Chapter 7

(1)

Chapter 7

(2)

What is Equilibrium



A chemical reaction process where the forward and reverse reactions are

occurring at the same rate.



Shown using a double arrow



Example:

(3)

Types of Equilibrium

 Solubility

 Phase

 Chemical reaction 3H3H₂₂(g) + N(g) + N₂₂(g) (g) 2NH2NH₃₃(g)(g)

(4)

Note:

Equilibrium can have a position- how much

product and reactant there are at equilibrium.

Shown with the double arrow

Reactants are favored Products are favored

Catalysts speed up both the forward and reverse reactions so don’t affect equilibrium position.

(5)

How is Equilibrium chemistry different from other reactions you are familiar with?

In an ordinary reaction; all reactants end up as products; there is 100%

conversion

(6)

In an ordinary reaction; all reactants end up as products; there is 100% conversion

CONCENTRATION CHANGE IN A REACTION CONCENTRATION CHANGE IN A REACTION

As the rate of reaction is dependant on the concentration of reactants... the forward reaction starts off fast but slows as the reactants get less concentrated

FASTEST AT THE START

SLOWS DOWN AS REACTANTS

ARE USED UP

TOTAL CONVERSION TO PRODUCTS THE STEEPER THE

GRADIENT, THE FASTER THE REACTION

(7)

In an ordinary reaction; all reactants end up as products; there is 100% conversion

CONCENTRATION CHANGE IN A REACTION CONCENTRATION CHANGE IN A REACTION

As the rate of reaction is dependant on the concentration of reactants... the forward reaction starts off fast but slows as the reactants get less concentrated

FASTEST AT THE START

SLOWS DOWN AS REACTANTS

ARE USED UP

TOTAL CONVERSION TO PRODUCTS THE STEEPER THE

GRADIENT, THE FASTER THE REACTION

(8)

Initially, there is no backward reaction but, as products form, it speeds up and

provided the temperature remains constant there will come a time when the backward and forward reactions are equal and opposite; the reaction has reached equilibrium.

EQUILIBRIUM REACTIONS EQUILIBRIUM REACTIONS

In an equilibrium reaction, not all the

reactants end up as products; there is not a 100% conversion.

BUT IT DOESN’T MEAN THE REACTION IS STUCK IN THE MIDDLE

FASTEST AT THE START NO BACKWARD REACTION

FORWARD REACTION SLOWS DOWN AS REACTANTS ARE USED UP

BACKWARD REACTION STARTS TO INCREASE

AT EQUILIBRIUM THE BACKWARD AND FORWARD REACTIONS ARE

EQUAL AND OPPOSITE

(9)

(10)

(11)

Note:

a reversible chemical reaction is a dynamic process

• everything may appear stationary but the reactions are moving both ways

• the position of equilibrium can be varied by changing certain conditions

Trying to get up a “down” escalator gives an

excellent idea of a non-chemical situation

involving dynamic equilibrium.

(12)

Summary When a chemical equilibrium is established ...

• both the reactants and the products are present at all times

• the equilibrium can be approached from either side

• the reaction is dynamic - it is moving forwards and backwards

• the concentrations of reactants and products remain constant

(13)

Calculating Concentrations at Equilibrium

(14)

for an equilibrium reaction of the form...

aA + bB cC + dD

then (at constant temperature) [C]^c . [D]^d = a constant, (K_c) [A]^a . [B]^b

where [ ] denotes the equilibrium concentration in mol/ L

K_c is known as the Equilibrium Constant

Note: do not include solids or liquid reactants or products in the calculation.

THE EQUILIBRIUM CONSTANT K THE EQUILIBRIUM CONSTANT K_c_c

VALUE OF K_c

AFFECTED by a change of temperature

NOT AFFECTED by a change in concentration of reactants or products a change of pressure

adding a catalyst

(15)

Writing Equilibrium Expressions



Write the equilibrium expressions for the following reactions.



3H

₂

(g) + N

₂

(g) 2NH

₃

(g)



2H

₂

O(g) 2H

₂

(g) + O

₂

(g)

(16)

K K

_eq_eq

= = [NH [NH

₃₃

] ]

²²

[H [H

₂₂

] ]

³³

[N [N

₂₂

] ]

3H3H₂₂(g) + N(g) + N₂₂(g) (g) 2NH2NH₃₃(g)(g) 2H2H₂₂O(g)O(g) 2H 2H₂₂(g) + O(g) + O₂₂(g)(g)

K K

_eq_eq

= = [H [H

₂₂

] ]

²²

[O [O

₂₂

] ]

[H [H

₂₂

O ] O ]

²²

(17)

Calculating Equilibrium

Calculate the equilibrium constant for the following reaction.

3H

₂

(g) + N

₂

(g) 2NH

₃

(g)

At equilibrium at 25ºC there 0.15 mol of N

₂

,

0.25 mol of NH

₃

, and 0.10 mol of H

₂

in a 2.0

L container.

(18)

KK_eq_eq = = [NH[NH₃₃]]²² [H[H₂₂ ] ]³³ [N [N₂₂ ] ]

 [N[N₂₂] = 0.15 mol/2.0L = 0.075mol/L] = 0.15 mol/2.0L = 0.075mol/L

 [NH[NH₃₃] = 0.25 mol/L = 0.125mol/L] = 0.25 mol/L = 0.125mol/L

 [H[H₂₂] = 0.10 mol/ 2.0 = 0.05mol/L] = 0.10 mol/ 2.0 = 0.05mol/L

K K

_eq_eq

= = ( (

0.125mol/L0.125mol/L

) )

²²

( (

0.05mol/L )^0.05mol/L ³³

( (

0.075mol/L)0.075mol/L)

= 1.67 x10= 1.67 x10³³ (mol/L) (mol/L)^-2^-2 =1.7x10=1.7x10³³ (mol/L) (mol/L)^-2^-2

(19)

Writing Equilibrium Expressions for heterogeneous equilibria

Write the equilibrium expressions for the following reaction

2H

₂

O(l) 2H

₂

(g) + O

₂

(g)

If the reaction involves pure solids or pure liquids the concentration of the solid or the liquid doesn’t change.

As long as they are not used up they are not used up we can leave them out of the

equilibrium expression. K K

_eq_eq

= = [H [H

₂₂

] ]

²²

[O [O

₂₂

] ]

(20)

The magnitude of K

If K > 1 Products are favored

K=1

If K < 1 Reactants are favored

(21)

LeChatelier’s Principle LeChatelier’s Principle



When a system at equilibrium is When a system at equilibrium is

disturbed by a change in a property, the disturbed by a change in a property, the

system adjusts in a way to oppose the system adjusts in a way to oppose the

change

change.

.

(22)

Le Chatelier Translated:



When you take something away from a When you take something away from a system at equilibrium, the system shifts system at equilibrium, the system shifts

in such a way as to replace some of what in such a way as to replace some of what

you’ve taken away.



When you add something to a system at When you add something to a system at equilibrium, the system shifts in such a equilibrium, the system shifts in such a

way as to use up some of what you’ve way as to use up some of what you’ve

added.

(23)

Changing Concentration

 If you add reactants (or increase their concentration).

 The forward reaction will speed up.

 More product will form.

 Equilibrium “Shifts to the right”

 Reactants  products

(24)

Changing Concentration

 If you add products (or increase their concentration).

 The reverse reaction will speed up.

 More reactant will form.

 Equilibrium “Shifts to the left”

 Reactants  products

(25)

Le Chatelier Example Le Chatelier Example

A closed container of N

₂

O

₄

and NO

₂

at

equilibrium. NO

₂

is added to the container.

N

₂

O

₄

(g) + Energy  2 NO

₂

(g)

The equilibrium of the system shifts to

the _______ to use up the added NO left

₂

.

(26)

Effect of Concentration:

1.

If you add more reactant, it shifts to the right increasing the formation of product, using up the reactants.

2.

If you add product, it shifts to the left

3.

If you remove product, it shifts to the right, increasing the formation of product.

4.

If you remove reactant, it shifts to the left

(27)

LeChâtelier’s Principle

If something is changed in a system at

equilibrium, the system will respond to relieve the stress.

Example:

(28)

Example 2

(29)

Changing Temperature



Reactions either require or release heat.



Endothermic reactions go faster at higher temperature.



Exothermic go faster at lower temperatures.



All reversible reactions will be exothermic

one way and endothermic the other.

(30)

Effect of temperature:



Remember: Energy is treated as a reactant if endothermic equation and as a product if

exothermic equation.



If heating a system, it shifts so extra heat is used up.



If cooling a system, then it shifts so more heat

is produced.

(31)

Example Endothermic Reaction:

 Heating this reaction causes the system to

shift to the right ( more products) because you treat energy like a reactant (this reaction

needs energy to go forward)

2NaCl +H₂SO₄ + energy < -- > 2HCl + Na₂SO₄

 Cooling this reaction causes the system to shift to the left (less reactants) so need to make up more

(32)

Changing Temperature



As you raise the temperature the reaction proceeds in the endothermic direction.



As you lower the temperature the reaction

proceeds in the exothermic direction.

(33)

Example 3

(34)

Changes in Pressure



As the pressure increases the reaction will shift in the direction of the least gases.

 Count the # of moles of gas for each side and compare.



At high pressure

2H

₂

(g) + O

₂

(g)  2 H

₂

O(g)

 3 moles vs 2moles therefore shift towards least #



At low pressure

2H

₂

(g) + O

₂

(g)  2 H

₂

O(g)

 therefore shift towards greater #

(35)

What does not effect the Equilibrium concentrations ?

Adding an inert gas, partial pressures of reactants and product are not changed.

No effect on equilibrium position

Also adding a catalyst will not affect equilibrium concentrations

No effect on equilibrium position

(36)

Determine the effect on the equilibrium in the following:

Consider:

2SO

₂

(g) + O

₂

2SO

₃

(g) H <0

a) Adding SO

₂

b) temp. is increased

c) Volume of container is increased

d) Catalyst is added

e) An inert gas is added

(37)

a) Shifts to the right

b) Shifts to the left

c) Pressure decreases for all the gases;

affects left side more- shifts to the left

d) No effect; equilibrium is reached faster.

e) Total pressure is increased; partial pressures of the individual gases is unaffected.

(38)

Determine the effect on the equilibrium in the following:

Consider:

H

₂

O (g) + heat 2H

₂

(g) + O

₂

(g)

a) Adding H

₂

b) temp. is increased

c) Volume of container is decreased

d) Catalyst is added

e) An inert gas is added

(39)