For the equilibrium PCl5 (g) PCl3 (g) + Cl2 (g), Kc = 2.0 * 101 at 240°C. If pure PCl5 is placed in a 1.00-L container and allowed to come to equilibrium, and the equilibrium concentration of PCl3 (g) is 0.27 M, what is the equilibrium concentration of PCl5 (g) ?

the equilibrium concentration of [PCl₅] is 3.64*10⁻³ M

Explanation:

for the reaction

PCl₅ (g) → PCl₃ (g) + Cl₂ (g)

where

Kc = [PCl₃]*[Cl₂]/[PCl₅] = 2.0*10¹ M = 20 M

and [A] denote concentrations of A

if initially the mixture is pure PCl₅, then it will dissociate according to the reaction and since always one mole of PCl₃ (g) is generated with one mole of Cl₂ (g), the total number of moles of both at the end is the same → they have the same concentration → [PCl₃ (g) ] = [Cl₂]=0.27 M

therefore

Kc = [PCl₃]*[Cl₂]/[PCl₅] = 0.27 M * 0.27 M / [PCl₅] = 20 M

[PCl₅] = 0.27 M * 0.27 M / 20 M = 3.64*10⁻³ M

[PCl₅] = 3.64*10⁻³ M

the equilibrium concentration of [PCl₅] is 3.64*10⁻³ M