# Question b2748

Jul 8, 2017

["NO"] = "0.714 M"

#### Explanation:

The first thing that you need to do here is to figure out the equilibrium constant of the reaction by using the known equilibrium concentrations of the three chemical species.

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

By definition, the equilibrium constant for this equilibrium looks like this

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

You know that at a given temperature, the system is at equilibrium when

["N"_2] = ["O"_2] = "0.100 M " and $\text{ " ["NO"] = "0.500 M}$

Even without doing any calculations, you should be able to say that ${K}_{c} > 1$ because the reaction vessel contains more product than reactants at equilibrium, which implies that at this temperature, the forward reaction is favored.

So, plug in the values to find

${K}_{c} = \frac{{0.500}^{\textcolor{red}{2}}}{0.100 \cdot 0.100} = 25$

Now, you are told that the concentration of nitric oxide is being increased to $\text{0.800 M}$.

In response, the system will react by changing the position of the equilibrium in such a way as to reduce the added stress, i.e. to decrease the concentration of nitric oxide $\to$ think Le Chatelier's Principle here.

This means that the reverse reaction will be favored. The concentration of nitric oxide will decrease by a value $x$ $\text{M}$, which implies that the concentrations of nitrogen gas and oxygen gas will increase by $\left(\frac{x}{\textcolor{red}{2}}\right)$ $\text{M}$ $\to$this is due to the $1 : \textcolor{red}{2}$ mole ratio that exists between the two reactants and nitric oxide.

You can thus say that equilibrium constant will be equal to--keep in mind that the equilibrium constant is, well, constant as long as the temperature of the reaction remains constant!

${K}_{c} = {\left(0.800 - x\right)}^{\textcolor{red}{2}} / \left(\left(0.100 + \frac{x}{\textcolor{red}{2}}\right) \cdot \left(0.100 + \frac{x}{\textcolor{red}{2}}\right)\right)$

K_c = (0.800 -x )^color(red)(2)/(((0.200 + x)/color(red)(2)) * ((0.200 + x)/color(red)(2))

${K}_{c} = \frac{4 \cdot {\left(0.800 - x\right)}^{\textcolor{red}{2}}}{0.200 + x} ^ 2$

This will be equivalent to

$\sqrt{{K}_{c}} = \sqrt{\frac{4 \cdot {\left(0.800 - x\right)}^{\textcolor{red}{2}}}{0.200 + x} ^ 2}$

$\sqrt{{K}_{c}} = \frac{2 \cdot \left(0.800 - x\right)}{0.200 + x}$

which will give you

$\left(0.200 + x\right) \cdot \sqrt{25} = 2 \cdot \left(0.800 - x\right)$

$1 + 5 x = 1.600 - 2 x$

$7 x = 0.600 \implies x = \frac{0.600}{7} = 0.0857$

Therefore, you can say that when the equilibrium is reestablished, the concentration of nitric oxide will be

["NO"] = "0.800 M" - "0.0857 M" = color(darkgreen)(ul(color(black)("0.714 M")))#

Notice that the final operation is a subtraction, so you must round the answer to three decimal places, the number of decimal places you have for $\text{0.800 M}$.