Question 17.8: Determining Equilibrium Concentrations from Initial Concentr......

Plan We have to find the “composition” of the equilibrium mixture, in other words, the equilibrium concentrations. As always, we write the balanced equation and use it to write the reaction quotient. We find the initial [CO] and [H_2O] from the amounts (0.250 mol of each) and volume (0.125 L), use the balanced equation to define x and set up a reaction table, substitute into Q_c, and solve for x, from which we calculate the concentrations.

Solution Writing the balanced equation and reaction quotient:

                 CO(g)  +  H_2O(g) \xrightleftharpoons[]{}  CO_2(g)  +  H_2(g)              Q_c  =  \frac{[CO_2][H_2]}{[CO][H_2O]}
Calculating initial reactant concentrations:

Setting up the reaction table (Table 1), with x = [CO] and [H_2O] that react (note the 1/1 molar ratio among all reactants and products):

Substituting into the reaction quotient and solving for x:
               Q_c  =  \frac{[CO_2][H_2]}{[CO][H_2O]}  =  \frac{(x)(x)}{(2.00  −  x)(2.00  −  x)}  =  \frac{x^2}{(2.00  −  x)^2}
At equilibrium, we have

               Q_c  =  K_c  =  1.56  =  \frac{x^2}{(2.00  −  x)^2}
We can apply the following math shortcut in this case but not in general: Because the right side of the preceding equation is a perfect square, we take the square root of both sides:

A positive number (1.56) has a positive and a negative root, but in this case, only the positive root has any chemical meaning, so we ignore the negative root:^*

So          2.50  =  2.25  x         \text{therefore}            x  =  1.11  M

Calculating equilibrium concentrations:
             [CO]  =  [H_2O]  =  2.00  M  −  x  =  2.00  M  −  1 .11  M  =  0.89  M
             [CO_2]  =  [H_2]  =  x  =  1.11  M

Check Given the intermediate size of K_c (1.56), it makes sense that the changes in concentration are moderate. It’s a good idea to check that the sign of x in the reaction table is correct—only reactants were initially present, so x has a negative sign for reactants and a positive sign for products. Also check that the equilibrium concentrations
give the known K_c:  \frac{(1.11)(1.11)}{(0.89)(0.89)}  =  1.56.

^*The negative root gives −1.25  =  \frac{x}{2.00  −  x}, or −2.50 + 1.25x = x.
So              2.50 = 0.25x    and    x = 10. M
This value has no chemical meaning because we started with 2.00 M as the concentration of each reactant, so it is impossible for x to be 10. M. Moreover, the square root of an equilibrium constant is another equilibrium constant, which cannot have a negative value.