# 19.10: Le Châtelier's Principle and the Equilibrium Constant

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## Le Châtelier's Principle and the Equilibrium Constant

Occasionally, when students apply Le Châtelier's principle to an equilibrium problem involving a change in concentration, they assume that $$K_\text{eq}$$ must change. This seems logical since we talk about "shifting" the equilibrium in one direction or the other. However, $$K_\text{eq}$$ 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:

$\ce{A} \rightleftharpoons \ce{B}$

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

\begin{align} \ce{A} &= 0.50 \: \text{mol} \\ \ce{B} &= 1.0 \: \text{mol} \end{align}

The value of $$K_\text{eq}$$ is given by:

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

Now we will disturb the equilibrium by adding 0.50 moles of $$\ce{A}$$ to the mixture. The equilibrium will shift towards the right, forming more $$\ce{B}$$. Immediately after the addition of $$\ce{A}$$ and before any response, we how have $$1.0 \: \text{mol}$$ of $$\ce{A}$$ and $$1.0 \: \text{mol}$$ of $$\ce{B}$$. The equilibrium then shifts in the forward direction. We will introduce a variable $$\left( x \right)$$, which will represent the change in concentrations as the reaction proceeds. Since the mole ratio of $$\ce{A}$$:$$\ce{B}$$ is 1:1, as $$\left[ \ce{A} \right]$$ decreases by the amount $$x$$, the $$\left[ \ce{B} \right]$$ 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.

$\begin{array}{l|ll} & \ce{A} & \ce{B} \\ \hline \text{Initial} & 1.0 & 1.0 \\ \text{Change} & -x & +x \\ \text{Equilibrium} & 1.0 - x & 1.0 + x \end{array}$

At the new equilibrium position, the values for $$\ce{A}$$ and $$\ce{B}$$ as a function of $$x$$ can be set equal to the value of the $$K_\text{eq}$$. Then, one can solve for $$x$$.

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

Solving for $$x$$:

\begin{align} 2.0 \left( 1.0 - x \right) &= 1.0 + x \\ 2.0 - 2.0x &= 1.0 + x \\ 3.0x &= 1.0 \\ x &= 0.33 \end{align}

This value for $$x$$ is now plugged back in to the Equilibrium line of the table and the final concentrations of $$\ce{A}$$ and $$\ce{B}$$ after the reaction is calculated.

$\left[ \ce{A} \right] = 1.0 - x = 0.67 \: \text{M}$

$\left[ \ce{B} \right] = 1.0 + x = 1.33 \: \text{M}$

The value of $$K_\text{eq}$$ 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 $$K_\text{eq}$$ for a reaction is described.

## Contributors

• CK-12 Foundation by Sharon Bewick, Richard Parsons, Therese Forsythe, Shonna Robinson, and Jean Dupon.