Question #80c7b
1 Answer
Explanation:
So, you're dealing with an equilibrium reaction that features carbon monoxide,
The first thing to do here is take a look at the value of the equilibrium constant,
Since
Now, notice that the reaction mixture does not contain any water at first. This tells you that the reaction will proceed in the direction that will result in the production of water.
Since water is a reactant, the equilibrium will shift to the left, i.e. consume carbon dioxide and hydrogen gas and produce water and more carbon monoxide.
Since you're dealing with
So, use an ICE table to help you with the calculations
#" " "CO"_text((g]) " "+" " "H"_2"O"_text((l]) " "rightleftharpoons" " "CO"_text(2(g]) " "+" " "H"_text(2(g])#
By definition, the equilibrium constant for this reaction will be
#K_c = (["CO"_2] * ["H"_2])/(["CO"] * ["H"_2"O"])#
In your case, this will be equivalent to
#K_c = ((1.6-x) * (3.0 -x ))/(x * (2.2 + x)) = 0.297#
Rearrange to get
#0.297 * (x^2 + 2.2x) = x^2 - 4.6x + 4.8#
This is equivalent to
#0.703x^2 - 5.2534x + 4.8 = 0#
This quadratic equation has two solutions, both positive
#x_1 ~~ 6.41" "# and#" "x_2 ~~ 1.066#
Since you need the equilibrium concentrations of all species to be positive, you must take
Therefore, the equilibrium concentration of water will be
#["H"_2"O"] = color(green)("1.1 M") -># to one decimal place
SID NOTE Notice that the equilibrium indeed shifted to the left in accordance to the value of the equilibrium constant.
In fact, for the same starting concentrations, the reaction will always shift to the left, since that will result in the production of water.
However, for