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1222P.B. College of Engineering

Campus Address : Administration Office :Irungattukottai, Chennai – Bangalore National Highway 3/117, Brahmin Street, Karambakkam,Sriperumbudur Taluk, Kanchipuram Dist., Chennai – 602 117. Porur, Chennai – 600 116.Tel. : 044-27198033, 9940026861, 9600112733 Tel. : 044-24766386, Telefax : 044-24766941E-mail : [email protected] Website : www.pbce.in Cell : 9600113031, 9962206098

MODEL QUESTION PAPER – II

CHEMISTRY

Note : i) Answer all the question from Part – I 30x1=30

ii) Answer any fifteen questions from part – II

iii) Answer any Seven Questions from Part – IIICovering all sections and Choosing at least two questions from each section

iv) Question No.70 is compulsory Answer any three from the remaining question in part – IV

v) Draw diagrams and write equations wherever necessary.

PART – I

Choose the correct answer

1. Which of following pairs have almost equal radii ?

(a) MO,W (b) Y, La (c) Zr,Hf (d) Nb, Ta

Ans. : (c) Zr,Hf

2. The most common oxidation state of lanthanide is

(a) +2 (b) +1 (c) +3 (d) + 4

Ans. : (c) +3

3. The fuel used in nuclear power plant is

(a) Copper (b) Lead (c) Uranium (d) Radium

Ans. : (c) Uranium

4. The Co-ordination number of Ni II in (Ni (Cl) 4) 2- is

(a) 2 (b) 4 (c) 5 (d) 6

Ans. : (b) 4

5. Loss of – particles equivalent to

(a) Increase of one proton only (b) decrease of one neutron only

(c) both (a) and (b) (d) none of these.

Ans. :(a) Increase of one proton only

6. If the activation energy is high then the rate of the reaction is.

(a) high (b) moderate (c) low (d) cannot be predicted

Ans. : (c) low

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7. The migration of colloidal particles under the influence of an electric field is

known as

(a) electrosmosis (b) electrolysis (c) electrodialysis (d) electrophoresis

Ans. : (d) electrophoresis

8. For chemisorptions which is wrong.

(a) It is irreversible (b) It requires activation energy

(c) It forms multi molecular layers on adsorbate

(d) Surface Compounds are formed

Ans. : (d) Surface Compounds are formed

9. The catalyst used for the decomposition of KClO3 is

(a) MnO2 (b) Cl2 (c) V2O5 (d) pt

Ans. : (a) MnO2

10. When one coulomb of electricity is passed through an electrolytic solution, the

mass deposited on the electrode is equal to

(a) equivalent weight (b) molecular weight

(c) electrochemical equivalent (d) one gram

Ans. : (a) equivalent weight

11. The isomerism exhibited by CH3 – CH2- NO2 and CH3- CH2 – O – N= O is

(a) Position (b) Chain (c) functional (d) tautomerism

Ans. : (c) functional

12. When nitromethane is reduced with Zndust + NH4Cl we get

(a) CH3N2 (b) C2H5NH2 (c) CH3NHOH (d) C2H5COOH

Ans. : (c) CH3NHOH

13. Electrophile used in the nitration of benzene is

(a) Hydronium ion (b) sulphonic acid (c) nitronium ion (d) bromide ion

Ans. : (c) nitronium ion

14. The reducing sugar of the following is

(a) Sucrose (b) Cellulose (c) Starch (d) Glucose

Ans. : (a) Sucrose

15. The amino acid without chiral carbon is

(a) Glycine (b) Alanine (c) Proline (d) Tyrosine

Ans. : (a) Glycine

16. The bond order of oxygen molecule is

(a) 2.5 (b) 1 (c) 3 (d) 2

Ans. : (d) 2

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17. Molecular orbital with least energy is

(a) 1s (b) 1s (c) 2py (d) 2py

Ans. : (a) 1s

18. Among the following which has the maximum ionisation energy.

(a) Alkali metals (b) Alkaline earth metals (c) Halogens (d) Nobel gases

Ans. : (d) Nobel gases

19. The shape of PCl5 is

(a) Pyramidal (b) Trigonal bipyramidal (c) Linear (d) Tetrahedral

Ans. : (b) Trigonal bipyramidal

20. The general outer electronic configuration of d-block elements is

(a) (n-l)d1-10 ns0-2 (b) (n-1) d1-5ns2 (c) (n-l)d0 ns1 (d) (n-l)d1-10 ns1-2

Ans. : (d) (n-l)d1-10 ns1-2

21. The total number of atoms per unit cell in BCC is

(a) 1 (b) 2 (c) 3 (d) 4

Ans. : (b) 2

22. The network obtained from a system is given by.

(a) w-PV (b) W+ PV (c) –W + PV (d) –W-PV

Ans. : (a) w-PV

23. Entropy is a function.

(a) State (b) Path (c) exact (d) inexact

Ans. : (a) State

24. The fraction of total moles of reactant dissociated is called.

(a) dissociation equilibrium (b) degree of association

(c) degree of dissociation (d) dissociation constant

Ans. : (c) degree of dissociation

25. Two moles of ammonia gas are introduced into a previously evacuated 1.0 dm3

vessel in which it partially dissociates at high temperature at equilibrium 1.0

mole of ammonia remains. The equilibrium constant Kc for the dissociation is

(a) (mole dm-3)2 (b) (mole.dm-3)2 (c) (mole.dm-3)2 (d) none of these

Ans. : (c) (mole.dm-3)2

26. Order of reactivity of alcohol towards sodium metal is

(a) Primary <secondary> tertiary (b) Primary >secondary> tertiary

(c) Primary <secondary< tertiary (d) Primary >secondary< tertiary

Ans. : (b) Primary >secondary> tertiary

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278

-274

-274

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27. Diethyl ether can be decomposed with

(a) HI (b) KMnO4 (c) NaOH (d) H20

Ans. : (a) HI

28. Which of the following produces ether, when heated with conc. H2SO4 at 143 K ?

(a) organic acid (b) Aldehyde (c) Alcohol (d) Ketone

Ans. : (c) Alcohol

29. Methyl Ketones are characterized by

(a) Fehling’s solution (b) Iodoform test (c) Schiff’s test (d) Tollen’s reagent

Ans. : (b) Iodoform test

30. The acid which redues Tollen’s reagent is

(a) acetic acid (b) benzoic acid (c) formic acid (d) oxalic acid

Ans. : (c) formic acid

PART –II

Note : i) Answer any Fifteen questions 15x3=45

ii) Answer in one or two sentences

31. State Heisenberg’s uncertainty principle.

Heisenberg’s uncertainty principle “it is impossible to measure

simultaneously both the position and velocity (or momentum) of a

microscopic particle with absolute accuracy or certainty.”

32. Why electron affinity of fluorine is less than that of Chlorine ?

EA of Fluorine is less than that of Chlorine

Because of small size of fluorine atom, repulsion among electrons of the

valence shell and also with electron to be added takesplace. Further there occurs

large crowding of electrons around the nucleus, the effective nuclear charge gets

decreased. Thus, the electron is having less attraction during addition. Hence

electron affinity gets decreased.

33. Write a note on plumbosolvency.

Plumbo solvency :

Lead is not attacked by pure water in the absence of air, but water

containing dissolved air has a solvent action on it due to the formation of lead

hydroxide (a poisonous substance). This phenomenon is called Plumbo

solvency.

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34. H3PO4 is triprotic prove.

35. Why do d-block elements exhibit variable oxidation states ?

This property is due to the following reasons.

i) These elements have several (n – 1) d and ns electrons.

ii) The energies of (n – 1)d and ns orbital are fairly close to each other.

36. What is “Spitting of silver”? How is it prevented ?

Molten silver absorbs about twenty times its volume of oxygen which it

again expels on cooling. Globules of molten silver are thrown off. This is called

“spitting of silver”. This can be prevented by covering the molten metal with a

layer of charcoal.

37. Determine the average life of U238 having t½ = 140 days

38. How are glasses formed ?

When certain liquids are cooled rapidly there is no formation of crystals at

a definite temperature, such as occurring on slow cooling. The viscosity of the

liquid increases steadily and finally a glassy substance is formed.

39. Write Kelvin statement of second law of thermodynamics

“It is impossible to construct an engine which operated in a complete

cycle will absorb heat from a single body and convert it completely to work

without leaving some changes in the working system”.

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40. Degree of dissociation of PCl5 at 1 atom and 25°C to O.2 calculate the Kp Value

for equilibrium.

PCl5 PCl3 + Cl2 at 25°C

41. What is meant by consecutive reaction ? Give an examples.

Consecutive reactions

The reactions in which the reactant forms an intermediate and the

intermediate forms the product in one or many subsequent reactions are called

as consecutive or sequential reactions. In such reactions the product is not

formed directly from the reactant. Various steps in the consecutive reaction are

shown as below :

A = reactant ; B = intermediate ; C = product.

Initially only the reactant A will be present. As the reaction starts, A

produces an intermediate B through K1 rate constant. As and when B is formed,

it produces the product C through K2 rate constant. After the completion of

reaction only ‘C’ is present and concentrations of A and B will be zero.

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42. Give three examples for first order order reaction :

43. Define colloidal solution.

A colloidal solution is a heterogeneous system consisting of dispersed

phase where colloidal particles are present and a dispersion medium where the

colloidal particles are dispersed.

44. State Kohlraush’s law.

KOHLRAUSH’S LAW :

‘‘at infinite dilution where in the ionisation of all electrolytes is complete, each

ion migrates independently and contributes a definite value to the total

equivalent conductance of the electrolyte’’.

45. Distinguish enantiomers and diastereomers.

Enantiomer Diastereomer

1. Optical isomers having the same

magnitude but different sign of

optical rotation

Differ in the magnitude of

optical rotation.

2. They have configuration with

non-super impossible object

mirror image relationship.

They are never mirror images.

3. Enantiomers are identical in all

properties except the sign of

optical rotation.

Diastereomers differ in all

physical properties.

46. How are 1-propanal and 2 –Propanal distinguished by Oxidation method ?

1-proponal and 2-proponal can be best distinguished by oxidation by

heating with copper followed by reaction with Fehling solution.

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47. Write the conversion reaction of ethylene glycol to 1, 4 – dioxan.

48. How do you prepare urotropine ? Mention its Use.

49. Formic acid reduces Tollen’s Reagent while acetic acid does not, Why ?

Formic acid is unique because it contains both an aldehyde group and

carboxyl group also. Hence it can act as a reducing agent.

Aldehyde group carboxylic acid group

50. An aromatic hydrocarbon A on nitration gives B Which is Known as oil of

mirbane. B on Warming with cone H2SO4 gives Compound C.

Identity A,B and C.

A- Benzene, B- Nitro Benzene, C- Nitro Benzene Sulphuric acid

51. Write a short note on antacids.

Certain drug formulations provide relief from such burning sensation.

These are known as antacids. Antacids are available in tablet as well as

gel/syrup forms. These antacids contain magnesium and aluminium hydroxides,

in addition to flavouring agents and colour.

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PART – III

(SECTION –A)

Note : Answer any seven Questions choosing atleast two questions from each section

52. Explain the formation of O2 molecule using molecular orbital theory.

Oxygen molecule, O2. The electronic configuration of oxygen (Z = 8) in

the ground state is 1s22s22p4. Each oxygen atom has 8 electrons, hence, in O2

molecule there are 16 electrons. Therefore, the electronic configuration of O2 is

as follows.

The molecular orbital energy level diagram of O2 molecule is given in Fig.

O O2 O

Atomic orbitals Molecular orbitals Atomic orbitals

53. Explain how silver is extracted from its chief ore.

Extraction of silver from the Argentite ore

Silver is extracted from the argentite ore by the Mac-Arthur and Forrest’s

cyanide process.

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The various steps involved in this process are as follows.

1. Concentration

The crushed ore is concentrated by froth-floatation process.

2. Treatment of the ore with NaCN

The concentrated ore is treated with 0.4-0.6% solution of sodium cyanide for

several hours. The mixture is continuously agitated by a current of air, so

that Ag present in the ore is converted into soluble sodium argento complex.

Ag2S + 4NaCN 2Na [Ag(CN)2] + Na2S

Sodium argento cyanide (soluble)

3. Precipitation of silver

The solution containing sodium argento cyanide is filtered to remove

insoluble impurities and filtrate is treated with zinc dust, silver gets

precipitated.

2Na [Ag(CN)2] + Zn Na2 [Zn(CN)4] + 2Ag

4. Electrolytic refining

The impure silver is further purified by electrolytic refining. The impure silver

is made the anode while a thin sheet of pure silver act as the cathode. The

electrolyte is silvernitrate acidified with 1% nitric acid. On passing electricity

pure silver gets deposited at the cathode.

54. Write the consequences of Lanthanide contraction.

Important consequences of lanthanide contraction are given below.

i) Basicity of ions

Due to lanthanide contraction, the size of Ln3+ ions decreases regularly

with increase in atomic number. According to Fajan’s rule, decrease in

size of Ln3+ ions increase the covalent character and decreases the basic

character between Ln3+ and OH- ion in Ln(OH)3. Since the order of size of

Ln3+ ions are

La3+ > Ce3+ ............... >Lu3+

ii) There is regular decrease in their ionic radii.

iii) Regular decrease in their tendency to act as reducing agent, with increase

in atomic number.

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iv) Due to lanthanide contraction, second and third rows of d-block

transition elements are quite close in properties.

v) Due to lanthanide contraction, these elements occur together in natural

minerals and are difficult to separate.

55. In what way [FeF6]4- differs from [Fe(CN)6]4- ? Explain

[FeF6]4-

(i) Fe+2 ion in this complex is sp3d2 hybridised.

(ii) The species is paramagnetic corresponding to four unpaired electrons.

[ Fe(CN)6 ]4-

(i) Fe2+ ion in this complex is d2sp3 hybridised.

(ii) The species is dimagnetic as there are no unpaired electrons in the

complex.

(SECTION –B)

56. State the various statements of second law of thermodynamics.

Second law of thermodynamics can be stated in many ways:

i) “It is impossible to construct an engine which operated in a complete cycle

will absorb heat from a single body and convert it completely to work

without leaving some changes in the working system”. This is called as the

Kelvin – Planck statement of II law of thermodynamics.

ii) “It is impossible to transfer heat from a cold body to a hot body by a

machine without doing some work”. This is called as the clausius statement

of II law of thermodynamics.

iii) ‘A process accompanied by increase in entropy tends to be spontaneous”.

This statement is called as the entropy statement of II law of

thermodynamics.

iv) “Efficiency of a machine can never be cent percent”.

v) The heat Efficiency of any machine is given by the value of ratio of output

to input energies. Output can be in the form of any measurable energy or

temperature change while input can be in the form of heat energy or fuel

amount which can be converted to heat energy.

thus % efficiency = output/input x 100.

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57. Derive the expressions for Kc and Kp for the decomposition of pcl5

The dissociation equilibrium of PCl5 in gaseous state is written as

PCl5(g) PCl3(g)+ Cl2(g)

Let ‘a’ moles of PCl5 vapour be present in ‘V’ liters initially. If x moles of

PCl5 dissociate to PCl3 and Cl2 gases at equilibrium at constant ‘V’ liters, then

molar concentrations of PCl5, PCl3 and Cl2 gases at equilibrium will be a x/V,

x/V, x/V respectively.

x is also known as the degree of dissociation which represents the fraction

of total moles of reactant dissociated.

x = Number of moles dissociated / Total number of moles present initially

If initially 1 mole of PCl5 is present then

If the degree of dissociation is small compared to unity, then (1–x) is

approximately equal to 1.0.

where x is small, degree of dissociation varies inversely as the square root of

pressure (or) varies directly as the square root of volume of the system.

58. Explain the experimental determination of rate constant of acid hydrolysis of

methyl acetate.

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Initially to a definite volume of (100 ml) hydrochloric acid (0.5 N), 10 ml of

ester is added and the start of the reaction corresponds to time of addition of

ester. The rate of the reaction is followed by withdrawing a definite volume of the

reaction mixture consisting of the ester and acid at various time intervals and

arresting the further progress of reaction by adding ice. The whole cold mixture

is titrated with standard NaOH (0.1 N) using phenolphthalein as the indicator.

Let the volume of alkali consumed at t = 0 be Vo cc which is equivalent to

the amount of hydrochloric acid present in the definite volume of the reaction

mixture drawn out at regular intervals of time. If Vt cc is the volume of alkali

consumed for the same definite volume of the reaction mixture drawn out after

reaction time ‘t’, then (Vt - Vo) cc is equivalent to the acetic acid produced by the

hydrolysis of ester in time ‘t’.

A final titration is done after about 8 hours or after refluxing the solution

for 45 mins to complete the hydrolysis which is Vcc. (V– Vo) cc is equivalent to

acetic acid produced from complete hydrolysis of ester.

Calculations

The initial concentration of ester = a (V– Vo) cc

Concentration of ester reacted at ‘t’ = x (Vt – Vo) cc

Concentration of ester remaining at time ‘t’ = (a – x) (V– Vt)

By substituting Vt values for various ‘t’ values, k is determined. These

values are found to be constant indicating k as the rate constant of the reaction.

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59. Calculate the potential of a half – cell consisting of Zinc electrode in 0.01m

ZnSO4 solution at 25°C |E° = 0.763V|

(SECTION – C)

60. Give any three methods of preparation of diethyl ether.

1. Intermolecular dehydration of alcohol. When excess of alcohol is heated

with con. H2SO4 or H3PO4, two molecules condense losing a molecule of

water forming ether.

2. Williamson’s synthesis: When ethyl bromide is heated with sodium

ethoxide diethyl ether is formed.

3. By heating ethyl iodide with dry silver oxide diethyl ether is formed.

61. Write the differences between acetaldehyde and benzaldehyde.

1. Because of the presence of Benzene ring, Benzaldehyde undergoes

electrophilic substitution reactions like nitration, halogenation sulphonation

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etc. – ‘CHO’ group is meta directing group. Hence meta substituted products

are formed

Halogenation :

+ HCl

(FeCl3 is the Lewis acid catalyst that forms the electrophilic Cl+ )

2. The absence of α hydrogen atom indicates that a carbanion cannot be

generated at α carbon atom. Hence aldol type condensation reaction cannot

occur.

3. Benzaldehyde does not undergo polymerization.

62. Explain the mechanism of esterification.

Mechanism of Esterification:

Protonation of the –OH group of the acid, enhances the nucleophilic attack by

alcohol to give the ester.

Step 1. Protonation of carboxylic acid

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63. Explain chromophore and auxo chrome theory about dyes.

According to this theory,

(i) An organic compound appears coloured due to the presence of certain

unsaturated groups (the groups with multiple bonds) in it. Such groups

with multiple bonds are called chromophores.

(ii) The compounds containing the chromophore group is called chromogen.

The colour intensity increases with the number of chromophores or the

degree of conjugation. For example, ethene (CH2 = CH2) is colourless, but

the compound CH3 – (CH = CH)6 – CH3 is yellow in colour.

(iii) The presence of certain groups which are not chromophores themselves,

but deepen the colour of the chromogen. Such supporting groups are

called auxochromes. Auxochromes may be acidic (phenolic) or basic.

Some important auxochromes are –OH, –NH2, –NHR, NR2.

The presence of an auxochrome in the chromogen molecule is essential to make

it a dye. However, if an auxochrome is present in the meta position to the

chromophore, it does not affect the colour.

For example, in the compound p-hydroxyazobenzene (a bright red dye),

PART – IV

Note : Question No.70 is compulsory and answer 4x10=40

any three from the remaining questions.

64. (a) Explain the Pauling scale for the determination eletronegativity. Give the

disadvantage of Pauling scale.

Consider a bond A-B between two dissimilar atoms A and B of a molecule

AB. Let the bond energies of A-A, B-B and A-B bonds be represented as

EA-A, EB-B and EA-B respectively. It may be seen that the bond dissociation

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energy of A-B is almost higher than the geometric mean of the bond

dissociation energies of A-A and B-B bonds i.e.,

Their difference () is related to the difference in the electronegativities of A

and B according to the following equation

Here, XA and XB are the electronegativities of A and B respectively.

The factor 0.208 arises from the conversion of Kcals to electron volt.

Considering arbitrarily the electronegativity of hydrogen to be 2.1, Pauling

calculated electronegativities of other elements with the help of this equation.

Disadvantage of Pauling scale

The disadvantage of Pauling’s scale is that bond energies are not known

with any degree of accuracy for many solid elements.

(b) How does fluorine differ from other halogens ?

1. Fluorine is the most reactive element among halogen. This is due to the

minimum value of F–F bond dissociation energy.

2. Fluorine decomposes cold dilute alkalies liberating OF2 and with conc.

alkali, O2 is liberated. Under similar conditions, the other halogens will

give rise to the hypohalites and halates respectively.

3. It has the greatest affinity for hydrogen, forming HF which is associated

due to the hydrogen bonding. Hydrofluoric acid is a weak acid whereas

the other hydrohalic acids are strong acids.

...... H– F...... H– F..... H–F.

4. It differs markedly from the other halogens in that it can form two types

of salts with metals. NaF and NaHF2.

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5. The salts of HF differ from the corresponding salts of other hydracids.

AgF is soluble in water while the other AgX are insoluble.

6. Being strongly electronegative it can have only a negative oxidation state

while the other halogens can have negative as well as positive oxidation

state.

65. (a) Write the Postulates of Werner’s theory Coordination compounds.

Postulates of Werner’s theory

1) Every metal atom has two types of valencies

i) Primary valency or ionisable valency

ii) Secondary valency or non ionisable valency

2) The primary valency corresponds to the oxidation state of the metal ion.

The primary valency of the metal ion is always satisfied by negative ions.

3) Secondary valency corresponds to the coordination number of the metal

ion or atom. The secondary valencies may be satisfied by either negative

ions or neutral molecules.

4) The molecules or ion that satisfy secondary valencies are called ligands.

5) The ligands which satisfy secondary valencies must project in definite

directions in space. So the secondary valencies are directional in nature

whereas the primary valencies are non-directional in nature.

6) The ligands have unshared pair of electrons. These unshared pair of

electrons are donated to central metal ion or atom in a compound. Such

compounds are called coordination compounds.

(b) Explain nuclear reactions taking place in sun.

NUCLEAR REACTIONS TAKING PLACE IN SUN (STARS)

It has been estimated that the sun is giving out energy equally in

all possible directions at the rate of 3.7x1033 ergs/sec. The energy of the

sun is supposed to arise from the fusion of hydrogen nuclei into helium

nuclei which in going on inside it all the time.

The various fusion reactions taking place in the sun are as follows:

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66. (a) Explain Schottky defect and Frenkel defect

Schottky defects

This defect is caused if some of the lattice points are unoccupied.

The points which are unoccupied are called lattice vacancies. The number

of missing positive and negative ions is the same in this case and thus,

the crystal remains neutral. The existence of two vacancies, one due to a

missing Na+ ion and the other due to a missing Cl- ion in a crystal of

NaCl is shown in Fig.

Schottky defects appears generally in ionic crystals in which the

positive and negative ions do not differ much in size.

Frenkel defects

This defect arise when an ion occupies an interstitial position

between the lattice points. This defect occurs generally in ionic crystals in

which the anion is much larger in size than the cation. AgBr is an

example for this type of defect.

One of the Ag+ ion occupies a position in the interstitial space

rather than its own appropriate site in the lattice is shown in Fig.

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(b) Write briefly the adsorption theory of catalysis

Adsorption of reactant molecules

The reactant molecules A and B strike the surface of the catalyst.

Formation of Activated complex

The particles of the reactants adjacent to one another join to form an

intermediate complex (A-B). The activated complex is unstable.

Decomposition of Activated complex

The activated complex breaks to form the products C and D. The

separated particles of the products hold to the catalyst surface by partial

chemical bonds.

Desorption of Products

The particles of the products are desorbed or released from the surface.

67. (a) Explain the buffer action of a basic buffer with an example.

The buffer NH4OH / NH4Cl can also be explained on the same lines

as of an acid buffer upon addition of HCl the H+ ions combine with

NH4OH to form NH4 + and H2O. pH is retained. Similarly when NaOH is

added, the OH– ions combine with NH4 + ions present in the buffer

solution to give NH4OH and hence pH is maintained.

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(b) Derive Nernst equation.

Nernst equation : Suppose the reaction occurring in a reversible cell is

represented by the equation

The decrease in free energy, – G, accompanying the process is

given by the well known thermodynamic equation

– G = – Go – RT ln J

where –Go is the decrease in free energy accompanying the same process

when all the reactants and products are in their standard states of unit

activity and J stands for the reaction quotient of the activities of the

products and reactants at any given stage of the reaction.

Substituting the value of J, we have

If E is the E.M.F. of the cell in volts and the cell reaction involves the

passage of ‘n’ faradays (i.e.,) F coulombs, the electrical work done by the

cell is in nFE volt-coulombs or Joules. Hence free energy decrease of the

system, –G, is given by the expression

where Eo is the E.M.F. of the cell in which the activity, or as an

approximation, the concentration of each reactant and each product of

the cell reaction is equal to unity. Eo is known as the standard E.M.F. of

the cell.

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is often referred to as the Nernst equation Replacing activities by

concentrations, as an approximation, the Nernst equation may be written

as

where the quantities in parentheses represent the concentration of the

species involved. Replacing [C] [D]/[A] [B] as equal to K, the equilibrium

constant in the molar concentration units,

where Eo = standard electrode potential

R = gas constant

T = Kelvin temperature

n = number of electrons transferred in the half-reaction

F = Faraday of electricity

K = equilibrium constant for the half-cell reaction as in equilibrium law.

68. (a) Discuss cis – trans isomerism with a suitable example (organic

compounds)

Geometrical Isomerism :

Isomerism that arises out of difference in the spatial arrangement of

atoms or groups about the doubly bonded carbon atoms is called

Geometrical isomerism. These isomers are not mirror images of each

other. Rotation about C=C is not possible at normal conditions and hence

the isomers are isolable.

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If different atoms or groups are bonded to the ‘C=C’ bond in a molecule,

more than one spatial arrangement is possible. For example, 2-butene

exists in two isomeric forms.

The isomer in which similar groups lie on the same side is called ‘cis

isomer’ (I). The other in which similar groups lie in opposite direction is

called ‘Trans isomer’ (II). This isomerism is called ‘Cis-Trans’ isomerism.

The two groups attached to the carbon atoms need not be same, it may be

different also. e.g.,

2-pentene

This isomerism arises out of the hindrance to rotation about the C=C bond in

such molecules.

(b) Account for the reducing nature of formic Acid.

Formic acid is unique because it contains both an aldehyde group

and carboxyl group also. Hence it can act as a reducing agent. It reduces

Fehling’s solution, Tollens reagent and decolourises pink coloured KMnO4

solution.

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COUNSELLING CODE


69. (a) Write any three methods of preparing Benzylamine.

1. Reduction of benzonitriles or benzamides

By the reduction of benzonitrile or benzamide catalytically or with

Lithium Aluminium Hydride.

2. From benzylbromide

By the action of alcoholic ammonia on benzyl bromide.

(b) Explain the function of lipids in Biosystems.

Functions of lipids in biosystems :

Fats and oils act as storage of energy in plants and animals. In the

animal body, fats are stored in fatty tissues, which are almost pure fat,.

Fat give about 2¼ times as much energy as carbohydrates or proteins.

Thus, as far as weight is concerned, storage of fat is the most economical

way for the body to maintain a reserve energy supply. Fat is a poor

conductor of heat. Hence, fat layer under skin serves to prevent losses of

heat from the body.

Wax acts as a protective agent on the surfaces of animals and

plants. Waxy coating on the surface of plants and fruits protects them

from excessive loss of moisture and becoming infected with fungi and

bacteria.

25

COUNSELLING CODE


Phospholipids like lecithins, and cephalins play a greater role in

biosystem. The lecithins are required for normal transport and utilisation

of other lipids, especially in the liver. lecithin aids in the organisation of

the cell structure. Cephalins are found in the brain. Cephalins have been

implicated in the process of blood coagulation.

Galactolipids occur in considerable amount in the white matter of

the brain and of all nervous tissue. The presence of galactose in the

glycolipids suggests the importance of milk sugar in the diet of infants

and children during the development of the brain and nervous system.

70. (a) Compound (A) with molecule formula. C6H6O gives violet colour with

neutral FeCl3, (A) reacts with CHCl3 and NaOH gives two isomers (B) and

(C) with molecular formula C2H6O2 compound (A) reacts with ammonia at

473K in the presence of ZnCl2 gives compound (D) with molecular formula

C6H7N. Compound (D) undergoes Carbylamine test Identify (A) , (B), (C)

and explain the reactions.

Solution :

When phenol is refluxed with chloroform and sodium hydroxide, a

formyl group –CHO is introduced at the ortho or para position to –OH

group.

Reaction with Ammonia

When heated with NH3 under pressure in presence of catalysts like

anhydrous zinc chloride or calcium chloride, hydroxyl groups of phenols

are replaced by amino groups.

(b) An element A present in period No.4 and Group No.12 on treatment with

dill. HNO3 froms B with the liberation of N2O. A when heated with air at

773K gives C which is Known as Philosophers’ wool Identify A.B and C

Explain the reaction involved.

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COUNSELLING CODE


The element in Period No.4 and Group No. 12 is Zinc. Hence “A” is

identified as Zinc.

Action of HNO3

Zinc reacts with HNO3 at various concentrations and it gives different

products.

Action of air

When heated in air at 773 K, it burns to form a white cloud of Zinc oxide

which settles to form a wooly flock called philosopher’s wool.

(c) Compound (A) with molecular formula C2H4O reduces Tollen’s reagent (A)

on treatment with HCN gives Compound (B) Compound (B) on hydrolysis

with an acid gives compound (C) with molecular formula C3H6O3 which is

an optically active compound . Compound (A) on reduction with

N2H4/C2H5ONa gives a hydrocarbon (D) of molecular formula C2H6 identify

(A), (B), (C) and (D) and explain the reactions.

Reduction under strongly basic condition :

A D (Ethane)

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COUNSELLING CODE


(d) An electric current is passed through three cells in series containing

solutions of CuSO4, AgNO3 and Kl respectively what weights of silver and

iodine will be liberated while 1.25g of copper is being deposited ?

(At weight of Ag = 108

Cu = 63.4

I = 127.1

28

COUNSELLING CODE


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