LeChâtelier’s Principle and the Equilibrium Constant

Learning Objectives

Describe the relationship between Le Châtelier’s principle and the equilibrium constant.

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LeChâtelier’s Principle and the Equilibrium Constant

Occasionally, when students apply LeChatelier’s principle to an equilibrium problem involving a change in concentration, they assume that K_eq must change. This seems logical since we talk about “shifting” the equilibrium in one direction or the other. However, K_eq is a constant, for a given equilibrium at a given temperature, so it must not change. Here is an example of how this works. Consider the simplified equilibrium below:

$A rightleftarrows B$

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

$A &=0.50 text{ mol} \B &=1.0 text{ mol}$

The value of K_eq is given by:

$K_{eq}= frac{left [B right ]}{left [A right ]}=frac{1.0 text{ M}}{0.50 text{ M}}=2.0$

Now we will disturb the equilibrium by adding 0.50 mole of A to the mixture. The equilibrium will shift towards 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, as [A] decreases by the amount x, the [B] increases by the amount x. We set up an analysis called ICE, which stands for Initial, Change, and Equilibrium. The values in the table represent molar concentrations.

$& qquad qquad qquad underline{;; A qquad qquad B ; ; ; ; ; ; ; ;;;} \& text{Initial} qquad qquad 1.0 qquad quad 1.0 \ & text{Change}qquad quad -x qquad quad +x \ & text{Equilibrium} qquad 1.0 -x qquad 1.0 +x$

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 K_eq. Then, one can solve for x.

$K_{eq}=2.0=frac{left [ B right ]}{left [ A right ]}=frac{1.0 + x}{1.0 - x}$

Solving for x:

$2.0(1.0-x) &=1.0 + x \2.0 - 2.0 x &=1.0+x \3.0 x &=1.0 \x &=0.33$

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 is calculated.

$left [ A right ]&=1.0-x=0.67 text{ M} \left [ B right ]&=1.0+x=1.33 text{ M}$

The value of K_eq has been maintained since $frac{1.33}{0.67}=2.0$ . This shows that even though a change in concentration of one of the substances in equilibrium causes a shift in the equilibrium position, the value of the equilibrium constant does not change.

Summary

Maintenance of the constant K_eq for a reaction is described.

Practice

Read the material at Chem Purdue and answer the following questions:

What concentration units should be used?
What quantities should you use for equilibrium problems?
What must the change in each quantity agree with?
What is “x”?

Review

Does K_eq change for a given reaction at a given temperature?
What does ICE stand for?
Will the equilibrium position change if materials are added to or removed from the reaction?
How does addition or removal of materials affect the K_eq?

Glossary

Initial Change Equilibrium (ICE): A calculation that looks at the initial conditions, change, and new equilibrium in a reaction.