Question #e2b8b

1 Answer
Jun 16, 2015

Answer:

#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")#