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# 19.3: The Effects of Applying Stress to Reactions at Equilibrium

Difficulty Level: At Grade Created by: CK-12

## Lesson Objectives

The student will:

• state Le Châtelier’s Principle.
• demonstrate on specified chemical reactions how Le Châtelier’s Principle is applied to equilibrium systems.
• describe the effect of concentration on an equilibrium system.
• demonstrate with specific equations how Le Châtelier’s Principle explains the effect of concentration.
• describe the effect of pressure as a stress on the equilibrium position.
• describe the pressure effect in Le Châtelier’s Principle.
• describe the effect of temperature as a stress on an equilibrium system.
• explain how Le Châtelier’s principle explains the effect of temperature.
• understand how a catalyst works in equilibrium reactions.
• explain the effect of a catalyst in equilibrium reactions.

## Vocabulary

• Le Châtelier’s Principle

## Introduction

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, and the macroscopic properties remain constant.

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.

Le Châtelier's principle does not provide an explanation of what happens on the molecular level to cause the equilibrium shift. Instead, it is simply a quick way to determine which way the reaction will run in response to a stress applied to the system at equilibrium.

## 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:

$A_{(aq)} + B_{(aq)} \rightleftharpoons C_{(aq)} + D_{(aq)}$

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 exact opposite is true when either a reactant or product is removed.

Example:

For the reaction $\mathrm{SiCl}_{4(g)} + \mathrm{O}_{2(g)} \rightleftharpoons \mathrm{SiO}_{2(s)} + 2 \ \mathrm{Cl}_{2(g)}$, what would be the effect on the equilibrium system if:

1. $[\mathrm{SiCl}_4]$ increases
2. $[\mathrm{O}_2]$ increases
3. $[\mathrm{Cl}_2]$ increases

Solution:

1. The equilibrium would shift to the right. $[\mathrm{Cl}_2]$ would increase, more $\mathrm{SiO}_2$ would be produced (but that does not increase its concentration since its a solid), and $[\mathrm{O}_2]$ would decrease.
2. The equilibrium would shift to the right. $[\mathrm{SiCl}_4]$ would decrease, more $\mathrm{SiO}_2$ would be produced (but again no change in concentration), and $[\mathrm{Cl}_2]$ would increase.
3. The equilibrium would shift left. $[\mathrm{SiCl}_4]$ and $[\mathrm{O}_2$] would increase, and $\mathrm{SiO}_2$ would be used up but not change its concentration.

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:

For the reaction $\mathrm{PCl}_{3(g)} + \mathrm{Cl}_{2(g)} \rightleftharpoons \mathrm{PCl}_{5(g)}$, which way will the equilibrium shift if:

1. $[\mathrm{PCl}_3]$ decreases
2. $[\mathrm{Cl}_2]$ decreases
3. $[\mathrm{PCl}_5]$ decreases

Solution:

1. left
2. left
3. right

Example:

Here's a reaction at equilibrium. Note the phases of each reactant and product.

$A_{(aq)} + B_{(s)} \rightleftharpoons C_{(aq)} + D_{(aq)}$
1. Which way will the equilibrium shift if you add some $A$ to the system without changing anything else?
2. After $A$ has been added and a new equilibrium is reached, how will the new concentration of $D$ compare to the original concentration of $D$?
3. After $A$ has been added and a new equilibrium has been established, how will the new concentration of $A$ compare to the original concentration of $A$?
4. After $A$ has been added and a new equilibrium has been established, how will the new concentration of $B$ compare to the original concentration of $B$?
5. Which way will the equilibrium shift if you add some $C$ to the system without changing anything else?
6. After $C$ has been added and a new equilibrium has been established, how will the new concentration of $D$ compare its original concentration?
7. After $C$ has been added and a new equilibrium has been established, how will the new concentration of $A$ compare its original concentration?
8. Which way will the equilibrium shift if you add some $B$ to the system without changing anything else?

Solution:

1. The equilibrium will shift toward the products.
2. The new concentration of $D$ will be higher than the original.
3. The new concentration of $A$ will be higher than the original, but lower than the concentration right after $A$ was added.
4. Since $B$ is a solid, its concentration will be the same as the original. There will be less of it since some was used in the equilibrium shift, but the concentration will be the same.
5. The equilibrium will shift toward the reactants.
6. The new concentration of $D$ will be lower than the original.
7. The new concentration of $A$ will be higher than the original.
8. Since $B$ is a solid, adding $B$ will not change its concentration and therefore has no effect on the equilibrium. It is possible that adding some $B$ will increase the surface area of $B$ and therefore increase the forward reaction rate, but it will also increase the reverse reaction rate by approximately the same amount, hence no shift in equilibrium.

### The Haber Process and Concentration Change Effects

The reaction between nitrogen gas and hydrogen gas can produce ammonia, $\mathrm{NH}_3$. 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.

$\mathrm{N}_{2(g)} + 3 \ \mathrm{H}_{2(g)} \rightleftharpoons 2 \ \mathrm{NH}_{3(g)} + \text{energy}$

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 product. Haber used Le Châtelier’s principle to solve this problem in other ways as well. These will be discussed in the following sections.

### Summary

• 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.

### Questions

1. What is the effect on the equilibrium position if the [reactants] is increased?
2. What is the effect on the equilibrium position if the [reactants] is decreased?
3. Which of the following will cause a shift in the equilibrium position of the equation: $2 \ \mathrm{C}_8\mathrm{H}_{18(l)} + 25 \ \mathrm{O}_{2(g)} \rightleftharpoons 18 \ \mathrm{H}_2\mathrm{O}_{(l)} + 16 \ \mathrm{CO}_{2(g)}$? i. add $\mathrm{C}_8\mathrm{H}_{18}$ ii. add $\mathrm{O}_2$ iii. remove $\mathrm{CO}_2$ iv. remove $\mathrm{H}_2\mathrm{O}$
1. i and ii only
2. ii and iii only
3. ii and iv only
4. i, ii, iii, and iv
4. Which of the following will cause a shift in the equilibrium position of the equation: $\mathrm{CaCO}_{3(s)} \rightleftharpoons \mathrm{CaO}_{(s)} + \mathrm{CO}_{2(g)}$? i. add $\mathrm{CaCO}_3$ ii. add $\mathrm{CO}_2$ iii. remove $\mathrm{CaO}$ iv. remove $\mathrm{CO}_2$
1. ii only
2. i and iii only
3. ii and iv only
4. i, ii, iii, and iv only
5. For the reaction: $\mathrm{N}_2\mathrm{O}_{5(s)} \rightleftharpoons \mathrm{NO}_{2(g)} + \mathrm{O}_{2(g)}$, what would be the effect on the equilibrium if:
1. $\mathrm{N}_2\mathrm{O}_5$ is added?
2. $\mathrm{NO}_2$ is removed?
3. $\mathrm{NO}_2$ is added?
4. $\mathrm{O}_2$ is added?
6. Answer the following questions when $[\mathrm{CO}]$ is increased in the following system at equilibrium: $\mathrm{Fe}_2\mathrm{O}_{3(s)} + 3 \ \mathrm{CO}_{(g)} \rightleftharpoons 2 \ \mathrm{Fe}_{(s)} + 3 \ \mathrm{CO}_{2(g)}$.
1. Write the equilibrium constant expression.
2. Which direction will this equilibrium shift?
3. What effect will this stress have on $[\mathrm{CO}_2]$?
7. For the reaction $\mathrm{C}_{(s)} + \mathrm{H}_2\mathrm{O}_{(g)} \rightleftharpoons \mathrm{CO}_{(g)} + \mathrm{H}_{2(g)}$, what would be the effect on the equilibrium system if:
1. $[\mathrm{H}_2\mathrm{O}]$ increases?
2. mass of $\mathrm{C}$ decreases?
3. $[\mathrm{CO}]$ increases?
4. $[\mathrm{H}_2]$ decreases?

## Effect of Changes of Pressure

A second type of stress studied by Le Châtelier was the effect of changing the pressure of a system at equilibrium. Reactants and products that are in solid, liquid, or aqueous states are not compressible, so raising or lowering the pressure on a system that contains no gases will have no effect on the equilibrium. If, however, the reaction at equilibrium contains at least one gas in either the reactants or products, altering the pressure will alter the equilibrium.

To get a better understanding of how pressure changes are affecting gaseous reaction components, it may help to consider the relationship between pressure and concentration for a gas at a given temperature. Recall that for an ideal gas, $PV = nRT$, where $R$ is a constant. Since temperature is also held constant in this case, $R$ and $T$ can be combined to form a new constant, which we’ll call $x$. Dividing both sides by volume now gives us $P = x \cdot (n/V)$. Remember that $n$ is in moles and $V$ is in liters; moles/liters = molarity. Thus, the pressure of a gas at a constant temperature is directly proportional to its concentration. When we are talking about the partial pressure of a gas, keep in mind that we are essentially talking about its concentration. This will help you to grasp some of the potentially confusing effects pressure can have on the equilibrium position.

Suppose we have the following reaction at equilibrium.

$A_{(aq)} + B_{(aq)} \rightleftharpoons C_{(aq)} + D_{(g)}$

Since the reaction has at least one gaseous substance involved in the reaction, its equilibrium will be affected by a change in the partial pressure of that gas. In order for this reaction to reach equilibrium, it would have to be in a closed reaction vessel so that the gaseous product do not escape. The gas must stay in contact with the solution for the reverse reaction to occur.

There are at least three ways to increase the pressure in the area above the liquid in the cylinder: 1) add some other gas not involved in the reaction, 2) add some gaseous $D$ into the space above the liquid, and 3) push the piston down so the space above the liquid in the cylinder is decreased (see figure below).

1. Increasing the pressure in the cylinder by adding a gas not involved in the reaction will increase the total gas pressure in the cylinder, but it will not affect the partial pressure of $D$, so there will be no effect on the equilibrium of this reaction. This concept might be easier to understand if we think about it in terms of concentration rather than pressure. If we have the same amount of gas in the same amount of space, the concentration of the gas and thus its partial pressure will not change. The fact that some other gas was added does not change either of these key values. This is why only changes to the partial pressure of the reaction component, not the total pressure, are able to affect the equilibrium position.
2. Adding some gaseous $D$ to the cylinder is the same as adding a reactant or product, which has already been discussed earlier.
3. Lowering the piston in the cylinder pushes the $D$ molecules into a smaller space, increasing both the partial pressure of gaseous $D$ and its concentration in the space above the liquid. Since the concentration of $D$ is increased, the reverse reaction rate will increase and the equilibrium will shift toward the reactants.

The reverse of this is also true. If you expand the volume of the cylinder by raising the piston, the partial pressure and concentration of $D$ will decrease. When the concentration of $D$ decreases, the reverse rate slows and the forward reaction rate will drive the equilibrium toward the products. This equilibrium shift increases the concentration and partial pressure of $D$, once again counteracting the stress you applied.

### Increasing Pressure Shifts Equilibrium Toward Fewer Moles of Gas

When you have a reaction at equilibrium with gaseous substances on both sides of the equation, the explanation for what happens is more complicated. Consider the following equation:

$A_{(aq)} + B_{(g)} \rightleftharpoons C_{(aq)} + D_{(g)}$

In this case, there are gaseous reactants and gaseous products. It should be easy to see that if we reduce the volume above this reaction to half of its previous volume, the partial pressures and the concentrations of these two gases will be doubled. Therefore, both reaction rates will increase because concentrations on each side have been increased. If we assume that we are still dealing with single collision reactions (so that the forward and reverse rates can be expressed as $R_F = k_F [A][B]$ and $R_R = k_R [C][D]$), then we can see that doubling the concentration of B doubles the forward rate and doubling the concentration of D doubles the reverse rate. Both rates will increase, but since they increase by the same factor, the equilibrium will not shift.

Now consider the reaction below:

$A_{(aq)} + B_{(g)} \rightleftharpoons C_{(g)} + D_{(g)}$

This time, we see that there is one gas in the reactants and two gases in the products. If we once again reduce the volume of the gases by half, the partial pressures and the concentrations of the gases will double. Therefore, we double the forward reaction rate because we doubled the concentration of $B$. In terms of the products, we are doubling the concentrations of $C$ and $D$. The new reverse rate is:

$Reverse \ Rate_{NEW} = k(2[C])(2[D]) = 4k[C][D]$

Therefore, the new reverse rate will be four times the original reverse rate. If we double the forward reaction rate and quadruple the reverse rate, the equilibrium shift will be toward the reactants.

When you increase the pressure (by reducing volume) on a reaction at equilibrium, the equilibrium shift will be toward the side that has fewer moles of gas. A decrease in pressure due to a volume expansion will shift the equilibrium to the side with more moles of gas. Once again, you should note that Le Châtelier's principle predicts this result. If the equilibrium shift is converting $2$ moles of gas to $1\;\mathrm{moles}$ of gas, then the shift is reducing the number of moles of gas and the total pressure will decrease. You applied a stress by increasing the pressure, and the equilibrium shift tends to counteract that stress by reducing pressure.

Example:

For the reaction $2 \ \mathrm{SO}_{2(g)} + \mathrm{O}_{2(g)} \rightleftharpoons 2 \ \mathrm{SO}_{3(g)}$, what would be the effect on the equilibrium position when the pressure is increased by reducing the volume?

Solution:

$2 \ \mathrm{SO}_{2(g)} + \mathrm{O}_{2(g)} \rightleftharpoons 2 \ \mathrm{SO}_{3(g)}$
$\text{3 moles gas} \rightleftharpoons \text{2 moles gas}$

If the pressure increases, the reaction would shift to the side with the least number of moles of gas. Since there are $3 \ \mathrm{moles}$ of gaseous reactants and $2 \ \mathrm{moles}$ of gaseous products, the equilibrium would shift right, producing more products.

Example:

For the reaction $4 \ \mathrm{NH}_{3(g)} + 5 \ \mathrm{O}_{2(g)} \rightleftharpoons 4 \ \mathrm{NO}_{(g)} + 6 \ \mathrm{H}_2\mathrm{O}_{(g)}$, what would be the effect on the equilibrium system if the pressure decreases?

Solution:

$4 \ \mathrm{NH}_{3(g)} + 5 \ \mathrm{O}_{2(g)} \rightleftharpoons 4 \ \mathrm{NO}_{(g)} + 6 \ \mathrm{H}_2\mathrm{O}_{(g)}$
$\text{9 moles gas} \rightarrow \text{10 moles gas}$

If the pressure decreases, the reaction would shift to the side with the greater number of moles of gas. Since there are nine moles of gaseous reactants and ten moles of gaseous products, the equilibrium would shift right, producing more products.

Example:

For the reaction $\mathrm{PCl}_{5(g)} \rightleftharpoons \mathrm{Cl}_{2(g)} + \mathrm{PCl}_{3(g)}$, what would be the effect on the equilibrium system when (1.) the volume increases, and (2.) the volume decreases

Solution:

1. When the volume is increased, pressure decreases, so the equilibrium will shift toward the side with more moles of gas. Therefore, $[\mathrm{PCl}_5]$ will decrease, and $[\mathrm{Cl}_2]$ and $[\mathrm{PCl}_3]$ will increase.
2. When the volume is decreased, pressure increases, so the equilibrium will shift toward the side with fewer moles of gas. Therefore, $[\mathrm{PCl}_5]$ will increase, and $[\mathrm{Cl}_2]$ and $[\mathrm{PCl}_3]$ will decrease.

### The Haber Process and the Effect of Pressure Change

Earlier, we discussed the reaction that produces ammonia, $\mathrm{NH}_3$:

$\mathrm{N}_{2(g)} + 3 \ \mathrm{H}_{2(g)} \rightleftharpoons 2 \ \mathrm{NH}_{3(g)}$

We also stated that the German chemist Fritz Haber used Le Châtelier’s principle to develop a method that produces more product. Previously, we saw that when ammonia is immediately removed from the reaction container, thus decreasing the concentration, the equilibrium shifts to make up for the stress and produces more ammonia.

Now we can add another improvement: Since there are $4 \ \mathrm{moles}$ of gas molecules in the reactants and $2 \ \mathrm{moles}$ of gas molecules in the product, increasing the pressure (by decreasing the volume) will shift the equilibrium to the right, producing more ammonia. Later, we will discuss one more factor to complete our discussion of the Haber process.

### Summary

• A decrease in volume will cause an increase in pressure, shifting the equilibrium to the side with fewer moles of gas.
• An increase in volume will cause a decrease in pressure, shifting the equilibrium to the side with more moles of gas.

### Questions

1. What is the effect on the equilibrium position if the pressure is increased?
2. What is the effect on the equilibrium position if the pressure is decreased?
3. Use Le Châtelier’s Principle to predict what will happen to the following reaction at equilibrium if the pressure is increased: $2 \ \mathrm{NH}_{3(g)} \rightleftharpoons \mathrm{N}_{2(g)} + 3 \ \mathrm{H}_{2(g)}$. Mark all that apply.
1. equilibrium position shifts right
2. equilibrium position shifts left
3. $[\mathrm{N}_2]$ will decrease
4. $[\mathrm{NH}_3]$ will increase
4. Use Le Châtelier’s principle to predict what will happen to the following reaction at equilibrium if the pressure is decreased: $2 \ \mathrm{NO}_{(g)} + 2 \ \mathrm{H}_{2(g)} \rightleftharpoons \mathrm{N}_{2(g)} + 2 \ \mathrm{H}_2\mathrm{O}_{(g)}$. Mark all that apply.
1. equilibrium position will not shift
2. equilibrium position shifts left
3. $[\mathrm{N}_2]$ will increase
4. $[\mathrm{NO}]$ will increase
5. Use Le Châtelier’s principle to predict what will happen to the following reaction at equilibrium if the volume is decreased: $2 \ \mathrm{NCl}_{3(g)} \rightleftharpoons \mathrm{N}_{2(g)} + 3 \ \mathrm{Cl}_{2(g)}$. Mark all that apply.
1. equilibrium position shifts right
2. equilibrium position shifts left
3. $[\mathrm{N}_2]$ will increase
4. $[\mathrm{NCl}_3]$ will decrease
6. For the reaction $2 \ \mathrm{N}_2\mathrm{O}_{(g)} + \mathrm{O}_{2(g)} \rightleftharpoons 4 \ \mathrm{NO}_{(g)}$, what would be the effect on the equilibrium system if the pressure increases (or the volume decreases)?
7. For the reaction $2 \ \mathrm{IBr}_{(g)} \rightleftharpoons \mathrm{I}_{2(g)} + \mathrm{Br}_{2(g)}$, what would be the effect on the equilibrium system if the pressure decreases (or the volume increases)?
8. For the reaction $\mathrm{H}_{2(g)} + \mathrm{CO}_{2(g)} \rightleftharpoons \mathrm{H}_2\mathrm{O}_{(g)} + \mathrm{CO}_{(g)}$, what would be the effect on the equilibrium system if
1. the pressure increases?
2. the volume decreases?
9. For the reaction $3 \ \mathrm{NO}_{(g)} \rightleftharpoons \mathrm{N}_2\mathrm{O}_{(g)} + \mathrm{NO}_{2(g)}$, what would be the effect on the equilibrium system if
1. the pressure increases?
2. the volume decreases?

## Effect of Changing Temperature

As we saw in the chapter “Chemical Kinetics,” 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:

$\mathrm{A}_{(g)} + \mathrm{B}_{(g)} \rightarrow \mathrm{C}_{(g)} + \mathrm{D}_{(g)}$

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. In the chapter “Chemical Kinetics,” we used an energy distribution curve to show the percentage of reactant particles that had sufficient activation energy to react. The figure below shows the energy distribution curves for $A$ and $B$ in the forward reaction and for $C$ and $D$ in the reverse reaction.

As you can see in the drawing, a much larger percentage of reactant particles than product particles have the activation energy required for a successful collision. Suppose we were to increase the temperature $10^\circ\mathrm{C}$. The activation energy requirements remain the same for both groups of particles, but the curves will shift right to reflect the additional energy these particles possess.

Both reaction rates will increase because a higher percentage of both reactants and products have enough energy to overcome the activation barrier. Additionally, the percentage of particles that exceed the activation energy is still higher for the reactants than the products, so the forward reaction will still be faster. However, if we look at the increase in sufficiently energetic particles, the product side has more of a change than the reactants. Thus, the reverse rate will increase more than the forward rate. The equilibrium will shift left, producing more reactants until a new equilibrium is established.

For nearly every reaction, either the forward or reverse reaction will require more activation energy than the other. The addition of energy to a reaction will increase both reaction rates, but it increases the rate of the slower reaction more. In an exothermic reaction, the products are lower in energy than the reactants, so they have a higher barrier to climb in order to reach the same transition state. Accordingly, the reverse rate will be slower for an exothermic reaction, so increasing the temperature will shift the equilibrium to the left. The reverse is true for an endothermic reaction. 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:

$2 \ \mathrm{SO}_{2(g)} + \mathrm{O}_{2(g)} \rightleftharpoons 2 \ \mathrm{SO}_{3(g)} \ \ \ \ \ \triangle H = -191 \ \text{kJ}$

This could also be written as:

$2 \ \mathrm{SO}_{2(g)} + \mathrm{O}_{2(g)} \rightleftharpoons 2 \ \mathrm{SO}_{3(g)} + 191 \ \text{kJ}$

When changing the temperature of a system at equilibrium, energy can be thought of as just another product or reactant. For this reaction, $191 \ \mathrm{kJ}$ of energy is produced for every mole of $\mathrm{O}_2$ and 2 moles of $\mathrm{SO}_2$ 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 Table below.

The Effect of Temperature on an Endothermic and an Exothermic Equilibrium System
Exothermic $(- \triangle H)$ Endothermic $(+ \triangle H)$
Increase Temperature Shifts left, favors reactants Shifts right, favors products
Decrease Temperature Shifts right, favors products Shifts left, favors reactants

Example:

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

1. $\mathrm{H}_{2(g)} + \mathrm{CO}_{2(g)} \rightleftharpoons \mathrm{CO}_{(g)} + \mathrm{H}_2\mathrm{O}_{(g)} \ \ \ \ \ \triangle H = + 40 \ \mathrm{kJ/mol}$
2. $2 \ \mathrm{SO}_{2(g)} + \mathrm{O}_{2(g)} \rightleftharpoons 2 \ \mathrm{SO}_{3(g)} + \mathrm{energy}$

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 Value of K Changes When Temperature Changes

The rate constant, $k$, for a reaction is determined for a particular temperature and is correct only at that temperature. The $k$ value is unaffected by changes in concentration, volume, or pressure. Recall that the equilibrium constant, $K$, is the ratio of $k_F$ to $k_R$, so if the values for the $k$'s don't change, neither will $K$. Therefore, $K$ is also unaffected by changes in concentration, volume, or pressure. However, changing the temperature will alter the values of $k_F$, $k_R$, and $K$.

Temperature is the only stress, of all those studied by Le Châtelier (and others), that has the effect of increasing or decreasing the value of the equilibrium constant. The Dutch chemist Jacobus Henricus van 't Hoff derived the relationship for how equilibrium constants change with temperature shortly after Le Châtelier had reported his finding. The qualitative results of his work are presented in the following table (Table below).

The Effect of Temperature on an Endothermic and an Exothermic Equilibrium System
Exothermic $(- \triangle H)$ Endothermic $(+ \triangle H)$
Increase Temperature Shifts left, favors reactants, $K \downarrow$ Shifts right, favors products, $K \uparrow$
Decrease Temperature Shifts right, favors products, $K \uparrow$ Shifts left, favors reactants, $K \downarrow$

Example:

How does an increase in the temperature affect the value of K and the concentration of the substance boldfaced in each of the following reactions?

1. $\mathbf{SO_{2(g)}} + \mathrm{energy} \rightleftharpoons \mathrm{S}_{(s)} + \mathrm{O}_{2(g)}$
2. $\mathrm{P}_{4(s)} + \ 10 \ \mathrm{Cl}_{2(g)} \rightleftharpoons 4 \ \mathbf{PCl_{5(g)}} + \mathrm{energy}$

Solution:

1. An increase in temperature for an endothermic reaction causes a shift to the right, so the equilibrium constant $K$ increases, and the products are favored. This means that the concentration of $\mathrm{SO}_2$ will decrease.
2. An increase in temperature for an exothermic reaction causes a shift to the left, so the equilibrium constant $K$ decreases, and the reactants are favored. This means that the concentration of $\mathrm{PCl}_5$ will decrease.

### The Haber Process and the Effect of Temperature Change

We have already seen that the maximum amount of ammonia produced in the Haber process can be improved by decreasing the concentration of the ammonia (by removing it from the reaction container) and increasing the pressure.

$\mathrm{N}_{2(g)} + 3 \ \mathrm{H}_{2(g)} \rightleftharpoons 2 \ \mathrm{NH}_{3(g)} + \text{energy}$

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 $550^\circ\mathrm{C}$ and 250 atm of pressure, with the ammonia being continually removed from the system.

### Summary

• For an endothermic reaction, an increase in temperature shifts the equilibrium toward the products, whereas a decrease in temperature shifts the equilibrium toward the reactants.
• For an forward exothermic reaction, an increase in temperature shifts the equilibrium toward the reactant side, whereas a decrease in temperature shifts the equilibrium toward the products.
• Increasing or decreasing the temperature causes the $K$ value to change.

### Questions

1. Why does temperature affect the value of the equilibrium constant?
2. Which direction will a system at equilibrium shift for each of the following?
1. adding energy to a forward exothermic equilibrium system
2. adding energy to a reverse exothermic equilibrium system
3. adding energy to a forward endothermic equilibrium system
4. adding energy to a reverse endothermic equilibrium system
3. At a certain temperature, the equilibrium constant for the reaction $\mathrm{Ag}^+_{(aq)} + \mathrm{Cl}^-_{(aq)} \rightleftharpoons \mathrm{AgCl}_{(s)} \ (\triangle H = -112 \ \mathrm{kJ/mol})$ is $7.43$. What would the value of the equilibrium constant be if the reaction were allowed to come to equilibrium at a higher temperature?
1. less than $7.43$
2. greater than $7.43$
3. the same, $7.43$
4. not enough information is available
4. With an increase in temperature, the equilibrium constant for a certain reaction was found to increase. Consequently, what can be said about the concentrations of the reactants and products?
1. They both increase.
2. They both decrease.
3. The [reactants] increase and the [products] decrease.
4. The [products] increase and the [reactants] decrease.
5. Predict the effect of an increase in temperature on the equilibrium position for each of the following.
1. $\mathrm{H}_{2(g)} + \mathrm{I}_{2(g)} \rightleftharpoons 2 \ \mathrm{HI}_{(g)} \ \ \ \ \ \triangle H = +51.8 \ \mathrm{kJ}$
2. $\mathrm{P}_4\mathrm{O}_{10(s)} + 6 \ \mathrm{H}_2\mathrm{O}_{(l)} \rightleftharpoons 4 \ \mathrm{H}_3\mathrm{PO}_{4(aq)} + \mathrm{heat}$
6. How does an increase in temperature affect the value of $K$ and the concentration of the products for each of the following?
1. $\mathrm{NO}_{2(g)} + \mathrm{NO}_{(g)} \rightleftharpoons \mathrm{N}_2\mathrm{O}_{(g)} + \mathrm{O}_{2(g)} \ \ \ \ \ \triangle H = -43 \ \mathrm{kJ}$
2. $4 \ \mathrm{NH}_{3(g)} + 5 \ \mathrm{O}_{2(g)} \rightleftharpoons 4 \ \mathrm{NO}_{(g)} + 6 \ \mathrm{H}_2\mathrm{O}_{(g)} + \mathrm{heat}$
7. Predict the effect on the equilibrium position if the temperature is increased in each of the following.
1. $\mathrm{C}_2\mathrm{H}_{2(g)} + \mathrm{H}_2\mathrm{O}_{(g)} \rightleftharpoons \mathrm{CH}_3\mathrm{CHO}_{(g)} \ \ \ \ \ \triangle H = - 151 \ \mathrm{kJ/mol}$
2. $\mathrm{HC}_2\mathrm{H}_3\mathrm{O}_{2(g)} + \mathrm{H}_2\mathrm{O}_{(g)} + \mathrm{energy} \rightleftharpoons \mathrm{CH}_3\mathrm{CH}_2\mathrm{OH}_{(g)} + \mathrm{O}_{2(g)}$

Use this information to answer questions $8 - 11$. The reaction below is at equilibrium in a closed container at $25^\circ\mathrm{C}$.

$2 \ \mathrm{NOBr}_{(g)} \rightleftharpoons 2 \ \mathrm{NO}_{(g)} + \mathrm{Br}_{2(g)} \ \ \ \ \ \triangle H = +16.1 \ \mathrm{kJ/mol} \ \mathrm{Br}_2$

1. What will happen to the concentration of NO if the temperature is increased?
1. increase
2. decrease
3. remain unchanged
2. What will happen to the concentration of $\mathrm{NOBr}$ if the temperature is increased?
1. increase
2. decrease
3. remain unchanged
3. What will happen to the concentration of $\mathrm{Br}_2$ if the temperature is increased?
1. increase
2. decrease
3. remain unchanged
4. What will happen to the value of $K_e$ if the temperature is increased?
1. increase
2. decrease
3. remain unchanged

## Effect of a Catalyst

We have now studied the effect of temperature, pressure, and concentration changes on a system in equilibrium. A summary of how these stresses affect the equilibrium position are found in Table below.

Summary of Le Châtelier Stresses and the Effect on Equilibrium
Stress Increase Decrease
Temperature Endothermic, shift right Endothermic, shift left
Exothermic, shift left Exothermic shift right
Pressure Shift to side with fewer moles of gas Shift to side with more moles of gas
Concentration Increase [reactants], shift right Decrease [reactants], shift left
Increase [products], shift left Decrease [products], shift right

Le Châtelier also studied the effect of catalysts on the equilibrium position. If the catalyst acted as a stress, would it shift the equilibrium position, and if so, how?

### A Catalyst Increases Both Reaction Rates Equally

Remember that a catalyst is a substance that increases the rate of a chemical reaction but is not consumed or destroyed by the reaction. A catalyst provides an alternate pathway between reactants and products that requires less activation energy. It will not have an effect on the equilibrium position because it increases both the forward and the reverse reactions equally. Adding a catalyst to a system in equilibrium will not cause any macroscopic change. Adding a catalyst to a reaction that is not in equilibrium will cause the system to reach equilibrium faster, but the end result (the final concentrations of reactants and products) will be exactly the same with or without a catalyst.

Example:

Predict the effect on the chemical equilibrium when each of the indicated changes are made to the following reaction at equilibrium.

$4 \ \mathrm{NH}_{3(g)} + 5 \ \mathrm{O}_{2(g)} \rightleftharpoons 4 \ \mathrm{NO}_{(g)} + 6 \ \mathrm{H}_2\mathrm{O}_{(g)}$
1. $[\mathrm{H}_2\mathrm{O}]$ is increased
2. Pressure is decreased
3. $[\mathrm{NH}_3]$ is decreased
5. Volume is decreased

Solution:

$4 \ \text{NH}_{3(g)}$ $5 \ \text{O}_{2(g)}$ $\rightleftharpoons$ $4 \ \text{NO}_{(g)}$ $6 \ \text{H}_2\text{O}_{(g)}$
1. $\uparrow$ $\uparrow$ $\leftarrow$ $\downarrow$ $\uparrow$
2. $\downarrow$ $\downarrow$ $\rightarrow$ $\uparrow$ $\uparrow$
3. $\downarrow$ $\uparrow$ $\leftarrow$ $\downarrow$ $\downarrow$
4. no effect
5. $\uparrow$ $\uparrow$ $\leftarrow$ $\downarrow$ $\downarrow$

A Khan Academy electronic lecture on Le Châtelier's Principle (9a) is available at http://www.youtube.com/watch?v=4-fEvpVNTlE (14:43).

### Summary

• A catalyst affects the rate of a reaction such that both the forward and reverse reactions would be changed equally.
• A catalyst has no effect on the equilibrium position.

### Questions

1. What is a catalyst? Give an example of a catalyst.
2. How does a catalyst work in an equilibrium reaction?
3. What will the effect of adding a catalyst be for the following reaction: $4 \ \mathrm{NH}_{3(g)} + 5 \ \mathrm{O}_{2(g)} \rightleftharpoons 4 \ \mathrm{NO}_{(g)} + 6 \ \mathrm{H}_2\mathrm{O}_{(g)} \ (\triangle H = -905 \ \mathrm{kJ})$?
1. the [products] will increase at equilibrium
2. the [reactants] will increase at equilibrium
3. the equilibrium constant will increase for the forward reaction
4. there will be no effect on the equilibrium concentrations
4. The reaction between nitrogen monoxide and carbon monoxide is represented as follows: $\mathrm{NO}_{(g)} + \mathrm{CO}_{(g)} \rightleftharpoons \frac{1} {2} \ \mathrm{N}_{2(g)} + \mathrm{CO}_{2(g)} \ (\triangle H = -374 \ \mathrm{kJ})$. If a catalyst was added to the system, all of the following would be affected in the system except:
1. an increase in $\triangle H$
2. an increase the rate of the reverse reaction.
3. an increase the rate of the forward reaction.
4. a change in the reaction path
5. Of the four factors listed below, which factors would not be determined by the value of the equilibrium constant for the following equation: $2 \ \mathrm{NaNO}_{3(s)} \rightleftharpoons 2 \ \mathrm{NaNO}_{2(s)} + \mathrm{O}_{2(g)}$? i. concentration of $\mathrm{NaNO}_{3(s)}$ ii. concentration of $\mathrm{NaNO}_{2(s)}$ iii. concentration of $\mathrm{O}_{2(g)}$ iv. addition of a catalyst
1. i and ii only
2. iii and iv only
3. i and iii only
4. i, ii and iv only
6. In the reaction $\mathrm{C}_{(s)} + \mathrm{O}_{2(g)} \rightleftharpoons \mathrm{CO}_{2(g)} \ (\triangle H = -393.5 \ \mathrm{kJ/mol}$), what would Le Châtelier’s principle suggest as a way to increase the concentration of $\mathrm{CO}_2$?
2. increase $\mathrm{O}_2$
3. increase the pressure
4. increase the temperature
7. Predict the effect on the chemical equilibrium when each of the following changes is made to this reaction at equilibrium: $\mathrm{H}_2\mathrm{O}_{(g)} + \mathrm{CO}_{(g)} \rightleftharpoons \mathrm{H}_{2(g)} + \mathrm{CO}_{2(g)}, \ \triangle H = -42 \ \text{kJ}$. What will the effect be on the amount of product produced?
1. Temperature is increased.
2. Pressure is increased.
3. $\mathrm{CO}_2$ decreases.
8. Predict the effect on the chemical equilibrium when each of the following changes is made to this reaction at equilibrium: $2 \ \mathrm{NaHCO}_{3(s)} \rightleftharpoons \mathrm{Na}_2\mathrm{CO}_{3(s)} + \mathrm{CO}_{2(g)} + \mathrm{H}_2\mathrm{O}(g), \ \triangle H = - 635.33 \ \mathrm{kJ}$. What will be the effect on the amount of product produced?
1. Temperature is decreased.
2. Pressure is decreased.
3. $[\mathrm{Na}_2\mathrm{CO}_3]$ decreases.
9. Define Le Châtelier’s principle.
10. When can Le Châtelier’s principle be applied?
11. Give an example of how Le Châtelier’s principle can be used to generate more products for a reaction at equilibrium.
12. Predict the effect on the chemical equilibrium when each of the following changes is made to this reaction at equilibrium: $2 \ \mathrm{SO}_{3(g)} + \mathrm{heat} \rightleftharpoons 2 \ \mathrm{SO}_{2(g)} + \mathrm{O}_{2(g)}$. What will the effect be on the amount of product produced?
1. Temperature is increased.
2. Pressure is increased.
3. $[\mathrm{O}_2]$ decreases.
13. Predict the effect on the chemical equilibrium when each of the following changes is made to this reaction at equilibrium: $2 \ \mathrm{N}_2\mathrm{O}_{4(g)} + \mathrm{heat} \rightleftharpoons 2 \ \mathrm{NO}_{2(g)}$. What will the effect be on the amount of product produced?
1. Temperature is decreased.
2. Pressure is decreased.
3. $[\mathrm{N}_2\mathrm{O}_4]$ decreases.

Feb 23, 2012

Mar 26, 2015