# What is the equilibrium concentration of Cl2in a 100 liter vessel containing o.235 mole of PCI5 and 0.174 mole of PCI3?

## The compound $P C {I}_{5}$ decomposes into $C {l}_{2}$ and $P C {I}_{3}$. The equilibrium of $P C {I}_{5} \left(g\right) r i g h t \le f t h a r p \infty n s C {l}_{2} \left(g\right) + P C {I}_{3} \left(g\right)$ has a ${K}_{e q}$ of $2.24 x {10}^{-} 2$ at 327 "^oC.

Aug 10, 2017

$\left[{\text{Cl}}_{2}\right] = 0.00303$ $M$

#### Explanation:

We're asked to find the equilibrium concentration of ${\text{Cl}}_{2}$ in a given reaction.

Let's first write the chemical equation for this reaction:

${\text{PCl"_5(g) rightleftharpoons "PCl"_3(g) + "Cl}}_{2} \left(g\right)$ " "(which is given to us)

The equilirium constant expression for this reaction is thus

K_"eq" = (["PCl"_3]["Cl"_2])/(["PCl"_5"]) = 2.24xx10^-2

We're given the number of moles of two species:

• $0.235$ ${\text{mol PCl}}_{5}$

• $0.174$ ${\text{mol PCl}}_{3}$

And since the vessel has a volume of $100$ $\text{L}$, the molar concentrations of each are

• ["PCl"_5] = (0.235color(white)(l)"mol")/(100color(white)(l)"L") = color(red)(ul(0.00235color(white)(l)M

• ["PCl"_3] = (0.174color(white)(l)"mol")/(100color(white)(l)"L") = color(green)(ul(0.00174color(white)(l)M

Plugging these values into the equilibrium constant expression equation:

K_"eq" = ((color(green)(0.00174color(white)(l)M))["Cl"_2])/(color(red)(0.00235color(white)(l)M)) = 2.24xx10^-2

Solving for $\left[{\text{Cl}}_{2}\right]$ yields

color(blue)(ulbar(|stackrel(" ")(" "["Cl"_2] = 0.00303color(white)(l)M" ")|)