+

Equilibrium

Topic 7

+7.1 Equilibrium

Reversible rxn a rxn that can go either way

A double arrow shows that it is reversible

CaCO3(s) CaO(s) + CO2(g)

+Equilibrium

Decomposition of CaCO3 is reversible, but what if we influence the system such that it is kept in a container at a constant high temp?

Goes to CaO and not back to CaCO3

Why?

All the CO2 escapes and the reaction will not go in reverse

+Con’t

If the same experiment is run, but in a closed container, what will the outcome be?

After the same amount of times as the first exp. There is still CaCO3

Why?

The rxn reaches equilibrium and amounts of product and reactant remain constant

+Dynamic equilibrium

Macroscopic properties are constant (concentrations of all reactants and products remain constant) and the rate of the forward rxn is equal to the rate of the reverse rxn

Forward reactants going to products

Reverse products going to reactants

+Equilibrium and rate of rxn

Consider the reaction:

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

If we begin with just hydrogen and iodine in a closed container, how will their concentrations change?

+Reaching Equilibrium

+Con’t

What must the rates of the forward and reverse rxns of a reversible rxn equal to be at equilibrium?

Each other

+Characteristics of the equilibrium rate

1. Macroscopic properties are constant at equilibrium

[products] and [reactants] remain constant

2. At equi. The rate of forward rxn = rate of reverse rxn

3. Equi. can only be obtained in a closed system

If in an open system, products can escape and have no opportunity to recombine

+Con’t

4. All species in the equation are present in the equi. Rxn mixtures

All reactants and all products

5. Equi. can be obtained from either direction

Does not matter if beginning with reactants or with products

+7.2 Position of equilibrium

What does the position of equi. refer to?relative amounts of reactants and products

present at equi.

Equi. does not imply 50% reactants and 50% products

Some reactions favor one over the other

+Ex.

2NO(g) N2(g) + O2(g) ΔH= +10 kJ mol-1

At 700 K the position of the equi lies mostly to the right

There is a large amount of products and not very much of the reactantThere is about 1.5 million times as many

product

+The effect of changing conditions on the position of equilibrium

Le Chatelier’s princliple: If a system at equilibrium is subjected to some

change, the position of equilibrium will shift in order to minimize the effect of the change

This allows us to predict which direction equi will shift

+Effect of temperature

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

At room temp, there is about 28 times more of the reactant than products (96.6% and 3.4%)

At 700 K there is only about 7 times more of the reactants than products (88% and 12%)

Which way did the equilibrium shift?

To the right

+Use Le Chatelier’s with temp.

Looking at the first rxn with HI, the enthalpy is positive, meaning what?

The reaction is endothermic

When the temp is increased the rxn is going to shift to minimize the effects

For this rxn, the equi shifts more toward the endothermic direction to take in the added heat, thus the equi shifted to the right

+Summed up

Heat reaction mixture position of equi is

shifted in the endothermic direction

Cool reaction mixture position of equi is

shifted in the exothermic direction

+Effects of pressure

2NO2(g) N2O4(g)

Brown colorless

If this rxn is performed in a sealed syringe, as the pressure increases the color initially gets darker and then becomes colorless as a new equi position is established

Which way did equi shift?

To the right

+Con’t

At higher pressure the equi shifted from 2 gaseous molecules to the side where there is just 1

One molecule takes up less space than two

Which way would the equi for this rxn shift?

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

left

Which direction will this equi shift?

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

neither

+Effects of concentration

2[CrO4]2-(aq) + 2H+(aq) [Cr2O7]2-(aq) + H20(l)

yellow orange

+Con’t

Looking as the first yellow solution, which direction does the equi lie?

To the left

When acid is added, the color changes to orange, meaning the equi is shifted which direction?

To the right

+What this means

Adding the acid increases the H+ ions, thus shifting the equi to the right to minimize the effects

If OH- is added, it does what to the H+?

Neutralizes it to form water and the equi shifts back to the left

+In general

If the concentration of one of the species in an equilibrium mixture is increased, the position of the equi shifts to the opposite side ot reduce the concentration of this species

+7.3 The equilibrium constant

CH3COOH(l) + C2H5OH(l) CH3COOCH2CH3(l) + H2O(l)

Ethanoic acid ethanol ethyl ethanoate

Various amounts of ethanoic acid and ethanol are reacted and allowed to come to equi at the same temp.

The equilibrium concentrations of each is determined and a ration is found (this is the constant):

[CH3COOCH2CH3(l)][H2O(l)]

[CH3COOH(l)][C2H5OH(l)]

+Constants

All concentrations are measured at equilibrium

Equilibrium law for the rxn

aA + bB cC + dD

Where all reactants are in the same phase, the following ratio is constant at a particular temp

Kc= [C]c[D]d c=in terms of concentration

[A]a[B]b

+Purpose of equilibrium constant

Provides info about how far a rxn proceeds at a particular temp

Kc >> 1 the rxn proceeds almost totally towards the products

Kc << 1 the rxn hardly proceeds toward the products

Allows for preduction about which direction the equi will lie for a rxn

+How changed conditions affects the value of the constant

Effect of pressure NOT affected by pressure

Effect of concentration NOT affected by concentration

If either of these two are changed, we can often conclude that the system is not at equilibrium

+Effect of temperature

The value of the constant for a particular rxn is affected only by a change in temp

For exo rxns, the value of the equi constant decreases as the temp is increased

For an endo rxn, the value of the equi constant increases as the temp is raised

+Catalysts and the equi constant

The presence of a catalyst does NOT affect the position of equi or the value of the constant, it simply reduces the time required to reach equi

Increases the rate of the forward and reverse rxns equally

+Constant and rate

The equi constant give info about how far a rxn goes towards completion (about the extent of the rxn)

It gives us NO info about how quickly the rxn occurs

Kinetic data, such as rate constant, indicates how quickly equi is attained by privies no info about the position of equi and how far the rxn proceeds

+

HL 2

+7.4 Calculations with the equilibrium constants

Calculating units:

CH3COOH(l) + C2H5OH(l) CH3COOCH2CH3(l) +H2O(l)

Kc= [CH3COOCH2CH3(l)][H2O(l)]

[CH3COOH(l)][C2H5OH(l)]

All units are mol dm-3

+Con’t

All units will cancel out

Thus, Kc has no units

Always the case when you have the same # of species on the reactant and product side

+Another Ex

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

Kc= [NH3]2

[N2][H2]3

Δn= total power on top- total power on bottom

What do the units equal?

Kc= (mol dm-3)-Δn

Kc= (mol dm-3)-2 = mol-2 dm6

+Calculating equi constants

Generally as simple as plugging in knows to solve for an unknown

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

Substance

H2 I2 HI

Equi concentration/ mol dm-3

0.18 0.39 1.95

+Con’t

Kc= [HI]2

[H2][I2]

Plug in values: 1.952 = 54

0.18*0.39

Kc has no units for this rxn

+Two different values of the constant for the same rxn under the same conditions

For the rxn: 2NO2(g) N2O4(g)

Set up the constant equation

Kc= 0.69 mol dm-3 at 400 K

However, the rxn could also be written as:

N2O4 (g) 2NO2 (g)

The constant (Kc’) for this is 1.46 mol dm-3 at 400 K

Kc’ = 1/ Kc

+Another Ex

The rxn: H2(g) + I2(g) 2HI(g)

Can also be written: ½ H2(g) + ½ I2(g) HI(g)

If both are set up in the equi constant equation, the answers will be different

Thus the constant is dependent upon how the equation is written

+

SL & HL 1/2

+7.5 Industrial Processes

Applying these ideas to industrial processes considering these two factors:

1. How much product is obtained?

2. How quickly is the product obtained?

+Haber process

What is the product of this process?

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

The N is obtained from the air and one of the sources on H is methane

Yield of ammonia is about 15- 20% per pass of the process

+

+Pressure and Temperature

+Contact process

What is the product of this process?

Sulfuric acid

It involves the following rxns:

S(s) + O2(g) SO2(g) ΔH=-297 kJ mol-1

2SO2(g) + O2(g) 2SO3(g) ΔH=-197 kJ mol-1

SO3(g) + H2O(l) H2SO4(l) ΔH=-130kJ mol-1

+Temperature considerations

This process is run at a relatively low temperature that causes a lower rate of rxn

Why would industries run this process at a lower temp rather than a higher temp?

The higher temp is not cost-affective due to the higher costs of fuel and the quicker corrosion of the rxn vessels

+Increasing pressure

Increasing the pressure would increase the yield of SO3 and would increase costs because a plant that is designed for high pressures would have to be built

+7.6 Phase equilibria

Physical equilibria equilibria involving change of state; specifically that between liquid and its vapor

Evaporation when a liquid is left in an open container, it evaporates

On the molecular level, the particles overcome the forces that hold them together and escape into the gaseous phase

+

+Liquid- vapor equilibrium

Consider a volatile liquid in a closed container

+Rate of vaporization

When rate of vaporization and condensation are equal, the system is at dynamic equilibrium

Dynamic two opposing processes occurring at the same rate

+

HL 2

+HL 2 Vapor pressure

The pressure exerted by the vapor in equilibrium with a liquid (or a solid)

Depends on nature of liquid

The higher the temp, the more energy the molecules have to break free from intermolecular forces

Thus, depends ONLY on the nature of liquid and temp

+Con’t

Independent of:Volume of containerVolume of the liquid Volume of the vaporShape of the containerSurface area of liquid

+Variation of vapor pressure and temp

+Boiling point of a liquid

A liquid boils when its vapor pressure equals the external pressure

Thus the normal boiling point of a liquid is the temp at which the vapor pressure of the liquid is 1 atm

+Bp and pressure

Water boils at 70°C on top of Mount Everest

Pressure decreases as the altitude increases

Boiling point is dependent on the external pressure

+Vapor pressure and intermolecular forces

At all temps 1-butanol has lower vapor pressure than ethoxyethane

This is due to the stronger intermolecular forces of 1-butanol

They have about the same masses and Van der Waal’s forces, but the molecules are bonded together differently

1-butanol has hydrogen bonding and ethoxyethane has mainly dipole-dipole forces

+Enthalpy of vaporization

The standard enthalpy change of vaporization (ΔHvap) is the energy needed to convert 1 mol of a liquid to vapor under standard conditions

Intermolecular forces

Vapor pressure

Boiling point

Enthalpy of vaporizatio

n

Weak High Low Low

Strong Low High high