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    Equilibria - Importance of equilibria in industrial processes - Topic 3.

    - be able to apply these concepts to given chemical processes
    - be able to predict qualitatively the effect of temperature on the position of equilibrium from the sign of DH for the forward reaction
    -understand why a compromise temperature and pressure may be used
    -know about the hydration of ethene to form ethanol and the reaction of carbon monoxide with hydrogen to form methanol as important industrial examples where these principles can be applied
    - know the importance of these alcohols as liquid fuels
    Application of Le Chatelier's principle to industrial production of ethanol.

    H2C=CH2(g) + H2O(g) -----------------------> CH3CH2OH(g) DH= - 42KJmol^-1


    - Le Chatelier's principle says high yield can be attained if High pressure and Low temperature.

    - In practice, 300C is a compromise temperature to give increased rate of reaction with Phosphoric acid catalyst allowing T to be no higher.

    - In practice, high pressure (6.5MPa) is used but no higher than that bc of costs and problems with polymerisation.

    - Produces 5% yield which rises to 95% with recycling.


    ~


    Application of Le Chatelier's principle to industrial production of Methanol.

    CO(g) + 3H2(g) ---------------------------------> CH3OH(g) DH= -91KJmol^-1


    - It's basically the same arguments as those with ethanol EXCEPT this time it's a compromise temperature of 230C and a compromise pressure of 10Mpa which produces 5-10% yield.
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    (Original post by metrize)
    New spec or old spec?
    Old.
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    Redox reactions - Oxidation and reduction - Topic 4.

    -know that oxidation is the process of electron loss
    -know that oxidising agents are electron acceptors
    -know that reduction is the process of electron gain
    - know that reducing agents are electron donors
    Oxidation
    Is
    Loss of electrons.
    Reduction
    Is
    Gain of electrons.

    Therefore, oxidising agents are electron acceptors and reducing agents are electron donors
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    Redox reactions - oxidation states - Topic 4.

    -know and be able to apply the rules for assigning oxidation states in order to work out the oxidation state of an element in a compound from its formula
    -understand oxidation and reduction reactions of s and p block elements
    The rules are basically:

    - All elements have an oxidation state of 0.
    - Add up to 0 in a neutral compound.
    - Add up to charge on an ion.

    And the elements that you need to know are:

    F = -1
    O = -2 (except in peroxides - H2O2 and fluorine)
    Cl, Br, I = -1 (except in compounds with oxygen/fluorine)
    H = +1 ( except in Metal hydrides- NaH)

    THEN

    Group 1 all = +1
    Group 2 all = +2
    Group 3 all = +3
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    Redox reactions - redox equations - Topic 4

    -be able to write half-equations identifying the oxidation and reduction processes in redox reactions when the reactants andproductsarespecified
    -be able to combine half-equations to give an overall redox equation
    When you're constructing half equations in acidic conditions, you need to remember the following steps:

    - Balance reduced or oxidised atoms.
    - Balance O with H2O
    - Balance H with H+
    - Balance the charge with e-

    When you're combining half equations to produce a full redox equation, you need to remember the following:

    - Make sure electrons will cancel.
    - Simplify the overall equation if possible.. OTHERWISE YOU WILL LOSE MARKS.


    When an element is oxidised, it's oxidation number is increased and when an element is reduced it's oxidation number is decreased.


    REDUCED IS DECREASED.

    Then you can just remember that oxidised is the opposite.
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    Group 7(The Halogens) - Trends in physical properties + oxidising/reducing abilities - Topic 5.

    ..I cba quoting the spec anymore.


    Trend in electronegativity as you go down the group.

    DECREASES.

    The reason is because the number of shells increase therefore there is an increasing distance and shielding from the nuclear attraction.


    Trend in boiling point as you go down the group.

    INCREASES.

    The reason is because, as you go down the group the number of shells increase which means the molecules become larger and as we know from unit 1, the larger the molecule the greater the strength of Van der Waalsforces therefore more energy is required to break these intermolecular forces hence a high boiling point.


    Trend in oxidising power as you go down the group.

    DECREASES

    The reason is bc the ability to gain electrons decreases as there are more shells and therefore an increasing distance and shielding from the attraction of the nucleus.


    Trend in reducing ability as you go down the group.

    INCREASES.

    The reason is bc the ability to lose electrons increases as there are more shells and therefore an increasing distance and shielding from the attraction of the nucleus.


    ~

    So overalll:

    Electronegativity, boiling point, oxidising power, reducing ability = decreases, increases, decreases, increases.
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    Group 7(The Halogens) - Identification of halide ions using silver nitrates - Topic 5.

    Reactions of solid NaCl with concentrated H2SO4

    NaCl + H2SO4 ----------------------------------------------> HCL + NaHSO4

    - Produces steamy fumes because of the HCL, white fumes with Ammonia.


    ~

    Reactions of solid NaBr with concentrated H2SO4

    NaBr + H2SO4 ----------------------------------------------> HBr + NaHSO4

    - Produces steamy fumes + white fumes with Ammonia agaaain.

    - Sulphuric acid oxidises Br- to Br2 and it itself is reduces to SO2.

    So the equations we need to know are:

    H2SO4 + 2H+ + 2e- -----------------------------> SO2 + 2H2O

    2Br- ----------------------------------> Br2 + 2e-

    If they ask for the full redox equation for this reaction, then remember it just means you add them together.

    - SO2 is a choking, colourless gas that reduces orange dichromate paper to green Cr3+ ions.

    - Br2 is an orange/brown liquid/gas.


    ~

    Reactions of solid NaI with concentrated H2SO4

    Okay so for this it's the same as above but then also:

    Sulphuric acid is reduced as follows:

    H2SO4 + 6H+ + 6e- -----------------------> S + 4H2O

    H2SO4 + 8H+ + 8e- -------------------------> H2S + 4H2O


    I- ions are oxidised to I2.

    So:

    2I- ------------------> I2 + 2e-

    - S is a yellow solid.

    - H2S is a 'bad eggs' gas.

    - Iodine is seen as a black solid, and upon heating it gives a purple vapour.
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    Group 7(The Halogens) - Identification of halide ions using silver nitrates - Topic 5

    Continuing with this same sub- topic( New post bc the other one was getting a bit long and it would've got kinda messy)

    Okay so the test for Halides is addition of HNO3(aq) followed by AgNO3(aq) + then I'll also do the trends in solubility with NH3..


    So let's start with Fluoride.

    Fluoride is basically pointless but I'm going to write it down anyway.

    Upon adding HNO3(aq) followed by AgNO3(aq). = NVR

    Upon adding dilute NH3(aq) = NVR

    Upon adding conc NH3(aq) = NVR

    So overall, you just need to remember that fluorine produces NVR for all 3.

    ~


    Okay, now let's do Chloride.

    Upon adding HNO3(aq) followed by AgNO3(aq). = White precipitate

    Upon adding dilute NH3(aq) = Precipitate dissolves

    Upon adding conc NH3(aq) = Precipitate dissolves

    So overall, Chloride is a white ppt which dissolves in both dilute AND concentrated ammonia solution.

    ~


    Okay, for Bromide.

    Upon adding HNO3(aq) followed by AgNO3(aq). = Cream precipitate

    Upon adding dilute NH3(aq) = Precipitate does not dissolve

    Upon adding conc NH3(aq) = Precipitate dissolves

    So overall, Bromide is a cream ppt which dissolves ONLY in concentrated ammonia solution.

    ~




    Lastly Iodide.

    Upon adding HNO3(aq) followed by AgNO3(aq). = Pale yellow precipitate

    Upon adding dilute NH3(aq) = Precipitate does not dissolve

    Upon adding conc NH3(aq) = Precipitate does not dissolve

    So overall, Iodide is a pale yellow ppt which DOES NOT DISSOLVE in either dilute or concentrated ammonia solution.
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    Okay, so not sure where this goes under but it's only a little bit more info + pretty much in the same bracket as all this other stuff so I'm just going to add it here:

    Colours of halogen solutions:

    Chlorine: Pale green solution.
    Bromine: Yellow/brown solution.
    Iodine: Brown solution + black ppt.

    Reactions of halogens with halide ions in aqueous solutions:

    Cl2(aq) + 2Br-(aq) ----------------> 2Cl- (aq) + Br2(aq)

    Cl2(aq) + 2I-(aq) ----------------> 2Cl- (aq) + I2(aq)

    Br2(aq) + 2I-(aq) ----------------> 2Br- (aq) + I2(aq)
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    Group 7(The Halogens) - Uses of chlorine and chlorate - Topic 5

    Chlorine with water:

    Cl2 + H2O ----------------------------------------------> HCl + HClO

    Okay so it's used for water treatment + the main point is that it kills bacteria; prevents disease. The health benefits of adding chlorine to water outweigh it's toxic effects.

    ~

    Chlorine in sunlight:

    2Cl2 + 2H2O ----------------------------------------------> 4HCl + O2


    ~


    Chlorine with cold, dilute NaOH:

    2NaOH + Cl2 -----------------------------------------> NaCl + NaClO + H2O


    So NaClO is used as a bleach+ therefore kills bacteria.
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    Group 2(Alkaline Earth Metals) - Trends in physical properties- Topic 6

    Trend for atomic radius as you go down the group.(Mg - Ba)

    INCREASES.

    The reason is bc there are more electrons in energy levels which are further from the nucleus and shielded from the increasing attraction of the nucleus.

    ~


    Trend for first ionisation energy as you go down the group.(Mg - Ba)

    DECREASES.

    The reason is bc the outer electron is more shielded from the attraction of the nucleus and is further away.

    ~


    Trend for melting points as you go down the group.(Mg - Ba)

    DECREASES.

    The reason is bc the metallic bond strength decreases because ionic radii increase, decreasing the attraction between cations + delocalised electrons.

    ~

    So overall:

    Atomic radius, 1st Ionisation energy, Melting points = Increases, decreases, decreases
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    Group 2(Alkaline Earth Metals) - Trends in chemical properties- Topic 6

    Reactions of Group II elements with cold water:

    - Mg reacts very slowly with cold water, even if powdered.

    - Ca produces effervescence, metal dissolves away + there is a slow development of a white precipitate.

    - Sr and Ba react in the same way as Ca, but with increasing vigour.

    The general equation for these reactions which you should be able to apply to any group II element is as follows:

    M + 2H2O ----------------------------> M(OH)2 + H2


    ~

    Reaction of Mg with steam:

    - Magnesium must be hot.
    - Glows with a bright white light.
    - White solid is formed + also a flammable gas.

    The equation for Mg with steam is:

    Mg(s) + H2O(g) ----------------------------> MgO(s)+ H2(g)
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    Group 2(Alkaline Earth Metals) - Trends in chemical properties- Topic 6

    Solubility trends of the Group II hydroxides and uses:

    - Hydroxides increase in solubility as you go down the group.

    - Mg(OH)2 is sparingly soluble and is used as an antacid - Milk of Magnesia.

    - Ca(OH)2 is used in agriculture to raise the PH of soil.

    ~


    Solubility trends of the Group II sulfates and uses:

    - Sulfates decrease in solubility as you go down the group.

    - BaSO4 is v insoluble and used as a contrast agent for x- raying the stomach or intestinal tract. (Given to patients as Barium meal)
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    Group 2(Alkaline Earth Metals) - Trends in chemical properties- Topic 6

    Test for sulphate ions.

    - Add dilute HCl(aq)

    (This is to remove any CO32- present + prevent precipitation of BaCO3.)

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


    - Add solution of BaCl2
    - If sulphate is present then a white precipitate is formed.

    Ba2+(aq) + SO42- (aq) ------> BaSO4(s)
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    Extraction of metals - Principles of metal extraction- Topic 7

    So, the main starting points for this are that:

    Metals are usually found as oxides or sulfide ores.

    Sulfide ores are usually converted to oxides by roasting in air.

    The SO2 produced after the roasting can cause acid rain if it escapes, but can also be used in the production of sulphuric acid.


    ~

    All extraction methods involve reduction.

    The choice of reduction method depends on several factors:

    - Cost of reductant
    - Energy requirements
    - Required purity of metal
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    Extraction of metals - Principles of metal extraction- Topic 7

    Extraction using carbon or carbon monoxide.

    Carbon + carbon monoxide are cheap + effective reducing agents used in the extraction of 3 ELEMENTS.

    Those 3 are IRON, MANGANESE + COPPER.

    I-M-C.

    Carbon monoxide is used for Iron and then Carbon is used for Manganese + Copper.

    Okay so the examples are as follows:

    ~

    IRON

    Fe2O3 + 3CO -----------------------------------> 2Fe + 3CO2

    Conditions: Blast furnace, high temperature + it is a continuous process.


    ~

    MANGANESE + COPPER.


    MnO2 + 2C --------------------> Mn + 2CO

    CuO + C -----------------------> Cu + CO


    Conditions: Both at high temperature.
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    Extraction of metals - Principles of metal extraction- Topic 7

    However there is a general limitation of C reduction.

    Therefore it cannot be used for Titanium, Tungsten + Aluminium.

    The reason why is bc for Ti and W, the carbide is formed e.g.


    TiO2 + 4C -----------------------------------> TiC + 2CO

    WO3 + 2C -----------------------------------> WC + 3CO


    For Al, Carbon is simply not reactive enough for reduction.
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    Extraction of metals - Principles of metal extraction- Topic 7

    Right, so since we know that Carbon cannot be used for extraction we've got to use other methods.

    Firstly we'll start with Aluminium.

    Extraction of Al from purified bauxite.

    Electrolysis of molten Al2O3 is dissolved in cryolite at high temperatures.

    The electrode equations are as follows:

    Al3+ + 3e- -------------> Al

    2O2- ---------------> O2 + 4e-

    The material of the cathode + anode is graphite but the anode needs replacing frequently bc..

    C + O2 ---------------------> CO2

    It is a continuous process + the energy costs are high.
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    Extraction of metals - Principles of metal extraction- Topic 7

    Alright, so now we can move onto Titanium.

    Extraction of Ti from TiO2.

    TiO2 + 2Cl2 + 2C --------------------------> TiCl4 + 2CO (high temperature required)

    TiCl4 is then purified by distillation.

    Either Na or Mg can be used as a reducing agent in this reaction, here are both of the equations:

    TiCl4 + 4Na --------------> Ti + 4NaCl

    OR

    TiCl4 + 2Mg --------------> Ti + 2MgCl2

    It is a batch process + the costs are high hence the limited use despite unique properties + high natural abundance.
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    Extraction of metals - Principles of metal extraction- Topic 7

    Okay, finally it's Tungsten.

    Extraction of W from WO3

    WO3 + 3H2 ------------> W + 3H2O

    Conditions: High temperature

    Risks: H2 is an explosive gas.
 
 
 
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