Question

A mixture containing 0.250 mol SbCl3, 0.250 mol Cl2, and 0.500 mol SbCl5 comes to equilibrium in a 5.00 L vessel at 648 deg celsius.

SbCl3 + Cl2 <----> SbCl5

Kc = 40.0

Determine the equilibrium concentration of each gas. So far I know that the reaction is in equilibrium and Q = K so there is no shift to the left or to the right. But because there is no shift, I am not able to figure out how to set up the ICE table. I have only been taught when there is a shift either to the left or to the right. Any help is appreciated. Thank you.

Answer 1

First we will calculate the Molarity of each species.

Molarity = Moles / Volume

Molarity of,

SbCl3 = 0.25 / 5 = 0.05 M

Cl2 = 0.25 / 5 = 0.05 M

SbCl5 = 0.5 / 5 = 0.1 M

Reaction: SbCl3 + Cl2 <----> SbCl5

Initial:        0.05......0.05............0.1

Change:.....-X.......-X................+X

Final: .....0.05-X...0.05-X........0.1+X

Kc for the reaction = [SbCl5] / [SbCl3] [Cl2]

=> 40 = (0.1 + X) / (0.05 - X)^2

=> X = 0 M

Therefore, the reaction is already at equilibrium and equilibrium conc. of each gas is

[SbCl3] = 0.05 M

[Cl2] = 0.05 M

[SbCl5] = 0.1 M