO vs CO2 is uneasy to estimate by hand because other compounds add their mess. Software is the only reasonable way to go, for instance Propep or RPA.

Because much CO is obtained, the "best" proportions of reactants are not 5:48 moles neither.

This is what Propep computes from 170:100 mass oxidizer:fuel to find 1454K

CH4                  2.4869e-008
CO                   2.1333e-001
CO2                  1.5372e-001
H                    3.6047e-006
HCN                  3.1352e-008
HNCO                 1.8723e-008
H2                   1.4424e-001
HCHO,formaldehy      1.5402e-008
HCOOH                3.1072e-008
H2O                  2.4909e-001
K                    7.5906e-003
KCN                  8.7922e-007
KH                   5.7913e-006
KOH                  3.8836e-002
K2                   7.9887e-007
K2CO3                4.7043e-005
K2O2H2               8.3765e-004
NH3                  1.4311e-006
N2                   1.0820e-001
OH                   1.5129e-007
Condensed species
K2CO3(L)             8.4100e-002

but 290:100 would attain 1635K.
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