Question

Phosphorus pentachloride decomposes according to the chemical equation. PCl5(g) yields PCl3(g)+Cl2(g) . A 0.141 mol sample of PCl5(g) is injected into an empty 2.10 L reaction vessel held at 250 °C. Calculate the concentrations of PCl5(g) and PCl3(g) at equilibrium. kc= 1.80 at 250 degrees celsius

Answer 1

PCl5(g) ----> PCl3(g) + Cl2(g)

Initial concentration of PCl5:

Moles = 0.141 mol

Volume = 2.10L

[PCl5] = moles/ Volume = 0.141mol / 2.10L = 0.067M

Constructing an ICE table:

PCl5 ----> PCl3 + Cl2

Initial(M): 0.067 0 0

Change: -x +x +x

Equilbrm: 0.067-x x x

Kc = [PCl3]*[Cl2] / [PCl5]

Keeping the values,

1.80 = x*x /0.067-x

1.80*(0.067-x) = x²

0.121 - 1.80x = x²

x² + 1.80x - 0.121 = 0

Solving the quadratic equation for x :-

x = 0.065M

So, [PCl3] = [Cl2] = 0.065M