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

- Page ID
- 53913

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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.