Solved question paper for CHEMISTRY May-2017 (B-TECH Electrical and Electronics Engineering 1st-2nd)

Engineering chemistry

Previous year question paper with solutions for Engineering chemistry May-2017

Our website provides solved previous year question paper for Engineering chemistry May-2017. Doing preparation from the previous year question paper helps you to get good marks in exams. From our CHEMISTRY question paper bank, students can download solved previous year question paper. The solutions to these previous year question paper are very easy to understand.

These Questions are downloaded from www.brpaper.com You can also download previous years question papers of 10th and 12th (PSEB & CBSE), B-Tech, Diploma, BBA, BCA, MBA, MCA, M-Tech, PGDCA, B-Com, BSc-IT, MSC-IT.

Print this page

Question paper 1

  1. SECTION - A

    1. Write briefly :
    (a) Discuss in brief various transitions shown by organic molecules in UV-Vis
    spectroscopy with examples.
     

    Answer:

    U.V.V spectroscopy known as ultra violet visible spectroscopy also known as electronic spectroscopy. It in valves transitions of different types of electrons presents in simple molecular from ground state to excited state.
    Various transition show by organic molecular in U.V.V spectroscopy is as follows:-
    σ → σ * Transition
    Due to having sigma (σ) bond higher in strength. It is a high energy process .i.e. short wavelength process this transition requires high energy i.e. short wavelength. The usual spectroscopy technique cannot be used below 200mm. since oxygen present in begins to absorb strongly in this region. To study such a transition, the entire path length must be evacuated. This region is less information.
    n → σ * Transition 
    This Transition takes place in hydrocarbons i.e. saturated in nature having one hetero atom with lone pair of electron 
    E.g. Halides, alcohols, ethers amines etc.

      Transition 
    Molecules i.e. unsaturated in nature show thin type of transition. It requires less energy as compared to n → σ * transition and hence absorption of longer wavelength takes place.
    E.g. Alkene, Alkyne, Aromatic etc.
     Transition
    It requires less amount of energy and hence absorption of the highest value of takes place.
    E.g.: Aldehydes, Ketones etc.

     Stands for non-bonding electrons i.e. lone pair of electrons

    Stands for electrons involved in formation of  bond.

    σ  Stands for electrons involved in formation of σ  bond.

     = anticoding.

  2. (b) Why a UV spectrum consists of bands and not of peaks?

    Answer:

    The electronic Transition is having Transition between vibrational and rotational states. So an electronic absorption band actually consists of closely spaced lines which merge to a give a broad band.

  3. (c) How will you distinguish between cyclohexarone and 3-methylcyclopentanone on the basis of IR spectroscopy?

    Answer:

    These are ketones. Due to having longer carbon chain in cyclohexane will show signal at lower vibration frequency on the other hand 3 methyl cyclopean tan one will show signal at higher vibration frequency.

  4. (d) Despite of the involvement of radiations & matter in both, what are the differences
    between spectroscopy and photochemistry?

    Answer:

    Spectroscopy:

    It is a study of transition occurring in a molecule when it interacts with electromagnetic radiation. Molecular spectroscopy deals with different of transition occurring various compounds.

    Photochemistry:

                That branch of science which deals with the study of absorption of light radiation (photos) by the molecules when electromagnetic radiations fall on the molecules, a part of the radiation is absorbed by the molecular and the remaining part is transmitted. The absorbed energy excited molecular from lower state to an excited state. In tow excited state, molecular are unstable and may undergo some rearrangement of bonds and lead to chemical reaction which is known as photo chemical reaction.

  5. (e) What is meant by photophysical processes? Give examples.

    Answer:

    That process which takes place in the presence of light but doesn’t involve any chemical change. Absorption of light by the substance is followed by the emission of absorbed light. If the absorbed light is emitted instantaneously the process is called fluorescence. If it is absorbed emitted after some time lag it is called phosphorescence 
    E.g. energy transfer, Electron transfer, Quenching etc.

  6. (f) What is nanochemistry?

    Answer:

     Nano chemistry:
    Utilization of synthetic chemistry to make Nano scale building blocks of different size, shape, concentration, composition. Surface structure and functionality. Basically, the spatial dimension of a nanoparticle is B/W 1-1Wnm.

  7. (g) Define Green Chemistry. What are its applications?

    Answer:

    Green chemistry:
           Whenever a branch of chemistry is concerned with study of chemical and chemical processor designed to reduced (-ve)environmental impacts that branch is known as Green chemistry.
    Application:- 
    (i) Designing safer chemicals  
    (ii) Reduction of derivation 
    (iii) Design for degradation
    (iv) Catalysis.

  8. (h) Differentiate between dry and wet corrosion.

    Answer:

    Dry corrosion:
              Whenever there is drivel chemical attack of atmospheric gasses on metal surface that type of corrosion results in Dry corrosion.  
    Wet corrosion:
             whenever metal is in the contact of moist air or any other liquid medium that results in wet corrosion.  
    Or
    Corrosion which occurs when a conducting liquid is in contact with metal or when two dissimilar metals or alloys are immersed or dipped partially in a solution.

  9. (i) Write a brief note on second generation petrochemicals. Give examples.

    Answer:

    Second generation petrochemical: 
           Derivation of first generation petrochemical is known as second generation petrochemical. It constitutes the intermediate chemical which serves as the raw material for consumer industries like as dyes, detergent, rubber, Tiber etc.
    E.g: Styrene derivative of benzene of ethylene
            Any lo nitrile propylene’s derivative.

  10. (j) What are supramolecular structures? Discuss their properties.

    Answer:

    Supermolecular structures:
           Large molecules formed by bonding or grouping of several smaller molecules are known as supermolecular structure. 
    Properties: 
    (i) These materials and devices have hierarchical structure. 
    (ii)These molecules/materials encode a lot of data.
    (iii)The supermolecular din dimers can hold the solvent and can trap smaller molecules and ions.

  11. SECTION - B

    2. (a) Compare the acid character each of phenol and benzoic acid in the ground and their excited states.

    (b) How can IR be used to distinguish between inter and intramolecular hydrogen bonding?
     

    Answer:

    (b) IR is used to distinguish between two types of hydrogen bonds I.e. intra molecular and inter molecular. H bond results into
    (i) Shift of J for the O-H bond, towards lowers value from the normal value 
    (ii) Intermolecular H-B bond result into formation of a broad band at lower value.
    (iii) Intermolecular H-bond result into a sharp band at lower J value than the J value of non-hydrogen bonded O-H group.
    Dilution reduces Intermolecular H-bond but it has no effect on Intermolecular H- bond. It results in that bond at lower J value because of Intermolecular H-bond But if after dilution. Sample does not show any effect on the band appearing at low j value. It results that bond is because of Intermolecular H-bond.

  12. 3. Describe in details about :
    (a) Desalination of water
    (b) Semiconductor photochemistry

    Answer:

    (a) Desalination of water 
     A process of removing common salt from the water is called Desalination of water. For this process freezing and Desalination, methods are used commonly. Freezing method is useful in old climatic condition. For Desalination process more fuel is consumed which is not environment friendly which makes them unfit for processing curtain new methods have been implemented which is as follow:
    Reverse osmosis:
        When two solutions having unequal concentration are separated from each by semipermeable member. A flow of solvent from low come to higher come of solution take place. This is known as osmosis. Semipermeable membrane is a selective membrane which does not allow the solvate to pass through if hydrostatic pressure is applied, provided that hydrostatic pressure is greater than atmospheric pressure. At this condition flow of solvent is revised i.e. from high come. To lower cone, this process is called reversed osmosis.
     Advantages:
    1. Useful for removing ionic, non-ionic, colloidal or high molecular mars matter along with nacre. 
    2. lifetime of membrane is 2 years so economical in nature.

    (b) Semi-conductor is the utilization of stable Semi-conductors as light absorbing species to drive synthetic reactions on their surface.
    Materials having conductivity between conductor and insulation is known as Semi-conductor Range of conductivity is 1 to 10-8 Ohm-1 cm-1 properties of Semi-conductor can be changed by adding some impurities of with pure Semi-conductor that is also known as intrinsic Semi-conductor. The process of adding impurities is known as doping and Semi-conductor obtained after doping is known as extrinsic Semi-conductor or doped Semi-conductor.
    Various application of Semi-conductor photochemistry as follows:
    (i) Photochemical Co2 reduction.
    (ii) Photo catalytic water splitting
    (iii)Artificial photosynthesis

  13. 4. What is meant by innocuous reagents? Write a brief note on microwave radiations in
    Green synthesis with examples.

    Answer:

    Innocuous reagent:-
             Chemical having little or no harmful effect i.e.those potentially toxic substance which cannot reach the ton get of tonicity due to some feature of structural design then for all practical purposes that chemical is treated as innocuous reagent.
    Microwave radiation in green synthesis:
               Microwave range from 1cm to 1m in wavelength. Frequency corresponds to J=2450 MHz. it is an environment friendly approach for synthesis. Microwave energy is nonionizing and does not alter the molecular structure of compounds being heated as it provides only thermal activation. Heating effect is due to polarization. When microwave radiation is given to a molecule them it aligns it’s with the applied field. Both conduction and dielectric. Polarization is a source of microwave heating. Ability of material convert E.M energy into thermal energy depends on its dielectric constant. The larger the dielectric constant, greater the coupling with microwave. 
    Microwave radiations are widely used as a source of heating in organic synthesis small molecular can be built in a fraction of time. Many reaction processes faster upon microwave irradiation than the tradition heating technique.
    Major application of microwave radiation:
    Due to having following advantages, it is widely used in the world of chemistry 
    (i) High efficiency of heating 
    (ii) No treat less to environment  
    (iii) Low operating cost
    (iv) Purity of final product 

  14. 5. (a) Discuss about anharmonicity in IR spectroscopy.
    (b) Discuss the specifications of boiler feed water and the significance of those
    specifications.

    Answer:

    (a)

    At higher vibration levels (hetero nuclear diatomic and polyatomic) Molecule doesn’t behave as S.H.O. But as anharmonic oscillator, because restoring force is no longer  to the displacement. Anharmonic oscillator shows additional transition (through of weak intensity) in addition of fundamental transition.
    For this type of escalated energy is represented as:
     Evib (anh)   =  (v+1/2) h∂  -  (v+1/2) h∂
    X= Anharmonic contt.
    Above equation show that energy of each individual is less and energy level are now more closely placed 
    Selection rule for vibration transition to take place in enharmonic order selection rule for vibration to take place in a harmonic oxalate is:-
    ∆v=±1,±2,±3………………….
    If transition is from V0 to V3 it is called fundamental transition (∆v=±1)
    If transition is from V0 to V3 it is called overtone or third and harmonic transition (OV=±3)
    Note: Absorption intensity of overtone is less than that of fundamental 
    Transition fromV1 TO V3 or V1TO V2 is known as hot bands they take place at high temperature.

    (b) Various specifications and their significance of boiler feed water are as follow:
    (a) It should be free from hardness as hard water causes sludge and scale formation 
    (b)It should be free from dissolved gases such as CO2, H2S, O2 to avoid corrosion
    (c)It should not and manganese contains iron as they get disported in boiler.

  15. SECTION – C

    6. (a) Write a note on Galvanic corrosion.
    (b) Discuss about tacticity of polymers and methods of synthesis of polymer with specific tacticity.
     

    Answer:

    (a) Galvanic corrosion: It is also known as bimetallic corrosion.as the name indicates, when two different metals are in contact with other in the presence of an electrolyte, the metal which is higher in rank in electrochemical series will act as anode and other will behave as cathode. Anode undergoes oxidation and gets corroded this process is known as galvanic corrosion. 

    (b)  Tacticity of polymer:
    Configuration of polymer is known as tactility of polymers. It influences the physical properties of that polymer. According to tacticity polymer can be classified as:
    Isotactic polymer:
    When function group are all on the same side of the chain that configuration is known as isotactic polymer. It is obtained by use of catalysts mounted on a crystalline solid.

               
    Syndiotactic polymer:
         If arrangement of functional groups is in alternating patties that type of polymer arrangement is known as syndotactic polymer 



    Atactic polymer:

    if a polymer shows random arrangement of functional groups around main chain. The polymer is said to be a tactic polymer. It can be obtained by use of catalysis on an amorphous solid.

  16. 7. What are the primary raw materials for petrochemical industry? How can the primary raw material be converted to usable secondary and tertiary petrochemical products?

    Answer:

    Primary raw materials for petro chemical industries are as follows:
    (i) Nature gas             (ii) Crude oil

    These raw materials can be converted into secondary and tertiary petrochemical products through different processing sachems like as thermal conversion processor catalytic conversion process, alkylation, hydrocracking, isomerization, vacuum distillation, solvent extraction adsorption process, absorption process, atmospheric distillation etc.

    (a) Thermal cracking:
              It refers to decomposition of bigger hydrocarbon molecule into low boiling hydrocarbon of lower molecular weight.
    E.g.: C10H22→C5H12+C5H10
    It can be carried out by following two ways 
    (i) Liquid phase thermal cracking 
    Heavy oil is cracked at suitable temp ranging from 470-520oC. the cracked products are separated in a fractionating column. The yield of this process is 50-60%
    (ii) Vapor phase thermal cracking: temp ranging 600-650oC. This requires less time and resulting full has better anti knock properties. It is only used those oils which can be readily vaporized.

    (b) Alkylation: 
    Is referring to chemical bonding of light hydrocarbon molecular with isobutene to from longer branched-chain molecule that makes high octane petrol.
     Polymerization: at high pressure and temperature, the light unsaturated hydrocarbon molecule, over an acidic catalyst react with each other to form larger hydrocarbon molecule.
    Above discussed all processes are used to product tertiary product.Following processes are used to produce secondary products
    (a) Vacuum distillation:
              The allows heavy hydrocarbon with B.P of 450o C and to be separated without partly cracking them into unwanted products such as coke and gas.
    (b) Adsorption process:
              It involves adsorbents such as silica gel and bleaching clay. The sulfur, O and N compounds; and hydrocarbon which readily undergo polymerization are adsorbed. Hence they are separated.
    (c) Absorption process:
             Purification of crude oil by absorption process involves the selective absorption of undesirable of crude oil by solvent the solvents used are nitrobenzene, dichloroethyl ether and SO2 liquid.

  17. 8. Discuss in details all the aspects including chemistry of production of ethylene and
    propylene.

    Answer:

    Production of ethylene 
                It is mainly produced by steam cracking in this method hydrocarbon are heated to 750-950oC. This result in many free radical reactions. This causes conversion of larger hydrocarbon to small ones and also unsaturation. Ethylene is separated from resulting complex mixture. It is energy intensive process, so heat is recovered from the gas leaving the funnel. The heat or steam is used to run the ethylene plant. So there is no need for import steam.

    Production of propylene:
              It is a byproduct obtained during the catalytic cracking. It can also be prepared during petroleum cracking. It can also be prepared from cracking of propane.
    Uses of propylene:
    →Production of glycerol
    →Alkyl alcohol production 
    →Production of propylene glycol

  18. 9. (a) Discuss in details the chemistry of various methods of corrosion control with
    chemical equations.
    (b) How will you classify the crude oil?

    Answer:

    (a) Various methods involved in corrosion control are as follow.
    (i) Cathodic protection:
    It is based on the principle that metal which is to be protected from corrosion has to be behave as cathode various cathodic protection method are as follow 
    (a) Sacrificial modicProduction method:
            In this method, the metal to be protected is connected through a wiv to a more anodic metal so that oxidation takes place on anodic metal and metal which is to be protected remaining unaffected. The more anodic metal than sacrifice itself on behalf of less anodic metal i.e. metal to be protected and it is called sacrificial mode. The sacrificial mode has to be replaced when completely eaten up. 
    It is basically used in the production of metal like as Mg1Zn1 that and AC and their Alloys. Marine structure water tank etc. can protect by this method.

    (b )Impressed current cathodic protection: 
              An impressed current is applied in aduiction opposite to diction of corrosion current. it tends to change in reactive end to unreactive end  i.e.anode to cathode. Hence the metal acting as anode can be protected from corrosion. Impressed current is applied to insoluble anode buried in soil and connected to Meta like structure to be protected. This type of protection is useful for water pipes, oil pipes etc.
     

    (ii) Modifying the environment:
    Corrosion can be controlled by modification of environment. Different steps involved as follow.
    (a) Deaeration: removal of O2 it decelerates the rate of corrosion this can be removed by adjusting the temperature and with mechanical agitation oxygen can be removed by using hydrogen, sodium,s sulphate etc.
    (b) Dehumidification: - Removal of Moisture content in air. This can be done by using Silica Gel alumina etc.
    (c) Neutralization of acidic character: Acidic character lead to oxidation of metal which can be controlled by neutralization of acidic character of metal with the help of NaOH, lime soap and NH3 like neutralizer

    (iii) Metal coating:- In this method the metal to be prevented from the corrosion is to be coated with a metal. Coating act as a behave between metallic surface and corroding environment it can be done in two ways
    (i) Anodic coating                         (ii) Cathodic coating
    In Anodic coating, the metal which is used for coating is more anodic than the metal which is to be protect i.e. base metal. The coating of Zn, mg and AC or iron is termed as Anodic coding as these metal lie above iron in galvanic serves 
    In cathodic coating, the metal which is used for coating is more cathodic than base metal because it is has more corrosion resistant the base metal coating of Sn, Ni, Cu on iron is cathodic coating.
    Other methods for protection of metal from corrosion are use of inhabitant’s organic coating in inorganic coating and purpose designing etc.

     

    (b) Crude oil is classified into three categories:
    (i) Paraffin base:
    If residue obtained after removal velocity compound of Crude oil is rich in waxParaffin. The Crude oil is called Paraffin base. This Crude oil has low Octane number.
    (ii) Asphalt and naphthalene base:
     When the obtained after removing velocity waste is rich in asphalt and naphthalene, then it is called Asphalt and naphthalene base. These have a high octane number.
    (iii) Intermediate Base:
     If residue contains Paraffins and naphthalene both after removing velocity compound. Then crude oil is called intermediate base.

Question paper 2

  1. SECTION-A

    1. Write briefly :
    a) Explain temporary and permanent hardness of water.
     

    Answer:

  2. b) Describe reason for cracking

    Answer:

  3. c) What is partition chromatography?

    Answer:

  4. d) What is the difference electrode potential and cell e. m. f.?

    Answer:

  5. e) Explain photosen itization.

    Answer:

  6. f) Write the definition of Eutectic point.

    Answer:

  7. g) In IR spectroscopy why the region below 1500 cm–1 is called as fingerprint region?

    Answer:

  8. h) Explain the selection rules of UV-vis Spectroscopy.

    Answer:

  9. i) What is metastable state?

    Answer:

  10. j) Which of the following will absorb at higher wave number for C=O stretching?

    ​​​​​​​

    Answer:

  11. SECTION-B

    2. a) Describe the methods for the treatment of municipal water.
    b) Calculate the amount of lime and soda required for softening 90,000 liters of water
    containing the following salts per liter : Ca(HCO3)2 = 162 mg, CaSO4 = 136 mg and
    NaCl = 56.1 mg. Purity of lime is 92 % and soda is 99%.

    Answer:

  12. 3. a) Explain the electrochemical mechanism of rusting of iron in humid atmosphere.
    b) Discuss protective measure for controlling corrosion.

    Answer:

  13. 4. a) Explain the need of chromatogram. Discuss various methods used for development
    and visualization of chromatogram.

    b) What is the significance of Rf? How it can be calculated experimentally?

    Answer:

  14. 5. a) What is over voltage? Discuss the factors affecting the overvoltage value.
    b) Discuss the conductometric titration involving a HCl and a KOH.

    Answer:

  15. SECTION-C

    6. Draw a well labeled Jablonski diagram and explain.
    a) Intersystem crossing
    b) Phosphorescence

    Answer:

  16. 7. a) Explain Principles of UV-Vis Spectroscopy.
    b) On the basis of IR spectroscopy, how can you distinguish between the following :
    (i) Alkane, alkene and alkyne
    (ii) Aldehyde and ketone

    Answer:

  17. 8. a) Discuss spin-spin relaxation in NMR spectroscopy
    b) Draw and explain the splitting pattern observed in the 1H NMR of CHCl2 CH2Cl

    Answer:

  18. 9. State and explain phase rule, Describe phase diagram of :
    a) Potassium iodide-water system
    b) Nicotine-water system

    Answer: