Question

Phosphorus trichloride reacts with chlorine to form phosphorus pentachloride.

PCl3(g) + Cl2(g)=
PCl5(g)
0.75 mol of PCl3 and 0.75 mol of Cl2 are placed in a 8.0L reaction vessel at 500K.
What is the equilibrium concentration of the mixture? The value of Kc at 500K is 49.
(6 marks)​

Answer 1

[PCl₃] = [Cl₂] = 0.035M

[PCl₅] = 0.059M

Step-by-step explanation:

Based on the chemical reaction, Kc is:

Kc = 49 = [PCl₅] / [PCl₃] [Cl₂]

Where [] are the concentrations in equilibrium of each species

The initial concentration of the reactants is:

[PCl₃] = [Cl₂] = 0.75mol / 8.0L = 0.094M

The reactants must be consumed whereas PCl₅ will be produced. That is:

[PCl₃] = [Cl₂] = 0.094M - X

[PCl₅] = X

Where X is the reaction coordinate

Replacing in Kc expression:

49 = [X] / [0.094 - X] [0.094 - X]

49 = [X] / 0.008836 - 0.188 X + X²

0.432964 - 9.212 X + 49 X² = X

0.432964 - 10.212 X + 49 X² = 0

Solving for X

X = 0.059M → Right solution

X = 0.15M → False solution. Will produce negative concentrations

Replacing:

[PCl₃] = [Cl₂] = 0.094M - 0.059M = 0.035M

[PCl₅] = 0.059M

[PCl₃] = [Cl₂] = 0.035M

[PCl₅] = 0.059M