Consider the reaction: NiO(s) + CO(g) rightleftharpoons Ni(s) + CO_2(g). K_c = 4.0 xx 10^3 (at 1500 K). If a mixture of solid nickel (II) oxide and 0.10M carbon monoxide comes to equilibrium at 1500 K, what is the equilibrium concentration of CO_2?

Oct 11, 2016

Here's what I got.

Explanation:

The trick here is to make sure that you do not include the concentrations of the two solids, nickel(II) oxide and nickel metal, in the expression of the equilibrium constant, ${K}_{c}$.

In this regard, the expression of the equilibrium constant will depend exclusively on the concentrations of carbon monoxide, $\text{CO}$, and of carbon dioxide, ${\text{CO}}_{2}$.

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

Now, you know that you're starting with $\text{0.10 M}$ of carbon monoxide and an unknown quantity of nickel(II) oxide. Notice that the reaction produces $1$ mole of carbon dioxide for every mole of carbon monoxide that takes part in the reaction.

If you take $x$ to be the concentration of carbon dioxide formed by the reaction, i.e. the concentration of carbon dioxide at equilibrium, you can say that

$\left[{\text{CO}}_{2}\right] = x$

$\left[\text{CO}\right] = 0.10 - x$

Plug this into the expression of ${K}_{c}$ and solve for $x$

$4.0 \cdot {10}^{3} = \frac{x}{0.10 - x}$

This gets you

$x = 4 \cdot {10}^{3} \cdot \left(0.10 - x\right)$

$x = 4 \cdot {10}^{2} - 4 \cdot {10}^{3} x$

$x \cdot \left(4 \cdot {10}^{3} + 1\right) = 4 \cdot {10}^{2}$

$x = \frac{4 \cdot {10}^{2}}{4 \cdot {10}^{3} + 1} = 0.099975$

Notice that the answer must be rounded to two sig figs, since that is how many sig figs you have for the concentration of carbon monoxide, so

["CO"_2] = color(green)(bar(ul(|color(white)(a/a)color(black)("0.10 M")color(white)(a/a)|)))

Keep in mind that you have

["CO"] = "0.10 M" - "0.10 M" ~~ "0.00 M"

because

["CO"] = "0.10 M" - "0.099975 M" = "0.000025 M"

and

$\text{0.000025 M" = "0.00 M" } \to$ rounded to two decimal places