Question

The equilibrium constant, Kc, for the following reaction is 83.3 at 500 K. PCl3(g) Cl2(g) PCl5(g) Calculate the equilibrium concentrations of reactant and products when 0.400 moles of PCl3 and 0.400 moles of Cl2 are introduced into a 1.00 L vessel at 500 K.

Answer 1

[PCl₅] = 0.336 M

[Cl₂] = [PCl₃] = 0.064 M

Step-by-step explanation:

We are given the equilibrium constant, kc , and the moles of reactants and products. Thus, our strategy here sholud be to express the quantities at equilibrium in terms of its constant by setting up our ICE table helper.

First lets start by writing the expression for the equilibrium constant:

PCl₃ (g) + Cl₂ (g) ⇄ PCl₅(g)

Kc = [PCl₅] / [Cl₂] / [PCl₃] =83.3

and setup the ICE table

We are given the moles introduced in the 1.00 L vessel, so we can calculate the molarities as M = mol/L

[Cl₂] [PCl₃] [PCl₅]

i 0.400 M 0.400 M 0

c -x -x +x

e 0.400 - x 0.400 - x x

x / (0.400 - x )² = 83.3

(0.16 - 0.8x + x²) x 83.3 = x

13.33 -66.66x + 83.3x² = x

13.33 - 67.66 x + 83.3 x² = 0

Solving the quadratic equation we have x₁ = 0.476 and x₂ = 0.336

The first solution is phisically impossible, since it will give us a negative quantity at equilibrim for the reactants.

With the second solution x = 0.336, the equilibrium concentrations are:

[PCl₅] = 0.336 M

[Cl₂] = [PCl₃] = 0.400 - 0.336 = 0.064 M