Equilibrium and Le Chatelier's Principle

In Lesson 7, we encountered the concept of a "reversible reaction" when we learned about weak acids and bases. A weak acid is called "weak" when its dissociation is reversible, meaning is can proceed in both the forward and reverse directions. Notationally, we represent a reversible reaction with double arrows for this reason.

Functionally, if a reaction is reversible, that means it will eventually reach an equilibrium in which both the reactants and products are present in a mixture. This idea, which was briefly touched on in Lesson 7, will be the subject of this section.

Reaching and Maintaining Equilibrium

Imagine a simple reaction: A + B → C + D

A and B are reactants, and C and D are products. If I take solutions of A and B and mix them together, they will start to react and form C and D. Let's think about how the rate of the reaction - in both the forward and reverse directions - will change over time.

We know from the previous section that concentration has a big effect on the rate of a reaction. Higher concentrations mean a faster reaction. Early on, after mixing A and B together, the concentrations of A and B are relatively high, and the concentrations of C and D are very low. This means that the forward reaction (between A and B) will be relatively fast, and the reverse reaction (between C and D, forming A and B) will be relatively slow. Over a given time period, we might expect 100 molecules of A and B to react with each other, whereas only 5 molecules of C and D might transform back into A and B. So, on balance, the reaction is progressing towards the products, at a rate of 95 molecules per unit time.

Any given set of atoms might proceed back and forth through this process numerous times, since the reaction is constantly happening in both directions.

However, because the forward reaction is happening faster than the reverse reaction, the concentrations of A and B will decrease over time, and the concentrations of C and D will rise. This will have the effect of lowering the forward rate of the reaction and raising the reverse rate. In the image at right, per the same time period as before, we might be seeing 70 molecules of A and B react, while 20 molecules of C and D complete the reverse reaction. So the net progress of the reaction has slowed to 50 molecules per unit time.

It's important to note that the forward and reverse reactions will have different activation energies and orientation factors. This means that they will not respond to concentration in the same way. In this image, the concentration of products has actually overtaken the concentration of reactants. But due to some quirk of the reaction, the forward process is still happening faster than the reverse.

At this point, the forward reaction might be going 40 molecules per unit time, with the reverse going 30 molecules per unit time, for a net progress of 10 molecules per unit time. We are approaching an important point.

With the forward reaction slowing and the reverse reaction accelerating, we must eventually arrive at a point where they will equal each other, shown at right. Once we reach this point, individual molecules of A and B are being churned into C and D, with the reverse process also happening, at the exact same rate.

As a result, the concentrations of all four components - A, B, C, and D - stop changing at this point. If left undisturbed, the mixture will continue shuttling molecules back and forth between reactants and products forever, with the overall concentrations never changing. This is known as a dynamic equilibrium. The plural of "equilibrium" is "equilibria."

In the real world, it is somewhat rare to have a chemical process reach equilibrium. In particular, any time organisms get involved, they will tend to disrupt equilibrium systems, because for an organism to be fully at equilibrium means being dead! However, as we have seen, concepts of equilibrium are very important to understanding the behavior of weak acids and bases, as well as the dissolution of solids (we have seen these equilibria in Lessons 7 and 3, respectively). Equilibria are also important in many reactions performed in chemical research laboratories, and in chemical manufacturing settings where those reactions are implemented. Scientists need to understand equilibria in order to successfully synthesize everything from fuels to pharmaceuticals to fertilizer.

Le Chatelier's Principle

When explaining how an equilibrium forms above, I told you that equilibria last forever, if left undisturbed. Obviously, this is unrealistic, and it will be useful for us to think about how we might disturb a reaction at equilibrium, and what would happen if we did so.

In general, the responses of equilibria to disturbance are governed by Le Chatelier's Principle. This principle states that "when a reaction at equilibrium is disturbed, it will respond in a way that counteracts the nature of the disturbance." We will see below what this means, and reason through why it is the case.

Disturbances of Concentration

In the diagram at right, reactants are in red and products are in blue. While we have separated the two to make it easier to visualize the changes that will take place, you should think of them as occupying the same space, with red molecules converting into blue and vice versa, with equal rates in both directions.

When additional reactant (red) is added to the system, the reactant concentration increases, with the product concentration staying the same. This "disturbs" the equilibrium, because the rate of the forward reaction will rise, putting it out of balance with the reverse reaction.

With the forward reaction rate increased, reactant (red) molecules are now being converted to products (blue) faster than blue molecules are converted to red. This means the concentration of products will rise, and the concentration of reactants will fall (relative to its post-disturbance value).

Eventually, the forward and reverse rates will once again equalize, with some of the added reactant converted to product. The product concentration is higher than it was in the previous equilibrium. The reactant concentration is higher than in the previous equilibrium, but lower than it was in the disturbed state.

Hopefully, you can see how this example illustrates Le Chatelier's principle. We had a reaction at equilibrium, which was disturbed by adding reactant. The reaction responded to this disturbance by turning reactants into products, re-establishing the equilibrium in a way that counteracted the addition of reactant. The shorthand we use to describe this response is to say that the reaction "shifted toward the products" after the disturbance.

There are four general ways in which equilibrium concentrations can be disturbed. The first is via the addition of a reactant, which we have just seen. It is equally easy to imagine disturbing a reaction by adding a product (blue in the example above). In that case, the higher blue concentration would speed up the reverse reaction, converting products to reactants, which we would call "shifting toward the reactants."

We can also disturb a reaction by removing reactants or products. This is usually accomplished by adding a separate substance that causes a "side reaction," consuming either a reactant or a product. In the case where a reactant was partially removed, the response would be to "shift toward the reactants," and where a product was partially removed, the response would be to "shift toward the products." See if you can reason through why this is the case, using the example above as a template.

Disturbances of Temperature

To explain how Collision Theory predicts the effects of temperature changes, there are two additional points we must make first.

The first is that for an endothermic reaction, the activation energy of the forward reaction must be greater than the activation energy of the reverse reaction. The opposite is true of an exothermic reaction. This comes simply from the requirement that the paths of the forward and reverse reactions are exactly the same, they are just followed in the opposite directions from each other, and the definitions of endothermic and exothermic reactions.

The second is that increasing the temperature has a greater effect on reaction rate for a reaction with a higher activation energy.

For example, raising the temperature from 20 to 30 ^oC for a reaction with an activation energy of 50 kJ/mole increases the fraction of collisions with enough energy to react by 97%. The same increase in temperature for a reaction with an activation energy of only 42 kJ/mol produces only a 75% increase in the fraction of collisions with enough energy to react.

Suppose, now, that we have an endothermic reaction in dynamic equilibrium and we increase the temperature. Because we’ve increased the temperature of both reactants and products (they are in the same container), both the forward and reverse reactions speed up. Because the reaction is endothermic, however, the forward reaction has a greater E_a than the reverse reaction, so the forward reaction rate increases by more than the reverse reaction rate does.

This means that we have disturbed the equilibrium, because the forward reaction rate must now be greater than the reverse rate.

The greater increase in the rate of the forward reaction causes a net formation of products, and a net loss of reactants. These changes in concentration cause the rate of the forward reaction to fall and the rate of the reverse reaction to rise until the two rates are again equal. This will occur with more product, and less reactant, than was present in the original equilibrium.

Thus, the reaction has shifted "toward the products" in response to a disturbance. Le Chatelier's Principle tells us that these responses always "counteract" the disturbance. In what sense is that true in this case?

The simplest explanation is that you can think of heat as a "reactant" if the reaction is endothermic and as a "product" if the reaction is exothermic. Raising the temperature means adding heat, and lowering the temperature means removing it. If you have an exothermic reaction (meaning heat is a "product") and you lower the temperature, that is equivalent to removing a product, meaning the reaction will switch toward the products.

Disturbances of Pressure

The effect of a change in pressure is subtler still, because whether changing the pressure affects the position of equilibrium or not depends on how we change the pressure.

In particular, if we change the pressure by adding a gas that does not participate in the reaction, there will be no shift. This actually makes sense if you think in terms of ideal gases. In such a mixture, each gas behaves as though it were the only one in the container. Adding an inert gas to an equilibrium has no effect on the concentrations or partial pressures of the gases in the reaction, so it has no effect on the rates of the forward or reverse reaction.

In general, there will be an effect only if we change the pressure by changing the volume of the container and then only if the equilibrium includes gaseous reactants and/or products.

Because liquids and solids are highly incompressible, they do not change their volume and therefore they do not change their concentration when the size of the container or the external pressure changes. Thus, if there are no gaseous reactants or products, a change in pressure has no effect on the reaction rates.

If both of these conditions are met, then reducing the volume squeezes the gaseous reactants and products into a smaller volume, increasing their concentration; while increasing the volume allows them to expand into a larger volume, decreasing their concentration.

Again, only reactants and products are affected. Thus, as you might suspect, the eventual effect of changing the volume depends on whether the gases are reactants, products, or both.

Suppose, for example, that the only gases present are reactants (red). Decreasing the volume increases the reactant concentration, increasing the rate of the forward reaction, but it has no immediate effect on the rate of the reverse reaction because none of the product concentrations have changed.

Since none of the products are gases, decreasing the volume does not change their concentrations appreciably. Because the products are either solids or liquids they are already condensed and they cannot easily be compressed into a smaller volume. And since their concentrations don’t change, neither does the reverse rate.

This increase in the forward rate only causes a net loss of reactant, which slows the forward rate back down, and a net gain in the concentration of product, which increases the reverse rate, until the two rates are once again equal.

The shift to the products uses up the gaseous reactant. With fewer moles of gas in the vessel, the pressure falls, countering the initial increase in pressure that resulted from the decrease in volume, just as Le Chatelier’s principle predicts.

If the only gases present are products, then the reverse is true. Increasing the pressure by reducing the volume will cause a shift to the left.

And this shift, because it causes a net reduction in the amount of gas in the container, causes the pressure to fall, in accordance with Le Chatelier’s principle.

As you might expect, if we lower the pressure by increasing the volume, then the situation in both cases would be reversed, and we would see the reverse shift.

As you may be noticing, the variety of possible situations (is the gas a product or reactant? is the pressure being increased or decreased) means that it will not really be possible to memorize the full set of possibilities and the resulting outcomes. You have to be able to reason about a given situation using Le Chatelier's principle and your knowledge of a reaction.

If both reactants and products include gases, the direction of the shift will depend on the relative amounts. In this reaction, since there are more moles of gaseous reactants than gaseous products, reducing the volume will cause the equilibrium to shift to the right.

IMPORTANT: You have to look at the number of moles based on the coefficients (4 vs 2 here) rather than the number of formulas you see (2 vs 1 here).

For each reaction shown here, let's think about which direction the equilibrium will shift when the pressure is increased by reducing the volume of the reaction vessel.

First Reaction: The left side of this equation has more moles of gas, so we can think of it as having "more pressure" than the right side. So if the pressure is increased, the equilibrium will compensate by moving toward the products, per Le Chatelier's principle.

Second Reaction: Only the right side has any gas, so when the pressure is increased the reaction will shift toward the reactants, as that will reduce the pressure, compensating for the disturbance.

Third Reaction: Only the left side has gas, so the way to compensate for an increase in pressure is to shift toward the products.

In this week's lab exercise, you will investigate how different types of disturbance affect the equilibrium of a reaction, including adding and removing reactants and changing the temperature. You will then think about how your observations support (hopefully!) Le Chatelier's principle.

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