# Question a76fd

Jan 3, 2018

Here's what I got.

#### Explanation:

The idea here is that the value of the equilibrium constant will actually tell you in which direction the equilibrium will shift.

In this case, you have

${K}_{c} = 1 \cdot {10}^{- 5} \text{ << } 1$

which means that, at equilibrium, the reaction vessel will oxygen gas and nitrogen gas than nitrogen oxide. In other words, at the temperature given to you, the equilibrium will lie to the left, meaning that you should expect the equilibrium concentrations of the reactants to exceed the equilibrium concentration of the product.

So, if you start with $\text{1 M}$ of oxygen gas, nitrogen gas, and nitrogen oxide, you should expect the equilibrium to shift to the left, i.e. the reverse reaction will be favored.

This means that the reaction will consume nitrogen oxide and produce oxygen gas and nitrogen gas.

Now, you can use the forward reaction of the equilibrium

${\text{O"_ (2(g)) + "N"_ (2(g)) rightleftharpoons 2"NO}}_{\left(g\right)}$

with

${K}_{c} = 1 \cdot {10}^{- 5}$

or you can use the reverse reaction of the equilibrium

$2 {\text{NO"_ ((g)) rightleftharpoons "O"_ (2(g)) + "N}}_{2 \left(g\right)}$

with

${K}_{c}^{'} = \frac{1}{K} _ c = \frac{1}{1 \cdot {10}^{- 5}} = 1 \cdot {10}^{5}$

Either way, you should expect the equilibrium concentrations to be

{(["NO"] < "1 M"), (["O"_2] > "1 M"), (["N"_2] > "1 M") :}

Let's use the reverse reaction of the equilibrium. The balanced chemical equation tells you that in order for the reaction to produce $1$ mole of oxygen gas and $1$ mole of nitrogen gas, it must consume $2$ moles of nitrogen oxide.

If you take $x$ $\text{M}$ to be concentration of nitrogen oxide that reacts to produce oxygen gas and nitrogen gas, you can say that, at equilibrium, you will have

["NO"] = "1 M" - 2 * xcolor(white)(.)"M"

["O"_2] = "1 M" + xcolor(white)(.)"M"

["N"_2] = "1 M" + xcolor(white)(.)"M"

By definition, the equilibrium constant for the reverse reaction of the equilibrium will be

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

In your case, this will be equivalent to

${K}_{c}^{'} = \frac{\left(1 + x\right) \left(1 + x\right)}{1 - 2 x} ^ 2$

You will thus have

$1 \cdot {10}^{5} = \frac{1 + 2 x + {x}^{2}}{1 - 4 x + 4 {x}^{2}}$

$1 \cdot {10}^{5} - 4 \cdot {10}^{5} \cdot x + 4 \cdot {10}^{5} \cdot {x}^{2} - 1 - 2 x - {x}^{2} = 0$

$\text{399,999"x^2 - "400,002"x + "99,999} = 0$

This quadratic equation will produce two solutions

${x}_{1} = 0.5024 \text{ }$ or $\text{ } {x}_{2} = 0.4976$

Now, notice that the equilibrium concentration of nitrogen oxide is equal to

["NO"] = (1 - 2x)color(white)(.)"M"

This means that, in order to avoid having a negative equilibrium concentration for this compound, you need to have

$x < 0.50$

Therefore, you can discard the first solution and say that

$x = 0.4976$

This means that, at equilibrium, the reaction vessel will contain

["NO"] = (1 - 2 * 0.4976)color(white)(.)"M" = "0.0048 M"

["O"_2] = (1 + 0.4976)color(white)(.)"M" = "1.4976 M"

["N"_2] = (1 + 0.4976)color(white)(.)"M" = "1.4976 M"#

I won't round the answers to one significant figure, but keep in mind that you should that because you have one significant figure for the initial concentrations.

So, as predicted, the reaction consumed nitrogen oxide and produced oxygen gas. I recommend redoing the calculations using the forward reaction of the equilibrium with

${K}_{c} = 1 \cdot {10}^{- 5}$

the values must come out the same! In that case, you have

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

which gets you

$1 \cdot {10}^{- 5} = {\left(1 - 2 x\right)}^{2} / \left(\left(1 + x\right) \left(1 + x\right)\right)$