Question

Phosphorus pentachloride decomposes according to the chemical equation PCl5(g)↽−−⇀PCl3(g)+Cl2(g)Kc=1.80 at 250 ∘C A 0.157 mol sample of PCl5(g) is injected into an empty 2.50 L reaction vessel held at 250 ∘C. Calculate the concentrations of PCl5(g) and PCl3(g) at equilibrium.

Answer 1

The concentration of PCl3 is 0.061 mol/L and for PCl5 is 0.002 mol/L

Step-by-step explanation:

The initial concentration of PCl5 is

n/V = 0.157/2.5 = 0.063 mol/L

During the reaction x mol/L is consumed, and x mol/L is formed of each product (the stoichiometry is 1 mol : 1 mol : 1 mol).

So, doing a table of reaction

PCl5 PCl3 Cl2

initial 0.063 0 0

reacted -x +x +x

equilibrium 0.063 - x x x

For a reaction aA + bB ↽−−⇀ cC +dD, the equilibrium constant is given by:

`Kc = ([C]^cx[D]^d)/([A]^ax[B]^b)`

All these concentration are the concentration of equilibrium. So, for the reaction:

`Kc = ([PCl3]x[Cl2])/([PCl5])`

`1.8 = (x*x)/(0.063 - x)`

x² = 0.1134 - 1.8x

x² + 1.8x - 0.1134 = 0

Using Bhaskara, with a = 1, b = 1.8, and c = -0.1134

Δ = b² - 4ac = (1.8)² - 4*1*(-0.1134) = 3.6936

`x = (-b +/- √(delta) )/(2a)`

x is a molar concentration, so it must be positive, then:

x = (-1.8 + √3.6936)/2

x = 0.061 mol/L

Which is the concentration of PCl3. The concentration of PCl5 is

0.063 - 0.061 = 0.002 mol/L