slide1 n.
Download
Skip this Video
Loading SlideShow in 5 Seconds..
Le Chatelier’s Principle PowerPoint Presentation
Download Presentation
Le Chatelier’s Principle

Loading in 2 Seconds...

play fullscreen
1 / 32

Le Chatelier’s Principle - PowerPoint PPT Presentation


  • 149 Views
  • Uploaded on

Le Chatelier’s Principle. Ways to disturb equilibrium:. 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

loader
I am the owner, or an agent authorized to act on behalf of the owner, of the copyrighted work described.
capcha
Download Presentation

PowerPoint Slideshow about 'Le Chatelier’s Principle' - nike


An Image/Link below is provided (as is) to download presentation

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.


- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -
Presentation Transcript
slide2

Ways to disturb equilibrium:

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

slide3

Concentration changes for

  • Reactants and Products
slide4

At equilibrium, the seesaw is balanced.

A + B

C + D

Think of equilibrium as a big seesaw.

To the eye, no changes are occurring

to the amount of reactants on the left

or to the amount of product on the right.

slide5

When you increase the concentration of

a reactant A, you are adding weight to the

left side of the seesaw.

C + D

A + B

How can you re-balance the seesaw? How

can you achieve equilibrium again?

slide6

C + D

A + B

By shifting some of the weight toward the right!

A + B

C + D

Equilibrium has been re-established when

concentrations stop changing.

slide7

Before A was added, the system was

at equilibrium.

A + B

C + D

slide8

C + D

A + B

At the moment that A was added, the [A]

went up. In our example, increased [A] is

symbolized as more weight on the left side

of the seesaw.

slide9

C + D

A + B

C + D

A + B

In the process of re-establishing equilibrium,

the concentration of C and D went up. In

our example, increased concentration is

symbolized as more weight on the right side

of the seesaw.

slide10

System was

at equilibrium.

C + D

[A] increased.

System not at

equilibrium.

A + B

A + B

A + B

C + D

C + D

System regains

equilibrium.

Let’s look at the overall process one more time.

slide11

New Equilibrium

Initial

Concentrations

First

Equilibrium

[C] & [D] go up

[A] up.

Conc.

Time

slide12

A + B

C + D

What would the seesaw look like if we

increased the [D]?

In which direction do we need to shift

“weight” in order to regain equilibrium?

Shift to the LEFT!

slide13

Caused by increasing [A].

Shift RIGHT to regain

equilibrium.

A + B

C + D

C + D

Result:

A + B

Caused by increasing [D].

Shift LEFT to regain

equilibrium.

Result:

[C] and [D] increase.

[A] and [B] increase.

slide14

C + D

A + B

What if we removed D as it was formed?

This would be the same as decreasing [D].

What would the seesaw look like?

When you decrease the [D], you are

removing “weight” from that side of the

seesaw.

slide15

C + D

A + B

How would you re-establish equilibrium?

Shift “weight” to the right. More products

will form.

This is a common way to make an equilibrium

reaction go to “completion.”

slide16

Remember this all ties into Keq.

Recall if:

Keq>1 are favored

Keq<1 are favored

products

reactants

slide18

What happens when you change the

temperature of a reaction?

It will depend on whether the reaction

is exothermic or endothermic.

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

reaction:

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

reaction:

slide19

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 change

in “weight” occurs on the product side.

If the reaction is endothermic, the change

in “weight” occurs on the reactant side.

slide20

A + B

C + D + heat

C + D + Heat

A + B

Exothermic Reaction with decreased temperature.

(want to produce MORE heat)

Exothermic Reaction with increased temperature.

(want to produce LESS heat)

Shift Left!

More Reactants

Shift Right!

More Products

slide21

C + D

A + B + heat

Endothermic Reaction with decreased temperature.

(need more heat)

A + B + heat

C + D

Endothermic Reaction with increased temperature.

(need less heat)

Shift Right!

More Products

Shift Left!

More Reactants

slide23

When you have gaseous reactants or products

and you change volume, you are changing

concentration:

PV = nRT

Pressure and volume are inversely related.

For gases, we use Kp and measure pressure

changes.

slide24

Let’s rearrange the ideal gas law to

show concentration:

P = n

RT V

n/V is concentration.

Pressure is directly related to

concentration and inversely related

to volume.

slide25

3 moles

2 moles

Let’s look at a reaction with gaseous

components:

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

There are THREE moles of gas on the

reactant side and TWO moles of gas

on the product side.

slide26

2 moles

3 moles

If we cut the volume in half, the pressure

will 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 the

increased concentration.

slide27

2 moles

3 moles

3 moles

2 moles

If we shift the reaction toward the side

with fewer moles of gas, the effect of

cutting the volume in half will be minimized.

For this reaction, cutting the volume in half

results in MORE product.

slide28

Cutting the volume in half, increases pressure.

Reaction will shift toward side with FEWER

moles of gas.

Doubling the volume, decreases pressure.

Reaction will shift toward side with MOST

moles of gas.

Is it always true that cutting the volume in

half will cause more products to form?

NO!

You have to examine each reaction with

gaseous components to see, first, which side

has more moles of gas.

slide29

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

1 mole of gas 2 moles of gas

2 moles

1 mole

If volume is increased, which direction

will reaction shift?

If volume increases, pressure decreases.

Shift to the side with the MOST moles of gas.

Reaction will shift to the right.

slide30

Let’s see if you can put

it all together !!!

slide31

How does a system at equilibrium respond to a stress?

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

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

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

slide32

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

1. In which direction will reaction shift if [A] is doubled?

First, note that the reaction is endothermic.

Second, note that the reactant side has more

moles of gas.

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 moles

loses “weight.” Shift left. More reactants formed.