# Question 80c7b

Dec 30, 2015

["H"_2"O"] = "1.1 M"

#### Explanation:

So, you're dealing with an equilibrium reaction that features carbon monoxide, $\text{CO}$, and water, $\text{H"_2"O}$, as the reactants and carbon dioxide, ${\text{CO}}_{2}$, and hydrogen gas, ${\text{H}}_{2}$, as the products.

The first thing to do here is take a look at the value of the equilibrium constant, ${K}_{c}$.

Since ${K}_{c} < 1$, you know that the equilibrium will lie to the left, i.e. the reaction mixture will contain more reactants than products at equilibrium.

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 $1 : 1$ mole ratios between all four chemical species, you can predict that the equilibrium concentration of hydrogen gas will be the lowest of the group.

${\text{ " "CO"_text((g]) " "+" " "H"_2"O"_text((l]) " "rightleftharpoons" " "CO"_text(2(g]) " "+" " "H}}_{\textrm{2 \left(g\right]}}$

color(purple)("I")" " " "2.2" " " " " " " " " "0" " " " " " " " " "1.6" " " " " " " "3.0
color(purple)("C")" "(+x)" " " " " "(+x)" " " " " " " "(-x)" " " " " "(-x)
color(purple)("E")" "2.2+x" " " " " " " "x" " " " " " " "1.6-x" " " " " "3.0-x#

By definition, the equilibrium constant for this reaction will be

${K}_{c} = \left(\left[\text{CO"_2] * ["H"_2])/(["CO"] * ["H"_2"O}\right]\right)$

In your case, this will be equivalent to

${K}_{c} = \frac{\left(1.6 - x\right) \cdot \left(3.0 - x\right)}{x \cdot \left(2.2 + x\right)} = 0.297$

Rearrange to get

$0.297 \cdot \left({x}^{2} + 2.2 x\right) = {x}^{2} - 4.6 x + 4.8$

This is equivalent to

$0.703 {x}^{2} - 5.2534 x + 4.8 = 0$

This quadratic equation has two solutions, both positive

${x}_{1} \approx 6.41 \text{ }$ and $\text{ } {x}_{2} \approx 1.066$

Since you need the equilibrium concentrations of all species to be positive, you must take

$x = 1.066$

Therefore, the equilibrium concentration of water will be

$\left[\text{H"_2"O"] = color(green)("1.1 M}\right) \to$ 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 ${K}_{c} > 1$, this shift will be less significant, meaning that the equilibrium concentration of water will be lower.