CK-12-Chemistry Intermediate

19.2. Le Châtelier’s Principle http://www.ck12.org

A⇀↽B

Let’s say we have a 1.0 liter container. At equilibrium, the following amounts are measured:

A = 0.50 mol

B = 1.0 mol

The value of Keqis given by:

Keq=

[B]

[A]

=

1 .0 M

0 .50 M

= 2. 0

Now we will disturb the equilibrium by adding 0.50 mole of A to the mixture. The equilibrium will shift toward
the right, forming more B. Immediately after the addition of A, and before any response, we now have 1.0 mol of A
and 1.0 mol of B. The equilibrium then shifts in the forward direction. We will introduce a variable (x), which will
represent the change in concentrations as the reaction proceeds. Since the mole ratio of A:B is 1:1, if [A] decreases
by the x moles, [B] increases by the same amount. We can now set up an analysis called ICE, which stands for
Initial, Change, and Equilibrium. The values listed in theTable19.4 represent molar concentrations.

TABLE19.4:Sample ICE table

A B

Initial 1.0 1.0

Change -x +x

Equilibrium 1.0-x 1.0+x

At the new equilibrium position, the values for A and B as a function of x can be set equal to the value of the Keq.
Then, one can solve for x.

Keq= 2. 0 =

[B]

[A]

=

1. 0 +x

1. 0 −x

Solving for x:

2.0(1.0 - x) = 1.0 + x

2.0 - 2.0x = 1.0 + x

3.0x = 1.0

x = 0.33

This value for x is now plugged back in to the Equilibrium line of the table, and the final concentrations of A and B
after the reaction are calculated.

A= 1.0 –x = 0.67 M

B = 1.0 + x = 1.33 M

The value of Keqhas been maintained since 1.33/0.67 = 2.0. This shows that even though a change in concentration
of one of the substances in an equilibrium causes a shift in the equilibrium position, the value of the equilibrium
constant does not change.