Calculate the equilibrium composition of a reactive system consisting of an equimolecular mixture of $\ce{CH4}$ and $\ce{H2O}$ at $\pu{900 K}$ and $\pu{1 atm},$ considered to be an ideal mixture of ideal gases. Under these conditions, the equilibrium constants of the two possible reactions are

$$ \begin{align} \ce{CH4 + H2O &<=> CO + 3 H2} &\quad K_{p,1} &= 1.307 \tag{R1}\\ \ce{CO + H2O &<=> CO2 + H2}&\quad K_{p,2} &= 2.210 \tag{R2} \end{align} $$

I have tried to solve it by expressing each $K_{p,i}$ independently, considering there are five constituents in this ideal mixture and that $\ce{CH4}$ and $\ce{H2O}$ are the starting compounds, with $n^0 = 1.$ I associated a different degree of advance, $\xi_i,$ to each of the reactions and solved the two-equation system, but it seems that this is not the proper solution because the reactions are not independent.

Can anybody figure out a solution?

  • 1
    $\begingroup$ Editing note: both $K_p$ values had a comma which I interpreted as a decimal separator, but it might as well be a thousands separator. I'm too lazy to look up a plausible $K_p$ value, so please be aware. $\endgroup$
    – andselisk
    Apr 23, 2021 at 17:48
  • 1
    $\begingroup$ Vandalism is not accepted behaviour even over own posts. $\endgroup$
    – Poutnik
    Apr 26, 2021 at 6:23

1 Answer 1



$$moles\ CH_4=1-e_1$$ $$moles\ H_2O=1-(e_1+e_2)$$ $$moles\ CO=e_1-e_2$$ $$moles\ CO_2=e_2$$ $$moles\ H_2=3e_1+e_2$$ where the $e_1$ is the number of moles of $CH_4$ destroyed and $e_2$ the number of moles of $CO_2$ that are produced.

So the equilibrium equations are going to read:$$\frac{(e_1-e_2)(3e_1+e_2)^3}{(1-e_1-e_2)(1-e_1)(2+2e_1)^2}=K_1$$and $$\frac{e_2(3e_1+e_2)}{(e_1-e_2)(1-e_1-e_2)}=K_2$$

  • $\begingroup$ @ShoubhikRMaiti. This is the start of how to solve it. I don't want to reveal too much. What part of this don't you understand? $\endgroup$ Apr 23, 2021 at 22:11
  • $\begingroup$ Ok I see, you are giving a hint answer. I think you should change the wording of number of times each of the two reactions has been executed, I had to read it a couple of times to understand. Maybe change it to extent of reaction or something like that. Also, hint answers are discouraged nowadays: chemistry.meta.stackexchange.com/questions/4287/… $\endgroup$
    – S R Maiti
    Apr 23, 2021 at 22:15
  • $\begingroup$ @ShoubhikRMaiti OK, I changed the wording to make things clearer. $\endgroup$ Apr 23, 2021 at 23:53
  • $\begingroup$ The reactions don't each have separate numbers of moles in the denominator to get the mole fractions for each. The total number of moles for both reactions is $(2+2e_1)$. And, the two equations are coupled, so you must solve these two non-linear algebraic equations simultaneously. $\endgroup$ Apr 24, 2021 at 11:39
  • $\begingroup$ –1 for improper notations and terms like "$moles\ CH_4$", "number of moles" and also, this is a hint, not an answer. Please post suggestions and hints in the comments section shall you decide not to reveal the complete solution. $\endgroup$
    – andselisk
    Apr 24, 2021 at 12:06

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