# Question #644c6

##### 1 Answer

#### Explanation:

You know that you have

#color(darkorange)(2)"NOBr"_ ((g)) rightleftharpoons color(blue)(2)"NO"_ ((g)) + "Br"_ (2(g))#

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

#K_c = (["NO"]^color(blue)(2) * ["Br"_2])/(["NOBr"]^color(orange)(2))#

You already know that **at equilibrium**, you have

#["Br"_2] = "0.178 M"#

Now, notice that **for every** **moles** of nitrosyl bromide that **react**, you get **moles** of nitric oxide and **mole** of bromine gas.

This means that **at equilibrium**, you will have

#["NO"] = color(blue)(2) * ["Br"_2]#

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

#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 **after** the reaction produced

Since you have a **mole ratio** between nitrosyl bromide and nitric oxide, you can say that in order to produce

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 **to the left**, i.e. the reaction vessel will contain more reactant than products at equilibrium.