# Calculating the amount of a product at the reaction equilibrium

$$\ce{Cr2O3}$$ 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. 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?

• You already have the equilibrium constants, so you don’t need to mess with the thermodynamics. – Chet Miller Jan 21 at 20:37
• The equilibrium constant values determined at 1500K tell you something about the partial pressures of the two gases in the final state. – Chet Miller Jan 21 at 20:41
• Yeah they do. But how can I use that to get some statement about the chromium here? – xotix Jan 21 at 20:46
• If you have the partial pressures, then you have the number of moles. – Chet Miller Jan 21 at 20:47
• 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? – xotix Jan 21 at 21:06