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 $CO_2$?

Md Ariful Islam

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Md Ariful Islam

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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 , 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

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