Equilibrium

Objectives

Outline the characteristics of chemical and

physical systems in a state of equilibrium

Deduce the equilibrium constant expression

from the equation for a homogeneous

reaction.

Dynamic EquilibriumWhen wet clothes are hung on a washing line to dry, the

water in them does not boil, because the temperature is still far below the normal boiling point of water.

Some of the water molecules will be traveling much faster than the average, and they will have enough kinetic energy to escape the surface of the water and enter the atmosphere.

The pressure they produce is known as the vapour pressure of water. The vapour pressure is very dependent upon the temperature.

As the vapour leaves the wet clothes it will be blown away by the wind, and more water will evaporate to replace it, until eventually the clothes become dry.

However what would happen if the water was held in a closed system?

Initially water will still vaporize, but once the air above the water becomes saturated with water vapour it can hold no more water.

A state of equilibrium is established between the liquid water and the water vapour.

Some water molecules will still be traveling fast enough to have enough kinetic energy to escape from the surface and enter the vapour phase.

At the same time some of the water vapour molecules will slow down and condense back into the liquid state.

Equilibrium is established when the rate of vaporization is equal to the rate of condensation:

H2O(l) H2O(g)

Vapour PressureVapour pressure is the pressure of a vapour in equilibrium with its non-vapour phases. All liquids and solids have a tendency to evaporate to a gaseous form, and all gases have a tendency to condense back into their original form (either liquid or solid). At any given temperature, for a particular substance, there is a pressure at which the gas of that substance is in dynamic equilibrium with its liquid or solid forms.

This is the vapour pressure of that substance at that temperature. The equilibrium vapour pressure is an indication of a liquid's evaporation rate. It relates to the tendency of molecules and atoms to escape from a liquid or a solid. A substance with a high vapour pressure at normal temperatures is often referred to as volatile.

This is an example of dynamic equilibrium. Can you think of another example of a

dynamic equilibrium?At equilibrium the process taking place has

not stopped, but the amounts on both sides of the process remain constant.

The rate of the forward process is equal to the rate of the backward

process

Explain how these diagrams demonstrate a dynamic equilibrium…

Closed SystemsA dynamic equilibrium

will only set up in a closed system.

In a completely closed system neither matter nor energy can be lost or gained.

When we use the term equilibrium we are usually referring to a closed system.

Vapour

Liquid

vaporisation

Water in a closed system

Reversible ReactionsSome reactions go virtually to completion,

however, many don’t.If a reaction does not go to completion it is

said to be reversible.Once some products have been formed they

start to react to re-form the reactants.The position of equilibrium can be reached

either by starting with the reactants or by starting with the products.

Whether we start with dinitrogen tetroxide or with nitrogen dioxide, once equilibrium has been attained the composition of the mixture is constant. We can tell when equilibrium is established as the graph becomes horizontal

00.20.4

0.60.81.01.2

1.41.6

1.82.0

00.20.4

0.60.81.01.2

1.41.6

1.82.0

N2O4 N2O4

NO2 NO2

At start 1.0 mol N2O4

At start 2.0 mol NO2

time time

Homogeneous Chemical EquilibriaA homogeneous reaction is one in which all the

reactants and products are in the same phase.A phase is similar to a state such as solid, liquid

or gas, but there is an important difference. There is a physically distinct boundary between two phases.

For example: two immiscible liquids such as oil and water, both are in the liquid state, but there are two different phases, as there is a clear boundary where the two liquids meet.

If all the reactants and products are in the same phase, such as all in the gaseous state, all in the liquid phase or all in aqueous solution, we can write an equilibrium expression for the reaction:

Consider the general reaction:

aA + bB cC dD+

At equilibrium the concentrations of A, B, C and D are written as [A]eqm’ [B]eqm’ [C]eqm’ and [D]eqm’

respectively.

Kc =[C]c

eqm x [D]deqm

[A]aeqm [B]b

eqmx

Kc is known as the equilibrium constant, and it is a constant at a stated temperature

Lets practicePut the following equations into equilibrium

expressions:C2H5OH(l) + CH3COOH(l) CH3COOC2H5(l) + H2O(l)

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

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

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

Activity:Read through pages 152- 155 of Calculations

in AS/A Level Chemistry by Jim Clark 2000Complete problem 1 on page 155, check your

answers on page297.

ObjectivesDeduce the extent of a reaction from the

magnitude of the equilibrium constant.Apply Le Châtelier’s principle to predict the

qualitative effects of changes of, pressure concentration and temperature on the position of equilibrium and on the value of the equilibrium constant.

State and explain the effect of a catalyst on an equilibrium reaction.

The magnitude of the equilibrium constantKc is essentially a measure of the amount of

products in an equilibrium mixture compared with the amount of reactants.

If Kc > 1 then the position of equilibrium lies very much to the right, the mixture contains virtually all products, the reaction has almost gone to completion.

If Kc < 1, then very little product will be formed, the mixture contains virtually all reactants, the equilibrium lies to the left.

Le Châteliers PrincipleHenri Le Châtelier (1850-1936) was a French

chemist who described what happens to the position of equilibrium if a change to the conditions is made.

Le Châtelier’s Principle states that

”if a dynamic equilibrium is disturbed by changing the conditions, the position of equilibrium moves to counteract the change”

An example:Consider the reaction between ethanoic acid

and ethanol to form ethyl ethanoate and water:

If we remove some of the water from the equilibrium mixture, Le Châtelier’s Principle enables us to predict that more of the acid and alcohol will react to replace it.

This concept of removing one of the products is a useful way to increase the amount of a desired product in an equilibrium reaction.

C2H5OH(l) + CH3COOH(l) CH3COOC2H5(l) + H2O(l)

The Conditions:Pressure in gaseous reactionsConcentration for reactions in the liquid

phaseTemperatureUsing catalysts.

The effect of changing the pressure on gaseous reactionsThis only applies to reactions involving gases.An example of a gas reaction that reaches

equilibrium is:

N2O4(g) 2NO2(g)

Dinitrogen tetraoxide

Nitrogen dioxide

1 mole

2 moles

colourless brown

Increasing the pressure of a gas means that there are more molecules of it in a given volume – it is the equivalent to increasing the concentration of a solution.

If we increase the pressure on this system, Le Châtelier’s principle tells us that the position of equilibrium will move left because fewer molecules exert less pressure.

In the same way if we decrease the pressure the equilibrium will move to the right.

So how can this equilibrium mixture demonstrate this?

Dinitrogen tetraoxide is a colourless gas and nitrogen dioxide is brown.

If we decrease the pressure which direction will the equilibrium lie, what would we see?

If we left the mixture for a while what would we see?

N2O4(g) 2NO2(g)

Dinitrogen tetraoxide

1 mole

colourless

Nitrogen dioxide

2 moles

brown

Equilibrium mixture

For a moment the mixture becomes paler because we have reduced the concentration of brown NO2

We can decrease the pressure by pulling out the syringe barrel.

After a few moments the mixture becomes darker as the equilibrium moves to the right and more brown NO2 is formed

However, if there are the same number of moles of gas on both sides of the equation, then pressure has no effect on the equilibrium mixture. For example:

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

2 moles 2 moles

The equilibrium position will not change in this reaction when the pressure is changed, so the proportions of the three gases will stay the same.

1. In which of the following reactions will the position of equilibrium be affected by changing the pressure? Explain your answers.

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

b. CH3CO2H(aq) CH3CO2-(aq) + H+

(aq)

c. H2(g) + CO2(g) H2O(g) + CO(g)

ConcentrationIf we increase the concentration of one of the

reactants, Le Chateliers principle says that the equilibrium will shift in the direction that tends to reduce the concentration of this reactant.

A(aq) + B(aq) C(aq) + D(aq)

Suppose we add some extra A. This would increase the concentration of A. The only way that this system can reduce the concetration of A, is by A reacting with B and moves the equilibrium to the right

A(aq) + B(aq) C(aq) + D(aq)

We end up with a greater proportion of products in the reaction than before we added A. The same thing would happen if we added more B.

We could also remove C as it was formed. The equilibrium would move to the right to produce more C (and D). The same thing would happen if we removed D s soon as it was formed.

TemperatureReversible reactions that are exothermic (give

out heat) in one direction are endothermic (take in heat) in the other direction.

Suppose we increase the temperature of an equilibrium mixture that is exothermic in the forward direction.

An example is:

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

The negative sign of ∆Hѳ means that heat is given out when sulphur dioxide and oxygen react to form sulphur trioxide.

Le Châteliers principle tells us that if we increase the temperature , the equilibrium moves in the direction that cools the system down.

To do this it will move in the direction which absorbs heat (is endothermic) i.e. to the left.

The equilibrium mixture will contain a greater proportion of sulphur dioxide and oxygen than before.

In the same way, if we cool the mixture the equilibrium will move to the right and increase the proportion of sulphur trioxide.

CatalystsCatalysts have no effect on the position of

equilibrium.

They work in such a way that they affect both

the forward and reverse reactions equally.

Therefore they speed up the rate at which

equilibrium is set up but do not alter the

composition of the equilibrium mixture.

ObjectivesApply the concepts of kinetics and equilibrium

to industrial processes

Haber Process

Contact process

Ammonia, NH3

Ammonia is an important chemical in industry.

World production is over 140 million tonnes each year.

Around 80% is used to make fertilisers like ammonium nitrate, ammonium sulphate and urea.

The rest is used to make synthetic fibres (including nylon), dyes explosives and plastics like polyurethane

Making ammoniaNitrogen and hydrogen react together by a

reversible reaction, which at equilibrium, forms a mixture of nitrogen, hydrogen and ammonia:

N2(g) + 3H2(g) 2NH3(g) ∆Hѳ = -92 kJ mol-1

The percentage of ammonia obtained at equilibrium depends on temperature and pressure.

The graph shows that high pressure and low temperature would give close to 100% conversion, while low pressure and high temperature would give almost no ammonia.

How could we use Le Châteliers Principle to predict the conditions for the best yield?

Le Châtelier’s PrincipleThe reaction is exothermic from left to right.

So, cooling the mixture will drive the equilibrium to the right (in the direction where heat is given out).

There will be a greater percentage of ammonia in the equilibrium mixture.

There are four molecules of gas on the left and two on the right. So, increasing the pressure will drive the equilibrium to

the right (in the direction where the pressure is lower).There will be a greater percentage of ammonia in the

reaction mixture.

The Haber ProcessIn industry Ammonia is made by the Haber

Process, in which the process of making ammonia is the key step.

The Raw Materials are:

AIR NATURAL GAS WATER

Provides the nitrogen (N2)

Provide the hydrogen by the following reaction

CH4(g) + H2O(g) → CO(g) + 3H2(g)

The nitrogen and hydrogen are fed into a converter in the ratio of 1 : 3 and passed over an iron catalyst.

Most plants run at a pressure of around 20000kPa (about 200 times atmospheric pressure) and a temperature of about 450K.

This is a lower pressure and a higher temperature than would give the maximum conversion.

These compromise conditions are used because:A high-pressure plant is expensive, both to build

and maintainA lower temperature would slow down the rate of

reactionAt lower temperatures the catalyst lasts longer

Nitrogen and hydrogen flow continuously over the catalyst so the gases do not spend long enough in contact with the catalyst to reach equilibrium – there is about 15% conversion to ammonia.

The ammonia is cooled so that it becomes liquid and is piped off.

Any nitrogen and hydrogen that is not converted into ammonia is fed back into the reactor.

The catalyst is iron in pea sized lumps (to increase the surface area). It lasts about 5 years before it becomes ‘poisoned’ by impurities in the gas stream and has to be replaced.

Natural gas CH4

Steam H2O

H2, CO2

Air

N2 + O2

H2, N2 CO2

Alkali

Carbon dioxide

by-product

Reactor, Fe catalyst

H2, N2

H2, N2 NH3,

Cooler

NH3,

Ammonia NH3

H2, N2 recycled

Raw Material

Product

Flow Chart of the Haber Process

The Contact ProcessThe manufacture of sulphuric acid in the

developed world has fallen in recent years.This is partly because the manufacture of

phosphate fertilizers, which uses sulphuric acid, has declined.

It is also because much of the industry has relocated to the developing world, where the plants are cheaper to build and operate, and where environmental restrictions can be less stringent.

The essential reaction is the conversion of sulphur dioxide to sulphur trioxide:


The sources of sulphur dioxide include the combustion of the element sulphur itself and the roasting (combustion) of sulphide ores in air.

As the reaction is exothermic and involves a reduction in volume, the same arguments as used for the Haber process can be applied.

A high yield will be favoured by a high pressure and a low temperature.

The rate at which equilibrium is reached will be favoured by a high pressure, a high temperature, and the use of a suitable catalyst.

As before, an optimum temperature must be used.In practice, plants operate at a temperature of about

450oC, and the catalyst is vanadium (V) oxide.Although in theory a high pressure is required, in

practice a pressure of only 2 atm is used as this will give a yield of over 98%.

Operating the plant at higher pressure than this would not be cost-effective.


V2O5(s), 450oC, 2atm

3 vol 2 vol

The reaction between sulphur trioxide and water then produces sulphuric acid:

SO3(g) + H2O(l) H2SO4(l)

The sulphur trioxide produced cannot be added directly to water, because the heat evolved is too great.

Instead it is absorbed by a concentrated solution of sulphuric acid to which small quantities of water are then added to keep the acid concentrated.

The sulphuric acid produced has many uses, ranging from a chemical feedstock to paints and pigments, fertilizers, detergents and soaps, and as the electrolyte in car batteries.

ObjectivesDescribe the equilibrium established between

a liquid and its own vapour and how it is affected by temperature changes.

Sketch graphs showing the relationship between vapour pressure and temperature and explain them in terms of the kinetic energy.

State and explain the relationship between enthalpy of vaporization, boiling point and intermolecular forces

Vapour PressureVapour pressure is the pressure of a vapour in equilibrium with its non-vapour phases. All liquids and solids have a tendency to evaporate to a gaseous form, and all gases have a tendency to condense back into their original form (either liquid or solid). At any given temperature, for a particular substance, there is a pressure at which the gas of that substance is in dynamic equilibrium with its liquid or solid forms.

This is the vapour pressure of that substance at that temperature. The equilibrium vapour pressure is an indication of a liquid's evaporation rate. It relates to the tendency of molecules and atoms to escape from a liquid or a solid. A substance with a high vapour pressure at normal temperatures is often referred to as volatile.

The vapour pressure of all liquids increases as the temperature increases.

All liquids show this kind of exponential – shaped graph, and the boiling point of the liquid will be the temperature at which the external pressure is equal to the vapour pressure above the liquid.It can be seen from the graph that the vapour pressure rises rapidly at higher temperatures. Using the Maxwell-Boltzmann distribution, try and explain this phenomena.

Consider the velocities of the particles in the liquid.

Not all particles are moving at the same velocity.If the temperature is increased

Temperature T1

Temperature T2

T2 >T1

Average kinetic energy at T1

Average kinetic energy at T2

Minimum energy for particles to form vapour

Particles traveling relatively slowly

Particles traveling close to the average velocity Particles

traveling much faster than the average velocity

At low temperatures only a few particles will be able to escape, and hence the vapour pressure is low.

At higher temperatures many more of the particles will have the necessary kinetic energy to escape, resulting in a much higher vapour pressure

If the system is open, then as the faster particles leave the system the average energy of the remaining liquid particles decreases, and the temperature of the liquid decreases.

This is why a small amount of a highly volatile liquid, such as ether, (ethoxyethane, C2H5OC2H5), on the back of your hand will feel cold to the touch.

Relationship between enthalpy of vaporization, boiling point and intermolecular forces.Boiling points are normally measured under a

pressure of 1atm. The boiling point of a substance is the temperature at which the vapour pressure is the same as atmospheric pressure.

Different liquids have different boiling points.When a liquid boils, the intermolecular forces of

attraction between particles in the liquid state are completely broken, and in an ideal gas there no forces of attraction between the gaseous particles.

Liquid(l) Vapour(g)

The stronger the forces of attraction, the

higher the boiling point.

The forces of attraction for liquids consisting

of simple molecules depend on the polarity of

the molecules: the more polar the molecule,

the stronger the forces of attraction, and the

higher the boiling point will be.

The strength of the forces of attraction also

affects the enthalpy of vaporization.

Water has a high enthalpy of vaporization due to the relatively strong hydrogen bonding that exists in liquid water.

H2O(l) H2O(g)∆Hѳ = +44 kJ mol-1

Compare this with the enthalpy of vaporization of silicon tetrachloride, which is non-polar.

The molecules of silicon tetrachloride in the liquid state will be attracted to each other by only weak van der Waals’ forces.

Even though molecules of silicon tetrachloride have a much greater mass, the enthalpy of vaporization is considerably lower.

SiCl4(l) SiCl4(g)∆Hѳ = +30 kJ mol-1

From the enthalpy of vaporization, we can deduce that the boiling point of silicon tetrachloride will be lower than the boiling point of water. The value is in fact 57.6oC.

Evidence of a close correlation between boiling point and enthalpy of vaporization is provided by comparing SiCl4 with Br2.

Both have a similar relative molecular mass, and both are non-polar and both the boiling points and their enthalpies of vaporization are almost identical

Compound Mr Bp (oC) Enthalpy of vaporization kJmol-1

SiCL4 170.09 57.6 30.0

Br2 159.80 58.8 30.7

Some compounds (e.g. 2-hydroxy propanoic

acid), also known as lactic acid,

CH3CH(OH)COOH) decompose below their

normal boiling points.

This could cause a problem if fractional

distillation is being used to purify the liquid or

separate the liquid from a mixture of liquids.

However, the problem can usually be easily

overcome by carrying out distillation under

reduced pressure.

ObjectivesThe equilibrium Law

Solve homogeneous equilibrium problems using

the expression for Kc

Calculating KcConsider the reaction between a moles of

ethanoic acid and b moles of ethanol to produce x moles of ethyl ethanoate and x moles of water at equilibrium. Let the total volume for the liquid mixture be V dm3

eqm conc /mol dm3

CH3COOH(l) + C2H5OH(l) CH3COOC2H5(l) H2O(l)+

Initial amount /mol

eqm amount /mol

a b 0 0

a - x b - x x x

a - x

V

b - x

V

x

V

x

V

Now comes the complicated part…..

[CH3COOC2H5(l)]eq

m

x [H2O(l)]eqm

[C2H5OH(l)]eqm x [CH3COOH(l)]eq

m

Kc =

=(x/V) x (x/V)

[(a – x)/V] [(b – x)/V]x

= x2

(a – x)

(b – x)

For exampleWhen 46g (1.0mol) of ethanol was reacted with

30.0g (0.50 mol) of ethanoic acid at 373K the equilibrium contained 37.0g (0.42 mol) of ethyl ethanoate. Calculate the value of Kc at 373K to the nearest integer. So we saw before that….

= x2

(a – x)

(b – x)

= (0.42)2

(1.0 – 0.42)

(0.50 – 0.42)

= 4

Kc now substitute the letters for the numbers

Working out the unitsThe units of Kc vary from case to case, and

you will have to work them out every time you calculate a value for Kc in a question. Write down the expression for Kc and then substitute units into it rather than numbers.

Example:

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

Kc =[NH3(g)]2

[N2 (g)] [H2(g)]3=

(mol dm-3)2

(mol dm-3) (mol dm-3)3=

(mol dm-

3)2

1

=mol2 dm-6

1= mol-2 dm6

ActivityRead through further examples of calculating

units on page 155 in Calculations in AS/A Level Chemistry by Jim Clark 2000

Complete problems 2 – 7 on pages 160 -161. Answers can be found on page 297 - 298

Calculating the composition of a reaction mixtureWe can use the equilibrium law to calculate

the composition of a reaction mixture that has reached equilibrium

Lets use this example again:


Suppose Kc = 4 at the temperature of the experiment and we want to know how much ethyl ethanoate we could produce by mixing 1 mol of ethanol and 1 mol of ethanoic acid.


Initial amount /mol

eqm amount /mol

1 mol

1 mol

0 0

(1 – x) mol

(1 – x) mol

x mol x mol

We do not know how many moles of ethyl ethanoate will be produced, so we call this x.

The equation tells us that x mol of water will also be produced and in doing so x mol of both ethanol and ethanoic acid will be used up.

So the amount of each of these remaining at equilibrium is (1 – x) mol

These figures are in moles, but we need concentration in mol dm-3 to substitute in the equilibrium law expression.


Initial amount /mol

eqm amount /mol

1 mol

1 mol

0 0

(1 – x) mol

(1 – x) mol

x mol x mol

Suppose the volume of the system at equilibrium Vdm-3.

Then……..

These figures may now be put into the expression Kc:

[CH3COOC2H5(l)]eq

m

= (1 – x)/V mol dm-3[C2H5OH(l)]eqm

[H2O(l)]eqm

[CH3COOH(l)]eq

m

=

=

=

(1 – x)/V mol dm-3

mol dm-3

mol dm-3

x /V

x /V

Kc =(x/V) x (x/V)

(1 – x)/V (1 – x)/Vx

Kc =(x/V) x (x/V)

(1 – x)/V (1 – x)/Vx

The V’s cancel, so in this case we do not need to know the actual volume of the system.

So….

4 =x2

(1 – x)2

Taking the square root for both sides

2 =x

(1 – x)So…

.x = 2 (1 –

x)x = 2 - 2x 2 = 3x x =

2

3

So 2/3 mol ethyl ethanoate and 2/3 mol of water are produced when the reaction reaches equilibrium, so the composition of the equilibrium mixture would be: ethanol 1/3mol, ethanoic acid 1/3mol, ethyl ethanoate 2/3mol, water 2/3mol


Initial amount /mol

eqm amount /mol

1 mol

1 mol

0 0

1/3 mol 1/3 mol 2/3 mol 2/3 mol

Calculating the amount of a reactant neededWe can also use Kc to find the amount of a

reactant needed to give a required amount of product.

For the following reaction in ethanol solution, Kc = 30.0 dm3 mol-1

CH3COCH3+ HCN CH3C(CN)(OH)CH3

propanone hydrogen cyanide

2-hydroxy-2-methylpropanenitrile

Suppose we are carrying out this reaction in 2.00dm3 of ethanol. How much of the hydrogen cyanide is required to produce 1.00mol of product if we start with 4.00mol of propanone? Let x be the number of moles of HCN required.

At equilibrium, we want 1 mol of product.

[CH3COCH3]eqm [HCN]eqm

[CH3C(CN)(OH)CH3]eqm

Kc = = 30.0dm3 mol-1

Equation CH3COCH3+ HCN CH3C(CN)(OH)CH3

At start 4.00 mol x mol 0 mol

At eqm (4.00 – 1.00) mol

(x – 1.00) mol 1.00 mol

3.00 mol (x – 1.00) mol 1.00 mol

x

These are the number of moles but we need the concentrations to put in the equilibrium law expression. The volume of the solution is 2.00dm3 and the units for concentration are mol dm-3 so we next divide each quantity by 2.00dm3

So at equilibrium….

3.00/2.00 mol dm-3[CH3COCH3]eqm

[HCN]eq

m


=

=

=

(x - 1.00)/2.00 mol dm-3

1.00/2.00 mol dm-3

Putting the figures into the equilibrium law expression and cancel any volumes:

3.00/2.00 mol dm-3[CH3COCH3]eqm

[HCN]eq

m


=

=

=

(x - 1.00)/2.00 mol dm-3

1.00/2.00 mol dm-3

30.0 dm3 mol-1

=1.00/2.00 mol dm-3

3.00/2.00 mol dm-3 x (x - 1.00)/2.00 mol dm-3

So...

Cancelling through and rearranging we have:

30 (3/2 ( x – 1 ) / 2 ) = 1/2

45 ( x - 1 ) = 1

45x = 46

x = 46/45 = 1.02

So, to obtain 1 mol of product we must start with 1.02 mol hydrogen cyanide, if the volume of the system is 2.0dm3.In this example the volume of the system does

make a difference, because this reaction does not have the same number of moles of products and reactants

30.0 dm3 mol-1

=1.00/2.00 mol dm-3

3.00/2.00 mol dm-3 x (x - 1.00)/2.00 mol dm-3

ActivityRead through the examples on pages 161 –

164 from Calculations in AS/A level Chemistry by Jim Clark 2000

Complete problems 8 – 10 on page 165. Answers can be found on page 299.

Equilibrium

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