4
$\begingroup$

I don't quite know how summarize my question in the title. I have a combustion process given by this equation: $$\ce{C10H22 + 15.5 O2 + 58.25 N2 -> 11 H2O + 10 CO2 + 58.25 N2}$$

This process occurs inside of a cylindrical combustion chamber of area A, but I think this is irrelevant for the purposes of my question. The combustion process is stationary, and the gas mixtures are ideal.

I wanted to know if, in general, this is true: $$\frac{\text{molar flow rate of }\ce{N2}}{\text{molar flow rate of }\ce{C10H22}} = \frac{\text{amount of substance of }\ce{N2}}{\text{amount of substance of }\ce{C10H22}}=\frac{58.25}{1}=58.25$$

With molar flow rate I want to mean the amount of substance that leave/enter the combustion chamber in 1 second.

I'm a beginner in chemistry, but I'm not sure if I can convert a ratio between molar flows like I did, using directly the stoichiometric coefficients in the balanced equation.

The idea behind my reasoning is that: $$\text{molar flow rate of }\ce{N2} = \frac{n_{\ce{N2}}(t+\Delta t) - n_{\ce{N2}}(t)}{\Delta t}$$ and: $$\text{molar flow rate of }\ce{C10H22} = \frac{n_{\ce{C10H22}}(t+\Delta t)-n_{\ce{C10H22}}(t)}{\Delta t}$$

Therefore, $$\frac{\text{molar flow rate of }\ce{N2}}{\text{molar flow rate of }\ce{C10H22}} = \frac{n_{\ce{N2}}(t+\Delta t) - n_{\ce{N2}}(t)}{n_{\ce{C10H22}}(t+\Delta t) - n_{\ce{C10H22}}(t)}$$

And, intuitively, this last equality should be proportional to $n_{\ce{N2}}/n_{\ce{C10H22}}=58.25/1=58.25$.

$\endgroup$
  • $\begingroup$ Welcome to Chem SE! If you haven't yet, take the tour [chemistry.stackexchange.com/tour] and if you have any other questions stop by the [help center] [chemistry.stackexchange.com/help]. Could you clarify in what context you have seen an integrated heat capacity? It is not a term I'm familiar with. $\endgroup$ – Another.Chemist Apr 15 '17 at 3:00
  • $\begingroup$ Hi, thank you! I'm studying the combustion chamber of a civil jet, and need to study the thermodynamics that happen in there. I'm currently doing an energy balance, which includes mass flow rates. But mass flow rates can be converted to molar flow rates, which in turn lead me to this question. $\endgroup$ – Jose Lopez Garcia Apr 15 '17 at 3:02
  • $\begingroup$ Looks like solid reasoning to me. $\endgroup$ – Martin - マーチン Apr 26 '17 at 12:12
  • $\begingroup$ Look, the molar flow rate is a production or consommation of a number of mole per unity of time. Your process is stationary so there is no variation of concentration with the time. What it says is if you do your reaction from 9:00 am to 10:00 am or from 9:10 am to 10:10 am your reaction will spend and produce exactly the same amount of each molecules in both cases. So there is no limitation to do what you did. $\endgroup$ – ParaH2 Apr 26 '17 at 18:18

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.