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Physical Chemistry Notes

MODULE P8: GASES.......................................................................................................................

INTRODUCTION.....................................................................................................................................KINETIC THEORY OF MATTER.................................................................................................................THE IDEAL GAS EQUATION....................................................................................................................USING THE IDEAL GAS EQUATION TO DETERMINE THE RMM OF GASES...................................................DALTON’S LAWS OF PARTIAL FRACTIONS..............................................................................................DEFINITION OF A MOLE FRACTION.........................................................................................................AVOGADRO’S LAW...............................................................................................................................

Molar volume...................................................................................................................................GRAPHICAL REPRESENTATION OF SPEEDS OF MOLECULES IN A GAS.........................................................

MODULE P9: THE SOLID STATE................................................................................................

METALLIC STRUCTURES........................................................................................................................The hexagonal close packed structure..............................................................................................The face centred cube structure.......................................................................................................The body centred cube structure......................................................................................................Metallic properties...........................................................................................................................

Hexagonal close packed - ABAB (hcp).......................................................................................................Cubic close packing - ABCABC (ccp).........................................................................................................Body centred cube (bcc)..............................................................................................................................

IONIC CRYSTALS...................................................................................................................................MOLECULAR CRYSTALS.........................................................................................................................GIANT COVALENT CRYSTALS.................................................................................................................

Diamond.....................................................................................................................................................Silicon dioxide............................................................................................................................................

THE STRUCTURE OF GRAPHITE...............................................................................................................POLYTHENE AND PLASTIC......................................................................................................................

MODULE P10: LIQUIDS AND SOLUTIONS.................................................................................

THE LIQUID STATE................................................................................................................................INTERMOLECULAR FORCES WITHIN A LIQUID..........................................................................................BINARY LIQUIDS SOLUTIONS - THE PROCESS OF DISSOLUTION.................................................................

Polar solvent --- Polar solute...........................................................................................................Non-polar solvent --- Polar solute....................................................................................................Polar solvent --- Non-polar solute....................................................................................................Non-polar solvent --- Non-polar solute.............................................................................................

ENERGY FACTORS IN THE DISSOLUTION OF AN IONIC SALT......................................................................SOAPS...................................................................................................................................................

MODULE P11: TWO COMPONENT SYSTEMS AND PHASE DIAGRAMS.............................

EQUILIBRIUM........................................................................................................................................MISCIBLE LIQUIDS.................................................................................................................................RAOULT’S LAW....................................................................................................................................EUTECTICS AND ALLOYS.......................................................................................................................COOLING CURVES FOR MIXTURES OF VOLATILE LIQUIDS.........................................................................

MODULE P12: THERMOCHEMISTRY.........................................................................................

INTRODUCTION.....................................................................................................................................THE LAW OF CONSERVATION OF ENERGY...............................................................................................ENTHALPY AND ENTHALPY CHANGES.....................................................................................................EXOTHERMIC AND ENDOTHERMIC REACTIONS AND THE SYMBOL OF DH.................................................STANDARD CONDITIONS AND STATES.....................................................................................................

PHYSICAL CHEMISTRY NOTES PART I I

HESS’S LAW.........................................................................................................................................LIMITATIONS OF DH VALUES.................................................................................................................BOND ENERGIES...................................................................................................................................

Bond dissociation and bond formation.............................................................................................Uses of bond energies......................................................................................................................Estimating enthalpy changes for reactions.......................................................................................

Determining the bond energy of methane....................................................................................................THE BORN-HABER CYCLE.....................................................................................................................

Representing the Born-Haber cycle on an energy diagram...............................................................Comparison of lattice energies for group 1 and 2 halides.................................................................

MODULE P13: REACTION KINETICS.........................................................................................

INTRODUCTION.....................................................................................................................................MEASURING REACTION RATES...............................................................................................................

Methods of determining reaction rates.............................................................................................Titrimeteric analysis...................................................................................................................................Colorimetric analysis..................................................................................................................................Pressure measurements...............................................................................................................................Conductrimetric analysis.............................................................................................................................

FACTORS AFFECTING REACTION RATES..................................................................................................The effect of concentration on rate of reaction.................................................................................

CALCULATING THE ORDER OF A REACTION FROM EXPERIMENTAL DATA..................................................USES OF REACTION RATE STUDIES.........................................................................................................ACTIVATION ENERGY AND TEMPERATURE..............................................................................................TWO-STEP REACTIONS...........................................................................................................................

MODULE P14: EQUILIBRIUM SYSTEMS....................................................................................

THE KINETIC MOLECULAR PICTURE OF EQUILIBRIUM.............................................................................EXAMPLES OF SYSTEMS IN EQUILIBRIUM................................................................................................

Physical systems..............................................................................................................................Chemical systems.............................................................................................................................

THE DISTRIBUTION OF A SOLUTE BETWEEN TWO IMMISCIBLE LIQUIDS.....................................................THE RELATIONSHIP BETWEEN THE CONCENTRATIONS OF THE SOLUTE IN SOLUTIONS................................THE EQUILIBRIUM LAW.........................................................................................................................

Calculations involving kc to remember.............................................................................................Gaseous mixtures.............................................................................................................................

FACTORS AFFECTING EQUILIBRIUM........................................................................................................Changing concentration by… (kc remains constant).....................................................................................Changing pressure in… (kp remains constant)............................................................................................Changing temperature in….........................................................................................................................Action of catalyst........................................................................................................................................

MODULE P15: ACID-BASE EQUILIBRIA....................................................................................

INTRODUCTION TO ACIDIC SYSTEMS (HYDROGEN ION THEORY)...............................................................BRONSTED-LOWRY THEORY..................................................................................................................

Conjugate pairs...............................................................................................................................THE PH SCALE......................................................................................................................................

How is pH calculated?.....................................................................................................................THE IONIC PRODUCT OF WATER.............................................................................................................STRENGTH OF ACIDS AND BASES............................................................................................................

Why does the double bonding increase the strength of an acid?.......................................................A-VALUES.............................................................................................................................................DISSOCIATION CONSTANTS....................................................................................................................CALCULATION OF PH VALUES FOR WEAK ACIDS AND BASES...................................................................ACID-BASE INDICATORS........................................................................................................................

The end point of titrations................................................................................................................SALT HYDROLYSIS................................................................................................................................BUFFER SOLUTIONS...............................................................................................................................

What are they?............................................................................................................................................Why does they work?..................................................................................................................................

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REDOX EQUILIBRIA......................................................................................................................

STANDARD ELECTRODE POTENTIAL........................................................................................................USING ELECTROCHEMICAL CELLS..........................................................................................................APPLICATIONS OF STANDARD ELECTRODE POTENTIALS...........................................................................

TablesTABLE 1: DIFFERENCES BETWEEN SOLIDS, LIQUIDS AND GASES.............................................................TABLE 2: DISSOCIATION ENERGIES OF BONDS.......................................................................................TABLE 3: RESULTS FROM EXPERIMENTAL DATA (CONCENTRATION EXPERIMENT)...................................TABLE 4: RESULTS FROM EXPERIMENT (PHASE DIFFERENCE).................................................................TABLE 5: FACTORS AFFECTING EQUILIBRIUM (CONCENTRATION)...........................................................TABLE 6: FACTORS AFFECTING EQUILIBRIUM (PRESSURE)......................................................................TABLE 7: FACTORS AFFECTING EQUILIBRIUM (TEMPERATURE)...............................................................TABLE 8: NUMBER OF DOUBLE BONDS IN VARIOUS ACIDS.....................................................................TABLE 9: RESULTS OF VARIOUS ACID/ALKALI COMBINATIONS IN SALT HYDROLYSIS...............................TABLE 10: STANDARD ELECTRODE POTENTIAL VALUES.........................................................................

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MODULE P8: GASES

IntroductionWe know that particles exist because we can be affected by their presence. If, for example, a person wears a string perfume or you spray air freshener in one corner of a room it is possible over a period of time to smell it anywhere in the room. This is due to the fact that small particles are moving in the air and spreading throughout the room. The same may be said when we put only a small pinch of salt and pepper into our food and it flavours the entire meal.

Particles of gases move by Brownian motion and diffusion occurs when the particles of one substance mix with the particles of another substance from high concentration to equilibrium. This mixing and moving of matter results from kinetic energy.

Brownian motion is the rapid random movement of particles. To observe this you can watch pollen grains in water or smoke specks in air through a microscope. The particles move due to the bombardment of other particles.

Kinetic theory of matter· All matter is comprised of tiny invisible particles.

· Particles of different substances are different sizes.

· In solids the particles are packed closely together and have only small amounts of energy preventing them from moving apart but allowing them to rotate in a fixed position.

· In liquids the particles are a little further apart and have greater amounts of energy and so they can flow; i.e. the particles can move over one another and take the shape of their container. Liquid particles have vibrational, rotational, and translocational motion.

· In gases, particles are widely separated and have large amounts of energy. They have enough energy to overcome their forces of attraction nearly completely. They move rapidly, randomly and haphazardly into all circulable space. i.e. The ideal gas can be considered to exist as a collection of point mass particles moving in random motion.

· Increased temperature = an increase in average kinetic energy of particles. Kinetic energy = vibrational, rotational, and translocational energy in gases and liquids and as vibrational and rotational energy in solids.

Gases differ from solids and liquids in that their particles are very far apart, they have low densities and are very compressible. The forces between gaseous molecules are very weak and the particles readily move away from each other to occupy their whole container.

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Solids Liquids Gases

Volume definite (fixed) definite take container’s

Shape definite take container’s shape but may not fill it

occupies whole container

Compressibility nil almost nil large

Density large large small

Table 1: Differences between solids, liquids and gases

The Ideal Gas equationClearly if the volume of a gas is decreased it's pressure increases. Mathematically this can be represented as:

V 1P

Boyle in 1962 stated this law as, “The volume of a fixed mass of gas is inversely proportional to it's pressure, provided the temperature remains constant”. Oddly enough this is known as Boyle's Law.

Also it is clear that when temperature is increased the volume increases. Mathematically, this is expressed as:

V TCharles’ Law states, “The volume of a fixed mass of gas is proportional to it’s absolute temperature (Kelvin) provided the pressure remains constant.”

COMBINING THESE RELATIONSHIPS:

V TP

or PV T

PV=kT

PVT

constant

so,P VT

P VT

P VT

=...= P VT

1 1

1

2 2

2

3 3

3

N N

N

Using the ideal gas equation to determine the RMM of gasesThe ideal gas equation states:

PV = nRTwhen n = number of moles of gas

However number of moles mass of gasRMM of gas

mM

So substituting PV = mRTM

,

5


rearranging, M = mRTPv

Also if the density of the has is known then = mv

Substituting density into previous gives, M = RTP

REMEMBER: When using R, all units must be standard, i.e. Nm-2 or Pa for pressure, m3 for volume (not dm3) and temperature in Kelvin.

Dalton’s Laws of Partial FractionsDalton considered the pressures applied by the constituents of a mixture of gases. E.g. Air is approximately 4/5

ths nitrogen and 1/5ths oxygen. \4/5

ths of the air pressure is due to nitrogen and 1/5

th due to the oxygen.

The contribution that one gas makes to the total pressure is known as it’s partial pressure. Partial pressure of a gas is the pressure that each gas would exert if it alone occupied the container.

Dalton summarised this by, ”In a mixture of gases which do not react chemically, total pressure of the mixture is the sum of partial pressures of the constituent gases.”.

i.e. P = P + P + P +...+PTOTAL 1 2 3 N

where P1...N are partial pressures.

E.g.: 3.0dm3 of Co2 at pressure 200kPa and 1dm3 of N2 at a pressure of 300kPa are introduced to a 1.5dm3 vessel. What is the total pressure in the vessel?

V1=3dm3 P1=200 kPa (For CO2)

V2=1.5 dm3 P2=?

P1V1=P2V2 => P2=400 kPa

V1=1 dm3 P1=300 kPa (For N2)

V2=1.5 dm3 P2=?

P1V1=P2V2 => P2=200 kPa

\ Total pressure in vessel = 600 kPa

Definition of a mole fractionIf two gases are present in a mixture then we can represent the number of atoms of each gas by saying:

Total number of atoms = n(a) + n(b)where n(a/b) is the number of atoms in gas a/b

A more accepted way of expressing this is in moles:

Total number of moles = m(a) + m(b)where m(a/b) is the number of moles in gas a/b

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The mole fraction of a gas A is the number of moles of A in the mixture divided by the total number of moles of all gases present in the mixture.

X = N(N + N )

AA

A BX = N

(N + N )B

B

A B

From this it follows that the partial pressure of a gas is the product of the mole fraction and the total pressure of the gas mixture.

I.e.: PX = PA A

where P = Total pressure of mixtureXA = Mole fractionPA = Partial pressure of A

Avogadro’s Law

Molar volumeThe French chemist Gay-Lussac studied the way in which gases reacted together. He found that “the volumes of the reacting gases measured at the same temperature and pressure are in a simple ratio to one another and to the volumes of any gaseous products”.

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

1 vol 3 vol 2 vol

Avogadro then suggested that “equal volumes of gases must contain equal numbers of molecules”. This is known as AVOGADRO’S LAW.

H2 (g) + Cl2 (g) -----> 2HCl (g)

1 vol 1 vol 2 vol1 mol 1 mol 2 mol

However, if equal volumes of gases contain equal numbers of molecules then it must be possible to work out the volume of 6x1023 molecules. Then it could be said that one mole of gas occupies a certain volume; this was found to be 22.414 dm3 at stp (0°C, 1 atm), or 24.056 dm3 at rtp (20°C, 1 atm).

E.g.: When 75 cm3 of the mixture of hydrogen and ethane was burned in an excess of oxygen, the volume of CO2 produced was 60 cm3 (all volumes stp). Find the percentage by volume composition of hydrogen and ethane.

C2H6 + 31/2 O2 -----> 2CO2 + 3H2O

60 cm3

1 mole of C2H6: 2 moles CO2

1 volumes of C2H6 : 2 volumes of CO2

30 cm3 of C2H6: 60 cm3 of CO2

so, ethane + hydrogen = 75 cm3, ethane = 30 cm3 \ hydrogen = 45 cm3

\ ethane = 40% hydrogen = 60%

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Graphical representation of speeds of molecules in a gasIf a sample of gas is analysed and the number of molecules with a specific speed recorded the graphs below are obtained. As temperature increases, speed increases.

8

T1


MODULE P9: THE SOLID STATE

Metallic structures

The hexagonal close packed structureIf atoms are thought of as small spheres then when they are packed together they line up as shown below:

It can be seen above that with the exception of the edges each sphere is surrounded by six others, thus forming a hexagonal shape. This is known as hexagonal close packing.

However, this is only one layer of the crystal. There are clearly millions of layers in one crystal. The subsequent atoms must sit in the gaps between the atoms in the first layer.

There is now, however, a problem, since the third layer can fit into two positions. Firstly, it can go directly above the first layer.

If the first layer is thought of as layer A and the second as B then the general pattern for all layers is ABABAB etc.

The general shape of this crystal is hexagonal and the crystal unit is known as hexagonal close packed.

9

T2

velocity


Each metal atom is surrounded by 12 other atoms and thus has a co-ordination number of 12.

The face centred cube structureSecondly, the third layer may be placed directly above the holes left by the first two layers. This structure has a different make-up of layers than the previous in that the third layer is not above the first. Thus the layers are built in an ABCABC fashion.

Careful examination (!!) of this structure shows that each unit cell is in fact a cube with an atom at each face.

This structure is referred to as a face centred cube close packed.

The body centred cube structureNot all metals have their atoms arranged in a hexagonal close packed arrangement.

The following diagram illustrates what happens. A cube with an atom at the centre is produced.

Metallic properties

Hexagonal close packed - ABAB (hcp)E.g. Zinc and MagnesiumHard tightly packed atomsHigh density and high melting point because of tightly packed atoms and strong forces between them.

Cubic close packing - ABCABC (ccp)E.g. Aluminium, copper and lead.

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% of mole-cules

Hexagonal closepacking


Good conductor because tightly packed atoms allow larger delocalised bonds to form.Hard, high density - tightly packed atoms.

Body centred cube (bcc)E.g. Iron and alkali metalsSoft - less tightly packed.Low density and low melting point because they are less tightly packed than the others.

Ionic crystalsOne of the most common ionic compounds is sodium chloride (common salt) and this has a simple cubic structure with each Na+ ion being surrounded by 6 Cl- ions.

If the sizes of the ions are similar then a slightly different arrangement is adopted. The shape is more open (due to the ion size) and is known as a body centred cube.

Ionic compounds are hard because strong electrostatic forces exist between the ions but they are also brittle because if the crystal is deformed the ions will repel each other and the crystal will break. The strong forces between the ions also give ionic substances high melting and boiling points.

They are good conductors when in solution or when molten because the lattice is broken down and the ions are free to move through the solution when a potential difference is applied. They are soluble in polar solvents (e.g. H2O) because the positive end of the solvent will be attracted to the negative ion and vice versa.

11

6:6 co-ordination number

Body centred cube

Face centred cube

Hexagonal close packed

Face centred cubeNa+ ion

Cl- ion


Molecular crystalsEven simple molecules can exist in crystal lattices. However the forces that hold them in the lattice are much smaller.

The forces holding the iodine atoms together are covalent bonds and the forces forming the crystal are Van der Waals bonds.

E.g.: Rhombic Sulphur - S8 White Phosphorous - P4

Another molecular crystal which is very common is that of ice. In this case the cohesive force is slightly stronger than the Van der Waal’s force. It is in fact a hydrogen bond.

Giant covalent crystals

DiamondEach carbon atom is bonded to four other carbon atoms.

This structure is a giant covalent molecule because each carbon is joined to four others by covalent bonds. The structure is a repeated tetrahedron.

Diamond is extremely hard - this is due to the very strong rigid structure of the crystal; it is known as a network of covalent bonds. The co-ordination number of this structure is 4.

12

Cs+

8:8 co-ordination number

C

CC

CC

CC

C

C

C

I2 molecules

Van der Waal’s bonds

Cl-


Silicon dioxideThis is formed from SiO2, but the basic building block of the crystal is a tetrahedral SiO4 unit. For this reason the structure is similar to diamond.

Si has a co-ordination number of 4

O has a co-ordination number of 2.

The structure of graphite

The carbon is joined to 3 others by covalent bonds shown as the unbroken lines. The layers are held together by Van der Waal’s bonds shown in dotted bold lines. In between the two layers is an area of delocalised electrons shown by the thin dotted lines.

Polythene and plasticThese substances exist as long chains and in both cases the chains are held together by Van der Waal’s forces. Since these forces are weak the structures may be easily deformed or stretched especially when heated.

13

Si

O OSiO

O

O

O O

OC C C


In sulphur the S8 rings break on heating and form tangled chains, when plunged into cold water the chains do not have time to reform into rings. As a result the plastic sulphur has a degree of elasticity due to the tangled chains being able to be stretched.

MODULE P10: LIQUIDS AND SOLUTIONS

The liquid stateGases are less dense than liquids and are more compressible than liquids. Liquids and solids are not very compressible. The molecules in a liquid have fairly strong bonds but not strong enough to hold them tightly. This is unlike a gas which has very weak or no bonds.

Gases and liquids both diffuse very quickly. Also both liquids and gases exhibit Brownian motion.

In close up the molecules are generally predictable; i.e. their bonds hold them tightly in roughly the same position but on zooming out the system is chaotic due to Brownian motion of the liquid.

Molecules in a solid state vibrate about their fixed positions. As the temperature increases molecules vibrate more and more until they break the bonds holding them together. The molecules are still close together but moving freely.

As the temperature increases, the molecules move around more and more quickly. Some molecules gain sufficient energy to escape from the liquid at the surface where they create a vapour pressure.

As temperature increases the kinetic energy of the molecule increases and more and more possess sufficient energy to escape thus increasing vapour pressure.

Eventually a point is reached when the vapour pressure equals the atmospheric pressure and the liquid boils.

This is known as the kinetic molecular picture of a liquid in an open vessel.

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temp +

O

Si O


For an enclosed vessel see 10.1.4. Note the existence of saturated vapour pressure.

The temperature at which a liquid boils is determined by the external pressure.

· If the external pressure is high (for example down a mine or in a pressure cooler) a greater vapour pressure will be required for the liquid to boil. Therefore it must be heated to a greater temperature for it to boil.

· If the external pressure is lower (for example on the top of a mountain or in a plane) then the liquid will boil at a lower temperature than normal as this will still generate sufficient vapour pressure to allow the disk to boil.

Intermolecular forces within a liquidNote the forces within a liquid determine many of its properties:

non-polar molecular liquids < polar molecular liquids < H-bonded liquids < molten salts or metals

The weaker the forces within the liquid are then the more volatile it will be and the lower boiling point it will have. As forces increase, so does surface tension and viscosity. The molar heat of vapourisation is greatest when the forces are strongest.

Binary liquids solutions - the process of dissolution

Polar solvent --- Polar soluteE.g.: NaCl + H2O

The main forces in the liquid are hydrogen bonds and ionic bonds in the solute.

When the solid is added to the liquid the cations and anions must first be separated from the lattice. This is possible because the water molecule has a positive and a negative end, and thus is able to solvate the ions and dissolve the solute.

Non-polar solvent --- Polar soluteE.g.: NaCl + Cyclohexane (C6H12)

In this case the ions still must be separated from the crystal lattice. However this is not possible since the solvent is not polar and cannot solvate the ions.

15

O

HH

O

O

H

H

H

H

H

H

HH

H HHH

HO

O O

O

Cl-

Atmospheric pressure

Vapour pressure

temp +


Polar solvent --- Non-polar soluteE.g.: I2 + H2O

A non-polar solute is insoluble in a polar solvent because the solute is not able to overcome the hydrogen bonding within the solvent and thus will not be able to dissolve in the solvent.

Non-polar solvent --- Non-polar soluteE.g.; I2 + Cyclohexane (C6H12)

The forces between the solvent molecules are now weak enough to be broken by the forces holding the solute together; thus solvation can occur and dissolution takes place.

In general : LIKE DISSOLVES LIKE

Energy factors in the dissolution of an ionic saltWhen considering energy factors it is important to ask what exactly happens in this process - as very often energy changes can be subtle and easily overlooked. For example NaCl in H2O.

· The ions in the NaCl must be released from their lattice. This involves energy input and is known as the crystal lattice energy.

· The ions must be solvated (hydrated) by the water. This involves a release of energy as the ions are being stabilised by the water. This is known as hydration (or solvation).

A very soluble salt will give out heat when dissolved (i.e. small lattice energy and high hydration output).

SoapsSoaps clean because they can emulsify fats and oils, i.e. it can convert them into a suspension of tiny droplets in water. Dirt is held to fabrics by a thin film of oil or grease which must be removed before the dirt can be washed away.

Soap owes its properties to its structure. One end is a highly polar carbohydrate ion which is hydrophilic and the other end is a long hydrocarbon chain which is hydrophobic and lipophilic.

E.g. Sodium Stearate CH3(CH2)16COO-Na+

When soap is dissolved in water the long hydrocarbon chain is attracted into the oil and fat in the dirt.

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H O Na+

C C C

C C

C

O-

Na+

OC


The hydrophilic ionic heads are attracted to the water and pulls the dirt out of the fabric. Because each of the heads is of the same charge the micelles repel each other and so can be washed away.

MODULE P11: TWO COMPONENT SYSTEMS AND PHASE DIAGRAMS

EquilibriumEquilibrium between phases is the study of systems in which there is more than one phase present.

E.g.: H2O (l) <-----> H2O (g)

Here there are two phases present - the liquid and the gaseous phase.

Equilibrium between phases simply means that at any given temperature there will be a certain amount of water vapour present and also a certain amount of liquid water present.

E.g.: CaCO3 (s) <-----> CaO (s) + CO2 (g)

Here there are three phases - two solid and one gaseous.

Miscible liquidsIf a mixture of two miscible liquids is studied, the total vapour pressure of the mixture is related to the amount of each component present.

If A’s molecules want to leave the surface they will be prevented from doing so since B’s molecules are also at the surface, and so realistically the vapour pressure of A will be reduced.

Let the total number of molecules present = 30Let the number of molecules of A = 20

\ A’s vapour pressure will be 2/3rds of its pure value

\ New vapour pressure of A = PoA x 2/3 ( Po

A = V.P. of pure A)

New vapour pressure = original vapour pressure x (molecules of A/total mols)

17

C

C


However real life does not work with molecules but with moles...

Vapour pressure of A in solution = moles of Atotal moles

pure vapour pressure of A

Vapour pressure of A = mole fraction of A P

Total vapour pressure = X P X PAo

A Ao

B Bo

\

\

This is known as Raoult’s Law.

Raoult’s LawThis may be shown graphically.

The graph shows that as the mole fraction of A approaches 1 the vapour pressure is directly proportional to the mole fraction.

When the composition A=0 and B=1 so B’s vapour pressure is decreasing because the amount of B is decreasing.

If a graph of boiling point of mixture is plotted against composition the graph slopes in the opposite direction than the vapour pressure composition graph.

This is for an ideal mixture. E.g. propan-1-ol and propan-2-ol

18

C

VP of A

Liquid B

Total VP of mixture

VP of B

Vapour pressure

Liquid A


However not all mixtures are ideal ( e.g. ethanol and cyclohexane)

The ethanol is polar and has stronger intermolecular forces than the cyclohexane. When mixed the process weakens the intermolecular forces and lowers the boiling point of the mixture so the line will curve downwards.

Eutectics and alloys

19

Po

A

Po

B

0

Mole fraction of A

---> 1

Ideal mixture

0

Mole fraction of A

---> 1

Boiling point

BP of mixture

BP of pure BBP of pure AC

OHpropan-1-ol

propan-2-ol

C

C

C OH

CC

C

C


The molecules in a liquid are held together by weak Van der Waals or hydrogen bonds. As they loose heat energy they loose kinetic energy and bonds become stronger and are held in a crystal lattice. The freezing point of a liquid is the point where freezing and melting are in equilibrium.

At A the liquid is cooling; at B it is freezing. The temperature does not change because all the energy is used in bond formation and at C the liquid is cooling further.

Cooling curves for mixtures of volatile liquidsIn winter salt is scattered on the roads to prevent the water freezing, this is because if an impurity is added (salt) to a pure liquid (water) then the freezing point of the liquid will be lower; i.e. the frost must be heavier to freeze the roads.

Similar depressions of freezing points occur with other mixtures e.g. solder. Solder is used in the electronics industry because of its low melting point and conductivity. Solder is an alloy of lead and tin.

Pure lead melts at 327oC and pure tin melts at 232oC. However a mixture of the two melts at a lower temperature than either of them. This is due to the atoms of both compounds preventing each other from forming an ordered lattice; i.e. pure solid. Thus the temperature needs to fall lower to attain more order.

Mixtures of lead and tin have cooling curves like X and Z .

X is a mixture of lead and tin which is richer in lead than tin and at T 1 there is a kink in the curve.

Since the composition is rich in Pb, at T1 crystals of pure lead begin to form. This shows the rate of cooling and gives a kink in the cooling curve.

Curve Z, which is richer in tin has a kink at T3 because tin is crystallising at this point and slowing the cooling down.

However curve Y has a curve which is characteristic of a pure substance. This would suggest that at Y’s composition both the lead and the tin are crystallising at the same time, this is true although they do not crystallise out as pure elements but as a mixture.

20

liquid richer in A than B

Boiling point

Non-ideal mixture

vapour

OH C C C

C

CC


This is because at this composition the atoms of lead and of tin are able to arrange themselves in such a way that they form a lattice of lead and tin crystals. This is known as a eutectic mixture and the plateau at T2 is the eutectic halt.Note curves X and Z also have a eutectic halt. This is because as the richer component separates the remaining liquids will eventually reach the eutectic composition and will crystallise showing the plateau.

SummaryT1 : crystals of lead separate

T2 : eutectic temperature

T3 : crystals of tin separate

Using cooling curves of various compositions a plot of temperature of crystallisation against composition may be achieved..

A: Melting point of pure lead

B: Melting point of pure tin

C: Eutectic point

T1 Temperature of lead separation in X

T2: Eutectic temperature

T3: Temperature of pure tin separation in Z

In region ADC, solid lead and liquid mixture are present. In region CEB, solid tin and liquid mixture are present. Region DC to axis solid lead and solid mixture present. Region CE to axis, solid tin and eutectic solid present.

21

X 25% Sn 75% Pb

Y 62% Sn 38% Pb

Z 87% Sn 13% Pb

T1

T2

T3

BP of pure A

liquid

0

Mole fraction of A

---> 1

BP of pure B

liquid richer in B than A


MODULE P12: THERMOCHEMISTRY

IntroductionThermochemistry deals with the relationship between chemical reactions and heat. Many chemical reactions are carried out solely to produce heat (e.g.: burning coal), the heat produced may also be used to power mechanical objects. Thus an understanding of thermochemistry is important as more and more emphasis is placed on the conservation of energy.

The law of conservation of energyThe work of many scientists and engineers has shown that energy, like matter, cannot be created or destroyed in chemical reactions. However energy may be transferred from one body to another and from one form to another. (E.g. a bunsen heating a beaker --- chemical to heat)

The law of conservation of energy states that “energy may not be created or destroyed but changed from one form to another”

Enthalpy and enthalpy changesIn a previous experiment temperature changes were recorded for the dissolution of various salts, lost changes in a lab take place under the approximately constant pressure of the earth’s atmosphere. In the experiment heat changes were being measured - these are also known as enthalpy changes.

The thermal energy gained or lost when a change takes place under constant pressure is called the enthalpy change (DH). The Greek letter D means change and H is the symbol for enthalpy (heat) therefore the symbol DH represents the difference between the final and initial enthalpy.

DH = H i - H f

Exothermic and endothermic reactions and the symbol of DHAn exothermic reaction is one which gives out heat, the enthalpy change is conventionally written as -DH (showing heat is lost to the surroundings). Secondly an endothermic reaction is one which absorbs heat, the enthalpy change is written as +DH (showing heat is taken in from the surroundings).

Standard conditions and statesThe values of changes of enthalpy that accompany physical and chemical changes depend upon the conditions under which the change takes place. Therefore to solve the problem a set of standard conditions has been defined so that results may be compared easily.

The International Union of Pure and Applied Chemistry (IUPAC) recommends using a pressure of 105 Pa (Nm2) or 1 bar as standard pressure although most tables are still using the old standard pressure of 1 atmosphere (105 Pa = 0.987 atm) so the difference is negligible.

22


The standard state of a substance is the state in which it exists under standard conditions. If all the reagents and products are in their standard states then the enthalpy change that accompanies a reaction is called a standard enthalpy change DHo for the reaction.

When writing chemical reactions in this way state symbols must be used.

Hess’s LawIn 1840 the Swiss born chemist Germain Henri Hess stated an empirical law. The thermal energy given off or absorbed in a given change is the same whether the change takes place in a single step or multiple steps.

Graphite + O2 -----> CO2

C (s) + O2 (g) ------> CO2 (g) DH = -393.5 kJ

This may also occur in two steps:

(i) C (s) + 1/2O2 (g) -----> CO (g) DH = -110.5 kJ

(ii) CO (g) + 1/2O2 (g) -----> CO2 (g) DH = -283.0 kJ

(i)+(ii) gives

C (s) + 1/2O2 (g) + CO (g) + 1/2O2 (g) ----> CO (g) + CO2 (g)

C(s) + O2 (g) -----> CO2 (g) DH= DH1+DH2 = -393.5 kJ

This may be summarised as follows:

Often certain reactions are difficult to carry out under lab conditions (unstable products) so to determine the enthalpy of that reaction other reactions are studied and the enthalpy changes combined as in the above example.

Often the most common enthalpy to be calculated is the enthalpy of formation.

H2 (g) + Ca (s) + O2 (g) -----> Ca(OH)2 (s) DH = -986 kJ

This would be very difficult and dangerous to carry out in a lab so Hess’s law is used.

23

T1

Temperature

327oC140oC Temperatu

re

Pure Pb


E.g.: Calculate DHrxn for the following:

Sn (s) + Cl2 (g) -----> SnCl2 (s)

using

Sn (s) + 2Cl2 -----> SnCl4 (l) DH = -545.2kJ

SnCl2 + Cl2 -----> SnCl4 (l) DH = -195.4kJ

Sn + 2Cl2 -----> SnCl4 DHrxn = -545.2

SnCl4 -----> SnCl2 + Cl2 DHrxn = 195.4

Sn + 2Cl2 + SnCl4 -----> SnCl4 + SnCl2 + Cl2 DHrxn = -349.8

Sn (s) + Cl2 (g) -----> SnCl2 (s)

If a reaction is exothermic, this means it is favourable and the product formed will be very stable.

H2 + ½O2 -----> H2O DHrxn = -286kJ

This reaction is actually explosive.

Limitations of DH values1. DH values give no indication of the reaction rate.

2. DH values only refer to the standard states and conditions, however, the reaction may behave differently with the addition of the catalyst or high pressure.

Bond EnergiesThis is the average energy needed to break a given type of bond. In methane, there are four C-H bonds. To calculate the bond energy of one C-H bond, the energy required to break all four bonds is measured and the result is divided by four. This gives the average bond energy for a C-H bond, or bond energy term, which is denoted by E.

\E(C - H) = 415.5kJ

Bond dissociation and bond formationBond dissociation is the energy required to dissociate a molecule into its atoms.

H2 (g) -----> 2H (g)

This is endothermic, DH is positive.

Bond formation is the energy released when bonds are formed.

2H (g) -----> H2 (g)

This is exothermic, DH is negative.

24


Uses of bond energiesBond energies may be used for:

1. comparing the strength of bonds.

2. understanding structure and bonding.

3. estimating enthalpy changes for reactions.

4. understanding the mechanisms or pathways of chemical reactions.

Estimating enthalpy changes for reactionsE.g.: Could the following reaction be used as rocket fuel?

N2H4 (g) + 2F2 (g) -----> N2 (g) + 4HF (g)

Bonds Broken F - N F - F N - N

E 340kJ 158kJ 163kJ

no. 4 2 1

åE 1560kJ 316kJ 163kJ åE = +2039kJ

Bonds Broken N º N H - F

E 945kJ 565kJ

no. 1 4

åE 945kJ 2260kJ åE = -3205kJ

Table 2: Dissociation energies of bonds

DHrxn = 2039 - 3205 = -1166kJ

This is exothermic, so the fuel may be used as rocket fuel.

Determining the bond energy of methaneC (s) + 2H2 (g) -----> CH4 (g) DH = -75kJ

C (s) -----> C (g) DH = +715kJ

½H2 -----> H DH = 218kJ

872 + 715 + 75 = 4(C - H)

4(C - H) = 1662

C - H = 417

E(C - H) = -416kJmol-1

25

Reagents

CO (g) + 1/2

O2 (g)

T3 C-

393.5 kJ

ZX-

283.0 kJ

DProducts

+ N FF

BAT2

0

% avg mass of tin

---> 100

HF

C (s) + O2 (g)

-110.5 kJ


Bond energies may also be unreliable.

The C -H bond in CHCl3 will be weaker than the one in CH4 because of the polarity of the molecules.

Thus, if a measurement were to be taken on one, the value would not agree with the accepted value.

This is also the case in benzene, C6H6.

Calculations would suggest that benzene contains three double bonds. Experimental values would suggest otherwise. In fact, the structure of benzene contains a mixture of double and single bonds which account for the discrepancy in the experimental values.

The Born-Haber cycleWhen sodium metal is added to chlorine gas, the reaction is spontaneous and often dangerous. Energy is released, but where does the energy come from? The equation for the formation of NaCl (s) is:

Na (s) + ½Cl2 (g) -----> NaCl (s)

This reaction may be broken down into five steps:

1. Vapourisation of sodium metal.Na (s) -----> Na (g) Enthalpy of sublimation DH = +91kJmol-1

2. Dissociation of Cl2 molecules into atoms.½Cl2 (g) -----> Cl (g) Enthalpy of dissociation DH = 121kJmol-1

3. Ionisation of sodium vapour.Na (g) -----> Na+ (g) + e- Ionisation energy/enthalpy of ionisation

DH = 496kJmol-1

26

F

FF

FH

FF

FF H Cl

H H

C H

+

N

N

HF H

N

ClC

HCl

HH C C

H H

C C


4. Addition of electron to Cl vapour.Cl (g) + e- -----> Cl- (g) Electron affinity DH = -349kJmol-1

5. Combination of sodium and chlorine.Na+ (g) + Cl- (g) -----> NaCl (s) Lattice energy DH = -771 kJmol-1

DHf = DHnnå

1

5

= 91 + 121 + 496 - 349 - 771 = -411kJmol-1

Representing the Born-Haber cycle on an energy diagram

DHf + 91 + 121 + 496 = 349 + 771

DHf = 441

Endothermic, so DHf = -441kJmol-1

Using the above relationship, it is possible to calculate any of the enthalpies of any step if DHf is known.

Comparison of lattice energies for group 1 and 2 halidesThe experimental value is calculated from the Born-Haber cycle. The theoretical value is obtained by considering the attractions and repulsions between the ions. The latter method assumes an even distribution of charge on each ion.

The theoretical values for alkali halides agree fairly well with the experimental value. This is because, in these compounds, the simple model of the ionic lattice (i.e. spherical ions and even distribution of charge) is indeed present.

In the silver halides, the picture is different. The experimental values show that a stronger bond is present. This is due to the reduced difference in electronegativity values, causing an uneven distribution of charge. This

27

H

H

CC

C

CH

H

C

C

H H

H

HC

C


leads to a decrease in ionic character and an increase in covalent character. The Ag+ ion has a greater ability to pull electrons towards itself - this strengthens the bond.

28


MODULE P13: REACTION KINETICS

IntroductionChemical reactions are all around us and very often we need to know or are told how long a reaction will take.

E.g.: Baking, dying, pizza making

The rate of a reaction can be studied in great detail and the information obtained is used to determine how a reaction proceeds, and thus the conditions for optimum production are obtained.

The rate of a reaction is defined as follows:

“The rate of change of amount or concentration of a particular reactant or product”

Rate of reaction =Change in concentration

Time takenUnits: moldm-3s-1

This equation gives the average rate over time taken. However the reaction may proceed very quickly over the first few seconds and then gradually taper off towards the end.

Measuring reaction ratesFor example, the decomposition of hydrogen peroxide (H2O2) using MnO2 as a catalyst.

The volume of O2 generated can be plotted against time giving the following graph.

To measure the reaction rate at any given time, take a tangent to the curve and measure It to the gradient.

REM rate = Change in concentrationTime

dCdt

gradient of curve

For the average rate the line is not a tangent but a line from the origin to the required amount of time.

29

Na+ (g) + Cl (g) + e-

DH (EA of Cl2) DH (IE of

Na)


Methods of determining reaction ratesThe reaction rate may be obtained by following some property which alters as the reaction proceeds.

Titrimeteric analysisE.g.: CH3COOH(aq) + I2(aq) ---(H+)--> CH2ICOOCH2 (aq) + I- (aq) + H+(aq)

E.g.: HCOOCH3 + OH- ----> HCOO- + CH3OH

In these cases the reaction may be followed by removing and analysing small portions of the reaction mixture at intervals. The removed portions must be added to a reagent to stop the reaction (i.e. quench it). This prevents further concentration changes before the analysis is carried out. Once the mixture is quenched it may be analysed by titrating the unreacted iodine against a standard solution.

Colorimetric analysisThis is especially useful for systems in which one of the substances is coloured. The colour intensity may be followed during the reaction using a photoelectric colorimeter. From these measurements the concentration of the coloured substance may be obtained at different times.

In a colorimeter a narrow beam of light passes through the solution under test and towards a sensitive photocell. In many colorimeters it is possible to select the most appropriate colour of light. The current generated by the photocell is proportional to the amount of light transmitted by the solution. This in turn depends on the depth of colour of the substance. Thus the current is greatest when the coloured substance is most dilute.

Pressure measurementsThis technique is particularly useful for gaseous reactions which involve changes in pressure.

30

Na(g) + ½Cl2 (g)

Na(g) + Cl (g)

DH (sub) DH

(lattice)

DH (diss Cl2)


Pressure may be measured at regular intervals and the reaction rate may be evaluated.

Conductrimetric analysisMany reactions in aqueous solutions involve changes in the ions present as the reaction proceeds. Consequently the electrical conductivity of the solution will change during the reaction and this may be used to determine the change in concentrations of reagents and products over time. This is done using two inert electrodes and a conductivity meter.

Factors affecting reaction ratesWhen thinking about reaction rates, remember:

For a reaction to occur, molecules must collide.

1 Surface area: More molecules collide if there is more surface area hence the reaction rate is increased.

2 Temperature: Molecules have more kinetic energy and move faster; hence there are more collisions and faster reactions.

3 Catalyst: A compound that speeds up the reaction rate with getting used up or being changed itself.

4 Concentration: More molecules per unit volume means that the collisions rate and hence reaction rate is increased.

5 Light: The addition of light often accelerates or initiates reactions. Examples of such reactions are:

5.1 Production of starch in plants.5.2 Darkening of AgCl in photographs.5.3 Activation of vitamin D in sunlight.5.4 Tanning in skins

The effect of concentration on rate of reactionBr2 (aq) + HCOOH (aq) ------> 2Br- (aq) + 2H+ (aq) + CO2 (g)

Since Br2 is a red-brown colour, the reaction can be followed by measuring the intensity of colour. The meter is first calibrated by measuring the intensity of known concentration, hence the intensities from the reaction can be converted easily to concentration.

Reaction rate = -d[Br ]

d t2

31

photocell

meter narrow beam of light

Average

Initial

Instantaneous

NaCl (s) Na+

(g) + Cl- (g)

77191 121 349 496

Na(s) + ½Cl2 (g)


The curve has a negative slope since the [Br2] is decreasing so the rate has an overall positive value.

If we then proceed to plot the rate of reaction against [Br2] then the following the following is obtained.

This graph shows that the reaction rate is proportional to the [Br2]

Rate = k [Br2]

We now have the following general equation.

Rate = k [x]However not all reaction give the previous graph of rate proportional to [Br2] so the equation must be modified by inserting an extra term.

Rate = k [x]n where k is rate constant and n is order of reaction

Reactions may be either 0,1,2 or 3 order reactions, the order tells us how the rate is related to the concentration.

When n=0, the reaction is said to be zero order.

When n=1, the reaction is said to be first order.

32

solution

filter of diffraction grating

CH3

(g)

H3C

H3CCH3

(g)(g)


Calculating the order of a reaction from experimental dataE.g.: 2NO (g) + 2H2 (g) -----> N2 (g) + 2H2O (g)

[NO] [H2] Rate

6x10-3 1x10-3 3x10-3

6 x10-3 2 x10-3 6 x10-3

6 x10-3 3 x10-3 9 x10-3

1 x10-3 6 x10-3 .5 x10-3

2 x10-3 6 x10-3 2 x10-3

3 x10-3 6 x10-3 4.5 x10-3

Table 3: Results from experimental data (concentration experiment)

To find the order wrt [NO] we must have the concentration of the H2

constant. Next we look at the different concentrations of NO and see how the rate is affected.

\relationship is 2nd order

The order of H2 is 1.

Rate = k[NO]2

Rate = k[H2]

\ overall rate = k[NO]2[H2]

Overall order of reaction is the sum of the separate orders = 3

Finding k --> [NO]=6 x10-3 [H2]=2 x10-3 Rate=6 x10-3

6 x10-3 = k (6 x10-3)2 (2 x10-3)

k = 8.33 x104 units = mol-2dm6s-1

Uses of reaction rate studiesVery often reactions occur in several steps.

The slowest step is known as the rate determining step. This step determines the rate of the reaction and very often the mechanism of the reactions.

This is crucial to synthetic chemists as the mechanisms can be used to work out synthetic routes to new chemicals.

Activation energy and temperatureAs the temperature of a reaction increases so does the number of collisions. However not all collisions result in a reaction. The molecule must collide with a certain amount of energy. This is known as the activation energy of a reaction.

33

Reaction rate

Time

[Br2]CH3

CCC

HI+

H3C


i.e. the amount of energy required to activate the reaction. This may be summarised by drawing energy diagrams.

The greater the activation energy is then the more effect a temperature change has on the reaction rate. This is because an increase in temperature leads to an increase in kinetic energy and thus the number of collisions. This increases the rate constant.

An increase of temperature increases the number of effective collisions and increases reaction rate.

The geometry of the collision must also be correct.

Two-step reactions

The rate determining step is from A to C. The activated complexes are found at B and D. The activation energy is AB and the enthalpy change is AE.

34

reagents

products

activation energy level

DH +ve for endothermic reaction

activated complex

energy

time

x9

x4

1x10-3

x2

x3

2x10-3

3x10-3

0.5x10-3

2x10-3

4.5x10-3

[x]

2nd order

1st order

zero orderRATE

[Br2]


The above reaction is a two-step reaction. The slowest step of the reaction is known as the rate determining step which determines the rate of the reaction and very often the mechanism of the reaction.

35


MODULE P14: EQUILIBRIUM SYSTEMS

The Kinetic Molecular Picture of equilibrium(Reference pg. 31 module book)E.g.: Br (l) Û Br (g)

This is a two-phase system (liquid, vapour)

At equilibrium the colour of the contents of the gas jar remains constant. Looking at the gas jar there seems to be nothing happening however a closer look at the surface of the liquid reveals something.

Experiments show that in this system molecules are in constant motion so at equilibrium:

THE RATE AT WHICH THE LIQUID BROMINE VAPOURISES EQUALS THE RATE AT WHICH THE GASEOUS BROMINE CONDENSES.

Because the molecules are moving the equilibrium is known as dynamic equilibrium.

Static equilibrium is slightly different in that once equilibrium has been reached then no further movements of the molecules occur.

Examples of systems in equilibrium

Physical systems1. Pure solid / pure liquid = ice in water

2. Pure liquid / pure vapour = mixture of steam and water

3. Pure solid / pure vapour = Iodine crystals / vapour

4. Pure solid in contact with a solution, crystallising and dissolving together = sugar crystals in a saturated sugar solution.

36

time

energy

activated complex

products

DH -ve for exothermic reaction

activation energy level


5. Solute distributed between two immiscible solvents = Iodine in trichloroethane and aqueous Iodine.

Chemical systems1. Chemical equilibria with reactants and products in same phase = Copper

sulphate and ammonium hydroxide.

2. Chemical equilibria with reactants and products in different phases = Calcium chloride and sodium carbonate.

The distribution of a solute between two immiscible liquidsE.g.: I2 dissolved in KI (aq) and mixed in CCl4

When iodine is dissolved in potassium iodide a dark brown solution forms. Once the CCl4 is added the iodine moves into the CCl4 layer ( this is due to the movement of a solute going from a high to low concentration ). CCl4

then turns pale purple and the potassium iodide layer lightens.

Eventually when equilibrium is reached the colour of both layers is constant. This is because dynamic equilibrium has been reached. We know this because:

n The macroscopic properties remain constant (no more colour change)

n At equilibrium the microscopic processes continue but these are balanced.

n

n

n

n

n

The relationship between the concentrations of the solute in solutions

E.g. Investigating the partition of butanedoic acid between water and ethoxyethane.

37

B

D

C

E

energy

time reagents

Liquid Bromine

Gaseous Brominereage

nts

products

A


N.B. Don’t be alarmed by the names for these; all that is needed is two immiscible solvents and a solute that dissolves in both solvents.

If 1g of the solute is added to equal volumes of the solvents, shaken, and settled, i.e. to reach equilibrium the following would be observed:

Both layers were analysed for acid, and the concentration of acid recorded. The experiment was then repeated five times, and with various masses of acids.

The results were as noted below:

Experiment # Organic layer Water layer

1 0.023 0.152

2 0.028 0.182

3 0.036 0.242

4 0.044 0.300

5 0.052 0.358

6 0.055 0.381

Table 4: Results from experiment (phase difference)

If these results are plotted on a graph organic vs. water then a straight line graph through the origin is obtained.

This means that y = mx + c

or concentration in organic concentration in H2O.

ka c ida c id

e th e r

w a te r

[ ][ ]

This constant k is known as the Partition co-efficient or the Distribution ratio.

The equilibrium lawThere is a relationship between the concentrations of reagents and products at equilibrium.

For any reaction if an equilibrium mixture contains substances A, B, C and D related by the chemical equation:

aA + bB ó cC + dD

it is found that:

[C] D]A] B]

kc d

a b c

[[ [

38

Liquid Br2 remaining

Liquid Br2 condensing

Gaseous Bromine


Kc is known as the equilibrium constant. Note its units change and they must be worked out carefully on a case-by-case basis.

E.g.: Write an expression for Kc for the following equation:

H2 + I2 ó 2HI

[HI]H ] I ]

k2

2 2c[ [

E.g.: What are the units of Kc in the following reactions at equilibrium:

(i) N2 + 3H2 ó 2NH3

kNH

N Hc3

2 2

[ ]

[ ][ ]

2

3 Units: mol-2dm6

(ii) 2NO + O2 ó 2NO2

kNO

NO Oc2

2

[ ]

[ ] [ ]

2

2 Units: mol-1dm3

Calculations involving kc to remember· The equilibrium law only applies to systems in equilibrium

· kc is a constant as long as temperature remains constant

· The numerical value of kc is unaffected by changes in concentrations of either reagents or products.

· The magnitude of kc gives an indication as to the extent of the reaction; i.e. if kc is large then the concentration of the products is large. If kc is very low then the concentration of reagents is high.

· kc is an indication of how far a reaction proceeds but not how fast; the rate constant determines this.

E.g. When 1 mole of HI is allowed to dissociate in a 1 dm3 vessel only 0.78 moles of HI are present in equilibrium. What is the equilibrium constant at this temperature for the reaction?

2HI ó H2 + I2

kH ] I ]

[HI]c2 2

2[ [

concentration HI : 1 concentration HI : 0.78at start I2 : 0 at equilibrium I2 : 0.11

H2 : 0 H2 : 0.11

k] ]

[0.78] (no units)c 2

[ . [ ..

011 0110 0199

ON NO ACCOUNT MUST NUMBERS BE FILLED INTO AN EQUATION WITHOUT FIRST WRITING THE EQUILIBRIUM EQUATION FOR THE REACTION.

E.g.: Suppose 2 moles of hydrogen and 1 mole of iodine are mixed together in a 1 dm3 vessel at 440oC. How many moles of HI, hydrogen, and iodine will be present at equilibrium if kc=0.02 ?

39


H2 + I2 ó 2HI

concentration H2 : 2 concentration H2 : 2-xat start I2 : 1 at equilibrium I2 : 1-x

HI : 0 HI : 2x

kH ] I ]

[HI]c2 2

2[ [

à0.02] ]

[2x]2 [ [2 1x x

à 0 08 2 32 2. x x x

à 0 92 3 2 02. x x

à \ x = 2.33 or 0.93

(but since x<1) 0.93 moles must react

\ 0.07 moles of iodine, 1.07 moles of hydrogen, 1.86 moles of HI

E.g.: The equilibrium constant for the reaction

2NO2 ó N2O4

is 200 mol-1dm-3 at 298K.

(i) What is the expression for kc?

(ii) If the concentration of N2O4 in the equilibrium mixture is 2x10-2 M, what is the concentration of NO2.

(i) kN O

[NO ]c4

22 200 2[ ]

(ii) 2002 10 2

[NO ]22

[NO ]2002

2

2 101 10

24

[NO ] moldm2-30 01.

Gaseous mixturesGaseous mixtures are rarely measured in concentrations (moldm-3). Instead they are measured in terms of partial pressures. For this reason equilibrium systems involving gaseous reagents and products have slightly different equilibrium equations:

aA (g) + bB (g) ó cC (g) + dD (g)

[pC] [pD][pA] [pB]

kc d

a b p

Note if any solids appear in the equilibrium equations ignore them as they do not exert a partial pressure.

40


Factors affecting equilibriumIf a system in equilibrium is subjected to change, processes will occur which tend to counter at the change imposed. This is known as Le Chateliers principle.

Changing concentration by… (kc remains constant)

partial removal of reagent product yield decreases

partial removal of product product yield increases

addition of more reagent product yield increases

Table 5: Factors affecting equilibrium (concentration)

Changing pressure in… (kp remains constant)

systems with mole increase pressure increase yield decrease

pressure decrease yield increase

systems with mole decrease pressure increase yield increase

pressure decrease yield decrease

systems with no change product unchanged yield unchanged

Table 6: Factors affecting equilibrium (pressure)

Changing temperature in…

exothermic reactions temperature increase decrease of kc yield decrease

temperature decrease increase of kc yield increase

endothermic reactions temperature increase increase of kc yield increase

temperature decrease decrease of kc yield decrease

Table 7: Factors affecting equilibrium (temperature)

Action of catalyst

Equilibrium will be reached more quickly but it has no effect on the product yield or on kc.

41


MODULE P15: ACID-BASE EQUILIBRIA

Introduction to acidic systems (hydrogen ion theory)Previously the definition of an acid was given as:

Acids provide hydrogen ions H+ (aq) in solutions.

For example, when a molecule like HCl dissolves in water, it dissociates into:

HCl + water à H+ (aq) + Cl+ (aq)

However studies show that the active ingredient of an acidic solution is not a simple H+ ions but in fact an OXONIUM or HYDRONIUM ion H3O+.

Thus when HCl dissolves in water, a reaction occurs:

HCl (aq) + H2O (l) à H3O+ (aq) + Cl- (aq)

Other acids behave in the same way.

H2SO4 (aq) + 2H2O (l) à SO42- (aq) + 2H3O+ (aq)

HNO3 (aq) + H2O (l) à NO3- (aq) + H3O+ (aq)

However not all reactions occur in aqueous mediums. Chemists often carry out reactions in non-aqueous solvents such as ether or liquid ammonia. In these cases, the above theory, although useful, does not hold; a new theory was therefore proposed to deal with the problem.

42

HH

O

H H

OH O

S

O OH

++Acid distributed between two layers

Ether layer

Water layer

I2 leaving CCl4

I2 entering CCl4

KI

CCl4 Su

rface interface

I2 in KI

Dark

Pale

KI

CCl

4

At start

At equilibrium

Liquid Bromine

H++

O HH

H

+ O

HH

H+

O-

O S O

O- O


Bronsted-Lowry theoryACIDS ARE PROTON DONORS

BASES ARE PROTON ACCEPTORS

This fits in with the earlier theory but can be applied to other areas of chemistry as well.

HCl (a) + NH3 (a) à NH4+ (a) + Cl- (a)

Note: In any acid-base reaction, there is always a transfer of protons.

Conjugate pairsLooking carefully at the change in the acid it can be seen that the acid is converted into a base and a proton.

H-Cl (a) ó H+ (a) + Cl- (a)Acid Proton donor Base Proton acceptor

In this way every acid has a conjugate base and equally every acid has a conjugate acid.

Example 1:CH3COOH ó CH3COO- + H+

Acid (proton donor) Conjugate base (proton acceptor)

Example 2:H2SO4 ó SO4

2- + 2H+

Acid (proton donor) conjugate base (proton acceptor)

Example 3:NH3 + H+ ó NH4

+

Base (proton acceptor) Conjugate acid (proton donor)

The pH scaleAcids and bases are often spoken of as being strong or weak and have corresponding pH values.

How is pH calculated?Acids which are strong have high degrees of dissociation:

i.e. HCl à H+ + Cl-

\They will have high concentrations of H+ ions. The pH scale is related to the concentration of H+ by the following equation:

pH log[ ]H

43

O

H

H

H

+ N

O

O

O- O H HN

O

O O


It would seem easier to refer to the strength of an acid by quoting its [H+], but acids vary from 10-15 M to 10 M. So with such a wide range chemists introduced a log scale. The scale was also made negative so that most pH’s would end up as positive values.

E.g.: Work out pH values for the following concentrations:

10-12 M [H+] : pH=12

10-11 M [H+] : pH=11

10-10 M [H+] : pH=10

Note that the concentration increases, so the pH value gets smaller. This is in keeping with the earlier theory that the strong acids have a low pH.

The ionic product of waterWater molecules take part in an equilibrium reaction:

H2O (l) ó H+ (aq) + OH- (aq)

kH ] OH ]

[H O]c

+ -

2

[ [, or k [H O] H ] OH ]c 2

+ - [ [

However, the equilibrium lies far to the left, which for our purposes means that the concentration of H2O is constant. Therefore if kc is a constant, and the [H2O] is constant, then the product must also be constant. The new product is labelled kw.

k H ] OH ]w+ -[ [

At 298oK, kw = 10-14 mol2dm-6,

but since [H+] = [OH-],

10-14 = [H+]2

10-7 = [H+]

\pH=+7

In the same way to pH,

pOH log[ ]OH

In the above example, kw=10-14 mol2dm-6. When writing equilibrium values, it is often more sensible to quote these as pkw;

pkw = -log(kw) = 14

kw = [H+][OH-]

-log(kw)= -log[H+] + (-log[OH-])

pkw = pH + pOHThis relationship is true for aqueous solutions as well as H2O.

Remember pH changes with concentration and so it is really only a rough guide to the strength of an acid.

44


Strength of acids and basesAcids are molecules containing one or more hydrogen atoms bonded to an atom of higher electronegativity. The stronger acids can gives off H+ ions more easily than the weaker ones.

A base’s strength depends on how easily it can attract protons and the number of lone pairs it has.

Acidic strengthH2O Strength increasing

CH3COOH

H2SO4

HClO4

Acid # double bonds

Water (H2O) 0

Ethanoic acid (CH3COOH) 1

Sulphuric acid (H2SO4) 2

Hypochloric acid (HClO4) 3

Table 8: Number of double bonds in various acids

THE MORE DOUBLE BONDS PRESENT, THE GREATER THE STRENGTH OF THE ACID

Why does the double bonding increase the strength of an acid?

The more double bonds present around the central atom (in the example to the left it is the second carbon), the more positive the central atom gets, and therefore more negative the oxygen from the OH becomes. This in turn makes the H more positive and so easier to dissociate.

45

H

H

H

Sulphuric acid

Hypochloric acid

Water

Ethanoic acid

H

Proton acceptor (acid)

Proton donor (acid)

Base

N

H H

H

+

N

HH

H

Cl

Cl- H

H +

+

O

O

O OH

OHO

C

C Cl

H

H

H


a-valuesIf an acid is strong then a large proportion of its molecules will dissociate into ions. If an acid is weak then only a small proportion will dissociate into ions. If a is termed as the degree of dissociation then a strong acid will have a large a value, and a weak acid a small a-value.

Since pH is a relatively feeble way of representing the strength of an acid, chemists had to come up with an alternative method of comparing the strengths of acids quantitatively.

Dissociation constantsThe acid HA will dissociate into the following ions:

HA ó H+ + A-

The equilibrium constant for this is:

kH ] A ]

[HA]c

+ -

[ [

Where the concentrations of the reagents changes the value of kc remains constant. Chemists now have a constant which is unaltered regardless of concentration. The constant kc was renamed ka to indicate the presence of acids and the equation became…

kH ] A ]

[HA]a

+ -

[ [

The constant ka is known as the dissociation constant.If ka is kept high then the concentration of H+ and A- is high and so the degree of dissociation, a, is high and this would refer to a strong acid.

If ka is low then the opposite is applied; a weak acid is present.

The stronger an acid is then the smaller the value of pka will be and the weaker it is, the higher the pka will be where:

pk a log[ ]ka

The same theory applies to bases.

BOH ó B+ + OH-

kB ] OH ]

[BOH]c

+ -

[ [

Kc remains unaltered when concentration changes.

kB ] OH ]

[BOH]b

+ -

[ [

A high kb indicates a strong base and a low kb denotes a weak base.

The stronger an acid is then the larger ka is and so the equilibrium is more to the right hand side of the equation.

E.g.: HNO3 ó H+ + NO3- ka = 40 moldm3 (dissociates a lot)

46


E.g.: CH3COOH ó H+ + CH3COO-

ka = 1.7 x 10-5 moldm-3 (dissociates a little - high pH)

Calculation of pH values for weak acids and basesIn order to calculate the pH value of a weak acid, two factors are needed:

(i) concentration of the solution, and(ii) dissociation constant of the acid.

CH3COOH ó H+ + CH3COO-

At start c

Used / produced x x x

At equilibrium (c-x) x x

kH ] CH COO ]

[CH COOH]a

+3

-

3

[ [

At equilibrium,

kx

c xa

2

But x = [H+], and \ x = ca

Substituting this into previous equation,

kc

c(1 x)a

2 2

a

kc1 xa

2

a

But since this is a weak acid (a << 1), the expression 1-a » 1 (slightly dodgy bit)

k ca2 a

But a 2+

2

Hc

[ ]

\ Substituting a in,

kH

ca

+

[ ]2

k c Ha+[ ]2

\ [ ]H k c+a (from this the pH can be calculated)

A similar arrangement holds for the calculation of the pH of a weak base.

Acid-Base indicatorsIndicators for acid-alkali titrations are simply weak acids which change colour on conversion to their conjugate base.

47


E.g.: HIn ó H+ + In-

acid colour conjugate base colour

or methyl orange

HMe ó H+ + Me-

red yellow

When H+ ions are added to the system the equilibrium shifts to the left producing a red colour. When alkali is added to the system, the equilibrium shifts to the right giving a yellow colour.

The end point of titrationsThe end point of a titration is the point at which the titration is stopped. The equivalence point for the two reacting solutions is the point when the reaction is complete; i.e. all H+ ions and OH- ions have been neutralised. At this point, [HIn]=[In-], and the colour produced is midway between the colour of HIn and In-.

kH ] In ]

[HIn]a (indicator)

+ -

[ [

At the equivalence point [HIn]=[In-]

\ Substituting this in gives:

k H ]a (indicator)+[

pk pHa (indicator)

Salt hydrolysisIons of certain salts react when dissolved in water to produce either an acidic or alkaline solution. This is known as Salt hydrolysis.

· In normal salts derived from strong acids and weak alkalis, the cations will hydrolyse to give acidic solution. E.g.: NH4Cl

· In normal salts derived from weak acids and strong alkalis, the anions will hydrolyse to give alkaline solutions. E.g.: Na2CO3

· In normal salts derived from weak acids and weak alkalis, both cations and anions will hydrolyse to give either acidic or alkaline solutions. E.g.: CH3CO2NH4

· In acid salts derived from strong acids and alkalis, the anions will hydrolyse to give acidic solutions. E.g.: NaHSO4

· In acid salts derived from weak acids and strong alkalis, the anions will hydrolyse to give alkaline solutions. E.g.: NaHCO3

Some of this is summarised in the table below:

Strong acid Weak acid

Strong alkali Acidic solution (NaHSO4) Alkaline solution (Na2CO3)

Weak alkali Acidic solution (NH4Cl) Either (CH3CO2NH4)

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Table 9: Results of various acid/alkali combinations in salt hydrolysis

Buffer solutions

What are they?A buffer solution consists of either:

(a) A solution of weak acid in the presence of one of its salts (e.g. CH3COOH and NaCH3COO). This is also represented as:CH3COOH ó CH3COO- + H+

Conjugate acid Conjugate base

(b) A solution of a weak base in the presence of one of its salts:NH3 ó NH4+Base Conjugate base

Why does they work?Take a for example a solution of a weak acid with it’s salt:

HA ó H+ + A- (The acid is partly dissociated)

MA à M+ + A- (The salt is fully dissociated)

The solution has a relatively high concentration of H+ (ionised acid) and A-

(base).

If acid is added to the solution, the H+ ions are mopped up by the A- ions, and thus the pH changed only very slightly.

If alkali is added to this system, the OH- ions combine with the H+ ions to form water. This reduces the concentration of H+ ions in the buffer and so to compensate the HA dissociates to restore the equilibrium. Thus the pH value is only slightly altered.

By having reserves of H+ & A-, changes in pH resulting from the addition of acid or alkali are minimised.

REDOX EQUILIBRIAIn a previous unit, it was concluded that in a redox reaction a transfer of electrons takes place. Redox equilibria is concerned with harnessing this transfer or movement of electrons and measuring it as a p.d. or voltage, i.e. electricity.

In a redox reaction, one species is losing electrons whilst the other is gaining them.

Zn (s) + Cu2+ (aq) à Zn2+ (aq) + Cu (s)

Zn (s) à Zn2+ (aq) + 2e- (oxidation)

Cu2+ (aq) + 2e- à Cu (s) (reduction)

In order to generate an electric current from this reaction, the two halves of the equation must be physically separated.

49

O

O


In order to keep both salts balanced in terms of their charge, a salt bridge is set up between the 2 beakers.

Standard electrode potentialVoltages of electrochemical cells are always measured between 2 layers. This poses a problem if a voltage is required which refers only to one half of the equation.

The problem is solved by keeping one of the layers constant and measuring the difference between the constant and the others. The standard electrode used for this purpose is the hydrogen electrode.

The reaction which takes place is:

H+ (aq) + 2e- ó H2 (g)

Under the conditions described the EMF or voltage is set at

E .SHE 0 000V

50

- -

-

--

-- -

- +++

++ +

+++++++

+++++

+++ +

+ ++ +

--- +

Oxidation

Reduction

OOCC

H

d+d-d

-

d+H

H

HO

S

O

O

-

-

-

-


Using electrochemical cellsIn order to standardise the way in which the cells are written, IUPAC set up a convention for doing so. Refer to the module book, page 46 part iii for further information.

Applications of standard electrode potentialsBelow is a list of standard electrode potentials for several elements:

Li+ (aq) | Li (s) Eo = -3.04 V Li+ + e- ó Li

K+ (aq) | K (s) Eo = -2.92 V K+ + e- ó K

Ca2+ (aq) | Ca (s) Eo = -2.80 V Ca2+ + 2e- ó Ca

Na+ (aq) | Na (s) Eo = -2.71 V Na+ + 2e- ó Na

Mg2+ (aq) | Na (s) Eo = -2.37 V Mg2+ + 3e- ó Mg

Al3+ (aq) | Al (s) Eo = -1.66 V Al2+ + 2e- ó Al

Zn2+ (aq) | Zn (s) Eo = -.076 V Zn2+ + 2e- ó Zn

Fe2+ (aq) | Fe (s) Eo = -0.44 V Fe2+ + 2e- ó Fe

Ni2+ (aq) | Ni (s) Eo = -0.27 V Ni2+ + 2e- ó Ni

Pb2+ (aq) | Pb (s) Eo = -.013 V Pb2+ + 2e- ó Pb

Cu2+ (aq) | Cu (s) Eo = +0.34 V Cu2+ + 2e- ó Cu

Ag+ (aq) | Ag (s) Eo = +0.80 V Ag+ + e- ó Ag

Table 10: Standard electrode potential values

For further notes on electrode potentials and calculations involving electrode potentials refer to the module book page 48.

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