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$\ce{Cr2O3}$ can be reduced to the elements chromium with high temperature and carbon by this two reactions:

$$ \begin{align} \ce{Cr2O3(s) + 3C(s) &-> 2Cr(s) + 3 CO(g)} \label{rxn:1}\tag{1}\\ \ce{Cr2O3(s) + \frac{3}{2}C(s) &-> 2Cr(s) + \frac{3}{2}CO2(g)} \label{rxn:2}\tag{2} \end{align} $$

The following image shows the equilibrium constants.

Equilibrium constants

Using the image, we can get $K_1$, $\Delta_rG^\circ_1$, $\Delta_rH^\circ_1$, $\Delta_rS^\circ_1$ of reaction $\eqref{rxn:1}$ as well as $K_2$ for $T = 1500\,\mbox{K}$.

Assume all of them are given. Now:

In a closed container with $V = 1.00\,\mbox{dm}^3$ inner volume, we put $5.00\,\mbox{g}$ of $\ce{Cr2O3}$ and $20.0\,\mbox{g}$ carbon. For $T = 1500\,\mbox{K}$ we have the reaction equilibrium $\eqref{rxn:1}$ and $\eqref{rxn:2}$.

Problem: What mass (in g) reduced chromium do we have in the equilibrium?

Question: Now I don't know how to solve the problem. I can do everything above like determining the Gibbs-Energy or the equilibrium constants, but what are my though now to get the amount of reduced chromium?

First, I don't get exactly what it means if they say "in the equilibrium". Does that refer to $\eqref{rxn:1}$ or $\eqref{rxn:2}$ or both?

Further, how do I actually calculate it?

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  • 2
    $\begingroup$ You already have the equilibrium constants, so you don’t need to mess with the thermodynamics. $\endgroup$ – Chet Miller Jan 21 '19 at 20:37
  • $\begingroup$ The equilibrium constant values determined at 1500K tell you something about the partial pressures of the two gases in the final state. $\endgroup$ – Chet Miller Jan 21 '19 at 20:41
  • $\begingroup$ Yeah they do. But how can I use that to get some statement about the chromium here? $\endgroup$ – xotix Jan 21 '19 at 20:46
  • $\begingroup$ If you have the partial pressures, then you have the number of moles. $\endgroup$ – Chet Miller Jan 21 '19 at 20:47
  • $\begingroup$ I think I might have seen my mistake. So I basically use $pV=nRT$ to get the moles of $CO_2$ and $CO$. I further know how much moles in total I put into the reaction. Now I can take the difference and get how much $Cr$ I got, right? $\endgroup$ – xotix Jan 21 '19 at 21:06

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