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

Question

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?

Answer 1

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.