Question

Question #e2b8b

Answer 1

`K_c = 1.3`

Explanation:

You're dealing with an equilibrium reaction in which phosphorus pentachloride, `PCl_5`, decomposes to produce phosphorus trichloride, `PCl_3` and chlorine gas, `Cl_2`.

Moreover, you know that 40% of the `PCl_5` undergoes this decomposition. This of course implies that 60% of the compound will remain unchanged.

From a concentration stand-point, this tells you that the equilibrium concentration of the reactant will be bigger than the individual concentrations of the reactants.

If you start with a total of 14.250 moles of `PCl_5`, and only 40% undergo decomposition, you'll end up with

`"14.250 moles" * 40/100 = "5.7 moles"` `->`react

`"14.250 moles" * 60/100 = "8.55 moles"` `->` do not react

Since the reactant produces 1 mole of `PCl_3` and 1 mole of `Cl_2` for every mole that reacts, you'll get (at equilibrium)

`n_(PCl_3) = "5.7 moles"`

`n_(Cl_2) = "5.7 moles"`

`n_(PCl_5) = "8.55 moles"`

Use the volume of the vessel to determine the concentration of the species involved in the reaction

`[PCl_5] = "8.55 moles"/"3 L" = "2.85 M"`

`[PCl_3] = [Cl_2] = "5.7 moles"/"3 L" = "1.9 M"`

So, for your reaction

`PCl_(5(g)) rightleftharpoons PCl_(3(g)) + Cl_(2(g))`

By definition, the equilibrium constant will be

`K_c = ([PCl_3] * [Cl_2])/([PCl_5])`

`K_c = (1.9 * 1.9)/2.85 = 1.267`

I'll leave the answer rounded to two sig figs, even though you only gave one sig fig for the volume of the vessel.

`K_c = color(green)("1.3")`