AQA A level Chemistry Unit 2 Notes

Group 2 Elements

Trend Reasons

Atomic Radius

Mg<Ca<Sr<Ba (Increases as you go down

group)

1) More Energy Levels 2) More Shielding Electrons

Ionisation Energy

Mg>Ca>Sr>Ba (Decreases as you go down

group)

1) Electron further from Nucleus 2) More Shielding Electrons

3) Whilst effective Nuclear Charge stays the same

Melting Points

Mg>Ca<Sr>Ba (Decreases as you go down

group but Sr has a higher melting point than Ca)

1) The ionic radius increases

2) Attraction between the delocalised electrons and the metal cations decreases.

WHY DOES A METAL HAVE A HIGHER MELTING POINT THAN THE OTHER METAL:

1) Delocalised electrons are closer to cations, therefore there is a stronger attraction between the cations and delocalised electrons.

Reactivity With Water

Mg<Ca<Sr<Ba (Increases as you go down

group)

1) Ions get bigger, therefore there is less force of an attraction

2) The outer-electron therefore is more easily lost.

Compared to group 1, metals are less reactive as ions get smaller across a period. Metal Reaction with Cold Water

Mg Reaction too slow to notice Ca Reaction quick but not violent Sr/Ba Reaction increasingly exothermic Example:

Ca(s) + 2H2O(l) → H2(g) + Ca(OH)2

Ca(s) + 2H2O(l) → H2(g) + Ca2+(aq) + 2OH-(aq)

Group 2 Sulphates Group Hydroxides (made by reacting with water)

Solubility

MgSO4>CaSO4>SrSO4>BaSO4

Decreases as you go down group. They are all ions and therefore

aqueous.

Mg(OH)2<Ca(OH)2<Sr(OH)2<Ba(OH)2

Increases as you go down group. They are all solids too.

Therefore: Ba(OH)2 is soluble butBaSO4 is insoluble.

Mg(OH)2 is insoluble but MgSO4 is soluble.

If anything is soluble, then it is easy to remove as it can be dissolved.

BaSO4 is used in medicine as a Barium meal as it is insoluble, therefore it doesn’t cause

harm.

Mg(OH)2 is used in medicine as an Indigestion Tablet as it is insoluble.

Ca(OH)2 is used in agriculture to neutralise soil acid.

Test for Sulphate Ions (SO42-) – AQUEOUS. Why

Add Dilute Hydrochloric Acid (HCl(aq))/

Add Dilute Nitric Acid (HNO3(aq))

(NOT sulphuric acid as this contains SO4

2-Destroys Carbonate ions, Barium Carbonate also forms a white precipitate.

How: 2H+ _{+ CO}

32- → H2O + CO2

Add Barium Chloride Solution (BaCl2(aq))/

Add Barium Nitrate Solution (Ba(NO3)2(aq))

Produces precipitate.

Positive Result White precipitate of Barium Sulphate (BaSO4)

Ba2+ _{+ SO}

42- → BaSO4 (white precipitate)

To distinguish between BaCl2 and MgCl2 we use Sodium Hydroxide, BaCl2 remains

Haloalkanes The following is called a Free Radical Substitution.

When an alkane reacts with a halogen under conditions of: 1) Ultraviolet Light 2) Heat

3) Other forms of energy

Then the U.V. Light provides the energy needed to break the covalent bond in the halogen or haloalkane molecule homolytically. Homolytic bond breaking creates radicals.

There are three stages: 1) Initiation: makes the radicals 2) Propagation: a) Take one atom

b) Pairs of Reaction

c) 2nd _{Reaction produces Radical of 1}st _Reaction

d) Makes the real products

e) Further steps occur if excess reactant 3) Termination: Destroys Radicals

Ozone is O3. It is formed by either: 1) Ultraviolet light on Oxygen light in the atmosphere

2) Electrical sparks at ground level

Polluting Ozone is a very toxic gas. The ozone layer in the atmosphere absorbs U.V. light limiting the amount reaching the surface of the Earth. At present, there are large holes in the North and South Pole.

Chlorine and chlorine compounds, which are found in the poles, destroy Ozone. First, U.V. light hits chlorine or chlorine based compounds:

Cl2 → 2Cl

∙

Then Cl

∙

reacts with Ozone (O3) in a pair of propagation steps. The Chlorine free radical is

called the reactive intermediate. Cl

∙

+ O3 → OCl

∙

+ O2

OCl

∙

+ O3 → Cl

∙

+ 2O2

This makes the overall equation: 2O3 → 3O2

Therefore, as you can see, Chlorine is a catalyst in turning Ozone (O3) into Oxygen (O2).

This is because the termination step Cl

∙

+ Cl

∙

→ Cl2 = makes Chlorine, which would once

again get hit by U.V. light. Therefore Chlorine provides an alternative route of lower activation energy.

Chlorine atoms are also formed from chlorine-containing organic compounds in the upper atmosphere due to the carbon-chlorine bond breakage in a homolytic way.

Therefore a C-Cl bond is not considered environmentally acceptable or friendly. The IUPAC name of CF3Cl is Chlorotrifluoromethane

Also remember that when a metal chloride reacts with Sulphuric Acid, it makes Hydrochloric Acid (HCl) and one other product.

Oxidation and Reduction Oxidation is the Loss of Electrons

Reduction is the Gain of Electrons Oxidation is the Loss of Hydrogen Reduction is the Gain of Hydrogen Oxidation is the Gain of Oxygen Reduction is the Loss of Oxygen

Oxidising Agents are REDUCED themselves and OXIDISE others. Reducing Agents are OXIDISED themselves and REDUCE others. General Rule:

If you see a substance that has lost electrons, then it is a Reducing Agent. If you see a substance that has gained electrons, then it is an Oxidising Agent Oxidation Numbers:

1) All elements are zero

2) All simple ions have an oxidation number equal to its charge

3) The SUM of the oxidation numbers of a complex ion (molecules with a charge) is equal to the charge on the ion

4) The SUM of the oxidation numbers of a molecule is zero Results:

1) It has been oxidised if the oxidation number gets more positive because it has lost electrons; it is therefore a Reducing Agent.

2) It has been reduced if the oxidation number gets more negative because it has gained electrons; it is therefore an Oxidising Agent.

Assume :- Oxygen atoms are -2 Hydrogen atoms are +1 FOR EXAMPLE:

CH4 – Hydrogen atoms are +1

There are 4 Hydrogen atoms, therefore 4 × +1 = +4

This is a molecule, therefore (rule 4 says) the SUM must equal zero That is how we know that C has oxidation number -4

CO2 – Oxygen atoms are -2

There are 2 Oxygen atoms, therefore -2 × 2 = -4

This is a molecule, therefore (rule 4 says) the SUM must equal zero That is how we know that C has oxidation number +4

CO32- – Oxygen atoms are -2

There are 3 Oxygen atoms, therefore -2 × 3 = -6

This is a complex ion (molecule with a charge), therefore (rule 3 says) the SUM must equal the charge on the ion

The charge on the ion is -2

That is how we know that C has oxidation number +4, because -6 +4 = -2 Redox Equations:

1) Write down the oxidation numbers

2) Identify the Oxidation equation and Reduction equation 3) Write Half-equations for both separately, use H2O, H+ and e

-4) Combine Reduction half equation and Oxidation half equation 5) Multiply equations appropriately so that electrons cancel

Nomenclature

Functional Group Word Endings Connection with

Carbon Chain

Conditions

Alcohol Anol 1) The isomer

will end with ‘ol’ if there is a number in the middle.

Aldehyde Anal 1) It is always

at the rear end of a chain. 2) Numbers begin from butanal

Ketones Anone 1) Isomer will

be in the middle of a chain

Carboxylic Acid Anoic 1) It is always

at the rear end of a chain. Amines Amine or begin with Amino

NO CONDITIONS

Nitriles Cyanide NO

CONDITIONS In a test for Alkanes and Alkenes, we use bromine water, it remains orange with Alkane, but turns colourless with an Alkene.

IMPORTANT: If it asks you for the displayed formula of any of the above, you must show the bonds. So instead of writing OH, you must O-H.

Geometrical Isomers take place at sights of restricted rotation.

If the both of the above connected chain are different to the bottom connected chain, then there will be E/Z isomerism.

Double bonded isomers and Rings isomers have restricted rotation. The E isomer looks like a Z shape.

The Z isomer looks like an E shape.

Type of isomer What Happens Priority

Functional Group Isomer

Ending changes

e.g. goes from ANE to ENE, etc.

1st

Chain Isomer First part of name changes

e.g. pentane to propane

2nd

Position Isomer

Numbers change

e.g. goes from 1-methylpentane to 2-methylpentane

3rd

Geometrical

Isomer e.g. goes from Z-1,2-dimethylcyclopropane to E-1,2 -dimethylcyclopropaneChanges from E to Z or vice versa 4

Alcohols

Primary Alcohols (Alcohols in which the Carbon connected to the OH is connected to 1 other Carbon) are oxidised (they are made to gain Oxygen or lose Hydrogen) first to Aldehyde, then to Carboxylic Acids.

The Reagent that will be the oxidising agent for the alcohol: Acidified Potassium Dichromate (H+_/Cr

2O72-/ [O])

EXAMPLE: Ethanol (primary alcohol)

CH3CH2OH + [O] → CH3CHO + [O] → CH3CO2H

Ethanol Ethanal Ethanoic Acid (Lost 2 Hydrogens) (Gained 1 Oxygen)

Aldehyde boils at a lower temperature than alcohols. Therefore we carry out the oxidation at a temperature greater than the boiling point of the Aldehyde so that the Aldehyde is

separated from the [O] and the alcohol.

By using Reflux, we can make the carboxylic acid.

Secondary Alcohols (Alcohols in which the Carbon connected to the OH is connected to 2 other Carbons) are oxidised (they are made to gain Oxygen or lose Hydrogen) to Ketones. EXAMPLE: Propan-2-ol (secondary alcohol)

CH3CHOHCH3 + [O] → CH3COCH3

Propan-2-ol Propanone (Lost 2 Hydrogens)

Tertiary Alcohols (Alcohols in which the Carbon is connected to 2 other Carbons) can’t be oxidised easily as the double bond can’t be made without breaking a C-C bond. Therefore, if it hints towards an alcohol that doesn’t react with Acidified Dichromate, then this alcohol is tertiary. Alcohols are a Homologous Series. In hydration, we use concentrated H2SO4

How each reagent helps distinguish between

Aldehydes (Anal) & Ketones (Anone)

Aldehyde Ketone

Acidified Dichromate Orange to Green Orange to Orange Tollen’s Reagent Colourless to Silver Mirror Colourless to Colourless Benedict’s or Feheiling Blue solution to Brick Red

Precipitate

Industrial preparation of Ethanol

Fermentation Industrial

Raw Material Glucose Ethene

Renewable Yes No

Other Requirements Yeast & Water Water & Phosphoric Acid catalyst

Temperature 37o_{C High Temperature, but Low Pressure}

Level of Technology Low High

Rate of Production Slow Very Fast

Purity Very impure Very pure

Purification Process Distillation – V.Expensive Not needed

We use the industry method. We don’t use fermentation because it is a slow reaction and is very expensive. The industry method involves Hydration/ Electrophilic Addition of ethene to from ethanol.

Brazil runs cars on alcohol through fermentation as it has lots of land; the sugar grows quickly and due to cheap labour.

Bio Fuels Fuel produced from biological or renewable sources

Fermentation is not Carbon Neutral: 1) Fuel used to distil impure ethanol releases CO2

2) Harvesting/planting sugar involved using fuel that releases CO2

For a complete combustion of a fuel to take place in air, you must have sufficient supply of oxygen. Fermentation: 1) Yeast 2) Sugar 3) Water 4) 37o_C Essential Conditions: 1) Yeast 2) 30o_{C ≥ T ≤ 42}o_C 3) No Air

It is an Oxidation reaction as respiration is Anaerobic. C6H12O6 → 2C2H5OH + 2CO2

Sugar (glucose) Ethanol Carbon Dioxide Can be claimed to be Carbon Neutral:

6 CO2 taken in, 6 CO2 given out due to 2 CO2 from fermentation and 4 CO2 from combustion.

Curly arrow begins at: 1) A Bonding Pair

Mechanism What Happens Effect

Nucleophilic Substitution Curly arrow goes to Substrate from Reagent

Makes a Saturated (single bond) reactant into a Saturated (single bond)

product Base Elimination Curly arrow goes to Hydrogen in the

Substrate from the Reagent ( which is called a Base)

Makes a Saturated (single bond) reactant into a Unsaturated (double bond)

product Electrophilic Addition Curly arrow goes to Reagent from

Double bond in Substrate

Makes a Unsaturated (double bond) reactant into a

Saturated (single bond) product

Acid Elimination Curly arrow goes to Hydrogen in the Acid from the Substrate

Makes a Saturated (single bond) reactant into a Unsaturated (double bond)

product 2) A Lone Pair

Nucleophile: Electron pair donor. Electrophile: Electron pair acceptor.

Reagent: Chemical/chemicals that react with an organic molecule. Substrate: The organic molecule that undergoes the reaction.

RULES:

1) First curly arrow goes to something positive

2) If curly arrow departs, it makes +1 charge 3) If curly arrow arrives, it makes -1 charge

4) Atoms have a max of 4 pairs, except Hydrogen which has a max of 1 pair

The substrate is an Alkene. It is the only mechanism that begins with an Alkene. It makes a Haloalkane.

Reagents:

1) Bromine (Br-Br) or Chlorine (Cl-Cl) : How: the double bond in the substrate is a centre of high electron density. This induces a DIPOLE in the Br-Br or Cl-Cl bond. 2) HCl or HBr: How: H-Cl and H-Br are polar bonds.

3) Sulphuric Acid (H-OSO3H): HOW: The H-O bond in Sulphuric Acid is highly polar.

IMPORTANT: The most stable intermediate connects to the Halogen forming more often and therefore giving the major product.

Types of Carbon What is it Stability

Tertiary Connected to 3 other Carbons Most Stable Secondary Connected to 2 other Carbons Less Stable Primary Connected to 1 other Carbon Least Stable

When an alkene becomes an alkane, but it does not involve curly arrows, then it is simply called Addition or Reduction or Hydrogenation.

Nucleophilic Substitution:

The substrate is a haloalkane. It can make an Alcohol, Nitrile or Amine. Reagents:

1) Aqueous Sodium/Potassium Hydroxide (HO¨-_{) Remember: We don’t use the Sodium}

Ion. This reagent makes an alcohol.

2) Alcoholic Potassium Cyanide (N C¨-_{) Remember: we don’t use the Potassium. This}

reagent makes a Nitrile.

3) Ammonia (H3N¨); more ammonia will be added because this makes an unstable

intermediate. This reagent makes an Amine.

If we have an excess of haloalkane, further reactions take place.

As we go down the Halogen group, the Haloalkane in the Nucleophilic Substitution increases the rate of the Nucleophilic Substitution (R-I reacts fastest with Reagent, R-Cl reacts slowest with Reagent). As we go down the group, the bond strength of the halogen with the alkane decreases even though polarity decreases (the more polar a bond, the weaker it is). Hence, the reason the rate of Nucleophilic reaction increases as you down the halogen group is due to polarisibility (the R-I bond is EASILY polarised).

The characteristic of the haloalkane that enables it to undergo the Nucleophilic Substitution is the fact that it has a polar bond between a Carbon and Halogen.

Base Elimination:

The substrate is a haloalkane. It makes an Alkene. Reagents (Base):

1) Alcoholic Potassium/Sodium Hydroxide (HO¨-_{) Remember: We don’t use the Sodium.}

Acid Elimination:

The substrate is saturated. It makes an alkene. Reagents:

1) H+ _{How: There is a lone pair in the substrate that goes to the positive H}+

Polymerisation

It is a reaction which turns many small molecules into long chain molecules. Hence:

MONOMER → POLYMER Addition Polymers: (Alkenes undergo Addition)

Made from alkenes or similar compounds EXAMPLE: Ethene to Poly(ethene)

Others:

Poly(propene) = plastic bottles

Poly(chloroethene)

(P.V.C) = drainpipes

Poly(phenylethene) (Polystyrene) = bubble wraps/packaging

If you are shown a polymer and asked what it was before, then count the number of Carbons, then this shall give you the alkene chain, then you need to work out where the double bond was.

Polymers are now recycled because: 1) It preserves raw materials 2) Prevents pollution

3) Saves energy

Polymers are non-biodegradable materials.

The Mass Spectrometer

Molecules can also be analysed by a mass spectrometer (not just elements). Cl2 + e- → Cl2+

∙

+ 2e

-The mass spectrometer can measure to at least 4 decimal places of an a.m.u.

Mr Mass Spectrometer Mr

CO 28 28.0104

N2 28 28.0134

C2H4 28 28.0536

Therefore, the mass spectrometer can distinguish between compounds of similar Mr.

Propane and Carbon Dioxide have the same Mr when it is to one decimal place.

Carbon has Mr 12.00000 by definition.

Some Chemicals have the same Mr EVEN TO 5 DECIMAL PLACES: Why:

They contain the same number of atoms of the same element, or another way of saying it is that they have the same molecular formula.

Some Chemicals have a different Mr when made to 5 DECIMAL PLACES: Why:

They contain different number of elements of different elements.

Trend Reasons

Melting Point

F2<Cl2<Br2<I2

(Increases as you go down group)

1) Van Der Waals forces increase as the size of the molecules increase.

Electronegativity

F2>Cl2>Br2>I2

(Decreases as you go down group)

1) Force decreases as size increases 2) Effective charge remains

approximately constant

Oxidising Ability (The ability to attract electrons to itself and take away electrons from

others)

F2>Cl2>Br2>I2

(Decreases as you go down group)

1) Small F- _{ion can attract extra}

electron strongly, therefore it forms more easily

2) Large I- _{ion can’t attract extra}

electron strongly, therefore it finds it more difficult to form

Order of Reactivity

F2>Cl2>Br2>I2

(Decreases as you go down group)

1) It needs to gain one electron 2)As the size increases, the nuclear force decreases

3) Therefore the nucleus finds it harder to attract an electron

Reducing Ability (The ability to give away

electrons to others)

Cl-_<Br-_<I

-(Increases as you go down group)

1) I- _{ion is very big, therefore it has a}

weak force from the nucleus and hence electrons are easily lost

2) Cl- _{ion is very small, therefore it has}

a strong force from the nucleus, hence electrons are difficult to remove

Strength of bond with C

F2>Cl2>Br2>I2

(Decreases as you go down the group)

1)The size of the molecules increases, therefore it becomes polar very easily

Halogens may or may not displace with halide ions. The colour given off is of the element of the displaced ion

Halides: F-_{, Cl}-_{, Br}-_{, I}-_{} colourless}

Halogens:

F2 – an aqueous solution of F2 cannot be made as it reacts with water

Cl2 – an aqueous solution of Cl2 has the colour (pale) green

Br2 – an aqueous solution of Br2 has the colour Orange/Brown

I2 – an aqueous solution of l2 has the colour Purple/Black

It takes less energy to break the weaker C-halogen bond; therefore the precipitate would form quicker.

Chlorine reacts slowly with water: It is used in swimming pools but as small doses as it is toxic. A Universal Indicator would turn pink in chlorine water.

Cl2(aq) + H2O →

HCl (oxidation number of Cl here is -1) + HOCl (oxidation number of Cl here is +1)(aq)

This is an example of disproportionation. When the same chemical is oxidised and reduced. Over time the HOCl (Chloric Acid) decomposes: 2HOCl(aq) → 2HCl(aq) + O2(g)

When Chlorine reacts with Sodium Hydroxide, Sodium Chloride and Sodium Chlorate (NaClO) is formed. When it says “give the formula” you must not write the name of the substance, you should write the actual chemical formula of it, even if it may be an element.

-Halide Ions Colourless Solution Colourless Solution Colourless Solution Add dilute HNO3

(Destroys Carbonate ions that could make Ag2CO3 which is

white and would give a false result

– HOW does it destroy it: 2H+ _{+ CO}

32- →

CO2 + H2O

Colourless Solution Colourless Solution Colourless Solution

Add Silver Nitrate (Forms The Precipitates)

White Precipitate

Ag+

(aq) + Cl-(aq) → AgCl(s)

Cream Precipitate

Ag+

(aq) + Br-(aq) → AgBr(s)

Yellow Precipitate

Ag+

(aq) + I-(aq) → AgI(s)

Add Dilute Ammonia (Dissolves AgCl(s))

Colourless Solution AgCl + 2NH3 → [Ag(NH3)2]+ + Cl

-Cream Precipitate Yellow Precipitate

Add Concentrated

Ammonia (Dissolves

AgBr(s))

Colourless Solution Colourless Solution AgCl + 2NH3 → [Ag(NH3)2]+ + Cl

-Yellow precipitate

Fluoride ions are not tested for in the test for Halide ions, this is because when we add the Silver Nitrate, it makes AgF, which is soluble and therefore forms no precipitates.

Add Concentrate Sulphuric Acid (H2SO4) Equation

NaCl (bleach) (kills bacteria )

Misty Acidic Fumes (HCl) Cl- _{can’t reduce SO} 4

2-NaCl + H2SO4 → HCl + NaHSO4

(Not redox)

The role of H2SO4 is simply an acid

NaBr

Misty Acidic Fumes (HBr) NOT redox reaction, it is a DISPLACEMENT REACTION Orange/Brown liquid fumes (Br2) H2SO4 + 2H+ + 2Br- → Br2 + SO2 + 2H2O

Acidic fumes (SO2)

NaI

Misty Acidic Fumes (HI) NOT redox reaction, it is a DISPLACEMENT REACTION Purple/Black solid fumes (I2) SO42- + 4H+ + 2I- → Br2 + SO2 + 2H2O

Unpleasant Smell (H2S - Hydrogen Sulphide) 10H+ + 8I- + SO42- → H2S + 4I2 + 4H2O

Reduction product: H2S

Oxidation product: 4I2

Yellow Precipitate/Solid (S8 –Sulphur) =

insoluble 6I- _{+ SO} 42- + 8H+ → 4H2O + S + 3I2 Reduction product: S Oxidation product: I2 Infrared Spectroscopy Infrared waves do not break bonds but makes them vibrate.

The frequency of the vibration depends on: 1) strength of the bond- single, double or triple 2) Masses of the bonded atoms (C, N, O=heavy.

More absorption (higher wave number) of an infrared wave makes that chemical a more effective greenhouse gas.

Bonds to Hydrogen 2500-3500 Triple Bond 2220-2260 Double Bond 1620-1680 Single Bond 750-1300 Alcohol Aldehyde and Ketone Carboxyli c Acid Nitrile Alkene

Finger print region unique to all compounds. It is compared to an authentic sample.

Metal Extractions

Metals are usually found in the ground as their positive ions in ionic compounds. Therefore extracting metal must involve reduction (gaining of electrons).

Positive Ion + Electrons → Metallic Elements e.g. Fe3+ _{+ 3e}- _{→ Fe}

Bonding Type Frequency

Bonds to Hydrogen 2500-3500

Triple Bond 2220-2260

Double Bond 1620-1680

Ores are usually found as sulphides or oxides.

Sulphides are roasted in air to convert to Oxides. (This makes SO2 – Acid Rain)

EXAMPLE:

WS2 + 3.5O2 → WO3 + 2SO2

The 1st _{choice reducing agent (it reduces others and oxidised itself) is Carbon (coke-made}

from coal).

Blast Furnace – we can use Iron, Copper or Manganese

1) Iron Ore (Fe2O3) – impure – main impurity is SiO2 – limestone (CaO) used to remove

it and this makes CaSiO3 (slag). The slag can be used to build roads.

2) Air 3) Coke 4) Limestone

Reduction Process in the Blast Furnace

Reducing Agent: Coke (Condition: high temperature of 1500o_{C); we are reducing Iron(III).}

Fe2O3 + C → Fe + CO

OR

Fe2O3 + C → Fe + CO2

The other main reducing agent is Carbon Monoxide (CO) (C + O2 → CO2

C + CO2 → 2CO) this (coke) is where the energy comes from in a blast furnace.

Fe2O3 + 3CO → 2Fe + 3CO2 (Iron is more likely to react with CO than C as CO is gas)

The iron still contains about 4% Carbon, this is too much Carbon and this makes iron brittle. However, Carbon (coke) does not work as a reducing agent for several metals, as it makes the product (metal) them too brittle after reducing them.

1) Only works for elements that are less reactive than Carbon, thus it doesn’t work for Mg (Magnesium), Na (Sodium) and Al (Aluminium).

2) Carbide remains brittle at very low amounts, even less than 4%. Thus it doesn’t work for Ti and W. It is not commercial in these metals to get the carbon out.

Metal Oxide How Reaction Extras

W (Tungsten) WO3

Tungsten is made by reducing (given electrons) WO3

with Hydrogen Gas.

WO3 + 3H2 → W + 3H2O Hydrogen is explosive;

therefore Water should not be used to put out a fire in which a metal is burning.

Ti (Titanium) TiO2

Titanium is made in a 2 stage process. First it is made into TiCl4, and then the

TiCl4 is then reduced

with a more reactive metal, either Na (Sodium) or Mg (Magnesium) and made into Titanium. An inert (no oxygen) atmosphere is required as Titanium can react with oxygen.

Stage 1:

TiO2 +2C + 2Cl → TiCl4 + 2CO

TiO2 +C + 2Cl2 → TiCl4 + CO2

Stage 2:

TiCl4 + 4Na → Ti + 4NaCl

TiCl4 + 2Mg → Ti + 2MgCl2

TiCl4 is a liquid with a low

boiling point. Therefore it can be distilled (separates TiCl4 from Carbon)

cheaply to purify. Also NaCl and MgCl are

soluble, hence they can be dissolved to get simply Na (Sodium) and Mg

(Magnesium) Therefore as you can see, Na and Mg are extracted from their oxides by reacting them with a more a reactive metal. Pure Titanium is expensive: Why: 1) Cl2 needed 2) Na needed 3) Argon needed Aluminium (Al) Al2O3 Purified Bauxite used in extraction (Very high melting point- this is

dissolved in molten CRYOLITE-lowers melting point). The melting point of mixture is much lower than pure Al2O3 much cheaper

Electrolysis is used to extract Al from Al2O3.

Cathode (negative electrode): Site of REDUCTION

Al3+ _{+ 3e}- _{→ Al}

Anodes: 2O2- _{→ O}

2 + 4e

-1) Electrolysis can only occur when we have ions that can move- that is why electrolysis must be of a molten mixture.

2) The cathode and anode are made of graphite, therefore oxygen formed burns away the electrodes – hence they are replaced frequently

3) The major cost in this process is electricity. However the product is very pure.

Copper (Cu)

Copper present in ground water (any dilute aqueous solution-low grade ore) can be captured by using scrap iron.

Cu2+

(aq)+ Fe(s) → Cu(s) + Fe2+(aq)

Or

CuO + C → Cu + CO

This is a low cost method of extracting Copper as scrap iron is cheap, and there is a low energy requirement.

Every time a metal is given and in the middle of its name there is a bracket with a symbol in it, then that symbol represents the positive charge on that metal ion.

FOR EXAMPLE: If you see Copper (II), it means Cu2+_{. Also the atom economy must be high}

when extracting.

Remember that Aluminium is expensive. Sometimes the cost of extracting a metal is high because of the energy required when heating it with another metal. Using enthalpy, we can find this high amount of energy. Al = Al3+_{, Zn = Zn}2+_{. Also Aluminium is recycled to save}

electricity, energy and conserve bauxite. Also, Titanium is stronger than Aluminium. When metal reacts with water, it makes an Oxide and Hydrogen.

Recycling 1) Preserves resources (Economic)

2) Reduces pollution (Environmental) 3) Saves energy (Economic)

4) Scrap Iron/Steel has higher iron content (Economic) Equilibrium Equilibrium is attained when there is a reversible reaction.

At equilibrium, the rate of formation of products from reactants is exactly balanced by the rate of formation of reactants from products.

Two features of a reaction at equilibrium: 1) Concentrations of reactants and products remain constant: WHY: Because the forward and backward reactions are proceeding at equal rates.

2) Forward and backward reactions proceeding at equal rates

However, according to how we change the conditions, we can alter the position of

equilibrium. However, whatever condition we change, the reaction moves so as to oppose the change.

We can change: 1) Temperature

2) Pressure (for gas reactions) 3) Concentration

EXAMPLE:

N2 + 3H2 2NH3

Pressure: If we increase the pressure, the reaction will attempt to decrease the pressure by converting the reactant (which is 4 moles) into the product (which is 2 moles). Therefore if we increase the pressure, the equilibrium moves to the side of fewest gaseous molecules. However if there are the same number of moles on each side, then the pressure shall have no effect.

Pressure increases Ammonia yield: There are more moles of reactant therefore an increase in pressure is opposed.

-92

N2 + 3H2 2NH3

+92

Temperature: The sign + means that it is endothermic, the sign - means that the reaction is exothermic. Therefore, if we increase the temperature, the reaction will attempt to decrease the temperature (try to cool down) by moving towards the endothermic reaction, therefore making the N2 and 3H2.

Temperature decreases Ammonia yield: The forward reaction is exothermic; therefore if we provide heat, the equilibrium will move to cool down by moving towards the endothermic reaction, therefore less Ammonia is formed.

N2 + 3H2 2NH3

Concentration: If we increase the amount of concentration of the product, the reaction will attempt to decrease the concentration by moving towards the reactants. Therefore, if we want a high yield of ammonia, the temperature must be kept low, and the pressure must be kept high:

Increase pressure

N2 + 3H2 2NH3

Increase temperature

Thus we use a compromise of both:

Temperature (450o_{C) Pressure (200atms)}

If we increased it further Less ammonia in mixture Too expensive equipment and pumping costs, and also risk of

explosion

If we lowered it further Reaction too slow Less ammonia and slow reaction A catalyst increases the rate of attainment of equilibrium and speeds up forward and

backward reactions. But it does not have an effect on the equilibrium amounts. The catalyst in ammonia manufacture is Iron.

Why does the catalyst have no effect on the position of equilibrium:

A catalyst increases the rate of the forward and backward reaction, and this increase in rate of forward and backward reaction is equal.

Equilibrium may also shift to oppose the loss of a substance.

Also, it is of paramount importance to remember that a negative ∆H (enthalpy has

decreased) is always an exothermic reaction. The actual sign - means exothermic reaction. Therefore the overall enthalpy of an exothermic reaction must have a - before it. FOR EXAMPLE: The following is an exothermic reaction:

If the enthalpy has decreased as shown above, it means the sum of the bond enthalpies of the product is lower than the sum of the bond enthalpies of the reactants.

Collision Theory For particles to react chemically, they must: 1) Collide

Only reactions with at least the activation energy can pass from reactants to products. Reacting particles have different activation energies; some high, some low.

Maxwell Boltzmann Distribution:

1) Increase temperature: The curve of the graph becomes flatter and distributes to the right.

2) Add a catalyst: The Ea on the graph moves to the left, however the distribution of the

graph is not affected, and the catalyst is usually hot: WHY: To provide the activation energy. If a catalyst remains hot during a reaction, it is because the reaction is exothermic.

3) If we double the pressure, the volume halves, therefore the concentration also doubles. Increasing Concentration/Pressure: The hump of the graph grows higher, but does not move right or left.

4) What does the area under the graph represent: Total number of molecules. The graph start at the origin because all molecules have some energy, therefore no molecules have no energy. However very few molecules have the Ea.

5) The Y axis is labelled ‘Number Of Molecules’ and the X axis is labelled ‘Energy’.

If we decrease the temperature: There is a decrease in the number of particles which have the minimum energy to react, therefore there are fewer successful collisions.

If we increase the temperature: There is an increase in the number of particles which have the minimum energy to react, therefore there are many more successful collisions. Also the Emp (the place where

the hump on the graph is) increases. This is more effective in increasing the rate than increasing the pressure

because increasing the pressure by a small amount increases the frequency of collisions by a small amount.

If we increase the concentration: The number of particles reacting is doubled (many more), therefore there is a greater collision frequency (higher chance of a successful collision).

If we decrease the concentration: The number of particles reacting is much fewer, therefore there is a lower collision frequency.

If we add a catalyst: There is an increase in the number of particles which have the minimum energy to react, therefore there are many more successful collisions. Also the graph distribution remains the same. Sometimes a metal catalyst is often used in the form of gauze or a powder to increase the surface area of the catalyst.

If we increase the surface area of reactants: There is a greater collision frequency.

Enthalpy

Once condition necessary for enthalpies of formation to be quoted as standard values at a specific temperature of 298K: All reactants and products are in their standard states.

There is a difference between the value calculated through mean bond enthalpies and the value given in a book because:

1) Mean bond enthalpies are not specific for this reaction 2) They are average values from many different compounds

An incomplete combustion can lead to a reaction being less exothermic. Questions about q = m c ∆T

Q = m c ∆T

(You must write this in the exam)

In Standard Enthalpy of reactions: ΔH = Σ ΔHf products − Σ ΔHf reactants

In Mean Bond enthalpy of reaction:

ΔH = Σ bonds broken (reactants) - Σ bonds made (products)

The calculated values of enthalpies of combustion of alcohols, when plotted against Mr

follow a straight line: WHY:

1) Each Alcohol increases by an extra CH2

2) Combustion of each alcohol in the series breaks one more C-C bond and two more C-H compared with the previous one and forms one more mol CO2 and more mol H2O.

Also sometimes the experimental values of enthalpy obtained by a student (through the above equation) are lower than the calculated values (using Hess’s law) due to one of: 1) Heat loss

2) Incomplete combustion

3) Reactants or products may not be in their standard states

Sometimes it is not possible to measure the enthalpy change directly for a combustion reaction as other products are also formed that are not showed in the equation.

Elements have standard enthalpy of formation of zero – by definition, but not when they are not in their standard states.

For example if gaseous Mg is formed, instead of solid Mg, then the enthalpy would be less negative.