Question

At some temperature, K for the reaction PCl₅ ⇌ PCl₃ + Cl₂ is 0.020. In one liter vessel, 0.10 bar PCl₅ and 0.020 bar PCl₃ are present initially. There is no Cl₂ present initially. When the system comes to equilibrium, how much Cl₂ will be present?

Answer 1

PCl₅  PCl₃ + Cl₂

Kc = [PCl₃] ∙ [Cl₂] / [PCl₅]

Set up ICE table

PCl₅      PCl₃ + Cl₂
I: 0.1 0.02 0
C: -x +x +x
E: (0.1-x) (0.02+x) x

Plug in the equilibrium concentrations to equilibrium condition and solve for x:

Kc = [PCl₃] ∙ [Cl₂] / [PCl₅]

0.02 = [ (0.02+x) x ] / (0.1-x)

0.02 (0.1-x) = (0.02+x) x

0.002 - 0.02x = 0.02x + x²

x² + 0.04x - 0.002 = 0

Solving the quadratic equation, we get:

x = 0.029 and x = - 0.069

The negative value of x is discarded, and hence x = 0.029

[Cl₂] = x = 0.029 M

Multiply by the volume and you get the number of moles.

So, moles of Cl₂ = 0.029 M x 1 L = 0.029 moles