READ: Equilibrium

READ: Equilibrium

Introduction to Equilibrium

Consider this generic reaction: A+B→C+D. Based on what we have learned so far, you might assume that the reaction will keep going forward, forming C and D until either A or B (or both) is completely used up. When this is the case, we would say that the reaction “goes to completion”. Reactions that go to completion are referred to as irreversible reactions – (reactions where products cannot be converted back into reactants).

Some reactions, however, are reversible reactions (reactions where products can also react to re-form the reactants). In our example, this would correspond to the reaction C+D→A+B. During a reversible reaction, both the forward and backward reactions are happening at the same time.

Equilibrium

As we learned earlier, the rate of a reaction depends on the concentration of the reactants. At the very beginning of the reaction A+B→C+D, we would not expect the reverse reaction to proceed very quickly. If only a few particles of C and D have been created in a large flask of A and B, then it is very unlikely that they will find each other because the concentration of C and D is just too low. If C and D cannot “find” each other and collide with the correct energy and orientation, no reaction will occur.

However, as more and more C and D are created, it becomes more and more likely that they will find each other and react to re-form A and B. Conversely, as A and B are being used up, the forward reaction slows down for the same exact reason. The concentration of A and B decreases over the course of a reaction because there are less A and B particles in the same size flask. At some point, the rates for the forward and reverse reactions will be equal, at which point the concentrations will no longer change. If A and B are being destroyed at the same rate that they are being created, the overall amount should not change over time. At this point, the system is said to be in equilibrium – (when the rate of the forward reaction is equal to the rate of the reverse reaction). A qualitative description of this process for the reaction between hydrogen and iodine to make hydrogen iodide is shown below.

Chemists use a double arrow to show that a reaction is in equilibrium. For the reaction above, the chemical equation would be:

H2+I2⇌2HI

Dynamic Equilibrium

When a reaction is at equilibrium, the concentration of each component is constant over time. As we saw before, both the forward and reverse reactions are still taking place, but since they are moving at the same rate, there is no change for the system as a whole. This condition is called dynamic equilibrium (a state where no overall change is taking place although both reactions are still occurring). Individual molecules are still being formed and broken down, but the system as a whole is not changing over time.

Changes to Equilibrium Systems

When a reaction has reached equilibrium with a given set of conditions, if the conditions are not changed, the reaction will remain at equilibrium forever. The forward and reverse reactions continue at the same equal and opposite rates.

It is possible, however, to alter the reaction conditions. For example, you could increase the concentration of one of the products, or decrease the concentration of one of the reactants, or change the temperature. When a change of this type is made in a reaction at equilibrium, the reaction is no longer in equilibrium. When you alter something in a reaction at equilibrium, chemists say that you put stress on the equilibrium. When this occurs, the reaction will no longer be in equilibrium, so the reaction itself will begin changing the concentrations of reactants and products until the reaction comes to a new position of equilibrium. How a reaction will change when a stress is applied can be explained and predicted and is the topic of this lesson.

Le Châtelier’s Principle

In the late 1800s, a chemist by the name of Henry-Louis Le Châtelier was studying stresses that were applied to chemical equilibria. He formulated a principle, Le Châtelier’s Principle, which states that when a stress is applied to a system at equilibrium, the equilibrium will shift in a direction to partially counteract the stress and once again reach equilibrium. For instance, if a stress is applied by increasing the concentration of a reactant, the equilibrium position will shift toward the right and remove that stress by using up some of the reactants. The reverse is also true. If a stress is applied by lowering a reactant concentration, the equilibrium position will shift toward the left, this time producing more reactants to partially counteracting that stress. The same reasoning can be applied when some of the products is increased or decreased.

Effect of Concentration Changes

Let's use Le Châtelier's principle to explain the effect of concentration changes on an equilibrium system. Consider the generic equation:

A(aq)+B(aq)⇌C(aq)+D(aq)

At equilibrium, the forward and reverse rates are equal. The concentrations of all reactants and products remain constant, which keeps the rates constant. Suppose we add some additional A, thus raising the concentration of A without changing anything else in the system. Since the concentration of A is larger than it was before, the forward reaction rate will suddenly be higher. The forward rate will now exceed the reverse rate. Now there is a net movement of material from the reactants to the products. As the reaction uses up reactants, the forward rate that was too high slowly decreases while the reverse rate that was too low slowly increases. The two rates are moving toward each other and will eventually become equal again. They do not return to their previous rates, but they do become equal at some other value. As a result, the system returns to equilibrium.

Le Châtelier's principle says that when you apply a stress (adding A), the equilibrium system will shift to partially counteract the applied stress. In this case, the reaction shifts toward the products so that A and B are used up and C and D are produced. This reduction of the concentration of A is counteracting the stress you applied (adding A).

Suppose instead that you removed some A instead of adding some. In that case, the concentration of A would decrease, and the forward rate would slow down. Once again, the two rates are no longer equal. At the instant you remove A, the forward rate decreases, but the reverse rate remains exactly what it was. The reverse rate is now greater than the forward rate, and the equilibrium will shift toward the reactants. As the reaction runs backward, the concentrations of C and D decrease slowing the reverse rate, and the concentrations of A and B increase, raising the forward rate. The rates are again moving toward each other, and the system will again reach equilibrium. The shift of material from products to reactants increases the concentration of A, thus counteracting the stress you applied. Le Châtelier's principle again correctly predicts the equilibrium shift.

The effect of concentration on the equilibrium system according to Le Châtelier is as follows: increasing the concentration of a reactant causes the equilibrium to shift to the right, using up reactants and producing more products. Increasing the concentration of a product causes the equilibrium to shift to the left, using up products and producing more reactants. The same reasoning can be applied products are removed from the system.

Example 1

For the reaction, what would be the effect on the equilibrium system if:

SiCl4(g)+O2(g)⇌SiO2(s)+2Cl2(g)

1. [SiCl₄] increases
2. [O₂] increases
3. [Cl₂] increases

Solution:

1. The equilibrium would shift to the right. [Cl₂] would increase, more SiO₂ would be produced, and [O₂] would decrease.
2. The equilibrium would shift to the right. [SiCl₄] would decrease, more SiO₂would be produced, and [Cl₂] would increase.
3. The equilibrium would shift left. [SiCl₄] and [O₂] would increase, and SiO₂ would be used up.

Let's take a moment to consider what happens to the concentration of a reactant or product that is changed. In our theoretical reaction, if you add A, the concentration of A will increase. The equilibrium shifts toward the products, and A is used. Where does the concentration of A end up, higher or lower than the original concentration? The concentration of A increases when you add more A, but it decreases as the equilibrium shifts. A new equilibrium, however, will be reached before the concentration of A gets back down to its original concentration. This is why Le Châtelier's principle says the equilibrium will shift to partially counteract the applied stress. The equilibrium shift will move toward returning the concentration to where it was before you applied the stress, but the concentration never quite gets back to the original value before a new equilibrium is established.

Example 2

For the reaction PCl3(g)+Cl2(g)⇌PCl5(g), which way will the equilibrium shift if:

1. [PCl₃] decreases
2. [Cl₂] decreases
3. [PCl₅] decreases

Solution:

1. left
2. left
3. right

Effect of Changing Temperature

In the previous section, you learned that one of the most important factors that determines reaction rate is temperature. Raising the temperature will increase the average speed of the individual particles, thus causing more frequent collisions. Additionally, this increase in energy means that more particles will have the energy necessary to overcome the activation barrier. Overall, a rise in temperature increases both the frequency of collisions and the percentage of successful collisions.

It should be clear that increasing the temperature of the reaction vessel will increase both the forward and reverse reaction rates, but will it increase both rates equally? Let's examine the potential energy diagram of a reaction to see if we can gain any insight there. Here is the potential energy diagram for our usual theoretical reaction:

A(g)+B(g)→C(g)+D(g)

As you can see, the forward reaction has a small energy barrier while the reverse reaction has a very large energy barrier. With the reactants and products at the same temperature, the forward reaction will be much faster that the reverse reaction if the concentration of reactants is equal to the concentration of products.

Suppose we were to increase the temperature of the reaction system by 10 degrees C. Both the forward and reverse reaction rates will increase because a higher percentage of reactants and products will have enough energy to overcome the activation barrier. However, the reverse reaction rate will be more drastically affected. The activation barrier for the reverse reaction is so large that very few molecules would be able to overcome the barrier at the lower temperature. When the temperature is raised, there is a significant increase in the number of product molecules that can overcome the activation barrier. On the other hand, the activation barrier for the forward reaction is so small that many of the molecules would be able to overcome the barrier at the lower temperature. When the temperature is raised, there is only a small increase in the number of reactant molecules that can overcome the activation barrier. Thus, the rate of the reverse reaction will increase more dramatically than the rate of the forward reaction. The equilibrium will shift to the left, producing more reactants until a new equilibrium is established.

All reactions are either endothermic or exothermic, so there will always be a difference in the activation energy for the forward and reverse reactions. Whenever the temperature is raised, it will add energy to the system and increase both the forward and reverse reaction rates. However, it will more significantly increase the rate of the slower reaction. In an exothermic reaction, the reverse reaction has the higher activation barrier, and is thus slower. When heat is added, the reverse reaction will speed up more than the forward reaction, and the equilibrium will shift to the left. In an endothermic reaction, the forward reaction has the higher activation barrier, and is thus slower. When heat is added, the forward reaction will speed up more than the reverse reaction, and the equilibrium will shift to the right. By looking at a potential energy diagram, you should be able to tell 1) whether the reaction is exothermic or endothermic, 2) whether the forward or reverse reaction would be slower, assuming equal concentrations of reactants and products, and 3) which direction the equilibrium would shift in respond to a change in temperature.

Following the same reasoning as above, we can see that decreasing the temperature of a reaction produces an equilibrium shift in the opposite direction. Cooling an exothermic reaction results in a shift to the right, and cooling an endothermic reaction causes a shift to the left. Le Châtelier's principle correctly predicts the equilibrium shift when systems are heated or cooled. An increase in temperature is the same as adding energy to the system. Look at the following reaction:

2SO2(g)+O2(g)⇌2SO3(g)ΔH=−191 kJ

This could also be written as:

2SO2(g)+O2(g)⇌2SO3(g)+191 kJ

When changing the temperature of a system at equilibrium, energy can be thought of as just another product or reactant. For this reaction, 191 kJ of energy is produced for every mole of O₂ and 2 moles of SO₂ that react. Therefore, when the temperature of this system is raised, the effect will be the same as increasing any other product. As the temperature is increased, the equilibrium will shift away from the stress, resulting in more reactants and less products. As you would expect, the reverse would be true if the temperature is decreased. A summary of the effect temperature has on equilibrium systems is shown in the table below:

Example 3

Predict the effect on the equilibrium position if the temperature is increased in each of the following:

1. H2(g)+CO2(g)⇌CO(g)+H2O(g)ΔH=+40 kJ
2. 2SO2(g)+O2(g)⇌SO3(g)+energy 

Solution:

1. The reaction is endothermic. With an increase in temperature for an endothermic reaction, the reaction will shift right, producing more products.
2. The reaction is exothermic. With an increase in temperature for an exothermic reaction, the reaction will shift left, producing more reactants.

The Haber Process

The reaction between nitrogen gas and hydrogen gas can produce ammonia, NH₃. However, under normal conditions, this reaction does not produce very much ammonia. Early in the 20th century, the commercial use of this reaction was too expensive because of the low yield.

N2(g)+3H2(g)⇌2NH3(g)+energy

A German chemist named Fritz Haber applied Le Châtelier’s principle to help solve this problem. Decreasing the concentration of ammonia by immediately removing it from the reaction container causes the equilibrium to shift to the right, so the reaction can continue to produce more products.

One more factor that will affect this equilibrium system is the temperature. Since the forward reaction is exothermic, lowering the temperature will once again shift the equilibrium system to the right and increase the ammonia produced. Unfortunately, this process also has a very high activation energy, so if the temperature is too low, the reaction will slow to a crawl. Thus, a balance must be struck between shifting the equilibrium to favor products and allowing products to be formed at a reasonable rate. It was found that the optimum conditions for this process (the ones that produce the most ammonia the fastest) are 550°C and 250 atm of pressure, with the ammonia being continually removed from the system.

Summary

• Irreversible reactions will continue to form products until the reactants are fully consumed.
• Reversible reactions will react until a state of equilibrium is reached.
• Dynamic equilibrium refers to an equilibrium where forward and reverse reactions are still occurring, but they are proceeding at the same rate, so there is no net change.
• In a dynamic equilibrium the concentrations of the reactants and products are constant.
• Increasing the concentration of a reactant causes the equilibrium to shift to the right, producing more products.
• Increasing the concentration of a product causes the equilibrium to shift to the left, producing more reactants.
• Decreasing the concentration of a reactant causes the equilibrium to shift to the left, using up some products.
• Decreasing the concentration of a product causes the equilibrium to shift to the right, using up some reactants.
• Changing the temperature of a reaction system will cause a shift in equilibrium based on the ΔH of the reaction. Heating an endothermic reaction causes a shift toward the products. Heating an exothermic reaction causes a shift toward the reactants.

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