Question #04928

1 Answer
Mar 24, 2017

Answer:

#["CO"_2] = "0.12997 M"#

Explanation:

The trick here is to realize that the concentrations of the two solids are not taken into account when calculating the equilibrium constant for this equilibrium reaction.

So, you know that at #"1500 K"#, the following equilibrium is established

#"NiO"_ ((s)) + "CO"_ ((g)) rightleftharpoons "Ni"_ ((s)) + "CO"_ (2(g))#

By definition, the equilibrium constant for this reaction is equal to

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

The concentrations of the two solids are not added to the expression of #K_c# because we take them as being constant.

That is the case because the concentration of a solid, or of a pure liquid, for that matter, depends exclusively on the density and on the molar mass of the solid (or liquid).

In other words, it doesn't matter how much solid you have, its concentration can always be taken as being constant.

Now, notice that #K_c > 1# at this temperature. This tells you that the equilibrium will lie significantly to the right, i.e. the forward reaction will be favored at this temperature.

You can thus expect the equilibrium concentration of carbon dioxide to be significantly higher than the equilibrium concentration of carbon monoxide.

In fact, you can say that the reaction will convert almost all the carbon monoxide to carbon dioxide.

You know that carbon monoxide and carbon dioxide exist in a #1:1# mole ratio. If you take #x# to be concentration of carbon monoxide consumed by the reaction to produce carbon dioxide, you can say that at equilibrium, you will have

#["CO"] = (0.13000 - x)color(white)(.)"M"#

#["CO"_2] = x color(white)(.)"M"#

Plug this into the expression of #K_c# to get

#4000.0 = x/(0.13000 - x)#

Rearrange to solve for #x#

#4000.0 * (0.13000 - x) = x#

#520 - 4000.0x = x#

#40001.0x = 520 implies x = 520/4001.0 = 0.12997#

Therefore, the equilibrium concentration of carbon dioxide will be

#color(darkgreen)(ul(color(black)(["CO"_2] = "0.12997 M")))#

The answer is rounded to five sig figs.

As predicted, the reaction consumed almost all the carbon monoxide to produce carbon dioxide, which is what you should expect when dealing with such a high value of #K_c#.