CHM 201

Physical Chemistry II

NOTA RINGKAS

CHAPTER 1

CHAPTER 2

CHAPTER 3

CHAPTER 4

CHAPTER 5

CHAPTER 6

 

JAWAPAN UJIAN

 

TEST 1

TEST 2

 

PEPERIKSAAN AKHIR

 

NOV 2006

MARKAH

 

TEST 1

TEST 2

 

 

Tahniah diucapkan kepada pelajar berikut kerana telah memperolehi keputusan yang cemerlang di dalam peperiksaan April 2007 :

 

Mohd Shazlan Mohd Isa                       [A]

Noorsya’ima Mohd Azlan                     [A]

Norsyafiqah Abdul Aziz                        [A]

Nuruljannah Zahraa’ Badrol Zaman       [A]

Siti Norlela Mohamad Radzi                 [A]

Siti Nur Atiqah Yahaya                         [A]

Nurul Ashikin Johari                             [A]

Nooraini Nasaruddin                            [A-]

 

Syllabus content

Chapter 1: Colligative properties

                        -Colligative properties

                        -Molecular mass determinations

 

Chapter 2: Phase equilibrium

                        -Definition and explanation

                        -One component system: ice-water-water vapour system and CO2 system

                        -Two component system

                        -Cooling curve

                        -Distribution of a solid between two solvents

 

Chapter 3: Solids

                        -Kinetic theory of solids

                        -Introduction to different types of solid

                        -Calculations

                        -Ionic solids

                        -Thermal properties of solids

 

Chapter 4: Colloid and surface chemistry

                        -Colloid

                        -Adsorption

                        -Surface chemistry

 

Chapter 5: Chemical kinetics

                        -Simple rate equation

                        -Concept of molecular collision in chemistry reactions

                        -Qualitative effects on rate of reaction

                        -Rate law and order of reaction

                        -Method to determine order of reaction

                        -Relation between rate constant and temperature in Arrhenius equation

                        -Catalytic kinetics

 

Chapter 6: Electrochemistry

                        -Electrode potential (IUPAC system)

                        -Nernst equation

                        -Electrolysis

 

Assessment

 

            Continuous assessment             40 %

            Test                                          30 %

            Laboratory work                      10 %

            Final Examination (3 hours)       60 %

 

 

 

CHAPTER 1

Colligative properties

These are the properties which depend only on the number of particles of solute present into the solution and are independent of the nature of the solute Ex:

  1. Elevation of boiling point
  2. Depression of freezing point
  3. Osmotic pressure
  4. Relative lowering of vapour pressure

The observed value of a colligative property is normal when the solute exists in solution in its normal state. It will be higher if the solute molecules undergo dissociation and lower than the normal value if the solute molecules undergo association in solution.

 

Raoult’s law (for miscible liquids): This states, “In a solution of two miscible and volatile liquids A and B, the partial pressure PA of liquid A is proportional to its mole fraction XA’.

Mathematically PA = PoA • XA similarly, PB = PoB • XB. Where PoA and PoB are the vapour pressures of pure components A and B respectively.

Roult’s law can also be stated, “for solution of volatile liquids, the partial pressure of any component is equal to the product of the vapour pressure of pure component and the mole fraction of that component in solutions”.

 

Raoult’s law (for solution of non-volatile solute): The relative lowering of vapour pressure for a solution is equal to the mole fraction of the solute when the solvent alone is volatile.

Since    PA = PoA • XA 

 Or       PA =  PoA(1-XB) because XA + XB =1

Or        PA = 1 - XB

            PoA

Or        XB = 1 - PA  =  PoA- PA

                         PoA        PoA

Or mole fraction of B = relative lowering of vapour pressure for dilute solution. PoA- PA = n

                                                                                                                             PoA        N

Solution which obeys Roult’s law for all concentrations is called  Ideal solution.

For ideal solution ∆V = 0, ∆H = 0, for such solution p = PA + PB + …….

Examples: 1. mixture of benzene and toluene

                2. A solution of chlorobenzene

                3. Mixture of ethyl bromide and ethyl iodide

 

A non ideal solution is one which does not obey Roult’s law. The deviation from Roult’s law may be positive or negative. For such solutions ∆V≠ 0, ∆H ≠ 0

Examples: positive deviation: ethanol – cyclohexane

                 Negative deviation = chloroform-acetone

 

 

 

CHAPTER 2

 

 

 

CHAPTER 3

 

Solids

Solids: Solids have a definite shape and definite volume. They possess regular arrangement of constituent particles (atoms, ions or molecules)

 

Main characteristics of solids: Solids possess one or more of  the under mentioned characteristics.

1) rigid

2) have definite volume

3) change to liquid on heating

4) have high densities

5) incompressible

6) have good thermal and electrical conductivity

7) possess definite shape

8) crystalline or amorphous in structure

 

Crystal structure: This refer to the regular and repetitive arrangement of atoms or molecules in crystals.

 

Space lattice (crystal lattice): The three dimensional pattern present in crystalline substance is known as crystal or space lattice. It is a network of line drawn in space such that the space is divided into equal volumes and the point of intersection of these lines are the lattice points upon which the atoms, ions or molecules make up the substance.

 

Unit cell: Unit cell is the smallest unit of lattice which when translated along the axes of the lattice produces the crystal lattice.

 

Different types of unit cells: Three types of unit cells are described below.

1) Body centered cubic structure:

 

 

Lattice points in a bcc unit cell are located at each of the eight corners and in the center of the unit cell. A body centered cubic unit cell contains 2 atoms

 

2) Face centered cubic structure:

 

 

Lattice points in a face centered cubic unit cell  are found at each of the eight corners and in the center of each face. A face centered cubic unit cell contains 1 atom from the eight corners plus 6*1/2=3 atoms from each face for a total of 4 atoms.

3) Simple cubic or primitive cubic

 

(a) A simple cubic unit cell showing the lattice points. (b) A portion of a simple cubic lattice. (c) Only a portion of each atom of a substance that forms a simple cubic lattice lies within a particular unit cell.

         Only 1/8 of each atom lies in a particular simple cubic unit cell

         Each simple cubic unit cell contains:

8 corners x 1/8 atoms = 1 atom

 

Coordination number: It is the number of nearest atom which are directly surrounding a given atom in a crystal. Coordination number in case of  simple cubic structure is 6, body centered cubic is 8 and face centered cubic is 12.

 

Atomic radius: It is the half of distance between adjacent atoms in a crystal of a pure element.

 

Calculations of atomic radius: If the structures of crystal and lattice parameters are known, the atomic radius can be calculated.

            In simple cubic cell r = a/2        where a is the edge of the cube and r is atomic radius.

            In body centered cubic r = a√3

                                                        4

            In face cubic centered r =   a         

                                                    2√2

 

Density of unit cell: =    mass of unit cell

                                 Volume of unit cell

 

Mass of  each particles: = molar mass of the particle

                                         Avogadro’s number

 

Volume of unit cell: = a3 where a is the edge

 

 

 

CHAPTER 4

Colloid and surface chemistry

Surface chemistry is the branch of chemistry which deals with the nature of surfaces and changes taking place on the surfaces.

Absorbent: It is the material on whose surface the absorption occurs

Adsorbate: The molecular species that gets adsorbed is known as adsorbate

Adsorption: The phenomenon of higher concentration of molecular species (liquids of gases) at the surface of a solid than in the bulk is called adsorption

Desorption: It refers to the process of removal of an adsorbed substance from the surface on which it is adsorbed.

Chemisorption (chemical adsorption):  If the adsorbate is held on surface by strong chemical bonding forces, the adsorption process is known as chemisorption.

Physiosorption (physical adsorption):  If the adsorbate is held on a surface by weak Van Der Waals forces, the adsorption process is known as physisorption

Adsorption: It is a phenomenon in which the molecules of a substance are uniformly distributed throughout the body of other substance

Positive adsorption: When the concentration of the adsorbate is more on the  surface of the adsorbent as compared to its concentration in the bulk then it is known as positive adsorption

Negative Adsorption: When the concentration of the adsorbate is less on the  surface of the adsorbent as compared to its concentration in the bulk then it is known as positive adsorption

 

Factors affecting adsorption of gas on solid:

  1. Nature of adsorbate (solid)
  2. Nature of gas
  3. Temperature
  4. Activation of adsorbate
  5. Pressure

 

Freundlich Isoterm: It is a graph between log x and log p. It is a straight line

                                                                    m

Adsorption isobar: It is a graph between the amount of substance adsorbed per gram (x/m) and temperature ‘t’ at constant pressure

Application of adsorption:

  1. In a gas mask to remove poisonous gases such as CH4, CO. (activated charcoal)
  2. For removing the hardness of water, in ion exchange resins
  3. In volumetric analysis as adsorption indicator (fluorescein)
  4. For removal of moisture and controlling humidity (silica)
  5. In decolourisation of sugar (animal charcoal)
  6. In chromatographic purification of compounds (silica gel)
  7. In creating high vacuum (charcoal cooled in liquid air)

 

Colloidal state: It is an intermediate state between true solutions and suspensions. It is heterogeneous and consists of two phases.

i)                    disperse phase (colloidal particles)

ii)                   dispersion medium (the medium in which colloidal particles are dispersed)

The size of colloidal particles ranges between 1 nm to 10 nm.

Lyophilic colloids: Colloids in which particles of dispersed phase have a great affinity for the dispersion medium are called lyophilic colloids ex: glue, gelatin

Lyophobic colloids: Colloids in which there is no affinity between particles of dispersed phase and the dispersion medium called lyophobic colloids Ex: Al(OH)3 solution

Multimolecular colloids: They consist of aggregates of particles of small molecules with diameter less than 1mm Ex: solution of gold particles

Macromolecular colloids: They consist of dispersed particles which are large molecules Ex: polymer

Classification of colloids

Disperse phase

Dispersion medium

Name of colloidal solution

Example

Gas

Liquid

Foam

Soda water, creams

Gas

Solid

Solid foam

Rubber, foam, bread

Liquid

Gas

Aerosol

Cloud, fog

Liquid

Liquid

Emulsion

Milk, medicines, emulsified oils

Liquid

Solid

Gels

Butter, cheese, jelly

Solid

Gas

Solid aerosol

Smoke, dust storm

Solid

Liquid

Solutions

Paints, muddy water

Solid

Solid

Solid solutions

Alloys, coloured glass

 

Micelles: Colloidal solutions which behave as strong electrolytes at low concentrations but exhibit colloidal properties at higher concentration are known as micelles. They also known as associated colloids

Demulsification: The decomposition of an emulsion into its constituent liquids is called demulsification

Dialysis: This the method used to separate impurities from the colloidal solution. It is based on the principle that colloidal particle cannot pass through a cellophane membrane or parchment while the ions of an electrolyte can pass through it.

Tyndall effect: If a strong beam of light is allowed to pass through a colloidal solution placed in dark, the path of the beam is illuminated by a bluish light. This is because of the scattering of light and this phenomenon is known as Tyndall effect

Electrophoresis: It refer to the movement of colloidal particles towards cathode or anode under the influence of electric field

Catalyst: Catalyst is a substance which increases the rate of chemical reaction without being used in the reaction

Homogeneous catalyst: When the catalyst mixes homogeneously with the reactants and forms a single phase, the catalyst is said to be homogeneous and the process is known as homogeneous catalysis

Heterogeneous catalyst: When the catalyst forms a separate phase with the reactants then it is called heterogeneous catalyst and the process is known as heterogeneous catalysis

Activity of catalyst: The ability of a catalyst to accelerate the rate of  a chemical reaction is known as activity of  catalyst. Sometimes it may be as high as 1010 times.

Selectivity of catalyst: It refer to the ability of catalyst to direct reactions to yield particular products Ex: n-heptane selectively gives toluene (C6H5-CH3) on platinum catalyst.

Shape selective catalyst: The catalytic reactions depending on the pore structure of catalysts are known as shape selective catalytic reactions and the catalysts are known as shape selective catalyst Ex: zeolite catalysts

 

 

CHAPTER 5

Chemical kinetics

Chemical kinetics is that branch of chemistry which deals with the speed or rate of reaction.

Homogeneous reactions: When all the reactants and products are in the same physical state, the reaction is said to be homogeneous reactions Ex: NaOH (aq) + HCl (aq) → NaCl (aq) + H2O

Heterogeneous reactions: Reactions, where reactants and products are in different physical states, are known as heterogeneous reactions. Ex: 2HCl (aq) + CaCO3 (S) → CaCl2 (aq) + CO2 (g) + H2O

Exothermic reactions: Reactions which are accompanied by evolution (release) of heat energy are known as exothermic reactions. ∆H is negative or heat of products is less than heat of reactants. Ex: O2 (g) + H2 (g) → H2O (g)  ∆H = -242.7kJmole-1

Endothermic reactions: These are the reactions which are accompanied by absorption of heat energy. ∆H is positive or heat of products is greater than heat of reactants. Ex: C (s) +H2O (g) → CO2 (g) + 2H2O (g)  ∆H = +131.8 kJmole-1

 

Different kinds of chemical reactions:

a)      Chemical combination

H2 + Cl2 → 2HCl

b)   Chemical decomposition

                        2KClO3 → 2KCl + 3O2

c)   Simple displacement reaction

                  Zn +CuSO4 → ZnSO4 +Cu

d)      Double displacement reaction

AgNO3 + NaCl → AgCl + NaNO3

 

Rate of reaction: It is defined as the decrease in concentration of the reactants or increase in the concentration of products per unit time. There occurs an increase in rate of reaction if the concentration of reactants is increased.

            For a reaction A → B

            Rate of reaction = - ∆A = + ∆B

                                        ∆t        ∆t

 

Reaction rate constant (K) or specific rate: For a reaction, it is defined as “the rate of reaction when the concentrations of the reactants are unity”. The value of K depends upon (i) Order of reaction (ii) temperature.

 

Average rate:

                        Average rate = Change in concentration

                                           Time involved for change

                                        = ∆A

                                           ∆t 

Instantaneous rate: It is the rate of the reaction determined at a particular moment of time. The average rate approaches the instantaneous rate as ∆t becomes smaller.

 

Factors which affect the rate of reaction:

a) Physical state of reactants ≈ rate of reaction increase in surface area.

b) Concentration of reactants ≈ rate of reaction increases with increase in concentration.

c) Catalyst ≈ it generally increase in presence of catalyst

d) Temperature  ≈ rate of reaction increases with increase in temperature.

 

Rate equation

It is a mathematical expression that shows the relationship between the rate of the reaction and the concentration of reactants.

            A + B → C + D

Rate = - d(A) = - d(B) = + d(C) = + d(D)

             dt          dt           dt           dt

This equation is called rate equation

 

Order of a reaction

It is equal to the total number of reacting species (ions, atoms or molecules) whose concentration actually changes during the course of a reaction.

For a general reaction;              A + B +C → products

If, rate = K[A]p.[C]q.[C]r  where K is rate constant, then order of the reaction = (p+q+r)

 

Molecularity

It is equal to the sum of coefficients of various reactants appearing in a balanced chemical equation

C12H22O11 + H2O → 2C6H12O6

For the above reaction, molecularity = 1+1=2

Molecularity is a theoretical concept and its value is always a whole number. If molecularity is one, it is unimolecular. If molecularity is two, it is bimolecular and so on.

 

First Order reaction

In case the rate of reaction is found to be directly proportional to the concentration (first power of concentration) of a single reactant, it is said to be first order reaction

A → products

d(A) = rate = K[A]  [rate equation for first order]

 dt

examples of first order reaction;

1)      rate of growth of bacterial culture

2)      radioactive disintegration process

 

Rate constant for the first order reaction:

            A → B

If ‘a’ is the initial concentration of A and ‘x’ of it disappears in time ‘t’. Then concentration of A after time ‘t’ is (a-x), therefore

 K =2.303 log  a     

         T        (a-x)

Units of K = sec-1

Second order reaction

This is a reaction whose rate is directly proportional to the square of concentration of a single reactant or equal to the concentration of two different reactants.

For such reaction;         A + B → products

Rate = K[A][B] or rate = K[A]2

K =    1      ln b(a-x)          where a, b are the initial concentrations of A and B respectively and (a-x), (b-x) are the concentrations of A and B after time ‘t’.

       t(a-b)     a(b-x)

units of K = L m-1 sec-1

Third order reaction

This is that reaction whose rate is found to be directly proportional to the cube of the concentration of a single reactant or equal to the square of concentration of one reactant or equal to the square of concentration of one reactant and concentration of another reactant.

Rate = K[A]3

Rate = K[A]2[B]

Rate = K[A][B]2

Rate = K[A][B][C]

 

Units of K = L2 mol-2sec-1

 

Zero order reaction

For such reaction;         Rate = k [concentration]0

Units of K = L mol-1 sec-1

 

Half-life of a reaction

It is the time taken for a reaction to reach the half of its initial concentration. It is denoted by t 1/2

For a first order reaction t ½  = 0.693    where K is the first order reaction constant.

                                                  K

Reaction mechanism

Collision theory: molecules react when they collide in a preferential direction.

Threshold energy: It is the minimum amount of energy from zero level, the reacting molecules must possess, in order to react. It is denoted by ET

Activation energy: It is defined as, “the extra energy over and above the average potential energy of the reactants which must be supplied to the reactants to enable them to cross over the energy barrier between the reactants and products.” It is denoted by Ea

            Ea = ET -ER      where ER is average energy of  reactants

For fast reactions activation energies are low. For slow reactions activation energies are high.

 

Theory of transition state: Chemical reactions occur through an intermediate stage which is known as transition state or activated state.

Particles in collision react only if, upon impact, they attain the energy of the transition state.

 

 

 

Catalyst’s effect on equilibrium constant: A catalyst does not alter the equilibrium constant of a reaction.

Effect of catalyst on activation energy: A catalyst lowers the activation energy

Effect of catalyst on free energy change: A catalyst does not alter the free energy change of the reaction.

 

CHAPTER 6

Electrochemistry

Electrochemistry is the branch of chemistry which concerned with the study of relationship between electrical energy and chemical energy and conversion of one form into another.

Electrochemical reactions: Reactions involving the flow of current are called electrochemical reactions. The following are different types of electrochemical reactions.

1)         Electrolysis: These reactions require electrical energy from external source to bring about a chemical change. (electrical

            energy is consumed)

2)         Electrochemical cells: In these cells the electrical energy is produced due to the chemical change taking place inside the cell.

 

Electrolytes: Substances which conduct electricity in the molten state, fused state or aqueous solutions are known as electrolytes.

Conductors: Those substances which allow the passage of current through them are called conductors.

Metallic conductors: Those metallic substances which allow the passage of electricity through them without undergoing any chemical change are called metallic conductors ex; silver, copper, aluminium.

Metallic conductance: It is the phenomenon of flow of electrons through a metallic substance. It occurs due to flow of electrons.

 

Voltaic or Galvanic cell: This cell is a simple device of producing electrical energy by chemical reactions. Such a cell is also known as an electrochemical cell.

 

 

Electrolytic cell: it is a device to convert electrical energy into chemical energy.

Electrode potential: It is the potential developed over an electrode when it is immersed into the solution of its own ions. It is denoted by E.

Oxidation potential:

 

Gambarajah yang terdapat di dalam nota ini diambil dari Chemistry matter and its changes, 4th ed.Brady and Senese.John Wiley and sons.

 

 

ANSWER SCHEME TEST 1

Answer

 

1. 100.0 mL eth.gly. x 1.15 g eth.gly.    x   1 mol eth.gly. = 1.86 mol eth. gly

                                     1.00 mL eth.gly.       62.0 eth.gly.

 

     100.0 mL water x 1.00 g water    x   1 mol water  = 5.56 mol water

                                 1.00 mL water       18.0 water

 

    mole fraction of water =      5.56         =  5.56    = 0.749

                                             5.56 + 1.86      7.42

   

    Raoult’s law : Pwater = (Xwater)(Powater) = (0.749)(525.8 mmHg) = 394 mmHg

 

2.  39.5 g x  1 mol   = 0.636 mol eth.gly.

                   62.07 g

     750 mL x 1.00g x 1kg      = 0.750 kg mass of solvent

                        ml       103g

 

    molality of solution = 0.636 mol  =  0.848 mol kg-1

                                        0.750 kg      

    The boiling point elevation = ∆Tb = (0.52 kg mol-1)( 0.848 mol kg-1)

                                                 = 0.44oC

    The solution boils at 100.44 oC

    The freezing point lowering = ∆Tb = (1.86oCkg mol-1)( 0.848 mol kg-1)

                                                  = 1.58 oC

    The solution freezes at -1.58 oC

 

3. c =  π    =            1.8 x 10-3 atm                      = 7.36 x 10-5 mol/L

          RT     (0.0821 L atm mol-1K-1)(298K)

 

 

     Molar mass =        5.0 g                 =  6.8 x 104 g/mol

                            7.36 x 10-5 mol

 

4. i)

Point

Degree of freedom

A

1

B

1

C

1

D

2

T

0

 

    ii) 

Point

Phases

A

2 phases (solid-liquid)

B

2 (liquid-vapor)

C

2 (liquid-vapor)

D

1 (vapor)

T

3 (solid-liquid-vapor)

 

     iii) T- triple point; C-critical point

 

5. i)

 

 

    ii) 

 

 

 

    iii) Fractional distillation yields pure hydrogen bromide and an azeotropic mixture (47% HBr and 53% water)

 

6. i) Let the mass extracted = a g

              a/100         = 5

               (5-a)/100

               a=4.17g

    ii) first extraction :       b/50    = 5

                                  (5-b)/100

                                    b=3.57g

         second extraction :     c/50           = 5

                                       (1.43-c)/100

                                     c=1.02g

         total mass extracted = 3.57 + 1.02 = 4.59g

 

7. Mass of Cd remains in solution = 40 g Bi x 40 g Cd = 26.67g Cd

                                                                          60 g Bi

    Mass of  Cd precipitated = 60-26.67 = 33.33 g

 

 

8. a) cubic closest packed spheres

    b) hexagonal closest packed spheres

 

9. Edge length = 2(133pm) + 2(181pm) = 628pm

 

 

 

ANSWER SCHEME TEST 2

 

1. Shine a beam of light on it. If light is scattered, the dispersion is colloid.

 

2. i) Size 1nm -1000 nm

              a) monodisperse : particles all the same size

              b) polydisperse: range of particles size

   

    ii) chemical properties

           a) Lyophilic – solvent loving

                                - stable towards flocculation by electrolytes

                                 - generally solubilized macromolecules

 

            b) Lyophobic – easily flocculated by electrolytes

                                  - generated mechanically

 

3. physisorption – molecules are noon covalently bound to the surface. Weak binding forces

      chemisorption – molecules are covalently bound to the surface. Strong binding forces.

 

4. Langmuir isotherm – Adsorption of ≤ 1 monolayer

    BET isotherm – adsorption > 1 monolayer

 

5. i)

           

     ii) The graph of rate against [H2O2] is a straight line. So the reaction is first order with respect to hydrogen peroxide concentration.

     iii) Rate = k [H2O2]

 

6. i) experiment 4 rate = <0.227 molL-1s-1 = 4 = 22

        experiment 2 rate = 0.057 molL-1s-1 = 1 = 1

         

          rate = k[O2][NO]2

      ii) for experiment 1

           0.028 molL-1s-1 = k (0.010 mol/L)(0.020 mol/L)2

                                 k = 7.0 x 103 L2mol-2s-1

 

7 i)

 

7 ii)

 

         

 

8 i) K = A e –Ea/RT

>

         K = 2.4 x 107 Lmol-1s-1

     ii) rate = k[NO][O3]

                = (2.4 x 107)( 0.0010)( 0.00050)

                 = 12 mol L-1s-1

9. i

      

     ii) Au3+ (aq) + 3e → Au(s)

     iii) 2Au3+ (aq) + 3 Zn(s)  → 2Au(s) + 3 Zn2+(aq)

 

 

 

 

 

 

ANSWER : FINAL NOV 06

 

Part A

 

1.B       2.C      3.D      4.A      5.C      6.A      7.D      8.C      9.B       10.B     11.D    12.A    13.C    14.D    15.A            16.B     17.B     18.C    19.A    20.D

 

Part B (hanya jawapan akhir sahaja diberikan selebihnya perlu cuba dan akan dibincangkan dalam kelas)

QUESTION 1

ai)        1,1,2,2

aii)        P = 3, C = 2, F = 1

bii)       57.14g

c)         5.565g

QUESTION 2

c)         0.149 m @ 14.9 cm

QUESTION 3

aii)        0.32V

aiii)       1.42V

c)         8.66 x 10-4 g

QUESTION 4

a)         121.6 g

b)         96.8 gmol-1

c)         293 g

QUESTION 5

b)         face centered cubic unit cell

QUESTION 6

aii)        78.0 sec

aiii)       57.8 sec

b)         2.0 x 105 Jmol-1

ci)        second order

cii)        6.14 x 10-2 M-1 s-1

ciii)       1.63 x 103 sec

 

ANSWER: FINAL NOV 2005

 

PART A

 

1.C      2.B       3.D      4.C      5.D      6 B       7.A      8.D      9.A      10.A    11.B     12.D    13.C    14.B     15.A            16.C    17.A    18. B    19.D    20.B

 

PART B

 

QUESTION 1

a) i) PA =XAPA0

    ii) benzene & toluene @ hexane & heptane

b) by using Henry’s Law ; 1.699 x 10-3

c) i) 101.60 oC

    ii) -5.82 oC

 

QUESTION 2

 a) ii) melting point – any point on BC line

     iii) If K is water the slope of BC line should be the negative because for water when pressure increase, the temperature decrease

     iv) solid, liquid and gas can exist together in dynamic equilibrium

c) to estimate the multilayer surface area.

 

QUESTION 3

a)i) The boiling point of HBr

   ii) negative . The boiling point composition graph shows maximum, vapor pressure graph shows a minimum

 

 T = boiling point of the mixture

 C = composition of the vapor

 

iv) 1st distillate : azeotropic mixture ; final distillate : azeotropic mixture

v) the mass of Y will be extracted into the ether is 1.872 g

 

QUESTION 4

a) i) Volume of cubic = 7.895 x 10-2 nm3

        volume occupied by one atom of P = 3.95 x 10-2 nm3

    ii) NA = zW            =  6.006 x 1023

                  ρV

b)ii) length = 4.04 x 10-8 cm

   iii) NA  = 6.02 x 1023

   iv) density = 2.27

 

QUESTION 5

a)i)

ii) Ea= 18kJmol-1

b) rate = k[NH4+]x[NO2]y  ; x = 1, y = 2

c) i) living organism operate efficiently in a very narrow temperature range. Enzymes as homogeneous catalysts that speed up desirable reactions without heating.

   ii) homogeneous-same phase; heterogeneous – different phase.

 

QUESTION 6

a) i)H + T → HT

    ii) R

   iii) step 2

   iii) Rate = K2[HR][T] -------- 1

        k1[H][R] = k-1[HR]

         [HR] = k1/ k-1[H][R]---------- 2

        substitute; rate = K2 (k1/ k-1) [H][R][T]

          Rate = k[H][R][T]

b) i) second order

    ii) k = 0.12 M-1 min-1

    iii) half life = 1/k [A]o = 8.3 min

 

QUESTION 6

a) i) anode Pb ; cathode Ag

    ii) Ag gain ; Pb lose

    iii) Pb(s)  + 2Ag+ → Pb2+ + 2Ag(s)

    iv) Eocell = 0.93 V

     v) Ecell = 0.893 V

b) 574.4 s

 

 

 

 

 

 

RESULT : TEST 1

 

 

 

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