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Zinc oxide is reduced at a constant temperature in a closed reactor using $\ce{ZnO(s)}$ and $\ce{C(s)}$ as the only starting materials. The following reactions are assumed to be at thermodynamic equilibrium. \begin{align} \ce{ZnO(s) + C(s) &-> Zn(g) + CO(g)}\\ \ce{2CO(g) &-> CO2(g) + C(s)} \end{align}

Assume ideal gas behaviour. Based on mole balance, the relationship applicable to the system at equilibrium is: \begin{align} \tag{A} P_{\ce{Zn}} &= P_{\ce{CO}} + 2P_{\ce{CO2}}\\ \tag{B} P_{\ce{Zn}} &= 2P_{\ce{CO}} + P_{\ce{CO2}}\\ \tag{C} P_{\ce{Zn}} &= P_{\ce{CO}} + P_{\ce{CO2}}\\ \tag{D} P_{\ce{Zn}} &= 0.5P_{\ce{CO}} + 2P_{\ce{CO2}} \end{align}

In this problem seeing that none of the given options consisted of $\ce{C(s)}$, so in order to get rid of that I just added the two equations and arrived at the equation $$\ce{ZnO(s) + CO(g) = Zn(g) + CO2(g)}.$$ Now just sort of comparing with the options I find that $\ce{CO(g)}$ and $\ce{CO2(g)}$ are on the opposite sides. It sort of gives me the intuition that may be the pressure term should have had a minus. Anyway I know that I have gone horribly wrong somewhere with regard to my concepts. Please tell me how to solve the problem henceforth.

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First of all, let me explain why your approach is a bit wrong... Addition of steps will only give you the net reaction, which is what you got. But in the process, the mechanism becomes blurred. For example, the net reaction which you have expressed, doesn't tell us the full story. One completely overlooks the fact the CO is also recombining to give CO2, and that is exactly where the fallacy arises.

Here is my solution to the problem. I hope it helps.

My solution to the problem. Hope it helps.

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  • $\begingroup$ Tell me something why are you equating number of moles of Zn(g) with CO(g) in the last step? $\endgroup$
    – user586228
    Commented Aug 15, 2019 at 15:06
  • $\begingroup$ From the first equation.. it is obvious that equal number of moles of Zn and CO are formed... The amount of CO consumed is inferred from the second equation... Now just add the results to get the residual moles of CO in the mix. $\endgroup$
    – Om Nom
    Commented Aug 15, 2019 at 16:16
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    $\begingroup$ Please note that the proper term for "number of moles" is amount of substance. The former would be the same as referring to the mass as "number of kilograms". A screenshot or picture of an exercise is not searchable. Please consider rewriting it, so that it can be of help for future visitors. $\endgroup$
    – Martin - マーチン ♦
    Commented Aug 22, 2019 at 14:39

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