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
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,
In this regard, the expression of the equilibrium constant will depend exclusively on the concentrations of carbon monoxide,
#K_c = (["CO"_2])/(["CO"])#
Now, you know that you're starting with
If you take
#["CO"_2] = x#
#["CO"] = 0.10 - x#
Plug this into the expression of
#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