Question

Kc = 1.8

PCl5 <==> PCl3 + Cl2

A 0.323 mol sample of PCl5(g) is injected into an empty 4.20 L reaction vessel held at 250 °C. Calculate the concentrations of PCl5(g) and PCl3(g) at equilibrium.

Answer 1

PCl₅(g)    PCl₃(g) + Cl₂(g)
I.C. : (0.323/4.20) 0 0
E.C.: (0.323/4.20)-x x x

where

I.C. refers to initial mole concentration (moles per liter),

E.C. refers to Equilibrium mole concentration and

x is degree of dissociation.

Equilibrium constant K_c,
K_c = [PCl₃][Cl₂]/[PCl₅]

Putting K_c=1.8, [PCl₃]=x, [Cl₂]=x and [PCl₅]= (0.323/4.20)-x

1.8 = [(x) (x)] / [(0.323/4.20)-x]
1.8 * (0.323/4.20 - x) = x²
x² = 0.1384 - 1.8x
x² + 1.8x - 0.1384 = 0
On solving this quadratic equation,
x = 0.07386 mol/L

So, concentration of PCl5 would be = ((0.323/4.20 - x)

= (0.07690 - 0.07386)

= 0.00304 mol/L

and that of PCl3 = x = 0.07386 mol/L.