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

- Page ID
- 53913

With online banking, management of your personal finances can become less complicated in some ways. You can automatically deposit paychecks, pay bills, and designate how much goes into savings or other special accounts each month. If you want to maintain \(10\%\) of your bank account in savings, you can set up a program that moves money in and out of the account when you get a paycheck or pay bills. The amount of money in savings will change as the money comes in and out of the bank, but the ratio of savings to checking will always be constant.

### 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.