Question

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`?

Answer 1

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, `"CO"`, and of carbon dioxide, `"CO"_2`.

`K_c = (["CO"_2])/(["CO"])`

Now, you know that you're starting with `"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

`["CO"_2] = x`

`["CO"] = 0.10 - x`

Plug this into the expression of `K_c` and solve for `x`

`4.0 * 10^3 = x/(0.10 - x)`

This gets you

`x = 4 * 10^3 * (0.10 - x)`

`x = 4 * 10^2 - 4 * 10^3x`

`x * (4 * 10^3 + 1) = 4 * 10^2`

`x = (4 * 10^2)/(4 * 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

`"0.000025 M" = "0.00 M" " ->` rounded to two decimal places