AP Chemistry

Ms. Grobsky CHEMICAL EQUILIBRIUM

The word equilibrium is commonly used in science

“Forces are in equilibrium”

Forces are equal

“Objects have reached thermal equilibrium”

Temperatures are equal

Chemical equilibrium occurs when a forward

reaction and its reverse reaction proceed at the

same rate in a closed system

To some extent, all reactions reversible (↔)

Some reactions have very little reversibility

Referred to as “product favored”

Other reactions have lots of reversibility

Referred to as “reactant favored”

WHAT IS EQUILIBRIUM?

Different types of arrows are used in chemical equations associated with equilibrium:

Single arrow

Assumes that the reaction proceeds to completion as written

Two single-headed arrows

Used to indicate a system in equilibrium

Two single-headed arrows of different sizes

May be used to indicate when one side of an equilibrium system is favored

DEPICTING CHEMICAL EQUILIBRIUM

For the general reaction:

A + B C

You can view the reaction as occurring in

three steps:

Initial mixing

Kinetic region

Equilibrium region

BASIC STEPS TO REACHING CHEMICAL

EQUILIBRIUM

When A and B are first brought together, there is no C present

The reaction proceeds as:

A + B C

This equation only represents the very start of the reaction

Things change as soon as some C is produced!

INITIAL MIXING STEP

As soon as some C has been produced, the reverse reaction is possible

A + B C

Overall, we still see an increase in the net concentration of C

As we approach equilibrium, the rate of the forward reaction becomes slower

KINETIC REGION

A point is finally reached where the

forward and reverse reactions occur at

the same rate

A + B C

There is no net change in the

concentration of any of the species

EQUILIBRIUM REGION

GRAPH FOR EXAMPLE C

on

ce

ntr

ati

on

Time

C

B

A

Equilibrium

Region

Kinetic

Region

ILLUSTRATING CHEMICAL EQUILIBRIUM C

on

ce

ntr

ati

on

Time

Kinetic Equilibrium

Region Region

Products

Reactants

• An equilibrium exists when no further change in

concentration occurs

• Note that the concentrations of products and

reactants do not have to be equal!

• The equilibrium

concentrations of the

reactants and

products are the

same regardless of

whether or not you

start with only the

reactants or only the

products!

No products

No reactants

ACHIEVING CHEMICAL EQUILIBRIUM

3H2(g) + N2(g) ↔ 2NH3

3H2(g) + N2(g) ↔ 2NH3

Equilibrium should not be viewed as a

static condition

While concentrations do not change, products

and reactants continue to interconvert at equal

rates

Rates are NOT zero even though you see no

visible changes

Therefore, chemical equilibrium is

dynamic

HOW TO VIEW CHEMICAL EQUILIBRIUM

A SUMMARY OF CHEMICAL EQUILIBRIUM

The forward and reverse reaction rates are equal

Macroscopically, the system (reactants and products)

appears to not be doing anything

Microscopically, the system (reactants and products)

is dynamic

The net change in concentrations of reactants and

products will remain unchanged

The equilibrium concentrations of reacts and products will not

usually be equal

HOW TO DEPICT THE

LEVEL OF COMPLETION

A CHEMICAL REACTION

REACHES

INTRODUCING THE EQUILIBRIUM EXPRESSION

AND THE EQUILIBRIUM CONSTANT

THE LAW OF MASS ACTION

Norwegian chemists Guldberg and Waage

proposed in 1864 that the rate of a chemical

reaction is directly proportional to the products of

the reactants

Called the Law of Mass Action

The Law of Mass Action is represented by an

equilibrium expression

To generalize this expression, consider the reaction :

The equilibrium expression for this reaction would be :

Keq is called the equilibrium constant

Defined as the ratio of the concentrations of the products compared to the concentrations of the reactants

It is a number with NO UNITS!

THE EQUILIBRIUM EXPRESSION

Keq = [C]c[D]d

[A]a[B]b

aA + bB cC + dD

SOME NOTES ON THE EQUILIBRIUM

EXPRESSION

Products always goes in numerator of

expression whereas reactants go in

denominator

Values are raised to the coefficient in the chemical

equation

ONLY aqueous solutions or gases go in the

equilibrium expression

The concentrations of solids and liquids are

essentially constant

Therefore, NO pure substances - solids or liquids –

appear in equilibrium expression

If K << 1 (K< 0.10), the

reaction doesn’t go very

far to completion

Reactant-favored

At equilibrium, [products] <<

[reactants]

If K = 1, there are

substantial amounts

of both product and

reactant at

equilibrium

MAGNITUDE OF K

WHAT DOES IT TELL ME ABOUT A CHEMICAL

REACTION?

MAGNITUDE OF K

WHAT DOES IT TELL ME ABOUT A CHEMICAL

REACTION?

If K >> 1 (K > 10), the

reaction goes to

completion

Product-favored

At equilibrium,

[products] >>

[reactants]

Equilibrium constant for a reaction is the

same REGARDLESS of initial reactant

concentrations and equilibrium concentrations

RATIO IS CONSTANT!

ONE MORE THING…

3H2(g)+N2(g) 2NH3(g) The equilibrium constant, Keq, for the

reaction above is determined by:

Keq = [ NH3 ] 2

[ H2 ] 3 [ N2 ]

At 472C, Keq = 0.105

EXAMPLE

MANIPULATING THE

EQUILIBRIUM CONSTANT

2NH3(g) 3H2(g)+N2(g) • Reversing a reaction will result in the new

equilibrium constant, :

Keq new =1 / Keq old

Keq new = [ H2 ] 3 [ N2 ]

[ NH3 ] 2

= (1 / 0.105) or 9.52

WHAT IF I REVERSE THE REACTION?

GENERAL SUMMARY

If the equation is reversed, the equilibrium

constant is inverted

NH3(g) 1.5H2(g)+0.5N2(g) • Changing the number of moles of reactants and

products will exponentially change the equilibrium

constant:

Keq new= (Keq old)^n

Keq new = [ H2 ]1.5 [ N2 ]

0.5

[ NH3 ]

= (9.52)^0.5 or 3.09

WHAT IF I CHANGE THE # OF MOLES IN

THE REACTION?

GENERAL SUMMARY

If the equation is multiplied by a factor,

the equilibrium constant is raised to the

same factor

WHAT IF THE REACTION IS A MULTI -

STEP PROCESS?

When chemical

equations are added,

their equilibrium

constants are

multiplied together to

get the overall

equilibrium constant

SUMMARY OF WHAT TO DO WHEN

MANIPULATING EQUILIBRIUM CONSTANT

THE DIFFERENT TYPES

OF EQUILIBRIUM

CONSTANTS

Different reactions have different symbols for

K

Kc The most used, general equilibria constant with molar concentrations [M]

in the expression

THE DIFFERENT TYPES OF K

Kc = [C]c[D]d

[A]a[B]b

For equilibria that involves gases, partial pressures

can be used instead of concentrations!

aA (g) + bB (g) eE (g) + fF (g)

Kp =

Kp is used when the partial pressures are expressed

in units of atmospheres (atm)

EQUILIBRIA INVOLVING GASES

pEe pF

f

pAa pB

b

TWO DIFFERENT K’S

For the reaction:

2SO3(g) 2SO2(g) + O2(g)

We can write two equilibrium expressions!

In general, Kp ≠ Kc

However, partial pressures are proportional to concentration at a constant temperature via:

PV = nRT

Where:

R is the gas law constant:

0.0821 atm·L/mol·K

T is the temperature, K

RELATIONSHIP BETWEEN

CONCENTRATION AND PRESSURE

CONVERTING BETWEEN KC AND KP

CONVERTING BETWEEN KC AND KP

The Δn is the change in the number of

moles of gasous products minus the

number of moles of gaseous reactants

For the following equilibrium, Kc = 1.10 x 107 at 700. oC.

What is the Kp?

PRACTICE!

atm L

mol K

2H2 (g) + S2 (g) 2H2S (g)

Kp = Kc (RT)Dng

T = 700 + 273 = 973 K

R = 0.0821

Dng = ( 2 ) - ( 2 + 1) = -1

PRACTICE!

Kp = Kc (RT)Dng

= 1.10 x 107 (0.0821 ) (973 K)

= 1.378 x105

atm L

mol K [ ] -1

Equilibria that involve more than one phase is called

heterogeneous equilibria

Example:

CaCO3 (s) ↔ CaO (s) + CO2 (g)

Equilibrium expressions for these types of systems do not

include the concentrations of the pure solids (or liquids)

because their concentrations do not vary!

Kc = [CO2]

Kp = PCO2

WHAT DO YOU DO IF THE REACTION HAS

MORE THAN ONE PHASE?

CALCULATING

EQUILIBRIUM

CONSTANTS

Chemical reactions tend to go to equilibrium

provided that the reaction takes place at a

significant rate

There is no relationship between the magnitude of

the equilibrium constant and the rate of a reaction!

Example:

2H2 (g) + O2 (g) 2H2O (g)

Kc = 2.9 x 1031 = [H2O]2

[H2]2 [O2]

This reaction is very product-favored

However, the reaction will take years to reach equilibrium at room

temperature!

EQUILIBRIUM AND RATE OF REACTION

Equilibrium constants can be found by experiment

If you know the equilibrium concentrations (or partial pressures) of the reactants and products, simply plug and chug into the expression!

Let us consider the following equilibrium:

H2 (g) + I2 (g) 2HI (g)

DETERMINING EQUILIBRIUM CONSTANTS

H2 (g) + I2 (g) 2HI (g)

At 425.4oC, it is determined that the concentration of all species at equilibrium is as follows:

H2 (g) 0.00022 M

I2 (g) 0.00772 M

HI (g) 0.00956 M

Calculate the equilibrium constant, Kc, for the system

PRACTICE!

The equilibrium expression for our system is:

Kc =

So, the equilibrium constant is calculated as:

Kc = = = 54

At equilibrium, are there mostly products, reactants, or a mixture of products and reactants in the system? Justify your answer.

PRACTICE!

[HI]2

[H2] [I2]

[HI]2

[H2] [I2] (0.00956)2

(0.00022)(0.00772)

The

Reaction

Quotient,

Q

DETERMINING WHETHER

A SYSTEM IS AT

EQUILIBRIUM

We can predict the direction of a reaction by

calculating the reaction quotient, Q

Expression uses using any set of concentrations

of substances rather than just equilibrium

concentrations

Always written like so for the general reaction:

aA + bB ↔ eE + fF

PREDICTING THE DIRECTION OF A REACTION

Q = [E]e [F]f

[A]a [B]b

By comparing Q to the Kc value, we can predict

the direction for the reaction!

Q < Kc - Net forward reaction will occur

Q = Kc - No change, at equilibrium!

Q > Kc - Net reverse reaction will occur

COMPARING Q AND K

At 472oC, Keq = .105 for the following reaction:

N2(g) + 3H2(g) 2NH3(g)

2 minutes after this reaction starts, you measure the

concentrations and find:

[N2] = .0020M

[H2] = .10M

[NH3] = .15M

Is the system at equilibrium? If not, how must the

system shift in order to reach equilibrium?

PRACTICE!

These are not

necessarily

equilibrium

concentrations

PRACTICE!

• The reaction quotient, Q, is calculated as

follows:

Q = [NH3]2 = (.15)2 = 1.1x104

[N2][H2]3 (.0020)(.10)3

PRACTICE - COMPARE “Q” TO “KEQ”

1x104 ≠ .105

So, this reaction is NOT at equilibrium

In this case, Q > Keq

1x104 > .105

So, Q must get smaller to reach equilibrium

What do you need to make “Q” smaller?

More products or more reactants?

COMPARE “Q” TO “KEQ”

If:

More reactants are needed!

Q = [NH3]2

[N2][H2]3

Making the denominator

bigger makes Q smaller

so you need more

reactants (shifts left)

PREDICTING

EQUILIBRIUM SHIFTS

WITH

LE CHȂTELIER’S

PRINCIPLE

Equilibrium concentrations are based on:

The specific equilibrium

The starting concentrations

Temperature

Pressure

Reaction specific conditions

Altering conditions will stress a system, resulting in an

equilibrium shift

PREDICTING SHIFTS IN EQUILIBRIA

If a system at equilibrium is disturbed by a change in temperature, pressure, or the concentration of one of the components, the system will shift its equilibrium position so as to counteract the effect of the disturbance

In other words, any stress placed on an equilibrium system will cause the system to shift to minimize the effect of the stress

A QUICK RUNDOWN OF LE CHȂTELIER’S

PRINCIPLE

• Adding a reactant or product shifts the equilibrium

away from the increase

• Removing a reactant or product shifts the

equilibrium towards the decrease

• To optimize the amount of product at equilibrium,

we need to flood the reaction vessel with reactant

and continuously remove product

– We illustrate the concept with the industrial preparation

of ammonia

LE CHATELIER’S PRINCIPLE

CHANGE IN REACTANT OR PRODUCT

CONCENTRATIIONS

N2(g) + 3H2(g) 2NH3(g)

N2(g) + 3H2(g) 2NH3(g)

EXAMPLE

Consider the Haber process:

If H2 is added while the system is at equilibrium, the system

must respond to counteract the added H2

That is, the system must consume the H2 and produce products until

a new equilibrium is established

Therefore, [H2] and [N2] will decrease and [NH3] increases.

EXAMPLE

N2 and H2 are pumped

into a chamber The pre-heated gases

are passed through a

heating coil to the

catalyst bed

The catalyst bed is

kept at 460 - 550 C

under high pressure

The product gas

stream (containing

N2, H2 and NH3) is

passed over a

cooler to a

refrigeration unit

In the

refrigeration unit,

ammonia liquefies

but not N2 or H2

ILLUSTRATION OF THE HABER PROCESS

THE HABER PROCESS

The unreacted nitrogen and hydrogen are recycled with the

new N2 and H2 feed gas

The equilibrium amount of ammonia is optimized because the

product (NH3) is continually removed and the reactants (N 2

and H2) are continually being added

LE CHATELIER’S PRINCIPLE

EFFECTS OF VOLUME AND PRESSURE

Note - Changing the pressure does not change the value of the

equilibrium constant at constant temperature

Solids and liquids are not affected by pressure changes

Changing pressure by introducing an inert gas will not shift an

equilibrium

Pressure changes only affect gases that are a portion of an

equilibrium

This is due to the relationship between volume and pressure for a

gas:

As volume is decreased, pressure increases

If pressure is increased the system will shift to counteract the

increase by producing fewer moles of gas

That is, the system shifts to remove gases and decrease pressure

An increase in pressure favors the direction that has fewer moles of

gas

In a reaction with the same number of product and reactant

moles of gas, pressure has no effect!

LE CHÂTELIER’S PRINCIPLE

EFFECTS OF VOLUME AND PRESSURE

EXAMPLE

Consider:

An increase in pressure (by decreasing the volume) favors the

formation of colorless N2O4

The instant the pressure increases, the system is not at

equilibrium and the concentration of both gases has

increased

The system moves to reduce the number moles of gas

The forward reaction is favored

A new equilibrium is established in which the mixture is

lighter because colorless N 2O4 is favored

N2O4(g) 2NO2(g)

The equilibrium constant is temperature dependent!

For an endothermic reaction, DH > 0 and heat can be

considered as a reactant

For an exothermic reaction, DH < 0 and heat can be

considered as a product

Adding heat (i.e. heating the vessel) favors away from the

increase

If DH > 0, adding heat favors the forward reaction

If DH < 0, adding heat favors the reverse reaction

Removing heat (i.e. cooling the vessel), favors towards the

decrease

If DH > 0, cooling favors the reverse reaction

If DH < 0, cooling favors the forward reaction

LE CHATELIER’S PRINCIPLE

EFFECTS OF TEMPERATURE CHANGES

Cr(H2O)6(aq) + 4Cl-(aq) CoCl42-(aq) + 6H2O(l)Co

EXAMPLE

Consider the following reaction:

DH > 0

Co(H2O)62+ is pale pink and CoCl 4

2- is blue

If a l ight purple room temperature equilibrium mixture is placed in a beaker of warm water, the mixture turns deep blue

Since DH > 0 (endothermic), adding heat favors the forward reaction, i.e. the formation of blue CoCl4

2-

If the room temperature equilibrium mixture is placed in a beaker of ice water, the mixture turns bright pink

Since DH > 0, removing heat favors the reverse reaction which is the formation of pink Co(H2O)6

2+

A catalyst lowers the activation energy barrier for the reaction

Therefore, a catalyst will decrease the

time taken to reach equilibrium

A catalyst does not

effect the composition of

the equilibrium mixture

LE CHATELIER’S PRINCIPLE

THE EFFECT OF CATALYSTS

“Any stress placed on an equilibrium system will cause the

system to shif t to minimize the effect of the stress”

LE CHATELIER’S PRINCIPLE

• How can you cause the color to change from

pink to blue?

• You can put stress on a system by adding or removing something from one side of a reaction

Co(H2O)62+ + 4Cl1- CoCl4

2- + 6H2O

LE CHATELIER’S PRINCIPLE

Co(H2O)62+ + 4Cl1- CoCl4

2- + 6H2O

Note: ∆H = + value

• What other stresses could be placed on this

system to change the color back and forth

between pink and blue?

Add heat

(removes H2O) Add Ag1+ (removes Cl1- ) Add ice

Add acetone

Add HCl

(adds Cl1-)