UNIT-I FUELS AND COMBUSTION PART A Howar efu lscas id? …

UNIT-I

FUELS AND COMBUSTION

PART A

1. Define a fuel.

Fuel is a combustible substance, which on combustion produces a large amount

of heat, which can be used for various domestic and industrial purposes. The fuels

commonly used contain carbon as the main constituent and some common fuels are

wood, charcoal, kerosene, diesel, producer gas etc.

2. How are fuels classified? Give examples

3. What is meant by Gross Calorific Value?

The quantity of heat evolved by the combustion of unit quantity of fuel is its gross

calorific value (GCV). Gross or higher calorific value is the quantity of heat liberated by

combusting unit mass of fuel in oxygen, the original material and the final product of combustion

being at a reference temperature of 25°C and the water obtained in the liquid state, represented

by GCV or HCV.

4. Define Net Calorific Value?

Net calorific value (NCV) is the quantity of heat evolved when a unit quantity of fuel is

burnt in oxygen, the original material and the final products of combustion being at a reference

temperature of 25°C and the water obtained from the fuel being at the vapor state. The net

calorific value is hence always less than the gross calorific value by the amount corresponding

to the heat of condensation of water vapours i.e., 587.0 kcal/kg.

NCV = HCV — Latent heat of water vap. formed

= HCV - Mass of hydrogen x a x Latent heat of steam

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since 1 part by mass of hydrogen produces 'a' part by mass of water.

5. What is a coke?

It is a carbonaceous residue obtained from the destructive distillation of coal,

petroleum and coal tar pitch. Petroleum yields coke during cracking processes. The main source

of coke is coal. Petroleum coke is used as metallurgical coke since it is pure.

6. What is an anti-knocking agent?

They are organometallic compounds that increase the octane number of gasoline when

added in low percentage to it. Most common is TEL (-tetraethyl lead). They can increase the

octane number over 100%.

7. What are Octane Number and Cetane Number?

Octane number

It is defined as the percentage of isooctane in isooctane and n-heptane

which has the same knocking as the gasoline sample when burnt in a standard test

engine under standard conditions.

ISOOCTANEN-HEPTANE

Cetane number

It is defined as the percentage of cetane in a mixture of cetane and 2-methyl

naphthalene which has the same ignition delay as the diesel oil when burnt in a

standard test engine under standard conditions.

2-METHYL NAPHTHALENE

CETANE



8. How will you improve the octane number?

Octane number of petrol is improved by adding additives like

tetraethyllead, TEL.

9. Give the composition of water gas.

48% H2; 44% CO. Rest CO2, N2 and methane.

10. Write the composition of producer gas.

50% N2; 30% CO; 10% H2. Rest CO2 and CH4

11. How is coke superior to coal.

i) coal dose not possess as much strength and porosity as coke

ii) by coking most of the undesirable sulphur is removed

iii) coke burns with short flame, due to expulsion of much of the volatile

matter during carbonization

12. (a) What is LPG? (b) What is LNG?

(a) LPG or Liquefied petroleum gas is obtained from 'Wet Natural gas' from

underground, by washing it with gas oil and fractionating the useful fraction.

(b) LNG is Liquefied Natural gas.

13. What are the advantages of a gaseous fuel?

• Can be distributed over a wide area by pipeline.

• Smooth combustion without smoke and ash.

• The temperature of heating can be controlled by controlling the gas flow to the burner.

• Higher calorific values.

14. Mention the disadvantages of a gaseous fuel.

• Danger of explosion

• Larger scale fire risk.



15. Why is calorific value of coal gas higher than that of producer gas?

Coal gas contains all the combustible gases like CH4, C2H4, C2H2, CO and H2, whereas

producer gas contains CO, H2 and N2. N2 acts as inert diluent. So calorific value of coal gas

is higher than that of producer gas.

16. Producer gas is made by passing air and steam through a thick bed of coal.

Why?

The primary purpose of steam is to use up the heat developed during exothermic

reaction of coal and O2 of air to maintain the temperature of producer.

17. Why is NCV greater than GCV?

Gross calorific value GCV includes the latent heat steam during combustion of a fuel,

but Net calorific value NCV excludes the latent heat of steam.

18. Why a good solid fuel must have low ash content?

Ash is inorganic in nature. So high ash content decreases the calorific value of a fuel.

19. What is CNG?

CNG is compressed natural gas used in motor engines now-a-days instead of gasoline fuel

causing less pollution.

20. What is the function of TEL?

1.0—1.5 ml of TEL is added per litre of petrol. TEL functions by being converted to a

cloud of finely divided lead oxide particles, which react with any hydrocarbon peroxide molecules

formed in the engine cylinder thereby solving down the chain oxidation reaction and preventing

knocking.

21. What is meant by the term fixed carbon?

It is the pure carbon present in coal. Higher the fixed carbon content of the coal, higher

will be its calorific value.

22. Name the different varities of coal.

1. Peat, 2. Lignite, 3. Sub-bituminous,

4. Bituminous, 5. Anthracite



PART-B

1. Proximate analysis includes the determination of moisture, volatile,

ash and fixed carbon content.

(i) Determination of Moisture content: 1 g of finely powdered air dried

sample is taken in a crucible and heated in an electrically heated hot air oven

at 105°C—110°C for 1 hr. After heating, the crucible is taken out side, cooled

in a dessicator and weighed.

Loss in weight of coal

Percentage of moisture content = ---------------------------×100

Weight of coal taken

(ii)Determination of Volatile matter: The dried sample left in the crucible

along with the lid is heated in a muffle furnace at 950 ± 25°C for 7 minutes

and then cooled in a dessiccator and weighed.

Loss in weight of coal

Percentage of volatile content = - -------------------------- ×100


(iii)Determination of Ash content: The residual sample after the two above

experiments in the crucible is heated in the furnace at 750°C for 30 minutes

without the lid. Then it is cooled in a desiccator and weighed.

Weight of ash left

Percentage of ash content = ---------------------- × 100


(iv) Determination of Fixed carbon: The fixed carbon content can be

determined indirectly by subtracting the percentage of moisture, volatile and

ash content from 100.

Percentage of fixed carbon =100 — % of (moisture + volatile matter + ash)

Significance:

a) Higher percentage of moisture content lowers the calorific value of



coal. Hence, lower the moisture content, better the quality of coal.

b) Higher percentage of volatile matter reduces the calorific value of coal.

Low volatile matter also reduces coking property of coal.

c) Ash being non - combustible reduces the calorific value of coal. Ash

deposition also causes problems in the furnace walls and the ultimate disposal

of ash is also a problem.

d) Higher the percentage of fixed carbon, higher is the calorific value

and better is the quality of coal.

2. Write notes on ultimate analysis of coal.

Ultimate Analysis: It refers the determination of weight percentage

of carbon, hydrogen, nitrogen, oxygen and sulphur.

(i) Carbon and hydrogen: A known amount of coal sample is taken and

burn in a current of O2 in combustion apparatus whereby CO2 and H2O

are formed. CO2 and H2O are absorbed by previously weighed tubes

containing KOH and anhydrous CaCl2. The increase in weight gives the

C and H content as follows:

C + O2 CO2

and

H2 +1/2 O2 H2O

2KOH + CO2 K2CO3 + H2O.

and CaCl2 + 7H2O CaCl2.7H2O.

Increase in wt. of

KOH tube

% of C= ------------------- ×12/44 ×100

Wt. of coal taken

Increase in wt. of CaCl2 tube

% of H = ---------------------------------- × 2/18 × 100

Wt. of coal taken

(ii)Nitrogen: About 1 g of accurately weighed coal sample is taken in a long

necked flask along with conc. H2SO4, K2SO4 and heated. Then it is treated



with excess of NaOH and the liberated NH3 is absorbed in known excess of

standard acid solution. The excess acid is back-titrated with standard NaOH

solution. From the volume of acid consumed N content is calculated as

follows:

Volume of acid consumed x Normality

% of N = ------------- —— ---- —— ------------------ . × 1.4

Wt. of coal taken

(iii) Sulphur: While determining the calorific value of a coal sample in a

bomb calorimeter, the S in the coal is converted to sulphate. Finally the

washings containing sulphate is treated with dil. HC1 and BaCl2 solution,

which precipitates BaSO4 which is filtered in a sintered glass crucible,

washed with water and heated to a constant weight.

Wt. of BaSO4

% of S= ------------------ × 32/233×100

Wt. of coal taken

( iv) Oxygen content = 100 - % of (C + H + S + N )

Significance:

a) Higher percentage of C and H increases the calorific value of coal and

hence better is the coal.

b) Higher the percentage of O2 lower is the calorific value and lower is

the coking power. Also, O2 when combined with Hydrogen in the coal, Hydrogen

available for combustion becomes unavailable.

c) S, although contributes to calorific value, is undesirable due to its

polluting properties as it forms SO2 on combustion.

2. Write notes on knocking.

Knocking:

An explosive sound produced due to rapid pressure rise in an internal combustion engine.

The knocking characteristics of petrol can be expressed in terms of octane number.



Octane Number:

Octane number is equal to the percentage by volume of iso-octane (2,2,4-

trimethyl pentane) in a mixture of n-heptane and iso-octane having the same knocking

tendency compared to the sample of gasoline being tested; iso-octane has the best

antiknocking properties and assigned an octane number of 100 whereas n-heptane has

poor antiknocking property and assigned an octane number of zero. The hydrocarbons

present influence the knocking properties of gasoline which vary according to the

series: straight chain paraffin > branched chain paraffin > olefin > cycloparaffin >

aromatics. The fuel which has same knocking tendency with the mixture having 80%

iso-octance has octane number 80.

The most effective antiknock agent added is tetraethyl lead (TEL) along with

ethylene dibromide which prevents the deposition of lead by forming volatile lead

halides. Other antiknocking agents are tetramethyl lead (TML), tertiary butyl

acetate, diethyl telluride.

1.0—1.5 ml of TEL is added per litre of petrol. TEL functions by being

converted to a cloud of finely divided lead oxide particles, which react with any

hydrocarbon peroxide molecules formed in the engine cylinder thereby solving down the

chain oxidation reaction and preventing knocking.

Cetane Number:

The knocking characteristics of diesel can be expressed in terms of cetane

number.

There is a delay period between the injection of diesel fuel and its ignition. This

delay period if becomes large, too much fuel accumulates in the cylinder and burn

very rapidly and causes "diesel knock". This delay period is connected to the type of

hydrocarbons present in the diesel. Increasing delay period occurs in the series: re-

paraffins < olefins < naphthenes < isoparaffins < aromatics. The order is the reverse for

gasoline anti-knock quality, n-hexadecane is given cetane number 100 and a-methyl

naphthalene given cetane number zero. If a given fuel matches in quality with the



blend having 40/60 blend of cetane and a-methyl naphthalene, it is assigned a cetane

number 40.

Cetane number is defined as the percentage of cetane present in a mixture of

cetane and 2- methyl naphthalene which has the same ignition lag as that of the fuel

under test.

The cetane number of diesel can be improved by adding amyl/butyl nitrate,

carbamates, ditertiary butyl peroxide and metal organic compounds.

3. How is metallurgical coke manufactured?

Otto-Hoffmanns method

Otto Hofmann developed a modern by product coke oven. Here, the heating is done by

a portion of coal gas produced during the process itself. It also utilizes the waste flue gases for

heating the checker work bricks. The oven consists of a number a narrow silica chambers, each

about 10-12 m long, 3-4 m tall and 0.4-0.45 m wide, erected side by side with vertical flues

between them to form a sort of battery. Each champer has a hole at the top to introduce the

charge, a gas off take and a refractory lined cast iron door at each end for coke discharge. The

oven works on heat regenerative principle i.e., the waste gas produced during carbonization is

utilized for hesting. The ovens are charged from the top and closed to restrict the entry of air.

Finely crushed coal is charged through the hole at the top and in a closed system it is

heated to 1200°C. The flue gases produced during combustion pass their sensible heat to the

checker brick-work, which is raised to 1000°C. The flow of heating gases is then reversed and

the hot checker bricks heat the inlet gas. This cycle continues till the volatile matter lasts.

Carbonization takes about 11 to 18 hours, after which the doors are opened and the glowing

coke mass is discharged by machine driven coke-pusher into coke-quencher. The hot coke is

quickly quenched by water spraying. 'Dry quenching' is also done by circulating flue gases

over hot coke and the hot gases are utilised to run waste heat boilers (Fig. 1).



Fig.1.

Recovery of byproducts

The gases coming out at 600°C-700°C get a spray of flushing liquor at the goose-neck

of the standpipe and the temperature is reduced to 80°C. Tar and steam get condensed. The

'Coke oven gas' is composed of NH3, H2O; tar contains naphthalene, benzene, moisture etc.

(i) Coal tar is condensed in the tank below.

(ii) Ammonia is recovered partly as aqueous solution and partly as sulfate.

(Hi) Naphthalene is recovered by passing the gas through a tower where water is sprayed

at very low temperature which condenses the naphthalene.

(iv) Benzene is recovered similarly by spraying petroleum.

(v) H2S is recovered by passing the gas through moist Fe2O3 as:

Fe2O3 + 3H2S --------- > Fe2S3 + 3H2O

Fe2S3 is again regenerated by exposing to atmosphere

Fe2S3 + 4O2 ----------------> 2FeO + 3SO2

4FeO + O2 -------------- > 2Fe2O3

4. Explain the cracking process?

Cracking Process

Cracking is the process in an oil refinery by which heavier fraction from the

fractional distillation converted into useful lighter fractions by the application of heat,

with or without catalyst, i.e., cracking is a process by which larger molecules break up

into smaller ones. The chief application of commercial cracking in all the refineries is

for the production of gasoline from gas oils. Other uses include the production of

olefins from naphthas and gas oils. The surplus of heavier petroleum fractions are also



cracked to get petrol. There are two methods of cracking.

1. Thermal Cracking

When cracking is carried out without any catalyst at high temperature from

450°C—750°C at pressures ranging from 1—70 atms, it is called thermal cracking.

The important reactions are decomposition, dehydrogenation, isomerization and

polymerization. The paraffins decompose to lower mol. wt. compounds, like paraffin

and an olefin.

2. Catalytic cracking:

The quality and yield of gasoline produced by cracking can be greatly improved by

using a suitable catalyst like aluminium silicate.

There are two main types of catalytic cracking:

(a) Fixed-bed catalytic cracking:

The catalysts are fixed in towers, through which the hot oil (500°C) flows

from the top and passes down. 40% of the charge is converted to gasoline and 2—

4% carbon is formed. This carbon deposits on the catalyst beds. The bottom liquid

is reboiled and recycled to the fractionating column and ultimately gas oil is

obtained having high octane value. The gasoline is stripped off dissolved gases and

purified. The carbon deposits on the catalysts are burnt by compressed air in one

chamber for reactivation while the other catalyst chambers are active (Fig. 18.7).

(b) Moving bed or fluidized bed catalytic cracking:

The finely powdered catalyst behaves as a fluid when suspended in gas or oil

vapour. The preheated heavy oil is forced through tower along with the fluidised

catalyst. At the top of the tower a cyclone separator is active to separate the cracked

oil vapour and passes it to the fractionating column. The catalyst powder is retained

and sent back. The catalyst becomes deactivated by a deposition, which is reactivated by

burning away the deposits with compressed air in a regenerator.



5. Discuss Bergius method of manufacture of Synthetic Petrol?

Bergius Process:

Destructive hydrogenation of coal in presence of a catalyst yields oil, but it is

not yet competitive with petroleum refining. Coal is ground and made into a paste with

a heavy recycle oil and a catalyst like iron, tin or molybdenum compound. The paste is

preheated and treated with H2 at 250—350 atm. pressure and 450°C—500°C temperature.

The unreacted coal is filtered-off and the liquid product distilled. Hydrogen combines

with coal to form saturated hydrocarbons, which decompose at high temperature

yielding low-boiling hydrocarbons. The crude oil is fractionated to get (i) gasoline (ii)

middle oil and (Hi) heavy oil which is recycled. Middle oil is hydrogenated in vapour

phase with catalyst to yield more gasoline. Yield of gasoline is 60% of the coal dust.



PART-C

1. Explain flue gas analysis by Orsat apparatus

Method of Analysis of Flue Gas

Analysis of flue gas will indicate the complete or incomplete combustion of a

fuel; which is very essential in respect of efficient utilization of the fuel. The

analysis is done in Orsat's apparatus and is based on the principle of absorption

of

• CO2 in KOH solution (500 g\ 500 ml)

• O2 in alkaline pyrogallic acid (25 g of pyrogallic acid in 400 g KOH solution, 500

ml)

• CO in ammoniacal cuprous chloride solution (100 g Cu2Cl2 + 125 ml liquor

ammonia

Rest water to make up the volume, 500 ml).

Description of the Apparatus

It consists of a water jacketted measuring burette, connected in series with the

absorption bulbs containing the above three solutions successively as depicted in the

Fig. 18.16. The water jacket maintains the temperature of the gas constant. The

absorption bulbs are filled with glass tubes for better absorption of the gases.

Procedure

• The apparatus is tested for its air-lightness.

• The flue gas (100 ml) is taken in the apparatus in the measuring burette by

adjusting the volume by water reservoir atmospheric pressure.

• The stopper (1) is opened for CO2 absorption and water reservoir is raised to

force thegas inside the bulb. The gas is finally taken in the burette and the

volume of the gas is measured by making the levels of water inside the burette

and water reservoir equal. The decrease in volume gives the percentage by

volume of CO2 in the flue gas.

• The stopper (2) is opened and provided as usual to get the percentage by volume

of O2 in the flue gas.

• The stopper (3) is opened and proceeded as usual to get the percentage by volume



of CO in the gas.

• The sequence of the bulbs should be strictly followed.

Implications of the Analysis

• If the flue gas contains greater percentage of CO it is implied that considerable

wastage

of fuel is taking place due to incomplete combustion and the O2 supply is

insufficient.

• The greater percentage of O2 in the flue gas indicates that O2 supply is in

excess.

• Result of the analysis will help to control the combustion process.



2. i) How is producer gas prepared ? Mention its uses

Producer Gas

It is a mixture of combustible gases like CO and H2 and large amount of non-

combustible gases like N2, CO2 etc. It is prepared by passing air and steam over an

incandescent bed of solid carbonaceous fuel in a reactor called "gas producer" or

simply a "producer". It is a fuel of low calorific value (1300 kcal/m3) but its advantage

is its cheapness and ease of production. The carbonaceous fuel used is generally coal or

coke, though wood waste, peat etc. can be used.

The gas producer consists of a steel vessel with inner lining made up of refractory

bricks. The dimension of the vessel is about 4 m height and at the top there is a cone

feeder and a side opening for producer gas outlet. At the base there is inlets for air and

steam and an outlet for the ash.

Uses:

1. It is a cheap, clean and easily producible gas and is used for heating open hearth

furnaces

2. as a reducing agent in metallurgical operations



ii) Discuss Fischer-Tropsch method of manufacture of Synthetic Petrol?

Fischer-Tropsch Process

The process is based on the catalytic hydrogenation of carbon monoxide

leading to the following reaction:

nCO + 2n H2→ CnH2n + n H2O

nCO + (2n+1) H2→ CnH2n+2 + n H2O

Olefin Water gas (CO + H2), produced by passing steam over heated coke is mixed

with hydrogen. The gas is purified by passing over Fe2O3 to remove H2S, then passing

over a mixture of Fe2O3.Na2CO3 to remove organic sulfur compounds. The purified gas

is compressed to 5 to 25 atm. at 200°C—300°C passed through a converter containing

catalyst which is a mixture of 100 parts cobalt, 5 parts thorium, 8 parts magnesium

and 200 parts keiselguh.

3. i) How is water gas prepared ? Mention its uses

Water Gas

Water gas is a gaseous fuel of medium calorific value (2800 kcal/m3)

generated by gasifying solid incandescent source of carbon in superheated

steam. The equipment is known as water gas generator and is more or less

similar to that shown in Fig. 2 where the following reaction takes place which

is endothermic.

Fig. 2. Preparation of Water gas



Water gas is obtained by the action of steam on a bed of coal heated to 1000°C.

C + H2O → CO + H2O

Since the above reaction is endothermic, the coal cools down after a few minutes

and the reaction proceeds in a different way to form CO2 and H2, instead of water gas.

C + 2 H2O → CO2 + 2 H2

In order to avoid the above undesirable reaction, the blow of air replaces the blow

of stream. The following reactions occur:

C + O2 → CO2 + 97 K cal.

2C + O2 → 2CO + 59 K cal

Due to exothermic reactions, the temperature of the bed rises and when the

temperature increases to 1000°C, air entry is stopped and stream is again passed. Thus,

stream and air are blown alternatively.

Uses:

Water gas is used:

As a source of hydrogen gas

In the manufacture of NH3 by hapers process

As an illuminating gas

As a fuel gas

For welding purposes

ii) A fuel contains C = 75%; H = 4%; O = 5%; S = 7% remaining ash.

Calculate the minimum quantity of air required for complete combustion of

1kg of fuel.

Given

Cup and cone feeder

Coke

Water gas outlet

Red hot coke at900°C-1000°C

Steam supply Refractory brick lining

:-4-Air supply Ash outlet

Ash



Weight of the fuel = 1 kg

Weight of C in the fuel = 0.75 kg

Weight of H in the fuel = 0.04 kg

Weight of O in the fuel = 0.05 kg

Weight of S in the fuel = 0.07 kg

Solution

CO + O2 → CO2

12 kg of carbon requires 32 kg of oxygen

0.75 kg carbon requires = (32/12) ×××× 0.75 = 2 kg

H2 + ½ O2 → H2O

2 kg of hydrogen requires 16 kg of oxygen

0.04 kg hydrogen requires = (16/2) ×××× 0.04 = 0.32 kg

S + O2 → SO2

32 kg of carbon requires 32 kg of oxygen

0.07 kg carbon requires = (32/32) ×××× 0.07 = 0.07 kg

Total weight of oxygen required = 2 + 0.32 + 0.07

= 2.39 kg

But weight of oxygen already present = 0.05 kg

Actual weight of oxygen required = 2.39-0.05 = 2.34 kg

Weight of air required = 2.34 × (100/23) = 10.17 kg of air

iii) Calculate the volume of air required for complete combustion of 1 litre of CO?

Consider the combustion reaction of CO

CO + ½ O2 → CO2

1 vol 0.5 vol

1 vol of CO requires 0.5 vol of oxygen for complete of combustion.

1 litre of CO requires 0.5 litre of oxygen for complete of combustion

The volume of air required = 0.5 ×××× 100/21 = 2.38 lit of air



4. (i) Discuss the manufacture of synthetic petrol by Bergius process.

Bergius Process:

Coal is ground and made into a paste with a heavy recycle oil and a catalyst like

tin oleate. The paste is sent along with H2 at 250—350 atm. Pressure into a converter

which is maintained at 450°C—500°C temperature. The unreacted coal is filtered-off and

the liquid product distilled. Hydrogen combines with coal to form saturated

hydrocarbons, which decompose at high temperature yielding low-boiling

hydrocarbons. The crude oil is fractionated to get (i) gasoline (ii) middle oil and (iii)

heavy oil which is recycled. Middle oil is hydrogenated in vapour phase with catalyst to

yield more gasoline. Yield of gasoline is 60% of the coal dust.

4.(ii)What is meant by cracking? Explain the process of catalytic cracking.

Cracking Process

Cracking is the process in an oil refinery by which heavier fraction from the

fractional distillation converted into useful lighter fractions by the application of heat,

with or without catalyst, i.e., cracking is a process by which larger molecules break up

into smaller ones. There are two methods of cracking.

1. Thermal Cracking: When cracking is carried out without any catalyst at

high temperature from 450°C—750°C at pressures ranging from 1—70 atms, it is

called thermal cracking. The important reactions are decomposition,

dehydrogenation, isomerization and polymerization. The paraffins decompose to

lower mol. wt. compounds, like paraffin and an olefin.



2. Catalytic Cracking. The use of catalyst during cracking accelerates

the reactions and at the same time modifies the yield and the nature of product.

Catalyst used are synthetic composition of silica and alumina, zeolites in the form

of beads or pellets.

There are two main types of catalytic cracking:

(a) Fixed-bed catalytic cracking: The catalysts are fixed in towers, through

which the hot oil (500°C) flows from the top and passes down. 40% of the

charge is converted to gasoline and 2—4% carbon is formed. This carbon

deposits on the catalyst beds. The bottom liquid is re boiled and recycled to

the fractionating column and ultimately gas oil is obtained having high

octane value. The gasoline is stripped off dissolved gases and purified. The

carbon deposits on the catalysts are burnt by compressed air in one

chamber for reactivation while the other catalyst chambers are active

(b) Moving bed or fluidized bed catalytic cracking: The finely powdered

catalyst behaves as a fluid when suspended in gas or oil vapour. The preheated

heavy oil is forced through tower along with the fluidised catalyst. At the top of

the tower a cyclone separator is active to separate the cracked oil vapour and

passes it to the fractionating column. The catalyst powder is retained and sent back.

The catalyst becomes deactivated by a deposition, which is reactivated by burning

away the deposits with compressed air in a regenerator



Advantages of catalytic cracking:

(i) The quality and yield of petrol is better.

(ii) External fuel is not necessary since the coal embedded in the catalyst supplies

the heat.

(iii) Operating pressure is lower.

(iv) Byproduct gas evolution being low, yield of petrol is higher.

(v) Due to higher aromatics content, the anti-knocking properties are higher.

(vi) Greater portion of S escapes as H2S so residual S content of the oil is low.

(vii) Gum-forming compounds are very low.

(viii) In presence of specific catalysts preferentially the cracking of naphthenic

materials takes place, so it becomes richer in paraffinic compounds.

(ix) Only the high-boiling hydrocarbons and the side chain of the aromatics are

decomposed preferentially.

5. Write notes on petroleum processing.

Petroleum Processing:

Drilling: Oil is brought to the surface by drilling holes upto the oil bearing

surface. By the hydrostatic pressure of natural gas the oil is pushed up or it is

pumped up by means of a pump. Two coaxial pipes are lowered to the oil reservoir,

through the outer pipe compressed air is forced, whereby the oil is forced out through

the inner pipe. This crude oil is sent to the refineries for further processing of the

crude oils.

Refining: After removal of dirt, water and natural gas, the crude oil is

separated into fractions by distillation and the fractions obtained are subjected to



simple purification procedures or complex treatments to yield different petroleum

products. All these steps are under petroleum refining which include:

i) Coltrell's process.

Crude oil is intimately mixed with water forming an emulsion. The water is

separated from the oil by passing the emulsion through Coltrell's electro static

precipitator.

ii) Removal of objectionable compounds:

Sulphur compounds have objectionable properties of pollution so they are

removed prior to distillation as copper sulfide by treatment

with copper oxide.

iii)Petroleum distillation.

1. The crude oil is subjected to distillation to about 400°C

temperature in an iron retort whereby all volatile components except the solid

residue are distilled out. These are separated in a fractionating column consisting of a

tall tower where thehigher boiling fractions condense first. This distillation is a

continuous process and the following fractions are obtained from refinery.

(a) Gasoline is obtained upto 200°C. The naphtha is condensed and

subjected to refining for the removal of sulfur, di olefins after refractionating.

(i) Petroleum ether boiling between 40°C-70°C

and (ii) Benzene boiling between 70°C-90°C and

(iii) Gasoline boiling between 90°C—200°C all are



obtained.

Its calorific value is 14,250 kcal/kg and is used as a fuel for internal

combustion engines in automobiles and aircrafts.

(b) Solvent naphtha is obtained as a side steam between 200°C-250°C.

This contains some gasoline, which is passed back to the main fractionating

column. Naphtha contains 6-10 carbon atoms.

(c) Kerosene oil is obtained between 250°C-300°C. The lower boiling

fraction mixed

with it is returned to the main column. Bottom liquid is refined and finally

can be used as

domestic fuel having calorific value of 1100 kcal/kg.

(g)Gas oil is obtained between 300°C-350°C. This is passed through a

cooler and then extracted with liquid SO2 to remove sulfur. It is used as a diesel

engine fuel with calorific value

of 11000 kcal/kg.

(h)The residual liquid coming out from the bottom on subsequent

treatment yield vaseline, grease, paraffin wax, asphalt-bitumen, petroleum,

coke etc.

Table 1: Common fractions from crude oil

Fraction Boiling range Composition Uses

Uncondensed gas Within 30°C C1-C4 As domestic or industrial

fuel under the name LPG

Petroleum ether 30°C-70°C C5 – C7 As a solvent.

Gasoline or

petrol

90°C-200°C C6 – C9 As a motor fuel solvent and

dry washing Naphtha 200°C-250°C C9 – C10 As a solvent

Kerosene oil 250°C-300°C C10 –C16 As fuel for domestic and

industrial uses Diesel oil 300°C-350°C C10-C18 As a fuel for diesel engine

Heavy oil 320°C-400°C C17-C30 For different fractions



UNIT-II

CORROSION AND ITS CONTROL

PART-A

1. What is corrosion?

The destruction of metals or alloys by the action of gaseous atmosphere,

water or any other reactive liquid medium is known as corrosion.

2. Mention the conditions for wet corrosion to takes place?

i) When two dissimilar metals are in contact with each other in presence of

aqueous solution or moisture, electrochemical or wet corrosion occurs.

ii) When a metal is exposed to an electrolyte with varying amount of

oxygen, then also wet corrosion takes place.

3. What is Pilling Bedworth ratio? Give its significance?

The ratio of the volume of oxide film formed to the volume of metal

consumed is called Pilling Bedworth ratio. It gives an idea about whether the

oxide film formed on the metal surface is protective or non-protective.

4. How do you classify the corrosion?

(a) Chemical corrosion => it involves direct chemical action between

metals and gases.

(b) Electrochemical corrosion => it involves deterioration of metal due to

flow of electric current from one point to another.

(c) Dry corrosion => it refers to the corrosion of metals involving direct

chemical action between metals and dry gases.

(d) Wet corrosion => it involves flow of electric current from one point to

another through some perceptible distance in the presence of liquid or

moisture in air.



5. What is dry corrosion? Give examples.

Dry corrosion is the direct chemical attack of metals by the gases such as O2, CO2,

SO2, H2O etc.

Eg: - i) Tarnishing of silver articles in H2O gas.

ii) Action of dry HCl on iron surfaces.

6. What is wet corrosion?

It is a type of corrosion which occurs when a conducting liquid is in contact with

metal (or) two dissimilar metals or alloys either immersed or partially dipped in a

solution.

7. What is galvanic corrosion?

When two metals are electrically connected and exposed to an electrolyte, the

metal higher in electrochemical series undergoes corrosion. Eg; Zn-Cu couple, Zn

gets corroded.

8. Distinguish between dry corrosion and wet corrosion.

Dry corrosion Wet corrosion

• It occurs in dry state

• It involves direct chemical

attack of the metal by gases

• It follows adsorption

mechanism.

• Corrosion products accumulate

on the same spot where

corrosion occurs.

It occurs in presence of moisture or

electrolyte

It involves setting up of large number of

tiny galvanic cells.

It follows electrochemical reaction

Corrosion occurs at anode while products

gather at cathode.



9. What is pitting corrosion?

Pitting is a localized attack which results in the formation of a hole around which

the metal is relatively unattached. The mechanism involves setting up of differential

aeration cell.

10. Name the factors which affect corrosion.

i) Air and Moisture

ii) Electrolyte in water

iii) pH

of the medium

iv) Presence of corrosive gases

v) Presence of impurities in a metal.

11. A steel screw in a brass marine hardware corrodes. Give reason.

This is due to galvanic corrosion. Iron which is higher in the emf series becomes

anodic and is corroded. Brass which is present lower in series acts as cathodic and is

not corroded.

12. Iron corrodes under drops of salt water. Explain.

This is due to differential aeration. Areas of iron covered by drops, having poor

access to oxygen, become anodic with respect to other areas which are freely

exposed to air. Due to electrochemical corrosion, the areas under drops undergo

corrosion, while the freely exposed parts remain unaffected.

13. The rate of metallic corrosion increases with increase in temperature. Give

reason.

With increase of temperature of the environment, the rate of reaction as well as

rate of diffusion increases, thereby corrosion rate increases.

14. Iron corrodes faster than aluminium, even though iron is placed below

aluminium in emf series. Why?

This is because aluminium forms a non-porous, very thin, tightly adhering

protective oxide film (Al2O3) on its surface and this film does not permit corrosion to

occur.



15. Rusting of iron is quicker in saline water than in ordinary water. Give reason.

The presence of sodium chloride in saline water leads to increased conductivity of

water layer in contact with the iron surface, thereby corrosion current increases and

rusting is speeded up.

15. What is cathodic protection? Mention its two applications.

Cathodic protection is the reduction or prevention of corrosion by making

metallic structure as cathode in the electrolytic cell. This can be done by either using

sacrificial anodic (or) impressed current cathodic method.

They give protection to cables, pipelines, ship hulls from marine corrosion.

16. What is impressed current cathodic protection?

It is a method in which an impressed current (greater than corrosion current) is

applied in opposite direction to nullify the corrosion, thereby converting the corroding

metal from anode to cathode.

17. What are corrosion inhibitors? Give examples.

Corrosion inhibitors are substances which are added to the corrosive environment

to decrease the corrosion rate.

Eg. Anodic inhibitors – phosphates & chromates.

Cathodic inhibitors – aniline and its derivatives.

18. What is the role of pigment in paint? Give examples.

i) Pigments are solid substances and they provide color to the paint.

ii) They provide capacity to the paint.

iii) They give protection to the paint film by reflecting harmful UV radiation.

Eg. Green – chromium oxide

Blue – Prussian blue.

19. What is meant by electroplating?

Electroplating is the process by which the coating metal is deposited on the base

metal by passing a direct current through an electrolytic solution containing the

soluble salt of the coat metal.



20. What is meant by electroless plating?

Electroless plating is the technique of depositing a noble metal from its salt

solution on a catalytically active surface of the metal to be protected by using a

suitable reducing agent without using electrical energy.

Metal ion + reducing agent � Metal + oxidized product.

21. What is meant by anodizing?

Anodizing is a process by which a thick oxide coating can be produced on the

base metal by making it as anode in the electrolytic cell. The electrolytes used to

assist this process are oxalic acid, chromic acid etc.

PART B

1. Explain the mechanism of oxidation corrosion.

This type of corrosion is brought about by the attack of oxygen gas on a free

metal surface, resulting in the formation of metallic oxide which is a corrosion product.

Mechanism:

Oxidation takes place at the surface of them metal forming metal ions, M2+

M M2+

+ 2e-

Oxygen is converted to oxide ion (O2-

due to the transfer of electrons from metal

½ O2 + 2e - →

O2-

Oxide ion reacts with the metal ions to form metal oxide film.

M2+

+ O2-

MO

The corrosion stops if the oxide film produced is impervious for oxygen thereby

preventing oxygen from penetration inside the film. On the other hand if the oxide film

produced is porous or volatile one, (or) direct contact of oxygen in its ionic form with the

metal ion to produce additional oxide film.

The nature of oxide film formed on the metal surface plays a significant role.

The different types of oxide films are



a) Porous & non-porous oxide film: (Pilling Bedworth rule)

In cash of alkali and alkaline earth metals (Na, K & Mg, La etc), the volume of

oxide film formed on the surface is less than that of the metal, consumed. Hence, the

oxide film formed is porous and non-protective.

In case of heavy metals such as Sn and Pb, the volume of oxide film formed on

the surface is more than the volume of the metal consumed. Hence, the oxide layer

formed is protective and non-porous is the ratio of the volume of oxide formed to the

volume of metal consumed is called Pilling Bedworth rule (or) Pilling-Bedworth ratio.

b) A stable layer is fine grained in structure and can get adhered tightly to the metal

surface. Such a layer is impervious in nature, and prevents the metal from further oxygen

coating, thereby shielding the metal surface.

Eg: The oxide films of Al, Sn, Pb etc are stable, adhering and impervious in nature.

c) Unstable oxide layer:

They are mainly produced on the surface of noble metals (Ag, Au & Pt) which

decomposes back to the metal and the oxide is liberated in the form of oxygen.

Metal oxide Metal + Oxygen.

d) Volatile oxide layer:

Volatile oxide layers volatilize as soon as they are formed, thereby leading to

further corrosion directly. Eg. Molybdenum Oxide (MoO3 ) is volatile.



2. Write notes on wet corrosion.

Electrochemical corrosion is a type of corrosion where the conducting

electrolytic liquid is in contact with metal or two dissimilar metals or alloys which

are dipped partially in a solution. Here one part of the metal becomes anode and

another cathode.

In anode, oxidation takes place with the liberation of electrons, where the

actual corrosion takes place. In cathodic metal, reduction takesplace with the

absorption of electron liberated from the anode.

Mechanism

At anote: M Mnt + ne- (oxidation)

In the cathodic part, reduction occurs which depends on the nature of the

corrosive environment.

i) If the corrosive environment is acidic, hydrogen evolution occurs at the cathode.

ii) If the corrosive environment is slightly alkalin (or) neutral, hydroxide ion form at the

cathodic part (or) absorption of oxygen takesplace.

e) Hydrogen evolution:

This type of corrosion takesplace when bare metals are in contact with acidic

solutions. For eg. Considering metal like iron, the anodic reaction is dissolution of iron

as ferrous ions with liberation of electrons.

Fe → Fe2+

+ 2e- (oxidation)

The liberated electrons flow from the anode to cathode. At cathode H+ ions are

liberated as hydrogen gas

2H+ + 2e

- → H2 ↑ (Reduction)

The overall reaction is

Fe + 2H+ Fe

2+ + H2 ↑



Absorption of oxygen (or) hydroxide ion formation:

This type of corrosion takes place when base metals are in contact with neutral

solutions like water with dissolved oxygen. A good example of this type of corrosion is

the rusting of iron.

Usually the surface of iron is coated with a thin film of iron oxide. Here the well

coated part acts as cathode and the cracked areas acts as anode.

In the anodic part iron dissolves as ferrous ions with the liberation of electrons.

Fe Fe 2+

+ 2e- (oxidation)

The liberated electrons flow from the anode to cathode through the metal, where

electrons to form OH– ions

½ O2 + H2O + 2 e- � 2 OH

– (Reduction)

The Fe2+

ions at anode and OH- ions at cathode diffuse and react to produce Fe (OH) 2

Fe2+

+2OH- � Fe(OH)

If enough oxygen is present, ferrous hydroxide is easily oxidized to forric

hydroxide (Fe (OH)3).

4 Fe (OH)2 + O2 + 2H2O 4 Fe (OH)3

This product is called yellow rust actually corresponds to Fe2O3. H2O

3. Explain galvanic type corrosion and differential aeration corrosion?

When two different metals are in contact with each other in presence of an

aqueous solution moisture, galvanic corrosion occurs. The more active metal (more

negative potential) acts as anode and the less active metal ( less negative potential ) act as

cathode.

Eg: i) when copper and zinc are kept in salt water, the more action metal zinc corrodes

while copper remains uncorroded.

Zn Zn2+

+ 2e- (oxidation)



Cu2+

+ 2e- Cu (reduction).

ii) Zn – Fe couple, in which Zinc (more active or higher in the emf series) dissolves in

preference to iron (less active metal) is Zn acts as anode and Fe acts as cathode.

iii) Steel screw in a brass marine hardware corrodes. This is due to galvanic corrosion.

Iron (higher in the emf series) becomes anodic and is corroded while brass (lower is emf

series) act as cathode and is not corroded.

iv) Steel pipe connected to copper plumbing corrodes.

Control of galvanic corrosion.

i) Selecting the metals as close as possible in the electrochemical series.

ii) Providing an insulating material between the two metals.

iii) Providing larger area for anode and smaller area for cathode.

Differential aeration corrosion:

This type of corrosion occurs when a metal is exposed to varying concentration of

oxygen or any electrolyte on the surface of the base metal. This causes a difference in

potential between aerated areas.

Eg. 1) Metal partially immersed in water (or) conducting solution. (water line

corrosion)

If a metal is partially immersed in a conducting solution, the metal part above the

solution is more aerated and hence become cathodic.

On the other hand, the metal part inside the solution is less aerated and thus,

become anodic and suffers corrosion.



A difference of potential is developed on the same metal and the flow of electrons

takes place from anode to cathode.

Zn Zn2+

+ 2e- (oxidation)

½ O2 + H2O + 2e- 2OH- (Cathode) (reduction)

Pitting corrosion:

It is a localized attack, resulting in the formation of a hole around which

the metal is relatively unattached. Eg. metal area covered by a drop of water, sand

dust etc.

The area covered by the drop of water acts as an anode due to less oxygen

concentration & undergoes corrosion. The uncovered area acts as cathode due to

high oxygen concentration. The rate of corrosion will be more, when the area of

cathode is larger and the area of anode is smaller. Therefore more and more

material is removed from the same spot. Thus, a small hole or pit is formed on

the surface of a metal.



At anode:

Fe Fe2+

+ 2e-

At cathode:

H2O + ½ O2 + 2e- → 2 OH

-

Fe2+

+ 2OH- Fe (OH)2 → Fe (OH)3

(o)

This type of intense corrosion is called pitting corrosion.

iii) Another example of differential aeration is a wire fence in which the areas

where the wire cross are less oxygenated than the rest of the fence and hence

corrosion takes place at the wire crossings which are anodic.

iv) It may also occur in different parts of pipeline is buried pipelines or cables

passing from one type of soil to another (from clay (less aerated) to sand (more

aerated)) get corroded due to differential aeration.

4. How do you control the corrosion.

Since both the cathodic and anodic steps must take place for corrosion to occur,

prevention of either one will stop corrosion. The most obvious strategy is to stop both

processes by coating the object with a paint or other protective coating. Even if this is

done, there are likely to be places where the coating is broken or does not penetrate,

particularly if there are holes or screw threads.

A more sophisticated approach is to apply a slight negative charge to the metal,

thus making it more difficult for the reaction M → M2+

+ 2 e– to take place.

Sacrificial coatings

One way of supplying this negative charge is to apply a coating of a more active

metal. Thus a very common way of protecting steel from corrosion is to coat it with a thin

layer of zinc; this process is known as galvanizing. The zinc coating, being less noble

than iron, tends to corrode selectively. Dissolution of this sacrificial coating leaves

behind electrons which concentrate in the iron, making it cathodic and thus inhibiting its



dissolution.

The effect of plating iron with a less active metal provides an interesting contrast.

The common tin-plated can (on the right) is a good example. As long as the tin coating

remains intact, all is well, but exposure of even a tiny part of the underlying iron to the

moist atmosphere initiates corrosion. The electrons released from the iron flow into the

tin, making the iron more anodic so now the tin is actively promoting corrosion of the

iron.

Cathodic protection

A more sophisticated strategy is to maintain a continual negative electrical charge

on a metal, so that its dissolution as positive ions is inhibited. Since the entire surface is

forced into the cathodic condition, this method is known as cathodic protection. The

source of electrons can be an external direct current power supply (commonly used to

protect oil pipelines and other buried structures), or it can be the corrosion of another,

more active metal such as a piece of zinc or aluminum buried in the ground nearby, as is

shown in the illustration of the buried propane storage tank below.



PART – C

1. What are paints? Mention its constituents. Explain the functions of various

constituents.

Paints:

Paint is a mechanical dispersion of one or more finely divided pigments in a

medium (thinner + vehicle). When a paint is applied to a metal surface, the

thinner evaporates, while the vehicle undergoes slow oxidation forming a

pigmented film.

Requisites or characteristics of a good paint:

1. It should spread easily on the metal surface.

2. It should have high covering power.

3. It should not crack on drying.

4. It should adhere well to the surface.

5. The colour of the paint should be stable.

6. It should be a corrosion and water resistant.

7. It should give a glossy film.



Constituents and their functions of a paint:

Constituents:

1. Pigments, 2. Vehicle or drying oil. 3. Thinner or solvents. 4. Fillers or

extendors. 5. Driers. 6. Plasticisers. 7. Anti-Skinning agents.

Functions:

1. Pigments:

Pigments are solid and colour producing substances in the paint.

i) It gives colour and opacity to the film

ii) It also provides strength to the film.

iii) It protects the film by reflecting the destructive U.V. rays.

iv) It gives an aesthetical appeal to the paint film.

Egs: i) White pigments � White lead, Zinc, litho phone etc.

ii) Coloured pigmens � Red – red lead, chrome red

Green – chromium oxide

Blue – Prussian blue

Yellow – Chrome yellow.

2) Vehicle (or) drying oil:

This is the film forming constituent of the paint. These are glyeryl esters of high

molecular weight fatty acids present in animal and vegetable oil. Eg. Linseed oil.

Soya been oil etc.

Functions:

i) The drying oil is the main film-forming constituents of the paint.

ii) It gives tough, adherent & glossy film coating.

iii) It makes the film impervious to water.

iv) It holds the pigment particles together on the metal surface.

3. Thinner (or) solvents:

Thinners are added to the paint to reduce the viscosity of the paints, so that they

can be easily applied to the metal surface. Thinners are highly volatile solvent.



Eg. Turpentine, Petroleum Spirit, Kerosene etc.

Functions:

i) They reduce the viscosity of the paint.

ii) They dissolve the oil, pigments etc and produce a homogeneous mixture.

iii) They increase the elasticity of the film.

iv) They increase the penetrating power of the vehicle.

4. Fillers or Extenders:

These are white or coloured pigments

i) Fillers are added to the paint to reduce the cost and increase the durability of the

paint. Eg. talc, carbonate, gypsum, china clay etc.

5. Driers:

Driers are used to accelerate the drying of oil film by oxidation, polmerisation and

condensation. Eg. Linoleates, tungstates and naphthenates of cobalt, lead,

manganese and zinc.

Functions:

i) They act as oxygen coarriers (or) catalysts

ii) The mainfunction of drier is to improve the drying quality of oil film.

6. Plasticisers:

They are chemicals added to the paint to give elasticity to the film and prevent

cracking of the film. Eg. tributyl phosphate, trycresyl phosphate etc.

7. Antiskinning agents:

They are chemicals added to prevent skinning of the paint film. Eg. polyhydroxy

phenols.

2. Write notes on i) Electroplating and ii) Electrolessplating.

Electro Plating:

Electroplating is the process by which the coating metal is deposited on the base

metal by passing a direct current through an electrolytic solution containing the



soluble salt of the coating metal. The base metal to be plated is made cathode of

an electrolytic cell, where a sthe anode is either made of the coating metal itself or

an inert material of good electrical conductivity.

Theory:

The anode is made of coating metal itself in the electrolytic cell (Cu). During

electolysis, the concentration of electrolytic bath (CuSO4) remains unaltered,

since the metal ions (Cu2+

) deposited from the bath on cathode are regenerated

continuously by the reaction of free anions with the anode.

CuSO4 ← Cu 2+

+ SO4 2-

On passing current, Cu2+

ions go to the cathode and get deposited there.

At cathode: Cu2+

+2 e- → Cu

The ree sulphate ions migrate to the copper anode and dissolve an equivalent

amount of copper to form CuSO4.

At anode: SO42-

+ Cu → CuSO4 + 2e-

Thus, there is a continuous replenishment of electrolyte during electrolysis.

So a thin layer of coating metal is obtained on the article (anode). To get a strong,

adherent and smooth deposit certain additives are added to the electrolytic bath. The

favorable conditions for a good electro deposit are optimum temperature, optimum

current density and low metal ion concentration.

Electro less plating:

It is technique of depositing a metal on a catalytically active surface of the metal

to be protected by using a suitable reducing agent with out passing electrical energy. The

reducing agent reduces the metallic ions to metal, which gets plated over the catalytically

activated surface, giving a uniform coating.

This plating involes

i) Preparation of active surface of the object to be plated

ii) Preparation of the plating bath



Procedure

i)Preparation of active surface of the object:

This is achieved by using any one of the following methods

a)Etching:Removal of unwanted particles by acid treatment

b)Electroplating:A thin layer of the metal is coated on the surface of the object.

ii)Preparation of the plating bath:

The plating bath is composed of

i)Coating metal - must be asoluable salt of a metal.

ii)Reducing agent -HCHO, sodium hypo phosphate.

iii)Exaltant -Succinate, fluoride etc which enhances the plating rate.

iv)Complxing agent -EDTA, citrate , tartrate etc which improves the quality of the

deposit.

v)Stabiliser - Cations of lead , cadmium which prevent the decomposition of plating

bath.

vi)Buffer solution - Sodium acetate, NaOH + Rochelle salt etc added to control the pH

of the bath.

Procedure:

The article to be plated is immersed in the bath containing the salt of the metal and

reducing agent. The metal oin from the solution is reduced to the corresponding metal

and gets plated over the surface of then object.

At anode:

2HCHO + 4OH- → 2HCOO

- + 2H2O+ H2 + 2e

-

At cathode:

M2+

+ 2e- → M

Not Reaction:

M2+

+ 2HCHO + 4OH- → 2HCOO

- + 2H2O + H2 + M

Applications of electroless Plating:

1. Plastic cabinets coated with Cu and Ni finds applications in printed circuit

boards.

2. Electroless nickel coated polymers are used in decorative as well as functional

purposes.

3. Electroless nickel plating is extensively used in electronic appliances.

3. Write a brief note on Surface conversion coatings.



These coatings are produced by covering the surface of a metal by chemical or

electrochemical methods. The metal is immersed in the solution of a suitable chemical

which reacts with the metal surface, producing an adherent coating. These coatings give

good protection to the base metal from corrosion.

1. Phosphate coating

2. Chromate coating

3. Anodization or Anodised coating.

1) Phosphate coating (or) phospating:

Phosphate coatings are produced by dipping base metal in a bath of aqueous

solution containing phosphoric acid and metallic phosphates (zinc, iron and manganese

phosphates are generally used).

Phosphate coating coatings are usually applied to iron, zinc, steel, cadmium and

aluminium. During phosphate coating, phosphating solution dissolves and reacts with the

surface of the base metal, forms an insoluble metal – phosphate compound. This metal –

phosphate compound forms an adherent deposit over the base metal. It is grey in colour.

It is used as a base for the paints, lacquers which increase the resistance of the

films to humidity.



2) Chromate coating:

Chromate coatings are produced by dipping the base metal in a bath of aqueous

solution containing acidic potassium chromate followed by the immersion is a bath of

neutral chromate solution.

Chromate coatings are usually applied to zinc, cadmium, aluminium and

magnesium. During chromate coating, chromating solution dissolves and reacts with the

surface of the base metal forms and in soluble metal – chromate compound containing

mixture of Cr (III) and Cr (VI). This metal – chromate compound is amorphous, non-

porous and more corrosion-resistant than phosphate coatings. It is golden brown or black

in colour.

It is used as a base for paints, lacquers and enamels.

3) Anodization:

Anodization or anodic oxidation is an electrolytic process in which a thick oxide

coating is produced on the base metal. Anodized coatings are generally produced on

non-ferrous metals like Al, Zn, Mg and their alloys by anodic oxidation process in which

the base metal is made as an anode. It is carried out by passing a moderate direct current

through a bath in which the metal or alloy is suspended from anode. The bath usually

contains sulfuric, chromic, phosphoric oxalic or boric acid.

Eg: Anodizing on aluminium:

Anodized coating on aluminium is done by making aluminium as an anode in an

electrolytic bath containing sulphuric, chromic (or) phosphoric acid at moderate

temeperatures (35 – 40 o C) and the cathode is a plate of lead or stainless steel. On

passing current oxidation starts at anode and oxygen combines with the anodic metal to

form the oxide. The oxide film, initially will be very thin and grows from the metal

surface outwards and increases in thickness as oxidation continues at aluminium anode.

The oxide film is very porous and soft & this can be sealed by immersing it in boiling



water. This converts porous alumina in to hydrated aluminium which occupies more

volume thereby the pores are sealed.

Advantages:

i) an insulating coat for the electrically conducting base metal.

ii) very good resistance to corrosion.

iii) thick oxide coating, hence it is more protective.

iv) used in automobile engine piston.



UNIT-III

ELECTROCHEMISTRY

PART A

1. What is electrode potential?

The tendency of an electrode to lose or gain electrons, when it is in contact

with its own ions.

2. What is an electrochemical cell?

A device used to convert the chemical energy produced in a redox reaction

into electrical energy.

3. Write the Nernst’s equation for the electrode reaction?

Mn+

(aq) + ne- → M(s)

EMn+/M = E0

Mn+/M + 2.303 RT/ nF log [Mn+

(aq)]

4. Define a reference electrode?

The tendency of an electrode to lose electrons, when it is in contact with

solution of its own ions.

5. Why glass electrode cannot be used for solution of pH above 9.0?

At pH above 9.0, the ions of the solution affect the glass interface and

render the electrode useless.

6. Glass electrode is preferred to quinhydrone electrode in measuring pH of a

solution. Give reason.

Glass electrode is simple, not easily oxidized and attains equilibrium

rapidly. It can safely be used up to pH of 10. On the other hand, quinhydrone

electrode can be used upto a pH of 8 only. Moreover, it cannot be used in solutions

containing redox system. Hence, use of glass electrode in pH measurement is

preferred over quinhydrone electrode.

7. Electrode potential of zinc is assigned a negative value; whereas that of



copper a positive value.

Zn electrode is anodic wrt to SHE, so its electrode potential is assigned a

negative sign. On the other hand, Cu electrode is cathodic wrt to SHE, so its

electrode potential is assigned a positive sign.

8. What is galvanic cell or voltaic cell?

It is a simple device of producing electrical energy by chemical reaction, e.g.,

Daniel cell. Such a cell is also known as electrochemical cell.

ZnSO

4 CuSO4

Zn -------- > Zn+2 + 2e~ (oxidation)

Cu2+ + 2e --------- > Cu (reduction)

In the above cell Zn-electrode is anode and Cu-electrode is cathode.

9. What is an electrolytic cell?

It is a device used for converting electrical energy into chemical energy.

10. What do you mean by electrode potential (E)?

It is the tendency of an electrode in a half cell to lose or gain electrons when it is

in contact with the solution of its own ions.

11. What are reduction and oxidation potentials?

Reduction potential is the tendency of an electrode in a half cell to gain electrons

and oxidation potential is the tendency of an electrode in a half cell to lose electrons.

12. What do you mean by standard electrode potential (E°)?

It is the electrode potential of a metal in contact with its ions when the

concentration of ions is 1 M (1 molar).



13. What is a salt bridge?

It is an inverted U-tube containing an electrolyte (e.g., KC1, KNO3). It connects

(acts as a bridge) the solutions of the two half cells.

14. What is the E.M.F. of a cell?

It is defined as the potential difference between the two terminals of the cell

when no current is drawn from it.

15. Define normal hydrogen electrode.

It is a reference or standard reference electrode. Its electrode potential is taken

as zero at all temperatures. A normal hydrogen electrode generally consists of a Pt-

foil coated with platinum, dipped in solution having 1 M H+ ion concentration

and hydrogen gas at 1 atmospheric pressure constantly bubbled over it. It can be

represented as:

16. What do you mean by potentiometry?

Potentiometry is an electrochemical method of analysis based on

measuring the potential difference (e.m.f.) between two half cells, one of which

is an indicating electrode and the other is a reference electrode.

17. What is an indicating electrode?

It is an electrode in balance with an redox couple, the potential of which is

given by Nernst equation.

18. What are reference electrodes?

These are the electrodes whose potential is constant and independent

of the composition of the contacting solution.

The most widely used are saturated calomel electrodes (G = + 0.246 V) and

the silver-silver chloride electrode (e = + 0.222 V).

19. Define the electrochemical series?

When the standard reduction potentials of the electrodes are arranged in an



increasing order, the series so obtained is known as electrochemical series.

20. How do you increase the value of reduction potential of an electrode?

By increasing the concentration of metal ions in contact with metal

electrode or increasing the temperature.

PART-B

1. How do you measure EMF of a cell?

The difference of potential which causes flow of current from an electrode of

higher potential to an electrode of lower potential is called electromotive force.

Measurement of EMF of a cell:

The emf of a cell can be measured using a potentiometer. This measurement is

based on poggendorff’s compensation principle. In this method, the emf to be measured

is opposed by the emf of another cell or battery until the two emfs become equal and

there is no net flow of current in the circuit.The electrical assembly used is known as

potentiometer.

It consists of a uniform wire AC of high resistance. A storage battery of constant

emf is connected at the ends A and C by the wire.

The cell X, the EMF of which is to be determined is included in the circuit by

connecting the positive pole to A and the negative pole to the sliding contact J through a

galvanometer. The sliding contact is moved along the wire AB till no current flows

through the galvanometer.The distance AD is noted. The emf of the cell Ex is

proportional to the distance AD,

i.e.,Ex α AD ------(1)

The cell X is replaced by a standard cell of known emf,Es. The position of the

sliding contact is readjusted by moving it the wire AB, as before ,till no current flows

through the galvanometer, i.e., the null point,is reached again at D’ as shown in the



figure.

Now the emf of the cell is proportional to the distance AD’,

i.e., Es α AD’ --------(2)

From (1) & (2)

Ex / Es = AD/ AD’

Ex= Es AD/ AD’

Hence the emf of unknown cell is measured.

2. Write notes on calomel electrode

This electrode consists of mercury, mercurous chloride and a solution of saturated

KCl. The cell is represented as;

Pt,Hg,Hg2Cl2 /KCl(sat)

Mercury is placed at the bottom of the electrode. Mercury is covered by a paste of

mercurous chloride and a solution of saturated KCl is introduced above the paste. A

platinum wire sealed in to a glass tube serves to make electrical contact of the electrode

with the circuit.

Calomel electrode

If the electrode acts as anode, the electrode reaction will be;

2Hg + 2Cl-

Hg2Cl2 +2e-

If the electrode acts as cathode the electrode reaction will be;

Hg2Cl2 +2e-

2Hg + 2Cl-

The electrode potential can be written as

E= Eº -0.0591/2 log[Cl- ]

2 or E= Eº -0.0591 log[Cl

- ]

The electrode potential depends on the concentration of chloride ion.



3. Discuss the construction of glass electrode.

It is an electrode, which produces current in response to a specific ion present in

the solution. This electrode is otherwise known as ion selective electrode.

Construction:

A glass membrane electrode consists of thin walled glass bulb containing AgCl

coated Ag electrode or platinum wire in 0.1M HCl. There is an equilibrium exist between

H+ ions of solution and Na

+ ions of glass. For a particular type of glass the electrode

potential depends on the concentration of H+ ions. This can be represented as;

AgCl, Ag, HCl (0.1M)/Glass or Pt, HCl (0.1M)/Glass

If this electrode acts as anode , the electrode reaction will be;

½ H2 H+ + e

-

If this electrode acts as cathode , the electrode reaction will be;

H+ + e

- ½ H2

The electrode potential can be written as;

E= Eºg -0.0591 log[H+] -----------(oxidation potential)

E = Eºg +0.0591pH

Determination of pH

with the help of glass electrode: The electrode is



dipped into the solution where pH is to be determined and is usually

combined with a reference electrode— generally calomel electrode—to

form a complete cell. The potential of the glass electrode varies linearly with

pH .

`

Solution of unknown pH

Ecell= ER - EL

Ecell= Ecalomel - Eglass

Ecell = +0.2422- ( Eºglass -0.0591log [H+]

Ecell = +0.2422- Eºglass -0.0591pH

pH

=+.2422 - EºG -Ecell /.0591

5. What is meant by EMF series? Explain the importance of EMF series.

The electrode potentials of various electrodes are arranged in the order of

increasing standard reduction potentials with respective to hydrogen scale.This

arrangement is known as electrochemical series

Importance of electrochemical series:

Saturated

calomel electrode

Glass

electrode



Calculation of standard EMF of a cell:

The standard EMF of a cell can be calculated using the standard reduction

potentials of right hand side and left hand side electrodes from the emf series.

Eºcell= EºR - EºL

Relative ease of oxidation and reduction:

Higher the value of standard reduction potential, greater is the tendency to get

reduced. For example, in the electrochemical series, fluorine has high positive value of

reduction potential (+2.87 V) and is reduced. Lower the value of standard reduction

potential, greater is the tendency to get oxidized. In the electrochemical series, lithium

has the lower negative reduction potential value (-3.05V) and is oxidized.

Predicting spontaneity of a reaction:

The spontaneity of a reaction can be determined from the standard emf of the cell.

If the emf of a cell is positive, the reaction is spontaneous. If it is negative, the reaction is

non spontaneous.

Eºcell= EºR - EºL = + ive (spontaneous)

Eºcell= EºR - EºL = - ive (non spontaneous)

Displacement behaviour of hydrogen:

If zinc electrode is immersed in H2SO4 solution, it displaces hydrogen from acid

solution.

Zn + H2SO4 ZnSO4 +H2

If silver electrode is immersed in H2SO4 solution, it will not displace

hydrogen from acid solution. Because it has higher reduction potential value.

Ag + H2SO4 No reaction



6. Derive Nernst equation for electrode potential.

Nernst equation for electrode potntial:

Consider the following redox reaction:

Mn+

+ ne- M

For such a redox reversible reaction, the free energy change (∆G) and its equilibrium

constant (K) are inter related as

∆G= - RTlnK + RTln [Product]/ [reactant]

∆G = ∆Gο + RTln [Product]/ [reactant] [ ∆G

ο = - RTlnK ]

Where ∆Gο =Standard free energy change

The above equation is known as Vant Hoff reaction isotherm. The decrease in free

energy in the above reversible reaction will produce electrical energy.

-∆G = nFE and ∆Gο =- nFE

ο

Where E is the electrode potential, Eο is the standard electrode potential, F = 96500

coulombs and n is the number of electrons transformed.

Comparing equations (1) and (2),it becomes

- nFE = -nFEο + RTln [ M]/ [M

n+] (consider the reduction reaction)

- nFE = -nFEο + RTln 1/ [M

n+] ([M] = 1)

Dividing eqn (3) by – nF



E = Eο - RT/nF ln 1/ [M

n+]

or

E = Eο + RT/nF ln [M

n+]

E = Eο + 2.303 RT/nF log [M

n+]

Where R =8.314 J/K/mole, F = 96500coulombs, T = 298K, the above equation becomes

E = Eο + .0591 /n log [M

n+] This the Nernst equation for reduction potential

For oxidation potential

E = Eο - .0591 /n log [M

n+]

Application:

i) It is used to calculate electrode potential of unknown metal.

ii) It can be used to calculate the emf of a cell.

7. Write notes on potentiometric redox titrations.

Potentiometric Titrations:

A titration in which concentration change is followed by potential change is called

a potentiometric titrations. Two electrodes are required to measure the potential of the

test solution. The potential of one electrode should be constant while that of the other

should change with concentration of the test species. The former is called reference

electrode and the latter is called an indicator electrode which is different for different

types of titrations. For a redox titration (Fe2+

Vs Cr6+

), calomel electrode is used as a

reference electrode and platinum is used as an indicator electrode. The two electrodes are

connected to a potentiometer.



Redox Titration:( Fe2+

Vs Cr6+

)

Titration of ferrous sulphate vs potassium dichromate is Fe2+

Vs Cr6+

.When

K2Cr2O7 is added to ferrous sulphate in acid medium, ferrous ion is oxidized to ferric ion

hexavalent Cr is reduced to trivalent state. The potential of this system depends on ratio

of ferrous to ferric.

trivalent Cr will be present. So the potential of this system will follow the concentration

ratio of hexavalent Cr to trivalent Cr. When the potential is plotted against the volume of

dichromate added, the following graph is obtained.



The potential is found to increase with the addition of K2Cr2O7. Initially the

increase in potential is gradual but at the end point, the increase is very steep because

almost all Fe2+

would have been converted to Fe3+

.Thereafter Fe2+

will not be present.

Beyond the end point , addtion of excess Cr6+

established equilibrium with Cr3+

present in

the solution. The potential now depends on Cr6+

/Cr3+

.The diffential graph is drawn for

getting more accurate end point.



The mid point of the curve is known as the end point. The cell is represented as

Pt, Hg, Hg2Cl2/ KCl(sat)//Fe2+

, Fe3+

(H+)/ Pt .

8. Write notes on conductivity Acid- Base Titration.

Strong Acid Vs Strong Base:

A known amount of acid (HCl) is taken in a beaker and alkali (NaOH) is taken in

the burette. A conductivity cell is introduced into the beaker solution. Initially the

conductivity of HCl is high, this is due to the presence of fast moving H+ ions. As NaOH

is added gradually, conductance will be going on decreasing until the acid has been

completely neutralized. This is due to yhe replacement of fast moving H+ ions by slow

moving Na+ ions. The point B indicates the complete neutralization of all H

+ ions.



HCl +NaOH NaCl + H2O

Further addition of NaOH will introduce fast moving OH- ions.Therefore the

conductance value will begin to increase. On plotting conductance against volume of

NaOH added, the two lines intersect at a point B which corresponds to the neutralization

point.

Advantages:

i) Conductometric titrations give more accurate end point.

ii) It can be used to titrate col,oured solutions where the colour change of the

indicator is not clear.

iii) It can be used for titratig dilute solutions.

iv) There is no need to have keen observation near the end point, as it is

determined graphically.



Problems

1. Calculate the emf of a cell at 25°°°°C, when the concentration of ZnSO4 and

CuSO4 are 0.001 M and 0.1 M, respectively. The standard potential of cell is

1.2 V.

Solution

Cell is: Zn(s) | Zn2+

(0.001 M) || Cu2+

(0.1 M) | Cu(s)

Ecell = Eo

cell + (0.0592 /2) log [Cu2+

]/ [Zn2+

]

= 1.2 + 0.0296 × 2

= 1.2 + 0.0592

= 1.2592 V

2. What is concentration of Ni2+ in the cell at 25°C, if the emf is 0.601 V?

Ni(s) | Ni2+

(a = ?) || Cu2+

(0.75 M) | Cu(s)

Given

Eo

Ni/Ni2+

= 0.25 V ; Eo

Cu2+

/Cu = 0.34 V

Solution

Ecell = Eo

cell + (0.0592 /2) log [Cu2+

]/ [Ni2+

]

0.601 = (0.34-(-0.25) + 0.0296 log [0.75/a]

0.011 = 0.0296 log 0.75/a

log 0.75/a = 0.011 / 0.0296

0.75/a = Antiliog 3.7162

a = 0.75 / 2,3529

= 0.3188 M.



UNIT-IV

PHASE RULE

PART-A

1. State Phase rule.

The phase rule is a generalization which explains the heterogeneous

equilibria. Mathematically, it is stated as

F = C-P +2

Where

F – the number of degrees of freedom for the equilibria

C – the number of components

P – the number of phases present in equilibrium

2. Define Phase?

A phase is defined as any homogeneous and physically distinct part of a

system which is mechanically separable from each other parts of the system by a

boundry surfaces. A phase may be solid, liquid and gas.

3. Define component?

Component is defined as the smallest number of independently variable

constituents in terms of which the composition of each phase can be

expressed in the form of a chemical equation.

4. Define Degrees of freedom (F)?

It is the minimum number of independent variables of a system,

which can completely define the equilibrium of a system, such as

temperature, pressure, concentration



5. What is an invariant system? Give an example.

A system in which the degree of freedom is zero is called an invariant

system. This signifies that no condition is required to define the system at that

point.

Example:

Triple point of water is where ice, water and water vapour are in equilibrium.

6. What is the significance of triple point?

A triple point for a system is invariant. It is constant for a substance at a set

of temperature and pressure conditions. If temperature or volume is altered, one

phase of the substance disappears and the system becomes univariant from

invariant.

7. State two important merits of phase rule.

(i) Phase rule is applicable to both physical and chemical equilibria.

(ii) Phase rule helps to predict the behaviour of a system under different

sets of conditions.

8. State limitations of phase rule.

(i) Phase rule can be applied only for a system which is in equilibrium.

(ii) Phase rule conditions that all phases of the system must be present

simultaneously under similar set of conditions.

9. What do you mean by eutectic? What is the eutectic composition of Pb-Ag

System? Give one application of eutectics.

Eutectic is a solid solution of two or more substances having the lowest

freezing point of all the possible mixture of those components. Eutectic

composition of Pb-Ag system —> 97.4 % Pb, 26% Ag. By suitable choice, very

low melting alloys can be selected for preparing safety devices like fire sprinklers,

plugs in automobiles etc.



10. How many degrees of freedom are there for the following systems? (i)

Water(l) →→→→ Water vapour(g) (ii) Gaseous mixture of H2 and O2

(i) For Water(l) →→→→ Water vapour(g) system either temperature or pressure is to be

specified to define the system completely. Hence degree of freedom is one.

(ii) For the gaseous mixture of H2 and O2 both the pressure and the temperature are to

be specified to define the system completely, so the system has two degrees of freedom.

11. For a two-component alloy system of Pb and Ag, state the degrees of freedom

(i) as per phase rule and (ii) as per condensed phase rule.

(i) For a two-component alloy system of Pb and Ag when P = 1, degree of

freedom according to phase has the highest value of F = C - P + 2

= 2 - l + 2 = 3.

(ii) As per condensed phase rule:

For solid-liquid equilibrium of alloy, practically no gas is present and hence

at atmospheric condit ion the pressure remaining constant, degrees of freedom

reduces by one and F becomes F = C - P + 1 = 2 - 1 + 1 = 2.

12. How many phases and components are present in the system,

CaCO3 (s) →→→→ CaO(s) + CO2(g)

The system consists of two solid phases and one gaseous phase. Hence,

P = 3. It is a two component system.

13. What do you meant by phase diagram?

A phase diagram is a graph obtained by plotting one degree of freedom

against another. If temperature is plotted against pressure, the diagram is called

temperature-pressure diagram.



14. What is the effect of pressure on the melting point of ice?

The melting point of ice decreases with increase of pressure.

15. Define Condensed phase rule?

For two-component alloy systems it is considered that there is practically

no gas phase and the effect of pressure is negligible. So the system is considered

to be under atmospheric pressure which reduces the degrees of freedom by one

and hence,

F=C-P+1

16. Define Triple point.

It is the point of a system at which the gaseous, liquid and solid phases of

a substance can co-exist in equilibrium. At triple point the system is non-variant.

17. What do you meant by Eutectic point?

A solid solution of two or more substances having lowest freezing point is

the eutectic mixture and the minimum freezing point attained is the eutectic point.

18. Is it possible to have a quadrupole point in phase diagram of one

component system?

Quadrupole point means 4 phases are present

P = 4; C = 1;

F = C – P + 2

F = 1 – 4 + 2

= - 1

Since, the degree of freedom, F = -1, it is not possible.

19. Define eutectic system?

A binary system consisting of two substances which do not react chemically but is

miscible in all proportions in liquid phase is called "eutectic system".



20. Define Ferous alloys and non-ferrous alloys.

Non-ferrous alloys do not contain iron as one of the constituents of the alloys,

e.g., Brass which is composed of Zn and Cu. Whereas ferrous alloys always contain

iron as one of the constituents of the alloys, e.g., Stainless steel contains Fe, Ni, Cr.

21. Give the Importance of alloy preparation

• to increase the hardness and tensile strength

• to improve the casting property

• to increase the resistance to corrosion

• to decrease the thermal and electrical conductivities.

• to lower the melting and boiling points of individual metal.

PART-B

1. Define Phase rule? Explain the terms involved?

Phase. "Phase is defined as any physically distinct portion of matter which itself is

homogeneous and uniform in composition, mechanically separable from other parts by

definite boundary surfaces".

(a) Thus, gases being mutually miscible in all proportions will constitute one

phase only. Thus mixture of H2 and O2 constitutes single phase.

(b) Mixture of two completely miscible liquids has single phase.

(c) Solution of a solute in a solvent constitutes single phase such as salt solution in

water.

(d) If two liquids are immiscible, they form two phases such as chloroform and

water.

(e) Except solid solutions all different kinds of solids form different phases.

(f) If we keep a mixture of miscible liquids in a closed vessel, above the liquid

mixture, there will be some vapors of the liquids, so the system will have two

phases.

(g) At freezing point, water consists of three phases:

Ice(s) → Water(f) → Water vapor(g)



(h) A heterogeneous mixture of the type

CaCO3(s) → CaO(s) + CO2(g)

It consists of three phases—two solid and one gaseous.

Component.

Component is defined as the smallest number of independently variable

constituents, in terms of which the composition of each phase can be expressed in the

form of a chemical equation.

This concept of component can be explained in connection with phase rule with

the help of the following examples.

(a) In ice-liquid water-water vapor system, the composition of each phase can be

expressed by a single component i.e., H2O. So, it is a one-component system.

(b) A salt solution will be a two-component system.

(c) In the thermal decomposition of MgCO3,

MgCO3(s) → MgO(s) + CO2(g).

The composition of each of the three phases can be expressed in terms of

at least any two of the three constituents, then the composition of any one

phase can be represented as

Phase: MgCO3 → MgCO3 + OMgO

Phase: MgO → MgCO3 + MgO

Phase: CO2 → MgCO3 - MgO

So it is a two-component system.

(d) Suppose a solid dissociates into a number of gaseous substances in

a closed vessel

NH4Cl(s) → NH4Cl(g) → NH3 (g) + HCl(g)

The overall composition of the dissociated vapour (NH3, HC1) is exactly the

same as that of the undissociated substance (NH4C1). Thus, the number of

component is one. However, if NH3 or HC1 is introduced in the system in excess,

the system becomes a two-component system.

(e) In the system of sodium sulfate-water, various phases may exist like

Na2SO4; Na2SO4, 7H2O; Na2SO4, 10H2O; ice; solution and vapor. The composition



of each phase can be expressed in terms of Na2SO4 and H2O, so the system is a

two-component system.

(f) In the equilibrium,

Fe(s) + H2O(l) → FeO(s) + H2

the minimum number of components required to express the composition is

three and hence is a three-component system.

Degrees of Freedom or Variance.

The number of degrees of freedom of a system is the minimum number of

the independent variables of a system, such as temperature, pressure and

concentration, which can completely define the equilibrium of a system.

(a) A system consisting of pure gas or gas mixture. It is a one component,

one-phase system. If the temperature and pressure are specified, the volume of

the gas is known. Hence the degree of freedom is two.

(b) The system ice(s) → Water(l) → vapour(g) is one-component,

three-phase system and all these three phases can co-exist at the freezing point

of water at a particular temperature and pressure. So this system will have no

degree of freedom i.e., it is a non-variant or invariant system. If temperature

or pressure is altered, one of the phases will disappear and the three phases

will not be in equilibrium.

(c) For the system, NaCl(s) →NaCl-Water(aq) →H2O(l) the solubility at

the saturation point is fixed by either temperature or pressure, so the system

is univariant.

(d) For the simple system water (l) → Water vapour (g) temperature or the

pressure is required to define the system, hence the system is univariant.

2. Draw a neat labeled phase diagram of water system and explain areas, curves

and triple point in it.

Water System

This system consists of three phases—ice, water, vapor and one-component



H2O. From Fig. 11.1, we find that curve BO is the vapor pressure curve of ice which

indicates that ice has a small but definite vapour pressure.

The point B has a natural limit of— 273°C, beyond which the two phases

merge. OA is the vapour pressure curve of water, also called the vaporisation curve.

The OA curve terminates at A, the critical temperature of +374°C above which

distinction between liquid and vapor vanishes.

For both of these curves OA and OB, they are one-component systems with two

phases (C = 1, P = 2). Since F = 1 there is only one degree of freedom i.e., if the

temperature is fixed, all other properties are also fixed.

The point of intersection of these two curves is point 0, called the triple

point, because the three phases, ice-water-vapour are in equilibrium, i.e., it is

actually the melting point of ice i.e., 0.0098°C under 4.579 mm pressure and at this

point C = 1, P = 3, Hence F = 0 i.e., at this point neither temperature nor pressure

can be altered without the disappearance of one of the three phases. The curve OC

represents the change of melting point of ice with pressure. The inclination of OC

towards the pressure axis shows that melting point of ice is decreased by

increasing pressure.



The curves OA, OB and OC divides the whole region into three portions

AOC, BOC and AOB in which only one phase is present and so the system

becomes bivariant and to locate any point in these areas, temperature as well as

pressure is to be known because the degree of freedom F = l - l + 2 = 2.

The dotted portion OD, which is the continuation of the curve AO below 0°C

is the vapor pressure curve of supercooled water below 0°C. This curve represents

a metastable system. This curve OD runs above the vapor pressure curve for ice

for the above reason and on slight disturbance the supercooled water immediately

changes to the stabler form i.e., ice.

3. Draw a neat phase diagram of Fe-C system and label it.

The equilibrium diagram of Fe-C is especially useful to understand the heat

treatment of steel. The phase-diagram of iron-iron carbide system is used upto 6.67%

weight of carbon. Pure iron exists in three allotropic modifications a, y and 8. C

atoms being small compared to iron atoms form only interstitial solid solutions and

the solubility of C in these allotropes is quite different.



In the phase diagram, the curve ABCD is the liquidus line and AEPGCH is the

solidus line. Above the liquidus there is only liquid phase whereas below the solidus

there is only solid phase and between the two lines both phases are present. On

dissolving C in molten iron (8 form at 1535°C) the freezing point is lowered until at

eutectic temperature of 1130°C corresponding to 4.3% C by weight. On further

addition of C, freezing point increases upto 6.67% C content and it is called cementite.

Pure iron changes from 8 to a form at a temperature of 910°C. The phase

diagram indicates that upto 0.088% C is commercially pure iron, whereas 0.088% to

2% C content is steel and C content above 2% represents cast iron.

From the curve, AEB area represents 8-iron and liquid. Area DCH represents

cementite and liquid. Area BCGP represents austenite and liquid and finally PGKI

represents 8-iron (i.e. austenite) only.

The transformation of one solid phase into another takes place at the critical

temperatures indicated on the diagram. For an iron-C alloy (cast iron) containing 3%

C, freezing begins at 1270°C. On further cooling, austenite separates out from the

liquid, then at 1130°C, the mixture is austenite containing 2% C and liquid has

eutectic composition.

Hence further removal of heat leads to total solidification of the liquid to

ledeburite, a mixture of austenite and cementite. Cooling further leads to the

decomposition of the eutectic mixture of cementite and finally at 723°C the residual

austenite transforms to pearlite and at temperature below 723°C, all the ledeburite is

transformed into pearlite and cementite mixture. The Fe-C phase diagram shows the

phases, which are formed when the cooling rate is slow i.e., when equilibrium is

attained. But with faster cooling rates metastable phases like martensite and bainite

are formed which are not shown in the phase diagram.



Phase diagram of Fe-C system



4. Explain the Pb-Ag system using phase diagram.

Lead-Silver-Sample (Eutectic System)

It is a two-component system with four possible phases—solid Ag, solid Pb,

solution of Ag + Pb and vapor. At constant atmospheric pressure, the vapour phase is

absent and the condensed phase rule is applicable as F = C — P + 1.

Curve AO represents the freezing point curve of Ag. A is the melting point

(961°C) of Ag. AO represents the curve along which the melting point of Ag falls

gradually on addition of Pb till the lowest point (303°C) is reached when the solution

gets saturated with Pb and hence any extra Pb added gets precipitated as solid phase

and so also melting point of Ag cannot be lowered any further. This point 0 is the

eutectic point (2.6 % Ag and 97.4 % Pb) and the whole mass at this point crystallizes

out with the above fixed composition.

Similarly, another curve BO is obtained starting from the melting point of pure

lead (327°C) at B, as Ag is added the melting point gradually falls till the lowest point

0 is reached with the above mentioned composition. According to reduced phase rule

equation, the system is univariant along AO and BO and at point 0, F=3 — P=3 — 3 =

0 i.e., invariant, which is the eutectic temperature and a lead-silver mixture can never

have a melting point below the eutectic temperature. The region above the curve AO and

BO represents solution of Pb and Ag and below the curve AO represents solid eutectic with

crystalline Ag and below the curve BO represents solid eutectic with crystalline Pb.

Phase diagram of Pb-Ag



Application:

The phase diagram can be used to separate Ag from argentiferrous lead ore.

This process is known as Pattinson’s process of desilverization of argentiferrous

lead.

Pattinson’s process of desilverization of argentiferrous lead.

The phase diagram of lead – silver is useful in the extraction of silver from

argentiferrous lead ore which has a very small percentage (.1%) of silver.This

process is known as Pattinson’s process. Let ´p’ represent the molten

argentiferrous lead containing very small amount of silver in it. It is a

homogeneous liquid and on cooling, the temperature falls with out change in

concentration till any point p on the curve OC is reached. On further cooling lead

begins to separate out and the solution becomes richer in silver. Further cooling

will shift the system along the line PO ,more of the lead separates out as solid till

the point O is reached when the percentage of Ag rises to 2.6%. This process of

increasing the relative proportion of silver in the alloy is known as Pattinson’s

process of desilverization of argentiferrous lead.

5. Discuss briefly the heat treatment of stainless steel?

Heat treatment of steel:

Heat treatment is defined as the process of heating and cooling of solid steel article

under carefully controlled conditions, thereby developing certain physical properties with

out altering its chemical composition.

The main characteristics and the relevant heat treatment processes are

i) Annealing ii)Hardening iii)Tempering iv) Normalising andv)Case –

hardening.

i)Annealing:

Annealing means softening. This is done by heating the metal to high temperature,

followed by slow cooling in a furnace. It increases machinability and removes the

imprisoned gases and internal streses. Annealing can be done in two ways:

i) Low temperature annealing



ii) High temperature annealing

ii)Hardening or Quenching:

Heating steel beyond the critical temperature and then suddenly cooling it either in oil

or brine solution or some other fluid It increases the hardness of steel. The faster the rate

of cooling, harder will be the steel produced. It increases its resistance to wear, ability to

cut other metals and strength.

iii)Tempering:

It is the process of heating the already hardened steel to a temperature lower than

its own hardening temperature and then cooling it slowly. In tempering, the temperature

to which the hardened steel is reheated is of great significance and controls the

development of the final properties. It removes any stress and strains that might have

developed during quenching. It reduces brittleness and also some hardness but toughness

and ductility are improved.

iv)Normalising:

It is the process of heating steel to a definite temperature and allowing it to cool

gradually in air. It recovers the homogeneity of the steel structure, removes internal

stresses and increases the toughness. This steel is suitable for engineering works.

v) Case hardening:

It is a process through which a wearing surface is produced on steel having a soft core

inside. This process is carried out in two stages:

i) Carburising ii)Hardening

i)Carburising:

The mild steel article is taken in a cast iron box containing small pieces of charcoal. It

is heated to about 900 to 950°C and allow to keep it as such for sufficient time, so that

the carbon is adsorbed to a required depth. The article is then allowed to cool slowly

within the iron box itself. Now the outer skin of the article is converted to high carbon

steel containing about .8% to 1.2% carbon.



a)Nitriding:

It is a process of getting super hard surface. In this, the metal alloy is heated in

presence of ammonia at a temperature of about 550°C. The nitrogen combines with the

surface constituents of the alloy to form extremely hard nitrides.

b)Cyaniding:

It produces a hard surface on low or medium carbon steels by immersing the metal in a

molten salt containing cyanide (KCN or NaCN ) at a temperature of about 870°C and

then ,quenching in oil or water. The hard surface is produced due to the absorption of

carbon and nitrogen by the metal surface.

6. Discuss the composition, properties and applications of the following

i)Nichrome ii)Stainless steel

i)Nichrome:

It is an alloy of nickel and chromium.

Nickel 60%

Chromium 12%

Iron 26%

Mangenese 2%

Properties:

i)It shows good resistance to oxidation and heat.

ii)Steels containing 16 to 20% chromium with low carbon content (.06-.15) possess

oxidation resistance upto 900°C.

iii)Steels containing 18% nickel with small amounts of chromium can withstand

temperature above 900°C.

iv)It possesses high melting point.

v)It can withstand heat up to 1000° to 1100°C.

vi) It posseses high electrical resistance.



Applications:

i) It is widely used for making resistance coils, heating elements in stoves.

ii) It is used in electric irons and other household electrical appliances.

iii) Used in making parts of boilers, gas turbines, aero engine valves etc.

iv) Used in making other machineries or equipments exposed to very high

temperatures.

Stainless steels or Corrosion resistant steels:

Composition

These are alloys containing chromium together with other elements such as Ni,

Mo, etc., Chromium is effective if its content is 16% or more.The carbon content in steel

ranges from .3-1.5%.

Type of stainless steels:

There are two main types of stainless steels:

i) Heat treatable stainless steels ii) Non heat treatable stainless steels

i) Heat treatable stainless steels:

These steels mainly contain up to 1.2% of carbon and less than 12-16% of

chromium.

Properties:

They are magnetic, tough and can be worked in cold condition.

Application:

i) They can be used up to 800°C.

ii) They have good resistance towards weather and water.

iii) They are used in making surgical instruments, scissors, blades etc.,

Non heat treatable stainless steels:

These steels possess less strength at high temperature. They are more resistance to

corrosion.

Types of non heat treatable steels:



Based on their composition, they are if two types.

i) Magnetic type ii) Non magnetic type.

Magnetic type

It contains 12- 22% of chromium and .35% of carbon.

Properties:

a) It can be rolled and machined by the use of specially designed tools.

b)It resists corrosion better than heat treatable stainless steels.

Application:

It is used in making chemical equipments and automobile parts.

Non magnetic type:

It contains 18-26% Cr, 8-21% Ni and .15% of carbon.Total percentage of Cr and Ni in

such steel is more than 23%.

Properties:

i)It exhibits maximum resistance to corrosion.

i) Corrosion resistance of which can be further increased by adding a little

quantity of molybdenum.

Application:

It is used in making household utensils, sinks. Dental and surgical instruments.

7. Write a brief note on non-freeous alloys?

Non ferrous alloys:

These alloys do not contain iron as one of the main constituent. The main constituents of

non ferrous alloys are Cu, AL, Pb, Sn etc., The melting points of non ferrous alloys are

lower than those of ferrous alloys.

Important non ferrous alloys:

i) Brass ii) Bronze

i) Brass:

Brass is an alloy of copper and zinc. They possess

a) greater strength, durability and machinability than copper.

b) Lower melting points than Cu and Zn.



c) Good corrosion resistance property.

Composition of Brass:

Cu - 60 –90%

Zn - 40 –10%

Important brasses and their properties:

Bronze:

Bronze is a combination of Cu and Sn.

Composition of bronze:

Cu =80%, Sn =20 –5%

They possess

i) lower melting point than steel and are more readily produced from their constituent

metals.

ii) better heat and electrical conducting property than most of the steels.

iii) Non oxidizing, corrosion resistance and water resistance property.

Types of brass

Composition Properties Applications

Guilding metal Cu=90%,Zn=10% Golden yellow in colour,

stronger and harder than

pure Cu.

In hard wares,

jewellery etc.,

Dutch metal Cu=80%,Zn=20% Golden yellow in

colour,suitable for all

drawing and forming

operations.

Cheap jewellery,

battery caps,

flexible hoses,

tubes etc.,

German silver Cu=25 -50%,

Zn=10-35%

Sn=5-35%

Ductile, malleable and

looks like silver

Utensils, bolts,

screws,

ornaments etc.,



Important bronzes and their properties:

Types of bronze Composition Properties Application

Coinage bronze Cu =89-92%,

Sn=11-8%

Soft, ductile and

durable.

Pumps, valves,

wires, utensils,

coins, statues etc.,

Gun metal Cu=85%

Zn=4%

SN=8%

Pb=3%

Hard, tough, strong

to resist the force of

explosion.

Foundary works,

heavy load bearing,

steam plants etc.,

Aluminium bronze Cu=90-93%

Al=10-7%

Golden yellow in

colour, strong,

readily fusible,

resistance to

corrosion etc.,

Bushes, jewellery,

utensils, coins and

photoframes.



UNIT-V

PART-A

1. What is atomic spectroscopy?

Ground state of an atom means it is with normal electronic configuration. In

this state the atom remains in its lowest energy state and this is the most stable state

of the atom. Excited state of an atom refers to the electronic configuration availed

by an atom after absorbing certain definite amount of energy. The valence electrons are

promoted to some higher permitted energy level by absorption of energy. In the excited

state, the atom is unstable and the excited electron tends to come back to the original

position i.e., ground state. After about 104 sec. the electron returns to the ground state

by emitting the amount of energy absorbed during excitation. The energy is emitted or

absorbed in the form of electromagnetic waves of definite frequency i.e., of definite

wavelengths.

2. What is atomic absorption spectroscopy?

This is the analytical technique based on the phenomenon of light absorption

(UV or visible). It is applicable both to qualitative and quantitative analyses.

3. What are the parameters for expressing the absorption?

Transmittance is the ratio of the intensity of light transmitted to the intensity of

incident light.

Where

A = Absorbance,

Io = Intensity of incident light,

I = Intensity of transmitted light



4. Calculate the concentration of a substance A in an ethanolic solution of which the

absorbance in a 1 cm cell at its A, λλλλmax 241 nm was found to be 0.890. The εεεε is 540 at

241 nm.

A= ε C l

0.890 = 540 x l x C

C = 0.00165 g/100 ml.

5. Mention the applications of UV.

(a) Qualitative:

Detection of conjugation

Detection of functional groups

Detection of geometrical isomers.

(b) Quantitative:

• Analysis of various samples (drugs, dyes etc.)

6. What are the different electronic transitions that take place on absorption of UV

light?

When a molecule absorbs UV radiations the electrons are excited to higher energy

levels. In the diagram below the electrons are represented

The following electron transitions take place:

σ→σ* , n→σ*, n→π* and π→π*.

These electronic transitions are responsible for UV absorption of a molecule.

7. How does molecular spectrum arise?

It arises due to the interaction of electromagnetic radiation with molecules. This

results in transition between rotational, vibrational and electronic energy levels.

8. State Beer-Lamberts law?

According to this law, when a beam of monochromatic radiation is passed

through a solution of an absorbing substance, the rate of decrease of intensity of radiation

‘dI’ with thickness of the absorbing solution ‘dx’ is proportional to the intensity incident

radiation (I) as well as the concentration of the solution (C). It is mathematically



represented as

- dI/ dx = k I C

Diagram

9. Name the important components of colorimeter?

1. Radiation source

2. Filter

3. Slits

4. Cell

5. Detector

6. Meter

10. Write the complexes formed in the colorimetric determination of Fe?

Fe3+

+ NH4CNS → blood red colour complex

11. What are the important processes that occur in the flame photometry?

1. It should evaporate the solvent from the sample solution

2. It should decompose the solid into atoms

3. It should excite the atoms and cause them to emit radiant energy

12. What is finger print region? Mention its important uses?

The vibrational spectral region at 1400-1700 cm-1 gives very rice and intense

absorption bands. This region is called as finger print region. It can be used to detect the

presence of functional group and also to identify and characterize the molecule just as a

finger print can be used to identify a person.



13. What are the principles of IR?

The atoms in a molecule bond are in a state of constant vibration and

rotation. They may be compared with two balls (atoms) joined by spring (bond). On

absorption of IR the bond may stretch, bend etc., as shown below. So stretching and

bonding of bonds are responsible for IR absorption.

14. How is an I.R. spectrum recorded?

IR source: A Nernst glower, a rod of an allow of Zirconium, Yttrium

and Erbium oxides. The rod is electrically heated to 1750 K

Rock salt disc or KBr disc is used as glass and quartz absorb I.R.

Sample preparation: Either the sample with KBr is made to pellet or

Nujol mull is used. Nujol is hydrocarbon in nature.

Recording of spectra: The sample is placed in Rock salt cell in the path of

I.R. The change of intensity of light transmitted draws a graph which is

IR spectrum.

15. How will you distinguish CH3COOH from CH3COCH3 with the help

of I.R. spectra?

C = O (str.) peaks will be observed in both the spectra in the region of 1700 cm-1

.

But an absorption bond at 2500—3000 cm-1

(broad) will be observed in spectrum of

CH3COOH due to dimeric association of CH3COOH molecules through hydrogen

bonding.

16. What happens to a molecule when it is irradiated with (a) UV-Vis., and (b) IR

light.

UV-Vis., light causes electronic transition

IR light causes vibrational and rotational transitions



17. How to calculate the number of vibrational modes for a different types of

compounds?

i) For a non-linear molecule containing ‘n’ atoms, the number of vibrational

modes (3n-6)

ii) For a linear molecule containing ‘n’ atoms, the number of vibrational modes

(3n-5)

18. Name two fuel gases used in flame photometry?

1. Acetylene

2. Propane

19. What is the main application of flame photometry?

Analysis of alkali metals, particularly in biological fluids and tissues.

20. How can you identify an unknown element from emission spectrum?

By comparing the unknown spectrum with the spectrum of a known

element.

21. Find out the absorbance of a solution if the transmittance of a solution is

18.5%.

Solution:

% Transmittance = 18.5 %

T = 0.185

Absorbance, A = - log T

A = - log 0.185

= 0.733



PART B

1. Explain flame photometry with a neat diagram?

Components

The various components of the flame photometer are described as follows:

Burner

The flame must possess the following functions:

(i) it should evaporate the solvent from the sample solution

(ii) it should decompose the solid into atoms

(iii) it should excite the atoms and cause them to emit radiant energy.

Mirror

The radiation from the flame is emitted in all directions in space. In order

to increase the amount of radiation reaching the detector, a concave mirror is used

which is set behind the burner.

Slits

Entrance slits: It is kept between the flame and monochromator. It permits only

the radiation coming from the flame & mirror.

Exist slits: It is kept between the monochromator and detector. It prevents the

entry of interfering lines.

Monochromator

It allows the light of the required wavelength to pass through but absorbs

the light of other wavelength.



Detector

The radiation coming from the filter is allowed to fall on the detector,

which measures the intensity of the radiation falling on it. Photo cell or photo

multiplier is used as detector, which converts the radiation into an electrical

current.

Amplifier & Recorder

The current coming out from the detector is weak, so it is amplified and

recorded.

2. Expalin the estimation of sodium by flame photometry.

The instrument is switched on. Air supply and gas supply are regulated.

First distilled water is sent and ignition is started. After the instrument is warmed

up for 10 min, the instrument is adjusted for zero reading in the disply. Since

sodium produces a characteristic yellow emission at 589 nm, the instrument is set

at λmax= 589 nm and the readings are noted.

A series of standard NaCl solution (1, 2, 3, 4, 5, ……………ppm) is

prepared and is sent one by one and the readings are noted. The calibration graph

is drawn between the concentration vs intensity of the emitted light. A straight

line is obtained.

Now the unknown sodium solution is sent and the reading is noted. Then

the concentration of sodium in the water sample is determined from the

calibration curve.



3. Draw a neat diagram of UV-Visible soectrophotometer and explain the each

component ? Explain the working of UV-Visible spectrophotometer.

Components

The various components of a visible UV spectrometer are as follows:

Radiation source

In visible UV spectrometers, the most commonly used radiation sources are

hydrogen or deuterium lamps

Monochromators

Monochromator is used to disperse the radiation according to the wavelength. The

essential elements of a monochromator are an entrance slit, a dispersing element and an

exit slit. The dispersing element may be a prism or filter.

Cell

The cells, containing samples or reference for analysis, should fillfil the following

conditions:

i) They must be uniform in construction

ii) the material of construction should be inert to solvents

iii) they must transmit the light of the wave length used.

Detectors

There are three common types of detectors used in UV-Vis., spectrometers. They

are barrier layer cell, photomultiflier tube, photo cell.

Recording system

The signal from the detector is finally received by the recording system. The

reading is done by recorder pen.



Working

The radiation from the source is allowed to pass through the monochromator unit.

The monochromator unit allows a narrow range of wavelength to pass through an exit

slit. The beam of radiation coming out of the monochromator is split into two equal

beams. One half of the beam is directed to pass through cell containing a solution of the

compound to be analysed. The another half is directed to pass through an identical cell

that contains only the solvent. The instrument is designed in such a way that it can

compare the intensities of the two beams.

If the compound absorbs light at a particular wavelength, then intensity of the

sample beam (I) will be less than that of the reference beam (I0). The instrument gives

output graph, which is a plot of wavelength vs absorbance of the light. This graph is

known as an absorption spectrum.

2. State Beer – Lamberts law and explain its applications.

According to this law, when a beam of monochromatic radiation is passed through a

solution of an absorbing substance, the rate of decrease of intensity of radiation ‘dI’

with thickness of the absorbing solution ‘dx’ is proportional to the intensity incident

radiation (I) as well as the concentration of the solution (C). It is mathematically

represented as

- dI/ dx = k I C

- dI/ I = kC dx

Integrating the above equation between limits.

I

- ∫dI/ I = kC ∫dx

Iº

-ln I/ Iº = kCx

Where k = Molar absorption coefficient

ln Iº /I =kCx

2.303 log Iº /I = kCx

log Iº /I = k/2.303 Cx

Where E = k/2.303 (Molar extinction coefficient )



A = ECx

Where A = log Iº /I

Application:

Beer – Lamberts law can be used to determine the concentration of unknown solution.

For that first we measure the absorbance of standard solution. The absorbance will be

As = ECsx --------(1)

Now measure the absorbance of unknown solution. The absorbance will be

Au = ECux ----------(2)

From 1 and 2

Au / As = Cu / Cs

Cu = Au / As × Cs

From the values of Au , As and Cs we can calculate the concentration of unknown

slolution.

5. How is nickel estimated by AAS?

The instrument is switched on. Air supply and gas supply are regulated.Now prepare

stock solution of nickel by dissolving one g of Nickel nitrate in dilute nitric

acid.When a blank solution is aspirated into the flame, the meter of the AAS

assembly is adjusted to zero absorbance.From the ststandard solution of Nickel, a

series of standard solutions are prepared by appropriate dilutions.Now these standard

solutions are aspirated one by one into the flame and absorbance are measured at 232

nm.Now a calibrated graph is plotted between absorbance and concentration of

nickel in ppm. The nickel solution of uknown concentration is now aspirated into the

flame and absorbance is measured under the same conditions as above. From the

absorbance, the concentration of the unknown nickel ion sample is determined.



PART C

1. Explain the principle, working of IR spectroscopy with a block diagram?

Principle

The atoms in a molecule bond are in a state of constant vibration and

rotation. They may be compared with two balls (atoms) joined by spring (bond). On

absorption of IR the bond may stretch, bend etc., as shown below. So stretching and

bonding of bonds are responsible for IR absorption.

Block diagram:

Components

Radiation source

The main sources of IR radiation

(a) Nichrome wire

(b) Nernst glower

When they are heated electrically at 1200 to 2000°C, they glow and produce IR

radiation.

Monochromator

It allows the light of the requires wave length to pass through, but absorbs the

light of other wavelength.

Sample cell

The cell, holding the test sample must be transparent to IR radiation.

Detector

IR detector convert thermal radiant energy into electrical energy. There are photo



conductivity cell, thermocouple, Pyroelectric detector.

Recorder

The recorder records the signal coming out form the detector.

Working of IR Spectorphotometer

The radiation emitted by the source the source is split into two identical beams

having equal intensity. One of the beams passes through the sample and the other

through the reference sample.

When the sample cell contains the sample, the half-beam travelling through it

becomes less intense. When the two half beams (one coming from the reference and the

other from the sample) recombine, they produce and oscillating signal, which is measured

by the detector. The signal from the detector is passed to the recording unit and

recorded.



2. a) Draw a neat diagram and explain the principle, working of flame photometry?

Principle

When a metallic salt solution is introduced into a flame, the following processes

will occur.

1. It should evaporate the solvent from the sample solution

2. It should decompose the solid into atoms

3. It should excite the atoms and cause them to emit radiant energy

The excited atoms which are unstable, quickly emit photons of different wave

lengths and return to the lower energy state. Then the emitted radiation is passed through

the filter, which permits the characteristic wave length of the metal under examination. It

is then passed into the detector, and finally into the recorder.

Block diagram

Components

The various components of the flame photometer are described as follows:

Burner

The flame must possess the following functions:

i) it should evaporate the solvent from the sample solution

(ii) it should decompose the solid into atoms

(iii) it should excite the atoms and cause them to emit radiant energy.

Mirror

The radiation from the flame is emitted in all directions in space. In order to

increase the amount of radiation reaching the detector, a concave mirror is used which is

set behind the burner.



Slits

Entrance slits: It is kept between the flame and monochromator. It permits only the

radiation coming from the flame & mirror.

Exist slits: It is kept between the monochromator and detector. It prevents the entry of

interfering lines.

Monochromator

It allows the light of the required wavelength to pass through but absorbs the light

of other wavelength.

Detector

The radiation coming from the filter is allowed to fall on the detector, which

measures the intensity of the radiation falling on it. Photo cell or photo multiplier is used

as detector, which converts the radiation into an electrical current.

Amplifier & Recorder

The current coming out from the detector is weak, so it is amplified and

recorded.

Working of the Flame photometer:

Air, at a given pressure, is passed into an atomiser. The suction so-produced

draws some solution of the sample into the atomiser. Air+sample solution is then mixed

with fuel gas in the mixing chamber. The Air+sample solution +fuel gas mixture is then

burnt in the burner. The radiation, emitted by the burner flame, is passed successively

through the lens, filter, detector, amplifier and finally into a recorder.

The above experiment is first carried out using a series of standard solutions and

the reading for each solution is noted. The graph is drawn between the concentration

against intensity of emitted light. The test solution is taken and similar experiment is

carried out. From the graph the concentration of the unknown sample can be determined.

b) The percentage transmittance of 0.0005 M solution of disodium fumarate in a

1 cm cell is 19.2 %. Caluculate (i) the absorbance (ii) the molar extinction coefficient

(εεεε).

Solution:

Given

% T = 19.2 %



T = 0.192

C = 0.0005 M

l = 1 cm

(i) Absorbance

A = - log T

= - log 0.192

= 0.717

(ii) Molar extinction coefficient

ε = A / C . l

= 0.717 / (0.0005 × 1)

= 1.434 × 103

mol dm-3

4. Explain the principle, working of colorimeter with a block diagram. How

will you estimate iron using colorimetry

Principle

This colorimteric method is convenient for the coloured substances. The intensity

of colour can be easily measured by using a photoelectric colorimeter, from which the

concentration of coloured solution can be obtained using Beer-Lamberts law.

If the substance is colourless, then a suitable complexing agent is added to the solution so

that a coloured complex is obtained which can absorb the light.

Working

In a colorimeter, a narrow beam of light is passed from radiation source through

the test solution towards a sensitive detector. Usually, colorimeter is provided with the

arrangement of filter and slits, which select the light of required wavelength. The detector

generates the current, which is proportional to the amount of light transmitted by the

solution.

The amount of light transmitted depends on the depth of colour of the test

solution. Thus the current from the photo cell will be more when the light transmitted is

more. This will be possible only if the coloured solution is most diluted.

Current α light transmitted α 1 / Concentration



The transmitted light is now a days are allowed to send through a meter, which is

calibrated to show not the fraction of light transmitted but the fraction of light absorbed.

The light absorbed is proportional to the concentration of the test solution.

Block diagram

Components:

Radiationsources :

The wavelength range of visible light les between 400 – 750 nm. In this region, a

tungsten-filament lamp is most widely used.

Filter or monochromator :

It is a instrument, which allows the light of the required wavelength to pass

through,but absorbs the light of the other wavelengths.

Slits:

(a) Entrance slit: It provide a narrow source of the light.

(b) Exit slit: It select a narrow band of dispersed spectrum for observation by the

detector.

Cell:

The cell, holding the test sample (usually a solution), should be transparent. For -

visible region the cell is made of colour-corrected fused glass.

Detector :

It is used for measuring the radiant energy transmitted through the sample.

Photosensitive devices are used to detect radiations. These detectors produce current,

which is directly proportional to the intensity of the incident radiation.

Meter:

It is used to measure directly the fraction of light absorbed.



Estimation of iron

Principle

Fe3+

is a colourless but Fe3+

forms blood red colour complex with KCNS or

NH4CNS.

Fe3+

+ KCNS → [Fe(CNS)6]3-

+ 6K+

Reagents required

Standard iron solution

0.865 gms of FAS is dissolved in distilled water 5-10 ml of conc. HCl is added

and the solution is diluted to 1 litre. 1 ml of this solution contains 0.1 mg of Fe.

KCNS solution

20 gms of KCNS is dissolved in 100 ml of water

1:1 HCl

50 ml of conc.HCl is added to 50 ml of distilled water

Procedure

A series of standard solution of Fe3+ are prepared by adding KCNS with small

amount of 1:1 con.HCl. then the colorimeter is set at zero absorbance using a blank

solution , with a proper filter. Now absorbance of each standard solution is then measured

using the same filter. A graph is plotted between absorbance vs concentration. This plot is

called as calibration curve and will be the straight line passing through origin. This is

according to Beer-Lamberts law:

A = ε C l

Where

A is absorbance

C is concentration

l is path length

ε is molar extinction coefficient



Similarly, the absorbance of test solution is measured using the same colorimeter. From

the calibration curve, the concentration of the unknown ferric iron solution can be

evaluated.



MODEL QUESTION PAPER

CH 102 ENGINEERING CHEMISTRY II

B.E / B.Tech. Degree Examination, Dec 2009 / Jan 2010

PART-A

Answer to ALL questions

1. Define octane number and cetane number?

2. What is meant by knocking?

3. What is Pilling – Bedworth ratio? Give its significance.

4. What is impressed current cathodic protection?

5. Define Standard electrode potential?

6. Zinc reacts with dilute H2SO4 to H2 but Ag does not. Why?

7. What is condensed phase rule.

8. What is number of phases when CaCO3 is heated?

9. Filters are invariably used in absorption spectroscopy. Why?

10. How can you identify an unknown element from emission spectrum?

PART-B

11. (a) Discuss any one method of manufacture of Synthetic Petrol?

OR

(b) Write briefly on proximate analysis of coal.

12. (a) Illustrate the reactions involved in differential aeration corrosion with reference to

iron material.

OR

(b) Write notes on electroless plating.

13. (a) Calculate the half- cell potential at 298 K for the reaction:

Zn2+

+2e- �Zn(s)

If [Zn2+

] is 5.0M and EZn2+

/Zn is -0.76 V

OR

(b) Explain acid-base titration conductometrically?

14. (a) State Gibb’s phase rule. Define and explain the terms involved in it by giving two

illustrative examples each.



OR

(b) Explain the Pb-Ag system using phase diagram.

15. (a) How do you estimate the concentration of a solution by calorimetry?

OR

(b) How do you estimate sodium using flame photometry?

PART-C

16. i) Discuss the Orsat method of flue gas analysis.

ii) Write a note on LPG.

OR

i) Write briefly on proximate analysis of coal.

ii) Outline the preparation and uses of water gas.

17. a) i) Discuss anyone cathodic protection methods. (4)

ii) Discuss the various types of corrosion inhibitors.

OR

b) i) What are paints? Mention its constituents.

Explain the functions of various constituents. (6)

ii) Write notes on phosphate & chromate coatings. (4)

18. a) Draw a neat diagram phase diagram and explain the lead –silver system? Briefly

write about pattinson’s process?

OR

b) Draw a neat phase diagram of Fe-C system and label it.

19. a) i) Derive Nernst equation?

ii) What are the applications of Nernst equation?

OR

b) i) Explain precipitation titration with an example

ii) Write any two difference between Electrolytic cells and Electrochemical cell

20. a) i)Explain with a block diagram, working of flame photometry?

ii) Explain the applications of calorimetry.

OR

b) i) Explain with a block diagram, working of IR Spectroscopy?

ii) Explain the estimation of iron by calorimetry.



Documents

UNIT-I FUELS AND COMBUSTION PART A Howar efu lscas id? …