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

Question

The compound $P C I_{5}$ $PCI_5$ decomposes into $C l_{2}$ $Cl_2$ and $P C I_{3}$ $PCI_3$ . The equilibrium of $P C I_{5} (g) ⇌ C l_{2} (g) + P C I_{3} (g)$ $PCI_5(g) rightleftharpoons Cl_2 (g) + PCI_3(g)$ has a $K_{e q}$ $K_(eq)$ of $2.24 x 10^{- 2}$ $2.24 x 10^-2$ at 327 $^{o}$ $"^o$ C.

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Answer 1 · 2017-08-10T04:12:18

$[{Cl}_{2}] = 0.00303$ $["Cl"_2] = 0.00303$ $M$ $M$

We're asked to find the equilibrium concentration of ${Cl}_{2}$ $"Cl"_2$ in a given reaction.

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

${PCl}_{5} (g) ⇌ {PCl}_{3} (g) + {Cl}_{2} (g)$ $"PCl"_5(g) rightleftharpoons "PCl"_3(g) + "Cl"_2(g)$ $" "$ (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:

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

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