Le Chatelier’s Principle

The 4 most commons changes to make for equilibrium reactions are:

1. Concentration changes for reactants

2. Concentration changes for products

3. Temperature changes for reaction

4. Volume/Pressure changes for gaseous reactions

Ways to disturb equilibrium:

Concentration changes forReactants and Products

Think of equilibrium as a big seesaw.

To the eye, no changes are occurringto the amount of reactants on the left or to the amount of product on the right.

At equilibrium, the seesaw is balanced.

A + B C + D

When you increase the concentration ofa reactant A, you are adding weight to theleft side of the seesaw.

A + BC + D

How can you re-balance the seesaw? Howcan you achieve equilibrium again?

By shifting some of the weight toward the right!

A + B

C + DA + B C + D

Equilibrium has been re-established whenconcentrations stop changing.

Before A was added, the system was at equilibrium.

A + B C + D

At the moment that A was added, the [A] went up. In our example, increased [A] issymbolized as more weight on the left sideof the seesaw.

A + BC + D

In the process of re-establishing equilibrium,the concentration of C and D went up. Inour example, increased concentration issymbolized as more weight on the right sideof the seesaw.

A + B

C + DA + B C + D

Let’s look at the overall process one more time.A + B C + D

System wasat equilibrium.

A + BC + D

[A] increased.System not atequilibrium.

A + B C + DSystem regainsequilibrium.

Conc.

Time

InitialConcentrations

First Equilibrium

[A] up.

New Equilibrium

[C] & [D] go up

What would the seesaw look like if we increased the [D]?

A + BC + D

In which direction do we need to shift “weight” in order to regain equilibrium?

Shift to the LEFT!

A + BC + D

A + BC + D

[C] and [D] increase.

[A] and [B] increase.

Caused by increasing [A].

Shift RIGHT to regainequilibrium.

Result:

Caused by increasing [D].

Shift LEFT to regainequilibrium.

Result:

What if we removed D as it was formed?This would be the same as decreasing [D].What would the seesaw look like?

A + BC + D

When you decrease the [D], you are removing “weight” from that side of theseesaw.

How would you re-establish equilibrium?

A + BC + D

Shift “weight” to the right. More products will form.

This is a common way to make an equilibriumreaction go to “completion.”

Remember this all ties into Keq.

Recall if:

Keq>1 are favored

Keq<1 are favored

products

reactants

Change in Temperature

What happens when you change the temperature of a reaction?

It will depend on whether the reactionis exothermic or endothermic.

Exothermic A + B <--> C + D + heat reaction:

Endothermic A + B + heat <--> C + D reaction:

Treat “heat” like a reactant or product.

If you increase the heat, you are adding“weight” to the seesaw.

If you decrease the heat, you are removing“weight from the seesaw.

If the reaction is exothermic, the changein “weight” occurs on the product side.

If the reaction is endothermic, the changein “weight” occurs on the reactant side.

Exothermic Reaction with increased temperature.(want to produce LESS heat)

A + BC + D + Heat

Exothermic Reaction with decreased temperature.(want to produce MORE heat)

A + BC + D + heat

Shift Left!More Reactants

Shift Right!More Products

Endothermic Reaction with increased temperature.(need less heat)

A + B + heatC + D

Endothermic Reaction with decreased temperature.(need more heat)

A + B + heatC + D

Shift Right!More Products

Shift Left!More Reactants

Changing Volume/Pressure

When you have gaseous reactants or productsand you change volume, you are changing concentration:

PV = nRT

Pressure and volume are inversely related.

For gases, we use Kp and measure pressurechanges.

Let’s rearrange the ideal gas law toshow concentration:

P = nRT V

n/V is concentration.

Pressure is directly related to concentration and inversely related to volume.

Let’s look at a reaction with gaseouscomponents:

2A(g) + B(g) <--> 2C(g)

There are THREE moles of gas on the reactant side and TWO moles of gason the product side.

3 moles 2 moles

If we cut the volume in half, the pressurewill double. This means that the concentration of ALL gases went up.

The side of the reaction with the most moles of gas, will be most disturbed by theincreased concentration.

3 moles2 moles

If we shift the reaction toward the sidewith fewer moles of gas, the effect ofcutting the volume in half will be minimized.

3 moles2 moles

For this reaction, cutting the volume in halfresults in MORE product.

3 moles 2 moles

Is it always true that cutting the volume inhalf will cause more products to form?NO!

You have to examine each reaction withgaseous components to see, first, which sidehas more moles of gas.

Cutting the volume in half, increases pressure.Reaction will shift toward side with FEWERmoles of gas.

Doubling the volume, decreases pressure.Reaction will shift toward side with MOSTmoles of gas.

If volume is increased, which directionwill reaction shift?

If volume increases, pressure decreases.Shift to the side with the MOST moles of gas.

Reaction will shift to the right.

3A(s) + B(g) <--> 2C(g)

1 mole of gas 2 moles of gas

1 mole2 moles

Let’s see if you can put

it all together !!!

• How does a system at equilibrium respond to a stress?

• What factors are considered to be stresses on an equilibrium system?

Changes in concentration, pressure (volume), and temperature.

If possible, the equilibrium shifts in the direction that relieves the stress.

2 A(g) + B(g) <--> C(g) DH = + 200 kJ

1. In which direction will reaction shift if [A] is doubled?Increase [A], shift right. More products formed.

2. In which direction will reaction shift if [C] is removed?

Decrease [C], shift right. More products formed.

3. In which direction will reaction shift if temp goes up?

Endothermic reaction. Heat is a reactant. Shift right. More products formed.

4. In which direction will reaction shift if volume goes up?

Increase V, decrease P. Side with most gas molesloses “weight.” Shift left. More reactants formed.

First, note that the reaction is endothermic.

Second, note that the reactant side has more moles of gas.