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#?

1 Answer
Oct 11, 2016

Answer:

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