Question #35e45
1 Answer
Explanation:
Start by writing the balanced chemical equation for this equilibrium reaction
#"N"_text(2(g]) + "O"_text(2(g]) rightleftharpoons color(red)(2)"NO"_text((g])#
Now, the equilibrium constant for a chemical reaction is calculated by using the equilibrium concentrations of the chemical species that take part in that reaction.
In your case, the problem provides you with moles, not with molarities. However, if you take into account the fact that all three gases share the same reaction vessel, you can use moles instead of molarities.
More on that later.
So, you know that you're mixing two gases, nitrogen gas,
You start by adding
However, after equilibrium is established, you find that the vessel contains
This means that you can use an ICE table to find the number of moles of each reactant present at equilibrium
#" ""N"_text(2(g]) " "+" " "O"_text(2(g]) " "rightleftharpoons" " color(red)(2)"NO"_text((g])#
Notice that the number of moles of nitric oxide increased by
This of course means that you have
#color(red)(2)x = 11.2 implies x = 11.2/2 = 5.6#
At equilibrium, the reaction vessel will thus contain, along with
#5.88 - 5.6 = "0.28 moles N"_2#
#16.2 - 5.6 = "10.6 moles O"_2#
Now, before doing any calculations, try to predict whether or not the equilibrium constant,
As you know,
In this case, the number of moles of each reactant decreased significantly. At the same time, the number of moles of the product increased significantly. This tells you that you can expect to see
By definition,
#K_c = (["NO"]^color(red)(2))/(["N"_2] * ["O"_2])#
Now, here is why I said that you can use the number of moles and not worry about molarity. Let's assume that the reaction vessel has a volume
#["N"_2] = "0.28 moles"/(V" L") = 0.28/V" M"#
#["O"_2] = "10.6 moles"/(V" L") = 10.6/V" M"#
#["NO"] = "11.2 moles"/(V" L") = 11.2/V" M"#
Plug these values into the expression for
#K_c = (11.2^2/color(red)(cancel(color(black)(V^2)))color(red)(cancel(color(black)("M"^2))))/(10.6/color(red)(cancel(color(black)(V))) color(red)(cancel(color(black)("M"))) * 0.28/color(red)(cancel(color(black)(V))) color(red)(cancel(color(black)("M")))) = 11.2^2/(10.6 * 0.28)#
Therefore,
#K_c = color(green)(3.31)#
The answer is rounded to three sig figs.
Indeed, our prediction that