1 2

COVENANT UNIVERSITY

ALPHA SEMESTER TUTORIAL KIT (VOL. 2)

P R O G R A M M E : C H E M I S T R Y

300 LEVEL

1 2

DISCLAIMER

The contents of this document are intended for practice and learning purposes at the undergraduate

level. The materials are from different sources including the internet and the contributors do not in

any way claim authorship or ownership of them. The materials are also not to be used for any

commercial purpose.

1 2

LIST OF COURSES

CHM312: Organic Chemistry

CHM315: Unit Operations/Heat Transfer

CHM317: Environmental Chemistry

CHM332: Industrial Inorganic Chemistry

CHM333: Instrumental Method of Analysis

CHM334: Applied Spectroscopy

CHM356: Metallurgy and Metal Fabrication

*Not included

1 2

COVENANT UNIVERSITY

CANAANLAND, KM 10, IDIROKO ROAD

P.M.B 1023, OTA, OGUN STATE, NIGERIA.

TITLE OF EXAMINATION: BSc EXAMINATION

COLLEGE: Science and Technology

DEPARTMENT: Chemistry

SESSION: 2015/2016 SEMESTER:

ALPHA

COURSE CODE: CHM 312 CREDIT UNIT: 3

COURSE TITLE: Organic Chemistry

INSTRUCTION: Attempt any FOUR (4) Questions TIME:

3 HOURS

PLEASE START THE RESPONSE TO EACH QUESTION ON A NEW PAGE OF THE

ANSWER BOOKLET

1. (a) Describe the Haworth synthesis of

Naphthalene.

(5 marks)

1 2

Haworth Synthesis

• Naphthalene can be synthesized by treating benzene with succinic anhydride in the presence in the presence of AlCl3.

O

O

O

AlCl3

CCOOH

Zn/Hg

HCl

COOH

O

Conc. H2SO4

OZn/Hg

HCl

Naphthalene 1,2,3,4-Tetrahydronaphthalene

3-Benzoylpropa-noic acid

3-Phenyllpropa-noic acid

Pd/C or

Seleniumdehydrogenation

BenzeneSuccinicanhydride

(b) Showing the appropriate equation, give the final products of sulfonation of anthracene

both at low and high

temperature.

(5 marks)

Sulphonation

excess Conc. H2SO4

Low Temperature

SO3H

Anthracene-2,7-disulphonic acid

HO3S

excess Conc. H2SO

4High Temperature

Anthracene-1,8-disulphonic acid

SO3HSO3H

(c) Give the IUPAC names of the following compounds:

1 2

CH3

CHCH2

CBr

Br O

CH3

CHCH2

OC

CH3

CH3

O

CH3

CO

CCH3

O O

CH3

CNH2

O(i) (ii) (iii) (iv)

(2 marks)

(i) 3-bromobutanoyl bromide; (ii) isobutyl acetate or 2-methylpropyl ethanoate or -

bromobutyryl bromide; (iii) acetic anhydride; (iv) acetamide

(d) Give the general products formed during the hydrolysis, alcoholysis, and aminolysis of

acid

halides.

(3 marks)

Hydrolysis, alcoholysis and aminolysis of acid chloride will give the products:

carboxylic acid, ester and amide respectively as shown in the figure below;

Chapter 21 16

Acid Chloride Reactions (1)

H2O

R'OH

R'NH2

R'COOH

R C

O

Cl

R C

O

OH + HCl

R C

O

OR'

R C

O

NHR'

R C

O

O C

O

R' + HCl

+ HCl

+ HCl

acid

ester

amide

acid anhydride =>

(e) Complete the following reactions:

CH3CH2COOH + NaOH ?(i)

CH3―CH2―COOH + NaOH CH3―CH2―COO Na+

+ H2O

sodium propanoate

1 2

COCH3

O

HCl

H2O

(ii) +

(2.5 marks)

[Total

17.5

marks]

2. (a) Write structures for the following bicyclic

alkanes:

(3 marks)

(i) bicyclo[1.1.0]butane; (ii) 2-chlorobicyclo[3.2.0]heptane; (iii) 7-

methylbicyclo[2.2.1]heptane

ClCH3

(b) (i) Draw the Newman projections for rotation about the C―C bond in

chloroethane and clearly label the most stable conformer and the least stable

conformer

(2 marks)

HH

H

H

Cl

H

H

H

HH

Cl

H H

Cl

H

more stable

The barrier to rotation about the C―C bond in chloroethane is 18.0 kJ/mol.

1 2

(ii) What energy value can you assign to an H―Cl eclipsing interaction given that

H―H eclipsing interaction is 4

kJ/mol?

(2 marks)

18 = 𝑥 + 4 + 4 = 𝑥 + 8

𝑥 = 18 − 8 = 10 kJ/mol

(iii) Sketch a diagram of potential energy versus angle of rotation about the C―C

bond of

chloroethane.

(2 marks)

E

Degrees of Rotation

30060 120 180 2400 360

18 kJ/mol

(iv) Label the energy barrier to rotation on your

diagram

(1 mark)

See diagram above

(c) (i) The relative basicities of some nitrogen-containing compounds are as follows:

C2H5NH2 > NH3 > C6H5NH2. Explain why these compounds behave as

bases.

(2 marks)

There is a lone pair of electrons on nitrogen that can be donated to a proton

(ii) Using diagrams, explain why aniline (C6H5NH2) is more acidic than ammonia

(NH3).

(2 marks)

The lone pair of electrons on nitrogen is delocalized into the benzene ring, which

means that the lone pair of electrons is not available for donation to a proton.

1 2

NH2NH2

+

(d) This question is about Diels-Alder reaction, a [4+2] cycloaddition reaction involving a

diene and a dienophile. Use curly arrows to describe the mechanism of the reaction

between 1,3-butadiene and propanal to give cyclohex-3-ene carbaldehyde.

(3.5 marks)

CH2

CH2

H

CH2

O O

H

[Total

17.5

marks]

3. (a) An aldol addition product loses water to form an aldol condensation product. Give the

mechanism of reaction for aldol condensation of propanal

(CH3CH2CHO).

(10 marks)

Aldol condensation is an organic reaction in which an enol or an enolate ion reacts with

a carbonyl compound to form a β-hydroxyaldehyde or β-hydroxyketone, followed by a

dehydration to give a conjugated enone.

(b) Briefly define the following terms showing the equation of reactions

(i). Claisen condensation

(ii). Malonic ester synthesis

1 2

(iii). Aldol condensation

(iv). Aldol addition reaction

(v). Enol

(7.5 marks)

[Total

17.5

marks]

The Claisen condensation is a carbon–carbon bond forming reaction that occurs between two esters or one ester and another

carbonyl compound in the presence of a strong base, resulting in a β-keto ester or a β-diketone.

Aldol condensation is an organic reaction in which an enol or an enolate ion reacts with a

carbonyl compound to form a β-hydroxyaldehyde or β-hydroxyketone, followed by a

dehydration to give a conjugated enone.

Aldol addition

In the aldol reaction, two molecules of an aldehyde or ketone react with each other in the

presence of a base to form a -hydroxy carbonyl compound.

Malonic Ester Synthesis

Four basic step in reaction

1. Enolate formation (strong base)

2. Enolate alkylation (alkyl halide)

3. Ester hydrolysis (aqueous acid)

4. Decarboxylation (heat)

Enols are isomers of aldehydes or ketones in which one alpha hydrogen has been removed

and placed on the oxygen atom of the carbonyl group. The molecule has a C═C and an ―OH

1 2

group, so it is called an ene/ol, i.e., an enol. Enols can be formed only from carbonyl

compounds which have alpha hydrogens. They can be formed by acid or base catalysis, and

once formed are highly reactive toward electrophiles, like bromine.

4. (a) Glycols undergo oxidative cleavage in presence of periodic acid. Give the products

resulting from periodate cleavage of the following

glycol.

(2 marks)

CH3CHOHCH 2OH ? ?+

O

CH3

H

O

H

H (b) An alkene can undergo addition reaction in the presence of HOX (X = halogen) to give

halohydrin, which on treatment with base gives an epoxide. Give the products of the

reaction.

(CH3)2C CH2

HOCl NaOH? ?

(2 marks)

CH2

CH3

CH3

CH3

OH Cl

H

CH3 H

O

CH3CH3

(c) Compound A, C7H8O is insoluble in aqueous NaHCO3 but dissolves in NaOH. When

treated with bromine water, (A) rapidly forms compound (B) C7H5OBr3. Give structures

for (A) and

(B).

(2 marks) OH

CH3

OH

BrBr

Br

CH3

A B

(d) How can C2H5OC2H5 and n-C4H9OH, which have the same molecular formula

(C4H10O) be distinguished by chemical

reactions?

1 2

(2 marks)

C2H5OC2H5 is an ether and cannot be oxidized by any oxidizing agent whereas n-

C4H9OH is an alcohol which can be oxidized by K2Cr2O7 to an acid or aldehyde.

(e) Starting from propane-2-thiol, suggest a Williamson ether synthesis for the following

compound

(CH3)2CH―S―CH3.

(2 marks)

(CH3)2CH―S―H + NaOH (CH3)2CH―S Na+

(+ CH3Br)

(CH3)2CH―S―CH3

(f) What do you understand by the term tautomersim?

(1.5 marks)

Tautomerism is the rapid equilibrium between two isomers.

(g) Give equations for the tautomersim in which each of the following compounds is the

more stable tautomer: CH3CHO;

C6H5COCH3

(4 marks)

O

CH3

CH3

CH2

OH

CH3

CH3

O

OH

CH2

(h) Certain alicyclic compounds can be prepared by carbene addition to an alkene, predict

the product(s) obtained from addition of singlet dichlorocarbene, ׃CCl2, to each of the

following

compounds:

(2 marks)

(i) (ii)

Cl

Cl

ClCl

[Total

17.5

marks]

1 2

5. Complete the following reactions and indicate the nomenclature of each of the products in

questions (a) to (c)

(d). The Scheme for the preparation of furan derivative is shown below using

cyclodehydration.

What is the reagent A and the name of product

B.

(3 marks)

A = ZnCl2/AcOH B = 2,5-dimethylfuran

1 2

(e). With the aid of equation of reaction only, explain Gattermann’s reaction in furan

chemistry.

(3 marks)

(f). Furan is a 5-membered aromatic heterocycle with only two C=C instead of three C=C

present in benzene. Account for the aromatic nature of furan.

(2.5 marks)

Since all monocyclic aromatic compounds must comply with Hückel formula

(4n+2)π to give 6 π electrons. Furan is aromatic because it is a planar (4n+2)π

electron molecule, hence, provided 6π electron for the ring system. One of the lone

pair of electron on oxygen is involved in the aromatic sextet in addition to two C=C

to give five resonant hybrids as shown below:

[Total

17.5

marks]

6. (a) When aqueous sodium hydroxide is added to phenol it gives a colorless solution, A.

On adding dilute hydrochloric acid, this solution becomes cloudy at first and then

separates to give colorless oil, B, as a lower layer. Identify A and B, and write equations

for the reactions

involved.

(4 marks)

(b) Complete the following reactions by identifying A, B and C.

(4.5 marks) OH OAc

A Br2 NaOH

H3O

B C

(c) The reaction below is a way of synthesizing alcohols. Complete the reaction by

identifying A, B and

1 2

C.

(4 marks)

CH3 Hg(OAc)2, H

2O NaBH

4

A B

(d) Write the IUPAC names for the following

compounds:

(2 marks)

O O CH3

Anesthesia Anisole

(i) (ii)

(e) The oxidation of butanethiol with nitric acid gives A. Identify A by structure and name.

(3 marks)

SH

HNO3 A

[Total

17.5

marks]

EQN:

(b)

1 2

(c)

(d) (i) Diethyl ether (ii) Methyl phenyl ether

(e)

Name: Butanesulfonic acid

1 2

COURSE CODE: CHM 312 COURSE TITLE: ORGANIC CHEMISTRY

1. Build a molecular model of ethane, and look at the interconversion of staggered and eclipsed forms. Measure the H―H distances in each case, and see if you can detect a difference Build a molecular model of propane, and look at the interconversion of staggered and eclipsed forms. Measure the H―H distances in each case, and see if you can detect a difference.

2. Draw the Newman projection for the staggered and eclipsed conformations of propane. Using the information of internal energies of rotation and the H―H and H―CH3 interactions draw a graph of potential energy versus angle of rotation for propane, and assign values to the energy maxima.

3. Look down the C2―C3 bond, and draw Newman projection formulas for the a. Most stable conformation of 2,2-dimethylbutane b. Two most stable conformations of 2-methylbutane. Which one is more stable? Why? c. Two most stable conformations of 2,3-dimethylbutane d. Sketch an approximate potential energy diagram for rotation about the carbon-

carbon bond in 2,2-dimethylpropane. e. Does the form of the potential energy curve of 2,2-dimethylpropane more closely

resemble that of ethane or that of butane? f. Repeat the problem for the case of 2-methylbutane.

4. The following questions relate to a cyclohexane ring depicted in the chair conformation shown.

4

5

6

1

2

3

a. Is a methyl group at C-6 that is “down” axial or equatorial? b. Is a methyl group that is “up” at C-1 more or less stable than a methyl group that is

up at C-4? c. Place a methyl group at C-3 in its more stable orientation. Is it up or down?

1 2

5. Draw the two chair conformers for each of the following, and indicate which conformer is more stable: a. cis-1-ethyl-3-methylcyclohexane b. trans-1-ethyl-2-isopropylcyclohexane c. trans-1-ethyl-2-methylcyclohexane d. cis-1,2-diethylcyclohexane e. cis-1-ethyl3-isopropylcyclohexane f. cis-1-ethyl-4-isopropylcyclohexane

6. Give the names for the following substituted alkanes.

7. Draw the structure and give the molecular formula for each of the following compounds.

a. 1-ethyl-3-methylcycloheptane

b. Isobutylcyclohexane

c. Cyclopropylcyclopentane

d. 3-ethyl-1,1-dimethylcyclohexane

e. 3-ethyl-2,4-dimethylhexane

f. 1,1-diethyl-4-(3,3-dimethylbutyl)cyclohexane

8. Give names for each of the following bicyclic alkanes.

9. Write structures for the following bicyclic alkanes:

a. 2-chlorobicyclo[3.2.0]heptane

b. 7-methylbicyclo[2.2.1]heptane

1 2

c. Bicyclo[2.1.0]pentane

d. Bicyclo[1.1.0]butane

10. Write the structures of a bicyclic compound that is an isomer of bicyclo[2.2.0]hexane and

give its name.

11. Name the following compounds:

12. Without drawing their structures, tell which of the following compounds is a fused bicyclic

compound and which is a bridged bicyclic compound, and how you know. a. Bicyclo[2.1.1]hexane

b. Bicyclo[3.1.0]hexane

13. Give IUPAC names for the following alkenes.

14. Draw the Kekulé structures of anthracene Propose a mechanism for the addition reaction shown below.

15. Give the names of the following compounds.

1 2

16. Draw the carbocation intermediate for attack at position 2 of naphthalene.

17. The table below shows that amine have higher boiling points than alkanes but have lower boiling points than alcohols with similar molecular mass. Explain the difference in boiling point between the amines and the alkanes and between the amines and the alcohols.

Compound Formula Molecular

mass, Mr

Boiling point, o

C

Propylamine CH3CH

2CH

2NH

2 59 48

Methylethylamine CH3NHC

2H5 59 37

Trimethylamine NH(CH3)3 59 3

Butane CH3CH

2CH

2CH

2CH

3 58 −0.5

Propan-1-ol CH3CH

2CH

2OH 60 65

18. The relative basicities of some nitrogen-containing compounds are as follows:

C2H5NH2> NH3> C6H5NH2

19. Explain why these compounds behave as bases.

20. Outline a preparation of benzylamine using the Gabriel synthesis.

21. Outline the synthesis of each of the following arylamines from benzene: a. o-Isopropylaniline

b. p-Isopropylaniline

c. 4-Isopropyl-1,3-benzenediamine

d. p-Chloroaniline

e. m-Aminoacetophenone

22. Give the structure of the major alkene formed when the hydroxide of each of the following quaternary ammonium ions is heated.

1 2

23. What amine and what diazonium salt would you use to prepare chrysoidine?

24. Outline the synthesis of orange II from 2-naphthol and p-aminobenzenesulfonic acid.

25. Chloramphenicol X is a polyfunctional antibiotic which was obtained naturally from

Streptomyces venezuelae and later synthesized based on functional group identification.

What is the expected product(s) when X is treated with:

(i) Sn/HCl (ii) PCl5 (iii) KMnO4/H+(excess) (iv) ZnCl2/HCl 6 marks

1 2

26. Acetylation of salicylic acid in acid medium, due to variation in reaction conditions, could

either give o-acetyl salicylic acid or compound X as shown in the scheme below:

(i) Itemise three methods you can use to ascertain that you have actually prepared acetyl salicylic acid and not compound X.

(ii) Carefully observe the structure of o-acetylsalicylic acid and write out the names of two functional groups present therein.

(iii) Give one test to distinguish between o-acetylsalicylic acid and X. (iv) What is the relevance and branding name of o-acetylsalicylic acid?

27. The halogenation of furan depends on the molar equivalent of chlorine gas involved. Write

the structures and names of the expected product(s) when:

(i) furan is treated with Cl2 gas at -40oC.

(ii) furan is treated with 1.0M equivalent of Cl2 gas at 40oC.

(iii) furan is treated with 5.0M equivalent of Cl2 gas at 40oC.

28. Write short notes on the following with required equation and reaction conditions to back

it up as example

a. Nitration of Pyrrole

b. Chlorination of Pyrrole

c. Sulphonation of Pyrrole

d. Acetilation of Pyrrole

e. Reduction of Pyrrole

1 2

29. Itemize five applications of pyrrole derivatives

30. Draw and name the products of the following reactions.

CH3 CH3

O O

CH3 O O CH3

O O

2 CH3CH

2OH

H+

diethyl propanedioate

31. Draw the complete mechanism for Fischer esterification of benzoic acid with methanol.

32. Draw the reagents that will react to produce the following ester.

33. In a tubular form list 15 sources of carboxylic acid in IUPAC and common names.

34. Name the following compounds

i) ii) iii) iv)

35. Acid chlorides react with alcohols producing esters and by product HCl by hydrolysis. Draw

and name the products of the following reaction.

36. Esters are among the most widespread of all naturally occurring compounds. Most have

pleasant odors and are responsible for the fragrance of fruits and flowers. Write structural

formulas for the following naturally occurring esters;

Flavour Name Structure

1 2

pineapple methyl butanoate

bananas isopentyl acetate

Apple isopentyl pentanoate

Rum isobutyl propanoate

oil of wintergreen

methyl salicylate [methyl 2-hydroxybenzoate)

nail polish remover

ethyl acetate

new car smell (plasticizer for PVC)

dibutyl phthalate

37. Write equations showing how the BELOW transformation can be carried

38. Predict the major product(s) of the following reactions:

C

OH

O

O CH3

C

O

O (CH2)3CH3

O

O (CH2)3CH3C

1 2

O CH3OH

c. H2SO

4, 25 oC

?

O

O H2

H

O H2 (2 moles), Pt

Ethanol?

?

?

?+(a)

(b)

(c)

(d)

39. (a) Draw the structures of the following compounds:

(i) benzyl alcohol (ii) ethylene glycol (iii) hydroquinone

(iv) 2-mercaptoethanol. (4 marks)

(b) Complete the following reaction which explains the synthesis of a typical aromatic ether:

OH

CH2

Br

NaH

0 oC, THF 27 oC, CH3OCH

2CH

2OCH

3

? ? (8 marks)

(c) Complete the oxidation reaction of the following thiol.

SH c. HNO3

? (5 marks)

40. Acid-catalyzed hydration of alkynes in the presence of Hg+2 yields an intermediate enol, which rapidly equilibrates with the corresponding carbonyl compound. The regiochemistry of the reaction is "Markovnikov"; starting with the equation below explain enol interconversion mechanism that will yield a corresponding carbonyl compound in this case.

C CHH

3O+/Hg2+

OH

H

H

(10 marks)

41. (a) List three methods each for the preparation of aldehydes and ketones. (3 marks)

(b) Complete the following reactions:

CH3CH2CH2CH2CH20H ?K2Cr2O7,, special conditions

CH3CH2CH2CH2CH20H ?K2Cr2O7,+

OH?

PCC

i)

ii)

iii)

C5H5NHCrO3Cl

1 2

(c) Explain what happens when benzoyl chloride is reacted with lithium aluminum hydride tri-tert-butoxide(LiAl(O-t-Bu)3

(d) Using equations of reactions explain the synthesis of 2-Methyl-3-pentanone viaCoupling of R2CuLi with acid chloride

1 2

ANSWERS TO SOME SELECTED QUESTIONS

Answer to 25:

Answer to 27:

Question 38

O OH

1 2

a) (5 marks)

b)

(5 mks)

c) (8 marks)

d) (5 marks)

Question 39

a)

b)

(10 mks)

c)

(5 mks)

CH2OH

OH

OH + C

i) OH ii)

OH OH

iii) OHHO iv) HSOH

270C, CH3OCH2CH2OCH3

NaH

O0 C, THF

Br

OH O- O

S

O

O

OH

1 2

Question 40

(a)

(10 mks)

Question 41

(a)

Aldehydes, syntheses:

1. Oxidation of 1o alcohols

2. Oxidation of methylaromatics

3. Reduction of acid chlorides

Ketones, syntheses:

1. Oxidation of 2o alcohols

2. Friedel-Crafts acylation

3. Coupling of R2CuLi with acid chloride

1 2

(b)

(c) DIY

(d) DIY

CH3CH2CH2CH2CH2OH

+ K2Cr2O7 CH3CH2CH2CH2CO2H

1-pentanol

pentanoic acid

K2Cr2O7, special conditions!CH3CH2CH2CH2CH=O

pentanalvaleraldehyde

CH2OHC5H5NHCrO3Cl

pyridinium chlorochromate

CH=O

benzaldehydebenzyl alcohol

CH3CH2CH2CH2CH2OH

1-pentanol

1 2

COVENANT UNIVERSITY

CANAANLAND, KM 10, IDIROKO ROAD P.M.B 1023, OTA, OGUN STATE, NIGERIA

TITLE OF EXAMINATION: B.Sc. Degree Examination

COLLEGE: Science & Technology

DEPARTMENT: Chemistry

SESSION: 2015/2016 SEMESTER: Alpha

COURSE CODE: CHM315 UNITS: 3

COURSE TITLE: Unit Operations/Heat Transfer

INSTRUCTION: Answer any FOUR (4) questions TIME: 3 hours

1. a Explain briefly the three modes of heat transfer (4.5

marks)

b. Calculate the thermal conductivity and nusselt number of a flow of gas (pr = 0.71, μ = 4.63

× 10-5

kg/ms, Cp = 1175 J/KgK) Over a turbine blade of chord diameter 20,000 μm, where

the heat transfer coefficient is 1000 W/m2K. (2

marks)

c. Calculate the appropriate Reynolds number and state if the flow is laminar or turbulent for

the following:

i. A 10 m (water line length) long yacht sailing at 13 km/hr in sea water density 1000 kg/m3

and μ = 1.3 × 10-3

kgm/s (2 marks)

ii. A roof of a coach 6 m long travelling at 100 km/hr in air density 1.2 kg/m3 and μ = 1.8

× 105 kg/ms (2 marks)

d. The walls of a house in a cold region consist of three layers, an outer brickwork of 15 cm

thickness and an inner wooden panel of 1.2 cm thickness. The intermediate layer is made

of an insulating material 7 cm thick. The thermal conductivities of the brick and the wood

used are 0.70 W/moC and 0.18 W/moC, respectively. The inside and outside temperatures

of the composite wall are 21 oC and -15 oC, respectively. If the layer of insulation offers

twice the thermal resistance of the brick wall, calculate (a) the rate of heat loss per unit area

of the wall and (b) the thermal conductivity of the insulating material. (7

marks)

[17.5

marks]

1 2

2. a. Explain the following terms:

i. Black body

ii. Natural Convection (3 marks)

b. A hot chamber has an 8 cm thick inner layer of firebrick (k = 1.04 W/moC) and a 13 cm

outer layer of ordinary brick (k = 0.69 W/moC). The inside and outside surface

temperatures of the composite wall are 400 oC and 75 oC, respectively. Considering that the

outer surface temperature is too high, it is decided to apply a 5 cm thick layer on the outer

surface. On doing this, the outer skin temperature is reduced to 60 oC and the rate of heat

loss decreased by 250 W/m2 of the wall area. Calculate the thermal conductivity of the layer

of plaster.

(6.5 marks)

c. Two identical bodies radiate heat to each other. One body is at 30 oC and other at 250 oC.

The emmisivity of both is 0.7. Calculate the net transfer per square meter.

(4 marks)

d. Hot air at 80 °C is blown over a 2 m by 4 m flat surface at 30 °C. If the average convection

heat transfer coefficient is 55 W/m2 °C. Determine the rate of heat transfer from the air to

the plate. (4

marks)

[17.5

marks]

3. a. Briefly define the terms extraction, salting out, extract and extent of reaction (7 marks)

b. Highlight the basic conditions necessary for unit extraction process selectivity (3 marks)

c. Differentiate between a batch and continuous reactors (4.5

marks)

d. Draw a schematic representation of a double pipe heat exchanger (3 marks)

[17.5

marks]

4. a. State the three (3) common types of heat exchanger (3 marks)

b. The temperature difference between the hot and cold fluids, varies along the length of

the heat exchanger, in the light of this, derive the Log Mean Temperature Difference

(LMTD) for a co-current flow. (7

marks)

c. State the mathematical expression for the overall heat transfer coefficient with respect to

fouling

(2 marks)

d. Oil having a specific heat capacity of 6.4 KJkg-1K enters a single-pass counter flow heat

exchanger at the rate of 4.5 kg/s and a temperature of 410 K. It is to be cooled to 350 K.

Water (Cpc = 3.5 KJkg-1K) is available to cool the oil at a rate of 5.1 kg/s and a temperature

1 2

280 K. Determine the surface area required if the overall heat-transfer coefficient is 230 Wm-

2K.

(8.5

marks)

[17.5

marks]

5. a. (i) What is dimension? (3 marks)

(ii) Write the symbol for expressing dimensions. (2.5

marks)

b. Derive the dimension for the following:

(i) Force (ii) Heat-transfer coefficient (iii) Power

(iv) Energy (v) Momentum (5 marks)

c. What is the difference between fluid statics and fluid dynamics? (2 marks)

d. State the fundamental equation of fluid pressure and explain each terms (2 marks)

e. Convert the following:

An early viscosity unit in the cgs system is the poise (abbreviated as P), or g/(cm . s),

named after J. L. M. Poiseuille, a French physician who performed pioneering

experiments in the year 1840 on water flow in pipes. The viscosity of water (fresh or

salt) at 20°C is approximately 0.01 P. Express this value in SI units. (3 marks)

[17.5

marks]

6. a. Define the following terms:

(i) Filtration (ii) Drying (5 marks)

b. Differentiate between surface and depth filtration (3 marks)

c. A food containing 80% water is to be dried at 100°C down to moisture content of 10%. If the

initial temperature of the food is 21 °C, the latent heat of vaporization of water at 100 °C and

at standard atmospheric pressure is 2257 kJ kg-1. The specific heat capacity of the food is 3.8

kJ kg-1 °C-1 and of water is 4.186 kJ kg-1 °C-1. Calculate:

(i) the quantity of heat energy required per unit weight of the original material, for drying under

atmospheric pressure. (4.5

marks)

(ii) Determine also the energy requirement/kg water removed. (2.5

marks)

(iii) Using the same material as specified above, if vacuum drying is to be carried out at 60 °C

under the corresponding saturation pressure of 20 kPa abs. and latent heat of vapourization at

the temperature is 2358 kJkg-1 (or a vacuum of 81.4 kPa), determine the heat energy required

to remove the moisture per unit weight of raw material.

(2.5 marks)

[17.5

marks]

1 2

COVENANT UNIVERSITY

CANAANLAND, KM 10, IDIROKO ROAD P.M.B 1023, OTA, OGUN STATE, NIGERIA

TITLE OF EXAMINATION: B.Sc. Degree Examination

COLLEGE: Science & Technology

DEPARTMENT: Chemistry

SESSION: 2015/2016 SEMESTER: Alpha

COURSE CODE: CHM 315 UNITS: 3

COURSE TITLE: Unit Operations/Heat Transfer

COURSE LECTURERS: Prof. Ajanaku K.O., Dr. Siyanbola T. O. and Mr. Aladesuyi O.

MARKING SCHEME

No 1

Conduction

Conduction heat transfer is an atomic or molecular process. It occurs in the presence of a temperature

difference and its not accompanied by any macroscopic or bulk motion in the medium. Conduction

is the only mode of heat transfer in a solid medium. It may also occur in a stagnant liquid or a gaseous

medium. Conduction has a role to play in virtually all of the heat transfer equipment. 1.5 marks

Convection

Existence of motion or a velocity field in a liquid or gaseous medium greatly enhances the rate of heat

transfer.Convection means the transport of heat energy by way of displacement of fluid elements from

one point to another which is at a different temperature. Convection may be of two types, forced

convection and free or natural convection. 1.5 marks

Radiation

Heat transfer by radiation does not require a material medium. A body at a temperature of absolute

zero always emits energy in the form of electromagnetic waves. The rate of release of such energy is

proportional to the fourth power of the absolute temperature of the body. This phenomenon is called

radiation and the basic law governing it is known as stefan-Boltzmann law. 1.5 marks

B.

i. pr = 𝜇𝐶𝑝

𝑘

1 2

0.71 = 4.63 ×10−

5𝑘𝑔

𝑚𝑠 ×1175

𝑘

K = 0.0766 W/Mk (1 mark)

To calculate Nusselt no (Nu) = ℎ𝐷

𝑘

Nu = 0.02 ×1000

0.0766 = 2610.96 (1 mark)

C

i Reynold number (Re ) = DV𝑝

𝜇 =

10 ×13 ×1000×10000

36000 ×0.0013 = 2.77 ×107

It’s a Turbulent flow…………………………………………………………….2 marks

ii. Reynold number (Re) = DV𝑝

𝜇 =

6 ×100 ×1000×1.2

36000 ×180000 = 1.1 × 10

-3

(Laminar) (2 marks)

D.

Q = ∆T

𝑇𝑜𝑡𝑎𝑙 𝑡ℎ𝑒𝑟𝑚𝑎𝑙 𝑟𝑒𝑠𝑖𝑠𝑡𝑎𝑛𝑐𝑒

Thermal Resistance (R) = R brick + Rwood + R insulator …………………………… 1 marks

R total = Lbick

𝐾𝑏𝑟𝑖𝑐𝑘 +

Lwood

𝐾𝑤𝑜𝑜𝑑 + 2(

Lbick

𝐾𝑏𝑟𝑖𝑐𝑘)

R total = 0.15

0.70 +

0.012

0.18 + 2(

0.15

0.70) ……………………………2 marks

R total = 0.214 + 0.066 + 0.4286 = 0.708 m2

o

C/ W

Q = ∆T

𝑇𝑜𝑡𝑎𝑙 𝑡ℎ𝑒𝑟𝑚𝑎𝑙 𝑟𝑒𝑠𝑖𝑠𝑡𝑎𝑛𝑐𝑒 =

21−(−15)

0.708 = 50.8 W/ m

2

…………………………….2 marks

R insulator = 2(R brick)

0.4286 = Linsulator

𝐾𝑖𝑛𝑠𝑢𝑙𝑎𝑡𝑜𝑟 =

0.07

𝐾𝑖𝑛𝑠𝑢𝑙𝑎𝑡𝑜𝑟 ………………..1 mark

= 0.4286 × 0.07 = kinsulator

= 0.03 W/ mo

C…………………………..2 marks

2.

Black Body: The black body is defined as a body that absorbs all radiation that falls on its surface. A

black body is a hypothetical body that completely absorbs all wavelengths of thermal radiation incident

on it. Such bodies do not reflect light, and therefore appear black if their temperatures are low enough

so as not to be self-luminous. All blackbodies heated to a given temperature emit thermal radiation.

1.5 marks

Natural (or free) convection: If the fluid motion is caused by buoyancy forces that are induced by

density differences due to the variation of temperature in the fluid.

1.5 marks

B

1 2

Q = ∆𝑇

𝑅 0.5 mark

R total = 0.08

1.04 +

0.13

0.69 = 0.26532 m

2 o

C/W 1 mark

Q = 325

0.26532 = 1224 W /m

2

1 marks

If the the rate of heat loss is decreased by 250 W /m2

therefore the outer layer rate of heat loss

provided by plaster.

1224 W /m2

- 250 W /m2

= 974. 9 W /m2

1 mark

Q = ∆𝑇

𝑅

974.9 = 340

𝑅 R plaster =

340

974.9 = 0.3487 m

2 o

C/W 1 mark

Thermal conductivity K of plaster = 𝑙

𝑅 =

340

0.3487 = 0.1434 W/m

o

C 2 marks

C.

Ф = 𝜎𝜀A (T1

4

– T2

4

) (1 mark)

Ф = 0.7 × 56.7 × 10-9

× (5234

- 3034

) (1 mark)

Ф = 2635 (2 marks)

D

Q = hA∆T 1 mark

A = 4 m × 2 m = 8 m2

1 mark

∆T = 50o

C, h = 55 W/m2o

C

Q = 55 × 8 × 50

Q = 22,000 W 2 marks

3a.

Extraction is a process that selectively dissolves one or more of the mixture components into an

appropriate solvent (2 Mks)

Salting out is a process whereby an inorganic salts, such as NaCl, are dissolved in water to reduce the

solubility of the organic compound in the aqueous layer. This makes the organic compound to

preferentially dissolve in the non-polar layer (2 Mks)

Extract is the solution of the dissolved compound of interest (1 Mks)

Extent of Reaction: The extent of a reaction (x) is defined as the change in the number of moles of

any reaction compound (reactant or product) due to the chemical reaction divided by its

stoichiometric coefficient (2 Mks)

1 2

b. For optimum selectivity of compound of interest the following are necessary:

i. the compound must be more soluble in the second solvent than in the first solvent (1 Mk)

ii. the impurities must be insoluble in the second solvent (1 Mk)

iii. the two selected solvents must be immiscible, or not soluble in one another (1 Mk)

c. A batch reactor has no stream continuously flowing into or out of it and is charged with the

reactants in a pure form or in a solution, heated to the reaction temperature and left for the

reaction to occur for a specific time while in continuous reactors an inlet and an outlet stream

flow continuously. Also, continuous reactors normally operate at steady state (4½ Mks)

d. Schematic representation of a double pipe heat exchanger (3 Mks)

4a. Types of heat exchanger:

(i) Shell and tube

(ii) Flat plate

(iii) Finned tubes (3Mks)

b. The heat transfer rate for a differential length is ∂q= U(Th-Tc) ∂A (1) (½ Mk)

Having assumed that the outer surface of the exchanger is perfectly insulated, the energy gained by

the cold fluid is equal to that given up by the hot fluid.

Energy given up by hot fluid: ∂q= -GhCph ∂Th (2) (½ Mk) Energy gained by cold fluid: ∂q=GcCph∂Tc (3) (½ Mk) Making ∂Th and ∂Tc subject of the formula in both eq (2) & (3)

∂Th= -∂q

GhCph

∂Tc= ∂q

GcCpc

Taking their differences,

(4) (½ Mk)

1 2

Substituting eq (1) into eq (4) gives:

(5) (½ Mk)

Rearranging eq (5)

(6) (½ Mk)

Letting ∂A tend to zero and integrating from one end of the heat exchanger to the other we obtain:

(7)

Solving the energy balance on each fluid for Cp gives:

(8) (1Mk)

Substituting eq (8) into eq (7)

(9) (1Mk)

Or, in terms of the difference in end temperatures

(10)

Thus, the heat transfer rate is equal to ( i.e. q)

(11) (1Mk)

Where the log mean temperature difference is

(12) (1Mk)

c. Mathematical expression:

d. Hot fluid Cold fluid

Thi = 220o

C Tci = 100o

C

Tho = 170o

C Tco = 140o

C

1 2

= ( Tho- Tco) – (Thi- Tci)

ln [(Tho-Tco)/(Thi-Tci)]

= (170-140)-(220-100)

ln[(170-140)-(220-100)

= 30-120

ln(30/120)

= - 90

ln(0.25)

= -90

-1.3862

= 64.93o

C

= (Tho-Tci)-(Thi-Tco)

ln[(Tho-Tci)/Thi-Tco)]

= (170-100)-(220-140)

ln[(170-100)/(220-140)]

= -10

-0.1335

=74.91 o

C

(8½ Mks)

5. a. (i) What is dimension?

It has been found from experience that everyday engineering quantities can all be expressed in

terms of a relatively small number of dimensions. These dimensions are length, mass, time and

temperature. For convenience in engineering calculations, force is added as another dimension.

(3 marks)

(ii) Write the symbol for expressing dimensions.

Dimensions are represented as symbols by: length [L], mass [M], time [t], temperature [T]

and force [F]. Note that these are enclosed in square brackets: this is the conventional way of

expressing dimensions. (2.5 marks)

b. Derive the dimension for the following:

Force = [F]

Heat-transfer coefficient = [F] [L]-1 [t]

-1 [T]

-1

Power = [F] [L] [t]-1

1 2

Energy = [F] [L]

Momentum = [M] [L] [t]-1

(5 marks)

c. What is the difference between fluid statics and fluid dynamics?

The study of fluids in motion is called fluid dynamics while the study of fluids at rest is called

fluid statics. (2 marks)

d. State the fundamental equation of fluid pressure and explain each terms

P = Zρg where P is pressure; Z is depth; ρ is density and g is acceleration due to gravity (2 marks)

e. Convert the following:

µ = [0.01 g/cm.s] 1kg/1000g (100 cm/m)

= 0.001 kg/m.s (3 marks)

[17.5

marks]

6. a. Define the following terms:

(i) Filtration

It may be define as a process of separation of solids from a fluid by passing the same

through a porous medium that retains the solids but allows the fluid to pass through.

(ii) Drying

Drying is one of the oldest methods of preserving food. Today the drying of foods is

still important as a method of preservation. Dried foods can be stored for long periods

without deterioration occurring. The principal reasons for this are that the

microorganisms which cause food spoilage and decay are unable to grow and multiply

in the absence of sufficient water and many of the enzymes which promote undesired

changes in the chemical composition of the food cannot function without water.

(5 marks)

b. Differentiate between surface and depth filtration

Surface Filtration

The size of particles retained is slightly higher than the mean pore size of medium.

Mechanical strength of filter medium is less, unless it is made of stainless steel.

It has low capacity.

The size of particles retained is more predictable.

Equipment is expensive because ancillary equipment such as edge clamps is required.

Ex. Cellulose membrane filter.

Depth Filtration

The size of particles retained is much smaller than the mean pore size of medium.

1 2

Mechanical strength of filter medium is high.

It has high capacity.

The size of particles retained is less predictable.

Equipment is cheaper because ancillary equipment is not required.

Ex. Ceramic filters and sintered filters.

(3 marks)

c. (i)

(4.5 marks)

(ii) (2.5 marks)

(iii) (2.5 marks)

[17.5

marks]

1 2

COURSE CODE: CHM 315 COURSE TITLE: UNIT OPERATIONS / HEAT TRANSFER

Section 1

1a. State the two (2) forms of heat exchanger. (2 Mk)

b . With the aid of annotated diagram briefly explain the temperature profile of a counter-

current flow heat exchanger. (3 Mk)

c. The temperature difference between the hot and cold fluids, varies along the length of

the heat exchanger, in the light of this, derive the Log mean temperature difference for a co-

current flow. (7 Mk)

d. Consider the following parameters given below:

Hot fluid Cold fluid

Thi= 220oC Tci= 100oC

Tho= 170oC Tco= 140oC

Calculate the for both counter-current flow and co-current flow. (8 Mk)

2a. Oil having a specific heat capacity of 6.4 KJ kg-1 K enters a single-pass counter flow heat

exchanger at the rate of 4.5 kg/s and a temperature of 410 K. It is to be cooled to 350 K. Water

(Cpc = 3.5 KJ kg-1 K) is available to cool the oil at a rate of 5.1 kg/s and a temperature 280 K.

Determine the surface area required if the overall heat- transfer coefficient is 230 Wm-2. K.

(10 Mks)

b. The composition of air often given in terms of only the two principal species in the gas

mixture

O2 , YO2 = 0.21

N2 , YN2 = 0.79

Determine the mass fraction of both Oxygen and Nitrogen. 3 moles of the gas mixture is

considered as basis for the calculation.

1 2

(Molecular weight of O2 and N2 are 0.032Kg/mol and 0.028Kg/mol respectively). (7 Mks)

c. State the three (3) common types of heat exchanger. (3 Mks)

3a. Types of heat exchanger:

(i) Shell and tube

(ii) Flat plate

(iii) Finned tubes 3Mks

3b.

According to the counter-current diagram above both the Thi and Tci flows in opposite direction in the heat

exchanger. The expressions for and respectively differs from that of co-current exchanger.

3 Mks

3c. The heat transfer rate for a differential length is ∂q= U(Th-Tc) ∂A (1) ½ Mk

Having assumed that the outer surface of the exchanger is perfectly insulated, the energy gained by

the cold fluid is equal to that given up by the hot fluid.

Energy given up by hot fluid: ∂q= -GhCph ∂Th (2) ½ Mk

Energy gained by cold fluid: ∂q=GcCph∂Tc (3) ½ Mk

Making ∂Th and ∂Tc subject of the formula in both eq (2) & (3)

1 2

∂Th= -∂q

GhCph

∂Tc= ∂q

GcCpc

Taking their differences,

(4) ½ Mk

Substituting eq (1) into eq (4) gives:

(5) ½ Mk

Rearranging eq (5)

(6) ½ Mk

Letting ∂A tend to zero and integrating from one end of the heat exchanger to the other we obtain:

(7)

Solving the energy balance on each fluid for Cp gives:

(8) 1 Mk

Substituting eq (8) into eq (7)

(9) 1Mk

1 2

Or, in terms of the difference in end temperatures

(10)

Thus, the heat transfer rate is equal to ( i.e. q)

(11) 1 Mk

Where the log mean temperature difference is

(12) 1 Mk.

d. Hot fluid Cold fluid

Thi = 220oC Tci = 100oC

Tho = 170oC Tco = 140oC

= ( Tho- Tco) – (Thi- Tci)

ln [(Tho-Tco)/(Thi-Tci)]

= (170-140)-(220-100)

ln[(170-140)-(220-100)

= 30-120

ln(30/120)

= - 90

ln(0.25)

= -90

-1.3862

= 64.93oC

1 2

= (Tho-Tci)-(Thi-Tco)

ln[(Tho-Tci)/Thi-Tco)]

= (170-100)-(220-140)

ln[(170-100)/(220-140)]

= -10

-0.1335

=74.91oC

4a.

4b. Solution:

1 2

q = -4.5 × 6.4 ( 350- 410)

q = -28.8(- 60)

q = 1728W

1728 = 5.1 × 3.5 (Tco – 280)

1728 = 17.85 (Tco- 280)

1728 = 17.85Tco – 4998

6726 = 17.85Tco

Tco = 6726

17.85

Tco = 376.81oC

= (350 – 280) – (410 – 376.81)

ln (70/33.19)

= 70 – 33.19

2.1091

= 36.81

2.1090

= 17.45oC

= 1728

230 × 17.45

1 2

= 1728

4013.5

A = 4.305 m2 10 Mks

4c. Solution:

O2 present = (3 mol) (0.21)= 0.63 mol

= 0.63 mol) × (0.032 kg)

mol

= 0.0201 kg ½ Mk

N2 present = (3 mol) × (0.79)= 2.37 mol

= (2.37 mol) × (0.028 kg)

mol

= 0.0663 kg ½ Mk

Total mass present= 0.0201 + 0.0663

= 0.0864 kg 1 Mk

wO2= 0.0201 kg = 0.23 2 Mk

0.0864 kg

wN2= 0.0663 kg = 0.77 2 Mk

0.0864 kg

Section 2

1. (a) With the aid of an annotated diagram, briefly explain the operation of a counter current

heat exchanger. Hence, define the term temperature cross. (5 Marks)

1 2

(b) A heavy lubricating oil (Cp = 3090 J/KgK) is cooled by allowing it to exchange energy

with water in a small heat exchanger. The oil enters and leaves the heat exchanger at 560K

and 510K respectively, and flows at a rate of 0.9 kg/s. Water at 292K is available in

sufficient quantity to allow 0.304 kg/s to be used for cooling purposes. Determine the

required heat transfer area for (a) counter flow (b) parallel flow operation.

The overall heat transfer coefficient may be taken as 350 W/m2K. (10 Marks)

(c) Which of the exchangers in question 1 (b) above is more efficient; state your reason?

(2 Marks)

(d) Explain the term mass transfer and state the two modes of mass transfer. (3 Marks)

2. (a) The temperature difference between the hot and cold fluids , varies along the length

of the heat exchanger, in the light of this, derive the Log mean temperature difference for a

co-current flow. (8 Marks)

(b) State the three (3) common types of heat exchanger. (3 Marks)

(c) A counter-flow double-pipe heat exchanger is used to cool a hot process fluid using

water. The process fluid flows at 20 kgs-1 and is cooled from 125 oC to 65 oC. The water

flows counter-currently to the process fluid, entering at 38 oC and leaving at 70 oC.

Assuming no heat loses. Calculate the required flow rate for the cooling water. Neglecting

the tube wall curvature, calculate the required area for heat exchange.

The specific heat for water is 4.2 kJ kg K-1 and that of the process fluid is 3.4 kJ kg K-1.

The process fluid side film heat exchanger coefficient is 2500 Wm-2 K-1 and the cooling

water side heat transfer coefficient is 1200 Wm-2 K-1. The tube wall thickness is 3mm and

the thermal conductivity 220 Wm-1 K-1. (9 Marks)

Solution to Section 2 1. (a)

1 2

(2 Mks)

Figure 1: schematic representation of a counter-current heat exchanger

In a counter-current-flow heat exchanger the fluids flows in opposite direction as shown

in figure 1 above, the fluids (i.e cold and hot) are forced to flow along the heat

exchanger by either pumps or fans. (1.5 Mk)

Temperature cross: this occurs in counter-current-flow configuration, when the outlet

temperature of the cold fluid exceeds the outlet temperature of the hot fluid outlet.

Tco>Tho. (1.5 Mk)

(b) First calculate the heat transfer rate for the hot fluid (i.e. oil)

q = -GhCph (Tho-Thi)

= - (0.9) (3090) (510-560)

= - (0.9) (3090) (-50)

= 139050 W (1 Mk)

Calculate the temperature of water (i.e. cold fluid) at the outlet region

q = GcCpc (Tco-292)

139050 = 0.304 × 4177 × (Tco-292)

1

139050 = 12969.808 (Tco-292)

1269.808 1269.808

1

Tco-292 = 109.504744

Tco = 109.504744 + 292

Tco = 401.504744 K (2 Mk)

1 2

For a counter-flow configuration, lm is calculated as

lm = (510-292) – (560-401.5)

ln ( 218 ÷ 158.5)

lm = 218 – 158.5

ln ( 218 ÷ 158.5)

lm = 59.5

ln 1.3754

lm = 59.5

0.3187

lm = 186.6959523

lm ≈ 186.7 K (2 Mks)

q =UA lm

139050 = 350 × A × 186.7

1

139050 = 65345 A

65345 65345 1

A = 2.1279

A ≈ 2.13 m2 (1 Mk)

For a parallel-flow configuration lm is calculated as

1 2

lm = (510-401.5) – (560-292)

ln (108.5 ÷ 268)

lm = 108.5-268

ln (0.404850746)

lm = - 159.5

- 0.904236807

lm = 176.3918464

≈ 176.4 K (2 Mks)

q = UA lm

139050 = 350 × A × 176.4

1

139050 = 61740 A

61740 61740 1

A = 2.252186589

A ≈ 2.25 m2 (1 Mk)

Drawings showing the temperature profile for each exchanger in question 1 (b). (2 Mks)

(c) The counter-current flow is more efficient because of its higher lm value. This is in

response to the fact that the exchanger performed better as far as cooling is concern also, the area

for the parallel flow is higher than that for a counter-current flow. (2 Mks)

(d) Mass transfer: In a situation whereby a system contains two or more components whose

concentrations vary from point to point, there is a natural tendency for mass to be transferred

thereby minimizing the concentration differences within the system. (2 Mks)

Modes of mass transfer transports:

1 2

(i) Molecular mass transfer

(ii) Convective mass transfer. (1 Mk)

2(a) The heat transfer rate for a differential length is ∂q= U(Th-Tc) ∂A (1) ½ Mk

Having assumed that the outer surface of the exchanger is perfectly insulated, the energy gained by

the cold fluid is equal to that given up by the hot fluid.

Energy given up by hot fluid: ∂q= -GhCph ∂Th (2) ½ Mk

Energy gained by cold fluid: ∂q=GcCph∂Tc (3) ½ Mk

Making ∂Th and ∂Tc subject of the formula in both eq (2) & (3)

∂Th= -∂q

GhCph

∂Tc= ∂q

GcCpc

Taking their differences,

(4) ½ Mk

Substituting eq (1) into eq (4) gives:

(5) ½ Mk

Rearranging eq (5)

1 2

(6) ½ Mk

Letting ∂A tend to zero and integrating from one end of the heat exchanger to the other we obtain:

(7) 1 Mk

Solving the energy balance on each fluid for Cp gives:

(8) 1 Mk

Substituting eq (8) into eq (7)

(9) 1Mk

Or, in terms of the difference in end temperatures

(10)

Thus, the heat transfer rate is equal to ( i.e. q)

(11) 1 Mk

Where the log mean temperature difference is

(12) 1 Mk.

(b). Types of heat exchanger:

(i) Shell and tube

(ii) Flat plate

(iii) Finned tubes 3Mks

1 2

(c). The heat given up by hot process fluid is equal to

q = -GhCph (Tho-Thi)

= -20 × 3.4 × 103 × (65-125)

= -20 × 3.4 × 103 × (-60)

= 4080000 W (1 Mk)

Since the heat given up by hot fluid is equal to the heat gained by the cold fluid, thus, the flow rate

required by cold water: q = GcCpc (Tco-Tci)

Gc = q

Cpc (Tco-Tci)

Gc = 4080000

4200 × (70-38)

Gc = 4080000

134400

Gc = 30.35714286 kg s-1

Gc ≈ 30.36 kg s-1 (2 Mks)

(1 Mk)

lm = (65-38) – (125-70)

ln (27÷ 55)

lm = 27-55

ln (0.49090909)

lm = - 28

1 2

- 0.711496319

lm = 39.35368217

lm ≈ 39.35 oC (1 Mk)

Neglecting the effect of fouling, the overall heat transfer coefficient is

U = 1 = 1

Rtot (1/hf) + (x/k) + (1/hw) (1 Mk)

U = 1

(1/2500) + (0.003/220) + (1/1200)

U = 802 W m-2 K-1 (1 Mk)

The heat transfer area is therefore equal to

A = q

U lm

A = 4080000

802 × 39.35

A = 4080000

31558.7

A = 129.2828919

A ≈ 129.28 m2 (2 Mk)

Section 3

1. Define the following terms (a) solid extraction (b) Liquid extraction (4 Mks)

2. Explain the differences between leaching and the washing of filtered solid. (4 Mks)

3. List four (4) different types of extraction equipments. (4 Mks)

1 2

4. Explain the following terms (a) Extraction battery (b) Diffusion battery. (8 Mks)

5. Describe the batchwise extraction operation of a mixer-settler. (6 Mks)

6. Show how the solvent phase and the solid phase go from one stage to another, in ideal stage

continuous counter-current leaching cascade. (4 Mks)

7. Species A and B are examples of two close-boiling mixture, suggest the best separation

technique for the separation of these species. State the reason for your answer.

(2 Mks)

8. Briefly explain the factors necessary for equilibrium to be attained. (3 Mks)

Solution to Section 3

H. (i) Solid extraction: This is used to dissolve soluble matter from its mixture with an

insoluble solid.

(ii) Liquid extraction: Is used to separate two miscible liquids by the use of a solvent

that preferentially dissolves one of them. (4 Mks)

I. In leaching, the amount of soluble material removed is often greater than in ordinary

filtration washing, and the properties of the solids may change considerably during the

leaching operation. Coarse, hard, or granular feed solids may disintegrate into pulp or

mush when their content of soluble material is removed. (4 Mks)

J. (i) Mixer-settler (ii) Sprayed and packed extraction tower (iii) Baffled towers

(iv) Agitated tower extractors (v) Perforated-plate towers. Any four (4) (4 Mks)

K.

(a) Extraction battery: When the rate of solution is so rapid that one passage of solvent

through the material is sufficient, fresh solvent is fed to the tank containing the solid

that is most nearly extracted; it flows through the several tanks in series and is

finally withdrawn from the tank that has been freshly charged. Such a series of tanks

is called extraction battery.

(b) Diffusion battery: In some solid-bed leaching the solvent is volatile, necessitating

the use of closed vessels operated under pressure. Pressure is also needed to needed

to force solvent through beds of some less permeable solids. A series of such

pressure tanks operated with countercurrent solvent flow is known as a diffusion

battery. (8 Mks)

1 2

4a. Batchwise extraction of mixer-settler: The mixer and settler may be the same unit. A

tank containing a turbine or propeller agitator is most common. At the end of the mixing

cycle the agitator is shut off, the layers are allowed to separate by gravity, and extract and

raffinate are drawn off to separate receivers through a bottom drain line carrying a sight

glass. The mixing and settling times required for a given extraction can be determined only

by experiment; 5 min for mixing and 10 min for settling are typical, but both shorter and

much longer times are common. (6 Mks)

4b. The ideal stage counter-current leaching cascade:

Solution of entering solid = Xa

Solution of leaving solid = Xb

Fresh solvent entering the system = Yb

Concentrated solution leaving the system = Ya

(4 Mks)

4c . Extraction method of separation is best used for species having close boiling temperature. Due

to the closeness in their boiling point, distillation method becomes inadequate because the distillate

at a particular temperature will constitute a mixture of the two species to be separated.

(2 Mks)

4d. Provided sufficient solvent is present to dissolve all the solute in the entering solid and there is

no adsorption of solute by the solid, equilibrium is attained when the solute is completely dissolved

and the concentration of the solution so formed is uniform.

(3 Mks)

Section 4

1. A furnace wall is constructed of a 3-cm thick flat steel plate, with a firebrick insulation 30-cm

thick on the inside, and rock wool insulation 6-cm thick on the outside. The inside surface

temperature of the firebrick insulation is 700℃. If the temperature of the outer surface of the rock

wool insulation is 50℃, what is the heat flux through the wall? Thermal conductivities are as follows:

firebrick, 0.1 W m_1 K_1; steel, 40 W m_1 K-1; rock wool, 0.04 W m_1 K_1.

2. Calculate the specific heat of a substance which absorbs 4.6 joules of heat when its mass of 2.0

kg increases with respect to the change in temperature of 60℃?

1 2

3. 1.50g of ethanol was burnt in a combustion apparatus, 500g of water vapour raised the

temperature from 20 0C to 39.5 0C. Calculate the enthalpy of combustion of ethanol. Given the

specific heat capacity of solution as 4.18 kJ/g.

4(a). The level of water in a storage tank is 4.7 m above the exit pipe. The tank is at atmospheric

pressure and the exit pipe discharges into the air. If the diameter of the exit pipe is 1.2 cm, what is

the mass rate of flow through this pipe? (g = 9.8 ms-2, ƿH2O = 1000 Kgm-3)

(b) List the classes of grinding equipment.

(c) Derive the two equations for size reduction.

5(a). Water for a processing plant is required to be stored in a reservoir to supply sufficient working

head for washers. A constant supply of 1.2 m3min-1 is pumped to the reservoir which is 22 m above

the water intake. The length of the pipe is about 120 m and the available galvanized iron piping is 15

cm. What will be the nature of the flow? (Density of water at 20 0C = 998 Kgm-3; viscosity of H2O

= 0.001 NSm-2)

(b) Differentiate between a hammer mill and a plate mill.

(c) State the basic assumption for the theories of size reduction.

6. Calculate the pressure drop along 170 m of 5 cm diameter horizontal steel pipe through which

olive oil at 20°C is flowing at the rate of 0.1 m3 min-1.

SOLUTION

1. Using: 𝑞𝑦 = 𝑇0− 𝑇3

(𝑏1

𝑘1 ⁄ )+ (

𝑏2𝑘2

⁄ )+ (𝑏3

𝑘3⁄ )

= T0− T3

∑ k)(b/ =

∆T

∑ /(b

k)

𝑞𝑦 = 700 − 50

(0.30.1⁄ ) + (0.03

40⁄ ) + (0.060.04⁄ )

= 973 − 323

4.501=

∆T

∑ /(b

k)

= 144.4 W m_2

To calculate temperature of the steel

From equation 2, 𝑇(1)= 𝑇0 − 𝑞𝑦 (

𝑏1𝑘1

⁄ )

𝑇(1)=700 − 144.4(0.30.1⁄ )

= 266.74℃

This is the inner surface temperature of the steel. The steel outer surface temperature is then given

by 𝑇(2)= 𝑇1 − 𝑞𝑦 (

𝑏2𝑘2

⁄ )

𝑇(2)= 266.74 − 144.4 (0.0340⁄ ) = 266.63 ℃

1 2

There is only a small temperature difference across the steel wall. This is due to the relatively small

thermal resistance offered by this layer (b2/k2), approximately 0.02% of the total resistance. The

wall is effectively at a uniform temperature of 267℃.

2. Quantity of Heat Needed (Q) = 4.6 joules

Mass of the substances (m) = 2.0 kg

Change in Temperature (ΔT) = 60℃

Specific Heat(c) = ?

Solution

Specific Heat(c) = 𝑄

𝑚Δ𝑇

Specific Heat(c) = 0.0383333333 J/kg. ℃

Specific Heat(c) = 0.0383 J/kg/℃

3. Q = MCΔT

= 500 𝑋 4.18 𝑋 19.5

= 40.8 𝑘𝐽

No of moles of ethanol, 𝜂 = 𝑚𝑎𝑠𝑠

𝑚𝑜𝑙𝑎𝑟 𝑚𝑎𝑠𝑠

= 1.5

46 = 0.0326 moles

Molar enthalpy of combustion = 𝑄

𝜂

= 40.8

0.0326

= 1250 kJ/mol

Since the enthalpy of combustion is exothermic, ∆𝐻𝑐= −1250 𝑘𝐽/𝑚𝑜𝑙

4. Z = 4.7m, Diameter = 1.2cm = 0.012m

V = (2gz) ½

V = (2 x 9.81 x 4.7) ½

= 9.6ms-1

Volume of flow = velocity x area

Area=(∏/4)D2

= (∏/4)(0.012)2

1 2

= 1.13 x 10-4m2

Volumetric flow rate = 9.61 x 1.13 x 10-4

= 1.08 x 10-3m3s-1

Mass flow rate = volumetric flow rate x density

= 1.08 x 10-3 x 1000

= 1.08Kgs-1

(b). Crushers and grinders

(c). Kick’s law

dE/dL = KLn

where dE = differential energy required,

dL = change in a typical dimension,

L= magnitude of a typical length dimension and

K, n = Constants.

Kick assumed that the energy required to reduce a material in size was directly proportional

to the size reduction ratio dL/L. This implies that n is equal to -1. If

K = KKfc

where KK is called Kick's constant and fc is called the crushing strength of the material, we

have:

dE/dL = KKfcL-1

which, on integration gives:

E = KKfc loge(L1/L2) (Kick’s equation)

5 (a) Diameter = 15cm = 0.15m

Cross sectional area in pipe = 𝜋/4(𝐷)2

= 𝜋/4(0.15)2

= 0.0177m2 Volume of flow V = 1.2 m3min-1

= 1.2/60 = 0.02 m3S-1

Velocity = 𝑣𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑓𝑙𝑜𝑤

𝑎𝑟𝑒𝑎

= 0.02/0.9177

=1.13 m/s

To determine the nature of flow, Reynolds number will be determined;

Re = Dvƿ/μ

0.15 ×1.13 998

0.001

= 169161

1 2

The nature of flow is turbulent since the Reynolds number is > 4000

(5b) In a hammer mill, swinging hammer heads are attached to a rotor that rotates at high speed

inside a hardened casing. The principle is illustrated below;

Grinders: (a) hammer mill, (b) plate mill

(5c). These theories depend upon the basic assumption that the energy required to produce a

change dL in a particle of a typical size dimension L is a simple power function of L.

Diameter of pipe = 0.05 m,

Area of cross-section A Type equation here.

= (Type equation here. /4)D2

= /4 x (0.05)2

= 1.96 x 10-3 m’

6. Viscosity of olive oil at 20°C = 84 x 10-3 Ns m-2 and density = 910 kg m-3,

and velocity = (0.1 x 1/60)/(1.96 x 10-3) = 0.85 m s-1,

(Re) = (Dv / )

= [(0.05 x 0.85 x 910)/(84 x 10-3)]

= 460

= 4615

f = 0.03.

1 2

f = 16/(Re) = 16/460 = 0.03 (from chart)

And so the pressure drop in 170 m, from eqn. (3.17)

Pf = (4f v2/2) x (L/D)

= [4 x 0.03 x 910 x (0.85)2 x 1/2] x [170 x 1/0.05]

= 1.34 x 105 Pa

= 134 kPa.

= 13.4kpa

1 2

COURSE CODE: CHM 317 COURSE TITLE: ENVIRONMENTAL CHEMISTRY What is soil pollution?

Answer: It is the introduction of substances, biological organisms, or energy into the soil, resulting in

a change of the soil quality, which is likely to affect the normal use of the soil or endangering public

health and the living environment. Worded differently, soil pollution or contamination is the presence

of man-made chemicals or other alteration to the natural soil environment.

Q. (2) What are some of the factors that may affect the rate of bioremediation?

Answer:

a) Temperature favorable for organisms

b) Water available (near field capacity)

c) Nutrients (N, P, K) in adequate supply

d) C:N ratio of material < 30:1

e) Material added is similar to naturally occurring organic material

f) Oxygen in sufficient quantity.

Q. (3) Define Chapman cycle and write the four reactions for the chapman cycle?

Answer:

Chapman cycle describes the 4 steps by which O2 is photolysed to produce O and O3, which can be

further photolysis to recover O2

Chapman Chemical Reactions:

O2 + hv 2O (< 242nm)

O + O2 + M O3 + M* (generates heat)

O3 + hv O2 + O (1D) (usually followed by rapid deactivation to 3P)

O + O3 2O2

Q. (4) a) What efforts have been implemented to counter the effects of ozone depletion?

1 2

Are there any signs that these efforts are working?

b) Advise on ways of reducing exposure to UV- radiation.

Answer:

a) Actions to Counter the Effects of Ozone Depletion

In 1987 a treaty known as the “Montreal Protocol” was signed by 46 countries, including the United

State of America to halt ozone depletion. The treaty was enforced in 1989. By 1996, the use of CFCs,

halons and CCl4 was phased out in developed countries while developing countries had up to 2010 to

do same. Also developed countries were scheduled to phase out production of HCFCs by 2030, while

developing countries were given up to 2040 to follow suit.

Signs of Recovery

There have been some signs of recovery

In 1997 satellite showed a decline of several known ozone-depleting gases

Satellite images show some slowing down of ozone loss But recovery is very slow.

b) Ways of reducing exposure to UV- radiation.

Stay out of the sun, especially between 10 a.m. and 3 p.m.

Do not expose yourself to the sun if you are taking antibiotics or birth control pills.

When in the sun, use a sunscreen with a protection factor of at least 15.

When in the sun, wear protective clothing and sunglasses that protect against UV-A and UV-

B radiation.

Protect the skin against the solar radiation using skin creams with sun protection factor (SPF).

When properly used, an SPF of 15 protects the skin from 93 percent of UVB radiation

Use lip balm with SPF.

Sunglasses with 100% UV block.

Q. (5) a) Provide the chemical formula for CFC–12.

b) List 3 known examples each of the following:

i) Volatile Organic Compounds (VOCs)

ii) Organic Hazardous Air Pollutants (OHAPs) iii) Impacts from climate change

iv) Inorganic Air Pollutants

1 2

v) Ozone-depleting chemicals.

Answer:

(a) CFC-12= Using the rule of 90, 12+90 = 102 Therefore, the number of carbon atom is 1,

the number of hydrogen atom is 0, and

the number of fluorine atom is 2.

To find the number of chlorine atom present, use the formulae: 2n+2= no of non- carbon atoms

Where n= no of carbon atoms

Therefore 2 (1) +2 = 0+2

4 =2

No of chlorine atoms= 4-2= 2

Therefore the chemical formulae for CFC-12= CF2Cl2

b) i) Volatile Organic Compounds: methane, benzene & chlorofluorocarbons

ii) Organic Hazardous Air Pollutants (OHAPs): formaldehyde, ethylene glycol & methanol,

methyl bromide iii) Impacts from climate change: temperature rise, massive biodiversity loss, economic and social

instabilities

iv) Inorganic Air Pollutants: lead, arsenic, mercury

v) ozone-depleting chemicals. Chlorofluorocarbons, methyl bromide & halocarbons

Q. (6) With the aid of a diagram shows the following:

i) The layers of the atmosphere

ii) Pollutants exerting local and global effects

1 2

Q. (7) Discuss the sources, effects and remediation of the following criteria pollutants

in a tabular form:

a) Particulate matter b) Nitrogen dioxide c)Lead

1 2

Pollutants/ Symbols

1.ParticulateMatter

PM 10 /PM 2.5

2. Nitrogen Dioxide

(NO2) Reddish-

brown gas.

3. Lead (Pb)

Sources

Power plant boilers,

steel mills, chemical

plants, unpaved roads

wood-burning stoves,

fireplaces, automobiles &

others.

Fuel combustion in motor

vehicles & industrial

sources.

High temperature burning

combining nitrogen &

oxygen in the air.

Smelters, lead-acid battery

manufacturing, electric arc

furnaces, incineration of

garbage containing lead

products, & use of leaded

gasoline.

Human health effects

(i) Aggravates respiratory

effects like asthma.

(ii) Impair visibility

(iii) heavy metal/ organic

chemical particles, may

cause cancer/ tissue

damage

(i)Respirator irritant.

(ii)Aggravates lung &

heart problems.

(iii)Precursor to ozone &

acid rain.

(iv) Causes brown

discoloration of

atmosphere.

Toxic to the nervous

system, organs, & most

levels of body function

Control method

(i) Pollution control

equipment (wet &

dry scrubbers &

cartridges)

(ii) Electrostatic

precipitators.

i) Exhaust gas

recirculation in

cars, reduction of

combustion.

(ii) Catalytic

converters reduce

90% of NOx.

(i)Phase-out of

leaded gasoline (ii)

use of pollution

control equipment

in industrial plants.

1 2

Q. (8). Highlight the physical properties of water.

Answer

The following are physical properties of water

Dielectric Constant 80

Density (g/cm3) 1.0

Boiling Point (oC) 100

Melting Point (oC) 0

1 2

Specific Heat (J/g) 4.18

Heat of Evaporation (KJ/g) 2.26

Heat of Fusion of Ice (KJ/g) 0.33

Q. (9). Identify the various sources of impurities in wastewater.

Answer

The sources of impurities in wastewater are as listed below:

(i) Floating or suspended solids e.g paper, rags, soil particles

(ii) Colloidal solids eg organic compounds, micro-organisms

(iii) Dissolved solids eg organic compounds, inorganic salts

(iv) Dissolved gases eg hydrogen sulphide

(v) Immiscible liquids eg oils and greases

Q. (10). Discuss the principles involved in aerobic and anaerobic oxidation of

wastewater.

Answer

Most organic waste oxidation by bacteria occurs in the presence of oxygen and this

process is called aerobic oxidation. Bacteria oxidize wastes to provide themselves

with sufficient energy needed for the synthesis of complex molecules. When

molecular oxygen is used as the terminal oxidizing agent the waste oxidation is

aerobic. The oxidation process can be represented by this equation:

aerobic

Wastes + Oxygen → Oxidized wastes + New bacteria

Bacteria

1 2

Oxidation can also proceed in the absence of air but in the presence of oxygen- containing compounds

such as SO42-

and SO32- and this process is termed anaerobic oxidation. For example, biological sludge

treatment is an anaerobic process. The solids in processed wastewaters are degraded in two stages by two

groups of anaerobic bacteria. In the first case, the organic compounds are oxidized to fatty acids, mainly

ethanoic acid, by acid forming bacteria. Next, methane producing bacteria convert the acid to methane.

Q. (11). Define the following terms i) mineralization ii) eutrophication

Q. (12). Explain the primary process involved in the chemical treatment of wastewater.

Q. (13). Explain the secondary process involved in the chemical treatment of wastewater.

Q. (14). What are the advantages of chemical methods of wastewater treatment over

biological treatment?

Q. (15) Write short notes on the following:

a) Hydrogen Chlorofluorocarbons (HCFCs)

b) Chlorine in the stratosphere

Q. (16) Describe the origin of stratospheric ozone and the role it plays in protecting life

on Earth.

Q. (17) What are CFCs? Give the chemical formulae of CFC-12, CFC-113 and

HCFC-142b

Q. (18) Explain the potential consequences of ozone depletion and propose three ways

for slowing these changes.

Q. (19) Briefly describe the structure of the atmosphere being sure to include the spheres

and the boundaries between each set of layers.

Q. (20) (a) List six types of water pollutants, and give an example of each.

(b) List the layers of the ocean in order of increasing depth.

1 2

COURSE CODE: CHM 332 COURSE TITLE: INDUSTRIAL INORGANIC CHEMISTRY Tutorial Questions

1. Write the chemical formula of the following ores:

epsom salt, siderite, cryolite, anglesite, corundum, calamine, chalcocite, rutile,

malachite, talc.

Ans.

Epsom salt MgSO4.7H2O

Siderite FeCO3

Cryolite Na3AlF6

Anglesite PbSO4

Corundum Al2O3

Calamine ZnCO3

Chalcocite Cu2S

Rutile TiO2

Malachite CuCO3.Cu (OH)2

Talc. Mg2(Si2O5).Mg(OH)2

2. Describe extraction of sodium form its ore.

3. Differentiate between pyrometallurgy and hydrometallurgy.

Ans.

Pyrometallurgy uses high temperatures to convert ore into raw metals, while

hydrometallurgy employs aqueous chemistry for the same purpose. The methods used

depend on the metal and their contaminants.

4. (a) Write short note on the following terms:

(i) Metal (ii) Mineral (iii) Ore (iv) Gangue (v) Open-cut mining (vi) Shaft Mining

(vii) Dredging.

Ans.

1 2

Ore - a large deposit of a mineral which is economically viable to mine and refine (e.g. iron

ore or haematite)

Gangue - the waste material of an ore from the crushing process

Open – cut mining: This mining involves digging a huge hole in the ground (e.g. Mining

of iron, copper, uranium)

Shaft Mining: involves mining in tunnels (e.g. coal, gemstones)

Dredging is the mixing large amounts of water with the crushed ore to allow the heavier

minerals to settle to the bottom (e.g. tin, mineral sands)

5. (a) With the aid of chemical equations, describe Contact process.

(b) Following wet acid process method, describe preparation of H2SO4

Ans.

a. 1. S (s) + O2 (g) → SO2 (g)

2. 2 SO2 (g) + O2 (g) 2 SO3 (g)

(in presence of V2O5)

3. H2SO4 (l) + SO3 (g)→ H2S2O7 (l)

4. H2S2O7 (l) + H2O (l) → 2 H2SO4 (l)

b. S(s) + O2(g) → SO2(g)

2H2S + 3O2 → 2H2O + 2SO2 (−518 kJ/mol)

2SO2 + O2 → 2SO3 (−99 kJ/mol)

SO3 + H2O → H2SO4(g) (−101 kJ/mol)

H2SO4(g) → H2SO4(l) (−69 kJ/mol)

6. Based on the equations of the reaction, sketch industrial layout for the production of H2SO4

through Contact process.

7. a. Give detailed account in the smelting process to extract lead from its mineral ore.

b. Explain the reason why gallium is significantly smaller than expected from its position

within the Group of elements. Explain.

Ans.

1 2

a. In the smelting process to extract lead from its mineral ore (Galena PbS),

The galena is first roasted or reacted with oxygen to remove the sulphur.

Lead Sulphide (Galena) + Oxygen →Lead Oxide + Sulphur Dioxide

PbS + O2 →PbO + SO2

Then the lead oxide is reacted with the carbon in coke to obtain the refined lead.

Lead Oxide + Carbon

Lead + Carbon Dioxide

PbO + C → Pb + CO2

b. The rationale for this may be attributed to an analogous effect in the lanthanide

contraction observed for the lanthanides and the 3rd row of transition elements. In multi-

electron atoms, the decrease in radius brought about by an increase in nuclear charge is

partially offset by increasing electrostatic repulsion among electrons. In particular, a

“shielding effect” results when electrons are added in outer shells, electrons already

present shield the outer electrons from nuclear charge, making them experience a lower

effective charge on the nucleus.

8. (a) List major chemical component of Portland cement and their source

(b) Use sketch diagram to represent cement manufacturing process

9. Describe the following process in manufacturing of Portland cement.

i. Grinding ii. Pyroprocessing

Ans.

i. Grinding.

The feed to the grinding process is proportioned to meet a desired chemical composition.

Typically, it consists of 80% limestone, 9% silica, 9% flyash, and 2% iron ore. These

materials are ground to 75 micron in a ball mill. Grinding can be either wet or dry. The

“raw meal” from dry milling is stored in a homogenizing silo in which the chemical

variation is reduced. In the wet process, each raw material is fed with water to the ball mill.

This slurry is pumped to blending tanks and homogenized to correct chemical composition.

The slurry is stored in tanks until required.

ii. Pyroprocessing

In the preheater, the raw meal from the mill is heated with the hot exhaust gas from the kiln

before being fed into the rotary kiln to form a semi-product known as clinker. The ash from

1 2

fuel used is also absorbed into the clinker. The particle size range for clinker is from about

2 inches to about 10 mesh.

10. (a) What do you understand by the term “Dye”

(b) Write chemical equations to justify the manufacture of phosphate fertilizers from

H2SO4.

11. Write the basic chemical reactions in the kiln during manufacturing of Portland

cement.

Ans.

Basic chemical reactions are: evaporating all moisture, calcining the limestone to produce free

calcium oxide, and reacting the calcium oxide with the minor materials (sand, shale, clay, and

iron). This results in a final black, nodular product known as“clinker” which has the desired

hydraulic properties.

T oC Reaction

100 Evaporation of water

>500 Evolution of combined water from the clay

900 Crystallization of

amorphous dehydration

products,

Carbon dioxide evolution from CaCO3

900 -1200 Main reactions between lime and clay to form clinker

1 2

1250 - 1280 Beginning of liquid formation

1280 -1550 Further liquid formation and final cement formation

12. (a) i. Describe Harber process in detail.

ii. List six stages involve in the modern method of preparing ammonia.

13(a) The catalyst used in the reforming reaction is deactivated (poisoned) by sulphur.

Describe how desulphurization can be used to solve this problem.

(b) List five uses of ammonia.

Ans.

a. Desulphurization

Hydrocarbon feedstocks contain sulphur in the form of H2S, COS, CS2 and mercaptans. The

catalyst used in the reforming reaction is deactivated (poisoned) by sulphur. The problem is solved

by catalytic hydrogenation of the sulphur compounds as shown in the following equation:

H2+RSH = RH + H2S(g)

The gaseous hydrogen sulphide is then removed by passing it through a bed of zinc oxide where it

is converted to solid zinc sulphide:

H2S+ZnO = ZnS+H2O

b. Uses of ammonia

Ammonia is the basis from which virtually all nitrogen-containing products are derived. The main

uses of ammonia include the manufacture of:

• Fertilizers ((ammonium sulfate, diammonium phosphate, urea)

• Nitric acid

• Explosives

• Fibres, synthetic rubber, plastics such as nylon and other polyamides

• Refrigeration for making ice, large scale refrigeration plants, air-conditioning units in buildings

and plants

• Pharmaceuticals (sulfonamide, vitamins, etc.)

1 2

• Pulp and paper

• Extractive metallurgy

• Cleaning solutions

14. Explain Primary (Steam) Reforming process in the manufacture of ammonia

15. Draw a flow chart to show Modern Method of Manufacturing Ammonia

Ans.

16. a List the major raw materials for manufacturing of ammonia

b. List the 10 uses of H2SO4

17. Write short note on each of the following stage of manufacturing ammonia

i. Purification. Ii. Ammonia Converter. Iii. Ammonia Separation iv. Ammonia Storage

Ans.

Purification.

1 2

The carbon dioxide is removed either by scrubbing with water, aqueous monoethanolamine

solution or hot potassium carbonate solution. CO is an irreversible poison for the catalyst used in

the synthesis reaction, hence the need for its removal The synthesis gas is passed over another

catalyst bed in the methanator, where remaining trace amounts of carbon monoxide and dioxide are

converted back to methane using hydrogen.

CO+3H2 = CH4+H2O

CO2+4H2 = CH4+2H2O

O2 + 2H2 2H2O

Note that the first equation is the opposite of the reformer reaction.

Ammonia Converter.

After leaving the compressor, the gaseous mixture goes through catalyst beds in the synthesis

converter where ammonia is produced with a three-to-one hydrogen-to-nitrogen stoichiometric

ratio. However, not all the hydrogen and nitrogen are converted to ammonia. The unconverted

hydrogen and nitrogen are separated from the ammonia in the separator and re-cycled back to the

synthesis gas compressor and to the converter with fresh feed. Because the air contains argon

which does not participate in the main reactions, purging it minimizes its build up in the recycle

loop.

Ammonia Separation

The removal of product ammonia is accomplished via mechanical refrigeration or

absorption/distillation. The choice is made by examining the fixed and operating costs. Typically,

refrigeration is more economical at synthesis pressures of 100 atm or greater. At lower pressures,

absorption/distillation is usually favoured.

Ammonia Storage

Ammonia is stored in tanks as a refrigerated liquid. Some ammonia is used directly as a fertilizer.

Most ammonia is converted in downstream processes to urea (46% nitrogen) or ammonium nitrate

(34% nitrogen) for use as fertilizer.

18. a. Write 5 Chemical properties of H2SO4

b. Use equation to represent the production of phosphate fertilizer from H2SO4

19. a. State the composition of Portland Cement

b. With the aid of chemical equations, describe reaction of N2 with

i. H2 ii. O2 iii. Mg iv. CaC2

Ans.

a. Portland cement are mainly calcium silicates and calcium aluminates. It has this

composition, 65% CaO, 20% SiO2, 5% Al2O3.

1 2

b.

20. a. Describe extraction of iron from its ore

b. List two examples each of Low Car

1 2

COURSE CODE: CHM 333 COURSE TITLE: INSTRUMENTAL METHOD OF ANALYSIS

1. Differentiate between qualitative and quantitative methods of analysis.

2. Explain 4 advantages that instrumental methods of analysis have over the

classical methods of analysis.

3. Justify the continued use of the classical methods of analysis.

4. Identify 7 factors that should be considered in the choice of analytical methods.

5. Explain the three types of atomic excitation (change) that could result when an atom absorbs a photon.

6. Write short notes on the following:

(i) ultraviolet spectroscopy (ii) infra-red spectroscopy.

(iii) Write short notes on X-ray methods.

7. A zinc solution of unknown concentration gave a polarographic diffusion current of 5.00 μA. A 5.00

ml portion of 0.0010 M Zn2+ was added to 10.00 ml of the unknown solution, and the polarogram was

taken again giving a diffusion current of 12.00 μA. What is the concentration of Zn2+ in the unknown

solution?

8. The polarographic diffusion current of a 3.0 x 10-3 M NiCl2 solution is 15 μA. If the volume of the solution

is 15 ml, how long will be required to reduce 1% of the Ni2+ ions in the cell?

9. Phenol can be titrated coulometrically by anodic generation of bromine as a titrant. The titration

reaction is

How many mg of phenol is present in a sample which requires 300 s to titrate with

a current of 25 mA?

10. Compare and contrast fluorescence and phosphorescence spectroscopy.

11. State the Franck-Condon Principle, and illustrate it with a named compound.

12. Illustrate fluorescent quenching with the Jablonski diagram, state the mathematical expression

involved and clearly define all the terms.

1 2

13. What are the basic principles of the following electroananlytical methods?

(i) coulometry (ii) voltammetry

14. Briefly describe the basic instrumentation of the following with a diagram

(i) controlled potential coulometry (ii) dropping mercury electrode

15. What are the parameters necessary for consideration in order to minimize electrolysis time in

controlled potential coulometry?

16. Identify all the annotated parts of the polarogram

17. Briefly describe with the aid of a diagram

(i) differential refractometer (ii) critical angle refractometer

18. What are the basic requirements of optical rotatory dispersion (ORD) instrument?

19. Briefly explain the following voltammetric methods

(i) Potential step voltammetry (ii) Normal pulse voltammetry

(iii) Differential pulse voltammetry

20. Silver is to be deposited from a 0.010 M AgNO3 solution which also contains 2.0 M HNO3. The

electrolysis cell includes two smooth platinum electrodes, and its resistance is 0.25 ohm. The

overvoltages at the anode and cathode are 0.85 and 0.05 V respectively. What voltage must be

applied in order to obtain an initial current of 0.75 A?

1 2

TUTORIAL ANSWERS

QUESTION 1

Qualitative method of analysis identifies the components or constituents of a sample while quantitative

method of analysis determines the amount of each of the components present in a sample.

QUESTION 2

1. Instrumental methods are usually faster than the classical methods of analysis.

2. Instrumental analyses are normally applicable at minute concentrations that may not be determined

by classical methods.

3. The instrumental methods are usually automated unlike the classical methods.

4. Instrumental methods can analyze a large number of samples compared to the classical methods.

QUESTION 3

Despite the many advantages of instrumental methods, the continued use of classical methods is justified

by these four major factors:

1. Apparatus required for classical procedures are cheaper and readily available in laboratories but many

instruments are expensive.

2. Instrumental methods require calibrations which could be time-consuming.

3. Although an instrumental method is ideal for a large number of routine determinations, an occasional,

non-routine analysis is often simpler by classical methods.

4. To obtain accurate results with instrumental methods, the reagents still need careful weighing,

measuring and preparation of standard solutions. Classical analysis provides the essential training and

experience.

1 2

QUESTION 4

1. The type of analysis required: elemental, routine or occasional.

2. Problems arising from the nature of the material to be investigated e.g. radioactive substances,

corrosive substances, substances affected by water.

3. Possible interference from components of the material other than those of interest.

4. The concentration range to be investigated.

5. The accuracy required.

6. The facilities available, particularly the instruments.

7. The time required to complete the analysis

8. The number of analysis of similar type (duplicate or triplicate) which have to be performed.

9. The necessity for destructive or non-destructive methods before analysis.

QUESTION 5

The three types of atomic excitation (changes) that could result when an atom absorbs a photon are:

1. Electronic excitation (change) - where electrons are moved from one energy

level to another.

2. Vibrational excitation (change) - where there is a change in the average space

separating the nuclei of neighbouring atoms.

3. Rotational excitation (change) - where there is the rotation of a chemical dipole

in a molecule.

QUESTION 6

(i) Ultraviolet Spectroscopy (UV)

1 2

When a molecule absorbs radiant energy in the UV region, the valence electrons in the molecule are raised

to higher energy orbits. The result is that fairly broad absorption bands are normally observed in the UV

region. Many substances, mainly organics, which do not absorb in the visible range and are therefore

colourless do absorb in the UV range of 190 – 400. The UV region is of more limited general usage, although

it is particularly suitable for the selective measurement of low concentrations of organic compounds such as

benzene ring containing compounds or unsaturated straight-chain compounds containing a series of double

bonds. Major application is in the determination of a single component of various samples at a single

wavelength. The amount of radiation absorbed by a substance is a function of its concentration or thickness

and the length of the path of the radiation through the sample. A plot of absorbance, A, against

concentration, C, gives a straight line. The system is governed by Beer’s law. At high concentration, the law

is not obeyed and a bend is observed.

(ii) Infrared Spectroscopy (IR)

The use of infrared spectra for quantitative determinations depends upon measuring the intensity of either

the transmission or absorption of the infrared radiation at a specific wavelength. Beer’s law also applies to

IR. The infrared region of the electromagnetic spectrum may be divided into three main sections:

1. Near-infrared 0.8-2.5 µm (12 500 - 4 000 cm-1)

2. Middle-infrared 2.5 - 50 µm (4000 - 200 cm-1)

3. Far-infrared 50 – 1000 µm (200 - 10 cm-1)

Infrared absorption spectra can be used to identify pure compounds or impurities. The IR instrument can

either have a single or a double beam. IR radiation is of low energy and its absorption by a molecule causes

changes in the vibrational or rotational energy of the molecule. As a result of the complexity of IR spectra, it

is highly unlikely that two different compounds will have identical curves.

(iii) X-ray Methods

X-ray method is a special technique in quantitative analysis. X-rays are produced when high-speed electrons

collide with a solid target. These X-rays are called primary X-rays. They arise because the electron beam may

1 2

displace an electron from the inner electron shells of an atom in the target; the lost electron is then replaced

by an electron from an outer shell and energy is emitted as X-rays. When a beam of short-wavelength

primary X-rays strikes a solid target, a similar mechanism as earlier described will cause the target material

to emit X-rays at wavelengths characteristic of the atoms involved. This is called secondary or fluorescence

radiation. X-ray fluorescence analysis is a rapid process which finds application in metallurgical laboratories,

the processing of metallic ores and in the cement industry.

QUESTION 7

𝐶𝑢 =𝑖1𝑣𝐶𝑠

𝑖2 + (𝑖2 − 𝑖1)𝑉

where i1 is the diffusion current of the wave obtained with V ml of unknown solution, and i2 is the diffusion

current obtained after addition of v ml of a known solution whose concentration (in the desired units) is Cs.

i1= 5.00 μA; V = 10.00 ml; 𝑣 = 5.00 ml; Cs = 0.0010 M; i2 = 12.00 μA; Concentration of Zn2+ in the unknown,

Cu= ?

∴ 𝐶𝑢 =(5×10−6)5.00×0.0010

12×10−6 + (12×10−6 − 5×10−6)10.00

𝐶𝑢 =2.5×10−8

8.2×10−5

= 3.05×10−4 M

QUESTION 8

id = 15 μA; V of Ni2+ = 15 ml; C of Ni2+ = 3.0 X 10-3 M; t = ? for reducing 1% of Ni2+

𝑁𝑁𝑖(𝐼𝐼) = (3×10−3)15

1000= 4.5×10−5 𝑚𝑜𝑙

1% 𝑜𝑓 4.5×10−5𝑁𝑖2+ = 4.5×10−7 𝑚𝑜𝑙

𝑄 = 𝐼𝑡; 𝑄 = 𝑛𝐹𝑁

1 2

𝐼𝑡 = 𝑛𝐹𝑁

15×10−6𝑡 = 2×96,500×4.5×10−7

𝑡 =2×96,500×4.5×10−7

15×10−6

𝑡 = 5.79 𝑠

QUESTION 9

Given that I = 25 X 10-3 mA; t = 300 s; 1F = 96,500 Cmol-1; mg of phenol = ?

The half reaction for the anodic generation of bromine

3𝐵𝑟− − 3 − 3𝑒− → 3𝐵𝑟

𝐼𝑡 = 𝑛𝐹𝑁

25×10−3×300 = 3×96,500×𝑁

𝑁 =25×10−3×300

3×96,500

𝑁 =7.5

289,500= 2.59×10−5 𝑚𝑜𝑙

𝑓𝑜𝑟𝑚𝑢𝑙𝑎 𝑚𝑎𝑠𝑠 𝑜𝑓 𝐶6𝐻5𝑂𝐻 = 94 𝑔𝑚𝑜𝑙−1; 𝑁 =𝑔

𝑀

2.59×10−5 =𝑔

94 ∴ 𝑔 = 2.44×10−3 ≡ 2.44 𝑚𝑔 (4½ marks)

QUESTION 10

Fluorescence: is self-emission of photon at singlet excited state. The emission arises from excited state

orbital electron having opposite spin to a ground state electron with which it is paired. Therefore, relaxation

of the excited state electron is spin-allowed. The emission is usually rapid and the life span of fluorescence

1 2

is about 10-8 s-1 or 10 ns. So it requires sophisticated electronics and optical device to sense and analyse this

emission.

Meanwhile, phosphorescence is self-emission from an excited triplet state orbital in which the electron has

the same orientation as the paired ground state electron, and as such transition to the ground state is spin

forbidden. That is why phosphorescence emission is usually slow about 103-10 s-1. So that phosphorescence

lifetimes are typically milliseconds to seconds. Even longer lifetimes are possible, as is seen from "glow-in-

the-dark" toys.

QUESTION 11

Franck-Condon Principle: All electronic transitions are vertical, that is they occur without change in the

position of nuclei. When transition of S1- S0 takes place in the absorption spectrum of molecule, the emission

spectrum is usually a mirror image of the absorption spectrum. Eg. polynuclear aromatic hydrocarbon, PAH.

1 2

QUESTION 12

Fluorescent Quenching

Collisional quenching: quenching occurs as a result of decrease in the intensity of fluorescence e.g Collisional

quenching

1 2

Decrease in intensity is given by

𝐹0

𝐹= 1 + 𝐾[𝑄] = 1 + 𝐾𝑞𝜏0[𝑄]

K is the Stern-Volmer quenching constant; Kq is the bimolecular quenching constant;

𝜏0 is the unquenched lifetime; [Q] is quencher’s concentration.

QUESTION 13

(i) Coulometry

Coulometric methods of analysis are based on an exhaustive electrolysis of the analyte. By exhaustive we

mean that the analyte is quantitatively oxidized or reduced at the working electrode or reacts quantitatively

with a reagent generated at the working electrode. There are two forms of coulometry: controlled-potential

1 2

coulometry, in which a constant potential is applied to the electrochemical cell, and controlled-current

coulometry, in which a constant current is passed through the electrochemical cell.

(ii) Voltammetry

Voltammetric techniques allow the determination of those components in a solution that can be

electrochemically oxidized or reduced. In these methods a potential is applied to the sample via a conductive

electrode, called the working electrode, which is immersed in the sample solution. The potential, which

serves as the driving force, is scanned over a region of interest. If at a particular potential a component of

the solution is oxidized or reduced, then a current will flow at the working electrode. The potential at which

this occurs identifies the component, and the amount of current produced is proportional to the

concentration of that component in the solution.

QUESTION 14

(i) Instrumentation of Controlled Potential Coulometry

The potential in controlled-potential coulometry is set using a three-electrode potentiostat. Two types of

working electrodes are commonly used: a Pt electrode manufactured from platinum-gauze and fashioned

into a cylindrical tube, and Hg pool electrode. The large overpotential for reducing H3O+ at mercury makes

it the electrode of choice for analytes requiring negative potentials. For example, potentials more negative

than –1 V versus the SCE are feasible at Hg electrode (but not at a Pt electrode), even in very acidic solutions.

The ease with which mercury is oxidized, however, prevents its use at potentials that are positive with

respect to the SHE. Platinum working electrodes are used when positive potentials are required. The

auxiliary electrode, which is often a Pt wire, is separated by a salt bridge from the solution containing the

analyte. This is necessary to prevent electrolysis products generated at the auxiliary electrode from reacting

with the analyte and interfering in the analysis. A saturated calomel or Ag/AgCl electrode serves as the

reference electrode.

1 2

(ii) Instrumentation of dropping mercury electrode

The mercury working electrode has found widespread use due to the high overpotential for hydrogen-ion

reduction (1.2 V) and the reproducibility of the metal surface. The original mercury electrode was the

dropping-mercury electrode (dme) in which the growth of the drop and the drop lifetime are controlled by

gravity. A stream of mercury is forced through a glass capillary (0.05 to 0.08 mm i.d.) under the pressure of

an elevated reservoir of mercury connected to the capillary by flexible tubing. Mercury issues from the

capillary at the rate of one drop every 2 to 5 s. The constant renewal of the drop surface (as it expands)

eliminates poisoning effects. Mercury forms amalgams with many metals and thereby lowers their reduction

potentials. The diffusion current assumes a steady value immediately and is reproducible.

QUESTION 15

Factors to be considered in minimizing electrolysis time in controlled potential coulometry

𝑡𝑒 = −1

𝑘ln(10−4) =

9.21

𝑘

From this equation we can see that increasing k leads to a shorter analysis time. For this reason, controlled-

potential coulometry is carried out in small-volume electrochemical cells, using electrodes with large surface

areas and with high stirring rates. A quantitative electrolysis typically requires approximately 30–60 min,

although shorter or longer times are possible

QUESTION 16

I = electrode reaction begins; II = half wave potential, E1/2;

1 2

III = diffusion current, id; IV = limiting current;

V = residual current; VI = discharge of the supporting electrolyte

QUESTION 17

(i) The differential refractometer, a light beam is transmitted through a partitioned cell that refracts the

beam at an angle that depends on the difference in refractive index between the sample liquid in one part

and a standard liquid in the other.

(ii) In the critical-angle refractometer, the light incident on the surface of the solution changes sharply from

reflected to transmitted light at a critical angle.

DIFFERENTIAL REFRACTOMETER

1 2

QUESTION 18

Basic requirements of Optical rotatory dispersion instrument

When ordinary white light, which is vibrating in all possible planes, is passed through a Nicol prism (the

polarizer), two polarized beams of light are generated. One of these beams passes through the prism, while

the other beam is reflected and does not interfere with the plane polarized beam, the one that is coincident

with the propagation axis of light. If the beam of plane-polarized light is also passed through a second Nicol

prism (the analyzer), it can be transmitted only if the second Nicol prism has its axis oriented so that it is

parallel to the plane-polarized light. If its axis is perpendicular to that of the plane-polarized light, the light

will not pass through. The sample cell is placed between the two Nicol prisms. If an optically active substance

is in the sample tube, the light is deflected. The analyzer prism is rotated to permit maximum passage of

light and is then said to be aligned. The angle of rotation (in degrees) is measured. The standard wavelength

is that of the green mercury line at 546.1 nm.

CRITICAL ANGLE REFRACTOMETER

1 2

QUESTION 19

(i) Potential Step Voltammetry

Potential step methods are based on the measurement of current as a function of time after applying a

potential to the working electrode. These methods seek to optimize the ratio of faradaic to charging current

by applying a sudden change (pulse) in applied potential and sampling the faradaic current just before the

drop is detached but after the capacitative current has largely decayed. The potential step methods

discriminate against the charging current by delaying the current measurement until close to the end of the

pulse.

(ii) Normal Pulse Voltammetry

In normal pulse voltammetry a series of square-wave voltage pulses of successively increasing magnitude is

superimposed upon a constant dc voltage signal. Near the end of each pulse (perhaps 50 ms in duration)

and before the drop is dislodged, the current is sampled for perhaps 17 ms.

(iii) Differential Pulse Voltammetry

In differential pulse voltammetry a series of potential pulses of fixed but small amplitude (10 to 100 mV) is

superimposed on a constant dc voltage ramp near the end of the drop life and after the drop has attained

the bulk of its growth. The current is sampled immediately before applying the potential pulse (perhaps 17

ms) and again (for 17 ms) just before the drop is dislodged.

QUESTION 20

Applied Voltage = overvoltages + IR

𝐴𝑝𝑝. 𝑉 = 0.85 + 0.05 + 𝐼𝑅

𝐴𝑝𝑝. 𝑉 = 0.9 + (0.75×0.25)

𝐴𝑝𝑝. 𝑉 = 1.09 𝑉 (4 𝑚𝑎𝑟𝑘𝑠)

1 2

COURSE CODE: CHM 334

COURSE TITLE: APPLIED SPECTROSCOPY 1. What is the wavelength (in metres) of an electromagnetic wave whose frequency is 1.61 x 1012 s1?

2. What is the frequency (in reciprocal second) of an electromagnetic radiation with a wavelength of 1.03 cm?

3. Calculate the energy of one photon of yellow light with a wavelength of 598 nm.

4. Calculate and compare the energy of a photon of wavelength 3.3 m with that of wavelength of 0.154 nm. Use the electromagnetic spectrum to identify the region to which each belongs.

5. Ethylene glycol, the substance used in automobile antifreeze, is composed of 38.7% C, 9.7% H, and

51.6% O by mass. Its molar mass is 62.1 g mol1. What is the empirical formula of ethylene glycol?

What is its molecular formula?

6. Use the formula to calculate the double bond equivalent for the following: C8H10

C6H12O6

C3H4Cl2

C8H7N

C13H10

C4H9NO

7. A 5.0 x 104 mol dm3 solution of Br2 in CCl4 absorbed 64% of the incident light when placed in a 2 cm cell at a wavelength where CCl4 does not absorb. Calculate the molar absorption coefficient of Br2.

8. Naphthalene (C10H8) absorbs around 310 nm in solution. The absorbance of a 1.0 x 103 mol dm3 solution is 0.29 in a 1.0 cm cell. Find the molar absorption coefficient.

1 2

At what concentration will the absorbance be 1.00?

9. Identify the auxochromes in the following compound:

O

O

Br

O2N

OH

OH

COOH

10. Four derivatives of cholesterol, A, B, C, and D have the following spectral properties:

IR: strong absorption near 1680 cm1

UV: max 230 nm, 10,700

IR: strong absorption near 1680 cm1

UV: max 241 nm, 16,600

IR: no absorption near 1680 cm1

UV: max 234 nm, 20,000

IR: no absorption near 1680 cm1

UV: max 315 nm, 19,800

Which structure corresponds to which derivative?

11. A solution was prepared using 0.0010 g of an unknown steroid (of molecular mass around 255) in 100 cm3 of ethanol. Some of this solution was placed in a 1-cm cell, and the UV spectrum was measured.

1 2

The solution was found to have max 235 nm, A = 0.74.

Compute the value of the molar absorptivity at 235 nm.

Which of the following compounds might give this spectrum?

12. Calculate the max for compounds with the following structures:

13. Which has a lower characteristic stretching frequency, the C―H or C―D bond? Explain briefly.

14. Which has a lower characteristic stretching frequency, the C═O bond or the C―O bond? Explain briefly.

15. The region of the IR spectrum which contains the most complex vibrations(600 – 1400 cm1) is called the _________________ region of the spectrum.

16. What effect does conjugation typically have on the frequency at which absorption by C―C occurs?

A) Conjugation decreases the frequency at which absorption occurs.

1 2

B) Conjugation increases the frequency at which absorption occurs.

C) Conjugation does not affect the frequency at which absorption occurs.

17. How does the O―H stretch in the IR spectrum of a carboxylic acid differ from the O―H stretch of an alcohol?

18. How could IR spectroscopy be used to distinguish between the following pair of compounds?

i. CH3OCH2CH3 and CH3CH2CH2OH ii. CH2═CHCH2CH(CH3)2 and CH3CH2CH2CH(CH3)2

19. Which compound would be expected to show intense IR absorption at 1680 cm1?

OH

OCH3

O

A B C

20. Predict and match which compound has the following IR:

Cyclohexanemethanol, Butanal, Propanoic acid, Hexane.

1 2

B.

21. How many double bond equivalents are there in each of the following molecules? (a) C6H12O6 (b) C8H9NO2 (c) C27H46O (d) C5H5N5

22. a. Explain how IR spectroscopy could be used to distinguish between these two compounds. Be as specific as possible.

b. Explain how 1H-NMR spectroscopy could be used to distinguish between the two compounds in problem a. Be as specific as possible.

23. a. Predict the approximate chemical shift position for each of the different hydrogens in the 1H-NMR spectrum of this compound.

b. Predict the multiplicity of each of the signals in the 1H-NMR spectrum of the compound in problem a.

c. Predict the integral for each of the signals in the 1H-NMR spectrum of the compound in problem a.

1 2

24. How many peaks would appear in the 13C-NMR spectrum of this compound? Give the approximate positions of these peaks.

25. Write short note on the phenomenon “Nuclear Overhauser Effect (NOE)”

26. The spectrum of a dibutyl phthalate ester is as shown in the Figure below. Assign the various protons to the appropriate chemical shift

27. Show a structure for the compound (C6H10) that has the following 13C-NMR spectrum and explain how you arrived at your answer.

1 2

28. The IR and 1H-NMR spectra of an unknown compound with the formula C5H10O2 are shown below. Interpret the IR spectrum, explaining which peaks indicate which functional groups. Explain the chemical shift, integral and multiplicity of each peak in the NMR spectrum. Finally, show the structure of the unknown compound. It should be clear from your explanations how you arrived at your answer.

29. The IR and 1H-NMR spectra of an unknown compound with the formula C5H10O are shown below. Interpret the IR spectrum, explaining which peaks indicate which functional groups. Explain the chemical shift, integral and multiplicity of each peak in the NMR spectrum. Finally, show the structure of the unknown compound. It should be clear from your explanations how you arrived at your answer.

1 2

30. a. Show the structure of the ion that is responsible for the peak at m/z = 43 in the mass spectrum of 2-heptanone.

b. Show the structure of the ion that is responsible for the peak at m/z = 99 in the mass spectrum of 2-heptanone.

31. Explain how these isomers could be distinguished by mass spectroscopy.

1 2

ANSWERS TO SOME SELECTED QUESTIONS

13. C―D bond; heavier atoms vibrate more slowly.

14. Stronger bonds are generally stiffer, thus requiring more force to stretch or compress them. The

C―O bond is theweaker of the two and hence has the lower stretching frequency.

15. Fingerprint region.

16. (A) Conjugation decreases the frequency at which absorption occurs.

17. Because of the unusually strong hydrogen bonding in carboxylic acids, the broad O―H stretch is

shifted to about3000 cm1 centered on top of the usual C―H absorption. This broad O―H absorption gives a characteristic overinflated shape to the peaks in the C―H region.

18. (i) O―H stretch at 3300 cm1

(ii) C―C stretch around 1640 cm1; vinylic C―H stretch above 3000 cm1.

19. Compound C

20. (a) Propanoic acid

(b) Cyclohexanemethanol.

21. (a) 1DBE (b) ……………….. (c) 10 DBE (d)……………………

23. (a). i. somewhat downfield from 3.0 because it is a CH2 and is further shifted downfield by the nearby carbonyl group

ii. somewhat downfield from 2.0 because it is a CH2 and is further shifted downfield by the nearby Cl

iii. slightly downfield from 3.7 because it is a CH2 group

iv. slightly downfield from 0.9 because of the nearby oxygen

1 2

(b).

i. split by 2 so triplet

ii. split by 2 so triplet

iii. split by 3 so quartet

iv. split by 2 so triplet

25. Nuclear Overhauser Effect (NOE): A third type of information available from NMR comes from the nuclear overhauser enhancement or NOE. This is a direct through-space interaction of two nuclei. Irradiation of one nucleus with a weak radio frequency signal at its resonant frequency will equalize the populations in its two energy levels. This perturbation of population levels disturbs the populations of nearby nuclei so as to enhance the intensity of absorbance at the resonant frequency of the nearby nuclei. This effect depends only on the distance between the two nuclei, even if they are far apart in the bonding network, and varies in intensity as the inverse sixth power of the distance. Generally the NOE can only be detected between protons (1H nuclei) that are separated by 5 ˚A or less in distance. These measured distances are used to determine accurate three-dimensional structures of proteins and nucleic acids.

29. The DU for C5H10O is 1.

From the IR spectrum we see sp3C―H bonds (3000 - 2850 cm-1) and a C═O near 1710

cm1. This fits for a ketone. The carbonyl group of the ketone accounts for the one degree of unsaturation.

In the 1H-NMR spectrum, the triplet at 2.4 (area 2) is due to two Hs split by two Hs. The

two Hs doing this splitting must be the two at 1.6. There are six lines in this signal, so

these two Hs must split by five, the two Hs at 2.4 and the three Hs at 0.9. Based on this

we can identify the presence of a propyl group. The singlet (area 3) at 2.1 must be due to an isolated methyl group. Putting these fragments together, the compound is

1 2

COVENANT UNIVERSITY

CANAANLAND, KM 10, IDIROKO ROAD

P.M.B 1023, OTA, OGUN STATE, NIGERIA.

TITLE OF EXAMINATION: B.Sc DEGREE EXAMINATION

COLLEGE: SCIENCE & TECHNOLOGY

DEPARTMENT: CHEMISTRY

SESSION: 2015/2016 SEMESTER: ALPHA COURSE

CODE: CHM 356 CREDIT UNIT: 2

COURSE TITLE: METALLURGY & METAL FABRICATION

INSTRUCTION: ANSWER ALL QUESTIONS IN SECTION A AND ANY TWO FROM

SECTION B TIME: 2 HOURS

SECTION A

A

a) What is a 'die' in metal forming operation. (2 Marks)

b) Differentiate between 'bulk deformation' and 'sheet metalworking' in metal

forming operations. (4 Marks)

c) What is 'abrasive processes' in material removal processes. (2 Marks)

d) Machining is important commercially and technologically for several reasons. Discuss

four (4) of the reasons. (8 Marks)

e) Give two (2) disadvantages associated with machining and other material

removal processes. (4 Marks)

B

a) Discuss the following as they relate to conventional machining:

1 2

(i) turning (ii) drilling (iii) milling (6 Marks)

b) Discuss the following as they pertain to sheet metalworking processes:

(i) bending (ii) drawing (iii) shearing [NOTE: Discuss with appropriate

DIAGRAM]. (9 Marks)

c) List five (5) methods of corrosion prevention that you know. (5 Marks)

Question Two

SECTION B

a) Explain physical metallurgy. (5 Marks)

b) Briefly discuss the five (5) basic steps in making sand castings. (10 Marks)

Question Three

a) Explain the following: (i) cutting tool (ii) machine tool and (iii) cutting fluid.

(6 Marks)

b) Discuss three (3) important properties required in a cutting tool material. (6 Marks)

c) List three (3) cutting tool materials. (3 Marks)

1 2

Question Four

a) Show with the aid of a ‘family tree’, the classification of metal forming operations.

(5 Marks)

b) Explain the following as they pertain to drilling:

(i) reaming (ii) tapping (iii) counterboring

(iv) countersinking (v) centering. (10 Marks)

Question Five

a) Differentiate between 'Bessemer converter' and 'Thomas converter' (4 Marks)

b) Discuss two (2) types of drill presses that you know. (5 Marks) c)

Give three (3) disadvantages each of:

(i) hot forming (ii) warm forming (iii) cold forming (6 Marks)

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COVENANT UNIVERSITY

CANAANLAND, KM 10, IDIROKO ROAD

P.M.B 1023, OTA, OGUN STATE, NIGERIA.

TITLE OF EXAMINATION: B.Sc DEGREE EXAMINATION

COLLEGE: SCIENCE & TECHNOLOGY

DEPARTMENT: CHEMISTRY

SESSION: 2015/2016 SEMESTER: ALPHA

COURSE CODE: CHM 356 CREDIT UNIT: 2

COURSE TITLE: METALLURGY & METAL FABRICATION (MARKING SCHEME)

INSTRUCTION: ANSWER ALL QUESTIONS IN SECTION A AND ANY TWO FROM

SECTION B TIME: 2 HOURS

SECTION A

A

a) What is a 'die' in metal forming operation. (2 Marks)

Die is a tool that brings about deformation in metal workpiece by applying stresses that

exceed the yield strength of the matal.

b) Differentiate between 'bulk deformation' and 'sheet metalworking' in metal forming

operations. (4 Marks)

Bulk deformation processes are generally characterized by significant deformation and

massive shape changes, and the surface area-to-volume of the work is relatively small

whereas sheet metal working processes are forming and cutting operations performed

on metal sheets, strips and coils. The surface area-to-volume ratio of the starting metal is

high.

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c) What is 'abrasive processes' in material removal processes. (2 Marks)

Abrasive process is a type of material removal processes which mechanically remove

material by the action of hard, abrasive particles. The process includes grinding, honing,

lapping and superfinishing.

d) Machining is important commercially and technologically for several reasons. Discuss

four (4) of the reasons. (8 Marks)

1. Variety of work materials. Machining can be applied to a wide variety of work

materials. Virtually all solid metals can be machined. Plastics and plastic composites can

also be cut by machining.

2. Variety of part shapes and geometric features. Machining can be used to create any

regular geometries, such as flat planes, round holes, and cylinders. By introducing

variations in tool shapes and tool paths, irregular geometries can be created, such

as screw threads and T-slots. By combining several machining operations in sequence,

shapes of almost unlimited complexity and variety can be produced.

3. Dimensional accuracy. Machining can produce dimensions to very close tolerances.

Some machining processes can achieve tolerances of +0.025 mm (+0.001 in), much more

accurate than most other processes.

4. Good surface finishes. Machining is capable of creating very smooth surface finishes.

Roughness values less than 0.4 microns (16 μ-in.) can be achieved in conventional

machining operations. Some abrasive processes can achieve even better finishes.

e) Give two (2) disadvantages associated with machining and other material removal

processes. (4 Marks)

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1. Wasteful of material. Machining is inherently wasteful of material. The chips generated

in a machining operation are wasted material. Although these chips can usually be

recycled, they represent waste in terms of the unit operation.

2. Time consuming. A machining operation generally takes more time to shape a given part

than alternative shaping processes such as casting or forging.

B

a) Discuss the following as they relate to conventional machining:

(i) turning (ii) drilling (iii) milling (6 Marks)

1. Turning: In turning, a cutting tool with a single cutting edge is used to remove material

from a rotating workpiece to generate a cylindrical shape. The speed motion in turning

is provided by the rotating workpart, and the feed motion is provided by the cutting tool

moving slowly in a direction parallel to the axis of rotation of the workpiece. -(2 marks)

2. Drilling: This is used to create a round hole. It is accomplished by a rotating tool that has

two cutting edges. The tool is fed in a direction parallel to its axis of rotation into the

workpart to form the round hole. -------------(2 marks)

3. Milling: In milling, a rotating tool with multiple cutting edges is moved slowly relative

to the material to generate a plane or straight surface. The direction of the feed motion

is perpendicular to the tool’s axis of rotation. The speed motion is provided by the rotating

milling cutter. -------------(2 marks)

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b) Discuss the following as they pertain to sheet metalworking processes:

(i) bending (ii) drawing (iii) shearing [NOTE: Discuss with appropriate

DIAGRAM]. (9 Marks)

1. Bending: It involves straining of a metal sheet or plate to take an angle along a

straight axis.

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2. Drawing: Drawing refers to the forming of a flat metal sheet into a hollow or concave

shape, such as a cup, by stretching the metal. A blankholder is used to hold down the

blank while the punch pushes into the sheet metal.

3. Shearing: Shearing process involves cutting rather than forming. The operation cuts the

work using a punch or die.

c) List five (5) methods of corrosion prevention that you know. (5 Marks)

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1. Proper material selection (1mk) 2. Alteration of environment (1mk) 3. Proper Design of material/structure (1mk) 4. Cathodic protection (1mk) 5. Anodic protection (1mk) 6. Coatings (1mk)

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Question Two

SECTION B

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a) Explain physical metallurgy. (5 Marks)

The physical metallurgist studies the behavior of metal and their alloy, in order to characterize

their internal structure, or microstructure; to understand how the microstructure influences the

properties of the metal: and to develop new and improved alloys. Physical metallurgists are

responsible for developing new aluminum alloys that reduce weight and improve the fuel

efficiency of aircraft, automobile steels that offer excellent properties without expensive heat

treatments, nickel super alloys that operate safely at high temperature.

The physical metallurgist develops strengthening mechanisms based on the microstructural

features in the metal. Strength can be increased by controlling the grain size in the metal,

deforming the metal, or adding alloys. Heat treatment of alloyed metals can often significantly

improve its strength or lessen the possibility of METAL FATIGUE. In steels, heat treatment causes

the formation of tiny precipitates of an iron – carbon compound. In a number of metals systems,

heat treatments can produce complicated microstructures that prevent even very large cracks

from growing. These metals, which have high fracture toughness, may allow airplanes to safely

operate with small cracks until the cracks are discovered and repaired.

b) Briefly discuss the five (5) basic steps in making sand castings. (10 Marks)

1. Patternmaking: Patterns are required to make molds. This mold is made by packing some

readily formed plastic material, such as molding sand, around the pattern. When this

pattern is withdrawn, its imprint provides the mold cavity, which is ultimately filled with

metal to become the casting.

2. Coremaking: Cores are forms, usually made of sand, which are placed into mold cavity to

form the interior surfaces of castings. Hence, the void space between the core and mold-

cavity surface is what eventually becomes the casting. Coreboxes are required to make

cores.

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3. Molding: This consists of all operations necessary to prepare a mold for receiving molten

metal. It usually involves placing a molding aggregate around a pattern held within a

supporting frame, withdrawing the pattern to leave the mold-cavity, setting the cores

in the mold cavity and finishing and closing the mold.

4. Melting and Finishing: The preparation of molten metal for casting is referred to as

melting. It is done in a specifically designated area of the foundry, and the molten metal

is transferred to the molding area where it is poured into the mold cavity.

5. Cleaning: This refers to all operations necessary to the removal of sand, scale, and excess

metal from the casting. The casting is separated from the molding sand and transported

to the cleaning department. This is done to improve the surface appearance of the

casting. Defective castings may be salvaged by welding or other repair. The casting,

after inspection for defects and general quality control, is then ready for further

processing e.g. heat treatment, surface treatment or machining.

Question Three

a) Explain the following: (i) cutting tool (ii) machine tool and (iii) cutting fluid.

(6 Marks)

1. Cutting tool is used to accomplish machining operations. It has one or more sharp

cutting edges and is made of materials that are harder than the work materials; the

cutting edge serves to separate a chip from the parent work material.

2. Machine tool is used to hold the workpart, position the tool relative to the work, and

provide power for the machining process at the speed, feed and depth that have been

set. By controlling the tool, work and cutting conditions, machine tools permit part

to be made with great accuracy and repeatability.

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3. Cutting fluid is often applied to the machining operation to cool and lubricate the cutting

tool. It is any liquid or gas that is applied directly to the machining operation to improve

cutting performance. It addresses (i) heat generation at the shear zone and friction zone,

and (ii) friction at the tool-chip and tool-work interfaces.

b) Discuss three (3) important properties required in a cutting tool material. (6 Marks)

1. Toughness: This is the capacity of a material to absorb energy without failing and

usually characterized by the combination of strength and ductility in the material. Tool

materials must possess high toughness to avoid fracture failure.

2. Hot Hardness: It is the ability of a material to retain its hardness at high temperatures.

This is required because of the high temperature environment in which the tool operates.

3. Wear Resistance: All cutting tool materials must be hard. Hardness is the single most

important property needed to resist abrasive wear. However, because of the other tool

wear mechanisms, wear resistance in metal cutting may depend on more than just tool

hardness to include other characteristics such as surface finish on the tool, chemistry of

tool and work materials.

c) List three (3) cutting tool materials. (3 Marks)

1. Plain carbon tool steel

2. High speed steel

3. Cast cobalt alloys

4. Cemented carbides

5. Cermets

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6. Ceramics

7. Synthetic diamonds

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Question Four

a) Show with the aid of a ‘family tree’, the classification of metal forming operations.

(5 Marks)

b) Explain the following as they pertain to drilling:

(i) reaming (ii) tapping (iii) counterboring

(iv) countersinking (v) centering. (10 Marks)

1. Reaming: Reaming is used to slighlty enlarge a hole, to provide a better tolerance on its

diameter, and to improve its surface finish. The tool is called a reamer, and it usually has

straight flutes.

2. Tapping: This operation is performed by a tap and is used to provide internal screw

threads on an existing hole.

3. Counterboring: This operation provides a stepped hole, in which a larger diameter

follows a smaller diameter partially into the hole. A counterbored hole is used to seat bolt

heads into a hole so the heads do not protrude above the surface.

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4. Countersinking: This is similar to counterboring, except that the step in the hole is cone-

shaped for flat head screws and bolts.

5. Centering: This operation drills a starting hole to accurately establish its location for

subsequent drilling with the used of a tool called center drill.

Question Five

a) Differentiate between 'Bessemer converter' and 'Thomas converter' (4 Marks)

Bessemer converter is an acid-lined converter whereas Thomas converter is a basic-

lined converter

b) Discuss two (2) types of drill presses that you know. (5 Marks)

1. Upright drill: This drill type stands on the floor and consists of a table for holding the

workpart, a drilling head with powered spindle for the drill bit, and a base and column

for support.

2. Radial drill: This is a large drill press designed to cut holes in large parts. It has a radial

arm along which the drilling head can be moved and clamped. The head therefore can

be positioned along the arm at locations that are a significant distance from the column

to accommodate large work. The radial arm can also be swivelled about the column to

drill parts on either side of the worktable.

3. Gang drill: It consists basically of two to six upright drills connected together in an in-

line arrangement. Each spindle is powered and operated independently, and they share

a common worktable, so that a series of drilling and related operations can be

accomplished in sequence simply by sliding the workpart along the worktable from one

spindle to the next.

4. Multiple-spindle drill: It comprises several drill spindles connected together to drill

multiple holes simultaneously into the workpart.

c) Give three (3) disadvantages each of:

(i) hot forming (ii) warm forming (iii) cold forming (6 Marks)

i) Hot Forming ---------------(3 marks in all)

1. Unlike in cold working, the shape of the workpart can be significantly altered

2. Lower forces and power are required to deform the metal

3. Metals that usually fracture in cold working can be hot worked

4. No strengthening of the part occurs from work hardening

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5. Strength properties are generally isotropic because of the absence of the oriented

grain structure typically created in cold working

(ii) Warm forming ---------------(3 marks in all)

1. Lower forces and power.

2. More intricate work geometries possible.

3. Need for annealing may be reduced or eliminated.

iii) Cold Forming ---------------(3 marks in all)

1. Better accuracy meaning closer tolerances

2. Better surface finish

3. Strain hardening increases strength and hardness of the part

4. Grain flow during deformation provide the opportunity for desirable

directional properties to be obtained in the resulting product

5. No heating of the work is required which saves on furnace and fuel costs

6. It permits higher production rate to be achieved