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# LeChateliers Principle and the Equilibrium Constant

## Predicting the effect of a change in conditions on a chemical equilibrium.

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LeChâtelier'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

Questions

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

#### Review

Questions

1. Does  $K_{eq}$ change for a given reaction at a given temperature?
2. What does ICE stand for?
3. Will the equilibrium position change if materials are added to or removed from the reaction?
4. How does addition or removal of materials affect the $K_{eq}$ ?