Chemistry Study Notes Form 5 Chemistry Study Notes Form 6 Chemistry Study Notes Msomi Maktaba All Notes

ADVANCED CHEMISTRY QUESTIONS & ANSWERS – SOLVED PROBLEMS

ANSWERS:

1.   11.  (a) (i) Partition law:

               “When a solute is dissolved in a mixture of two immiscible solvents, it distributes itself in a constant ratio of concentration”.

         (ii)   Conditions

1.The two solvents must be completely immiscible.
2.  Temperature should be constant.
3.  There should be no change in molecular state of the solute i.e. should not dissociate or associate in any of the solvents.

     (b)   (i)  Standard molar enthalpy of formation:

       Is the heat change when 1 mole of a substance is formed from its elements in their standard states?

              www.msomimaktaba.com (g) + www.msomimaktaba.comwww.msomimaktaba.com (g)  www.msomimaktaba.com Nwww.msomimaktaba.com (g)

             Heat of solution of a substance:

   Is the enthalpy change when 1 mole of a substance is added to so much water that further addition of water produces no farther change.

   Ionization energy:
Is the energy required to remove the most loosely held electron from the shell of an atom or ion.

(c)    (i)  Hess’s law:

The enthalpy change for a particular reaction is same whether the reaction takes place in one step or in a series of steps.

         (ii)     Solution:
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  Applying Hess’s law

   www.msomimaktaba.comHC   = -393 KJ mol-1

      www.msomimaktaba.comHC   = +285 KJ mol-1

 www.msomimaktaba.com

                www.msomimaktaba.com

 K.D  = www.msomimaktaba.com

              www.msomimaktaba.com

 0.0243a = 10-4   -a

1.0243a = 10-4

 a = 9.76×10-5  moles

     Fraction of Iodine extracted by Cwww.msomimaktaba.com will be
= 9.76 x 10-5

1 x 10-4
=   0.976

(iii)   Solution:

        www.msomimaktaba.com

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12.(a)  (i)  Nucleophilic addition reaction is the reaction in which nucleophile is added

         First followed by electrophile.

 Elimination reaction is the reaction whereby an atom or group of atoms are removed from a compound

   e.g:   Cwww.msomimaktaba.com Cwww.msomimaktaba.com Cl  +  NaOHwww.msomimaktaba.comwww.msomimaktaba.com   Cwww.msomimaktaba.com  +  NaCl  +  www.msomimaktaba.comO

   Cwww.msomimaktaba.com   www.msomimaktaba.comCwww.msomimaktaba.com and Cwww.msomimaktaba.com Cwww.msomimaktaba.comCwww.msomimaktaba.com OH

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  By Lucas Test

Propanol will not form cloudness when it undergoes Lucas test but propan -2-ol will form cloudness after 5 minutes upon Lucas test.

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(ii)  Cwww.msomimaktaba.com   CH – Cwww.msomimaktaba.com and Cwww.msomimaktaba.com Cwww.msomimaktaba.com Cwww.msomimaktaba.com Cl
Using KOH followed by AgNwww.msomimaktaba.com in the presence of HNwww.msomimaktaba.com, yellow ppt will be observed for Cwww.msomimaktaba.com Cwww.msomimaktaba.com Cwww.msomimaktaba.comCl while there is

                                 Cl

www.msomimaktaba.comNo reaction for Cwww.msomimaktaba.com CH Cwww.msomimaktaba.com

(iii)  Cwww.msomimaktaba.com Cwww.msomimaktaba.com CHO and Cwww.msomimaktaba.com Cwww.msomimaktaba.com www.msomimaktaba.com– Cwww.msomimaktaba.com Cwww.msomimaktaba.com
Cwww.msomimaktaba.com Cwww.msomimaktaba.com CHO will react with Tollen’s reagent [Ag (Nwww.msomimaktaba.com]OH to form a white ppt of silver at the bottom of the test-tube while
Cwww.msomimaktaba.comCwww.msomimaktaba.com C  www.msomimaktaba.com www.msomimaktaba.com– Cwww.msomimaktaba.com Cwww.msomimaktaba.com will not react.

(c) (i)  Propanoic acid from ethane

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Cwww.msomimaktaba.com Cwww.msomimaktaba.comN  +  www.msomimaktaba.comO 4
www.msomimaktaba.com  Cwww.msomimaktaba.com Cwww.msomimaktaba.com  www.msomimaktaba.com –  Nwww.msomimaktaba.com www.msomimaktaba.comCH3 Cwww.msomimaktaba.comwww.msomimaktaba.com– OH

Propanoic acid

(ii)  Propan-2-ol from propan-1-ol

Cwww.msomimaktaba.comCwww.msomimaktaba.com Cwww.msomimaktaba.com OH  +  www.msomimaktaba.comSwww.msomimaktaba.com   www.msomimaktaba.comCwww.msomimaktaba.com  CH  =  Cwww.msomimaktaba.com

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(iii) Propyne from Ethyne

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λ m  =  www.msomimaktaba.com

www.msomimaktaba.com  = 133 S cm3 mol

www.msomimaktaba.com = 6.4×10-3 g cm-1

C = 5 X 10 -2 M

www.msomimaktaba.com =   www.msomimaktaba.comwww.msomimaktaba.com

1 dm3 = 10 3 cm3

x =1cm3

             = 10-3

=www.msomimaktaba.com            www.msomimaktaba.com  =www.msomimaktaba.com

= www.msomimaktaba.com  www.msomimaktaba.com

www.msomimaktaba.com  = 128 S cm-2 mol-1

∝  = O.96  www.msomimaktaba.com

13. Salt hydrolysis is the reaction between salt and water to form an acidic or alkaline solution:

Order of reaction:                                                                        Molecularity:

–   It cannot be determined                                                – Can be determined from the equation

From equation (determined experimentally).

–   It can be a whole number or fraction .                        – It can only be a whole number

–   It is the sum of concentrations terms on                     – It is the number of ions atoms or

Which the rate of reaction actually depends                molecules that must collide so as to

(It is sum of exponents of the concentrations              results in a chemical reaction.

in the rate law).

–   It cannot be obtained from the stoichiometry             – It can be obtained from the

Of an unbalanced equation.                                           Stoichiometry of the equation.

www.msomimaktaba.com = 30 x 60 = 1800s at 27www.msomimaktaba.com

www.msomimaktaba.com = 10×60 = 600s at  47www.msomimaktaba.com

Log www.msomimaktaba.com  =  www.msomimaktaba.comwww.msomimaktaba.com

n log www.msomimaktaba.com  =  www.msomimaktaba.com www.msomimaktaba.com

log www.msomimaktaba.com =  – 1.088 x 10-5 Ea

www.msomimaktaba.comwww.msomimaktaba.com     Ea  = 4.3833

K = www.msomimaktaba.com
K =  www.msomimaktaba.com

K = 3.85×10-4

Tollen’s reagent (silver mirror test)

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Acyl halides (Acid halide)

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Derivatives of carboxylic acids upon hydrolysis go back to carboxylic acids

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Transesterification is the reaction between ester and alcohol to form another ester

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Conditions for a chemical reaction to take place

i.  The collision of the molecules must be energetic enough to break the bonds between the molecule

ii. The colliding molecule must have right orientation (geometry)

Log K1 =     log A – www.msomimaktaba.com
Log K2 =     log A – www.msomimaktaba.com

Log K1 –log K2 = www.msomimaktaba.comwww.msomimaktaba.com

Log www.msomimaktaba.com= www.msomimaktaba.comwww.msomimaktaba.com

In www.msomimaktaba.com= Kt           www.msomimaktaba.com= www.msomimaktaba.com

www.msomimaktaba.comwww.msomimaktaba.com= Kt         www.msomimaktaba.com=www.msomimaktaba.com

Homolytic bond cleavage.

Is the type of bond cleavage in which the covalent bond breaks symmetrically so that are electron moves to each end.

Heterolytic bond cleavage.

Here the covalent bond breaks unsymmetrical and all bonding electrons are taken by the more electronegative atom

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www.msomimaktaba.com (g) +www.msomimaktaba.com     →      2Nwww.msomimaktaba.com  Slow

R= www.msomimaktaba.com [www.msomimaktaba.com] [www.msomimaktaba.com] this slow step will determine rate law

www.msomimaktaba.com = www.msomimaktaba.com [NO] [NO]

www.msomimaktaba.com= [NOwww.msomimaktaba.com

www.msomimaktaba.com=www.msomimaktaba.com [www.msomimaktaba.com]

At equilibrium www.msomimaktaba.com= www.msomimaktaba.com

www.msomimaktaba.com [NO] 2 = www.msomimaktaba.com [www.msomimaktaba.com]

www.msomimaktaba.com[NO] 2 = [www.msomimaktaba.com]

R= www.msomimaktaba.com  www.msomimaktaba.comwww.msomimaktaba.com[NO]2 [www.msomimaktaba.com]

R= K [NO]2 [www.msomimaktaba.com]

www.msomimaktaba.com

The slow step will determine the rate law

R = www.msomimaktaba.com[Br] [www.msomimaktaba.com]

Consider the equilibrium reaction

www.msomimaktaba.com
www.msomimaktaba.com

www.msomimaktaba.com= www.msomimaktaba.com [Bwww.msomimaktaba.com]

www.msomimaktaba.com = www.msomimaktaba.com [Br] 2

At equation:

www.msomimaktaba.com = www.msomimaktaba.com

www.msomimaktaba.com[Bwww.msomimaktaba.com] = www.msomimaktaba.com [Br]2

www.msomimaktaba.com[Bwww.msomimaktaba.com] =[ Br]2

[Br] = www.msomimaktaba.com[Br]1/2

R= K0www.msomimaktaba.comBr1/2
www.msomimaktaba.com

R= K [Brwww.msomimaktaba.com [www.msomimaktaba.com]

Kh = www.msomimaktaba.com

Cwww.msomimaktaba.comCOONa   →  Cwww.msomimaktaba.comCOwww.msomimaktaba.com + Nwww.msomimaktaba.com

www.msomimaktaba.com

Start    0.1                 0          0

0.1 – x           x          x

Kh = www.msomimaktaba.com

α = www.msomimaktaba.com× 100

[H COONa] = www.msomimaktaba.com

Let added volume be x

[H COONa] = www.msomimaktaba.com

0.1= www.msomimaktaba.com

n=0.1x

[H COONa] = www.msomimaktaba.com

[HCOOH] = www.msomimaktaba.com

14. Explain the meaning and significance of colloids

15. Discuss the properties of soil colloids

               –  Surface area

               –   Electric charge

               –   Ion exchange (diagram)

16. Explain the mechanism of ion exchange in soil

17. Calculation of percentage base saturation of a soil sample. With worked examples (how to calculate)

18. Aluminum has high polarizing power hence the compound formed between Al and carbon will be unstable therefore it will decompose immediately after formation.
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The chloride outside is not part of the complex meaning that it is ionization. Therefore Ag will react. In [ p(NH3)2 cL4)] the chloride is part of the complex hence there is no ionizable chloride ion and there no reaction

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19. (a)(i)  CH3 – C-CH3 + CH3 Mg CL www.msomimaktaba.comCH3 –C- CH3

(ii) CH3 –C- C-CH3 +NaOH + I2     www.msomimaktaba.com   CHI3 + NaOOC – CONa + NaI
(iii) CH3 CH2 C – NH2
www.msomimaktaba.comCH3 CH2 NH2 + KBr +CO2 +H2O
(iv) CH3 CH2 NH2 + CHcL3
www.msomimaktaba.comCH3 CH2 N =C + HCL

20. Distinguish between
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When Br2/H2O is added to phenol, white ppt is formed

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When Br2/H2O is added, no reaction

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(iv) CH3 CH2 OH +I2 + NaOH    www.msomimaktaba.com   No reaction

22. With the aid of chemical equations explain the following

(i)   Mercury(ii) iodide solution but not in potassium iodide solution but in water

(ii)  The pink solution of cobalt (ii) chloride turns blue when conc. HCL is added gelatinous  ppf of

(iii)  Copper (ii) hydroxide turns deep blue in excess ammonia

(iv)  ZnO and Pbo dissolves in hot conc sodium hydroxide solution

23. using relevant balanced chemical equation describe the process of extracting copper from copper pyrite under the following heading.

 (i) Concentration

(ii) Roasting

 (iii) Removal of ion impurities

(iv) Self- reduction reaction

24.  Cobalt copper, Iron and manganese are d-block elements

a.            What is meant by the term d- block element

b.            Write E.C of Cu,  Fe2+, Mn2+

 (i)   Explain in terms of E.C why Fe2+ ions are readily oxidized to Fe3+ ions but manganese (ii) ions are not readily oxidized to Mn3+

25.  2.5 x 10-3 moles of a compound with a formula was dissolved in 0.1M a silver nitrate solution. 50 cm3 were required for complete precipitation of the chloride ions present.

      (i) Deduce the ionic formula of the compound

      (ii) Draw the structure of the complex ion present and name it

26.  Write the formula of the following complexes

     (i) Tetra ammine copper(ii)  Sulphate mono hydrate

     (ii)Potassium

27.  Write down a balanced chemical equation for the following

       (i)  Adding of excess ammonia solution to aluminium ion

       (ii)  Ion(iii) oxide is heated with aluminium power

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