AQA A level Chemistry Unit 2 Notes

Group 2 Elements Trend Atomic Radius Ionisation Energy

Melting Points

Reactivity With Water

Reasons

MgBa (Decreases as you go down group) Mg>CaBa (Decreases as you go down group but Sr has a higher melting point than Ca)

1) More Energy Levels 2) More Shielding Electrons 1) Electron further from Nucleus 2) More Shielding Electrons 3) Whilst effective Nuclear Charge stays the same 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. 1) Ions get bigger, therefore there is less force of an attraction 2) The outer-electron therefore is more easily lost.

MgBaSO4 Decreases as you go down group. They are all ions and therefore aqueous.

Group Hydroxides (made by reacting with water) Mg(OH)2I2 (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

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

Cl-Br2>I2 (Decreases as you go down the group)

Strength of bond with C

Reasons

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

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. Cl-

Br-

I-

Halide Ions 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+ + CO32- → CO2 + H2O Add Silver Nitrate (Forms The Precipitates) Add Dilute Ammonia (Dissolves AgCl(s)) Add Concentrated Ammonia (Dissolves AgBr(s))

Colourless Solution

Colourless Solution

Colourless Solution

Colourless Solution

Colourless Solution

Colourless Solution

White Precipitate Ag+(aq) + Cl-(aq) → AgCl(s)

Cream Precipitate Ag+(aq) + Br-(aq) → AgBr(s)

Yellow Precipitate Ag+(aq) + I-(aq) → AgI(s)

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

Cream Precipitate

Yellow Precipitate

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

Misty Acidic Fumes (HCl)

Cl- can’t reduce SO42-

NaCl (bleach) (kills bacteria )

NaCl + H2SO4 → HCl + NaHSO4 (Not redox) The role of H2SO4 is simply an acid Misty Acidic Fumes (HBr)

NOT redox reaction, it is a DISPLACEMENT REACTION

Orange/Brown liquid fumes (Br2)

H2SO4 + 2H+ + 2Br- → Br2 + SO2 + 2H2O

NaBr Acidic fumes (SO2) Misty Acidic Fumes (HI)

NaI

Purple/Black solid fumes (I2)

NOT redox reaction, it is a DISPLACEMENT REACTION SO42- + 4H+ + 2I- → Br2 + SO2 + 2H2O

Unpleasant Smell (H2S - Hydrogen Sulphide)

10H+ + 8I- + SO42- → H2S + 4I2 + 4H2O

Yellow Precipitate/Solid (S8 –Sulphur) = insoluble

Reduction product: H2S Oxidation product: 4I2 6I- + SO42- + 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. H=light)

More absorption (higher wave number) of an infrared wave makes that chemical a more effective greenhouse gas. Bonding Type Frequency Bonds to Hydrogen

2500-3500

Triple Bond

2220-2260

Double Bond

1620-1680

Single Bond

750-1300

Bonds to Hydrogen 2500-3500

Triple Bond 2220-2260

Double Bond 1620-1680

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

Single Bond 750-1300

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 1500oC); 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

W (Tungsten)

Oxide

WO3

How Tungsten is made by reducing (given electrons) WO3 with Hydrogen Gas.

Reaction WO3 + 3H2 → W + 3H2O

Extras

Hydrogen is explosive; therefore Water should be used to put out a fir which a metal is burnin

Ti (Titanium)

Aluminium (Al)

TiO2

Al2O3

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:

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.

TiO2 +2C + 2Cl → TiCl4 + 2CO TiO2 +C + 2Cl2 → TiCl4 + CO2 Stage 2: TiCl4 + 4Na → Ti + 4NaCl TiCl4 + 2Mg → Ti + 2MgCl2

Cathode (negative electrode): Site of REDUCTION Al3+ + 3e- → Al Anodes: 2O2- → O2 + 4e-

TiCl4 is a liquid with a l boiling point. Therefore can be distilled (separa TiCl4 from Carbon) cheaply to purify. Also NaCl and MgCl are soluble, hence they ca dissolved to get simply (Sodium) and Mg (Magnesium) Therefor you can see, Na and M are extracted from thei oxides by reacting them with a more a reactive metal. Pure Titanium is expensive: Why: 1) Cl2 needed 2) Na needed 3) Argon needed 1) Electrolysis can only occur when we have io that can move- that is w electrolysis must be of molten mixture. 2) The cathode and an are made of graphite, therefore oxygen forme burns away the electro – hence they are repla frequently 3) The major cost in th process is electricity. However the product is very pure. This is a low cost meth of extracting Copper as scrap iron is cheap, an there is a low energy requirement.

Copper present in Cu2+(aq) + Fe(s) → Cu(s) + Fe2+(aq) Copper (Cu) ground water (any dilute aqueous Or solution-low grade ore) can be captured CuO + C → Cu + CO by using scrap iron. 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 = Zn2+. 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. 1) 2) 3) 4)

Recycling Preserves resources (Economic) Reduces pollution (Environmental) Saves energy (Economic) 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:

If we increased it further

Temperature (450oC)

Pressure (200atms)

Less ammonia in mixture

Too expensive equipment and pumping costs, and also risk of

If we lowered it further

Reaction too slow

explosion 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 2) With at least the activation energy

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.