# How is the equilibrium constant for the water-gas shift reaction calculated after it is disturbed?

Consider the following equilibrium process at $$\pu{686 ^\circ C}$$: $$\ce{CO2(g) + H2(g) <=> CO(g) + H2O(g)}$$

The equilibrium concentrations of the reacting species are $$[\ce{CO}] = \pu{0.050 M}$$, $$[\ce{H2}] = \pu{0.045 M}$$, $$[\ce{CO2}] = \pu{0.086 M}$$, and $$[\ce{H2O}] = \pu{0.040 M}$$. (a) Calculate $$K_c$$ for the reaction at $$\pu{686 ^\circ C}$$. (b) If we add $$\ce{CO2}$$ to increase its concentration to $$\ce{0.50 mol/L}$$, what will the concentrations of all the gases be when equilibrium is reestablished?

I've answered (a) already, and I got $$K_p$$ and $$K_c$$ both equal to $$0.52$$; in other words, I'm certain that the system is in equilibrium. However, I'm having trouble answering (b). Do just add $$\pu{0.50 mol/L}$$ to all the other concentrations?

So at equilibrium, we can list the concentration of each species:

\begin{align} [\ce{CO2}] &= 0.50 - x = 0.500 - x,\\ [\ce{H2}] &= 0.045 - x,\\ [\ce{CO}] &= 0.050 + x,\\ [\ce{H2O}] &= 0.040 + x, \end{align} where $$x$$ is the change in concentration that is given to all species.

Therefore, we get the following expression at the new equilibrium: $$\frac{[\ce{H2O}][\ce{CO}]}{[\ce{CO2}][\ce{H2}]} = \frac{(0.050+x)(0.040 + x)}{(0.045 - x)(0.500 - x)} = 0.5168$$

We could plug the equation into WolframAlpha and solve for $$x$$ but if you can't do that, you can make some reasonable approximations.

First, since $$0.500-x$$ is likely a small change, we can approximate it to be only $$0.500$$, giving us a new expression of $$\frac{(0.050 + x)(0.040 + x)}{(0.045 - x)(0.500)} \approx 0.5168.$$

At this point, you can't really approximate anymore. Honestly, at this point, either brute force it by solving the quadratic or plug it into an online calculator. You eventually get a value of $$x = \pu{0.0251 M}$$. That means the concentration of each species at equilibrium are: \begin{align} [\ce{CO2}] &= \pu{0.4749 M},\\ [\ce{H2}] &= \pu{0.0199 M},\\ [\ce{CO}] &= \pu{0.0751 M},\\ [\ce{H2O}] &= \pu{0.0651 M}. \end{align}

It's good practice to plug these values back into the equilibrium expression to see if we made bad approximations, but in our case, it matches.

• I've fixed your MathJax, please use \ce{...} only for actual chemical expressions. Here's some boilerplate with links: On Chemistry mathematical and chemical expressions can be formatted using MathJax (and LaTeX Syntax). If you want to know more, please have a look here and here. We prefer to not use MathJax in the title field, see here for details. Aug 6 at 23:16
• The question says "add CO2 to increase its concentration to 0.50mol/L," so I interpret that to mean the new CO2 is .50. Change the denominator a bit to have (.500 -x). My old graphing calculator (TI83 Plus) gives x=.02516. Aug 7 at 21:35