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Experiment 6

Le Chatelier’s Principle

Introduction

Many chemical reactions do not go to completion. That is, the reactants do not fully transform into products. The

reason for this is that many reactions can proceed both from reactants to products (forward reaction) and from products

to reactants (reverse reaction):

A → 2 B

and

2 B → A

When this is the case, rather than the reaction going to completion, an equilibrium is established where the rates of the

forward and reverse reactions are equal. While the reaction has not stopped or gone to completion, the concentrations

of the chemical species will no longer be changing – as fast as reactant is converted to product an equal amount of

product is converted back to reactant. Rather than saying the reaction is finished, we say that it has reached a state of

equilibrium. Reactions that can establish equilibrium are written with a double arrow:

A ⇌ 2 B

It is established that for systems at equilibrium, the ratio of products to reactants raised to their coefficients from the

chemical equation will be a constant at a specific temperature. This is called the law of mass action. For the general

reaction

aA + bB ⇌ cC + dD

Using the law of mass action, we can establish the equilibrium expression

Kc = concentration equilibrium constant

The subscript (eq) denotes that the values in this equation can only be equilibrium values.

Kc is a constant for a chemical reaction at a given temperature. Initial concentrations or pressures, adding or removing

reactants or product, or adding a catalyst will have no effect on the value of K. At a given temperature, the

concentrations will always adjust themselves (stoichiometrically of course) to establish the appropriate value of K. This

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may take time if a reaction proceeds slowly, but given enough time, equilibrium will be established and the ratio of

reactants to products raised to their coefficients will be the K value.

LeChatelier’s Principle:

Since K is a constant at a given temperature, any change to the reaction system such as concentrations changes or

temperature changes will result in the reaction adjusting to offset the change and reestablish the equilibrium. This is

known as Le Chatelier’s Principle. Another way Le Chatelier’s principle is often stated is: When a stress is applied to an

equilibrium system causing the system to no longer be at equilibrium, the reaction will proceed in the direction to

minimize this stress and reestablish a new equilibrium. Common “stresses” are adding or removing a species in the

reaction, changing volume, or changing temperature. When one of these stresses is applied to a system at equilibrium,

it will cause the values in the equilibrium expression to no longer be equal to the K at that given temperature. The

reaction will then proceed in the appropriate direction until equilibrium has been reestablished.

Stresses that affect Equilibrium

Recall that the equilibrium constant is only affected by temperature. Aside from temperature, any stress that is applied

to a system will affect the concentration of the system.

Adding Reactants

Consider the general reaction and equilibrium conditions:

A (aq) + 2 B (aq) ⇌ C (aq)

At equilibrium, let’s make it simple and say that all concentrations are equal to 1M:

Substance Equilibrium / M

A 1

B 1

C 1

Therefore,

If 1 M equivalent A is added to the equilibrium mixture, then

Substance After more A is added / M

A 1 + 1

B 1

C 1

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Since the reaction is not at equilibrium, we cannot use the equilibrium expression (K = …). Instead we use the same

expression replacing K with Q. Q stands for reaction quotient and is the ratio of product to reactant (raised to their

coefficients) under non-equilibrium conditions.

When Q is not equal to K, the reaction is not at equilibrium and will proceed in the appropriate direction to establish

equilibrium. If Q is less than K, the reaction will proceed in the forward direction converting the reactants to products to

establish equilibrium. When Q is greater than K, it will proceed in the reverse direction converting products into

reactants to establish equilibrium.

In this circumstance, since the temperature was not changed, K is still equal to 1. Q is less than K, so the reaction will

proceed to make more product. Adding reactant will cause greater amounts of product to be produced. In terms of Le

Chatelier’s principle, adding one of the reactants was the stress applied to the system. To reduce this stress (i.e. reduce

the amount of the added reactant), the reaction will “remove” some of the reactant by converting it to product.

Removing Product

Going back to our original equilibrium mixture what if we removed some of the product.

After 0.5 M C is removed

Substance After C is removed / M

A 1

B 1

C 1 – 0.5

Once again, we see that Q is less than K causing the reaction to proceed to make more product. The stress caused by

removing C is reduced, by producing some C to take its place. The same principles apply if one or more products are

added or one or more reactants are removed. In these cases Q will be greater than K and the reaction will proceed

toward the reactants side to reestablish equilibrium.

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Volume of Solution

When adding or removing solvent to a reaction system, the concentrations of all species present will change. If we take

our original sample in aqueous solution and double the volume by adding more water, the concentrations will change as

follows:

Concentrations after adding additional water

Substance Concentration after dilution / M

A 0.5

B 0.5

C 0.5

In this case, Q is greater than K = 1 meaning the reaction will proceed in the reverse direction. The Le Chatelier’s

explanation here is that if you dilute the reaction media, then it will want to do the opposite – get more concentrated.

To get to a more concentrated media, the reaction will proceed toward the reactants in this case, because there are 3

moles of species in the reactants and only 1 mole in the products (according to the balanced chemical equation). When

diluting a reaction mixture, the reaction will proceed in the direction of greater numbers of dissolved species.

Volume of Gas

Changing the volume of a gas is similar to changing the volume with solutions. Let’s take the same reaction but assume

we had gases.

A (g) + 2 B (g) ⇌ C (g)

Kp is set-up the same as Kc; however, pressures are used instead of concentrations.

Staying simple we will make K = 1 and all equilibrium pressures 1 atm. If we compress the system to one half its original

volume, then the pressure of each component will double. In which direction will the above reaction proceed if the

temperature is held constant?

Halving the volume will double the pressures:

Substance Pressure after volume change / atm

A 2

B 2

C 2

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Q is less the K=1, so the reaction proceeds toward product. In this case, decreased volume increased the pressures (or

concentrations) of all species. We can see mathematically that this will cause the reaction to proceed toward the side of

the reaction with less number of moles of gas – in this case the products as there is only 1 mole of C for every 3 moles of

reactant (1 A and 2 B). With respect to LeChalelier’s principle, with decreased volume, pressures go up, so the reaction

will shift to reduce the pressure. Fewer moles of gas will have lower pressure, so the reaction shifts to the side with

fewer moles of gas – in this case the product.

Catalyst

While a catalys increases the rate at which a reaction will reach equilibrium, it will have no effect on the concentrations

of reactant and product once equilibrium is established. A catalyst lowers the activation energy subsequently increasing

the rate of the reaction for both the forward and reverse directions. So, a catalyst has no effect on equilibrium.

Temperature

Temperature affects equilibrium differently than all of the previous stresses that have been considered. Changing the

temperature does not change the concentrations of the substances; rather, it changes the K for the reaction.

A + 2 B ⇌ C

Recall that K = 1 for this reaction at 25 oC. Additionally, it is important to know that the reaction is exothermic. If the

temperature is increased, in what direction will the reaction proceed if it is at equilibrium at 25 oC? Knowing it is

exothermic, the reactants are at a higher energy than the products. Alternatively, you can consider that as the reaction

proceeds it releases energy as a product:

A + 2 B ⇌ C + energy

So, increasing the temperature puts more energy into the system (a product), so to reduce the energy, the reaction

proceeds toward the reactants.

At equilibrium at 25 oC

Substance Equilibrium / M

A 1

B 1

C 1

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If the temperature is increased to 35 oC, some of the product will be converted to reactant and equilibrium will establish

at the following new concentrations.

Substance Equilibrium / M

A 1.2

B 1.4

C 0.8

In terms of Le Chatelier’s principle, the added energy of the higher temperature will be reduced if it can be converted to

potential energy in the system. In this exothermic process, converting products to reactants requires an input of energy,

so it will proceed in this direction absorbing energy to reduce the added kinetic energy that has been added.

Another way of looking at it is that at higher temperatures, a reaction will favor the higher energy side. At lower

temperatures, the lower energy side is favored.

Surface Area or Amount of a Solid reactant or Product

One interesting aspect of equilibrium is that neither the amount nor surface area of a solid substance in a chemical

reaction will influence the equilibrium. Consider the dissolving of calcium fluoride in water:

CaF2 (s) ⇌ Ca2+ (aq) + 2F- (aq)

Kc = [Ca2+][F-]2

While it might seem unusual here to leave out the solid calcium fluoride reactant, this makes sense if you have ever

considered the dissolving of table salt, NaCl, in water. When you first place some salt in water, it dissolves. If you keep

placing more salt in, the solution becomes saturated and the excess salt sits on the bottom of the container. If more salt

is added, it will not dissolve; rather, it will sit on the bottom with the rest of the excess salt. The understanding of this is

similar to understanding why surface area does not affect vapor pressure. In this case, the solid is in equilibrium with

the dissolved ions. While putting more solid in or increasing the solid’s surface area will provide more area for the ions

to leave the solid and enter the solution, it provides an equal amount of area for the ions in the solution to recrystallize

in the solid phase.

This will apply to all solids in equilibrium systems – they are not included in the equilibrium expression.

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Experiment

In the experiments to be performed, Le Chatelier’s principle will be investigated by applying different stresses to

reactions at equilibrium. The first reaction studied is the equilibrium of a saturated solution with a solid, in this case

sodium chloride. To a saturated solution, chloride ion will be added to observe the affects it has on the equilibrium.

NaCl (s) ⇌ Na+ (aq) + Cl- (aq)

The second and third experiments will involve the equilibrium of a cobalt chloride complex ion with a cobalt water

complex ion:

[CoCl4]2- (aq) + 6H2O (l) ⇌ [Co(H2O)6]

2+ (aq) + 4 Cl- (aq)

For this reaction, both temperature and concentration of species will be examined. The unique aspect of this

equilibrium is that the [CoCl4]2- ion is blue in solution, while the [Co(H2O)]2+ ion is pink in solution. Any change in the

equilibrium can be observed by a change in the color of the solution.

The final reaction studied is the dimerization of nitrogen dioxide.

2NO2 (g) ⇌ N2O4 (g)

As this is a gas phase reaction, the effects of volume and pressure will be investigated. Changes in equilibrium can be

observed since the NO2 is a brown gas and N2O4 is colorless.

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Procedure

Concentration with Saturated solution:

1. Obtain approximately 4 mL of a saturated NaCl solution.

2. Add 5 drops of 12M HCl solution. (caution: 12M HCl is not your friend!)

3. Record Observations

Temperature:

1. Place approximately 3 mL of a 0.1M CoCl2 solution in methanol into a test tube.

2. Place the test tube in a hot water bath (hot but not boiling) in the fume hood and record your observations.

3. Remove the test tube from the hot water bath and allow to cool to near room temperature.

4. Place the test tube in an ice bath and record your observations.

5. Allow the test tube and solution to return to room temperature for the next part.

Concentration:

1. To your CoCl2 solution add the provided 12M HCl solution dropwise until a change is observed. Mix well

between drops.

2. To this solution add water dropwise until an observation is observed. Mix well between drops.

3. Dispose of the cobalt solution in the beaker in the hood. This will be placed in the heavy metal waste container

by the TA.

Pressure and Volume:

1. In the fume hood there are syringes with 30 mL of NO2 in them. Keep them in the hood for the experiment as

NO2 is a toxic gas.

2. Before you begin take note of the color of the gas in the syringe. You will be compressing it to one half its

original volume. If no reaction occurs, then the color should be twice as dark. If more NO2 is formed when it is

compressed, it will be more than twice as dark, while if some of the NO2 is converted to N2O4 then it will not be

twice as dark. If you look quickly, you will see that the gas initially gets dark due to the compression, then turns

lighter or darker due to the shift in equilibrium.

3. Quickly push on the plunger to compress the gas to 15mL– make observations.

4. Return the plunger to the 30mL mark.

5. Place the syringe in the ice water bath keeping the volume at 30 mL – make observations.

6. Leave the syringe for the next group.

Last Name:___________________________________ First Name:_____________________________________

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Data

Last Name:___________________________________ First Name:_____________________________________

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Questions

For these questions you are asked to give explanations. The explanations should be intelligent, thoughtful explanations

that demonstrate your understanding of the principles involved in this experiment. One thing to be careful about is that

when explaining a phenomenon, you should not just say that Le Chatelier’s principle says that it would happen. For

example, if you add a reactant to an equilibrium mixture, it will convert it to more product. While Le Chatelier’s

principle says that if you add the reactant, it will want to reduce it, this is not the reason it converts to product. Just

because Le Chatelier said it would is not an explanation. (If I say the sun will rise tomorrow, does it rise because I said it

would or is there a better explanation?) You need to discuss the effects that the added reactant has on the

concentrations and whether or not the concentrations will give the appropriate value of the equilibrium constant. If

they do not, what direction will the reaction proceed so that they will result in the appropriate K value.

1. Upon adding the HCl to the saturated sodium chloride solution, how did it affect the equilibrium? Explain why

the added HCl shifted the equilibrium.

NaCl(s) ⇌ Na+ (aq) + Cl- (aq)

2. In which direction did the following equilibrium shift when it was placed in the hot water bath?

[CoCl4]2- (aq) + 6H2O (l) ⇌ [Co(H2O)6]

2+ (aq) + 4 Cl- (aq)

3. Is the above reaction (from question 2) endothermic or exothermic? Give an explanation based on your results.

Last Name:___________________________________ First Name:_____________________________________

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4. Upon addition of the water, explain the reasoning for the shift in equilibrium. (Remember that liquids do not

appear in the equilibrium expression, so you cannot say that it is increasing reactants making Q<K. Water does

not directly appear in Q or K .

5. When the NO2 was compressed to 15mL was the color more or less than twice as dark as the color at 30mL?

6. Based on your answer to question 5, in which direction did the reaction shift with the reduced volume? Explain

why this occurred.

2NO2 (g) ⇌ N2O4 (g)

7. Is the above reaction (from question 6) endothermic or exothermic? Give an explanation based on your results.

Last Name:___________________________________ First Name:_____________________________________

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8. A catalyst could be added to speed up the dimerization of NO2. If a catalyst is added to the syringe with 30 mL

of the brown gas (which is an equilibrium mixture of NO2 and N2O4) , what change would be observed and why?