# Question 35e45

Dec 26, 2015

${K}_{c} = 3.31$

#### Explanation:

Start by writing the balanced chemical equation for this equilibrium reaction

${\text{N"_text(2(g]) + "O"_text(2(g]) rightleftharpoons color(red)(2)"NO}}_{\textrm{\left(g\right]}}$

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, ${\text{N}}_{2}$, and oxygen gas, ${\text{O}}_{2}$ at ${2000}^{\circ} \text{C}$, until equilibrium is established with nitric oxide, $\text{NO}$.

You start by adding $5.88$ moles of $\text{N} 2$ and $16.2$ moles of ${\text{O}}_{2}$ to the reaction vessel. Initially, the reaction vessel does not contain any nitric oxide.

However, after equilibrium is established, you find that the vessel contains $11.2$ moles of nitric oxide.

This means that you can use an ICE table to find the number of moles of each reactant present at equilibrium

${\text{ ""N"_text(2(g]) " "+" " "O"_text(2(g]) " "rightleftharpoons" " color(red)(2)"NO}}_{\textrm{\left(g\right]}}$

color(purple)("I")" " " "5.88" " " " " " "16.2" " " " " " " " " " "0
color(purple)("C")" "(-x)" " " " " "(-x)" " " " " " "(+color(red)(2)x)
color(purple)("E")" "5.88-x" " " "16.2-x" " " " " " "11.2

Notice that the number of moles of nitric oxide increased by $\textcolor{red}{2} x$, where $x$ represents the number of moles of each reactant consumed to make the product.

This of course means that you have

$\textcolor{red}{2} x = 11.2 \implies x = \frac{11.2}{2} = 5.6$

At equilibrium, the reaction vessel will thus contain, along with $11.2$ moles of $\text{NO}$

$5.88 - 5.6 = {\text{0.28 moles N}}_{2}$

$16.2 - 5.6 = {\text{10.6 moles O}}_{2}$

Now, before doing any calculations, try to predict whether or not the equilibrium constant, ${K}_{c}$, is smaller than $1$ or bigger than $1$.

As you know, ${K}_{c} < 1$ implies that the reaction favors the reactants, and ${K}_{c} > 1$ means that the raection favors the products.

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 ${K}_{c} > 1$.

By definition, ${K}_{c}$will be equal to

${K}_{c} = \left(\left[{\text{NO"]^color(red)(2))/(["N"_2] * ["O}}_{2}\right]\right)$

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 $V$ liters. The equilibrium concentrations of the the chemical species will be

["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 ${K}_{c}$ to get

${K}_{c} = \left({11.2}^{2} / \textcolor{red}{\cancel{\textcolor{b l a c k}{{V}^{2}}}} \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{M"^2))))/(10.6/color(red)(cancel(color(black)(V))) color(red)(cancel(color(black)("M"))) * 0.28/color(red)(cancel(color(black)(V))) color(red)(cancel(color(black)("M}}}}\right) = {11.2}^{2} / \left(10.6 \cdot 0.28\right)$

Therefore,

${K}_{c} = \textcolor{g r e e n}{3.31}$

The answer is rounded to three sig figs.

Indeed, our prediction that ${K}_{c} > 1$ turned out to be true.