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    Anyone doing this exam?

    I'm retaking it which means learning the mechanisms again and all the metal extraction stuff again!! :mad:

    But as Unit 5 is a lot more synoptic so it will help me which is good.
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    Hey; yep, I'm doing it.. Hoping to build on 92/100 in Unit one and 46/50 (so far on ISA).
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    *Refers to the 7th June 2010 exam*

    Papers, Questions, Notes, Doubts..etc

    BASICALLY EVERYTHING on AQA Unit 2 Chemistry- new specification

    this thread contains the latest paper for this unit

    ALL AFTER EXAM DISCUSSIONS CAN ALSO BE POSTED HERE!! :yes:
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    Do you have any doubts or stuff? I can help if you need some assistance!
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    How Is every1 findind Unit 2 chemistry??????????????????????? ????


    specification: http://www.a-levelchemistry.co.uk/AQ...20-W-SP-10.pdf
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    It's hard, freaking hard :eek3:
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    (Original post by CHEM1STRY)
    Do you have any doubts or stuff? I can help if you need some assistance!
    thanx!!!!

    What website do you think is most useful for AQA for notes n papers???
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    Whats Enthalpy all about?

    dont have a clue as we only just started the topic before easter and havent gone through it yet.:confused:
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    (Original post by chanella)
    thanx!!!!

    What website do you think is most useful for AQA for notes n papers???
    I use the AQA endorsed books - they are BRILLIANT for learning stuff!
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    (Original post by Toddy1806)
    It's hard, freaking hard :eek3:
    I know!!!
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    (Original post by CHEM1STRY)
    I use the AQA endorsed books - they are BRILLIANT for learning stuff!

    Ok thanx!!
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    (Original post by Bobby_Friction)
    Whats Enthalpy all about?

    dont have a clue as we only just started the topic before easter and havent gone through it yet.:confused:
    Its about heat change, and also involves lots of calculations..
    This site: http://www.a-levelchemistry.co.uk/ has quite good notes on that
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    (Original post by Bobby_Friction)
    Whats Enthalpy all about?

    dont have a clue as we only just started the topic before easter and havent gone through it yet.:confused:
    ENTHALPY CHANGES

    1. Exothermic and endothermic reactions

    When a chemical reaction takes place, the products and reactants have different stabilities and thus there is a change in potential energy. However since total energy is always conserved, any change in potential energy must be balanced by an equal and opposite change in kinetic energy.

    The lower the chemical potential energy of a given chemical species, the more stable it is. This means that stable species have a lower potential energy than unstable species.

    In some reactions, the products are more stable than the reactants. The products therefore have less potential energy than the reactants, and the potential energy of the reacting species decreases.

    Since the total energy is always conserved, it follows that the kinetic energy of the species must increase. The particles thus move faster and the temperature increases. Reactions in which the products are more stable than the reactants thus involve a transfer of energy from potential to kinetic and an increase in temperature.

    PE  KE

    Such reactions give out heat and are thus said to be EXOTHERMIC.

    In other reactions, the reactants are more stable than the products. The products therefore have more potential energy than the reactants, and the potential energy of the reacting species increases.

    Since the total energy is always conserved, it follows that the kinetic energy of the species must decrease. The temperature of the system thus decreases. Reactions in which the reactants are more stable than the products thus involve a transfer of energy from kinetic to potential and a decrease in temperature.

    KE  PE

    Such reactions absorb heat and are said to be ENDOTHERMIC.


    2. Standard enthalpy changes

    The change in chemical potential energy during a chemical reaction is known as the enthalpy change for that reaction. It is given the symbol H.

    By convention, if the reaction is exothermic (ie heat is given out) the enthalpy change is said to be negative: H = -ve. If the reaction is endothermic (ie heat is absorbed) the enthalpy change is said to be positive: H = +ve.

    The enthalpy change of a reaction depends on the reaction conditions. It is therefore necessary to specify standard conditions for the measurement of enthalpy changes. These are taken to be atmospheric pressure (1 atm) and room temperature (298K). Enthalpy changes measured under standard conditions are known as standard enthalpy changes and are given the symbol Ho. During these chemical changes, the pressure should be kept constant.

    The enthalpy change for a reaction is the heat energy change measured under conditions of constant pressure.

    The standard enthalpy change for a reaction is the heat energy change measured under standard conditions: 100 kPa and a stated temperature (usually 298K).

    The enthalpy change also depends on the amount of substance used. It is therefore necessary to specify the amount of reactants used. Enthalpy changes are conventionally measured in kJmol-1.

    So H = heat energy change/no. of moles

    Given a reaction: A + 3B  2C + 4D
    The standard enthalpy change for this reaction is taken to be the enthalpy change under standard conditions when one mole of A reacts with three moles of B to give two moles of C and four moles of D.

    Eg If 0.2 moles of A react with 0.6 moles of B and 200 kJ of energy are released, the enthalpy change is 200/0.2 = -1200 kJmol-1



    3. Standard enthalpies of formation and combustion

    The term "standard enthalpy of reaction" can be used to describe any chemical reaction, but there are some important reactions which have special names:

    The standard enthalpy of formation of a substance is the enthalpy change when one mole of that substance is formed from the most stable allotropes of its elements in their standard states under standard conditions.
    It is given the symbol Hof.

    Eg C(s) + 2H2(g)  CH4(g), H = -74.8 kJmol-1.
    The standard enthalpy of formation of methane is -74.8 kJmol-1.

    Eg H2(g) + 1/2O2(g)  H2O(l),H = -285.8 kJmol-1
    The standard enthalpy of formation of water is -285.8 kJmol-1.

    Eg N2(g) + 2H2(g) + 3/2O2(g)  NH4NO3(s),H = -365.6 kJmol-1
    The standard enthalpy of formation of ammonium nitrate is -365.6 kJmol-1.

    Eg 1/2N2(g) + 1/2O2(g)  NO(g),H = +82.0 kJmol-1
    The standard enthalpy of formation of nitrogen monoxide is + 82.0 kJmol-1.

    The standard enthalpy of formation of all elements in their standard states is zero.

    The standard enthalpy of combustion of a substance is the enthalpy change when one mole of that substance is burned in an excess of oxygen under standard conditions.
    It is given the symbol Hoc.

    Eg H2(g) + 1/2O2(g)  H2O(l),H = -285.8 kJmol-1
    The standard enthalpy of combustion of hydrogen is -285.8 kJmol-1.

    Eg CH4(g) + 2O2(g)  CO2(g) + 2H2O(l),H = -890.3 kJmol-1
    The standard enthalpy of combustion of methane is -890.3 kJmol-1.

    Eg C2H5OH(l) + 3O2(g)  2CO2(g) + 3H2O(l),H = -1367.3 kJmol-1
    The standard enthalpy of combustion of ethanol is -1367.3 kJmol-1.

    Burning a substance in oxygen is almost always exothermic, so standard enthalpies of combustion almost always have negative values.

    Substances which do not support combustion, like water, carbon dioxide and most other oxides, have zero enthalpy of combustion.

    4. Calculating enthalpy changes from temperature changes

    Enthalpy changes are generally measured by carrying out a reaction under controlled conditions in a laboratory and measuring the temperature change.

    The amount of heat required to change the temperature of a system by 1K is known as the heat capacity of a system (Hc). It is measured in JK-1.

    The heat energy change for a given reaction can therefore be calculated from the equation:
    q = T x Hc.

    The specific heat capacity (c) is the amount of heat required to heat 1g of a substance by 1K. So: heat capacity = specific heat capacity x mass
    Mass = volume x density
    So: heat capacity = specific heat capacity x volume x density

    q = VρcT
    or q = mcT

    If a reaction is taking place in solution (and therefore water is the main species present) it is reasonable to assume that the solution behaves as if it were pure water.
    The density of water is 1.0 gcm-3 and the specific heat capacity of water is 4.18 Jg-1K-1.

    So the calculation is q = (volume of solution) x 4.18 x T. The enthalpy change can then be calculated by dividing the energy change by the number of moles of reactants.

    If the temperature goes up the enthalpy change is negative. If the temperature goes down the enthalpy change is positive.

    MEAN BOND ENTHALPIES

    During a chemical reaction, the bonds in the reactants are broken. This is an endothermic process; energy is required to do this. After the bonds have been broken, however, the bonds in the products are formed. This is an exothermic process; energy is released when this happens.

    The enthalpy change for a chemical reaction can be deduced from consideration of the energy required to break bonds in the reactants and the energy released when the bonds in the products are formed. It can be calculated from the following equation:
    H = Energy required to break bonds in reactants - Energy required to break bonds in products.

    This method can be used to calculate the enthalpy changes for any reaction which does not involve ionic bonds. The breaking and making of ionic bonds involves a more complicated sequence of energetic processes and thus cannot be considered in this way.

    a) Molecules in the gaseous state

    In gas phase reactions, the bonds which are broken and formed are covalent bonds. The energy required to separate completely the atoms in one mole of covalent bonds is known as the bond dissociation enthalpy of that bond (Hb).

    A-B(g)  A(g) + B(g)

    Bond H /kJmol-1
    C-H +413
    O-H +464
    C-C +347
    C=C +612
    C=O +805
    H-F +568
    H-Cl +432
    Cl-Cl +243
    Br-Br +193
    O=O +498

    These bond enthalpies are mean values; the exact strength of a bond depends on its environment. Even the strengths of the same type of bond in the same molecule may vary:
    Eg in water: H-O-H(g)  H-O(g) + H(g); H = +502 kJmol-1
    H-O(g)  H(g) + O(g); H = +427 kJmol-1
    It is necessary therefore to take an average = ie (502 + 427)/2 = +465 kJmol-1.


    Thus bond enthalpies calculated from different reactions may vary slightly. They also do not take intermolecular forces into account. Mean bond enthalpies thus only give you approximate values for enthalpy changes.

    b) Solids and liquids

    In giant covalent substances, all the covalent bonds have to be broken before free gaseous atoms can be formed.
    In either case, the energy required to separate completely the atoms in one mole of the substance is known as the atomisation energy (Hat).
    M(s)  M(g), or AxBy(s)  xA(g) + yB(g)

    Substance Hat/kJmol-1
    C(s)  C(g) +717
    Si(s)  Si(g) +377
    SiO(s)  Si(g) + 2O(g) +1864

    Using mean bond enthalpies to calculate approximate enthalpy changes:

    The approximate enthalpy change can be calculated by considering the mean bond enthalpies of the bonds broken and bonds formed

    Eg Consider the combustion of methane:

    CH4(g) + 2O2(g)  CO2(g) + 2H2O(l)


    Broken: 4 C-H bonds, 2 O=O bonds: total energy = (4 x 413) + (2 x 498)
    = +2648 kJmol-1
    Formed: 2 C=O bonds, 4 O-H bonds: total energy = (2 x 805) + (4 x 464) = +3466 kJmol-1

    So approximate ∆H = 2648 – 3466 = -818 kJmol-1
    This answer is only approximate because it is using mean bond enthalpies, not molecule-specific ones, and because it is ignoring changes in intermolecular forces. The correct value of ∆H for the reaction is -890 kJmol-1.


    Determining mean bond enthalpies from given data:

    If the enthalpy change for a reaction is known, and most of the bond enthalpies are known, it is possible to calculate the mean bond enthalpy of a particular bond:

    Eg The enthalpy of formation of methane is known to be -76 kJmol-1. The bond enthalpy of a H-H bond is +436 kJmol-1, and the enthalpy of atomization of carbon (C(s)  C(g)) is +713 kJmol-1.

    Using this information, the mean bond enthalpy of a C-H bond can be calculated.

    C(s) + 2H2(g)  CH4(g)
    The bonds broken are the bonds in carbon (C(s)  C(g)) and 2 H-H bonds:
    Total energy = 713 + 2(436) = 1585
    The bonds formed are four C-H bonds:
    Total energy = 4x (x = mean C-H bond enthalpy)

    So 1585 – 4x = -76
    So 4x = 1585 + 76 = 1661 kJmol-1
    So x = 415 kJmol-1 = mean bond enthalpy of C-H

    HESS' LAW

    Hess' Law states that "the enthalpy change for a chemical reaction depends only on the initial and final states and is independent of the path followed".

    In other words whichever route, however direct or indirect, by which the reaction proceeds, the overall enthalpy change for the reaction will be same. It is an application of the principle of conservation of energy.

    Hess' Law can be used to calculate many enthalpy changes which cannot be measured directly.

    Many enthalpy changes can be calculated simply by using enthalpies of formation and combustion and applying Hess' Law.

    1. Calculations using enthalpies of formation

    Consider the reaction CH4(g) + Br2(l)  CH3Br(g) + HBr(g)
    One way by which this reaction could proceed would be to convert all the reactants into their constituent elements in their standard states and then to convert the elements into products.

    Since the enthalpy of formation of CH4 is -74.8 kJmol-1 then the enthalpy change when one mole of CH4 is converted into its elements will be +74.8 kJmol-1. Given that the enthalpies of formation of CH3Br and HBr are -37.2 kJmol-1 and -36.4 kJmol-1 respectively, then the enthalpy change for the reaction by this route can be calculated:



    An energetic cycle such as this is known as a Hess' Law cycle.

    H = -(-74.8) + (-37.2) + (-36.4) = +1.2 kJmol-1


    The enthalpy change of any reaction can be calculated if the enthalpies of formation of all the reactants and products are known. The enthalpy of formation of elements in their standard states is always zero.

    The calculation can be condensed into a simple formula:

    ∆H = Σ[∆Hf(products)] – Σ[∆Hf(reactants)]

    2. Calculations using enthalpies of combustion

    All organic compounds burn in air to give carbon dioxide and water.

    Consider the reaction C(s) + 2H2(g)  CH4(g)
    One way by which this reaction could proceed is to convert the reactants to carbon dioxide and water by combustion, and to convert carbon dioxide and water to methane and oxygen (ie the reverse of combustion).

    If the enthalpy of combustion of methane is -890 kJmol-1, then the reverse must be +890 kJmol-1. The enthalpy of combustion of hydrogen needs to be doubled because there are two moles of hydrogen in the equation.

    If all the relevant enthalpies of combustion are known, the enthalpy change of the reaction by this route can be deduced.



    H = -394 + (2 x -286) - (-890) = -76 kJmol-1

    The enthalpy change for a reaction can always be calculated if the enthalpy of combustion of reactants and products are known.

    The calculation can be condensed into a simple formula:

    ∆H = Σ[∆Hc(reactants)] – Σ[∆Hc(products)]
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    (Original post by CHEM1STRY)
    I use the AQA endorsed books - they are BRILLIANT for learning stuff!
    there are a few disastrous mistakes like p171, the second sodium iodide equation
    its a good book but you have to watch out for mistakes like that
    it should be worked through with the specification too as there is a lot of stuff u dont need to know
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    PM me if you want the past papers + ms.

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    (Original post by chanella)
    Its about heat change, and also involves lots of calculations..
    This site: http://www.a-levelchemistry.co.uk/ has quite good notes on that
    Thanks for the link and the notes on the other post. ill be sure to read them when my brains awake
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    (Original post by CHEM1STRY)
    I use the AQA endorsed books - they are BRILLIANT for learning stuff!
    Do you know where to get a few Unit 3( practical) papers??
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    (Original post by LukethePianoMan)
    there are a few disastrous mistakes like p171, the second sodium iodide equation
    its a good book but you have to watch out for mistakes like that
    it should be worked through with the specification too as there is a lot of stuff u dont need to know

    ok!!
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    ny1 done the Jan 2010 Paper?? How was it??
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    (Original post by Bobby_Friction)
    Whats Enthalpy all about?

    dont have a clue as we only just started the topic before easter and havent gone through it yet.:confused:
    Enthalpy change is the heat energy change of a system measured under constant pressure. We give it the symbol \Delta H. If \Delta H is a positive number, the chemical system has gained heat energy, in other words it has taken it in from its surroundings. This is called 'endothermic'. If the \Delta H is negative, heat energy has been given out from the chemical system. This is called 'exothermic'.

    If you have a look at any reaction there will always be some form of energy change going from products to reactants. This is because we need to BREAK the bonds of the reactants and FORM the bonds of the products.



    This diagram shows a reaction. Reaction coordinate means the progress of the reaction, and energy is the energy of the substances. As you can see, the energy of reactants is > energy of products, so heat energy has been GIVEN OUT i.e. it is EXOTHERMIC. \Delta H is negative as a result. Initially energy must be put in to break the bonds. Ea is called the activation energy - imagine it like a hill that molecules have to get over before they can roll down (start reacting). It is the minimum energy required for a reaction to proceed. The top of the curve is called the transition state or activated complex; here some bonds have formed and some have been broken. Just like a ball at the top of a hill it's very unstable, and so the energy is given off.



    This is an endothermic reaction. The molecules take in energy in this case, and the energy of products is higher than reactants. The \Delta H value is positive because the net energy change is such that the system has taken in energy.

    I hope this has given you a brief introduction to energetics, you will learn more as you go on.
 
 
 
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