# Question #e2b8b

Jun 16, 2015

${K}_{c} = 1.3$

#### Explanation:

You're dealing with an equilibrium reaction in which phosphorus pentachloride, $P C {l}_{5}$, decomposes to produce phosphorus trichloride, $P C {l}_{3}$ and chlorine gas, $C {l}_{2}$.

Moreover, you know that 40% of the $P C {l}_{5}$ 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 $P C {l}_{5}$, and only 40% undergo decomposition, you'll end up with

$\text{14.250 moles" * 40/100 = "5.7 moles}$ $\to$react

$\text{14.250 moles" * 60/100 = "8.55 moles}$ $\to$ do not react

Since the reactant produces 1 mole of $P C {l}_{3}$ and 1 mole of $C {l}_{2}$ for every mole that reacts, you'll get (at equilibrium)

${n}_{P C {l}_{3}} = \text{5.7 moles}$

${n}_{C {l}_{2}} = \text{5.7 moles}$

${n}_{P C {l}_{5}} = \text{8.55 moles}$

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

$\left[P C {l}_{5}\right] = \text{8.55 moles"/"3 L" = "2.85 M}$

$\left[P C {l}_{3}\right] = \left[C {l}_{2}\right] = \text{5.7 moles"/"3 L" = "1.9 M}$

$P C {l}_{5 \left(g\right)} r i g h t \le f t h a r p \infty n s P C {l}_{3 \left(g\right)} + C {l}_{2 \left(g\right)}$

By definition, the equilibrium constant will be

${K}_{c} = \frac{\left[P C {l}_{3}\right] \cdot \left[C {l}_{2}\right]}{\left[P C {l}_{5}\right]}$

${K}_{c} = \frac{1.9 \cdot 1.9}{2.85} = 1.267$

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

${K}_{c} = \textcolor{g r e e n}{\text{1.3}}$