Question

Phosphorus pentachloride decomposes according to the chemical equation. PCl5(g) ightleftharpoons PCI3(g) + Cl2(g) Kc = 1.80 at 250 �C A 0.363 mol sample of PCl5(g) is injected into an empty 4.60 L reaction vessel held at 250 �C. Calculate the concentrations of PCl5(g) and PCl3(g) at equilibrium.

Answer 1

The KC expression

K = [PCl3][Cl2] / [PCl5]

initially

[PCl5] = molV = 0.363/4.6 = 0.07891 M

[PCl3] = 0

[Cl2] = 0

in equilibrium

[PCl5] = 0.07891 - x

[PCl3] = 0 + x

[Cl2] = 0 + x

substitute in K

K = [PCl3][Cl2] / [PCl5]

1.8 = x*x/(0.07891 -x)

1.8* 0.07891 - 1.8x = x^2

x^2 + 1.8x - 0.142038 =0

x = 0.0757

substitute

[PCl5] = 0.07891 - 0.0757 = 0.00321

[PCl3] = 0 + x = 0.0757

[Cl2] = 0 + x = 0.0757

proof:

Q = (0.0757*0.0757)/(0.00321) = 1.7851, which is pretty near to 1.8, Kc value reported