Question

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 `PCI_5` decomposes into `Cl_2` and `PCI_3`. The equilibrium of `PCI_5(g) rightleftharpoons Cl_2 (g) + PCI_3(g)` has a `K_(eq)` of `2.24 x 10^-2` at 327 `"^o`C.

Answer 1

`["Cl"_2] = 0.00303` `M`

Explanation:

We're asked to find the equilibrium concentration of `"Cl"_2` in a given reaction.

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

`"PCl"_5(g) rightleftharpoons "PCl"_3(g) + "Cl"_2(g)` #" "#(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` `"mol PCl"_5`

• `0.174` `"mol PCl"_3`

And since the vessel has a volume of `100` `"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 `["Cl"_2]` yields

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