# Question 644c6

May 19, 2017

["NOBr"] = "4.64 M"

#### Explanation:

You know that you have

$\textcolor{\mathrm{da} r k \mathmr{and} a n \ge}{2} {\text{NOBr"_ ((g)) rightleftharpoons color(blue)(2)"NO"_ ((g)) + "Br}}_{2 \left(g\right)}$

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

${K}_{c} = \left({\left[\text{NO"]^color(blue)(2) * ["Br"_2])/(["NOBr}\right]}^{\textcolor{\mathmr{and} a n \ge}{2}}\right)$

You already know that at equilibrium, you have

["Br"_2] = "0.178 M"

Now, notice that for every $\textcolor{\mathrm{da} r k \mathmr{and} a n \ge}{2}$ moles of nitrosyl bromide that react, you get $\textcolor{b l u e}{2}$ moles of nitric oxide and $1$ mole of bromine gas.

This means that at equilibrium, you will have

$\left[{\text{NO"] = color(blue)(2) * ["Br}}_{2}\right]$

That happens because the reaction produces twice as much nitric oxide than bromine gas. You can thus say that, at equilibrium, you have

["NO"] = color(blue)(2) * "0.178 M" = "0.356 M"

Rearrange the equation for the equilibrium constant to solve for $\left[\text{NOBr}\right]$, the equilibrium concentration of nitrosyl bromide

K_c = (["NO"]^color(blue)(2) * ["Br"_2])/(["NOBr"]^color(orange)(2)) implies ["NOBr"] = color(darkorange)(sqrt(color(black)( (["NO"]^color(blue)(2) * ["Br"_2])/K_c)))

Plug in your values to find

["NOBr"] = color(darkorange)(sqrt(color(black)( (0.356^color(blue)(2) * 0.178)/(1.05 * 10^(-3))))) = color(darkgreen)(ul(color(black)("4.64 M")))

The answer is rounded to three sig figs, the number of sig figs you have for the equilibrium concentration of bromine gas.

If you want to find the initial concentration of nitrosyl bromide, use the fact that you are left with $\text{4.64 M}$ after the reaction produced $\text{0.356 M}$ of nitric oxide and $\text{0.178 M}$ of bromine gas.

Since you have a $\textcolor{\mathrm{da} r k \mathmr{and} a n \ge}{2} : \textcolor{b l u e}{2}$ mole ratio between nitrosyl bromide and nitric oxide, you can say that in order to produce $\text{0.356 M}$ of the product, the reaction must consume $\text{0.356 M}$ of the reactant.

Therefore, the initial concentration of nitrosyl bromide was

["NOBr"]_ 0 = "4.64 M" + "0.356 M" = "5.00 M"#

Finally, does this result make sense?

Notice that you have ${K}_{c} < 1$. This tells you that the equilibrium will lie to the left, i.e. the reaction vessel will contain more reactant than products at equilibrium.