Question

A 0.268 mol sample of PCl5(g) is injected into an empty 4.05 L reaction vessel held at 250 °C. Calculate the concentrations of PCl5(g) and PCl3(g) at equilibrium. 1.80 @ 250 C

Answer 1

The chemical reaction is as follows:

PCl5(g)<=====>PCl3(g)+Cl2(g)

Given that; Kc = 1.80 at at 250 °C

Now write the expression for the equilibrium constant of this reaction:

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

Molarity of PCl5(g) = moles / volume in L= 0.268 mol/ 4.05 L

=0.066 M

PCl5(g)<=====>PCl3(g)+Cl2(g)

I                       0.066                                     0         0

C                      -X                                +X       +X

E                      0.066-X                     X          X

At equilibrium

Kc= X*X/0.066-X

1.80 = X^2 / (O.066-X)

X^2-0.1188 +1.80X=0

I solved the above by the quadratic formula
X = (-b ±√ (b² - 4ac) ] / (2a)

X= 0.0637

PCl5(g)= 0.066-X= 0.066-0.0637=0.0023M

PCl3(g)= Cl2(g) =X= 0.0637