Question 6558e

Feb 27, 2016

$6.2 \cdot {10}^{- 4}$

Explanation:

The idea here is that you can find the equilibrium concentration of atomic bromine, $\text{Br}$, by using the percent of molecular bromine, ${\text{Br}}_{2}$, that gets converted by the reaction.

Once you know the equilibrium concentrations of both chemical species, you can solve for the equilibrium constant of the reaction.

Calculate the concentration of molecular bromine by using the given number of moles and volume of the reaction vessel.

$\textcolor{b l u e}{c = \frac{n}{V}}$

["Br"_2] = "1.05 moles"/"0.98 L" = "1.0714 M"

Now, you know that 1.2% of the amount of molecular bromine present in the reaction vessel will get converted to atomic bromine.

This tells you that out of every $1000$ molecules of bromine, $12$ will form atomic bromine and $988$ will remain as molecular bromine.

Right from the start, this should tell you that the equilibrium constant is significantly smaller than $1$. The reaction vessel contains significantly more molecules of bromine than atoms of bromine, which means that the equilibrium lies to the left.

${\text{Br"_text(2(g]) rightleftharpoons color(red)(2)"Br}}_{\textrm{\left(g\right]}}$

the equilibrium constant, ${K}_{c}$, will take the form

${K}_{c} = \left(\left[{\text{Br"]^color(red)(2))/(["Br}}_{2}\right]\right)$

It's important to remember that the expression of the equilibrium constant uses equilibrium concentrations.

Now, when 1.2% of the molecular bromine forms atomic bromine, you're left with 98.8% of the initial concentration of molecular bromine.

["Br"_2] = 98.8/100 * "1.0714 M" = 105.85/100"M"

Since $1$ molecule of bromine forms $\textcolor{red}{2}$ atoms of bromine, you can say that the equilibrium concentration of atomic bromine will be

["Br"] = color(red)(2) xx 1.2/100 * "1.0714 M" = 2.57/100"M"

Plug these values into the expression for ${K}_{c}$ to find

K_c = (2.57^color(red)(2) * 10^(-4))/(105.85 * 10^(-2)) = color(green)(6.2 * 10^(-4)#

The answer is rounded to two sig figs, the number of sig figs you have for the volume of the reaction vessel.

SIDE NOTE It's worth noting that ${K}_{c}$ is not unitless, but for simplicity I'll leave it without the corresponding units.