# 6 moles of "Cl"_2 are placed in a 3L flask at 1250K. At this temperature, "Cl"_2 begins to dissociate into "Cl" atoms. What is the value for K_c if 50.0% of the "Cl"_2 molecules dissociate when equilibrium has been achieved?

## 6 moles of Cl2 are placed in a 3L flask at 1250K. At this temperature, Cl2 begins to dissociate in to Cl atoms. What is the value for Kc if 50.0% of the Cl2 molecules dissociate when equilibrium has been acheived?

Jun 3, 2016

${K}_{c} = 4$

#### Explanation:

The idea here is that you need to write a balanced chemical equation for the dissociation of chlorine gas, ${\text{Cl}}_{2}$, into chlorine atoms, $\text{Cl}$, then use the mole ratio that exists between the two species to figure out the two equilibrium concentrations.

So, you will have

${\text{Cl"_ (2(g)) rightleftharpoons color(red)(2)"Cl}}_{\left(g\right)}$

Notice the every mole of chlorine gas that dissociates produces $\textcolor{red}{2}$ moles of chlorine atoms.

You know that the reaction vessel initially contained $6$ moles of chlorine gas and that 50% of these molecules dissociate to form chlorine atoms.

6 color(red)(cancel(color(black)("moles Cl"_2))) * ("50 moles Cl"_2)/(100color(red)(cancel(color(black)("moles Cl"_2)))) = "3 moles Cl"_2

You can thus say that at equilibrium, the reaction vessel will contain

${n}_{C {l}_{2}} = {\text{6 moles" - overbrace("3 moles")^(color(blue)("50% of the initial amount")) = "3 moles Cl}}_{2}$

and

${n}_{C l} = \text{0 moles" + color(red)(2) xx "3 moles" = "6 moles Cl}$

Use the volume of the vessel to calculate the concentrations of the two species

["Cl"_2] = "3 moles"/"3 L" = "1 M"

["Cl"] = "6 moles"/"3 L" = "2 M"

By definition, the *equilibrium constant for this reaction, ${K}_{c}$, will be

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

Plug in your values to find

K_c = ("2 M")^color(red)(2)/"1 M" = ("4 M"^color(red)(cancel(color(black)(2))))/(1color(red)(cancel(color(black)("M")))) = "4 M"

The equilibrium constant is usually given without added units, but keep in mind that, as you can see in this example, that is not always the case

${K}_{c} = \textcolor{g r e e n}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{4} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

The answer is rounded to one sig fig, the number of sig figs you have for the number of moles of chlorine gas added to the reaction vessel and for the volume of the vessel.