List of abbreviations - Newcastle University
A2 TOPICS INDEX
0. Organic Isomers – new for A2 p2
1. Aromatic Chemistry p3
2. Amines p9
3. Carbonyl Compounds p12
4. Carboxylic Acids p18
5. Esters p19
6. Acid Chlorides & Anhydrides p20
7. Amino Acids p22
8. Organic Spectroscopy p23
9. Acids, Bases & pH p34
10. Equilibria p48
11. Kinetics p54
12. Thermodynamics p56
13. Periodic Trends in Chemical Properties p66
14. Transition Elements p71
15. Redox p90
16. Electrochemistry p93
Appendix: Post-16 Learning Resources Flyer - CCDC after p95
Appendix: Post-16 Learning Resources Flyer - PDBe after p95
ORGANIC CHEMISTRY
TYPES OF ISOMERISATION
• STRUCTURAL (including positional) (covered in AS)
• STEREOISOMERISM
- only ever two isomers possible for each stereochemical centre (as stereo = two).
- stereochemistry is concerned with the 3-dimensional arrangement of atoms in space.
- stereoisomers are two different geometric arrangements of atoms in space for the
same compound.
1. GEOMETRIC (E-Z) (covered in AS)
- structural feature: two different groups attached to each side of a C=C bond.
Occurs because there is no rotation about a C=C bond (due to the geometry of the π–bond), which gives rise to two possible structures for the alkene;
2. OPTICAL (new for A2)
- structural feature: four different groups attached to the same carbon atom – this
carbon is called an asymmetric or chiral carbon atom and can be indicated by an *.
Occurs because this arrangement of groups gives rise to two possible structures which are mirror images; e.g. 2-chloropentane:
mirror
[pic]
Optical isomers have identical chemical and physical properties as they have the same
functional groups arranged in a similar way in space.
They only differ in that a solution of the same concentration of each isomer rotates plane polarised light by the same amount but in opposite directions.
Each of the two optical isomers is called an enantiomer.
An equimolar mixture of the two optical isomers is called a racemic mixture and does not rotate plane polarised light at all, since the two equal and opposite effects cancel each other out. Synthetic chemical reactions in the laboratory or industry usually produce racemic mixtures (see later – carbonyl chemistry) whereas reactions in biological systems usually only produce one enantiomer (stereoselectivity).
1. AROMATIC CHEMISTRY
1.1 THE STRUCTURE OF BENZENE
The structure contains a ring of delocalised electrons above and below the plane of the molecule. The benzene ring structure is usually represented as a regular hexagon:
[pic]
Using enthalpies of hydrogenation, the extra stability of benzene can be calculated. The delocalisation energy is found to be about 150 kJ mol-1.
If benzene reacted by addition, this energy would be lost. Therefore, in its reactions, benzene keeps the electron cloud complete by undergoing substitution reactions.
The electron cloud attracts positive species, which are known as electrophiles. The most common reactions of benzene are therefore electrophilic substitutions.
2. ELECTROPHILIC SUBSTITUTION
[pic]
(1) (2) (3) (4)
1.2.1 Nitration: forms a nitro compound (1), e.g. nitrobenzene, C6H5NO2
1.2.2 Friedel Crafts Alkylation (Edexcel, OCR & WJEC ONLY): forms an alkylbenzene (2), e.g. methylbenzene, C6H5CH3
1.2.3 Friedel Crafts Acylation (AQA, Edexcel & OCR ONLY): forms a ketone (3), e.g. phenylethanone, C6H5COCH3
1.2.4 Halogenation (Edexcel, OCR & WJEC ONLY): forms a halo compound (4),
e.g. bromobenzene, C6H5Br
1.3 MECHANISM OF ELECTROPHILIC SUBSTITUTION OF BENZENE
[pic]
In each case an electrophile is generated in a preliminary step. This is attracted to the electron-rich ring and a bond is formed from a carbon in the ring to the electrophile. This process disrupts the delocalised ring and forms an unstable intermediate, which rapidly loses a proton to reform the delocalised ring of electrons.
1.3.1 Nitration
Here the electrophile is the nitronium ion, NO2+ formed from conc. HNO3 and conc. H2SO4
[pic]
Benzene is heated to 50oC with a mixture of conc. nitric acid and conc. sulphuric acid.
The mechanism can be represented as:
[pic]
Nitration is an important step in:
(i) the synthesis of explosives, e.g. TNT or nitroglycerine or
(ii) the production of aromatic amines, such as phenylamine, C6H5NH2, which can be used to form dyes (see later).
The overall reaction is:
[pic]
or: [pic]
1.3.2 Friedel-Crafts Alkylation (Edexcel, OCR & WJEC ONLY)
The electrophile is the carbocation, R+ (where R is an alkyl group). R+ is formed by the reaction of a chloroalkane with an electron deficient species (a Lewis acid) such as aluminium chloride:
[pic]
Aluminium chloride is regenerated by reaction of the proton lost with the AlCl4- ion:
[pic]
The overall reaction:
[pic]
For example:
[pic]
ethylbenzene
The mechanism is the same as nitration, with R+ (CH3CH2+) in place of NO2+.
Alkylation reactions are also possible with alkenes in the presence of HCl and AlCl3.
The electrophile is the carbocation formed according to:
[pic]
Ethylbenzene is formed as below:
[pic]
Such a route is favoured industrially because ethene is cheaper than chloroethane.
With propene, the secondary carbocation ion is formed in preference to the primary so the major product is (1-methylethyl)benzene (cumene).
[pic]
Uses of alkylation:
Ethylbenzene and 1-methylethyl)benzene are made on a large scale industrially via alkylation.
• Ethylbenzene is used in the manufacture of phenylethene (styrene), the monomer for making polystyrene, and
• (1-methylethyl)benzene (cumene) is oxidised to make phenol and propanone.
1.3.3 Friedel-Crafts Acylation (AQA, Edexcel & OCR ONLY)
This reaction produces a ketone when the electrophile RCO+ reacts with benzene. The electrophile, the acylium ion, is formed by the reaction of an acyl chloride with AlCl3
[pic]
The overall reaction:
[pic]
e.g.
[pic]
1.3.4 Halogenation (Edexcel, OCR & WJEC ONLY)
When benzene reacts with chlorine gas or liquid bromine in the presence of a halogen carrier (aluminium chloride or iron(III) bromide) the electrophile Cl+ or Br+ is generated as follows:
[pic]
When the overall reaction is:
[pic]
or [pic]
When the overall reaction is:
[pic]
The catalyst, e.g. iron(III) bromide is regenerated by reaction of the FeBr4- ion with the proton lost:
[pic]
Compare the difficulty of these reactions of benzene with the rapid reaction of alkenes with ‘bromine water’.
1.4 Phenol C6H5OH (Edexcel, OCR & WJEC ONLY)
The lone pair on the oxygen atom is delocalised into the aromatic ring, increasing its electron density. This makes substitution in phenol much easier than in benzene. Bromine water will form 2,4,6-tribromophenol without the need for a catalyst.
[pic]
The equivalent chlorine compound is trichlorophenol (TCP), which is used in antiseptics.
Phenol also reacts as a weak acid (weaker than carbonic acid). This is due to the attraction of the electron density into the ring which weakens the O-H bond.
[pic]
Phenol is also less reactive than aliphatic alcohols. For example, in reactions with acid chlorides and anhydrides to form esters (see later), it is usually first converted into the phenoxide ion.
[pic]
2. AMINES
Amines are organic derivatives of ammonia. There are 3 types:
Primary – one hydrogen replaced, e.g. CH3NH2 methylamine
Secondary – two hydrogens replaced, e.g. (CH3)2NH dimethylamine
Tertiary – three hydrogens replaced, e.g. (CH3)3N trimethylamine
2.1 REACTIONS
2.1.1 As Bases (NOT OCR)
• Ammonia and primary amines behave as BASES (proton acceptors) by donating the lone pair of electrons on the nitrogen to an incoming proton.
[pic]
• Aliphatic (alkyl) amines are stronger bases than ammonia due to the electron donating (inductive) effect of the alkyl groups which make the lone pair of electrons on the nitrogen atom ‘more available’. Hence, the above reaction occurs more readily.
• Phenylamine is a weaker base than ammonia due to the lone pair of electrons on the nitrogen atoms being situated in a p-orbital which overlaps with the orbital of the delocalised electrons in the aromatic ring. This makes the lone pair of electrons ‘less available’ and so the following reaction is less likely to occur:
[pic]
2.1.2 With Nucleophiles
Ammonia and amines can react as NUCLEOPHILES in substitution reactions with:
(i) Haloalkanes
[pic]
[pic]
Overall equation:
[pic]
The product primary amine is a better nucleophile than the original ammonia (for the same reasons as it’s a better base) so if an excess of haloalkane is used, further reactions occur producing a mixture of amines and a quaternary ammonium salt.
To prevent further reaction and thus produce a high yield of the primary amine, a large excess of ammonia is used.
With two long chain alkyl groups, these quaternary ammonium salts are used as cationic surfactants in fabric softening.
2.1.3 A specific reaction of phenylamine (Edexcel & WJEC ONLY)
Phenylamine reacts with nitrous acid in HCl to form a diazonium salt.
[pic]
These salts are used in solution below 5oC to form diazo compounds which are used as dyes, e.g. benzenediazonium chloride couples with phenol in alkaline solution to form the dye:
[pic]
(ii) Acyl chlorides and anhydrides (see later section – AQA, Edexcel & WJEC ONLY)
2.2 PREPARATION OF AMINES
2.2.1 Primary aliphatic amines are prepared either by:
(i) The reaction of ammonia with a haloalkane (not a good synthetic route – see above) or
(ii) The reduction of nitriles. The reducing agent is lithium tetrahydroaluminate in ether (LiAlH4/ether) or catalytic hydrogenation.
e.g.
[pic]
2. Aromatic amines are prepared by:
The reduction of nitro compounds. Tin and conc. HCl are used in the laboratory while iron and conc. HCl are used in industry as the reducing agent.
[pic]
3. CARBONYL COMPOUNDS
Aldehydes and ketones, e.g.
3.1 PREPARATION VIA OXIDATION OF ALCOHOLS
Alcohols can be classified as primary, secondary or tertiary depending on the carbon skeleton to which the hydroxyl group is attached (R = any alkyl group).
[pic]
Although many reactions of the OH functional group are the same in all alcohols, the three types of alcohol differ in their reactions with oxidising agents such as acidified potassium dichromate(VI).
1. Oxidation of alcohols
Although many reactions of the OH functional group are the same in all alcohols, the three types of alcohol differ in their reaction with oxidising agents such as acidified potassium dichromate(VI).
Acidified potassium dichromate(VI) turns from orange to green when it acts as an oxidising agent.
1. Primary alcohols are oxidised, first to aldehydes, e.g. ethanol is oxidised to ethanal:
[pic]
An aldehyde still has one hydrogen atom attached to the carbonyl carbon, so it can be oxidised further to a carboxylic acid:
[pic]
If an aldehyde is required, a primary alcohol is added to warm acidified potassium dichromate(VI) and the aldehyde is immediately distilled off to prevent further oxidation.
If a carboxylic acid is required, a primary alcohol is heated with the oxidising agent under reflux conditions to prevent excape of the aldehyde before it can be oxidised further.
2. Secondary alcohols
Are oxidised to ketones which have no hydrogen atoms attached to the carbonyl carbon so cannot easily be oxidised further:
[pic]
3. Tertiary alcohols are not oxidised by acidified dichromate(VI)
No colour change occurs with tertiary alcohols – would have to break a C-C bond.
3.2 DISTINGUISHING BETWEEN ALDEHYDES AND KETONES
Because of the hydrogen atom attached to the carbonyl group, aldehydes are easily oxidised to carboxylic acids, but ketones are not easily oxidised.
Mild oxidising agents are used to distinguish between them, e.g.
(i) Tollen’s reagent contains the complex ion [Ag(NH3)2]+ “ammoniacal silver nitrate”
It is prepared by adding an excess of aqueous ammonia to silver nitrate solution. Aldehydes reduce this ion and produce a silver mirror on the walls of a test tube; or silver ppt or black ppt or grey ppt; ketones cannot reduce this ion and thus do not form a silver mirror.
(ii) Fehling’s or Benedict’s solution contains a deep blue copper(II) complex ion.
This is reduced by aldehydes but not ketones to form an orange/red/brick red/yellow precipitate of copper(I) oxide, Cu2O.
The oxidation of aldehydes in these reactions can be represented by the equation:
[pic]
(iii) with 2,4-dinitrophenylhydrazine (2,4-DNP or 2,4-DNPH)
An alternative method to identify the presence of a carbonyl group is to prepare the 2,4-dinitrophenylhydrazine (1) derivative (Edexcel, OCR & WJEC ONLY).
[pic]
(1)
These are brightly coloured (often orange or yellow) crystalline solids. Furthermore, the 2,4-DNP derivative can be used to identify a specific aldehyde or ketone, as each derivative has a unique melting point depending on the parent carbonyl compound - data tables of these values are available.
3.3 REDUCTION OF CARBONYL COMPOUNDS
3.3.1 Hydrogenation
(i) Catalytic
Both C=O and C=C bonds are saturated by reaction with H2 with a nickel catalyst, e.g.
[pic]
(ii) by reaction with NaBH4 sodium tetrahydridoborate(III) in methanol (or LiAlH4 in ether)
Aldehydes can be reduced to primary alcohols and ketones to secondary alcohols. These are nucleophilic addition reactions (see later) so only C=O bonds are affected, not C=C
[pic]
4. NUCLEOPHILIC ADDITION TO CARBONYL COMPOUNDS
3.4.1 Addition of HCN
[pic]
Note the optical isomerism of the final product (see below)
The mechanism involves nucleophilic attack by the cyanide ion on the δ+ carbon. The anion formed then accepts a proton to complete the overall addition:
[pic]
Since the carbonyl group is planar, the nucleophile can attack from either side of the plane with equal likelihood, so both optical isomers (enantiomers) are formed in equal quantity – a racemic mixture:
[pic]
(cf. biological reactions, which tend to produce only one of the isomers – they are said to be stereoselective).
3.4.2 Reaction with NaBH4 or LiAlH4 (mechanism AQA & OCR ONLY)
The addition of H- is followed by H+ from an acid
[pic]
The anion formed then accepts a proton to complete the overall addition.
Note that in this example, a racemic mixture of optical isomers is not produced.
However, consider this reaction:
[pic]
Since this leads to a product in which a carbon atom has four different substituents, it does produce a mixture of optical isomers.
Since the carbonyl group is planar, the nucleophile can attack from either side of the plane with equal likelihood, so both optical isomers (enantiomers) are formed in equal quantity – a racemic mixture:
[pic]
3.4.3 Other reactions of carbonyl compounds (Edexcel, OCR & WJEC ONLY)
(i) with 2,4-dinitrophenylhydrazine (see section 3.2 (iii))
The formation of an orange precipitate when 2,4-dinitrophenylhydrazine is added to an aldehyde or a ketone confirms the presence of a carbonyl group.
The melting point of the product can be compared with data tables to identify the carbonyl compound exactly.
4. CARBOXYLIC ACIDS
Carboxylic acids contain the functional group –COOH or –CO2H
4.1 AS ACIDS
|Acid |Formula |pKa |
|Methanoic acid |HCOOH |3.75 |
|Ethanoic acid |CH3COOH |4.76 |
|Propanoic acid |CH3CH2COOH |4.87 |
[pic]
Carboxylic acids are weak acids. They react with metals, alkalis, or carbonates to form salts, although the reactions are less vigorous than with strong acids.
When carboxylic acids react with sodium hydrogen carbonate, carbon dioxide is evolved. This reaction can be used as a test for the carboxylic acid group.
[pic]
With alkalis, such as sodium hydroxide, a soluble salt is formed.
[pic]
4.2 OTHER REACTIONS OF CARBOXYLIC ACIDS
4.2.1 With alcohols to form esters
Esters can be formed by the reaction of carboxylic acids with alcohols in the presence of strong acid catalysts, e.g. concentrated sulphuric acid.
[pic]
5. ESTERS
Esters contain the functional group RCOOR’ where R & R’ can be an alkyl or an aryl group.
They are formed by the reaction of alcohols with either
i) carboxylic acids in the presence of conc. sulphuric acid (catalyst)
[reversible reaction] or
ii) acid chlorides or anhydrides [not reversible]
1. HYDROLYSIS OF ESTERS
When esters are heated with alkali, hydrolysis occurs and an alcohol and a carboxylate salt are formed, e.g.
[pic]
This hydrolysis reaction is used widely with naturally occurring esters, such as oils and fats, to produced useful products including soaps and glycerol.
Oils and fats are esters of propane-1,2,3-triol (glycerol) with 3 long chain carboxylic acids. When these fats are boiled with sodium hydroxide, glycerol is formed and a mixture of the sodium salts of the three acids. These salts are soaps, e.g.
[pic]
6. ACID CHLORIDES & ANHYDRIDES
Acid chlorides are colourless liquids which fume in moist air (steamy acid fumes of HCl). They react readily with compounds containing the –OH or –NH2 groups in nucleophilic addition/elimination reactions.
[pic]
The carbon of the carbonyl group is easily attacked by nucleophiles, so reactions of these compounds are rapid with acid chlorides, but slower with anhydrides.
Anhydrides are often preferred in industry for this reason. They are cheaper, less susceptible to hydrolysis by moisture in air and less dangerous as they don’t give off HCl fumes. A major use of ethanoic anhydride is in the preparation of aspirin.
1. REACTIONS (AQA & Edexcel only; mechanisms AQA ONLY)
6.1.1 with water (WJEC – not mechanism)
Carboxylic acids are formed, e.g. ethanoic acid:
[pic]
[pic]
[pic]
2. With alcohols
Esters are formed, e.g. methyl ethanoate:
[pic]
[pic]
6.1.3 With NH3
Amides are formed, e.g. ethanamide:
[pic]
6.1.4 With amines
Substituted amines are formed, e.g. N-methylethanamide:
[pic]
[pic]
[pic]
6.2 POLYAMIDES
such as Nylon(6,6) –[CO(CH2)4CONH(CH2)6NH]-
are formed by a similar reaction between dicarboxylic acids such as HOOC(CH2)4COOH and diamines such as H2N(CH2)6NH2
NB the linkage between the monomers in the Nylon-6,6 polymer is an AMIDE and NOT a peptide (since the monomers are not amino-acids!).
7. AMINO ACIDS
α-Amino acids have the general structure RCH(NH2)COOH
Apart from the first member glycine, CH2(NH2)COOH, they are all optically active.
They react with both acids and alkalis to form salts:
[pic]
pH < 7 pH > 7
In neutral solutions amino acids exist as internal salts called zwitterions:
[pic]
pH = 7
Note that this does not occur via internal proton transfer – the geometry of the molecule does not permit the acid group to donate H+ to its own amino group – it is an inter-molecular process between two amino-acids!
Proteins are polymers of amino acids joined together by amide bonds. This is often called the peptide link:
[pic]
Hydrolysis of peptides and proteins in the presence of strong acids or enzymes breaks this peptide link and liberates the component amino acids, which can be identified by thin-layer chromatography (tlc) of the mixture.
8. ORGANIC SPECTROSCOPY
Modern methods of chemical analysis use the instrumental methods of:
1. Mass spectrometry
2. Infra-red spectroscopy
3. Nuclear magnetic resonance spectroscopy and
4. Ultra-violet and visible spectroscopy (not AQA)
A chemist will often need to piece together information from all of these techniques to be completely sure of the identity of an organic molecule.
8.1 MASS SPECTROMETRY
8.1.1 HOW A MASS SPECTROMETER WORKS
A mass spectrometer uses the principle that heavier ions are deflected less than lighter ones when passed through a magnetic field.
The five processes which occur in the spectrometer are:
1) vaporisation, (2) ionisation, (3) acceleration, (4) deflection, (5) detection.
8.1.2 INTERPRETING A MASS SPECTRUM
8.1.2.1 Finding Mr
As long as the molecular ion is not completely fragmented, the peak with the highest m/z value corresponds to this ion, M+, the mass of which is equal to the relative molecular mass, Mr.
If isotopes are present, such as 35Cl and 37Cl or 79Br and 81Br, two molecular ion peaks will occur at M and M+2, e.g. 1-bromopropane, CH3CH2CH2Br, has two molecular ion peaks;
m/z = 122 and 124, with equal intensity since the two isotopes of Br are of equal abundance, whereas 1-chloropropane, CH3CH2CH2Cl, has two molecular ion peaks; m/z = 78 and 80, with relative intensities 3:1 since the two isotopes of Cl are in 75%:25% relative abundances.
8.1.3 FRAGMENTATION PATTERNS
On electron impact, molecular ions break up, so the mass spectrum will contain many lighter particles formed by fragmentation. Study of the mass of these fragments or the difference in mass from the molecular ion will often provide clues to the molecular structure.
|Mass loss |Possible group |
|15 |CH3 |
|17 |OH |
|18 |H2O |
|29 |C2H5 |
|43 |CH3CO or C3H7 |
|45 |COOH |
|77 |C6H5 |
Note: the molecular ion is a radical ion [pic] so when writing an equation for its fragmentation the products will be one ion and one radical. This can occur in two possible ways:
[pic]
i.e. the overall equation [pic] is unbalanced in charges and therefore wrong although both X+ and Y+ will be produced and detected, but not from fragmentation of a single molecular ion.
• The peak with the maximum intensity is called the base peak and intensities of the other peaks are shown as percentages of the intensity of this peak.
• Small peaks which occur at M+1 are due to the presence of carbon-13 isotopes which have a natural abundance of 1.1%. The ratio of the heights of the M : M+1 peaks may be approximated to 100 : (n x 1.1) where n is the number of carbon atoms in the molecule.
• More stable fragment species give rise to higher peaks, especially acylium ions [RCO]+ and carbocations, R+, e.g. the major fragment peaks in the spectrum of butanone, CH3COCH2CH3 correspond to the acylium ions [CH3CO]+ and [COCH2CH3]+ and also the carbocations +CH2CH3 and +CH3 (see next page).
[pic]
8.2 INFRA-RED SPECTROSCOPY
Covalent bonds in molecules absorb infra-red radiation at characteristic frequencies, which depend on both the bond strength and on the masses of the two atoms involved:
• Weak covalent bonds and high masses absorb at low frequencies
• Strong covalent bonds and low atomic masses absorb high frequencies.
Absorptions are usually quoted in wavenumbers (the reciprocal of wavelength) i.e. the number of wavelengths in one cm (with unit cm-1)
In most molecules, covalent bonds may stretch, rotate or bend in many ways absorbing many different amounts of energy and producing very complicated spectra. It is not necessary to identify each absorption; only certain characteristic absorptions for particular functional groups.
Information can be deduced from absorptions which are absent as well as those present.
8.2.1 INTERPRETATION OF IR SPECTRA
8.2.1.1 Regions
The infra-red spectrum is conveniently divided into four regions:
|Region / cm-1 |4000 – 2500 |2500 – 2000 |2000 – 1500 |1500 – 400 |
|Bonds |C-H, O-H |C=C, C=N |C=C, C=O |C-C, C-O, C-X |
8.2.1.2 Fingerprinting (AQA & WJEC ONLY)
The part of the spectrum below 1500 cm-1 is called the fingerprint region. Here i.r. spectra are complicated because of the large number of absorptions due to single bond vibrations. Each molecule has a unique pattern in this region and this can be used to identify it completely. This is done (usually by computers) which compare the spectrum of an unknown molecule with those of known molecules until an exact match is found.
Even if the absorptions by two different molecules in other regions of a spectrum are very similar, the fingerprint region is unique to each molecule so the two can be distinguished.
8.2.1.3 Table of frequencies
The frequency for a particular covalent bond does not change much from molecule to molecule, so it is possible to draw up a table of typical absorption frequencies. Each bond is shown absorbing over a range of wavelengths because of the effect of adjacent groups, e.g. the carbonyl group C=O absorbs in the region 1680 – 1750 cm-1.
|Bond |Wavenumber / cm-1 |
|C-H |2850 – 3300 |
|C-C |750 – 1100 |
|C=C |1620 – 1680 |
|C=O |1680 – 1750 |
|C-O |100 – 1300 |
|O-H in alcohols |3230 – 3550 |
|O-H in acids |2500 - 3000 |
8.2.1.4 Intensity of absorptions
Some absorptions are intense while others are weak. The difference depends on the polarity of the bonds and the size of the change in polarity as a bond stretches, etc. Thus polar bonds such as C=O or C-O have intense absorptions and large peaks, whereas non-polar bonds such as C-C or C=C have weak absorptions. So note that peak size does NOT matter for IR spectroscopy (contrast with peak areas in NMR spectroscopy later).
Examples (see next page for the actual i.r. spectra)
(i) butan-1-ol and butan-2-ol (example spectra A & B)
The intense broad absorptions at about 3350 cm-1 are typical of an OH group. Note the absence of strong absorptions around 1700 cm-1 showing that C=O bonds are not present.
The spectra are almost identical above 1500 cm-1 but the fingerprint region below 1500 cm-1 can be used to distinguish between these isomeric alcohols.
(ii) methyl propanoate (example spectrum C)
The absorption at about 1750 cm-1 is due to a C=O; that at 1200 cm-1 is due to C-O and there is no O-H absorption at about 3300 cm-1. The peak at just under 3000 will be due to C-H bonds.
(iii) propanoic acid (example spectrum D)
The absorption at 1700 cm-1 indicates C=O, that at 3000 cm-1 shows O-H is present (partly obscuring the C-H absorption). 1230 cm-1 shows C-O present.
Without other information, these i.r. spectra alone would not allow complete identification of the compounds.
[pic]
8.3 NUCLEAR MAGNETIC RESONANCE
NMR gives information about the relative number and position of hydrogen or carbon atoms in a molecule.
Proton (1H) NMR spectra are usually recorded in solution in a proton-free solvent to avoid unwanted absorptions. Such solvents include CCl4 or deuterated compounds such as CDCl3.
The signals detected are described as having chemical shift relative to a standard frequency.
The standard is tetramethylsilane, (CH3)4Si which gives a single intense peak, upfield, i.e. to the right of most others. TMS is non-toxic, inert, and has a low boiling point and so is easily removed from the sample.
Chemical shifts for hydrogen (δH) are expressed as parts per million and have values between 0 and 12. By convention the scale increases from right to left.
Hydrogens close to electronegative atoms or π bonds will be further downfield (to the left).
Tables of chemical shifts can be used to identify specific types of proton.
8.3.1 INTERPRETING THE SPECTRUM
8.3.1.1 Chemical Structure
1H NMR spectra give the following four pieces of information about chemical structure:
• The number of absorptions indicates how many types of non-equivalent protons are present (different “chemical environments”).
• The position of the absorptions (the chemical shift, δ) gives clues to the environment of each kind of proton, see the table below.
• The intensities (area or integration) of the absorptions show how many equivalent protons are associated with each absorption peak. This can be shown as an integration trace, i.e. the spectrometer measures the areas of the peaks and shows this as a curve. The heights of the integration curve at each peak show the relative numbers of protons associated with the peak.
• The splitting of an absorption provides information about neighbouring protons (by applying the n+1 rule).
NB for 13C NMR spectra only the first three apply – there is NO splitting for 13C peaks.
|TYPICAL PROTON SHIFTS | |
|Type of proton |/ppm |
|RCH3 |0.8 – 1.0 |
|R2CH2 |1.2 – 1.4 |
|R3CH |1.4 – 1.6 |
|RCOCH1-3 |2.1 – 2.6 |
|RCH1-2OR |3.3 – 3.9 |
|RCH1-2Br |3.4 – 3.6 |
|RCH1-2Cl |3.6 – 3.8 |
|R2C=CH1-2 |4.6 – 5.0 |
|ArH |6.0 – 8.5 |
|RCHO |9.5 – 9.9 |
8.3.1.2 Spin-spin coupling
The absorption peak of one proton can be split by ‘coupling’ with the spin of neighbouring protons. The signal due to the first proton is no longer a single line but is split into 2, 3, 4 or more.
The amount of splitting is given by the n + 1 rule:
If there are no adjacent protons the signal can’t be split and appears as a singlet, S.
If there is one adjacent proton the signal is split into 2 – a doublet, D (1:1).
If there are two adjacent protons the signal is split into 3 – a triplet, T (1:2:1).
If there are three adjacent protons the signal is split into 4 – a quartet, Q (1:3:3:1).
Note that the presence of both a triplet (T) and a quartet (Q) in a spectrum indicates the presence of an ethyl group, CH3CH2-
8.3.1.3 Use of D2O to identify OH groups (OCR ONLY)
Deuterium 2H does not absorb radio frequency radiation as it has an even mass number. The exchange of the proton in an OH group with deuterium from D2O will cause the peak due to the OH proton to disappear from the spectrum – and thus can be used to identify the presence of (an) OH group(s) in a molecule.
Examples (see next page for actual 1H n.m.r. spectra)
(i) ethanol (example spectrum A)
CH3CH2OH, has six protons in three different environments
CH3 (δ = 1.2 ppm), CH2 (δ = 3.7 ppm), and OH (δ = 5.7 ppm).
The low resolution spectrum shows three peaks with the areas in the expected ration of 3:2:1.
The normal (high) resolution spectrum shows splitting of the signals:
The CH3 peak is split into a triplet by the two adjacent hydrogens in CH2
The CH2 peak is split into a quartet by the three adjacent hydrogens in CH3
Note: the OH singlet is further downfield than CH2 which is further downfield than CH3 reflecting their proximity to the electronegative oxygen atom.
(ii) methyl propanoate (example spectrum B)
CH3CH2COOCH3 has 8 protons in three different environments
A singlet (δ = 3.7 ppm) for 3 protons corresponds to CH3 (or OCH3) not adjacent to other protons but furthest downfield so nearest to the electronegative oxygen.
A quartet (δ = 2.4 ppm) for 2 protons corresponds to CH2 adjacent to 3 other protons
A triplet (δ = 1.2 ppm) for 3 protons corresponds to CH3 adjacent to 2 other protons
(iii) ethoxyethane (example spectrum C)
CH3CH2OCH2CH3 has 10 protons in two different environments since the molecule is completely symmetrical.
A quartet (δ = 3.5 ppm) for 2 protons corresponds to CH2 adjacent to 3 other protons
A triplet (δ = 1.2 ppm) for 3 protons corresponds to CH3 adjacent to 2 other protons
[pic]
9. ACIDS, BASES & pH
1. BRONSTED-LOWRY THEORY
- applies to species in aqueous solution:
An acid is a proton donor.
A base is a proton acceptor.
e.g. full equation: HCl + NaOH → NaCl + H2O
ionic equation: H+ + OH- → H2O
HCl (a Bronsted-Lowry acid) donates a proton to the hydroxide ion (a Bronsted-Lowry base).
Also: HNO3 + H2SO4 → H2NO3+ + HSO4-
Concentrated nitric acid accepts a proton from concentrated sulphuric acid, i.e. concentrated nitric acid acts as a Bronsted-Lowry base in the presence of a stronger acid, sulphuric. This reaction is important in the nitration of benzene (see Aromatic Chemistry section).
All acid-base reactions involve the transfer of protons.
Note that water can act as both an acid and a base (this is called amphoteric).
As an acid: H2O + NH3 → NH4+ + OH-
acid base acid base
As a base: H2O + HCl → H3O+ + Cl-
base acid acid base
As both: H2O + H2O → H3O+ + OH-
acid base acid base
2. STRONG ACIDS AND BASES
Are completely/totally/fully/100% dissociated into ions.
In aqueous solution, hydrochloric, nitric and sulphuric acids are strong, i.e. completely split up into ions, e.g. HCl (aq) → H+ (aq) + Cl- (aq)
0% 100%
Alkali metal hydroxides, e.g. NaOH or KOH are strong bases as the ionic lattice is completely split up releasing all the hydroxide ions when the solids dissolve into a large volume of water.
pH is a logarithmic scale used to simplify the way the concentration of H+ is expressed.
pH = –log10[H+] or simply –log[H+] [ ] means concentration in mol dm-3
pH can be measured using Universal Indicator (a mixture of several indicators) or using a pH meter which must first be calibrated using a buffer solution (see section 9.6 later).
In ALL calculations, pH MUST be given to 2 decimal places! ( (check your Exam Board’s requirements!)
a) Calculating pH from [H+]
e.g. (i) calculate the pH of a 0.200M solution of hydrochloric acid
The acid is fully ionised, so [H+] = 0.200M
Hence pH = -log[0.200] = 0.70 (note answer to 2dp!)
(ii) calculate the pH of a 1.00M solution of hydrochloric acid
[H+] = 1M hence pH = -log[1.00] = 0.00
(iii) calculate the pH of a 2.00M solution of hydrochloric acid
[H+] = 2M hence pH = -log[2.00] = -0.30 (again note answer to 2dp!)
Note that for solutions of monoprotic strong acids with concentrations greater than 1.00M, the pH will be negative.
a) Calculating [H+] from pH
e.g. calculate the concentration of hydrochloric acid when a solution has a pH of 3.20
pH = -log[H+] hence log[H+] = -3.20 so [H+] = 10-3.20 = 6.31 x 10-4 mol dm-3
so [HCl] = 6.31 x 10-4 mol dm-3 (note default calculation accuracy of answer – 3 sig. figs.)
Ionic product of water, Kw
The equation: [pic]
Can be simplified to: [pic]
The equilibrium constant for this, Kc = [H+] [OH-]
[H2O]
Since [H2O] is huge compared to the concentration of the ions, [H2O] is effectively a constant in pure water and so can be incorporated into a new equilibrium constant, Kw, called the ionic product of water:
Kw = [H+] [OH-] and has the value 1.00 x 10-14 mol2 dm-6 at 25oC. LEARN THIS!! (
In a neutral solution, [H+] = [OH-] so Kw = [H+]2 = 1.00 x 10-14 at 25oC
Hence [H+] = 1.00 x 10-7 mol dm-3 and thus pH = -log[1.00 x 10-7] = 7.00
For the process: [pic]
ΔH = +57.6 kJ mol-1
At higher temperatures, Le Chatelier’s Principle predicts that the equilibrium moves to the right to act to oppose the change, so that [H+] rises and thus pH falls, and the opposite if the temperature is reduced.
This does NOT mean that pure water becomes acidic when it is heated since [OH-] also increases by the same amount! It DOES however mean that the 0-14 scale of pH only applies at 25oC!
For simplification of pH calculations, it will always be assumed that the solutions used are at 25oC.
Using Kw, we can calculate the pH of strong bases (alkalis)
a) calculating pH from known a concentrations of a strong alkali
(i) calculate the pH of a 0.200M solution of sodium hydroxide
The base is fully ionised, so [OH-] = 0.200M
[H+] = Kw / [OH-] = 1.00 x 10-14 / 0.200 = 5.00 x 10-14 so pH = -log[5.00 x 10-14] = 13.30
(ii) calculate the pH of a 1.00M solution of potassium hydroxide
[OH-] = 1.00M so [H+] = Kw / [OH-] = 1.00 x 10-14 / 1.00 = 1.00 x 10-14 so pH = 14.00
Note that for solutions of strong bases more concentrated than 1.0M, the pH will be greater than 14.00!
b) calculating the concentration of a strong base given the pH of the solution
e.g. calculate the concentration of NaOH when a solution of it has a pH of 12.50
pH = 12.50 hence [H+] = 10-12.50 = 3.16 x 10-13 mol dm-3
so [OH-] = Kw / [H+] = 1.00 x 10-14 / 3.16 x 10-13 = 0.0316 thus [NaOH] = 0.0316 mol dm-3
Calculating pH of mixtures of strong acids and strong alkalis
e.g. (i) calculate the pH of the resulting solution produced by mixing 25.0cm3 of 0.100M sodium hydroxide with 15.0cm3 0.120M hydrochloric acid
No. moles NaOH = MV / 1000 = 0.100 x 25.0 / 1000 = 2.50 x 10-3 = moles OH-
No. moles HCl = MV / 1000 = 0.120 x 15.0 / 1000 = 1.80 x 10-3 = moles H+
Since H+ + OH- → H2O
No. moles OH- in excess = 2.50 x 10-3 – 1.80 x 10-3 = 0.700 x 10-3 = 7.00 x 10-4 mol
These are present in 25.0 + 15.0 = 40.0cm3 of solution
So [OH-] = 1000n / V = 1000 x 7.00 x 10-4 / 40.0 = 1.75 x 10-2 mol dm-3
[H+] = Kw / [OH-] = 1.00 x 10-14 / 1.75 x 10-2 = 5.71 x 10-13
so pH = -log[5.71 x 10-13] = 12.24
e.g. (ii) calculate the pH of the resulting solution produced by mixing 25.0cm3 of 0.100M sodium hydroxide with 35.0cm3 0.120M hydrochloric acid
No. moles NaOH = MV / 1000 = 0.100 x 25.0 / 1000 = 2.50 x 10-3 = moles OH-
No. moles HCl = MV / 1000 = 0.120 x 35.0 / 1000 = 4.20 x 10-3 = moles H+
Since H+ + OH- → H2O
No. moles H+ in excess = 4.20 x 10-3 – 2.50 x 10-3 = 1.70 x 10-3 mol
These are present in 25.0 + 35.0 = 60.0cm3 of solution
So [H+] = 1000n / V = 1000 x 1.70 x 10-3 / 60.0 = 0.028 mol dm-3
pH = -log[0.028] = 1.55
3. WEAK ACIDS AND BASES
Are not completely/fully/ 7.
Finally, note that a weak acid – weak base combination does not have an easily measureable equivalence point and thus this type of titration is not carried out.
The pH at half-equivalence for a weak acid, HA, is the point at which [HA] = [A-].
Thus Ka = [H+] [A-] / [HA] simplifies to Ka = [H+] or pH = pKa.
This provides an experimental method to measure pKa and hence Ka for a weak acid:
1. titrate the HA solution with strong base (e.g. NaOH) and record the pH curve;
2. determine equivalence point from pH curve – thus the volume (V) of NaOH needed to reach equivalence point;
3. determine the 1/2V value and read off the pH for this volume – this pH = pKa
4. hence calculate Ka since pKa = -logKa (so K = 10-pKa).
pH curve for weak acid (HA) titrated against a strong base (NaOH) – Ka determination
[pic]
Volume of 0.100M sodium hydroxide added (cm3)
[Image edited from ]
9.4a pH curves
[pic]
[pic]
5. INDICATORS
An acid-base indicator is simply a weak acid where the colours of the acid (HIn) itself and its ion (In-) are noticeably different.
e.g. [pic]
colour 1 colour 2
At low pH, the unionised form HIn predominates and thus the solution appears colour 1.
At high pH, the ionised form In- predominates and thus the solution appears colour 2.
Complete colour change from one colour to the other usually takes place over around 1 - 2 pH units.
Hence it is very important to choose the correct indicator for each combination of acid-alkali titration.
The indicators colour change pH range MUST lie completely on the steep part of the titration curve (i.e. around the equivalence point) so one drop from the burette will change the pH sufficiently to change the indicator colour completely.
The two most common indicators are methyl orange and phenolphthalein:
|Indicator |Colour 1: acid (HIn) |pH range |Colour 2: base (In-) |
|Methyl orange |Red |3.2 – 4.4 |Yellow |
|Phenolphthalein |Colourless |8.2 – 10.0 |Pink |
Now look back at the pH curves and decide which is the best indicator from these two to use for each of the four types of acid-base titration?
6. BUFFER SOLUTIONS
A buffer solution resists a change in pH when small amounts of acid or alkali are added or when it is diluted.
An acid buffer maintains a pH below 7. Acid buffers are prepared by mixing equimolar quantities of a weak acid and its salt, e.g. ethanoic acid and sodium ethanoate (an alternative way of achieving this is to half-neutralise some ethanoic acid with a strong alkali).
[pic]
LARGE small small
[pic]
small LARGE
When a small amount of H+ is added to this buffer mixture, the equilibrium moves slightly to the left – the large [CH3COOH] rises slightly and the large [CH3COO-] falls slightly.
Ka = [H+] [CH3COO-] i.e. [H+] = Ka x [CH3COOH]
[CH3COOH] [CH3COO-]
Both [CH3COOH] and [CH3COO-] are large relative to H+ and so their ratio remains almost constant – as Ka is a constant and so [H+] and thus the pH remains almost constant.
If the buffer solution is simply diluted with water, the ratio of [CH3COO-] / [CH3COOH] again remains constant, so [H+] and thus pH remain constant.
An alkaline buffer maintains a pH above 7. This consists of an equimolar mixture of a weak base together with its salt, e.g. ammonia and ammonium chloride:
[pic]
LARGE small small
[pic]
small LARGE
When H+ is added, the equilibrium moves to the right; [NH4+] rises slightly and [NH3] falls slightly, but [OH-] remains almost constant and so pH remains almost constant.
Buffers are very important in the control of pH both industrially in the manufacture of medicines, dyes, shampoos, etc. They are also found in biological systems, where they ensure that the pH of aqueous solutions in living creatures is maintained within narrow ranges, e.g. the blood in our bodies is kept within the pH range 7.35 – 7.45 to ensure that enzymes within our bodies work correctly and most efficiently.
Quantitative treatment of acid buffers
(i) A buffer was 0.100M with respect to both HA and A- (Ka = 1.80 x 10-5 mol dm-3). Calculate the pH of this buffer
Ka = [H+] [A-] / [HA] so Ka = [H+] hence pH = -log[1.80 x 10-5] = 4.74
This is known as the half-equivalence point, and can be used experimentally to find the value of Ka (since Ka = [H+] at this point, pH = pKa, which can be read off the pH curve – see also the pH curves section 9.4, p40).
(ii) calculate the change in pH when 10.0cm3 of 1.00M HCl are added to 1000cm3 of this buffer
No. moles H+ added = MV / 1000 = 1.00 x 10.0 / 1000 = 0.0100 mol
The extra H+ reacts with A- to make more HA, so:
No. moles HA present after addition = original + created = 0.100 + 0.010 = 0.110 mol
New concentration of HA = 0.110 / V
No. moles A- present after addition = original – removed = 0.100 – 0.010 = 0.090 mol
New concentration of A- = 0.090 / V
Ka = [H+] [A-] / [HA] = 1.80 x 10-5
[H+] = Ka x [HA] / [A-] = 1.80 x 10-5 x 0.110 / 0.090 (the volumes V cancel out)
[H+] = 2.20 x 10-6 so pH = 4.66
Note pH has only decreased by a very small amount = 4.74 – 4.66 = -0.08 units
(iii) calculate the change in pH when 10.0cm3 of 1.00M NaOH are added to 1000cm3 of this buffer
No. moles OH- added = MV / 1000 = 1.00 x 10.0 / 1000 = 0.010 mol
The extra OH- reacts with HA to make more A-, so:
No. moles HA present after addition = original - removed = 0.100 – 0.010 = 0.090 mol
New concentration of HA = 0.090 / V
No. moles A- present after addition = original + created = 0.100 + 0.010 = 0.110 mol
New concentration of A- = 0.110 / V
Ka = [H+] [A-] / [HA] = 1.80 x 10-5
[H+] = Ka x [HA] / [A-] = 1.80 x 10-5 x 0.090 / 0.110 (again, the volumes V cancel out)
[H+] = 1.47 x 10-5 so pH = 4.83
Note pH has only increased by a very small amount = 4.74 – 4.83 = +0.09 units
(iv) calculate pH of the buffer formed when 25.0cm3 of 0.100M NaOH are added to 75.0cm3 of 0.200M HA (for which Ka = 1.80 x 10-5 mol dm-3)
Reaction which occurs is: HA + OH- → A- + H2O
No. moles HA used = 75.0 x 0.200 / 1000 = 0.0150
No. moles OH- added = 25.0 x 0.100 / 1000 = 0.00250
No. moles A- formed = no. moles OH- used = 0.00250
Concentration of A- in the buffer = 0.00250 / V
No. moles HA removed = 0.00250
No. moles HA in the buffer = 0.0150 – 0.00250 = 0.01250
Concentration of HA = 0.01250 / V
Ka = [H+] [A-] / [HA] = 1.80 x 10-5
[H+] = Ka x [HA] / [A-] = 1.80 x 10-5 x 0.0125 / 0.00250 (again, the volumes V cancel out)
[H+] = 9.00 x 10-5 so pH = -log[9.00 x 10-5] = 4.05
10. EQUILIBRIA
Knowledge and understanding of the following AS Equilibria topic is assumed when A2 is studied:
Many chemical reactions reaction go to completion
e.g. [pic]
Other reactions do not go to completion and are reversible:
e.g. [pic]
Left to right in the equation is called the forward reaction.
Right to left in the equation is called the backward or the reverse reaction.
At equilibrium the concentrations of reactants and products remain constant.
10.1 The effects of changing reaction conditions
Factors to be considered:
• Change in concentration.
• Change in pressure.
• Change in temperature.
• Addition of a catalyst.
The qualitative effect of changing reaction conditions can be predicted using Le Chatelier’s principle:
“a system at equilibrium will react to oppose any change imposed upon it” (o.w.t.t.e.)
10.1.1 The effect of a change in concentration
Consider the reaction:
[pic]
If more water is added, the equilibrium position is displaced to the right and the equilibrium
yields of CH3COOH (l) and C2H5OH (1) are increased.
If more CH3COOH (l) or more C2H5OH (l) is added the equilibrium is displaced to the left
and the equilibrium mixture contains more CH3COOC2H5(l) and H2O(l).
NB. If the reactants or products are gases, a change in the pressure of any gaseous species
is equivalent to a change in the concentration of that species.
10.1.2 The effect of a change in total pressure
Pressure only has a significant effect if the reaction involves gases.
If there are more moles of gaseous reactant than there are moles of gaseous product, an increase in total pressure will displace the reaction to the right.
The converse of this statement also applies
e.g. [pic]
Total number of moles of gaseous reactant is 2: total number of moles of gaseous product is 4:
Thus when the total pressure is increased the system responds by moving to the left.
10.1.3 The effect of a change in temperature
A change in temperature changes the rate of the forward and backward reactions by different amounts and hence changes the equilibrium position.
e.g. [pic]
NB. the ΔH value quoted is always assumed to be for the forward reaction as written.
Exothermic reactions Applying Le Chatelier’s principle;
Heat energy is being evolved; an increase in temperature moves reaction in the direction of
the endothermic reaction, i.e. to the left; the equilibrium yield of product is reduced.
N.B. For an exothermic reaction, an increase in temperature reduces the time needed to
reach equilibrium but decreases the equilibrium concentration of products.
Endothermic reactions Applying Le Chatelier’s principle;
e.g. [pic]
Heat energy is being absorbed; an increase in temperature moves reaction in the direction of
the endothermic reaction, i.e. to the right; the equilibrium yield of product is increased.
N.B. For an endothermic reaction, an increase in temperature also reduces the time needed to
reach equilibrium, but this time increases the equilibrium concentration of products.
10.1.4 The effect of a catalyst
A catalyst has no effect on the composition or position of the equilibrium mixture.
A catalyst increases the rate of both the forward and the backward reactions equally. Hence
the equilibrium position is achieved more quickly.
10.2 Application to an industrial process (
Example: [pic]
10.2.1 The operating pressure
Four moles of gaseous reactant form two moles of gaseous product hence the synthesis of
ammonia is favoured by high pressure.
The pressure used is often around 2.0 x 104 kPa. Higher pressures produce greater
equilibrium yields of ammonia but the high cost of generating higher pressures makes this uneconomic. 2.0 x 104 kPa is a good compromise between cost and equilibrium yield of
ammonia.
10.2.2 The operating temperature
As the ammonia synthesis reaction is an exothermic process; ΔH = - 92 kJ mol-1 the best
equilibrium yield of ammonia is obtained at a low temperature.
However, at low temperatures the rate of reaction will be slow. A compromise is necessary.
The usual operating temperature is in the range 650 – 720 K.
3. The use of a catalyst
The rates of both the forward and backward reactions are increased to the same extent by the
use of a catalyst and hence the time taken for the reaction to reach equilibrium is reduced.
The catalyst usually employed is based on iron with potassium hydroxide added to promote
its activity.
2. Equilibrium constants
10.3.1 Definition
The equilibrium constant is a numerical measure of the position of the equilibrium, i.e. if it lies to the left-hand side (LHS) or the right-hand side (RHS) or indeed somewhere in the middle.
It is defined as a mathematical expression simply from the balanced equation for the reaction.
So for the general equilibrium:
[pic]
where A, B, C & D are reagents and
a, b, c & d are the stoichiometric quantities (no. of moles) of each species which react together.
10.3.2 In terms of concentration, Kc
Kc = [C]c [D]d
[A]a [B]b
where [X] represents the concentration of each species in mol dm-3.
You can see that simplistically Kc is a ratio of RHS : LHS and thus the magnitude (size) of Kc is a measure of the position of the equilibrium.
i.e. if [RHS] >> [LHS] then the “fraction” is top-heavy and Kc is a large number (>1) – the eqm. lies on the RHS.
Conversely, if [RHS] [RHS] so the eqm. constant expression is bottom-heavy and thus should be a fraction (> [LHS] so expression is top-heavy and thus Kc is large (>1).
10.3.4 In terms of pressure, Kp
For the general equilibrium:
[pic]
where A, B, C & D are reagents and
a, b, c & d are the stoichiometric quantities (no. of moles) of each species which react together.
Kp = p(C)c p(D)d
p(A)a p(B)b
where p(A) represents the partial pressure of each species in Pa or kPa or MPa.
Note that unlike the Ideal Gas Equation at AS level, the units of pressure do NOT have to be converted into the SI unit of Pa – they can be used as supplied in the question and the units of Kp will simply be a power of those pressure units (see the examples later).
You can see that simplistically Kp is also a ratio of RHS : LHS and thus the magnitude (size) of Kp is a measure of the position of the equilibrium.
10.3.5 Mole fractions and calculation of partial pressures
The partial pressure of a species, p(A), depends only on its proportion in the gaseous mixture, represented by its mole fraction (ΧA) and the total pressure (PTOT).
The mole fraction of a species, A, is defined as ΧA = nA / nTOT where nA is the number of moles of A and nTOT is the total number of moles of all gaseous species present. As a self-check, Σ (XN) = 1.
Thus p(A) = (nA / nTOT) x PTOT and as a self-check when you’ve calculated them all, Σ p(X) = PTOT
10.3.6 Calculation of Kp
e.g. 1 The following equilibrium was established at 500 K:
[pic]
At equilibrium, the number of moles of A, B and C were 1.2, 1.7 and 0.8 mol respectively at a total pressure of 300 kPa. Write an expression for Kp and calculate the value at 500 K, clearly stating its units?
Kp = p(C)
p(A) p(B)2
p(A) = 1.2 / (1.2 + 1.7 + 0.8) x 300 = 97.3 kPa
p(B) = 1.7 / 3.7 x 300 = 137.8 kPa
p(C) = 0.8 / 3.7 x 300 = 64.9 kPa (OR p(C) = 300 – 97.3 – 137.8)
Self-check: Σ p(X) = 97.3 + 137.8 + 64.9 = 300 kPa !
Thus Kp = 64.9 = 3.513 x 10-5 kPa-2
97.3 x 137.82
With Kp you can also usually roughly check if the answer is sensible – in this case, the p(LHS) >> p(RHS) so the eqm. constant expression is bottom-heavy and thus should be a fraction (> p(LHS) since the p(H2) is cubed, so the eqm. constant expression is top-heavy and thus should not be a fraction (>1).
11. KINETICS
Knowledge and understanding of the following AS Kinetics topics are assumed when A2 is studied:
• Collision Theory
• Definition of Activation Energy, Ea
• Explaining qualitatively, in terms of the Collision Theory, the effect of concentration or pressure, temperature, surface area or catalyst on the rate of reaction.
11.1 Rate Equations
These define the relationship between the concentrations of individual reactants and the overall rate of the reaction. The expression for the rate equation is derived from experimental data (contrast with the expression for the equilibrium constant, Kc, which is derived purely from the balanced chemical equation for the reaction (see section 10.3).
11.1.1 Definition
The rate equation is of the form: rate = k [X]a [Y]b [Z]c
where [X] is the concentration of each reactant species in mol dm-3 and
k = the rate “constant” – dependent on both the reaction and temperature and
a, b & c are the orders of reaction – determined from the experimental data – and limited for all A-level specifications to values of 0, 1 or 2.
11.1.2 Order of reaction
If doubling the concentration of [A] has no effect on the rate – 0 (zero) order.
If doubling the concentration of [A] doubles the rate – 1st (first; 1) order.
If doubling the concentration of [A] quadruples the rate – 2nd (second; 2) order.
To summarise the effect of each order on the rate:
|order |[A] x 2 |[A] x 3 |[A] x 4 |
|0 |none |none |none |
|1st |x 2 |x 3 |x 4 |
|2nd |x 4 (22) |x 9 (32) |x 16 (42) |
11.1.3 Deducing the Rate Equation
e.g. 1 The rate data in the following table were obtained at 298 K for the reaction:
[pic]
|Experiment |[X] mol dm-3 |[Y] mol dm-3 |Rate mol dm-3 s-1 |
|1 |0.15 |0.24 |3.50 x 10-2 |
|2 |0.45 |0.24 |1.05 x 10-1 |
|3 |0.45 |0.12 |1.75 x 10-2 |
Deduce the order of reaction with respect to X and Y. Write an expression for the rate equation. Using the data for experiment 1, calculate a value for the rate constant and state its units.
a. For X – compare expts. 1 & 2: [X] x 3, rate x 3 – 1st order.
b. For Y – compare expts. 1 & 3: [Y] / 2, rate / 2 – 1st order.
c. Rate = k [X] [Y]
d. k = rate / [X] [Y] = 3.50 x 10-2 / (0.15 x 0.24) = 0.972 dm6 mol-2 s-1
e.g. 2 The rate data in the following table were obtained at 325 K:
|Experiment |[E] mol dm-3 |[F] mol dm-3 |Rate mol dm-3 s-1 |
|1 |2.60 x 10-3 |1.60 x 10-2 |2.20 x 10-4 |
|2 |2.60 x 10-3 |3.20 x 10-2 |2.20 x 10-4 |
|3 |5.20 x 10-3 |1.60 x 10-2 |8.80 x 10-4 |
Deduce the order of reaction with respect to E and F. Write an expression for the rate equation. Using the data for experiment 1, calculate a value for the rate constant and state its units.
a. For E – compare expts. 1 & 3: [E] x 2, rate x 4 – 2nd order.
b. For F – compare expts. 1 & 2: [F] x 2, rate no change – 0 order.
c. Rate = k [E]2 [F]0 or more simply rate = k [E]2
d. k = rate / [E]2 = 2.20 x 10-4 / (2.60 x 10-3)2 = 32.5 dm3 mol-1 s-1
e.g. 3 In the presence of the catalyst rhodium, the reaction between NO and H2 occurs
according to the following equation:
[pic]
The kinetics of the reaction were investigated and the rate equation was found to be
rate = k [NO]2 [H2]
The initial rate of reaction was 6.20 x 10–6 mol dm–3 s–1 when the initial concentration of NO was 2.90 x 10–2 mol dm–3 and the initial concentration of H2 was 2.30 x 10–2 mol dm–3.
i. Calculate the value of the rate constant under these conditions and give its units?
ii. Calculate the initial rate of reaction if the experiment is repeated under the same
conditions but with the concentrations of NO and of H2 both doubled from their original values.
i. k = rate / [NO]2 [H2] = 6.20 x 10-6 / (2.90 x 10-2)2 x (2.30 x 10-2) = 0.321 dm6 mol-2 s-1
ii. rate = k [NO]2 [H2] = 0.321 x (5.80 x 10-2)2 x (4.60 x 10-2) = 4.97 x 10-5 mol dm-3 s-1
12. THERMODYNAMICS
Knowledge and understanding of the following AS topics is assumed when A2 is studied:
• Enthalpy change, ΔH
• Calorimetry
• Hess’s Law
• Bond enthalpies (AQA ONLY for A2 – already covered at AS)
12.1 Definitions – these need to be LEARNT! (
Enthalpy change, ΔHo; the heat energy change measured at constant pressure.
12.1.1 Standard conditions
Pressure: 100 kPa; Temperature: 298 K
State symbols must be given with equations, e.g.
[pic]
12.1.2 The Enthalpy of Formation, ΔHfo
The standard enthalpy change when one mole of a compound is formed from its elements under standard conditions, 100 kPa and 298 K, with all reactants and products in their standard states, e.g.
[pic]
NB This definition means that ΔHfo for any element is always ZERO
12.1.3 Ionisation enthalpy, ΔHio
The standard enthalpy change when an electron is removed from one mole of a species in the gas phase to form a positive ion and an electron also in the gas phase.
First ionisation enthalpy, e.g.
[pic]
Second ionisation enthalpy, e.g.
[pic]
12.1.4 Enthalpy of atomisation of an element, ΔHato
The standard enthalpy change when one mole of gaseous atoms is formed, e.g.
[pic]
[pic]
12.1.5 Electron affinity, ΔHeao
The standard enthalpy change when an electron is added to one mole of isolated gaseous atoms to form a negative ion
First electron affinity, e.g.
[pic]
Second electron affinity, e.g.
[pic]
12.1.6a Lattice Dissociation enthalpy, ΔHlatto
The standard enthalpy change when one mole of a solid ionic lattice dissociates into gaseous ions, e.g.
[pic]
12.1.6b Lattice Formation enthalpy, ΔHlatto
The standard enthalpy change when one mole of a solid ionic lattice is formed from gaseous ions, e.g.
[pic]
Note that these latter two have the same numerical value but opposite signs – one is endothermic (dissociation) and the other is exothermic (formation) (cf. BDE’s from AS)
12.1.7 Enthalpy of solution, ΔHsolno
The standard enthalpy change when one mole of an ionic solid dissolves in a large enough amount of water that the dissolved ions are well separated and do not interact with each other, e.g.
[pic]
12.1.8 Enthalpy of hydration, ΔHhydo
The standard enthalpy change when one mole of ions are hydrated, e.g.
[pic]
[pic]
12.2 Conservation of Energy
Energy cannot be created or destroyed but can only be changed from one form into another.
Working in open vessels at a fixed pressure we are only concerned with changes in heat energy, i.e. enthalpy changes.
12.2.1 Hess’s Law – met previously at AS
This applies to Enthalpy changes:
“The enthalpy change in a reaction is independent of the route by which the reaction occurs and depends only on the initial and final states of the reactants” (o.w.t.t.e.)
12.2.2 Born-Haber cycles
The enthalpy of formation of an ionic compound can be broken down into a number of separate enthalpy steps – just a more complicated application of Hess’s Law! (
The enthalpy change for any one step can be calculated provided all other values are known.
Example Calculate the second electron affinity of oxygen using the following data:
[pic]
The Born-Haber cycle for the formation of MgO:
[pic]
[pic]
Cycling clockwise from the elements, Mg (g) + 1/2O2 (g);
+ 148 + 738 + 1451 + 249 + (– 142) + 2nd ΔHeao + (– 3845) = – 602
Hence 2nd ΔHeao = + 798 kJ mol-1
Born-Haber cycle – Workshop Question 1
A Born–Haber cycle for the formation of calcium sulphide is shown below. The cycle includes enthalpy changes for all Steps except Step F (the cycle is not drawn to scale).
[pic]
Born-Haber cycle – Workshop Question 2
A Born–Haber cycle for the formation of magnesium(II) chloride is shown below.
[pic]
|Name of standard enthalpy change |Substance to which enthalpy change refers |Value of enthalpy change/ kJ mol-1 |
|Enthalpy of atomisation |Chlorine |+121 |
|Enthalpy of atomisation |Magnesium |+150 |
|Enthalpy of formation |Magnesium Chloride |–642 |
|First ionisation enthalpy |Magnesium |+736 |
|Electron affinity |Chlorine |–364 |
|Enthalpy of lattice formation |Magnesium Chloride |–2493 |
3. Enthalpy of Solution
Example Given that ΔHlatto (NaBr) = +/- 735 kJ mol-1, ΔHhydo (Na+) = - 405 kJ mol-1 and ΔHhydo (Br-) = - 335 kJ mol-1, calculate the enthalpy of solution of NaBr(s)
Again, construct the Hess’s Law cycle:
[pic]
Note ΔHlatto defined in terms of forming the ionic lattice from gaseous ions, i.e. exothermic.
[pic]
ΔHsolno + (– 735) = (– 405) + (– 335)
Hence ΔHsolno = – 5 kJ mol-1
OR in general:
[pic]
Note this time ΔHlatto is defined in terms of dissociation of the ionic lattice to gaseous ions, i.e. endothermic (+ve).
ΔHsolno = + 735 + (– 405) + (– 335) = - 5 kJ mol-1
12.3 Free-energy change and entropy change
For an endothermic reaction to go there must be an additional energy factor.
Entropy, S This can be thought of in terms of disorder.
Increasing disorder, increasing entropy.
Increasing temperature increases molecular movement, increases entropy.
[pic]
| |H2O (s) |H2O (l) |H2O (g) |
|So / J K-1 mol-1 |+ 48 |+ 70 |+ 188 |
If ΔSo is positive, disorder is increasing, when in a reaction there is an increase in the number of moles of particles.
Entropy increases when a unidentate ligand is replaced by a multidentate ligand,. e.g.
[Cu(H2O)6]2+ + EDTA4- → [Cu(EDTA)]2- + 6 H2O Δn = 7 – 2 = + 5
Entropy also increases when solids dissolve in solvents, e.g.
[pic]
NH4NO3 is very soluble in water, even though both ΔHsolno and ΔSsolno are positive.
12.3.1 Feasible/Spontaneous Change
• In thermodynamics, a Feasible reaction is one for which the Gibb’s free energy change, ΔGo, is < 0, i.e. –ve. That is, it is possible (feasible) for it to occur according to thermodynamics.
• A Spontaneous reaction is a feasible one which actually happens under the given conditions for the reaction.
• It is thus perfectly possible for a reaction to be feasible but not spontaneous, e.g. combustion; ΔGo is –ve, but the reaction does not occur spontaneously – why not? Spontaneity depends largely on kinetic factors, such as the size of the Ea, etc.
12.3.2 Gibbs free-energy change
This links entropy change, temperature and enthalpy change.
ΔGo = ΔHo – TΔSo
When ΔGo = 0 equilibrium has been established
When ΔGo > 0 reaction not feasible
When ΔGo < 0 reaction is feasible
12.3.3 Equilibria When ΔGo = 0 the system is at equilibrium.
Consider the equilibrium [pic] established at 273 K
ΔGo = 0 = ΔHo – TΔSo
Hence for this equilibrium ΔHfuso = TfusΔSfuso
and ΔSfuso ’ ΔHfuso / Tfus = 6.0 / 273 = +0.0219 kJ K-1 mol-1
i.e. ΔSfuso ’ +21.9 J K-1 mol-1
NB. For the following equilibrium [pic] established at 373 K
ΔSvapo ’ +118.0 J K-1 mol-1 showing that much more ‘disorder’ is created when water vaporises than when it melts.
12.3.4 Calculating the temperature at which a reaction becomes feasible
When data is available it is possible to calculate when a reaction either becomes feasible or ceases to be feasible.
Example 1 Use the data given below to calculate the lowest temperature at which the following reaction is feasible
[pic]
| |H2O (g) |C (s) |H2 (g) |CO (g) |
| ΔHfo / kJmol-1 |– 242 |0 |0 |– 111 |
|So / J K-1 mol-1 |+ 189 |+ 6 |+ 131 |+ 198 |
[pic]
ΔHo = [0 + (–111)] – [0 + (– 242)] = + 131 kJ mol-1
[pic]
ΔSo = (131 + 198) – (189 + 6) = + 134 J K-1 mol-1
Reaction feasible when ΔGo = ΔHo – TΔSo = 0; so T = ΔHo / ΔSo
Hence T = 131 x 1000 / 134 or 131 / (134 x 10-3) = 978 K
NB. Note the conversion factor of 1000 to convert ΔH and ΔS into the same units! (
12.3.5 Qualitative use of ΔG = ΔH – TΔS
Example 1 Explain why ice does not melt below 273K under standard conditions?
Since ΔG = ΔH – TΔS = 0 for minimum feasibility.
For the process: [pic]
ΔH = +ve; ΔS = +ve; hence 0 = (+ve) – T x (+ve);
Thus T = 273 will be the minimum value for which TΔS = ΔH and thus ΔG = 0
Example 2 Explain why NH4NO3 is very soluble in water even though ΔHsolno is +ve?
Again, ΔG = ΔH – TΔS = 0 for minimum feasibility.
ΔHsolno is +ve but ΔS is also +ve since [pic] involves a large increase in disorder (1 mole solid to 2 moles aqueous ions).
Thus 0 = (+ve) – T x (+ve), so at a minimum T, TΔS > ΔH and solubility becomes feasible. This T must be < 298 K and thus ammonium nitrate dissolves readily in cold water!
A qualitative summary
Since ΔG = ΔH – TΔS = 0 for minimum feasibility:
|ΔH |ΔS |TΔS |ΔG |Feasibility |
|+ve |+ve |+ve |–ve at or > T |minimum Tfeas |
|–ve |+ve |+ve |always –ve |always feasible |
|+ve |–ve |–ve |always +ve |never feasible |
|–ve |–ve |–ve |–ve at or < T |maximum Tfeas |
13. PERIODIC TRENDS IN CHEMICAL PROPERTIES
13.1 Trends in the reaction of elements with water
Sodium with cold water:
Floats; melts; moves; gas (might burn with yellow flame); colourless solution formed.
[pic]
Magnesium (i) with cold water:
Very slow reaction, gas produced
(ii) with steam:
Burns after ignition, brilliant white flame; white solid formed and gas
[pic]
13.2 Trends in the reaction of elements with oxygen
The elements Na to S all burn in oxygen, or air, after ignition to form oxides.
Sodium burns with a yellow flame; Sulphur with a blue flame, others with a white flame.
All, except sulphur, form oxides with their Group number oxidation state. Sulphur forms SO2.
[pic]
13.3 Acid-base properties of Period 3 oxides
Hydration: Water molecules held by an attraction between an ion and polar water
molecules
Hydrolysis: A reaction with water producing new species.
Factors involved in ionic compound hydrolysis:
• The charge on the ion
• The size of the ion
13.4 The reactions with water of the oxides formed by Period 3 elements
Na Mg Al Si P S [Cl] [Ar]
Oxides Na2O MgO Al2O3 SiO2 P4O10 SO2 [Cl2O] [none]
SO3
13.4.1 Oxides of Metals
For simple ionic compounds, the enthalpy of solution depends on the balance between the magnitude of the lattice enthalpy and the sum of the hydration enthalpies of the ions (see earlier Thermodynamics section 12.2.3; p59).
Entropy changes must also be considered when total energy changes are calculated.
As a guide to understanding, we can predict lattice enthalpies and solubilities using ionic radii.
Na2O is soluble water (all simple sodium compounds are soluble in water)
[pic]
In the solution formed:
Na( ions are hydrated; i.e. Na+ (aq)
O2( ion are hydrolysed; i.e. [pic]
What you need to know about Na2O and its reaction with water is:
Bonding Ionic
Structure Lattice
Melting point 1548 K
Observation Dissolves in water forming a colourless solution
pH of solution 14
[pic]
MgO predict a higher lattice enthalpy than Na2O (this is correct!).
[pic]
Predict MgO is sparingly soluble in water; any oxide ions are hydrolysed.
What you need to know about MgO and its reaction with water is:
Bonding Ionic
Structure Lattice
Melting point 3125 K
Observation Sparingly soluble
Observation White suspension produced
pH of solution 9
[pic]
Al2O3 predict an even higher lattice enthalpy than MgO (this is also correct!).
[pic]
Predict A12O3 will be insoluble in water. It is insoluble!
What you need to know about A12O3 and its reaction with water is:
Bonding Ionic-covalent
Structure Lattice
Melting point 2345 K
Observation Insoluble in water, no reaction.
pH That of water, 7; i.e. no change
13.4.2 Covalent Oxides
Silicon dioxide SiO2
[pic]
Does silicon dioxide dissolve in, or react with, water?
Factors to consider
• The strength of the Si – O covalent bonds.
• The polarity of the Si – O covalent bonds
• How strongly water molecules will be attracted
• Will the enthalpy of attraction exceed the Si – O and O – H covalent bond energies?
What you need to know about SiO2 and its reaction with water is:
Bonding Covalent
Structure Macromolecular
Melting point 1883 K
Observation Insoluble in water, no reaction
pH That of water, 7; i.e. no change
Reactions of other covalently bonded Period 3 oxides
• All are molecular
• All contain X = O covalent bonds (X = a non-metal element)
• Bond polarity important
• All react with water.
What you need to know about P4O10 and its reaction with water is:
Bonding Covalent
Structure Molecular
Melting point 573 K
Observation Reacts violently with water
Observation Reaction exothermic
Observation Colourless solution formed
pH 0, a strong acid formed
[pic]
Sulphur dioxide, SO2
Knowing the structure of SO2 will help you to understand the reaction of SO2 with water.
[pic]
What you need to know about SO2 and its reaction with water is:
Bonding Covalent
Structure Molecular
Melting point 200 K
Observation Moderately soluble in water
Observation Colourless solution formed
pH of solution 3; H2SO3 is a weak acid.
[pic]
Sulphur trioxide, SO3
Knowing the structure of SO3 will help you to understand the reaction of SO3 with water.
[pic]
i.e. the strong acid H2SO4
What you need to know about SO3 and its reaction with water is:
Bonding Covalent
Structure Molecular
Melting point 290 K
Observation Very soluble in water
Observation Colourless solution formed
pH of solution 0, H2SO4 is a strong acid
[pic]
14. TRANSITION ELEMENTS
Transition Series
|Transition Series |Elements | |
|First Series |Sc – Zn |3 d |
|Second Series |Y – Cd |4 d |
|Third Series |La – Hg |5 d |
14.1 General Characteristics of Transition Metals:
1. Form complex ions
2. Form coloured ions
3. Act as catalysts
4. Form compounds with the transition element in different oxidation states
These properties are linked to the presence of an incomplete d-sub-shell in an ion.
Consider:
Sc 1s2 2s2 2p6 3s2 3p6 3d1 4s2
Cu 1s2 2s2 2p6 3s2 3p6 3d10 4s1
Cu+ 1s2 2s2 2p6 3s2 3p6 3d10 4s0
Cu2+ 1s2 2s2 2p6 3s2 3p6 3d9 4s0
14.2 Complex Formation Definitions
|Complex |A central metal atom or ion surrounded by ions or molecules called ligands |
|Ligand |An ion or molecule which donates an electron pair to form a co-ordinate bond, |
| |i.e. a Lewis Base or nucleophile |
|Coordinate bond |A covalent bond formed using two electrons from one atom (the ligand) |
|Coordination number |The number of coordinate bonds formed to the central metal atom (NOT the number |
| |of ligands – see later) |
14.3 Co-ordination Numbers / Shapes of Complexes
If the coordination number is 6, the shape is octahedral:
[pic]
If the coordination number is 4, the shape is often tetrahedral:
[pic]
If the coordination number is 2, the shape is linear:
[pic]
14.4 Types of ligands
14.4.1 Unidentate ligands
These are “one-bite” ligands – one lone pair of electrons donated per ligand.
Examples include ligands with a period 2 donor atom: H2O, NH3, OH-, CN-
or ligands with a period 3 donor atom: Cl-, S2O32-
Examples:
|Metal Ion |Ligand |Coordination Number |Complex |Shape |Colour |
|Cr3+ |H2O |6 |[Cr(H2O)6]3+ |octahedral |green |
|Co3+ |NH3 |6 |[Co(NH3)6]3+ |octahedral |brown |
|Cu2+ |Cl- |4 |[CuCl4]2- |tetrahedral |blue |
|Ag+ |S2O32- |2 |[Ag(S2O3)2]3- |linear |colourless |
|Ag+ |NH3 |2 |[Ag(NH3)2]+ |linear |colourless |
|Ag+ |Cl- |2 |[AgCl2]- |linear |colourless |
14.4.2 Bidentate Ligands
These are “two bite” ligands - two lone pairs of electrons donated per ligand.
Examples:
Ethylenediamine (en):
[pic]
Ethanedioate anion:
[pic]
14.4.3 Multidentate ligands
Examples: a hexadentate ligand - six lone pairs of electrons donated per ligand.
Ethylenediaminetetraacetate ion, EDTA4-
[pic]
Example 1 [Fe(C2O4)3]3-
Example 2 [Pt(NH3)2Cl2]
14.5 Formation of coloured ions
14.5.1 Colour depends upon:
1. The transition metal.
2. The oxidation state of the transition metal.
3. The number of ligands surrounding the transition metal ion.
4. The nature of the ligands surrounding the transition metal ion.
14.5.2 Origin of colour:
• In presence of ligand, the 5 x 3d orbitals of a transition metal ion split to form 2 slightly different energy levels
• The electrons in the lower 3d sub-level absorb visible light
• Electrons move to a higher, more excited energy state (d → d transitions)
• Difference in energy between levels is:
E2 – E1 = ΔE = hν ’ hc / λ
(where h = Planck’s constant, ν is the frequency of light absorbed, c = speed of
light and λ is the wavelength).
• We see the colours which are not absorbed, i.e. the complementary colour
14.5.3 An application
• Measurement of transition metal ion concentration using a colorimeter/ spectrophotometer
• Note that an intense colour is needed for accurate results and this can be achieved by use of a suitable ligand, e.g.
Fe2+ (aq) absorption weak: add 2,2-bipyridyl (intense red complex)
Fe3+ (aq) absorption weak; add thiocyanate (intense red complex)
14.6 Catalytic activity
The transition metal as a catalyst has the following effect on the reaction:
• Provides an alternative route.
• Lowers the activation energy.
• Hence more molecules can react.
• Remember the catalyst remains unchanged in amount and chemical nature.
There are 2 types of catalysts: homogeneous (same state) and heterogenous (different state):
14.6.1 Examples of Homogeneous catalysis include
NB These two examples are specifically mentioned in both the AQA & Edexcel specifications! (
Example 1:
Overall: S2O82- (aq) + 2 I- (aq) → 2 SO42- (aq) + I2 (aq);
This reaction is catalysed by Fe2+
(note this reaction has a high Ea due to a reaction between two like charged ions)
Step 1: S2O82- (aq) + Fe2+ (aq) → 2 SO42- (aq) + Fe3+ (aq)
Step 2: 2 I- (aq) + Fe3+ (aq) → I2 (aq) + Fe2+ (aq)
NB Fe2+ does not appear in the overall equation and is recovered unchanged at the end of the reaction.
Example 2:
MnO4- / C2O42- : catalysed by Mn2+ (NB this is an example of autocatalysis – the catalyst is generated in the reaction and thus the reaction speeds up as it proceeds).
2 MnO4- (aq) + 5 C2O42- (aq) + 16 H+ (aq) → 2 Mn2+ (aq) + 5 CO2 (g) + 8 H2O (l)
Mechanism:
Step 1: 4 Mn2+ + MnO4- + 8 H+ → 5 Mn3+ + 4 H2O
Step 2: 2 Mn3+ + C2O42- → 2 Mn2+ + 2 CO2
NB Mn2+ does not appear in the overall equation and is recovered unchanged at the end of the reaction.
14.6.2 Examples of Heterogenous catalysis
These are reversible reactions
The 5 steps in heterogenous catalysis and a simple analogy:
|1. Diffusion to surface |Beach party to ice cream seller |
|2. Adsorption on surface |Arriving at the counter |
|3. Reaction on surface |Making friends |
|4. Desorption from surface |Leaving the counter |
|5. Diffusion from surface |Moving away |
Example 1: Vanadium(V) oxide, V2O5, used in the Contact Process:
[pic]
Step 1: SO2 + V2O5 → SO3 + V2O4
Step 2: 2 V2O4 + O2 → 2 V2O5
Example 2: Iron on silica support, used in the Haber process to manufacture ammonia:
[pic]
Example 3: Nickel, very finely divided, used in the hydrogenation of vegetable oil:
[pic]
Hydrogenation causes hardening of vegetable oil forming fat. It is an important step in the manufacture of margarine. The extent of hydrogenation is linked to hardness.
14.7 Acidity reactions of metal aqua complexes (hydrolysis) (AQA ONLY)
Consider separate solutions containing the ions:
[M(H2O)6]3+
[M(H2O)6]2+
[M(H2O)6]+
Example 1: for a solution containing approximately 1 mol dm-3 of [M(H2O)6]3+
[pic]
[pic]
Approximately 1000 n 1 1
Hence: [H+] ~10-3 and pH ~3
Example 2: for a solution containing approximately 1 mol dm-3 of [M(H2O)6]2+
[pic]
[pic]
Approximately 100 000 n 1 1
Hence: [H+] ~10-5 and pH ~5
14.8 Reactions of metal aqua complexes with other bases
Consider the relative base strengths of the following:
OH-; NH3; CO32-
Example 1: The addition of a 2.0 mol dm-3 solution of NaOH to a solution containing, e.g. Al3+ (aq)
[pic]
[pic]
NB The [Al(H2O)2(OH)4]- complex ion is often more simply written as [Al(OH)4]-
Summary:
• OH- acting as a base.
• Precipitation when no repulsion; species hydrogen bond together.
• Excess OH- (aq) forms negatively-charged species.
• Repulsion causes separation, precipitate re-dissolves.
Example 2: Addition of concentrated NH3 (aq) to a solution containing Cr3+ (aq)
[pic]
[pic]
Summary:
• NH3 is a weaker base than OH-.
• NH3 acts first as a base than as a ligand.
• Precipitation when no repulsion; species hydrogen bond together.
• Excess NH3 (aq) forms positively charged species.
• Repulsion casues separation, precipitate re-dissolves.
Example 3: In concentrated NH3 (aq) the ratio H2O: NH3 is 2:1
Substitution reactions are in equilibria and sometimes not all H2O replaced
[pic]
[pic]
Example 4: Addition of solution containing CO32- to solution containing M3+ (aq)
CO32- acts only as a base, e.g.
[pic]
[pic]
The following INCORRECT stepwise equations aid understanding of this reaction:
[pic]
NB Aqueous complex ions of metals in oxidation state (II) form simple carbonate
precipitates on the addition of aqueous Na2CO3. NO CO2 is produced.
e.g. Fe2+ (aq) + CO32- (aq) → FeCO3 (s)
14.9 Redox reactions
Useful rules of thumb:
• High oxidation state to low oxidation state:
Reagents: acid (e.g. dilute H2SO4) + reducing agent (e.g. Zn or SO2)
• Low oxidation state to high oxidation state
Reagents: base (e.g. NaOH (aq) ) + oxidising agent (e.g. H2O2)
Explanations:
High oxidation states are commonly found in complexes containing oxygen
Consider the reduction: MnO4- → [Mn(H2O)6]2+
Oxidation States: VII II
Consider the oxidation: [Cr(H2O)6]3+ → CrO42-
Oxidation States: III VI
14.10 Oxidation states of vanadium (AQA & Edexcel only)
Summary table:
|Oxidation state |Formula of species |Colour of aqueous solution |
|+5 |VO3- |yellow |
|+5 |VO2+ |orange |
|+4 |VO2+ |blue |
|+3 |V3+ |green |
|+2 |V2+ |violet |
There are two aqueous species which are both V(V) – converted in presence of acid:
VO3- + 2 H+ → VO2+ + H2O
V V
Note that this is NOT a redox reaction as there are no electrons in the overall equation.
V(V) can be reduced stepwise through the other oxidation states down to V(II) with an appropriate reducing agent, i.e:
VO2+ + 2 H+ + e- → VO2+ + H2O SO2 will reduce V(V) to V(IV)
V IV
VO2+ + 2 H+ + e- → V3+ + H2O SO2 will reduce V(IV) to V(III)
IV III Sn will reduce V(IV) to V(III)
V3+ + e- → V2+ Zn will reduce V(III) to V(II)
III II
Each of these reductions can be predicted using the appropriate half-equations and Eo values listed below:
VO2+ + 2 H+ + e- → VO2+ + H2O Eo = +1.00 V
VO2+ + 2 H+ + e- → V3+ + H2O Eo = +0.34 V
SO42- + 4 H+ + 2e- → H2SO3 + H2O Eo = +0.17 V
Sn2+ + 2e- → Sn Eo = - 0.14 V
V3+ + e- → V2+ Eo = - 0.26 V
Zn2+ + 2e- → Zn Eo = - 0.76 V
V2+ + 2e- → V Eo = - 1.18 V
14.11 Substitution reactions
14.11.1 Substitution by chloride ligands
1. Chloride ions are larger than water and ammonia molecules (period 3 vs. period 2).
2. Substitution of all water ligands by chloride ligands results in a change in coordination number and shape.
3. The colour of the complex changes.
4. Since chloride ions are charged, the charge on the complex changes.
5. Substitutions are equilibria; conc. HCl (aq) gives chloro complexes (NaCl is not very soluble and does not have a sufficiently high chloride ion concentration).
e.g. [Co(H2O)6]2+ + 4 Cl- → [CoCl4]2- + 6 H2O
pink solution blue solution
[Cr(H2O)6]3+ [Cr(H2O)5Cl]2+ [Cr(H2O)4Cl2]+
green/blue pale green dark green
14.11.2 Substitution of monodentate ligands by bidentate ligands:
e.g. [Co(H2O)6]2+ + 3 en → [Co(en)3]2+ + 6 H2O
e.g. [Co(H2O)6]2+ + 3 C2O42- → [Co(C2O4)3]4- + 6 H2O
These complexes are octahedral.
NB: bidentate ligands can also bond two metal ions together acting as a bridge.
14.11.3 Substitution of monodentate ligands by the hexadentate ligand EDTA4-
1. EDTA4- is a hexadentate ligand and forms a 6 coordinate complex.
2. A colour change ususally occurs when water is replaced by EDTA4-.
3. The complexes formed are octahedral.
4. There is a large entropy change when a monodentate ligand is replaced by EDTA4-
e.g. [Co(H2O)6]2+ + EDTA4- → [Co(EDTA)]2- + 6 H2O
2 ions form 7 species
14.12 The stability of complexes
Complexes in which bidentate or multidentate ligands are bonded to one central metal ion are known as chelates. Chelate stability depends on:
1. Size of ring formed by ligand with metal ion, i.e. strain.
2. Metal-ligand bond energy.
3. Number of bonds to be broken when ligand replaced, e.g. 6 for EDTA4-.
4. A reaction is feasible when Gibbs Free Energy, ΔG, is negative.
ΔG = ΔH – TΔS.
If the change in entropy, ΔS, is large and positive; and change in energy ΔH is small when bond energies are similar, then TΔS >>> ΔH. Hence, Gibbs Free Energy (ΔG) is large and negative.
NB: The stability of EDTA4- complexes is so high that the reactions go to completion so that, for example, no precipitate forms when NaOH (aq) is added. This is known as the chelate effect.
13. Summary Table of Aqueous Metal Ion Chemistry (AQA)
| |Al3+ |Fe2+ |Fe3+ |Cu2+ |
|Formula & colour | | | | |
|of species | | | | |
| | | | | |
|aqueous solution |[Al(H2O)6]3+ |[Fe(H2O)6]2+ Pale |[Fe(H2O)6]3+ Brown |[Cu(H2O)6]2+ |
| |Colourless solution |green solution |solution |Light blue solution |
| | | | | |
|dropwise NaOH |[Al(H2O)3(OH)3] |[Fe(H2O)4(OH)2] |[Fe(H2O)3(OH)3] |[Cu(H2O)4(OH)2] |
| |White ppt |Green ppt |Brown ppt |Blue ppt |
| | |turns to brown ppt in | | |
| | |air | | |
| | | | | |
|excess NaOH |[Al(H2O)2(OH)4]-Colour|No change |No change |No change |
| |less solution | | | |
| | | | | |
|dropwise NH3 |Al(H2O)3(OH)3] |[Fe(H2O)4(OH)2] |[Fe(H2O)3(OH)3] |[Cu(H2O)4(OH)2] |
| |White ppt |Green ppt |Brown ppt |Blue ppt |
| | |turns to brown ppt in | | |
| | |air | | |
| | | | | |
|excess NH3 |No change |No change |No change |[Cu(NH3)4(H2O)2]2+ |
| | | | |Deep blue solution |
| | | | | |
|excess Na2CO3 |[Al(H2O)3(OH)3] |FeCO3 |[Fe(H2O)3(OH)3] |CuCO3 |
| |White ppt & CO2 |Green ppt |Brown ppt & CO2 |Green ppt |
| |produced | |produced | |
| |X |X |X | |
|excess conc. HCl | | | |[CuCl4]2- |
| | | | |Green solution |
| | | | |Tetrahedral |
| | | | | |
|pH of aqueous |pH = 2 – 3 |pH = 7 |pH = 2 – 3 |pH = 5 – 6 |
|solution | | | | |
| |acidic |neutral |acidic |weakly acidic |
NB. Shapes of ALL complexes are octahedral unless otherwise stated in the table.
14.13a My Summary Rules for Learning Aqueous Metal Ion Chemistry (AQA)
1. With dropwise NaOH:
ALL give a ppt. of the metal hydroxide, [M(OH)n(H2O)6-n], where n = 2 or 3
2. With excess NaOH:
Al only dissolves and gives [Al(OH)4(H2O)2]-
Others don’t (still metal hydroxide ppt. above)
3. With dropwise NH3:
ALL give a ppt. of the metal hydroxide (same as with dropwise NaOH)
4. With excess NH3:
ALL dissolve except Fe2+ and Fe3+ (still metal hydroxide ppts.)
Cu gives [Cu(NH3)4(H2O)2]2+; Al gives [Al(NH3)6]3+
5. With Na2CO3:
M3+ ALL give ppt. of metal hydroxide and effervescence of CO2
M2+ ALL give ppt. of metal carbonate, MCO3
6. With c. HCl:
Cu only - gives tetrahedral [CuCl4]2- complex ion
7. Acidity:
M3+ ALL acidic
M2+ ALL neutral, except Cu2+ which is only slightly acidic
14. Summary Table of Aqueous Metal Ion Chemistry (OCR-A)
| |Cr3+ |Mn2+ |Fe2+ |Fe3+ |Cu2+ |
|Formula & colour| | | | | |
|of species | | | | | |
| | | | | | |
|aqueous solution|[Cr(H2O)6]3+ Purple |[Mn(H2O)6]2+ |[Fe(H2O)6]2+ |[Fe(H2O)6]3+ Brown |[Cu(H2O)6]2+ |
| |solution |Pale pink |Pale green solution |solution |Light blue solution |
| | |(colourless) solution| | | |
| | | | | | |
|dropwise NaOH |[Cr(H2O)3(OH)3] |[Mn(H2O)4(OH)2] |[Fe(H2O)4(OH)2] |[Fe(H2O)3(OH)3] |[Cu(H2O)4(OH)2] |
| |Grey/green ppt |Pale brown ppt turns |Green ppt |Brown ppt |Blue ppt |
| | |darker brown ppt in |turns to brown ppt | | |
| | |air |in air | | |
| | | | | | |
|excess NaOH |[Cr(OH)6]3- |No change |No change |No change |No change |
| |Green solution | | | | |
| | | | | | |
|dropwise NH3 |[Cr(H2O)3(OH)3] |[Mn(H2O)4(OH)2] |[Fe(H2O)4(OH)2] |[Fe(H2O)3(OH)3] |[Cu(H2O)4(OH)2] |
| |Grey/green ppt |Pale brown ppt turns |Green ppt |Brown ppt |Blue ppt |
| | |darker brown ppt in |turns to brown ppt | | |
| | |air |in air | | |
| | | | | | |
|excess NH3 |[Cr(NH3)6]3+ |No change |No change |No change |[Cu(NH3)4(H2O)2]2+ |
| |Purple solution | | | |Deep blue solution |
| |X |X |X |X | |
|excess conc. HCl| | | | |[CuCl4]2- |
| | | | | |Green solution |
| | | | | |Tetrahedral |
NB. Shapes of ALL complexes are octahedral unless otherwise stated in the table.
14.14a My Summary Rules for Learning Aqueous Metal Ion Chemistry (OCR-A)
1. With dropwise NaOH:
ALL give a ppt. of the metal hydroxide, [M(OH)n(H2O) 6-n] where n = 2 or 3
2. With excess NaOH:
Cr only dissolves and gives [Cr(OH)6]3-
Others don’t (still metal hydroxide ppt. above)
3. With dropwise NH3:
ALL give a ppt. of the metal hydroxide (same as with dropwise NaOH)
4. With excess NH3:
Cr & Cu only dissolve - Cr gives [Cr(NH3)6]3+ and Cu gives [Cu(NH3)4(H2O)2]2+
Others don’t (still metal hydroxide ppt. as above)
5. With c. HCl:
Cu only - gives tetrahedral [CuCl4]2- complex ion
15. Summary Table of Aqueous Metal Ion Chemistry (Edexcel)
| |Cr3+ |Fe2+ |Fe3+ |Co2+ |Cu2+ |
|Formula & colour| | | | | |
|of species | | | | | |
| | | | | | |
|aqueous solution|[Cr(H2O)6]3+ Purple |[Fe(H2O)6]2+ Pale |[Fe(H2O)6]3+ Brown |[Co(H2O)6]2+ |[Cu(H2O)6]2+ |
| |solution |green solution |solution |Pink solution |Light blue solution |
| | | | | | |
|dropwise NaOH |[Cr(H2O)3(OH)3] |[Fe(H2O)4(OH)2] |[Fe(H2O)3(OH)3] |[Co(H2O)4(OH)2] |[Cu(H2O)4(OH)2] |
| |Grey/green ppt |Green ppt |Brown ppt |Blue ppt |Blue ppt |
| | |turns to brown ppt in | |but leave in air forms| |
| | |air | |beige ppt | |
| | | | |[Co(H2O)3(OH)3] | |
| | | | | | |
|excess NaOH |[Cr(OH)6]3- |No change |No change |No change |No change |
| |Green solution | | | | |
| | | | | | |
|dropwise NH3 |[Cr(H2O)3(OH)3] |[Fe(H2O)4(OH)2] |[Fe(H2O)3(OH)3] |[Co(H2O)4(OH)2] |[Cu(H2O)4(OH)2] |
| |Grey/green ppt |Green ppt |Brown ppt |Blue ppt |Blue ppt |
| | |turns to brown ppt in | | | |
| | |air | | | |
| | | | | | |
|excess NH3 |[Cr(NH3)6]3+ |No change |No change |[Co(NH3)6]2+ → |[Cu(NH3)4(H2O)2]2+ |
| |Purple solution | | |[Co(NH3)6]3+ |Deep blue solution |
| | | | |Straw/yellow solution | |
| | | | |turns to brown | |
| | | | |solution in air | |
| |X |X |X | | |
|excess conc. HCl| | | |[CoCl4]2- |[CuCl4]2- |
| | | | |Blue solution |Green solution |
| | | | |Tetrahedral |Tetrahedral |
NB. Shapes of ALL complexes are octahedral unless otherwise stated in the table.
14.15a My Summary Rules for Learning Aqueous Metal Ion Chemistry (Edexcel)
1. With dropwise NaOH:
ALL give a ppt. of the metal hydroxide, [M(OH)n(H2O) 6-n], where n = 2 or 3
2. With excess NaOH:
Cr only dissolves to give [Cr(OH)6]3-
Others don’t (still metal hydroxide ppt. above)
3. With dropwise NH3:
ALL give a ppt. of the metal hydroxide (same as with dropwise NaOH)
4. With excess NH3:
ALL dissolve except Fe2+ and Fe3+ (still metal hydroxide ppts.)
Cr & Co give [M(NH3)6]n+; Cu gives [Cu(NH3)4(H2O)2]2+
5. With c. HCl:
Co & Cu only - give tetrahedral [MCl4]2- complex ions
15. REDOX
1. Redox reactions
You should already know from AS Chemistry;
• That oxidation and reduction involve transfer of electrons
• Recall the acronym “OILRIG” to remember that:
Oxidation Is Loss and Reduction Is Gain (of electrons)
• To use OILRIG to deduce if a species has been oxidised or reduced in a reaction
• To know the difference between oxidised and an oxidising agent and reduced and a reducing agent! (
• Be able to write balanced half-equations when reactant and product are specified
• Be able to add two balanced half-equations to produce the overall equation (electrons cancel).
1. Writing balanced half-equations:
1. write down the reactant and product as supplied.
2. balance the equation for atoms – remember, you can use H+ and.or H2O to balance the loss or gain of H or O in an equation – do NOT write H2 or O2 !!
3. finally, add the appropriate number of electrons to balance the charges.
Example Deduce the half equation for the oxidation, in acid solution, of SO2 to SO42-
SO2 → SO42- Two “oxygens” needed - add 2 H2O to LHS
2 H2O + SO2 → SO42- + 4 H+ Balance for atoms – add 4 H+ to RHS
2 H2O + SO2 → SO42- + 4 H+ + 2e- Balance for charges by adding 2e- to RHS
2. Writing the balanced full equation from the half-equations:
1. firstly, multiply up each half-equation as necessary to get the same number of electrons in each one.
2. add the two half equations together – the electrons will cancel out and you may find that other species such as H+ or H2O cancel down or out too.
3. finally, double check the final equation is balanced for the atoms of each element on each side and the charges too.
Example Deduce the full equation for the oxidation of SO2 to SO42- by MnO4- ?
Half equations: 1. 2H2O + SO2 → SO42- + 4H+ + 2e-
2. MnO4- + 8H+ + 5e- → Mn2+ + 4H2O
Multiply 1. by 5 and 2. by 2 to get same no. of electrons in each one:
1. 10H2O + 5SO2 → 5SO42- + 20H+ + 10e-
2. 2MnO4- + 16H+ + 10e- → 2Mn2+ + 8H2O
Add together: 10H2O + 5SO2 + 2MnO4- + 16H+ → 5SO42- + 20H+ + 2Mn2+ + 8H2O
Cancel H+ and H2O on each side to produce final equation:
2H2O + 5SO2 + 2MnO4- → 5SO42- + 4H+ + 2Mn2+
15.2 Redox titration calculations
15.2.1 Basic method
A reminder that most moles calculations involve the same three steps:
Step 1 – calculate the no. of moles (n) of the known substance;
Step 2 – use the balanced equation or stoichiometry for the reaction (always supplied)
to deduce the no. moles of the unknown (the one you’re asked to find in the
question);
Step 3 – convert the unknown no. of moles into the required answer (mass, Mr,
concentration, volume, % purity, etc)
We will use this method in all the calculations in this booklet.
15.2.2 Calculations
Again, these use the same basic three step method as for other moles calculations (see previous section).
– this time step 1 to calculate the known usually (but not always) involves the titre value.
e.g. 1 25.0 cm3 of an acidified solution of ethandioate, C2O42-, was found to react with exactly 22.7 cm3 of a 0.020 M solution of potassium manganate(VII).
Calculate the concentration of the ethanedioate solution?
1. No. moles MnO4- (known) = C x V / 1000 = 0.020 x 22.7 / 1000 = 4.54 x 10-4 mol
2. Using equation:
[pic]
2 mol MnO4- reacts with 5 mol C2O42-
1 mol MnO4- reacts with 5/2 mol C2O42-
4.54 x 10-4 mol MnO4- reacts with 4.54 x 10-4 x 5/2 = 1.14 x 10-3 mol C2O42-
3. Concentration (C) of C2O42- = 1000 x n / V = 1000 x 1.14 x 10-3 / 25.0 = 0.0454 M
e.g. 2 A 0.152g iron sample containing an insoluble impurity was found to react with exactly 17.3 cm3 of a 0.0160 M solution of Cr2O72-.
Calculate the % by mass of iron in the sample?
1. No. moles Cr2O72- (known) = C x V / 1000 = 0.0160 x 17.3/1000 = 2.77 x 10-4 mol
2. Using equation:
[pic]
1 mol Cr2O72- reacts with 6 mol Fe2+
2.77 x 10-4 mol Cr2O72- reacts with 2.77 x 10-4 x 6 mol Fe = 1.66 x 10-3 mol
3. Mass of Fe = n x Mr = 1.66 x 10-3 x 55.9 = 0.0928 g
4. % Fe = 0.0928 / 0.152 x 100 = 61.1 %
e.g. 3 Separate aqueous solutions of potassium manganate(VII) (KMnO4) and sodium manganate(VII) (NaMnO4) each contained 1.00 g of the compound in 250 cm3. A 25.0 cm3 sample of one of these solutions reacted with exactly 17.6 cm3 of a 0.100 M acidified solution of sodium ethanedioate, Na2C2O4.
1. No. moles C2O42- (known) = C x V / 1000 = 0.100 x 17.6/1000 = 1.76 x 10-3 mol
2. Using equation:
[pic]
5 mol C2O42- reacts with 2 mol MnO4-
1 mol C2O42- reacts with 2/5 mol MnO4-
1.76 x 10-3 mol C2O42- reacts with 1.76 x 10-3 x 2/5 mol MnO4- = 7.04 x 10-4 mol
3. Moles of MnO4– in 250 cm3 = 7.04 x 10-4 x 10 = 7.04 x 10-3 mol
4. Molar mass of MMnO4 salt, Mr = m/n = 1.00/7.04 x 10-3 = 142 g mol-1
5. Mr (KMnO4) = 158 g mol-1 & Mr (NaMnO4) = 142 g mol-1 so solution is NaMnO4
16. ELECTROCHEMISTRY
16.1 Electrode potentials
These are measured against a reference electrode, the HYDROGEN electrode, which is arbitarily assigned a potential of 0.00 V.
All other half-cells have their potentials measured against this hydrogen electrode – if this is done under the usual “standard conditions”, i.e. a pressure of 100 kPa, a temperature of 298 K and a 1M solution, the electrode potential becomes the standard electrode potential for that half-cell, given the symbol Eo.
We can use tables of Eo values to predict:
1. if two species are capable of reacting with each other in a redox reaction.
2. the likely reaction which will occur if two half-cell systems are brought together.
We will work through an example of each type of question.
A table of common Eo values is given below – you do NOT need to learn these – the half-equations and Eo values you will need to answer an examination question will always be supplied in the question paper! (
Table of some common electrode potentials
[pic]
The more +ve Eo values are systems which are good oxidisers – they consume electrons and would form the +ve right-hand electrode of a electrochemical cell. They do the forward reduction reaction as written in the table above.
The more –ve Eo values are systems which are good reducers – they produce electrons and would form the –ve left-hand electrode in an electrochemical cell. They do the reverse oxidation reaction to that written in the table above.
Tables of Eo values are usually written in this format, with the most +ve best oxidisers at the top (on the left hand side) and the most –ve best reducers at the bottom (on the right-hand side).
It thus follows that any species on the left-hand side (a potential oxidiser) can react with any species below it on the right-hand side (a potential reducer). It also follows that species with the reverse relationship in the table can NEVER react together.
16.2 The Rules:
1. write down the half-equation with the most +ve Eo – this is the RH electrode
2. write down the half-equation with the more –ve Eo in reverse – this is the LH electrode.
3. Multiply electrons and then add the two half-equations together to produce the overall reaction.
4. Calculate Ecello = ERighto - ELefto (DO NOT multiply the Eo values – they are independent of the number of electrons transferred).
e.g 1: predict if V2+ can react with I2. Write an overall equation for the reaction and calculate the overall Ecello.
1. Most +ve Eo: [pic]
2. Reverse more –ve Eo: [pic]
3. Add half-eqns together: [pic]
4. Calculate Ecell = ER - EL = + 0.54 – (- 0.26) = + 0.80 V
Since ΔG = - nEoF (you don’t need to remember this equation); when Eo is +ve,
ΔG will be –ve and thus the reaction is feasible. Of course, if the forward reaction has a –ve Eo, then the reverse reaction will have a +ve Eo and be the feasible one in this case.
e.g. 2: predict the cell reaction which will occur if the Pb2+/Pb and Fe2+/Fe3+ cells are connected together. Write the overall cell reaction and calculate the Ecell.
1. Most +ve Eo: [pic]
2. Reverse more –ve Eo: [pic]
3. Add half-eqns together: [pic]
4. Calculate Ecell = ER - EL = + 0.77 – (- 0.13) = + 0.90 V
16.3 An alternative approach? (
e.g. Deduce the overall cell reaction between Cr2O72- and NO3- and calculate the Ecell?
Half-equations & electrode potentials:
[pic]
Method:
1. Sketch a horizontal scale to represent the voltage.
2. Mark on the more +ve half-cell system – this is the RH electrode and thus the +ve.
3. Mark on the more –ve half-cell system – this is the LH electrode and thus the –ve.
4. Electrons flow from the –ve cathode to the +ve anode; thus oxidation takes place at the cathode and reduction takes place at the anode as the electrons flow to there.
5. Reverse the half-equation for the cathode (LH electrode) as it produces electrons.
6. Add to produce the full equation and thus overall cell reaction.
7. Ecell = ER – EL.
This is a snapshot of the Powerpoint slide I went through in the workshop session:
INTENTIONALLY BLANK PAGE! (
-----------------------
A2/Year 13 Chemistry
Revision Workshop
Spring 2020
Reviewing the A2 content of your Examinations
Participants Notes
NO2 CH3 COCH3 Br
3
Benzenecarbaldehyde
(benzaldehyde)
[pic]
ethanoic acid
[pic]
propanone
[pic]
H
H
CH3CH2
CH3CH2
CH3CH2
methyl ethanoate
sodium octadecanoate
(sodium stearate; SOAP)
pH
4
E2 – E1
= ΔE
3d
Energy
Area under curve
= total number of particles
Ea
No. of particles with given energy
Ea
(catalyst)
n
Overall:
Overall:
3+
7
6
5
4
3
2
1
................
................
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