ATOMIC STRUCTURE - Newcastle University



AS TOPICS INDEX

1. Atomic Structure p2

2. Bonding p7

3. Periodicity p13

4. Group II p15

5. Group VII p18

6. Redox p23

7. Energetics p26

8. Kinetics p31

9. Equilibria p34

10. Organic Chemistry - index p38

11. Quantitative - index p56

Appendix: Post-16 Learning Resources Flyer - CCDC after p64

Appendix: Post-16 Learning Resources Flyer - PDBe after p64

1. ATOMIC STRUCTURE

1.1 Fundamental Particles

Mass/kg Charge/C Relative mass Relative charge Location

Proton 1.6 x 10-27 1.6 x 10-19 1 +1 Nucleus

Neutron 1.6 x 10-27 0 1 0 Nucleus

Electron 9.1 x 10-31 1.6 x 10-19 5.45 x 10-4 -1 Cloud

1.2 Structure of the atom

Nucleus: all except hydrogen contain protons and neutrons

positive charge equal to number protons present

almost all the mass of the atom in the nucleus

nuclear radius very much smaller than the atomic radius nucleus very dense

Electrons: arranged in energy levels

in an atom, the number of electrons and protons are equal

1.3 Mass numbers and isotopes

Mass number (A): number of protons + number of neutons in the nucleus

Atomic (proton) number (Z): number of protons in the nucleus

Isotopes: same element;

same number of protons BUT

different numbers of neutrons

Since chemical properties depend on the number of protons and electrons, isotopes react chemically in the same way.

Example: Chlorine has two main isotopes: [pic]

35 means mass number (p + n) = 35

17 means atomic number (p only) = 17

Hence number of neutrons = 35 – 17 = 18

37 means mass number (p + n) = 37

17 means atomic number (p only) = 17

Hence number of neutrons = 37 – 17 = 20

1.4 The mass spectrometer

This is used to separate ions with nuclei of differing masses. Samples need to be in the gaseous state and at low pressure.

Example Chlorine, Cl2

[pic]

Deductions

(1) Ions present: m/z 35 37 70 72 74

Ion 35Cl+ 37Cl+ (35Cl+)2 (35Cl37Cl+) (37Cl+)2

(2) Relative atomic mass (Ar) of chlorine

This can be calculated from the m/z values and relative intensities, e.g.

(3 x 35) + (1 x 37) = 35.5

(3 + 1)

1.5 Relative molecular mass (Mr)

This is usually given by the maximum value of m/z (or the peak furthest to the right or peak with largest m/z value; NOT “the largest m/z peak” – this implies height) – this corresponds to the molecular ion M+.

1.6 Electron arrangements

Electrons are arranged in successive energy levels. The energy level nearest to the nucleus is filled first. The arrangement of elements in the Periodic Table can be used to deduce the electronic configuration of an atom. Within energy levels there are also sub-levels. These can be summarised as:

|Principal level (n) |Sub-levels |Orbitals in |Maximum e- |Maximum e- in level |

| | |sub-level | |(2n2) |

|1 |s |1 |2 |2 |

|2 |s |1 |2 | |

| |p |3 |6 |8 |

|3 |s |1 |2 |18 |

| |p |3 |6 | |

| |d |5 |10 | |

|4 |s |1 |2 | |

| |p |3 |6 | |

| |d |5 |10 | |

| |f |7 |14 |32 |

|etc. | | | | |

Each orbital can hold a maximum of 2 electrons. These electrons must have opposite spin and every electron can have a different electronic description.

The order of filling of the sub-shells can be determined using the Periodic Table which shows the relationship between chemical properties and electronic configuration.

Examples: Reading the electronic configurations from the Periodic Table gives:

Na 1s2 2s2 2p6 3s1 and Na+ 1s2 2s2 2p6

Cl 1s2 2s2 2p6 3s2 3p5 and Cl- 1s2 2s2 2p6 3s2 3p6

Co 1s2 2s2 2p6 3s2 3p6 3d7 4s2 and Co2+ 1s2 2s2 2p6 3s2 3p6 3d7

Note: Although the Periodic Table shows that the 3d sub-shell fills after the 4s, the 3d are actually slightly nearer to the nucleus. This explains why the 4s are lost when Co2+ is formed.

There are two deviations in the electron arrangements of two of the first row Transition Metals (d-block): chromium & copper:

e.g. the electronic configurations of chromium and its ions are:

Cr 1s2 2s2 2p6 3s2 3p6 3d5 4s1

Cr2+ 1s2 2s2 2p6 3s2 3p6 3d4 and Cr3+ 1s2 2s2 2p6 3s2 3p6 3d3

e.g. the electronic configurations of copper and its ions are:

Cu 1s2 2s2 2p6 3s2 3p6 3d10 4s1

Cu+ 1s2 2s2 2p6 3s2 3p6 3d10 and Cu2+ 1s2 2s2 2p6 3s2 3p6 3d9

1.7 Ionisation energies

First ionisation energy (enthalpy) is for the reaction

E (g) → E+ (g) + e- (g)

Second ionisation energy (enthalpy) is for the reaction

E+ (g) → E2+ (g) + e- (g) etc.

Variation in first ionisation energy across Period 3:

[pic]

Explanation: Outer electron enters same principal energy level

Nuclear charge increasing Na to Ar

Shielding (by inner full energy levels) constant

Electrons more strongly attracted

Hence general increase in 1st IE across period

Deviation 1: fall from Mg to Al due to electron entering a 3p sub-shell.

3p higher energy level (further away from nucleus) so easier to remove than from 3s.

Deviation 2: fall from P to S due to first electron pairing in 3p sub-shell.

mutual repulsion between electrons makes removal easier

Successive ionisation energies:

These can be used to indicate the number of electrons in the outer energy level of an atom

[pic]

The large increase after the removal of the fifth electron indicates that this element has 5 outer electrons and thus is in Group V of the Periodic Table.

2. BONDING

1. Types of bonding

There are three main types of chemical bonding:

(i) Ionic found in compounds of (usually) a metal with a non-metal.

(ii) Covalent found in compounds of a non-metal with another non-metal.

(iii) Metallic found in metal elements.

(i) Ionic Bonding: This involves a transfer of electrons, typically, but not always, from metals to non-metals.

Metal atoms lose electrons to form positive ions.

Non-metal atoms gain electrons to form negative ions.

The oppositely-charged ions are strongly attracted to each other by electrostatic forces, so ionic compounds are lattice structures, i.e. solids with high melting points.

(ii) Covalent Bonding: This involves sharing of one or more electron pairs.

A covalent bond is a shared pair of electrons between two atoms with one electron usually coming from each atom. Covalent bonds are formed between atoms within small molecules, e.g. CH4 or within giant macromolecular structures, e.g. in diamond or graphite.

Co-ordinate or dative covalent bonding:

In a co-ordinate bond, both electrons come from one atom, i.e. from a lone pair on the donor atom. Once formed, a co-ordinate bond is identical in every respect to any other covalent bond,

e.g. the ammonium ion from ammonia:

[pic]

iii) Metallic bonding

Metal elements exist as giant lattices of close packed metal ions in a sea of delocalised electrons. Metals usually have high melting and boiling points as a lot of energy must be supplied to overcome the attraction between the +ve metal ions and the delocalised electrons.

The number of delocalised electrons determines the melting point; if the number is small (e.g. one per atom as in Group 1) the lattice is held together less strongly and the melting point is quite low (less than 100oC). If the number is greater (e.g. 6 for Cr) then melting point can be very high (1890oC).

2.2 Bond polarity and the polarisation of ions

Electronegativity differences between atoms will mean that the pair of electrons in a covalent bond will rarely lie centrally between the two nuclei.

Electronegativity (the power of an atom to attract a bonding pair of electrons) increases from left to right across a period and from bottom to top in a group.

2.2.1 Polar covalent bonds

In covalent bond between atoms of different electronegativity, the electron density is displaced towards the more electronegative atom and the covalent bond is said to be polar, e.g. [pic]

where δ+ represents an electron deficient atom (less electronegative) and δ- represents an atom with excess of electron density (more electronegative).

Substances made up of molecules with an overall dipole will have higher boiling points than those consisting of non-polar molecules.

2.2.2 Polarisation of anions

Ions in lattices of purely ionic compounds are spherical, but if the positive ion is small and/or highly charged it will distort the electron cloud of the negative ion.

This polarisation of the negative ion leads to a build up of extra charge density between the two nuclei, ie. to partial covalency, e.g.

[pic] [pic]

Spherical ions in NaCl distorted ions in LiI

Larger negative ions are more easily polarised but the effect is only important when positive ions with charge of 3+, e.g. Al3+ are involved, e.g. pure A1C13 is a covalent molecule. However 2+ ions (Be2+) or even 1+ (Li+) show some polarising power because their sizes are so small, i.e. LiI is ionic but has some covalent character, whereas BeC12 is covalent.

Polarising power depends on the ratio charge of the ion, often called the charge density.

size

NaCl MgC12 A1Cl3 SiCl4

ionic ionic with a very little covalent with a little covalent

covalent character ionic character

2.3 The bonding spectrum

Purely ionic bonding (e.g. in NaCl) or purely covalent bonding (e.g. in H2) are the two extremes of bonding types – most bonds occur somewhere in between:

increasing polarisation

[pic]

IONIC POLAR COVALENT

[pic] [pic] [pic]

[pic]

increasing electronegativity difference

2.4 Intermolecular forces

Covalent molecules are attracted to each other by intermolecular forces of three types:

i) van der Waals/London forces (weakest) (0 – 5 kJ mol-1)

ii) dipole-dipole forces (5 – 15 kJ mol-1)

iii) Hydrogen “bonding” (strongest) (20 – 40 kJ mol-1)

(i) van der Waals forces (AQA); London/dispersion forces (OCR)

• present between ALL molecules and arise because a temporary dipole induces a temporary dipole on an adjacent molecule and they attract each other.

• increases with the size of the molecule, e.g. down group VII.

• depends on the surface area of the molecule, e.g. for a set of isomeric alkanes, the least branched has the greatest surface area and so the greatest van der Waals/London forces and thus the highest boiling point.

(ii) dipole-dipole forces

• forces between polar molecules

• A covalent bond between atoms with different electronegativities will be polar. If the dipoles within a molecule do not cancel then the overall molecule is polar and there will be dipole-dipole forces of attraction between molecules.

• Permanent dipoles are stronger than temporary ones, so dipole-dipole forces are stronger than van der Waals/London forces.

(iii) Hydrogen bonding

• An extreme and very specific form of dipole-dipole intermolecular forces.

• Between molecules which contain H atoms bonded to N, O or F atoms.

• These three most electronegative atoms attract electrons strongly from hydrogen which becomes very electron deficient or δ+. A strong intermolecular force then results from the electrostatic attraction of the lone pair of electrons on the N, O or F with the δ+ H.

• Hydrogen bonds are the strongest intermolecular force and cause high boiling points and high solubility in water for many compounds such as water, alcohols, ammonia or hydrogen fluoride.

2.5 States of matter: solid, liquid and gas

Solids have particles in fixed positions in a lattice, the particles can vibrate but not move about.

Hence a solid has a fixed shape.

The lattice may be held together by ionic attractions, covalent bonds, metallic bonding, hydrogen bonding, dipole-dipole forces or van der Waals forces.

If sufficient energy is supplied to overcome these forces, the particles can begin to move about; this is melting.

Liquids do not have a fixed shape because the particles can move about however they remain very close together.

If sufficient energy is supplied, the particles overcome the inter-particle forces almost completely and escape from the liquid. This is called evaporation or boiling. The energy required to boil a liquid is always more than that required to melt the same substance.

Gases are made up of particles which are widely spaced and move in rapid random motion. The forces between particles in the gas phase are negligible.

2.6 Types and properties of solids

There are four types of crystal depending on the type of bonding involved.

i) Metallic crystals (see section 2.1 (iii) earlier)

ii) Ionic crystals

These contain a lattice of oppositely-charged ions held together by strong electrostatic forces; hence ionic compounds have high melting points.

(iii) Molecular crystals

These contain covalent molecules which have strong covalent bonds within the molecule but only weak forces between the molecules, e.g. hydrogen bonding, dipole-dipole forces or van der Waals/London forces. Molecular solids such as iodine or sulphur have low melting points.

Analogy: a brick wall; the bricks = molecules (strongly held together); the mortar = intermolecular forces (weaker; break much more easily).

(iv) Macromolecular crystals

Macromolecules such as diamond contain covalent bonds between atoms throughout the whole structure. Diamond is therefore very hard and has a high melting point as many strong covalent bonds must be broken.

Graphite has carbon atoms covalently bonded in planes and so has a high melting point. However, there are no covalent bonds, only delocalised electrons, between the planes so these can slide over each other causing graphite to be a lubricant. Graphite is also the only non-metal element to conduct electricity because of the delocalised electrons between the planes of atoms.

NB Graphite does NOT have any type of intermolecular forces between the layers – if it did, it would be (simple) molecular and have a relatively low melting point (it actually sublimes at ~3500oC!).

7. Shapes of molecules

Ions attract equally in all directions; covalent bonds are directional.

Covalent bonds consist of pairs of electrons between the two atoms involved. Pairs of electrons repel each other so that molecules have particular shapes depending on the number of pairs of electrons surrounding the atoms.

2.7.1 Valence Shell Electron Pair Repulsion theory (VSEPR)

The electron pairs in the outer shell of an atom may also be non-bonding or lone pairs. The shape of a molecule or ion can be predicted by working out the number of electron pairs surrounding the central atom. The shape and bond angle will be slightly modified if some of the pairs are non bonding because lone pairs repel other electron pairs stronger than bonding pairs do.

Repulsive forces decrease in the order:

Lone pair–lone pair > lone pair–bonding pair > bonding pair–bonding pair

2.7.2 Rules

To determine the shape of a molecule or ion (having single bonds only):

1. Count the number of valence electrons from the central atom (group number)

2. Count electrons added by the bonding atoms (always 1 from each)

3. Add or subtract for negative or positive ions if necessary

4. Divide by two to get number of electron pairs

5. Shape is based on one of the following five basic shapes:

|No. of bonding pairs |Name of Base Shape |Bond angles |Example |

| 2 |Linear |1800 |BeC12 |

| 3 |Trigonal Planar |1200 |BF3 |

| 4 |Tetrahedral |109.50 |CH4 |

| 5 |Trigonal Bipyramidal |1200 and 900 |PF5 |

| 6 |Octahedral |900 |SF6 |

If lone pairs are present, other shapes and names will result will result, as the lone pairs are NOT included when determining the named shape of the molecule/ion.

Example 1: Ammonia

NH3 has three bonding pairs and one lone pair; the shape is based on a tetrahedral electron arrangement but the atoms are in a trigonal pyramid or simply pyramidal arrangement with bond angle 1070

Example 2: Water

H2O has two bonding pairs and two lone pairs; the shape is also based on a tetrahedral electron arrangement but the molecule is V–shaped or bent (some Exam Boards accept “non-linear” as an alternative name () with bond angle 104.50

VSEPR Molecular Geometries

[pic]

Image from bonding

3. PERIODICITY

1. The classification of elements into Periodic Table Blocks

[pic]

3.2 Physical properties of elements in Period 3

| |< ----- s-block ----- > |< ---------------------------------- p-block --------------------------------- > |

| |Na |Mg |Al |Si |P |

|4 |Be |[He] 2s2 |0.089 |899 |1551 |

|12 |Mg |[Ne] 3s2 |0.136 |737 |924 |

|20 |Ca |[Ar] 4s2 |0.174 |590 |1116 |

|38 |Sr |[Kr] 5s2 |0.191 |549 |1042 |

|56 |Ba |[Xe] 6s2 |0.198 |503 |998 |

4.1.1 Changes in atomic radius

[pic]

• Number of electron shells increases.

• Outer electrons more shielded from nuclear attraction.

4.1.2 Variation in first ionisation energy

[pic]

Explanation:

• Same number of outer electrons for each element.

• Radius of atom increases Be to Ba.

• Outer electron is further from the nucleus.

• More electron shells.

• Electron shielding increased Be to Ba.

• Hence attraction for outer electron falls.

• Ionisation energy falls.

4.1.3 Changes in melting point

[pic]

Explanation

• All form M2+ ion.

• All have two delocalised electrons.

• Size of the M2+ ion increases Mg2+ to Ba2+

• Charge/size ratio falls Mg2+ to Ba2+

• Expect melting point to decrease Mg to Ba.

• Discontinuity at Mg due to its different crystal structure.

4.2 The reactions of Group II metals with water

Reactivity increases Mg to Ba.

Magnesium: reacts very slowly with cold water, i.e.

Mg (s) + 2H2O (l) → Mg(OH)2 (aq) + H2 (g)

can be burnt in steam, i.e.

Mg (s) + H2O (g) → MgO (s) + H2 (g)

hydrogen produced

Others: react with cold water

reaction vigour increases Ca to Ba

hydrogen produced

hydroxide ions produced, e.g. Ca (s) + 2H2O (g) → Ca(OH)2 (aq) + H2 (g)

4.3 Changes in oxide / hydroxide solubility

Solubility increases Mg to Ba.

Mg(OH)2 sparingly soluble

if a solution containing hydroxide ions is mixed with a solution of Mg2+ ions, a white precipitate/suspension of the hydroxide will form (“Milk of Magnesia”).

Mg2+ (aq) + 2OH- (aq) → Mg(OH)2 (s)

Ca(OH)2 fairly soluble. Forms a solution known as “lime water” but will precipitate if concentrations of Ca2+ (aq) and OH- (aq) are high.

Sr(OH)2 and Ba(OH)2 are completely soluble and form strongly alkaline solutions.

4.4 Changes in sulphate solubility

Solubility decreases Mg to Ba.

MgSO4 very soluble (“Epsom salts”)

CaSO4 quite soluble (causes water to be “hard”)

will form a white precipitate if the concentrations of Ca2+ (aq) and SO42- (aq) are high

SrSO4 and BaSO4 are both completely insoluble.

4.4.1 The “Sulphate Test”

To the solution under test add:

• dilute HCl or HNO3 (NOT H2SO4 – why? ()

• then a solution of a soluble barium compound e.g. BaCl2 (aq) or Ba(NO3)2 (aq)

The formation of a white precipitate indicates that sulphate ions are present, i.e.

Ba2+ (aq) + SO42- (aq) → BaSO4 (s)

NB Acid must be added first when this test is carried out because e.g. BaCO3 is also insoluble.

The acid removed the carbonate ions from solution by converting them to CO2 (g)

2H+ (aq) + CO32- (aq) → H2O (l) + CO2 (g)

Use – as a “barium meal” – an aqueous suspension of the insoluble barium sulphate given to patients with suspected stomach/digestive tract/intestinal problems.

Barium sulphate is very dense thus strongly absorbs X-rays – can detect blockages etc.

Safe to use even though aqueous barium ions are toxic as barium sulphate is completely insoluble, so no danger of barium ions being absorbed by the patient!

5. GROUP VII

5.1 Trends in Chemical Properties of Halogens and Halide ions

|Element |Atomic Number |Outer Electrons |Atomic radius/ nm|Ionic radius/ nm |Boiling Point K |

|F |9 |2s22p5 |0.071 |0.133 |85 |

|Cl |17 |3s23p5 |0.099 |0.180 |238 |

|Br |35 |4s24p5 |0.114 |0.195 |332 |

|I |53 |5s25p5 |0.133 |0.215 |457 |

Graphical representation of data

[pic]

5.2 Trend in electronegativity of the halogens

• Electronegativity is the power of an atom to withdraw electron density from a covalent bond

• Electronegativity falls as the atomic number of the halogen increases

Explanation

• Increase in atomic number (expect the attraction to increase)

• Increase in number of electron shells (expect attraction to decrease due to repulsion)

• Increase in atomic radius of the atom (expect attraction to decrease)

On balance the increase in shielding and atomic radius more than compensate for the increase in nuclear charge and electronegativity falls

5.3 Trend in boiling point of the halogens

[pic]

Explanation

• Halogens exist as diatomic molecules, X2

• Weak intermolecular, van der Waals/London, attractive forces

• Van der Waals/London attractive force increases with molecular size

5.4 Trends in chemical properties

The oxidising power of the halogens decreases in the order F2 > Cl2 > Br2 > I2

The reasons for this trend are complex and include:

1. The enthalpy change for the process X2 (standard state) → X2 (g)

2. The strength of the X-X bond X2 (g) → 2 X (g)

3. The electron affinity of the atom X (g) + e- → X- (g)

4. The enthalpy change when X- (aq) is formed from a dissociated crystal lattice

5.5 The relative oxidising power of the halogens

5.5.1 The reactions of Cl2 (aq) with Br- (aq) and I- (aq)

|Halide Ion |Observation |Conclusion |Equation |

|Br- (aq) |Brown/Yellow |Br2 displaced |2 Br- + Cl2 → 2 Cl- + Br2 |

|I- (aq) |Brown colour and/or black |I2 displaced |2 I- + Cl2 → 2 Cl- + I2 |

| |precipitate | | |

5.5.2 The reactions of Br2 (aq) with Cl- (aq) and I- (aq)

|Halide Ion |Observation |Conclusion |Equation |

|Cl- (aq) |No change |Cl2 not displaced | |

|I- (aq) |Brown colour and/or black |I2 displaced |2 I- + Br2 → 2 Br- + I2 |

| |precipitate | | |

5.5.3 The reactions of I2 (aq) with Cl- (aq) and Br- (aq)

|Halide Ion |Observation |Conclusion |Equation |

|Cl- (aq) |No change |Cl2 not displaced | |

|Br- (aq) |No change |Br2 not displaced | |

5.6 Trends in reducing power of halide ions

F- < Cl- < Br- < I-

5.6.1 The reactions of SOLID sodium halides with CONCENTRATED sulphuric acid (not OCR(A); WJEC covered in A2)

A logical approach:

[pic]

Which of these ions will lose electrons most easily?

Oxidation states of Sulphur

[pic]

Which of these elements, in the oxidation state shown, will accept electrons most readily?

What is the likely product when an H+ ion meets an X- ion?

Predict the products of the reaction of each solid sodium halide with concentrated H2SO4 and state what is observed in each reaction.

| |+6 (SO3; SO42-) |+4 (SO2; SO32-) |0 (S) |-2 (H2S) |

|F- | |X |X |X |

|Cl- | |X |X |X |

|Br- | | |X |X |

|I- | | | | |

|NaX |Observations |Products |Type of reaction |

|NaF |Steamy fumes |HF |Acid-base (F- acting as a base) |

|NaCl |Steamy fumes |HCl |Acid-base (Cl- acting as a base) |

|NaBr |Steamy fumes |HBr |Acid-base (Br- acting as a base) |

| | | | |

| |Colourless gas |SO2 |Redox (reduction product of H2SO4) |

| | | | |

| |Brown fumes |Br2 |Redox (oxidation product of Br-) |

|NaI |Steamy fumes |HI |Acid-base (I- acting as a base) |

| | | | |

| |Colourless gas |SO2 |Redox (reduction product of H2SO4) |

| | | | |

| |Yellow solid |S |Redox (reduction product of H2SO4) |

| | | | |

| |Smell of bad eggs |H2S |Redox (reduction product of H2SO4) |

| | | | |

| |Purple fumes |I2 |Redox (oxidation product of I-) |

Summary of oxidation state changes

• Under these conditions neither F- nor Cl- can reduce H2SO4

• Br- can reduce H2SO4 from oxidation state +6 to +4

• I- can reduce H2SO4 from oxidation state +6 to +4, 0 and to –2

A self-test check sheet

|Halide |Acid-base products |H2SO4 reduction products in oxidation states of |

| | |+4 |0 |-2 |

|NaF (s) | | | | |

|NaCl (s) | | | | |

|NaBr (s) | | | | |

|NaI (s) | | | | |

5.7 Using silver nitrate and ammonia solution as a test for halide ions in solution:

|Halide |Precipitate |Colour |Solubility in ammonia solution |

|F- |None | | |

|Cl- |AgCl |White |Soluble in dilute NH3 (aq) |

|Br- |AgBr |Cream |Sparingly solube in dilute NH3 (aq) |

| | | | |

| | | |Soluble in concentrated NH3 (aq) |

|I- |AgI |Yellow |Insoluble in concentrated NH3 (aq) |

Example equations:

AgNO3 (aq) + NaCl (aq) → AgCl (s) + NaNO3 (aq)

or Ag+ (aq) + Cl- (aq) → AgCl (s)

Note: the addition of dilute nitric acid is essential to prevent precipitation of other insoluble compounds, e.g. with CO32-

6. REDOX

6.1 Introduction

Oxidation and Reduction

Initial definition:

• 2 Mg + O2 → 2 MgO

• Fe2O3 + 2 Al → Al2O3 + 2 Fe

• CuO + C → Cu + CO

Oxidation: addition of oxygen or removal of hydrogen

Reduction: removal of oxygen or addition of hydrogen

Consider these two reactions:

• SO2 + H2O + HgO → H2SO4 + Hg

• SO2 + 2 H2O + C12 → H2SO4 + 2 HCl

What is the oxidising agent in the second reaction? (

6.2 Oxidation state (or oxidation number)

6.2.1 Rules for assignment of oxidation state

Species Oxidation state

Elements not combined with others 0

Oxygen in compounds (except peroxides) -2

• Oxygen in peroxides -1

Hydrogen in compounds (except metal hydrides) +1

• Hydrogen in metal hydrides -1

Group I metals in compounds +1

Group II metals in compounds +2

Simple ions

The oxidation state of the element is simply the charge on the ion.

NB the sum of the oxidation states in compound = 0

Complex ions

The oxidation state of the central atom in a complex ion is the charge the element would have if it was a simple ion and not bonded to other species.

i.e. the sum of the oxidation states in a complex ion = the overall charge on the ion

Example

CO32- O is always -2 and overall Σ ox. states = 2- (charge on the ion), thus C = +4

Examples of oxidation states in simple ions

• Na+: K+: Ag+ have an oxidation state +1

• Mg2+: Ca2+: Ba2+ have an oxidation state +2

• F-: C1-: I- have an oxidation state -1

• O2-: S2-: have an oxidation state -2

6.2.2 Using change in oxidation state to deduce Oxidation or Reduction

[pic]

Oxidation state becoming more +ve = Oxidation

Oxidation state becomiong more –ve = Reduction

Oxidation Is the process of electron Loss

Reduction Is the process of electron Gain

Hence the acronym “O I L R I G” to aid memory!

Identifying redox reactions

Example 1 Zn + 2 HCl → ZnCl2 + H2

Ionic Equation : Zn + 2 H+ → Zn2+ + H2

Example 2 CuO + 2 HCl → CuCl2 + H2O

6.3 Writing redox equations

6.3.1 Rules for 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 (unless they are specified as reactants or products)!!

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. Rules for 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.

6.4 Summary

• 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 be able to assign an oxidation number to each element in a reaction

• To use the change in oxidation number to deduce if a species has been oxidised or reduced

• 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).

7. ENERGETICS

7.1 Enthalpy changes

When experiments are carried out at constant pressure, the heat energy change is called the

enthalpy change and given the symbol ΔH.

7.1.1 The sign of the enthalpy change

• Products have less energy than reactants; energy given out; exothermic; ΔHo is –ve

• Products have more energy than reactants; energy taken in; endothermic; ΔHo is +ve

NB. ALL ΔH values ALWAYS have a sign – do NOT assume no sign = +ve!

7.1.2 Standard conditions

100 kPa;

298 K

Standard states for elements and compounds i.e. as they are under these conditions

If ΔH is measured under these standard conditions, it is called the standard enthalpy change

and given the symbol ΔHo

NB. State symbols must be given when writing equations for enthalpies:

e.g. C2H5OH (l) + 3O2 (g) → 2CO2 (g) + 3H2O (l)

7.1.3 Definitions

1. the standard enthalpy of formation, ΔHfo

|The enthalpy change when 1 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

2. the standard enthalpy of combustion, ΔHco

|The enthalpy released when 1 mole of a compound is completely burnt in oxygen under |

|standard conditions, 100 kPa and 298 K, with all reactants and products in their standard |

|states |

e.g. C2H5OH (1) + 3O2 (g) → 2CO2 (g) + 3H2O (l) ΔHco = – 1367.3 kJ mol-1

7.1.4 Calorimetry

The heat energy change, q, can be calculated using the formula q = m x c x ΔT

When reactions are carried out in aqueous solution, we assume that the only substance

gaining or losing heat energy is the water.

When the heat energy evolved on combustion of a fuel is used to heat a metal can containing

water, we need to know the heat energy needed to raise the temperature of the can plus water by

one degree. This is called its heat capacity.

Example

The heat energy evolved when 0.30g of propane was burnt raised the temperature of a metal

can containing water by 16.4oC. the heat capacity of the can plus water was 0.760 kJ K-1.

Use this data to calculate the enthalpy of combustion of propane.

Heat energy absorbed by the can plus water q = 0.760 x 16.4 = 12.46 kJ

Moles of propane burnt = mass / Mr = 0.30 / 44.1 = 6.80 x 10-3

Enthalpy change = (heat energy, q / number moles, n) = 12.46 / 6.80x10-3 = – 1832 kJ mol-1

7.2 Conservation of Energy

Energy cannot be created or destroyed but can be changed from one form into another.

7.2.1 Hess’s Law 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.) |

Example

Use the following data to calculate the enthalpy of formation, ΔHfo, of ethanol C2H5OH

ΔHco C2H5OH = – 1367.3 kJ mol-1

ΔHco C = – 393.5 kJ mol-1

ΔHco H2 = – 285.8 kJ mol-1

[pic]

[pic]

Cycling anticlockwise we find the above relationship between the ΔH values.

Rearrangement and substitution gives ΔHfo = (2 x – 393.5) + (3 x – 285.8) – (– 1367.3)

Hence ΔHfo = – 277.1 kJ mol-1

Note 1: if you are provided with ΔHco values for all reactants and products, you can use the relationship, derived from the cycle:

|ΔHo (reaction) = Σ ΔHco (reactants) – Σ ΔHco (products) |

Note 2: if you are provided with ΔHfo values for all reactants and products, you can use the alternative relationship, derived from the cycle:

|ΔHo (reaction) = Σ ΔHfo (products) – Σ ΔHfo (reactants) |

7.3 Bond Enthalpies

The BOND ENTHALPY is the enthalpy change when a covalent bond is broken:

A – B (g) → A (g) + B (g)

It is given the symbol E(A–B) where A – B is the structure of the covalent bond.

Bond enthalpy in diatomic molecules depends on

• atomic size

• nuclear charge

• shielding

• bond length

Bond enthalpy is a measure of the strength of a covalent bond.

In more complex molecules it is necessary to consider the environment of the bond, e.g.

[pic]

Since the exact environment of a bond will change from molecule to molecule it is more

practical to use mean bond enthalpy values.

The average N-H bond enthalpy in ammonia is thus (448 + 368 + 356) / 3 = +391 kJmol-1

Mean bond enthalpies are average values of covalent bond enthalpies measured in a wide range of different compounds and are thus only approximate – as are any enthalpy values calculated from them! (

Example

Propane burns in excess oxygen as shown by the equation below:

C3H8 (g) + 5O2 (g) → 3CO2 (g) + 4H2O (g)

Use the mean bond enthalpies given below to calculate the enthalpy of combustion of propane,

which has the structure:

[pic]

You can construct a Hess’s cycle for this process in a similar way to those previously, i.e.

[pic]

OR

[pic]

[pic]

Heat energy to break bonds in reactants:

C-C x 2 = +347 x 2 = + 694

C-H x 8 = +413 x 8 = + 3304

O=O x 5 = +498 x 5 = +2490

Total = +6488

Heat energy to break bonds in products:

C=O x 6 = +805 x 6 = +4830

O-H x 8 = +464 x 8 = +3712

Total = +8542

Enthalpy change = + 6488 – 8542 = – 2054 kJ mol-1

NB. Data book value is – 2043 kJ mol-1. The difference of 11 kJ is from using mean BE values.

8. KINETICS

8.1 Collision theory

• Particles must collide for a reaction to occur.

• Not all collisions result in a reaction.

• Reaction only occurs if particles collide with sufficient energy.

Activation energy, Ea: the minimum energy necessary for a reaction to occur.

8.2 Maxwell-Boltzmann distributions

Curves can be drawn showing how molecular energies are distributed in a sample of gas:

[pic]

• The area under the curve gives the total number of molecules in the sample.

• No molecules have zero energy – graph passes through the origin (0,0).

• Very few molecules have very high energy.

• The curve is asymptotic at high energy – i.e. it NEVER touches the x-axis.

• Emp is the most probably energy.

• Ea is the activation energy.

8.3 Factors affecting reaction rate

8.3.1 Increasing concentration or pressure of gas or surface area of a solid at a fixed temperature

[pic]

• Area under the curve is increased.

• More particles more collisions.

• More collisions with E > Ea

• Rate increase

8.3.2 Increasing temperature at a fixed concentration

[pic]

i. Area under the curve unchanged.

ii. Particles have more energy.

iii. Distribution displaced to high energies.

iv. More collisions.

v. Many more collisions with E > Ea

vi. Rate greatly increased

8.3.3 Adding a catalyst

[pic]

• Unchanged chemically and

• Unchanged in amount at end of reaction.

• Provides an alternative route.

• Lower activation energy.

• Many more particles able to react.

• Rate increased.

• Equilibrium position unchanged.

9. EQUILIBRIA

Many chemical reactions 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.

9.1 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.)

9.1.1 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).

N.B. 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.

9.1.2 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.

9.1.3 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.

9.1.4 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.

9.2 Application to an industrial process (

Example: [pic]

9.2.1 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.

9.2.2 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 constant

9.3.1 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 from the balanced equation.

The Equilibrium Constant, in terms of concentration, Kc is defined as:

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. ORGANIC CHEMISTRY

1. IUPAC Nomenclature p39

2. Isomers p40

3. Alkanes p42

4. Alkenes p45

5. Haloalkanes p49

6. Alcohols p52

10.1 I.U.P.A.C. NOMENCLATURE FOR ORGANIC COMPOUNDS

These rules were designed by the International Union of Pure and Applied Chemistry (I.U.P.A.C.) so that you can work out the precise molecular structure of an organic compound simply from its name, and vice-versa, anywhere in the world!

Rule 1

Find the longest carbon chain in the molecule – this gives the STEM of the name.

Rule 2

Identify the PRINCIPAL FUNCTIONAL GROUP – this gives the SUFFIX (end) of the name.

Rule 3

Number the carbon chain to give the principal functional group the LOWEST possible number – this identifies where on the chain it is.

FOR MANY SIMPLE HYDROCARBONS, THIS IS AS FAR AS YOU NEED TO GO – FOR MORE COMPLEX ONES, CARRY ON ….. !

Rule 4

Identify other substituents or functional groups and name these as PREFIXES (in front of the STEM of the name).

Rule 5

Do NOT renumber the carbon chain, but give these groups the numbers they have from the existing numbering of the chain.

Rule 6

If there is more than one of a particular substituent group, identify how many by using a multiplier, i.e. two = di, three = tri, four = tetra, etc.

List the numbers of the carbons carrying each substituent in ascending numerical order in front of the prefix, separated by commas.

Rule 7

If there is more than one prefix, list them ALPHABETICALLY, NOT NUMERICALLY, in front of the name, IGNORING any multipliers.

e.g. CFCl3 is trichlorofluoromethane (since c comes before f).

10.2 ISOMERS

10.2.1 STRUCTURAL

- many isomers often possible for a given chemical formula.

- Structural – same molecular formula but different bonding arrangement of the atoms or different way/order the atoms are bonded together; but NOT simply “different arrangement of atoms” (could imply stereochemistry);

e.g. C2H6O; ethanol and methoxymethane (dimethyl ether):

[pic]

ethanol methoxymethane

Isomers have different chemical properties (different functional groups) and different physical properties (different intermolecular forces and/or shapes, etc.).

- Positional or chain – same formula, same carbon skeleton and functional group, but functional group is bonded in a different position in each molecule; e.g. C5H12O; pentan-1-ol, -2-ol & -3-ol (and others):

[pic]

pentan-1-ol pentan-2-ol pentan-3-ol

Isomers have similar chemical properties (same functional group) but slightly different physical properties (different shapes and thus strength of intermolecular forces, etc.).

2. 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.

GEOMETRIC (E-Z or cis-trans)

- 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;

e.g. 1-bromo-2-chloropropene:

[pic]

Z-1-bromo-2-chloropropene E-1-bromo-2-chloropropene

E = opposite side; Z = same side.

Where the two substituent groups are the same, an alternative naming can be used:

e.g. for but-2-ene:

[pic]

trans-but-2-ene cis-but-2-ene

cis = same side; trans = opposite side (across)

NB. cis-trans is a special case of E-Z stereoisomerism of alkenes where the two substituent groups are the same.

Geometric isomers have identical chemical properties (same functional group) but have slight differences in physical properties (due to the differing shapes/sizes of

the molecules).

10.3 CRUDE OIL

Alkanes and alkenes are hydrocarbons; compounds which contain hydrogen and carbon only.

Petroleum is a complex mixture of hydrocarbons, mostly alkanes. This mixture is separated into smaller mixtures of hydrocarbons (fractions) with similar boiling points and similar numbers of carbon atoms by fractional distillation (NOT OCR and WJEC)

|Name of fraction |Boiling range (oC) |Carbon atoms |Uses |

|LPG (gases) |Up to 25 |1 – 4 |Calor gas; camping gas |

|Petrol (gasoline) |40 – 100 |4 – 12 |Petrol |

|Naptha |100 – 150 |7 – 14 |Petrochemicals |

|Kerosine (paraffin) |150 – 250 |11 – 15 |Jet fuel |

|Gas oil (diesel) |220 – 350 |15 – 19 |Central heating fuel |

|Mineral (lubricating) oil |Over 350 |20 – 30 |Lubricating oil |

|Fuel oil |Over 400 |30 – 40 |Ships and power stations |

|Wax, grease |Over 400 |41 – 50 |Candles, grease |

|Bitumen |Over 400 |Above 50 |Roofing; road surfacing |

10.4 ALKANES

The homologous series of saturated hydrocarbons with the general formula CnH2n+2

|Number of carbon |1 |2 |3 |4 |5 |6 |

|atoms | | | | | | |

|Name |Methane |Ethane |Propane |Butane |Pentane |Hexane |

|Molecular formula |CH4 |C2H6 |C3H8 |C4H10 |C5H12 |C6H14 |

Alkanes contain only carbon-carbon and carbon-hydrogen bonds, which are strong and non-polar. Alkanes therefore do not react with acids, alkalis, electrophiles or nucleophiles. They have only three common reactions:

1. Combustion

2. Cracking

3. Free radical chlorination

1. Combustion

As all hydrocarbons, alkanes burn in air or oxygen in very exothermic reactions. Hence, their important use as fuels. In a plentiful supply of oxygen, complete combustion occurs to form carbon dioxide and water. For example:

CH4 + 2 O2 → CO2 + 2 H2O ΔHo = - 890 kJ mol-1

C4H10 + 6.5 O2 → 4 CO2 + 5 H2O ΔHo = - 2880 kJ mol-1

As the number of carbons increases, more oxygen is required per mole of hydrocarbon for complete combustion.

If insufficient oxygen is available, incomplete combustion occurs, e.g. in internal combustion engines where carbon monoxide is formed by the incomplete combustion of petrol vapour.

C8H18 + 8.5 O2 → 8 CO + 9 H2O

These engines also produce other pollutant gases, notably:

• oxides of nitrogen, NO and NO2 (known as NOx) formed by the direct reaction of nitrogen and oxygen at the spark;

• unburnt hydrocarbons;

• compounds of lead (from leaded petrol).

Much of the pollution can be removed by the use of catalytic converters. These contain the metals platinum, palladium and rhodium which catalyse reactions between the pollutants and help to remove up to 90% of the harmful gases, e.g.

2 CO + 2 NO → 2 CO2 + N2

C8H18 + 25 NO → 8 CO2 + 12.5 N2 + 9 H2O

2. Cracking (NOT OCR(A) and WJEC)

The supply of petrol and smaller fractions from crude oil does not meet the demand, so longer chain molecules are cracked to make smaller molecules. These are chemical processes as they involve breaking C-C and C-H bonds. Large alkanes are cracked to form smaller alkanes and alkenes and sometimes hydrogen:

High Mr alkanes → smaller Mr alkanes + alkenes (+ hydrogen)

When cracked, molecules may break up in different ways to give a mixture of products that can be separated by fractional distillation, e.g. two possible fragmentations of C14H30 are:

C14H30 → C7H16 + C3H6 + C2H4

C14H30 → C12H24 + C2H4 + H2

Two main processes are used; thermal cracking and catalytic cracking:

|Type |Mechanism |Conditions |Products |

|Thermal |Radical |High Temp (400 – 900oC) |High % of alkenes |

| | |High pressure (7000 kPa) | |

|Catalytic |Carbocation |Slight pressure (NOT low!) |Few alkenes, mainly branched alkanes |

| | |High Temperature (450oC) |(for motor fuels), |

| | |Zeolite (aluminosilicate) catalyst |cycloalkanes and aromatics |

3. Chlorination

In uv light or at T > 500oC methane reacts with chlorine to form hydrogen chloride and a mixture of chlorinated methanes. The overall reaction for the first stage is:

CH4 + Cl2 → CH3Cl + HCl

The mechanism is a free-radical substitution involving three stages:

1. Initiation:

[pic]

Propagation:

[pic]

Termination:

[pic]

Further substitution can also occur forming CH2Cl2, CHCl3, and finally CCl4

The likelihood of further substitution can be reduced if an excess of methane is used.

4. ALKENES

Unsaturated hydrocarbons, general formula CnH2n, e.g. ethene (C2H4), propene (C3H6), butene (C4H8 – remember structural and geometric isomers!)

[pic]

Ethene is a planar molecule; the double bond and the four atoms attached lie in one plane.

The double bond consists of a single bond and a π bond. This π bond consits of a cloud of electrons above and below the plane of the molecule and causes alkenes to be vulnerable to attack from electrophiles, i.e. electron deficient atoms or ions. These electrophiles add across the double bond forming saturated compounds in electrophilic addition reactions.

10.5.1 Electrophilic additions

(i) Hydrogen bromide

C2H4 + HBr → CH3CH2Br

bromoethane

[pic]

(ii) Bromine

C2H4 + Br2 → BrCH2CH2Br

1,2-dibromoethane

A solution of bromine in an organic solvent is decolourised by an alkene. This is used as a test for unsaturation, i.e. to distinguish alkenes from alkanes which do not react.

Mechanism:

[pic]

(iii) Concentrated sulphuric acid (mechanisms – AQA ONLY)

Reaction 1

Alkenes are absorbed by cold, concentrated sulphuric acid forming alkyl hydrogensulphates, e.g. ethene forms ethyl hydrogensulphate:

[pic]

Reaction 2

Warming ethyl hydrogensulphate in dilute sulphuric acid causes hydrolysis and produces ethanol

[pic]

These two reactions result in the overall addition of water to an alkene.

10.5.2 Electrophilic addition to unsymmetrical alkenes

If the alkene is unsymmetrical, e.g. propene, and the molecule added is also unsymmetrical, e.g. HBr or sulphuric acid (H-OSO2OH), two possible products can form, but the major product is the one formed via the more stable carbocation (carbonium ion).

The inductive electron-donating effect of alkyl groups means that stability increases in the order:

[pic]

Propene reacts with:

Hydrogen bromide to produce mostly 2-bromopropane via a secondary carbocation

CH3CH=CH2 + HBr → CH3CHBrCH3 (2-bromopropane)

[pic]

A little 1-bromopropane will also be formed via the less stable primary carbocation:

[pic]

The reaction with concentrated sulphuric acid occurs to produce mainly 2-propyl hydrogensulphate, via the secondary carbocation. Hydrolysis of this hydrogensulphate forms propan-2-ol:

[pic]

CH3CH(OSO3H)CH3 + H2O → CH3CH(OH)CH3 + H2SO4

propan-2-ol

3. Other additions

With hydrogen (hydrogenation) with a nickel catalyst at about 150oC form alkanes:

[pic]

Vegetable oils, which are polyunsaturated compounds, are “hardened” in this way during the manufacture of margarines.

With steam (hydration) at 300oC and 65 bar (= 6500 kPa or 6.5 MPa) in the presence of phosphoric acid as catalyst:

[pic]

Compare direct hydration with fermentation as a method to form ethanol (see alcohols section later).

10.5.4 Polymerisation (self-addition)

Ethene molecules link together in the presence of a catalyst to form an addition polymer which is saturated, poly(ethene). Other polymers such as poly(chloroethene) or PVC and poly(phenylethene) (polystyrene) can be formed where some or all of the hydrogens in ethene have been replaced, i.e.

[pic]

10.6 HALOALKANES

The homologous series of compounds with the general formula CnH2n+1X

Where X is a halogen, i.e. F. Cl, Br or I

e.g. CH3Cl (chloromethane) or CH3CHBrCH3 (2-bromopropane)

Halogen atoms are very electronegative so carbon-halogen bonds are polar.

The electrons in the C-X bond are attracted towards the halogen atom which becomes δ-.

The carbon atom is δ+ and therefore open to attack by nucleophiles, i.e. ions or molecules with a lone pair of electrons.

When nucleophilic attack occurs, the carbon-halogen bond breaks and a halide ion is released. The nucleophile replaces the halogen atom in a nucleophilic substitution reaction, i.e.

[pic]

Note: the conflict between the polarity of the bonds and their strength.

Because F is the most electronegative element, the C-F bond is the most polar, so the carbon attached to F is most vulnerable to attack by a nucleophile, but the C-F bond is also the strongest and thus less easily broken.

In practice it is the bond strength which is more important so the fluoroalkanes are the least reactive and the iodoalkanes the most reactive (remember washing your Teflon frying pan!).

| |C-F |C-Cl |C-Br |C-I |

|Polarity |Increasing |

|Strength |Increasing |

10.6.1 Nucleophilic substitution reactions

(i) Sodium or potassium hydroxide

When warmed with aqueous sodium or potassium hydroxide, hydroxide ions acts as nucleophiles and alcohols are formed, e.g.

CH3CH2Br + OH- → CH3CH2OH + Br-

ethanol

[pic]

Compare with alcoholic NaOH later – AQA only

(ii) Potassium cyanide

When haloalkanes are warmed with an aqueous/alcoholic solution of potassium cyanide, nitriles are formed, e.g.

CH3CH2Br + CN- → CH3CH2CN + Br-

propanenitrile

[pic]

Nucleophilic substitution with cyanide ions adds an extra carbon to the chain.

(iii) Ammonia

When halolakanes are warmed with an excess of ammonia in a sealed container, primary amines are formed, e.g. bromoethane forms ethylamine

CH3CH2Br + NH3 → CH3CH2NH2 + HBr

Or better as:

CH3CH2Br + 2 NH3 → CH3CH2NH2 + NH4Br

ethylamine

[pic]

The excess of ammonia minimises the chance of further reaction of the primary amines to form secondary or tertiary amines or quaternary ammonium salts:

[pic]

10.6.2 Elimination from haloalkanes – AQA, Edexcel & WJEC - mechanism AQA ONLY

As well as acting as a nucleophile, the hydroxide ion is also a base, so will remove a proton from a haloalkane in an elimination reaction in which an alkene is formed, e.g.

CH3CHBrCH3 + OH- → CH3CH=CH2 + H2O + Br-

2-bromopropane propene

[pic]

Note: in some cases more thane one alkene may be formed, e.g. elimination from 2-bromobutane produces both but-1-ene and but-2-ene (both cis-but-2-ene and trans-but-2-ene).

10.6.3 Concurrent displacement (i.e. substitution) and elimination

Whether substitution or elimination occurs depends on several factors, but simply:

• Subsitution is favoured by warm aqueous conditions. OH- acts as a nucleophile

• Elimination is favoured by hot ethanolic conditions, OH- acts as a base

10.7 ALCOHOLS

The homologous series with the general formula CnH2n+1OH

e.g. methanol, CH3OH and ethanol, CH3CH2OH

10.7.1 Ethanol production

(i) By fermentation of sugars:

[pic]

(ii) By direct hydration using steam, phosphoric acid catalyst at 300oC and 65 MPa:

C2H4 (g) + H2O (g) → CH3CH2OH (g)

Comparison of methods used to produce ethanol:

|Method |Rate of reaction |Quality of product |Raw material |Type of process |

|Fermentation |Slow |Impure |Sugars (a renewable |Batch (expensive on |

| | |(aqueous solution) |resource) |manpower) |

|Hydration |Fast |Pure |Ethene from oil (a finite |Continuous (cheap on |

| | | |resource) |manpower) |

10.7.2 Ethanol as a “carbon neutral” biofuel?

Ethanol is already used as a fuel in countries with no natural source of crude oil but have large areas of farmland and thus can grow sugar cheaply and easily, e.g. Brazil.

There is a current debate if ethanol produced from fermentation of sugars is a carbon-neutral fuel? – i.e. there is no net increase of carbon dioxide into the atmosphere.

Thus, production of sugars (glucose) in plants by photosynthesis:

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Fermentation of sugars to produce ethanol:

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Finally, combustion of ethanol as a biofuel:

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Thus it does appear to be carbon neutral – 6 CO2 removed in photosynthesis and a total of 6 (2+4) CO2 released back into the atmosphere from fermentation and combustion.

However, this does not consider other potentially significant sources of CO2 emissions such as production and transport costs?

3. Classifications and reactions

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).

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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).

4. Oxidation of alcohols

The three types of alcohol (primary, secondary and tertiary) differ in their reactions 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:

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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.

5. Distinguishing between aldehydes and ketones

Aldehydes are easily oxidised to carboxylic acids, but ketones are not easily oxidised. Observing whether or not further oxidation occurs can therefore be used to distinguish between them. Mild oxidising agents are used, e.g.

• Tollen’s reagent (ammoniacal silver nitrate) is reduced by aldehydes which produce a silver mirror on the walls of the test tube; ketones do not form a silver mirror.

• Fehling’s/Benedict’s solution (which containes a deep blue copper(II) complex ion) is reduced by aldehydes, but not by ketones, to form a brick red/red/yellow/orange precipitate of copper(I) oxide, Cu2O.

In these reactions, aldehydes are oxidised to carboxylic acids:

RCHO + [O] → RCOOH ([O] = oxidising agent)

10.7.6 Elimination

Alcohols can be dehydrated when heated to 180oC with conc. sulphuric or phosphoric acids.

The reaction is acid catalysed, e.g. the dehydration of propan-2-ol to propene:

The overall reaction is :

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Mechanism - AQA ONLY

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NB some alcohols may form more than one alkene on dehydration, e.g. butan-2-ol can form

but-1-ene and also but-2-ene (both cis-but-2-ene and trans-but-2-ene) by loss of H+ from either side of the intermediate carbocation, i.e.

[pic]

but-2-ene intermediate carbocation but-1-ene

Compare the two methods to form alkenes by elimination:

of water from alcohols using hot conc. H2SO4

of HBr from bromoalkanes using hot alcoholic KOH

11. QUANTITATIVE

1. Moles p57

2. Reacting Masses p58

3. Volumetric Calculations p58

4. Empirical and Molecular Formulae p60

5. Ideal Gas Equation p62

6. Mixed Method Calculations p64

11.1 Moles

11.1.1 Definition

A mole is a number - defined as the number of particles present in 1 mole of any substance – 6.02 x 1023 particles.

To measure out 1 mole of any substance you simply weigh out the relative atomic mass (RAM, Ar) of an element or the relative formula mass (RFM, Mr) of a compound.

e.g. 1 mole of sodium particles is 23.0 g (since the Ar of Na is 23.0);

1 mole of water molecules is 18.0 g (H2O; RFM = (2 x 1) + 16 = 18.0)

1 mole of calcium carbonate is 100 g (CaCO3 = 40 + 12 + (16 x 3) = 100)

11.1.2 Equation/calculations

The most generally applicable equation which links no. of moles (n) to the mass present (m) and the molar mass (Mr) of any substance is:

n = m / Mr

You need to be proficient in using this equation to calculate either the number of moles (n) directly or to be able to rearrange it to calculate either the mass (m) or the molar mass (Mr).

e.g. 1 how many moles is 6.50 g of Mg?

Using n = m / Mr = 6.50 / 24.0 (as 24.0 = Ar for Mg) = 0.271 mol

e.g. 2 what mass is 0.370 moles of NaCl?

Using m = n x Mr = 0.370 x 58.5 (as 58.5 is the Mr for NaCl) = 21.6 g

e.g. 3 what is the molar mass of a substance X if 1.25 x 10-2 moles has a mass of 0.744 g?

Using Mr = m / n = 0.744 / 1.25 x 10-2 = 59.5 g mol-1

11.2 Reacting masses calculations

11.2.1 Basic method

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, etc)

We will use this method in all the example calculations in this booklet.

11.2.2 Reacting masses example

What mass of water can be made from 20.0 g of hydrogen?

1. No. moles hydrogen (known) = m / Mr = 20.0 / 2 = 10.0 g

2. Using equation:

2 mol 2 mol

n mol n mol

10.0 mol 10.0 mol

3. Convert moles of water (n) into mass, m = n x Mr = 10.0 x 18 = 180 g

11.3 Volumetric calculations

11.3.1 Equation

There is a second equation which links no. moles (n) with the concentration of a solution (C) (sometimes alternatively called the Molarity, M) and the volume (V):

n = C x V / 1000

NB since V has units of cm3 but C has units of mol dm-3 (also known as Molar or M) the 1000 is a conversion factor between cm3 and dm3.

Again, you should be proficient in using this equation to calculate either n, C or V.

11.3.2 Volumetric calculations examples

e.g. 1 calculate the no. of moles (n) of HCl contained in 35.0 cm3 of a 0.0100 M solution?

n = C x V / 1000 = 0.0100 x 35.0 / 1000 = 3.50 x 10-4 mol

e.g. 2 calculate the volume of NaOH solution if it contains 0.790 mol with a concentration of 1.50 M?

V = 1000 x n / C = 1000 x 0.790 / 1.50 = 527 cm3

e.g. 3 calculate the concentration of a solution of AgNO3 if 1.36 x 10-3 mol were dissolved in 50.0 cm3 water?

C = 1000 x n / V = 1000 x 1.36 x 10-3 / 50.0 = 0.0272 M

e.g. 4 calculate the mass of AgNO3 needed to make the previous solution?

Since we know that n = 1.36 x 10-3 mol from the previous question, using m = n x Mr = 1.36 x 10-3 x 170 = 0.231 g

However, if you’d been asked to calculate the mass of AgNO3 to make 50.0 cm3 of a 0.0272 M solution in example 3, the approach could be slightly different:

Since n = C x V /1000 and n = m / Mr thus m / Mr = C x V / 1000

so m = C x V x Mr / 1000 = 0.0272 x 50.0 x 170 / 1000 = 0.231 g

(of course, it would be quite correct to do this in two steps by first using n = C x V / 1000 to determine the number of moles (n) of AgNO3 needed and then to substitute this into n = m / Mr to calculate m).

11.3.3 Titration calculations

Again, these use the same basic three step method as before (section 11.2.1) – this time step 1 to calculate the known usually (but not always) involves the titre value.

e.g. 1 25.0 cm3 of a solution of HCl was found to react with exactly 22.7 cm3 of a 0.200 M solution of NaOH.

Calculate the concentration of the HCl solution?

1. No. moles NaOH (known) = C x V / 1000 = 0.200 x 22.7 / 1000 = 4.54 x 10-3 mol

2. Using equation:

1 mol NaOH reacts with 1 mol HCl

4.54 x 10-3 mol NaOH reacts with 4.54 x 10-3 mol HCl

3. Concentration (C) of HCl = 1000 x n / V = 1000 x 4.54 x 10-3 / 25.0 = 0.182 M

e.g. 2 A solution containing 0.325 g of an unknown acid HA was found to react with exactly 17.2 cm3 of a 0.150 M solution of NaOH.

Calculate the Mr of the unknown acid?

1. No. moles NaOH (known) = C x V /1000 = 0.150 x 17.3 / 1000 = 2.58 x 10-3 mol

2. Using equation:

1 mol NaOH reacts with 1 mol HA

2.58 x 10-3 mol NaOH reacts with 2.58 x 10-3 mol HA

3. Mr of HA = m / n = 0.375 / 2.58 x 10-3 = 145 g mol-1

11.4 Empirical and Molecular Formulae

11.4.1 Empirical Formula

Definition: “the formula giving the simplest ratio (by moles) of number of atoms of each element in a compound”.

11.4.2 Calculating Empirical Formula

The data may be supplied from either of two sets of data:

- from the supplied masses of each element which react together

OR

- from % composition (by mass) of each element in the compound.

e.g. 1 From supplied masses of each element: 10.800g of Mg forms 18.000g of an oxide – calculate the empirical formula of magnesium oxide?

Mg O

1. Masses: 10.800 g 18.000 – 10.800 = 7.200 g

2. No. moles ( / Ar) 10.800 / 24 7.200 / 16

= 0.45 mol = 0.45 mol

3. / smallest moles 0.45 / 0.45 0.45 / 0.45

= 1 = 1

4. Empirical formula = MgO

NB you can round the ratio numbers up or down to the nearest whole number ONLY if they are within + 0.1 of a whole number – e.g. 2.9 can be rounded up to 3, 5.1 can be rounded down to 5 but e.g. 2.5 CANNOT be rounded up to 3 or indeed down to 2! Must multiply up all the ratios to produce whole numbers in this case!

e.g. 2 From % composition (by mass) of each element – given that they are 40.2% K, 26.9% Cr and 32.9% O, calculate the empirical formula of the compound?

K Cr O

1. % by masses 40.2% 26.9% 32.9%

2. No. moles ( / Ar) 40.2 / 39.0 26.9 / 52.0 32.9 / 16.0

= 1.031 mol = 0.517 mol = 2.056 mol

3. / smallest moles 1.031 / 0.517 0.517 / 0.517 2.056 / 0.517

= 1.99 = 1 = 3.98

4. Empirical formula = K2CrO4

11.4.3 Water of Crystallisation calculations

This can also be done using the basic empricial formula method above, except that instead of a column for each element, the two columns are for the anhydrous salt and water.

e.g. 1 From supplied masses: 0.869 g CuSO4.nH2O gave a residue of 0.556 g on heating – calculate the empirical formula of the hydrated salt?

CuSO4 H2O

1. Masses: 0.556 g 0.869 – 0.556 = 0.313 g

2. No. moles ( / Mr) 0.556 / 159.5 0.313 / 18

= 0.00348 mol = 0.0174 mol

3. / smallest moles 0.00348 / 0.00348 0.0174 / 0.00348

= 1 = 5

4. Empirical formula = CuSO4.5H2O

11.4.4 Molecular Formula

Definition: “the formula containing the actual number of atoms of each element (by moles) in a compound”.

It thus follows that the Molecular Formula must be a whole number (integer) multiple of the Empirical Formula, i.e. MF = n x EF.

Thus if the EF = X2Y3 then the MF could be X2Y3 or X4Y6 or X6Y9 etc.

To deduce the MF from the EF you also need to know the RMM (Mr) of the compound.

e.g. 1 A liquid X contains 72.0% C, 6.67% H & 21.33% O and has Mr = 150 g mol-1. Calculate the empirical formula of the compound?

C H O

1. % by masses 72.0% 6.67% 21.33%

2. No. moles ( / Ar) 72.0 / 12.0 6.67 / 1.0 21.33 / 16.0

= 6.00 mol = 6.67 mol = 1.33 mol

3. / smallest moles 6.00 / 1.33 6.67 / 1.33 1.33 / 1.33

= 4.5 = 5 = 1

Ratios not all < 0.1 so 4.5 x 2 = 9 5 x 2 = 10 1 x 2 = 2

4. Empirical formula = C9H10O2

5. EF mass = 150 g mol-1 so MF = Mr / EF mass = 150 / 150 = 1

6. Molecular formula = 1 x EF = C9H10O2

11.5 The Ideal Gas Equation

11.5.1 Equation and units

P V = n R T

where: P = pressure (in Pa or Nm-2)

V = volume (in m3)

n = no. of moles

R = Universal Gas Constant = 8.31 J K-1 mol-1

T = temperature (K)

The Ideal Gas Equation (IGE) is an expression derived from a combination of Boyles’ and Charles’ Laws for gases and is extremely useful as it links both the pressure (P), volume (V) and temperature (T) of a gas to the number of moles (n). The major thing to watch out for is that each quantity is in the correct units since the equation is a little unusual in that all the quantities must be in the standard SI units!

You should be proficient in rearranging the IGE to calculate any of the variables, i.e. n, P, V or T.

The two major units which usually require converting are Pressure and Volume:

- Pressure is usually measured in kPa, with 100 kPa being standard atmospheric pressure, but the IGE requires the pressure in Pa – conversion factor x 1000

- Volume is usually measured in cm3, but the IGE requires the volume in m3 – conversion factor is / 1,000,000 or x 10-6.

11.5.2 Examples of Ideal Gas Equation calculations

e.g. 1 Compound A is an oxide of sulphur. At 415 K, a gaseous sample of A of mass 0.304 g occupied a volume of 127 cm3 at a pressure of 103 kPa.

Calculate the no. of moles (n) of A in the sample and thus the Mr of A?

1. P V = n R T so n = P V / R T

2. Units conversions: P = 103 kPa = 103,000 Pa & V = 127 x 10-6 m3

3. So n = 103,000 x 127 x 10-6 / (8.314 x 415) = 3.79 x 10-3 mol

4. Mr = m / n = 0.304 / 3.79 x 10-3 = 80.2 g mol-1

e.g. 2 A 0.143 g gaseous sample of ammonia occupied a volume of 2.86 x 10-4 m3 at a pressure of 100 kPa at temperature T.

Calculate the temperature of the sample?

1. P V = n R T so T = P V / n R

2. n = m / Mr = 0.143 / 17.0 = 8.41 x 10-3 mol

3. Unit conversion: P = 100 kPa = 100,000 Pa

4. So T = 100,000 x 2.86 x 10-4 / (8.41 x 10-3 x 8.314) = 409 K

e.g. 3 2.32 kg of methane occupied a volume of 143 dm3 at a temperature of 298 K.

Calculate the pressure inside the container?

1. P V = n R T so P = n R T / V

2. Unit conversions: m = 2.32 kg = 2,320g & V = 143 dm3 = 143 x 10-3 m3

3. n = m / Mr = 2,320 / 16 = 145 mol

4. So P = 145 x 8.314 x 298 / 143 x 10-3 = 2,510,000 Pa = 2.51 MPa! (

11.6 Mixed methods calculations

These involve using two different equations in steps 1 and 3 (as opposed to most standard calculations which use the same equation in both steps). Steps 1 & 2 are identical to all the other moles calculations we have done earlier in this booklet.

e.g. 1 what volume of CO2 will be produced at 773 K and 100 kPa if 50.0 kg of calcium carbonate is decomposed in a lime kiln?

1. No. moles CaCO3 = m / Mr = 50 x 1000 / 100 = 500 mol

2. From equation:

1 mole CaCO3 produces 1 mole CO2

500 mol CaCO3 will produe 500 mol CO2

3. P V = n R T so V = n R T / P

4. Unit conversion: P = 100 kPa = 100,000 Pa

5. V = 500 x 8.314 x 773 / 100,000 = 32.1 m3

Tip: leave units in the ones the calculation puts them into UNLESS you are specifically asked for a particular unit in your answer.

e.g. 2 what mass of magnesium hydroxide would be needed to exactly neutralise

2.5 dm3 of 2M HCl which had accidentally been poured down a drain?

1. No. moles HCl = C x V / 1000 = 2.00 x 2500 / 1000 = 5.00 mol

2. From equation:

1 mol needs 2 mol

2.5 mol needs 5.0 mol

3. Mass = n x Mr = 2.5 x 58.0 = 145 g

e.g. 3 what is the concentration of the H2SO4 solution produced if 200 dm3 of gaseous SO3 are bubbled through 50.0 dm3 water at 298 K and 100kPa?

1. P V = n R T so n = P V / R T

2. Unit conversions: P = 100 kPa = 100,000 Pa & V = 200 x 10-3 m3

3. No. moles SO3 = n = 100,000 x 200 x 10-3 / (8.314 x 298) = 8.07 mol

4. From equation:

1 mole SO3 produces 1 mole H2SO4

8.07 mol SO3 will produce 8.07 mol H2SO4

5. n = C x V (since V in dm3) so C = n / V = 8.07 / 50.0 = 0.161 M

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AS/Year 12 Chemistry

Revision Workshop

Spring 2020

Reviewing the AS content of your Examinations

Participants Notes

T1

T1

T1

T1

2

ethanoic acid

propanone

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