First off, I am not a chemist. I'm an engineer with an interest in thermodynamics for engine simulation for the basis of emissions. So if I have asked something stupid, this will be why.

So I am trying to work out the partial pressure equilibrium constant for dissociation of CO2 to carbon monoxide and oxygen, to then move onto multiple reactions to calculate thermal NOx using the extended zeldovich mechanism. Now from engineering textbooks, $K_p$ is used as the partial pressure of the reactants, divided by the partial pressure of the products. However, from chemistry definitions, it is the other way round. (Partial pressure of products over reactants.)

My question is, can both be used, or is there a trick necessary to be able to this I have missed? The math works out a bit nicer the engineering way, but I want it to be correct obviously.

I have checked all my engineering books and found that indeed, it is defined as products divided by reactants. However, I found the bit that is confusing me, just it was hidden behind math.

For calculating the standard Gibbs free energy change, is that always sum of products minus sum of reactants?

I've done the calculations, and the only way I can get the math to agree is sum of reactants – sum of products:

From "Transport Fuels Technology"

$$K(\ce{CO2}) = 3.0549$$

From JANAF Tables

$$ \begin{align} ΔG_\mathrm{f}^\circ(\ce{CO2}) &= -395.461 \\ ΔG_\mathrm{f}^\circ(\ce{CO}) &= -367.826 \\ ΔG_\mathrm{f}^\circ(\ce{O2}) &= 0 \end{align} $$

$$ΔG^\circ = -RT\ln K$$ $$\frac{ΔG^\circ}{-RT} = \ln K$$

$$ \begin{align} K &= \exp{\left(\frac{ΔG^\circ}{-RT}\right)} \\ &= \exp{\left(\frac{27635}{-8.314\cdot 3000}\right)} \\ &= 0.330 \end{align} $$

Or reverse $ΔG^\circ$:

$$ \begin{align} ΔG^\circ &= \sum ΔG_\mathrm{f}^\circ (\text{reactants}) - \sum ΔG_\mathrm{f}^\circ (\text{products}) \\ &= -395.461 - (-367.826) \\ &= \pu{-27.635 kJ mol-1} \\ &= \pu{-27635 J mol-1} \end{align} $$

$$ΔG^\circ = -RT\ln K$$ $$\frac{ΔG^\circ}{-RT} = \ln K$$

$$ \begin{align} K &= \exp{\left(\frac{ΔG^\circ}{-RT}\right)} \\ &= \exp{\left(\frac{-27635}{-8.314\cdot 3000}\right)} \\ &= 3.028 \end{align} $$

which is in agreement with the book I am using.

  • 3
    $\begingroup$ I've never seen a chemistry engineering textbook where "reversed" $K_\mathrm{p}$ is used. Could you please add some references or context where exactly you've seen it? $\endgroup$
    – andselisk
    Mar 2 '19 at 6:56
  • 1
    $\begingroup$ As a chemical engineer, I have never seen this either. It is always products divided by reactants, as indicated in Fundamentals of Engineering Thermodynamics by Moran et al, Introduction to Chemical Engineering Thermodynamics by Smith and van Ness, and Chemical Engineers' Handbook by Perry et al. $\endgroup$ Mar 3 '19 at 13:33
  • $\begingroup$ Please don't provide links to the third-party cloud services/screenshots etc. I also would encourage not to use Word since it's not suited for scientific publishing at all. I re-typed the page from your Dropbox with MathJax, feel free to correct it if I made a mistake somewhere. Also, please visit this page, this page and this one on how to format your future posts better with MathJax and Markdown. $\endgroup$
    – andselisk
    Mar 7 '19 at 14:17
  • $\begingroup$ As for the question, it would be nice you'd added the reaction you are investigating the equilibrium for. Is it burning CO to CO2? $\endgroup$
    – andselisk
    Mar 7 '19 at 14:23
  • $\begingroup$ OK, i will have a look at MathJax. Have not come across that one. Thankyou for your patience. I have updated my question to show i am looking at dissociation of CO2 to CO + 0.5 O2 $\endgroup$
    – Ross Hanna
    Mar 7 '19 at 21:45

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.