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
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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.
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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
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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
Weight of coal taken
(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
Weight of coal taken
(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
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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
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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.
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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
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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).
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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
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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.
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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.
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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
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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.
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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
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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
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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
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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
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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.
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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
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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
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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
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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
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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.
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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.
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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.
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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.
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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
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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.
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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 ↑
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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)
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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.
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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.
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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
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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.
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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.
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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.
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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
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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
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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.
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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.
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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
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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.
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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
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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).
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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
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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
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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.
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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
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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
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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
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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
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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.
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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.
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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.
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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.
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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.
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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.
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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
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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.
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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.
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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".
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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)
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(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
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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
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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.
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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.
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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.
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Phase diagram of Fe-C system
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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
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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
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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.
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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.
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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:
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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.
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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.,
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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.
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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
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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
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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.
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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
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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
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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.
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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.
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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.
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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 )
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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.
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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
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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.
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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.
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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 %
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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
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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.
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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
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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.
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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.
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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.
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