C1.2  PHYSICAL CHEMISTRY.

                                                                                             C1.2.1

Q.18  Describe enthalpy changes associated with changes of state.

      Describe in qualitative terms the changes associated with the change of state.

        (cf.  Section C2. 1. I 0)

      Osmosis is NOT required.

Ans   In all three states particles have energy and are moving.  Gases have the highest enthalpy – the particles move fast enough to almost completely overcome the forces between them.  On cooling a liquid is formed and the gas particles attract each other more strongly, and have restricted movement.  Eventually when they become a solid, with the lowest enthalpy, the particles attract each other so strongly there is no translocational movement at all.

C 1. 2.2

The gaseous state, and pV=nRT.

Determination of relative molecular mass for gases and volatile liquids.

Q.19  Outline the basic assumptions of the ideal gas model. 

Ans   Assumptions of an ideal gas model:

·           the molecules are in constant random motion due to collisions between them.

·           all collisions, whether between molecules or between molecules and the walls, are perfectly elastic and therefore there is no loss of kinetic energy.

·           kinetic energy is directly proportional to the absolute temperature of the gas.

·           the molecules exert no attractive or repulsive forces on one another or the walls of the vessel.

·         the size of the molecules is negligibly small compared with the volume occupied by the gas.

Gases show behaviour closest to ideal behaviour at low pressure and high temperatures.

Q.20  Explain the symbols used in the ideal gas equation.

Ans   PV = nRT                IDEAL GAS EQUATION

            where:      P = pressure ( Pa )

            V = volume  ( m3 )

            n = number of moles

            R = gas constant ( 8.31 JK-1 mol-1  )

            T = temperature ( K )

      1 atmosphere pressure = 101325 Pa = 101.325 kPa = 760 mmHg

NB: R can also be given the value 0.082 atmdm3K-1 mol-1where: P = pressure ( Pa )

     V = volume  ( m3 )

      Practice simple calculations using the ideal gas equation.

      Derivation of the equation and deviations from ideality are NOT required.

Q.21  Do the following questions in ‘Advanced Chemistry calculations’ by Alec Thompson

Study worked example P.91 ( Advanced Chemistry Calculations / Alec Thompson )

Exercise 5.22, 5.23 + 5.24  P.92 ( Advanced Chemistry Calculations )

                       5.29  P.93 ( Advanced Chemistry Calculations )

                       5.30  P.93 ( Advanced Chemistry Calculations )

            5.31 + 5.32  P.93 ( Advanced Chemistry Calcuations ).

C1.2.3

Q.22  Explain the term ‘Conservation of energy’.

Ans   In a chemical reaction, the total energy is unchanged.  Energy may be exchanged between the materials and the surroundings but the total energy of the materials and the surroundings remain constant.  This important concept is known as the Law of Conservation of Energy and also as the First Law of Thermodynamics.

Q.23  Define Hess' law. 

Ans   Hess’s Law is simply an application of the more fundamental law of conservation of energy and is often called the Law of Constant Heat Summation.

      It says that the energy change in converting reactants, A+B, to products, X+Y, is the same, regardless of the route by which the chemical change occurs, provided the initial and final conditions are the same.

Q.24  Define standard molar enthalpy changes of –

(a)   Formation

Ans   The standard molar enthalpy change of formation of a substance is the heat evolved or absorbed when one mole of the substance is formed from its elements under standard conditions.

(b)            Combustion

Ans   The standard molar enthalpy of combustion of a substance is the amount of heat energy evolved or absorbed when one mole of the substance is fully burnt in oxygen under standard conditions.

(c)   Atomisation

Ans   The standard molar enthalpy of atomisation of a substance is the amount of heat evolved / absorbed when one mole of gaseous atoms is formed form the element under standard conditions.

(d)   Solution

Ans   The standard molar enthalpy of atomisation of a substance is the amount of heat energy evolved / absorbed when one mole of solute is dissolved in a solvent under standard conditions to form an infinitely dilute solution.

Q.25  Describe simple experiments which would determine enthalpy changes, including the bomb calorimeter.

Ans   Refer to ILPAC experiment A9

      & to the text for bomb calorimeter

      Essentially, the principle of bomb calorimetry and similar experiments to determine enthalpy changes, is to carry out the reaction in an energetically sealed environment and by considering the energy change of the surroundings, calculate the energy change taking place within the reaction chamber.  Eg.

      If the reaction takes place in solution, heat is taken from/given to the solvent during the course of the reaction.  If the temperature and hence energy change within the solvent is measured the enthalpy change can be determined.

      Often the energy changes are again simulated by placing a wire into the reaction chamber and by altering the electricity running through it matching the energy changes detected in the surrounding.  Since the electrical energy supplied by the wire can be calculated, and accurate estimate can be made of the enthalpy changes.  

Q.26  Calculations of molar enthalpy changes.

Ans   Do the Exercises 4.4, 4.5 & 4.6 in ‘Advanced Chemistry Calculations’ by Alec Thompson

C 1. 2.4

Q.27  Explain your understanding of  the terms ‘bond energy’ and ‘average bond enthalpy’.

Ans   The bond enthalpy is the energy required to break one mole of specified bonds in a gaseous molecule to form gaseous atoms.

      Average bond enthalpies are so called because they are an average of the energy for the particular bond as it appears in various compounds.  As a result, enthalpy changes determined by this method can only provide a close approximation.  Those enthalpies determined using a bomb calorimeter and Hess’ Law of constant Heat Summation are accurate.

Q 28        Describe how average bond enthalpies can be used  to predict enthalpy changes.

Ans   In chemical reactions bonds are broken in the reactants and new bonds are formed in the products.  Breaking bonds requires an energy input, whilst forming bonds releases energy.  The net energy change is given by the difference in these.

             Do the questions on pages 56-63 in ‘Advanced Chemistry calculations’ by Alec Thompson

C1.3  INORGANIC CHEMISTRY

C1.3.1

Q.29  Outline generally the properties of the main group elements of the Periodic Table.

Ans   GROUP I

Known as the alkali metals. They are all soft, white, lustrous but rapidly tarnish in air.  Good conductors of heat and electricity. Elements are too reactive to find uncombined and are normally extracted from their compounds by electrolysis.  Good reducing agents.  Have low electronegativity values ie. very electropositive therefore tend to lose their electrons relatively easily. High in activity (electrochemical) series. Usually stored under oil due to reactivity. React readily to form +ve ions with a 1+ charge All react with cold water to produce a metal hydroxide and hydrogen (reducing water to hydrogen). Have lower densities, m.p and b.p than normally associated with metals.

Have a giant metallic structure. Show only one oxidation number in their compounds: +1

Form stable involatile ionic compounds.

GROUP II.

Known as alkaline earth metals. Also good reducing agents. Low electronegativity values but higher than group 1 metal in the same period, due to increase in nuclear charge.

Less reactive than those in group 1

Show an oxidation number of +2 in their compounds. Form + ve ions with a 2+ charge.

GROUP VII.

Known as halogens (salt formers) because they combine readily with metals to form salts.

Most reactive group of non-metals. Strong oxidising agents.

All halogens exist as diatomic molecules. These molecules persist in gaseous, liquid and solid states. All coloured and depth of colour increases with increasing atomic number.

Have one electron less than the noble gas which follows them in the periodic table, therefore chemistry dominated by a tendency to gain a completely filled outermost electron shell.

React with metals to form ionic compounds eg. NaCl. With non-metals they tend to form simple molecular compounds. In these the halogen is linked by a single covalent bond (Hal ) to the other element.

GROUP O

Known as noble gases. Monatomic. Exist as single atoms in the gas phase at room temp. These symmetrical non-polar atoms have no permanent dipole and don't form any normal bonds. Will all condense to liquids and ultimately form solids if the temperature is low enough, suggesting the existence of intermolecular forces.  (INSTANTANEOUS DIPOLE – INDUCED DIPOLE)

These hold atoms together in the solid and liquid state.

Name implies, reluctance to form compounds. Have full outer shell and thus stable and very unreactive. Used for lighting, glow different colours.

C1.3.2

Q.30  Identifying and explaining trends, describe how the elements H to Ar vary with relation to the physical properties listed below,

        (a) melting points,

        (b) covalent and ionic radii,

        (c) standard molar first ionisation energies. 

Ans

C1.3.3

Q.31  Identify trends within Period 3 and describe how the elements Na to Ar vary with relation to the chemical properties listed below –

(a)            oxidation numbers (valencies)

Ans   The metals sodium, magnesium and aluminium have oxidation numbers of +1, +2 & +3 respectively, because the number of electrons in their outer shell matches the group number.

      The non-metals tend to have an oxidation number which equals that of the number of their group subtracted from eight, although they have a range of oxidation states both positive and negative.  In their covalent compounds, oxidation state is both positive and negative.  In ionic compounds, the oxidation state is always negative.

(b)            Bonding

Ans   The elements of groups I, II & III are metals and have therefore, giant metallic structures.  Silicon is giant covalent in structure and the structure of phosphorus to argon is simple covalent bonding.  Group VII’s element exist as diatomic molecules between which are weak Van der Waals forces.

(c)   Reactions with water.

Ans   Sodium reacts with water to produce its hydroxide and hydrogen:

Na   +   H2O   "   NaOH   +1/2H2

      It is a very vigorous reaction in which sodium floats on the surface and a yellow flame is produced.

      Magnesium reacts with steam to produce its oxide and hydrogen:

Mg   +   H2O   "   MgO   +   H2

      Aluminium does not react with water due to its oxide layer which forms rapidly from contact with atmospheric oxygen.

      Silicon, phosphorus and sulphur do not react with water.

      Chlorine dissolves slightly in water and also reacts very slightly to produce a mixture of acids (hypochlorous and hydrochloric):

Cl2   +   H2O   "   HCl   +   HClO

      Argon, as a noble gas with a full outer shell, does not react with water.

Q.32  Describe the acid/base characteristics of the oxides of Period 3.

Ans   Sodium and magnesium form alkaline oxides.

      Aluminium oxide is amphoteric.

      Silicon, phosphorus, sulphur and chlorine oxides are acidic.

C 1. 3.4

Q.33  Describe the trends in compounds of the elements Na to Ar with regard to -

(a)   how the chlorides are prepared.

Ans   All chlorides of the elements Na to P can be prepared by direct synthesis.  Sulphur chloride cannot.

      Na   +   1/2Cl2   "   NaCl

      Mg   +   Cl2   "   MgCl2

      2Al   +   3Cl2   "   2AlCl3

      Si   +   2Cl2   "   SiCl4

      P   +   11/2 Cl2   "   PCl3

      S2Cl2   +   Cl2   D   2SCl4

(b)   how the oxides are prepared

Ans   Oxides of sodium to phosphorus can be prepared by direct synthesis (except sulphur to chlorine)

      2Na   +   1/2O2   "   Na2O

      Mg   +    1/2O2   "   MgO  (also from decomposition of carbonate)

      4Al   +   3O2   "   2Al2O3

      Si   +   O2   "   SiO2

      4P   +   5O2   "   P4O10

SO2   +   O2   D    2SO3

      2HClO4   "   Cl2O7   +   H2O

      2Cl2   +   2HgO   "   Cl2O   +   HgO   +   HgCl2

 (c)  the structures and physical properties of the chlorides and oxides.

Ans   CHLORIDES

      NaCl

      AlCl3

      Al2Cl6

      SiCl4

      PCl3                      Liquid, covalent molecules

      SCl2

      Cl2                Gas, covalent molecule

      OXIDES

      Na2O               Basic

      MgO       Ionic

      Al2O3                   Amphoteric

      SiO2               Giant

      P4O10      Acidic   Simple            Covalent

         SO3                Simple

      Cl2O7              Simple

 (d)  how the chlorides and oxides react with water (including full and ionic equations properly balanced).

Ans   Reaction of chlorides with water

1.  Na        chlorides dissolve in H2O to form an ionic

2.  Mg        neutral solution

3.  [Al(H2O)6](aq    "   [Al(H2O)5OH](aq)    +   H+(aq)

4.  SiCl4(l)   +   2H2O(l)   "    SiO2(s)    +   4H+(aq)    +    4Cl-(aq)

5.  PCl3(l)   +   3H2O(l)   "   H3PO3(aq)   +   3H+(aq)   +   3Cl- (aq)

6.  2S2Cl2(l)   +   2H2O(l)   "  3S(s)   +   SO2(aq)   +   4H+(aq)   +   4Cl(aq)

      Reaction of oxides with water

      Na & Mg form strong alkaline solutions

      1. Na2O(s)   +   H2O(l)   "   2NaOH(aq)

      2. MgO   +   H2O   "   Mg(OH)2(aq)    sparingly soluble

      3.  Al2O3  - insoluble in H2O i.e. no reaction

      4.  SiO2  -  insoluble i.e. no reaction

        P & S  oxides both dissolve to form acidic solution

      5. P4O10   +   6H2O   "   4H3PO4   (phosphoric acid)

      6. SO2   +   H2O   "   H2SO3   (sulphurus acid)

        or

        SO3   +   H2O   "   H2SO4   (sulphuric acid)

      7. Cl2O   +   H2O   "   2HClO(aq)     (hydroclorous acid)

        or

        Cl2O7   +   H2O   "    2HClO4(aq)     (hydrochloric acid)

(e)   how sodium hydride (NaH) reacts with water.

Ans   Sodium Hydride.  Na H.

It's an ionic compound which decomposes at 800°C

It's a white crystalline solid.

PREPARATION  : direct synthesis

             Na(s)   +   1/2 H2(g) =  NaH (s)

REACTION WITH WATER :  soluble and produces a very alkaline solution.

                                                        pH = 12         ( strong base )

                   Na H + H2O    =    NaOH + H2

There's a quite vigorous reaction, with a hissing sound and effervescence observed.

Decomposes below melting point. Unstable to moisture in air.

(f)   how hydrogen chloride (HCl) reacts with water.

      Note – peroxides and superoxides are not required.

Ans   Hydrogen Chloride   HCl.

Simple molecular covalent compound,  HCl (g) exists as discrete molecules.

PREPARATION:   direct synthesis.

                                  H2(g)    +     Cl2(g)   =  2HCl (g)

                   or

                                  NaCl (s)    +   H2SO4(l)   =     NaHSO4(aq)    +    HCl(s)

REACTION WITH WATER:    soluble and very acidic pH = 2

when added to water HCl(g) dissolves into ions.

                                           HCl (g)   +   H2O(l)     =      H3O+(aq)  +    Cl-(aq)

                                                                                   hydrochloric acid.

                                          Hydroxonium ion

C 1.3.5

Q.34  Within Groups 1, 2, 7 & ) identify trends in

(a)   Atomic radii

Ans   The atomic radius increases as you descend the group because there are more electrons which are going into an additional energy level/shell.

(b)   Electron affinities

Ans   The electron affinity decreases as you descend the group.  Electron affinity is a measure of the desire and ability of the atoms of the element to gain electrons.  It is to do with how well the electron will be held in the ion that is subsequently formed.  For example, fluorine has the highest electron affinity because having formed the F- ion the additional electron is close to the attractive, positive charge of the nucleus and has only one shell between it and the nucleus to act as a shield.  Whereas Br has three shells between it and the nucleus to act as shields and more importantly, has the additional electron further from the attractive positive charge of the nucleus.

 (c)  Bonding

Ans   Two general observations can be made.

1.       Metallic character increases as you descend the group. 

                  E.g. Group IV – first two are non-metals & last two are metals

2.       The strengths of the bonds increase down the group in respect to Groups VII & VIII because of the increasing strength of the intermolecular forces between them.  However, the strength of metallic bonds in Groups I & II decrease as you descend the group because apart from Berylium, the melting points of Group II metals are less than 840°C  and all Group I metals melt below 200°C. Most transition metals melt above 1000°C.

As you descend the group the atoms become bigger and the outer electrons get more and more weakly held by the nucleus.  Thus the outer electrons can drift further from the nucleus than in most other atoms.  The larger atomic size results in weaker forces between neighbouring atoms because there is reduced attraction of the nuclear charge for the shared mobile outer electrons.

(d)   Oxidation numbers (valencies)

Ans   Group 0   =  0

      Group I    =   oxidation number  = +1

      Group II  =   +2

      Group VII         Element                  Oxidation number

                        F           -1

                        Cl          -1, +1, +3, +4, +5, +6, +7

                        Br          -1, +1, +5, +3, +4, +6, +7

                        I           -1, +1, +3, +5, +7

Q.35  Which member of the group may show properties atypical of the group as a whole? Give examples of such atypical properties.

Ans   Atypical behaviour is often shown by the first member of the group.

      E.g. F has only one oxidation state whereas others have different oxidation states.

E.g. The hydride of lithium does not decompose below its melting point but the others in group I metal hydrides do.

C 1.4.1

Carbon compounds containing chains, branched chains and rings.

Q.35  A key property of carbon is cantenation.  What do you understand by this term?

Ans   Cantenation describes the ability of an element to have its own atoms joined together many times in chains or rings.

Q.36  Explain why, together with the tetravalency of carbon, it gives rise to a large and wide variety of compounds.

Ans   Since carbon may form up to four bonds with any combination of C, N, H, O, P, S and halogens, a wide diversity of compounds is possible.  Chains may be limitless and the very stable rings add even more diversity in their connections and combinations.

Q.37  Identify and describe the three types of bonds formed in carbon compounds.

Ans   We find

      Single bonds       E.g. O – H             Triple bonds   E.g.      C=N

                        C – H

      Double bonds  E.g.      C=C

                              C=O

      We also find hydrogen bonds in some organic compounds – e.g. ethanol   

      Where hydrogen acts as a bridge between two electronegative atoms holding one by a normal covalent bond, and the other by electrostatic attraction a hydrogen bond is said to exist.

                        H           H

                        C     O                 H