--- title: "Au B Sc I Chemistry Iiiok" book: "test" category: "MA" publisher: "Ratan Prakashan Mandir Pvt. Ltd." type: "Educational Material" ---  According to Latest Syllabus For Dr. Bhimrao Ambedkar University (A.U.) Examination # VIKAS ## GUIDE BOOK B.Sc. I ### CHEMISTRY-III Km. Pankaj Published by Ratan Prakashan Mandir Pvt. Ltd. 2nd Floor, Centre Plaza, Parinay Kunj, Lajpat Kunj Marg, Agra-282002 Copyright Authors & Publishers Published by #### Ratan Prakashan Mandir Pvt. Ltd. 2nd Floor, Centre Plaza, Parinay Kunj, Lajpat Kunj Marg, Agra-282002 ISBN: 978-93-7940-158-8 Price 245.00 only Printed at : KIDS INTERNATIONAL PVT. LTD. C-60, 61, 62, 63, EPIP, Shastripuram, Agra - 282007 Ph. : +91 9719004921 B. Sc. I Chemistry III (Physical Chemistry) SYLLABUS Unit I - (A) Mathematical Concept : Logarithmic relations, curve sketching, linear graphs and calculation of slopes, differentiation of functions like ky, ex, nn, cos x, sin x, log x, maxima and minima, partial differentiation and reciprocity relations. Integration of some useful/relevant functions; permutations and combinations, factorials Probability. - (B) Computers : General introduction to computers, different components of a computer, hardware and software, input-output devices; binary numbers and arithmetic, introduction to computer languages operating system. Unit II Solid State : Definition of space lattice, unit cell. Laws of crystallography— (i) Law of constancy of interfacial angles, (ii) Law of rationality of indicies, (iii) Law of symmetry. Symmetry elements in crystals. X-ray diffraction by crystals, Derivation of Bragg equation. Determination of crystal structure of NaCl, KCl and CsCl (Laue's method and powder method). Unit III - (A) Liquid State : Intermolecular forces, structure of liquids (a qualitative description). Structural difference between solids, liquids and gases. Liquid crystals. Difference between liquid crystal, solid and liquid. Classification. Structure of nematic and cholestric phases. Thermography and seven segment cell. - (B) Colloidal State : Definition of colloids, Classification of colloids. Solids in liquids (Sols), properties-kinetic, optical and electrical, Stability of colloids, Protective action, Hardy-Schulze law, gold number. Liquids in liquids (emulsions), types of emulsions, preparation Emulsifier. Liquids in solids (gels), classification, preparation and properties, inhibition, general applications of colloids. Unit IV Gaseous State : Postulates of kinetic theory of gases, deviation from ideal behaviour van der Wall's equation of state. Critical Phenomena, PV isotherms of real gases, continuity of states, the isotherms of van der Wall's equation, relationship between critical constants and van der Wall's constants, the law of corresponding states, reduced equation of state. Molecular Velocities : Root mean square average and most probable velocities. Qualitative discussion of the Maxwell's distribution of molecular velocities collision number, mean free path and collision diameter Liquification of gases (Based on Joule-Thomson effect.) Unit V Chemical Kinetic and Catalysis : Chemical kinetics and its scope, rate of a reactions, factors influencing the rate of a reaction. Concentration, temperature, pressure, solvent, light, catalyst. Concentration dependence of rates, mathematical characteristics of simple chemical reactions—Zero order, first order, second order, pseudo order, half life and mean life. Determination of the order of reaction— differential method, method of integration, method of half life period and isolation method. Radioactive decay as a first order phenomena. Experimental methods of chemical kinetics : Conductometric, Potentiometric, optical methods, polarimetry and spectrophotometer. Theories of chemical kinetics : Effect of temperature on rate of reaction. Arthenius equation, concept of activation energy. Simple collision theory based on hard sphare model, transition state theory (equilibrium hypothesis), Expression for the rate constant based on equilibrium constant and thermodynamic aspects. Catalysis : Characteristics of catalysed reactions, Classification of Catalysis, Miscellaneous examples. Question Index S. No. P. No. Section ‘A’ Multiple Choice Type Questions (1 to 185) 11-31 Very Short Answer Type Questions (1 to 53) 32-38 Section ‘B’ Short Answer Type Questions 1\. A coin is tossed succesively three times. Find the probability of getting exactly one head or two heads. 38 2\. Prove that, log 15 = log 1 + log 3 + log 5 39 3\. Find the local maxima and local minima of f(x) = x3 ‒ 6x2 + 9x +15. Also find the local maximum or local minimum values as the case may be. 39 4\. How many four digit numbers can be formed with the digits 1, 2, 3, 4, 5, 6, when a digit may be repeated any number of times in any arrangement. 39 5\. - (i) Determine the differentiation coefficient of aex. - (ii) Determine the differential coefficient of ex2 + 4x2 + 6. 40 6\. How many 4-letter words with or without meaning can be formed out of the letter of word ‘LOGARITHMS’ if repeatation of letters is not allowed. 40 7\. Calculate the logarithms of 34∙56. 40 8\. Show the development of electronic computer in the form of a Table. 40 9\. Give main functions of computer. 40 10\. Give the Values of following : (i) log432 (ii) log7 (49) 41 11\. Three dice are thrown together. Find the probability of getting a total of at least 6. 41 12\. Evaluate, j(3x2 - 5x - 24)dx. 41 13\. Solve, 0-243 x 100 x 52 , , \_ x = 60 x 0-035 by logarithms. 42 6 Vikas, 2010 (A. U.) S. No. - 14\. Determine the differential coefficient of log x relative to x by first principal. - 15\. Write short notes on : - (i) Arithmetic and Logical Unit (ALU), - (ii) Central Processing Unit. - 16\. A bag contain 9 black 12 white balls. One balls is drawn at random, what is the probability that the ball drawn is black. P. No. 42 43 43 17\. 18\. 19\. 20\. If z = 16x3 ‒ 10x2y2 + 7y4 ‒ 8, then find the values of ∂∂z and ∂z x ∂y . Determine the maximum value of r for the curve c4 \_ a2 b2 r2 sin2 0 cos2 0 Draw a graph of function x2 ‒ 4x + 5 = 0. Evaluate, 44 44 44 21\. 22\. 23\. 24\. 25\. 26\. 27\. 28\. 29\. 30\. 31\. 32\. 33\. - (i) tan2 x dx (ii) sec x dx sec x + tan x Write a short note on Isotropy and Anisotropy. Write short notes on symmetry elements. Write short notes on the following : - (a) Law of constancy of Symmetry - (b) Law of rational intercepts. Write short note on Tyndall Effect. Write short note on electrophoresis. Give the application of liquid crystals ? Write a short note on Associated Colloids. Write short notes on the following : - (a) Source of electrical charge on colloidal particles - (b) Protection and Gold Number. What do you mean by liquid crystals ? Discuss its types. Give the kinetic molecular description of liquid state. Write short notes on : - (a) Thermography, - (b) Seven segment cell. What do you mean by coagulation ? Discuss Hardy-Schulze’s law. Prove that, 9 1 RTc a = RTcVc and b = 8 8 Pc 45 46 46 47 48 49 49 50 50 51 52 53 54 54 S. No. - 34\. Write short notes on collision frequency ? - 35\. What are the limitations of Van der Waal’s equations ? - 36\. Write explanatory note on law of corresponding states ? - 37\. State and explain law of equipartition energy. - 38\. What do you mean by mean free path ? - 39\. Explain homogeneous and heterogeneous catalysis with suitable example. - 40\. Optical rotation of a carbohydrate in 0∙5N‒HCl at different intervals of time as Time (minute) 0 10 20 30 40 ® Optical rotation +32∙4 +28∙8 +25∙5 +22∙4 +19∙6 ‒11∙1 Prove that the reaction is of first order. - 41\. Write short note on : - (i) Auto-Catalyst, (ii) Induced Catalyst, (iii) Catalytic Poisons. - 42\. Discuss in detail the mechanism of enzyme catalysed reaction. - 43\. Write short notes on Pseudo-molecular reaction. - 44\. 50% of a gas decomposed in second order way in 40 minutes. What would be the time taken for 75% of the gas to decompose? - 45\. Explain the rate of a reaction. Discuss the factors affecting the rate of a reaction. - 46\. 10 ml of ethyl acetate were added to a flask containing 100 ml of 0∙1 N-HCl placed in a thermostat maintained at 25°C. 5 ml of the reaction mixture were withdrawn at different intervals of time and after chilling titrated against standard alkali, the following data were obtained. Time (min) 0 75 119 183 ® Alkali used (ml) 9∙6 12∙10 13∙30 14∙75 21∙05 From the above data find out the order of reaction. - 47\. 50% of H2O2 decomposes in 10 minutes in an expe-riment. Find out its velocity coefficient if it is a reaction of first order (log102=0∙3010) - 48\. The density of hydrogen at 0°C and 760 mm pressure is 0∙00009 gm/cc. Determine it RMS velocity. Density of mercury is 13∙6 gm/cc. - 49\. Explain the molecularity and order of a reaction. Describe the difference between them. - 50\. The specific rate of reaction of a chemical reaction at 300Å is 0∙001 minute‒1 which becomes 0∙002 minute‒1 at 310Å. Calculate the energy of activation of the reaction. P. No. 55 56 56 57 57 58 59 60 61 61 61 62 63 64 64 65 65 S. No. P. No. Section ‘C’ Essay Type Questions 1\. 2\. What do you mean by operating system ? Explain in brief. 66 Integrate the following : dx (i) x cos xdx, (ii) 68 -L + sin x 3\. What do you mean by computer languages ? Explain merits and demertis of Machine Languages. 69 4\. One card is drawn from a pack of 52 cards, each of the 52 cards being equally likely to be drawn. Find the probability that : (i) the card drawn is red. - (ii) the card drawn is a king. - (iii) the card drawn is red and a king. - (iv) the card drawn is either red or a king. 72 5\. Differentiate the following : ex 2 x + 3 ex (i) (ii) 2 (iii) x x -5 1 + sin x (iv) sin x3 (v) sin3x (vi) esinx 73 6\. What do you mean by binary number system. How decimal fraction is written in binary fractions ? Write 0.700 in binary system. 74 7\. What is software ? Give the relationship between hardware and software. Discuss its types. 76 8\. What do you mean by assembly language ? Discuss its demerits and advantages. 77 9\. What is high level language ? Describing compiler and interpreter give the advantages of high level language. 79 10\. 11\. What are Weiss and Miller’s indices ? Explain in brief. 82 What do you mean by space lattice and unit cell ? Explain also simple cubic lattice, face-centered cubic space lattice and bodycentered cubic space lattice with suitable example. 83 12\. What do you mean by X-ray Crystallography ? Derive Bragg’s equation nX : 2d sin 0 for the reflection of X-ray by crystals. How internal structure of a crystal is determined by its application ? Describe rotating crystal method for the crystal analysis. 86 13\. 14\. Represent Crystal lattice of NaCl and KCl. 88 Give the types of liquid crystals ? Also give their properties and uses. 91 S. No. P. No. 15\. 16\. Throw light on the structure of liquids. 93 Give reasons for the following : - (i) Bleeding is stopped by adding alum. - (ii) Where river and sea water meet a delta is formed. - (iii) Alum is added to drinking water supply. - (iv) Colloidal particles move under and applied electric field. - (v) When As2S3 and Fe(OH)3 sols are mixed precipitation is occured. - (vi) Sols cannot filtered by ordinary filter paper. - (vii) Colloidal particles make zig-zag motion. 94 17\. What do you mean by colloidal state or sol ? Describe various types of colloidal dispersion. 95 18\. Discuss the Maxwell distribution law of molecular velocities. 97 What is the effect of temperature on velocity distribution ? 19\. - (a) Define gram molecular specific heat. Starting with kinetic equation of gas calculate the ratio of Cp and Cv. How does this vary with molecular complexity of the gas ? - (b) The Specific heat of a mono-atomic gas is 0∙075 at constant volume. Determine its molecular weight. 98 20\. Explaining the isotherm of CO2 gas, define the following : - (i) Critical temperature, - (ii) Critical pressure, - (iii) Critical volume. 101 21\. Discuss in brief the other equation of states for explaining the behaviours of real gases. 104 22\. - (a) Derive Van der Waals equation. Give units of van der Waals contants use in it. 105 - (b) Calculate the values ‘a’ and ‘b’ for CO2 if Tc = 31°C, Pc = 72∙8 atoms and R = 0∙082 lit atm. per deg. per mole. 109 23\. - (a) Explain enzyme catalysis with suitable examples. - (b) What are the general characteristics of enzyme catalysed reactions ? 109 24\. - (a) How does the period of half charge depends upon the initial concentrations of the first and second order. - (b) . To study the rate of decomposition of H2O2, a same amount of reaction mixture was titrated against standard KMnO4 solution. The different data obtained are as follows : S. No. P. No. 25\. 26\. 27\. 28\. Time (minute) 0 10 20 30 Vol. of KMnO4 used (ml) 46∙1 29∙8 19∙6 12∙3 Prove that the decompostion of H2O2 is a first order reaction. 111 (a) Derive the rate law constant for second order reaction. (b) In an experiment ethyl acetate is saponified by sodium hydroxide. The progress of the reaction is performed by titrating 25 c.c. of the reaction mixture with standard acid at different intervals. By taking equal concentrations of ester and alkali, the following data was obtained in the experiment : Reading 1 2 3 4 5 Time (minute) 0 5 15 25 55 Vol. of Acid 16 10∙24 6∙13 4∙32 2∙31 (a – x) c.c. Prove that the order of reaction is two. How much amount of ethyl acetate would have been decomposed after 35 minutes ? 113 - (a) Discuss an expression for the first order rate constant of a reaction. - (b) Give Integration and Equifraction methods employed for the determination of order of reaction. 115 - (a) What do you mean by temperature coefficient and activation energy of a reaction ? Derive an expression for determining the activation energy of a reaction. - (b) The value of rate constant for this reaction is 346 x 10“5 at 25ºC. N2O5 ^ N2O4 + {O2 and 447 x 10“3 at 65°C. Calculate the activation energy for the reaction. 116 Explain the modern Adsorption theory of Catalysis. 118 B. Sc. I, Chemistry III Section ‘A’ Multiple Choice Type Questions : 1\. Persons who assemble the computerised data are called : - (a) User (b) Computer Personal - (c) Hardware (d) a and b both. 2\. The brain of computer is : (a) Memory (b) C. U. (c) Peripherrals (d) C. P. U. 3\. Integrated circuit was used during : - (a) Ist generation (b) IIIrd generation - (c) IVth generation (d) Vth generation. 4\. Blaze’s adding machine is called : - (a) Pascaline (b) Difference engine - (c) Abacuos (d) Sensor calculator. 5\. Joystic is an example of : - (a) Input unit (b) Output unit - (c) Memory (d) All the above. 6\. Keyboard is an : - (a) Output Device (b) Storage Device - (c) Input Device (d) Processing Device. 7\. Central Processing unit (CPU) consisting : - (a) (i) and (iii) (b) (i) and (ii) - (c) (ii) and (iii) (d) all these. 8\. Which out of following is an output device ? - (a) Audio input unit (b) Light pen - (c) Tracker ball (d) Monitor. 9\. Linker is an example of : - (a) Operating system (b) System software - (c) Application software (d) Utility software. 10\. The unit of memory is : - (a) Bit (b) Byte - (c) Kilobite (d) All the above. 11\. If log3x = 4, then value of x will be : (a) 4 (b) 1/4 (c) 81 (d) 12. 12\. If log 2 = 0∙3010 then what will be the value of log 32 ? (a) 1.5050 (b) 10 (c) 0∙06020 (d) 0∙4000. 13\. In which quartile point (2, ‒4) will be situated ? - (a) first quartile (b) second quartile - (c) third quartile (d) fourth quartile. 12 Vikas, 2010 (A. U.) 14\. Choose the correct alternative if f (x, y) = x cos y + y cos x : a.) df=^f (b) f=f ' dx dy dy dx ôx gy a2f a2f (c) — \* — (d) None of these. ox dy dy dx 15\. The gradient of line parallel to straight line 2x + 3y ‒ 4 = 0 will be : (a) ‒2/3 (b) 2/3 (c) ‒3/2 (d) 3/2. 16\. dx Give the value of J ( a — x) : (a) ‒ log (a ‒ x) (b) log x (c) cos x (d) sin2x. 17\. The vaule of J tan x dx is : (a) sec x (b) ‒log (cos x) (c) ‒log (sec x) (d) sin x. 18\. What will be the value of P(A ) if P(A) = P(B) = 1/3 and P(A n B) = 1/6 ? (a) 2/3 (b) 5/6 (c) 1/4 (d) 3/5. 19\. The characteristic in log 0∙004837 is : (a) 2 (b) 3 (c) 2 (d) 3. 20\. The value of log 25 ‒ log 5 is : (a) log 25/log 5 (b) log 25 x log 5 (c) log 125 (d) log 5. 21\. af Iff (x, y) = x + y2 - 3axy, then find the value of dx : (a) 3x2 + 3ay (b) 3x2 ‒ 3ay (c) 3y2 ‒ 3ax (d) 3x2. 22\. Value of J e4 dx is : (a) 4 e4 x + C (b) 4 e3 x + C (c) 4 e3x + C (d) 3 e4x + C. 23\. The second derivative of y = x2 is : (a) 2 (b) 2x (c) 2x3 (d) 3x4. 24\. The value of nC + nC is : r r-1 (a) n–1Cr (b) nCr+1 (c) n-1Cr (d) n–1Cr–1 . 25\. The value of log 1000 is : (a) 4 (b) 3 (c) 2 (d) 1. 26\. If log x 16 = 2/3, then x will be : (a) 32 (b) 48 (c) 64 (d) None of these. - 27. ddx (x sin x) will be : - (a) cos x + sin x (b) cos x ‒ sin x (c) x cos2x (d) x cos + sin x. - 28. The value of ∫3 cos x dx is : - (a) 3 cos x (b) 3 sin x (c) sin x (d) ‒3 sin x. - 29. If y = loga x , then value of dxy will be : - (a) logae (b) 1 logae (c) x logae (d) 1 . xx - 30. What will be the value of f (‒3) if f(x) = 7 x2 +16 - (a) ‒ 19 (b) 7 (c) 5 (d) 6. - 31. Which of the following types of metals make the most efficient catalysts ? - (a) Alkali metal (b) Transition metal - (c) Alkaline earth metals (d) Radioactive metals - 32. Which one of the following statement regarding catalyst is not true ? - (a) A catalyst remains unchanges in composition and quantity at the end of reaction - (b) A catalyst can initiate reactions - (c) A catalyst does not alter the equilibrium in a reversible reaction - (d) Catalysts are sometime very specific in respect of reaction. - 33. Minimum energy of activation of an exothermic reaction is : - (a) zero (b) negative - (c) positive (d) either of the three. - 34. Which of the following has maximum value of fescarlating power ? - (a) Pb2+ (b) Pb4+ (c) Sr2+ (d) Na+ - 35. Gold numbers is the measure of the : - (a) protective action by a lyophilic colloid on a lyophobic colloid. - (b) protective action by a lyophobic colloid on a lyophilic colloid. - (c) number of milligrams of gold in a standard red gold sol. - (d) stability of gold sol. - 36. Which of the following terms is not related with colloids ? - (a) Brownian movement (b) Dialysis - (c) Ultrafiltration (d) Wavelength. - 37. The extra stability of lyophilic colloid is due to : - (a) Charge on their particles - (b) A layer of medium of dispersion on their particles. - (c) The smaller size of their particles - (d) The large size of their particles 38 39\. 40\. 41\. 42\. 43\. 44\. 45\. 46\. 47\. 48\. 49\. 50\. 51\. The base digit in binary number is : (a) 2 (b) 8 (c) 16 (d) 6 The closed packing of order AB, AB, ................ represents : - (a) Simple cubic complex (b) Body centered cubic complex - (c) Face centered cubic complex (d) Hexagonal closed complex. The example of 8 : 8 coordination among following is : - (a) CaF2 (b) ZnS (c) CaCl2 (d) KCl. Compound which has sharp melting point is : - (a) Plastics (b) Glass (c) Coke (d) Urea. Select correct statement among following : - (a) Glass is a crystalline solid - (b) Glass is an ideal solid - (c) Pure compound (solid) has certain melting point - (d) Iron is diamagnetic compound. Fog is an example of colloidal system of : - (a) Liquid dispersed in gas - (c) Solid dispersed in gas Light scattering takes place in : - (a) Solution of electrolyte - (c) Electrodyalysis (b) Gas dispersed in gas (d) Solid dispersed in liquid (b) Colloidal solutions (d) Electroplating. Which of the following electrolyte is least effective in causing flocculation of ferric hydroxide sol ? - (a) K3\[Fe(CN)6\] (b) K2CrO4 (c) KBr (d) K2SO4. The migration of Colloidal particles under the influence of an electric field is correctly known as : - (a) Cataphoresis (b) Electrophoresis (c) Electroosmosis (d) Electrodyalysis. Which out of following is an input device ? - (a) Mouse (b) Key board (c) Light Pen (d) All the above. In CsCl crystal neighbouring ions to Cs+ ions are : - (a) 6Cl‒ ions, (b) 8 Cl‒ ions, (c) 6 Cs+ ions, (d) 8 Cs‒ ions. The co-ordination number of Cs+ ion in CsCl is 8. What will be the co- ordination number of Cl‒ ion? - (a) 8, (b) 4, (c) 6, (d) 12. Which among the following is not a type of liquid crystal ? - (a) Nematic (b) Smectic (c) Cholestric (d) Amostropic. Hydrogen chloride is a gas whereas hydrogen fluoride is a liquid having low b. p. because ? - (a) H-F bond is strong - (b) H-F bond is weak - (c) Molecules are associated due to hydrogen bonding - (d) Hydrogen fluoride is a weak acid. - 52. The cytobectic group model for liquid structure was proposed by : - (a) Dalton (b) Steward - (c) Graham (d) Barnal and Scott - 53. Maximum hydrogen bonding is in : - (a) Triethyl amine (b) Ethanol - (c) Diethyl ether (d) Acetone - 54. One of the reasons for greater reactivity of finally divided Pt-catalyst is that it has : - (a) particles which are almost atomic in dimensions - (b) particle size which can spresd easily through whose reactants - (c) much larger surface area. - (d) a physical state only in which it can react quickly. - 55. Liquids diffuses slowly in comparison to gases because : - (a) Liquids have no definite shape - (b) Liquid molecules are heavy - (c) Liquid molecules are in rapid motion - (d) Molecules attracts each other by strong intermolecular forces. - 56. Liquids crystals are applicable in : - (a) Thermography (b) Digital watches - (c) Electronic calculator (d) All the above - 57. An Emulsifier : - (a) Accelerates the dispersion (b) Homogenesis the emulsion - (c) Stabilizes the emulsion (d) Aids of flocculation of emulsion. - 58\. A colloidal solution is subjected to an electric field the colloidal particles move towards the anode coagulation of the same colloidal solution is studied using NaCl, BaCl2 and AlCl3 solutions the coagulating power is : - (a) NaCl > BaCl2 > AlCl3 (b) NaCl > AlCl3 > BaCl2 - (c) BaCl2 > NaCl > AlCl3 (d) AlCl3 > BaCl2 > NaCl. - 59. Which compound among following is ferri magnetic ? - (a) Fe, (b) MnO2, (c) Fe3O4, (d) CO. - 60. In a crystal of solid how many type of lattices are possible ? - (a) 23, (b) 7, (c) 230, (d) 14. - 61. The Bragg’s law equation is : - (a) nπ = 2θ sin θ, (b) nλ = 2a sin θ, - (c) 2nλ = d sin θ, (d) n = θ/2sin θ. - 62. On the basis of geometry of atoms and ions crystals are divided into following classes : - (a) 3, (b) 7, (c) 14, (d) 4. 63\. 64\. 65\. 66\. 67\. 68\. 69\. 70\. 71\. 72\. 73\. 74\. 75\. 76\. 77\. The geometry of NaCl crystal is : - (a) Face centered cubic (b) Body centered cubic - (c) Hexagonal (d) Simple cubic. In NaCl crystal every Na+ ion is surrounded by how much Cl‒ ion having equal distance ? - (a) 2 (b) 4 (c) 6 (d) 8. How many molecules are in a unit cell of sodium chloride - (a) 2 (b) 4 The space lattice of CaF2 is : - (a) Face centered cubic (c) Simple cubic (c) 6 (d) 8. - (b) Body centered cubic - (d) Hexagonal closed packing. What types of lattice is found in the crystal of potassium chloride - (a) Face centered cubic (b) Body centered cubic - (c) Simple cubic (d) Simple tetragonal. The number of atoms in a common cubic unit cell is : - (a) 1, (b) 2, (c) 3, (d) 4. The number of ions surrounding Cs+ ion in CsCl crystal are : - (a) 4, (b) 6, (c) 8, (d) 12. The name of the crystalline arrangement having arrangement of the type ABC, ABC, ABC is : - (a) Octahedral closed packing - (c) Cubic closed packed It is impossible to liquefy a gas : - (a) Above its critical temperature - (c) If it is ideal The coordination number of Na+ (b) Tetrahedral (d) Octahedral. (b) Below its critical temperature (d) If it is non ideal. ion in cubic NaCl crystal is : (a) 4, (b) 6, (c) 8, (d) 12. In calcium fluoride co-ordination number of Ca2+ ion is 8. What will be the co-ordination number of F‒ ion ? - (a) 4, (b) 3, (c) 6, (d) 8. In molten state NaCl is good conductor of electricity because it has : - (a) Free ions (b) Free electrons - (c) Free molecules (d) Na and Cl atoms. For a lattice point in face centered cubic lattice number of neighbouring ions will be : - (a) 8, (b) 6, (c) 12, (d) 14. The limit of radius ratio for tetrahedral geometry is : - (a) 0∙155, (b) 0∙414 , (c) 0∙732, (d) 0∙225. In NaCl (Sodium chloride) crystal : - (a) Na+ and Cl‒ ions are present (b) Na and Cl atoms are present - (c) NaCl molecule is present (d) NaCl dimer is present. - 78. Crystalline substance from the following whose unit cell has only one atom is : - (a) Face centered cubic (b) Body centered cubic - (c) Simple cubic (d) None of these. - 79. In rock salt structure crystal number of formula units in per unit cell is : - (a) 2, (b) 4, (c) 6, (d) 1. - 80\. For the chemical reaction A —→ E, it is found that the rate of the reaction doubles when the concentration of A is increased four times. The order in A for this reactions is : - (a) Two, (b) One, (c) Zero, (d) Half. - 81. The dimensions of the rate constant of a second order reaction involves : - (a) Neither time nor concentration - (b) Time and concentration - (c) Time and square of concentration - (d) Time only. - 82. Proton accelerate the hydrolysis of esters. This is an example of : - (a) a heterogeneous catalysis (b) a promoter - (c) an acid-base catalysis (d) a negative catalyst - 83. Lyophilie solution are more stable than lyophobic solution because : - (a) The colloidal perticles have positive charge - (b) The colloidal particles have negative charge - (c) The colloidal particles are solivated - (d) There are strong electrostatic replusions between the negatively charged colloidal particles. - 84. Which of the following is not a properties of hydrophilic solution? - (a) High concentration of dispersed phase can be easily attained - (b) Coaguletion is reversible - (c) Viscosity and surface tensions are nearly same as that of water - (d) The change of the particle depends on the pH value of the medium, it may be +ve,-ve or even zero. - 85. In a reversible reaction, a catalyst : - (a) increase the rate of the forward reaction only - (b) increase the rate of the forward reaction to a greater extent than that of the backward reaction - (c) increase the rate of the forward reaction and decrases that of backward reaction - (d) increase the rate of the forward and backward reactions equally. - 86. If the reaction rate of a given temperature becomes slower, then : - (a) the free energy of activation is higher - (b) the free energy of activation is lower - (c) the entropy charges - (d) the initial concentration of the reactants remains constant - 87. Which of the following kinds of catalysis can be explained by the adsorption theory ? - (a) homogeneous catalysis (b) acid-base catalysis - (c) heterogeneous catalyst (d) enzyme catalysis - 88. Peptization denotes : - (a) Digestion of food - (b) Hydrolysis of proteins - (c) Breaking and depression into the colloidal state - (d) Precipitation of solid form colloidal dispersion. - 89. During dyalysis : - (a) All kinds of particles can diffuse through the semipermeable membrane - (b) Only solvent molecules can diffuse - (c) Solvent molecules and ions can diffuse - (d) Solvent molecules ions and colloidal particles can diffuse. - 90. The ion which has the highest ionic radius is : - (a) Na+, (b) Al3+, (c) Mg2+, (d) Si4+. - 91. If in a crystal the number of atom per unit is 2, the crystal structure will be : - (a) Octahedral (b) Face centered cubic - (c) Body centered cubic (d) Simple cubic. - 92. Which among following is covalent crystal ? - (a) Common salt,(b) Soda, (c) Diamond, (d) Copper oxide. - 93. In a reaction on 2A + B —→ A2B, the reactant A will disappeare at ? - (a) Half the rate that B will decrease - (b) The same rate that B will decrease - (c) Twice the rate that B will decrease - (d) The same rate that A2B will form. - 94. The effect of catalyst in a chemical reaction is to change the : - (a) activation energy (b) equilibrium concentration - (c) heat of the reaction (d) final product - 95. One As2S3 sol has a negative charge. Capacity to precipitate it is highest in : - (a) AlCl3, (b) Na3PO4, (c) CaCl2, (d) K2SO4. - 96\. A closed vessel contains equal number of oxygen and hydrogen molecules at total pressure of 740 mm. If oxygen is removed from the system the pressure : - (a) Becomes 1/9 of 740 mm (b) Becomes half of 740 mm - (c) Remain unchanged (d) Becomes double of 740 mm. - 97. The colour of the colloidal particles of gold obtained by different methods differs because of : - (a) Variable valency of gold - (b) Different concentration of gold particles - (c) Different types of impurities - (d) Different diameters of colloidal particles. - 98. The strongest force among intermolecular forces is : - (a) London force (b) Dipole-Dipole attraction - (c) Hydrogen bonding (d) All are strong. - 99. According to Kinetic theory of gases, for a diatomic molecule : - (a) The pressure exerted by the gas is proportional to the mean velocity of the molecule - (b) The pressure exerted by the gas is proportional to the root mean square velocity of the molecule - (c) The root mean square velocity of the molecule is inversely proportional to the temperature. - (d) The mean translational Kinetic Energy of the molecule is proportional to the absolute temperature. - 100\. An ideal gas can not be liquefied because : - (a) its critical temperature is always above 0°C - (b) its molecules are relatively smaller in size - (c) it solidifies before becoming a liquid - (d) forces operating between its molecule are negligible. - 101\. The catalyst iron employed in the Haber process, contains molebdenum, the function of which is : - (a) to increase the rate of combination of gases - (b) to counter balance for the presence of impurities in the gases - (c) to act as a catalyst promoter and increse its activity - (d) to make up for the adverse temperature and pressure conditions - 102\. A reactions rate constant is given be : 25000 K = 12 x 10"14 e—RT sec"1 It means : - (a) log K versus log T will give a straight line with slope as ‒25000 - (b) log K versus log T will give a straight line with slope as 25000 1 - (c) log K versus log T will give a straight line with slope as ‒25000 1 - (d) log K versus log T will give a straight line. - 103\. For an endothermic reaction where AH represents the enthalpy of the reaction in kJ/mole, the minimum value for the energy of activation will be : - (a) less than AH (b) zero - (c) more than AH (d) equal to AH - 104\. Which is not true in case of an ideal gas ? - (a) It cannot be converted into a liquid - (b) There is no interaction between the molecule. - (c) All molecules of the gas move with same speed - (d) At a given temperature PVi is proportional to the amount of gas - 105\. If C1, C2, C3.............. represent the speeds of n1, n2, n3 molecules then the root mean square speed is : 1/2 (a) (b) (c) (d) 222 nici + n2C2 + n3C3 +............ ni + n2 + n3 +............ i1/2 d222 nici + n2C2 + n3C3 +. ni+ n2 + n3+ ■ 2 1/2 2 1/2 2 1/2 dnici i + dn2C2 i + dn3C3 i + n1 n2 n3 2 1/2 bnici + n2C2 + n3C3 +............g ni + n2 + n3+............ - 106. For a first order reaction : - (a) the degree of dissociation is equal to (1 ‒ e‒kt) - (b) a plot of reciprocal concentration of the reactant Vs time gives a straight line. - (c) the time taken for the completion of 75% reaction is thrice the t 2 of the reaction - (d) the pre-exponential factor in the Anhenius equations has the dimension of time, 1T‒1. - 107\. An example of autocatalytic reaction is : - (a) the decompositon of nitro glycerine - (b) thermal decomposition of KClO3 + MnO2 mixture - (c) breakdown of 6C14 - (d) hydrogenation of vegetable oil using nickel catalyst - 108. A catalyst is used : - (a) only for increasing the velocity of the reaction - (b) for altering the velocity of the reaction - (c) only for decreasing the velocity of the reaction - (d) all (a), (b) and (c) are correct. - 109\. For the reaction H2(g) + Br2(g) ^ 2HBr(g) the experimental data suggests Rate = K = \[H2\] \[Br2\]1/2 The molecularity and order of reaction for the reaction is : (a) 2 and 2 respectively 1 (b) 2 and 1 2 respectively 1 (c) 1 2 and 2 respectively 11 (d) 1 2 and 1 2 respectively. - 110\. The correct relationship of Van der Waall’s equation is : - (a) PV = RT ###### F aI 1 - (c) G PJ b v + b g = RT (d) PV = ~ mnv. H v2K 3 - 111\. Which of the following statement is false : - (a) The product of pressure and volume of a fixed amount of a gas is independent of temperature - (b) The gas equation is not valid at high pressure and low temperature - (c) Molecules of different gases have the same Kinetic energy of a given temperature - (d) The gas constant per molecules is known as Boltzmann constant. - 112\. The critical temperature of a substance is defined as : - (a) The temperature above which a substance decomposes - (b) The temperature above which a substance can exist only in ages - (c) Melting point of the substance - (d) Boiling point of the substance. - 113\. Helium atom is two times heavier than a hydrogen molecule. At 298 K, the average K. E. of a helium atom is : - (a) two times that of hydrogen molecule - (b) same as that of a hydrogen molecule - (c) four times that of a hydrogen molecule - (d) half that of a hydrogen molecule. - 114\. In which of the following Tyndall effect is not observed ? - (a) Suspension (b) Emulsions - (c) Sugar solution (d) Gold sol. - 115. Lyophilic colloids are stable due to : - (a) Charge on the particles - (b) Large size of the particles - (c) Small size of the particles - (d) Layer of dispersion medium of the particles. - 116\. Van der Waals equation explains the behaviour of : - (a) Mixture of gases (b) Ideal gas - (c) Real gas (d) None of above. - 117\. The molecular weight of a gas is 44. The volume occupied by 22g of a gas at S. T. P. would be : - (a) 224 liters (b) 2∙24 liters (c) 11∙2 liters (d) 1∙12 liters. - 118\. When the temperature is increased, surface tension of water : - (a) increases (b) decreases - (c) remains constant (d) show irregular behaviour - 119\. Which one of the following explains that the reactions of high molecularity are rare : - (a) The greater the number of collisions between the colloiding molecules the more we be the weight. These molecules will have to cross the energy barrier. - (b) The activation energy of many collision becomes very high - (c) Many body collisions have low probability - (d) None of these. - 120\. A zero order reaction is one whose rate is independent of : - (a) The temperature of the reaction - (b) The concentration of the reaction - (c) The concentration of the products - (d) The material of the vessel in which the reaction is carried out. - 121\. Air can oxidize sodium sulphite in aqueous solution but can not do so in the case of sodium arsenite. If, however, air is passed through a solution containing both sodium sulphite and sodium arsenite then both are oxidised. This is an example of : - (a) positive catalysis (b) negative catalysis - (c) induced catalysis (d) auto catalysis - 122\. During hydrogenation of oil which of the following catalyst is commonly used ? - (a) Pd or CuCl2 (b) Ni (c) Fe (d) V2O5 - 123\. The ratio of root mean square velocity (Vrms) of molecular hydrogen and molecular oxygen are the same temperature is : - (a) 4 : 1, (b) 1 : 4, (c) 1: 2, (d) 1: 8. - 124. A zero order reaction is one : - (a) whose rate is affected by concentration - (b) in which concentration of the reaction does not charge with time - (c) in which reactant do not react - (d) in which one of the reactants is in large excess - 125\. If the concentration of reactants is increased the rate of reaction : - (a) Remain uneffected (b) Increases - (c) Decreases (d) May increase or decrease. - 126\. The efficiency of an enzyme in catalysing a reaction is due to its capicity : - (a) to form a strong enzyme-substance complex - (b) to decrease the bond energies of the substrate moelcule - (c) to change the shape of the substrate molecule - (d) to lower in activation energy of the reaction. - 127\. Regarding criteria of catalysis which one of the following statements is not true ? - (a) The catalyst in unchanged chemically at the end of the reaction - (b) A small quantity of catalyst is often sufficient to bring about a considerable amount of reaction - (c) In a reversible reaction the catalysis alters the equilibrium position - (d) The catalyst accelerates the reactions - 128\. An ideal gas obeying kinetic theory of gases can be liquefied if : - (a) Its temperature is more then critical temperature Tc. - (b) It can not be liquefied at any P and T. - (c) Its pressure is more than Pc at a temperature less then Tc. - (d) Its pressure is more then critical pressure. - 129\. The rate constant of reaction depends upon : - (a) Temperature - (b) Initial concentration of the reactants - (c) Time of the reaction - (d) Extent of reaction. - 130\. The rate expression for a chemical reaction is given by : Rate = k\[A\]m\[B\]n - (a) The order of the reaction is m - (b) The order of the reaction is n - (c) The order of the reaction is m + n - (d) The order of the reaction is m – n. - 131\. The compounds which have not resistance for flow of electricity are called : - (a) Semi conductors (b) Conductors - (c) Super conductors (d) Non-conductors. - 132. Gold number is associated with : - (a) Amount of pure gold in gold coin - (b) Purple of cassius - (c) Protective gold - (d) Electrophoresis.76. - 133\. Intermolecular forces present in liquids is due to : - (a) Orientation effect (b) Induction effect - (c) Dispersion effect (d) All the above - 134\. Which of the following compounds is used as a contact catalyst ? - (a) Boron (b) Germanium - (c) Nickel (d) Uranium - 135\. A catalyst is a substance which : - (a) at less the equilibrium in a reaction - (b) does not participate in the reaction but speed it up - (c) participates in the reaction and provides an easier pathway for the same - (d) is always in the same phase as the reactants. - 136\. When KClO3 is heated, it decomposes into KCl and O2, if some MnO2is added the reactions goes much faster because : - (a) MnO2 decomposes to give oxygen - (b) MnO2 provides head by reacting - (c) better contact is provide by MnO2 - (d) MnO2 acts as a catalyst - 137\. A catalyst is used in a reaction to : - (a) change the nature of reaction products - (b) increase the reaction yield - (c) decreases the need for reactants - (d) decrease the time required for the reaction - 138\. The van der Waal’s equation of stere is obeyed by real gases. For n molecules of real gas the expression will be : - (a) F P + na I F - - b I = n RT (b) n F P + ^2| b v - b g = nRT ###### H n v2 K H n K H v K F n2aI - (c) I P + J J bnv - bg = nRT (d) I P + —f bv - nbg = nRT . H v2 K H v K - 139\. In a reaction involving hydrolysis of an organic chloride in presencce of large excess of water. RCl + H2O> ROH + HCl - (a) Molecularity = 2, order of reaction = 2 - (b) Molecularity = 2, order of reaction = 1 - (c) Molecularity = 1, order of reaction = 2 - (d) Molecularity = 1, order of reaction = 1. - 140\. Which statement is wrong among the following ? - (a) Haber’s process of NH3 requires ion as catalyst - (b) Friedel-Crafts’s reaction uses anhydrous AlCl3 - (c) Hydrogenation of oil uses ion as catalyst - (d) Oxidation of SO2 to SO3 requires V2O5 - 141 Which of the following best explains the effect of a catalyst on the rate of a reversible reaction ? - (a) It provides a new reaction path with a lower activation energy - (b) It moves the equilibrium position to the right - (c) It increase the kinetic energy of the reacting molecules - (d) It decrease the rate of the reverse reaction. - 142\. Diazonium salt decomposes as : C6H5N2+ CL- ^ C6H5C1 + N2 At 0°C, the evolution of N2 becomes two times faster when the initial concentration of the salt is doubled. Therefore it is : - (a) a first order reaction - (b) a second order reaction - (c) independents on the initial concentration of the salt - (d) a zero order reaction. - 143\. Molecular velocities of two gases at the same temperature are u1 and u2. There masses are m1 and m2 respectively. Which of the following expression is correct [m1](#bookmark133) mm (b) ~ = ~ u1 u2 22 (d) m i u i = m 2 u 2 . - [(a) „2„2](#bookmark134) [u1](#bookmark135) - (c) m 1 u i = m 2 u 2 - 144\. In van der Wall’s equation of state for a non-ideal gas the term that accounts for inter molecular forces is : [a ^](#bookmark136) - (a) (V - b) (b) RT (c) I P + 772 I (d) — \\ V J RT - 145\. Which of the following volume (V)-temperature (t) plots represent the behaviour of one mole of an Ideal gas at 1 atmosphere pressure ? V (L) (b) (22.4 L, 273 K) ![](AU B. Sc. I Chemistry-III(OK)_files/AU20B.20Sc.20I20Chemistry-III(OK)-1.png) (28.6 L, 373 K) T (K) V (L) (d) (22.4 L, (142 L, 273 K) 373 K) T (K) - 146. Alum is added to muddy water because : - (a) It acts as a disinfactants - (b) It results in coagulation of clay and clear water - (c) Clay is soluble in alum, hence remove it - (d) It makes water alkaline, which is good for health. - 147\. A class of compounds which is anisotropic is : - (a) Gases (b) Liquids - (c) Liquid crystals (d) None of the above - 148\. According to Kinetic theory of gases : - (a) There are intermolecular attractions - (b) There are no intermolecular attractions - (c) The velocity of molecules decreases for each collision - (d) Molecules have considerable volume. - 149\. Which one of the following statements is incorrect in the case of heterogeneous catalysis ? - (a) The catalyst lowers the energy of activation - (b) The catalyst actually forms a compound with the reactant - (c) The surface of a catalyst plays a very important role - (d) There is no change in the energy of activation - 150\. The specific rate constant of a first order reaction depends on the : - (a) Concentration of the reactants - (b) Concentration of the products - (c) Time - (d) Temperature. - 151\. A catalystic poison acts by : - (a) getting absorbed at the active centers of the catalytic surface - (b) chemically combining with the catalyst - (c) getting absorbed into the catalyst - (d) chemically combining with any one of the reactants. - 152\. ('ll,/),, + HO dil^SO4 > cho + C,H.,O, . 12 22 11 2 6 12 6 6 12 6 Sucrose in this reaction dil H2SO4 is called : - (a) homogeneous catalyst (b) heterogeneous catalysis - (c) homogeneous catalysis (d) heterogeneous catalyst. - 153\. Following unit cell arrangement is found in tetragonal crystal group ? - (a) a = b = c (b) a = b # c a = p =y = 90° a = p = y = 90° - (c) a # b # c (d) a = b # c a = p = y = 90° a = p = 90°; y = 120°. O - 154\. CH3O— —N = N— —OCH3 is an example of : - (a) Crystalline solid (b) Amorphous solid - (c) Liquid crystal (d) None of the above. - 155\. A catalyst remain unchanges at the end of the reaction as regards : - (a) quantity and physical state - (b) chemical composition and physical stage - (c) chemical composition, quantity and physical state - (d) quantity and chemical composition. - 156\. What is the role of a catalyst in a catalysed reaction ? - (a) Lowers the activation energy - (b) Increases the activation energy - (c) Affects the free energy - (d) Affects the enthalpy change. - 157\. A catalyst : - (a) increases the free energy charge in the reaction - (b) decreases the free energy charge in the reaction - (c) does not increase or decrease the free energy charge in reaction - (d) can either decrease or increase the free energy charge depending on what catalyst we use. - 158\. Choose the correct statement(s) from the following : - (a) Enzymes are non-specific catalysts - (b) Nitric oxide acts as a homogeneous catalysts in the lead chamber process for the manufcature of sulphuric acid - (c) Aluminium chloride behaves as a Lewis acid catalyst in Friedel Crafts reaction - (d) Hydrolysis of ethyl acetate is catalysed by acids but not by alkalis. - 159\. The rate constant (K´) of one reaction is double the rate constant (K´´) of another reaction. Then the relationship between the two corresponding activation energies of the two reactions (Ea´ and Ea´´) will be : - (a) Ea´ > Ea´´, (b) Ea´ = Ea´´, (c) Ea´ < Ea´´, (d) Ea´ = 4Ea´´. - 160\. Under a given set of experimental conditions with increase in the concentration of the reactants the rate of a chemical reactions : - (a) Decreases (b) Increases - (c) Remains unaltered (d) First decrease and then increases. - 161\. In which of these processess in platinum used as a catalyst ? - (a) oxidation of ammonia to form nitric acid - (b) hardening of oils - (c) production of synthetic rubber - (d) synthesis of methanol. - 162\. Which one of the following statements is wrong in case of enzyme catalysis ? - (a) Enzyme work best at an optimum tempertaure. - (b) Enzymes work at an optimum pH - (c) Enzymes are highly specific for substances - (d) An enzymes raises activation energy. - 163\. Colloids are purified by : - (a) Brownian motion (b) Precipitation - (c) Dialysis (d) Filtration. - 164\. The blue colour in the sea is due to : - (a) Refraction of the blue light by the impurities in sea water - (b) Refraction of blue sky by sea water - (c) Secttering of blue light by water molecules - (d) Absorption of other colours except the blue colour by water molecule. 165. The migration of colloidal particles under influence of an electric field is known as : - (a) Electro-osmosis (b) Brownian movement - (c) Cata phoresis (d) Dialysis. - 166\. Diamond is an example of : - (a) Covalent solid - (b) Electrovalent solid - (c) Solid containing hydrogen bonds - (d) Glass. - 167\. Which among the following is not a property of solids ? - (a) Solids are always crystalline in nature - (b) Solids have high density and low compressibility - (c) Solids diffuses very slowly - (d) The volume of solids is fixed. - 168\. The half life period of a first order reaction is : - (a) Dependent on the rate constant - (b) Directly proportional to the rate constant - (c) Independent of the initial concentration - (d) Dependent on the initial concentration - 169\. The hydrolysis of ethyl acetate is reaction of CH3COOC2H5 + H2O ————→ CH3COOH + C2H5OH - (a) First order, (b) Second order, (c) Third order, (d) Zero order - 170\. When a catalyst is added to the system, the : - (a) equilibrium concentrations are increased - (b) equilibrium concentrations are unaffected - (c) equilibrium concentrations are decreased - (d) rate of forward reactions increases whereas the rate of backward reaction decreases. - 171\. Which of the following statements about a catalyst is/are true ? - (a) Acatalyst accelerates the reaction by bringing down the free energy of activation - (b) A catalyst does not take part in the reaciton mechanism - (c) A catalyst makes the reaction more fesible by making the ΔG° more negative - (d) A catalyst makes the equilibrium constant of the reaction more favourable for the forward reaction. - 172\. Kinetic theory of gases assumes that tiny particles called molecules : - (a) Contian average KE proportional to absolute temperature - (b) Exert no force - (c) Exert attractive force on each other - (d) Contain constant KE at all temperatures. - 173\. A chemical reaction is catalysed by a catalyst X. Hence X : - (a) does not affect equilibrium constant of the reaction - (b) decreases the rate constant of the reaction - (c) reduces the enthalapy of the reaction - (d) increase activation energy of the reaction. - 174\. Platirized asbestos is used as a catalyst in the manufacture of H2SO4. It is an example of : - (a) heterogeneous catalyst (b) auto catalyst - (c) homocatalyst (d) induced catalyst - 175\. At a given temperature the energy of activation of two reaction are same if : - (a) The specific rate constants for the two reactions are the same. - (b) The temperature coefficients of the specific rate constant for the two reactions are the same. - (c) ΔH for the two reactions are the same but not zero - (d) ΔH for the two reactions are zero. - 176\. If reaction A and B to give shows first order kinetics in A and second order in B, the equation can be written as : - (a) rate = K \[A\] \[B\]1/2 (b) rate = K \[A\]1/2\[B\] - (c) rate = K \[A\] \[B\]2 (d) rate = K \[A\]2\[B\] - 177\. Rate of reactions : - (a) Increases with increase in temperature - (b) Decrease with increase in temperature - (c) Does not depend on temperature - (d) Does not depend on concentration. - 178\. According to Kinetic theory of gases : - (a) there are intermolecular attractions - (b) there is no intermolecular attraction - (c) the velocity of molecules decrease is each collision. - (d) molecules have considerable volume. - 179. According to Kinetic theory of gases, for a diatomic molecule : - (a) the pressure exerted by a gas is proportional to the mean velocity of the molecule. - (b) the pressure exerted by the gas is proportional to the root mean square velocity of the molecule - (c) the root mean square velocity of the molecule is inversely proportional to the temperature - (d) the mean translational energy of the molecule is proportional to the absolute temperature. - 180\. In a reversible reaction the energy of activation of the forward reaction is 50 kcal. The energy of activation for the reverse reaction will be : - (a) 50 kcal - (b) > 50 kcal - (c) < 50 kcal - (d) either greater than or less than 50 kcal. - 181\. The decomposition of hydrogen peroxide can be slowed by addition of a smaller amount of acetamide. The latter acts as a : - (a) detainer (b) stopper (c) promoter (d) inhibitor. - 182\. A negative catalyst retards the reaction because : - (a) it decreases the speed of the reactant molecules - (b) it enhances the activation energy of the reaction - (c) it deactivases the reactant molecules - (d) none of these. - 183\. The enzyme which ptyatin used for the digestion of food is present in : - (a) Saliva (b) Blood (c) Intestine (d) Adrenal glands - 184\. According to adsorption theory of catalysis, the speed of the reaction increases because : - (a) the concentration of reactants molecules at the active centers of the catalysts becomes high due to adsorption - (b) In the process of adsorption the activation energy of the molecule becomes taste - (c) Adsorption produces heat which increases the speed of the reaction (d) All of the above. - 185\. Gaseous of ten deviates from ideal gas behaviour because their molecules : - (a) Possess negligible volume - (b) Are poly atomic - (c) Have forces of attraction between them - (d) Are not attracted to one another. - B. Sc. I, Chemistry III Answers 1\. (d) 2\. (a) 3\. (b) 4\. (a) 5\. (a) 6\. (c) 7\. (b) 8\. (d) 9\. (b) 10\. (d) 11\. (c) 12\. (a) 13\. (b) 14\. (a) 15\. (a) 16\. (a) 17\. (b) 18\. (a) 19\. (d) 20\. (d) 21\. (b) 22\. (a) 23\. (a) 24\. (a) 25\. (b) 26\. (b) 27\. (d) 28\. (b) 29\. (b) 30\. (c) 31\. (b) 32\. (b) 33\. (a) 34\. (b) 35\. (a) 36\. (d) 37\. (b) 38\. (a) 39\. (d) 40\. (c) 41\. (d) 42\. (c) 43\. (a) 44\. (b) 45\. (c) 46\. (b) 47\. (d) 48\. (b) 49\. (a) 50\. (d) 51\. (c) 52\. (b) 53\. (b) 54\. (c) 55\. (d) 56\. (d) 57\. (c) 58\. (d) 59\. (c) 60\. (d) 61\. (b) 62\. (c) 63\. (a) 64\. (c) 65\. (b) 66\. (a) 67\. (c) 68\. (a) 69\. (c) 70\. (c) 71\. (a) 72\. (b) 73\. (a) 74\. (a) 75\. (b) 76\. (c) 77\. (a) 78\. (c) 79\. (d) 80\. (d) 81\. (b) 82\. (c) 83\. (a) 84\. (b) 85\. (d) 86\. (a) 87\. (c) 88\. (c) 89\. (c) 90\. (a) 91\. (c) 92\. (c) 93\. (c) 94\. (a) 95\. (a) 96\. (a) 97\. (d) 98\. (c) 99\. (a) 100\. (d) 101\. (c) 102\. (d) 103\. (d) 104\. (c) 105\. (c) 106\. (a) 107\. (a) 108\. (b) 109\. (b) 110\. (b) 111\. (c) 112\. (b) 113\. (b) 114\. (d). 115\. (c) 116\. (c) 117\. (b) 118\. (b) 119\. (c) 120\. (b) 121\. (c). 122\. (b) 123\. (a) 124\. (b) 125\. (b) 126\. (d) 127\. (c) 128\. (a). 129\. (a) 130\. (c) 131\. (c) 132\. (c) 133\. (d) 134\. (c) 135\. (c). 136\. (d) 137\. (d) 138\. (d) 139\. (b) 140\. (c) 141\. (a) 142\. (a) 143\. (d) 144\. (c) 145\. (c) 146\. (b) 147\. (c) 148\. (b) 149\. (d) 150\. (c) 151\. (a) 152\. (c) 153\. (b) 154\. (c) 155\. (d) 156\. (a) 157\. (b) 158\. (b) 159\. (c) 160\. (b) 161\. (a) 162\. (b) 163\. (d) 164\. (c) 165\. (c) 166\. (a) 167\. (c) 168\. (c) 169\. (a) 170\. (b) 171\. (a) 172\. (b) 173\. (a) 174\. (a) 175\. (a) 176\. (c) 177\. (d) 178\. (b) 179\. (d) 180\. (d) 181\. (d) 182\. (c) 183\. (a) 184\. (c) 185\. (c). Very Short Answer Type Questions Q. 1. According to technology used, give the types of computer. Ans. According to technology used, computers are of three types : - (a) Analog Computer (b) Digital Computer (c) Hybrid Computer. Q. 2. Give the names of different components of a computer. Ans. Different components of a computer are : - (i) Input unit, (ii) Central Processing Unit (CPU), (iii) Output unit. Q. 3. What is the use of joystick ? Ans. The joystick is mostly used in children’s games. Q. 4. Give the classification of computer languages. Ans. The computer languages can be classified in the following three broad categories : - (i) Machine Language (Low level language) - (ii) Assembly Language (Middle level language) - (iii) High Level Language. Q. 5. Give two advantages of high level languages. Ans. (i) They are easier to learn. - (ii) They are easier to write, correct and modify. Q. 6. What is Software ? Ans. Software : Computer system is a dumb machine like a record player without a cassette. To make computer a meaningful machine we feed in a set of instructions known as software. But we can not touch it like hardware components. Rather you can feel it. On the other hand, software can also be defined as a set of programme, documents, procedures and routines associated with the operation of a computer system doveloped by the user. Q. 7. Work out the slope of the straight lines which are parallel and perpendicular to 2y = 3x + 1. Ans. Given that line is 2y = 3x + 1 y = 3 x +1 2 31 — x + — 22 3 Slope of the parallel line = 2 Slope of the perpendicular line -1 slope of the line -1 2 Q. 8. Differentiate x4 + 3x. d4 d4d Ans. —I x + 3 x ) = —x + —3 x dx dx dx = 4x3 + 3. Q. 9. Write 35 = 243 in logarithmic form. Ans. 35 = 243 log3 243 = 5 Q. 10. If log 2 = 0.3010, find out the value of log ^8. Ans. log ^8 = log 81/2 = 12 log8 = 12log23 - 3 3 = -log2 = -x 0.3010 = 0.4515 Q. 11. Simplify (a3b–5)– 5. Ans. (a3b‒5)‒5 = a‒15∙b+25 b25 = a15 Q. 12. Give the Braggs equation to determine the distance between constituent particles in a crystals. Ans. Bragg’s equation is n X = 2 d sin 0 where X ^ wavelength of X-rays d ^ distance between two parallel planes n ^ order of diffraction 0 ^ Angle between the incident X-rays and plane of the crystal. Q. 13. Give the law of rational indices. Ans. The intercepts that any face of the crystal makes with the crystallographic axes are either equal or small rational multiples of the intercepts made by unit face. Q. 14. Give the types of cubic crystal lattices. Give their names. Ans. Cubic crystal lattices are of three types : - (i) Simple Cubic Crystal - (ii) Body Centered Cubic Crystal - (iii) Face Centered Cubic Crystal. Q. 15. Write the number of total space lattices. Ans. No. of total space lattices is 14. Q. 16. Give the types of space lattices of tetragonal crystal lattices and give their numbers. Ans. Types of tetragonal space lattices are simple and body centered and their numbers are two. Q. 17. What are ionic crystal of AB types ? Give example. Ans. Ionic crystals in which ratio of cations and anions is 1 : 1 and whose co-ordination numbers are equal, are called ionic crystals of AB types. For example : NaCl crystal. Q. 18. Give the co-ordination number of Zn2+ and S2– ions in zinc blende crystal. Ans. The co-ordination of Zn2+ and S2‒ ions in zinc blende are 4-4 respectively. Q. 19. Give the examples of metals having BCC structure. Ans. Alkali metals of group IA like Na, K and Rb etc. has BCC structures. Q. 20. Give the types of crystalline solids. Ans. Crystalline solids are of four types : - (i) Ionic crystals - (ii) Molecular crystals - (iii) Covalent crystals (iv) Metallic crystals. Q. 21. What do you mean by unit cell ? Ans. That smallest unit which upon regular repetation in space form crystal lattice is called unit cell. Q. 22. How much is the size of colloidal particles ? Ans. The size of the colloidal particles is greater than 10‒7 cm but less than 10‒5 cm. Q. 23. What do you mean by a sol ? Give an example. Ans. When a solid substance gets dispersed in a medium to form a colloidal solution it is called a sol. Usually sol is called a colloidal solution, such as colloidal solution of arsenious sulphide, Fe(OH)3, gold etc. Q. 24. What is a get ? Give an example. Ans. When a liquid dispersed into a solid to form a colloidal system, it is called a get. For example jelly, cheese, butter etc. Q. 25. What do you mean by an emulsion ? Ans. When a liquid disperses in an immiscible liquid to form a colloidal system, it is called an emulsion. For example, cream, milk, rubber latek etc. Q. 26. What is peptization and peptizer ? Ans. The process through which a freshly prepared precipitate is converted into a colloidal solution is termed as peptization. Peptizer is a electrolyte by which the process of peptizaton occur. Q. 27. Give two examples of liquid crystals. Ans. p-azoxy anisol and cholesteryl benzoate are liquid cyrstals. Q. 28. Give the types of liquid crystals. Ans. Liquid crystals are of three types : - (i) Smectic liquid crystal - (ii) Nemactic liquid crystal - (iii) Cholesteric liquid crystal. Q. 29. Write down the van der Waals gas equation. Ans. (P + Va2)(V-b) = RT. Above equation is called van der Waals gas equation. Q. 30. What do you mean by most probable velocity ? Ans. It is the velocity gained by the maximum number of molecules in a gas. It is represented by Vmp. 2RT RT V = a---= 1A — mp M M Q. 31. Give the kinetic gas equation. Ans. On the basis of kinetic molecular theory Clausius and Chronig gave the following kinetic gas equation : 12 PV = — mnv where P ^ pressure of gas, V ^ volume of the gas m ^ molecular wt. of gas, n ^ no. of gas moleculars. Q. 32. Write a note on dialysis. Ans. The process of removal of soluble impurity from a colloidal solution through a semipermeable membrane is called dialysis. Q. 33. Arrange the power of ions in increasing order to coagulate As2S3 sol. Ans. As2S3 is a negative sol. It will be coagulated by positive ions. Hence the order will be Na+ < Mg2+ < Al3+ < Sn4+ Q. 34. Show that kinetic energy of n mole of a gas is proportional to absolute temperature. 12 Ans. Kinetic equation is PV = 3 mnv 12 (ii) (iii) But kinetic energy KE = mnv 1 21 PV = mnv = ∙ mnv 3 32 2 = 3 K. E. But PV = nRT 3 K. E. = nRT 2 Because in equation (iii), n and R are constants, ∴ K. E. ∝ T Hence, kinetic energy for gas molecules is proportional to absolute temperature. Q. 35. What do you mean by coagulation or floceulation ? Ans. Conversion of colloidal solution into precipitate by means of an electrolyte is called coagulation or floceulation. Q. 36. State Boyle’s law. Ans. Temperature remaining constant, the volume of a given mass of a gas is inversely proportional to the pressure. V ∝ P or PV = constant. Q. 37. What do you mean by an ideal gas ? Give ideal gas equation for one mole of a gas. Ans. A gas which obeys gas laws under all conditions of temperature and pressure is known as an ideal gas. Ideal gas equation for one mole of a gas is PV = RT Above equation is called ideal gas equation for 1 mole of a gas. Q. 38. Give the relation between average velocity, most probable velocity and root mean square velocity. Ans. Uave 2RT Ump M and Urms rms ave mp =3:^^ = 1.7 : 1.6 : 1.4 = 1.0 : 0.921 : 0.816 Q. 39. Give any two postulates of kinetic molecular theory. Ans. (i) A gas is composed of large number of particles, called molecules. - (ii) The volume of an individual molecule is negligible as compared with the total volume of the gas. Q. 40. Define critical temperature. Ans. The critical temperature may be defined as the temperature above which a gas cannot liquified by the pressure alone however big may be its magnitude. The pressure required to liquify a gas at its critical temperature is called the critical pressure and the volume occupied by a mole of a gas at the critical temperature and critical pressure is known as the critical volume. Q. 41. What is R in gas equation PV = nRT ? Give the value of R in litre atmosphere per kelvin per mole. Ans. R is gas constant. Its value is 0.0821 litre atmosphere per kelvin per mol. Q. 42. By which enzyme hydrolysis of urea occurs ? Ans. Urease enzyme hydrolyses urea : urease NH2CONH2 + H2O--> 2NH3 + CO2 Q. 43. Give name of two negative catalysts. Ans. (i) Glycerol in the decomposition of H2O2. - (ii) Alcohol in the oxidation of Na2SO3. Q. 44. What is velocity constant ? Ans. When concentration of all the reactants in any reaction becomes unity, the velocity of this time is called velocity constant. Q. 45. Give an example of pseudo molecular reaction. Ans. CH3COOC2H5 + H2O === CH3COOH + C2H5OH Ethyl acetate Acetic acid Ethanol Q. 46. Define activation energy. Ans. The additional amount of energy which the reactant molecule must acquires so that their collisions may result in an actual chemical charge is called activation energy. Q. 47. Give the types of catalysis on the basis of physical state of catalyst and reactants. Ans. Catalysis is of two types on the basis of physical state of catalyst and reactants : - (i) Homogeneous catalysis - (ii) Heterogeneous catalysis. Q. 48. Give a chemical equation of auto catalyst. COOH Ans. 5 | + 2KMnO4 + 3H2SO4 ^ K2SO4 + MnSO4 + 10CO2 + 8H2O COOH Here Mn2+ works as auto catalyst. Q. 49. Is positive catalyst increases or decreases activation energy ? Ans. Positive catalyst decreases activation energy. Q. 50. Give the expression for velocity constant for first order reaction. Ans. 2∙303 a K =----log----- t (a - x) ∙693 K = Q. 51. What is temperature coefficient ? Ans. The temperature coefficient may be defined as the ratio of the rate constants of a reaction of two different temperatures, which differ by 10°C. . K35°C Temperature coefficient = K25°C Q. 52. If concentration is expressed in mole/litre and time in minutes then calculate the specific velocity constant for second order reaction. Ans. mol‒1 litreminute‒1 Q. 53. What do you mean by order of reaction ? Ans. It is defined as the number of molecules or atoms whose molecular concentrations undergoes a measurable or appreciable change in the determination of the rate of reaction. Section ‘B’ Short Answer Type Questions Q. 1. A coin is tossed succesively three times. Find the probability of getting exactly one head or two heads. Sol. : Let S be the sample space then S = {HHH, HHT, HTH, THH, TTH, THT, HTT, TTT} Let E be the even of getting exactly one head or exactly two heads, Then, E = {HHT, HTH, THH, HTT, THT, TTH} Clearly, n (E) = 6, n (S) = 8 nE 6 3 ■• P (E) = n (S) = 8 = 4 Ans. Q. 2. Prove that, log 15 = log 1 + log 3 + log 5 Sol. : log 15 = log (1 x 3 x 5) = log 1 + log 3 + log 5 Ans. - Q. 3. Find the local maxima and local minima of f(x) = x3 -6x2 + 9 x +15. Also find the local maximum or local minimum values as the case may be. Sol. : Here, f(x) = x3 - 6 x2 + 9 x +15 ^ fr(x) = 3 x2 - 12 x + 9 For a local maxima or minima we must have f(x) = 0 New, fr(x) = 0 ^ 3(x2 - 4x + 3) = 0 = 3(x - 1) (x - 3) = 0 ^ x = 3, 1 When x = 3, In this case, when x is slightly less than 3, then f'(x) = 3(x - 3) (x - 1) is negative and when x is slightly more that three, then f'(x) is positive. Thus, f'(x) changes from negative to positives x increases through 3. - So, x = 3 is a point of local minima. Local minimum value = f 3) = 15 when x = 1, In this case when x is slightly less than 1, thenf'(x) = 3(x - 3) (x - 1) is positive and when x is slightly more than 1, then f'(x) is negative. Thus, f'(x) changes sign from positive to negative as x increases through 1. - So, x = 1 is the point of local maximum. The local maximal value =f'(1) = 19 Q. 4. How many four digit numbers can be formed with the digits 1, 2, 3, 4, 5, 6, when a digit may be repeated any number of times in any arrangement. Ans. Clearly the thousands place can be filled in six different ways as any one of the 6 digits may be placed at it. Since, there is no restriction on repitition of digits, each one of the hundreds, tens and units digits can be filled in 6 ways. - ■ Required number of numbers = 64 = 1296 Q. 5. (i) Determine the differentiation coefficient of aex. (ii) Determine the differential coefficient of ex2 + 4x2 + 6. Sol. : (i) Let y = aex dy = a.ex dx (Because the differentiation of ex is ex) (ii) Let y = ex3 + 4x2 + 6 :. yr- = 3 ex1+8x Ans. dx Q. 6. How many 4-letter words with or without meaning can be formed out of the letter of word ‘LOGARITHMS’ if repeatation of letters is not allowed. Sol. : The word ‘LOGARITHMS’ contains 10 different letters. : No. of required words = No. of arrangements of 10 letters, taken 4 at a time. = P (10, 4) = 10 x 9 x 8 x 7 = 5040 Ans. Q. 7. Calculate the logarithms of 34∙56. Sol. : Because the given number is greater than 1, the number of digits before decimal is 2. Hence mantissa will be 2 ‒ 1 = 1. Decimal number = 0∙5386 Hence, log1034∙56 = Mantissa + Decimal Number = 1.5386 Ans. Q. 8. Show the development of electronic computer in the form of a Table. Ans. Generations Period First Generation From 1946 to 1956 Second Generation From 1956 to 1964 Third Generation From 1964 to 1970 Fourth Generation From 1970 to 1985 Fifth Generation From 1985 to onward Q. 9. Give main functions of computer. Ans. Initially computer is used for calculating purposes but with the development of culture, now it is used in a very vast field. In present days it is used in following fields : - (i) To accepts data or input - (ii) To store data or storage - (iii) To analysis data or processing - (iv) To give the result of analysed date or output. A computer cannot only store and process data but also retrieve data i.e., take out data from the memory or storage as and when desired. Q. 10. Give the values of following : - (i) log4 32 (ii) log7 (419) Sol. : (i) log432 = x 4x = 32 >25 22x = 25 ^ 2 x = 5 or x = 2.5 Ans. - (il) log7( 49) = x - 7x - (49) = (7)‘ = 7-2 ^ x - -2 or i°g7 (49) = -2 Ans. Q. 11. Three dice are thrown together. Find the probability of getting a total of at least 6. Sol. : Since, three dices are thrown, the total number of points in the sample space S Is (6 x 6 x 6) - 216 Let E = Event of getting a total of at least 6 Then E = Event of getting a total of less then 6 i.e. 3, 4 or 5 a (E) - {(1, 1, 1), (1, 1, 2), (1, 2, 1), (2, 1, 1), (1, 1, 3), (1, 3, 1), (3, 1, 1), (1, 2, 2), (2, 1, 2), (2, 2, 1)} - •■• n (E)=10 [n(E)10](#bookmark155) [Sol. : P (nor E) = R ==](#bookmark156) [E n(S)](#bookmark157) - • P(E) - 1 - P(nor E) - {1- 2^} 206 = 216 Q. 12. Evaluate, 103 = 108 Ans. |(3x2 -5x-24)dx . Sol. : Here, J" (3x2 - 5x - 24) dx - 3 jx2dx - 5J" xdx - xj dx 3 2 = 3∙ x3 ‒ 5∙ x2 ‒ 24x = x3 ‒ 52x2 ‒ 24 Ans. Q. 13. Solve, 0∙243×100×5∙2 x = 60×0∙035 by logarithms. 0∙243×100×5∙2 Sol. : x = 60×0∙035 Taking log on both sides log x = log 0∙243 + log 100 + log 5∙2 ‒ log 60 ‒ log 0∙035 = 1∙3856 + 2∙0000 + 0∙7160 ‒ 1∙7782 ‒ 2 ∙5441 = 1 + 0∙3856 + 2.0000 + 0∙7160 ‒ 1.7782 + 2 ‒ 0∙5541 = 1∙7793 Ans. x = antilog 1∙7793 = 60∙16 Q. 14. Determine the differential coefficient of log x relative to x by first principal. - Sol. : Let y = logex and y + 5y = loge(x + 5x ) - ™ Sy = log e ( x + Sx ) —log x Then, Sx Sx So dy = lim dy = i,mlog e(x+Sx) —log x dx x ^0 dx Sx^0 Sx = lim Sx ^0 loge ( X ) Sx lim Sx ^0 Sx 1ÎSxV.1ÎSxV T "2( ) + 3( T ) — Sx Sx lim Sx ^0 = lim 1 Sx ^0 x 1 x 1 Sx . 1(Sx) 2 x2 + 3 x3 Sx 1 Sx+1(Sx)2 \_ 2 x2 3 x3 = 1x \[putting δx = 0\] Thus y = 1 or d (log x) = 1 . Ans. dx x dx x Q. 15. Write short notes on : - (i) Arithmetic and Logical Unit (ALU), - (ii) Central Processing Unit. Ans. (i) Arithmetic Logic Unit (ALU) : All calculations ae performed and all comparisons are made in the ALU. The data and instruction, stored in the primary memory prior to the processing, are transferred to the ALU where processing takes place. Intermediate results generated in the ALU are temporarily transferred back to the primary storage until needed again at a later time. After completion of processing, the final results which are stored in the primary memory are released to an output device. ALUs are designed to perform the four basic arithmetic operations—add, substract, multiply and divide—and logic operations or comparisons such as less than, equal to, or greater than. - (ii) Central Processing Unit : The control unit and the arithmetic logic unit of a computer system are jointly known as the Central Processing Unit (CPU). The CPU is the brain of any computer system. In a human body, all Program & Data ![](AU B. Sc. I Chemistry-III(OK)_files/AU20B.20Sc.20I20Chemistry-III(OK)-2.png)major decisions are taken by the brain and all other parts of the body function as directed by the brain. Similarly, in a computer system, all major calculations and comparisons are made inside the CPU and the CPU is also responsible for activating and controlling the operations performed by all other units of the computer systems. Q. 16. A bag contain 9 black 12 white balls. One balls is drawn at random, what is the probability that the ball drawn is black. Sol. : Total No. of balls = (9 + 12) = 21 Thus, if S is the sample space then n(S) = 21 and if E is the even of getting a black ball then n(E) = 9 n(E) 9 3 ∴ P (getting a black ball) = n(S) = 21 = 7 Ans. Q. 17. If z = 16x3 ‒ 10x2y2 + 7y4 ‒ 8, then find the values of ∂z ∂z ∂x and ∂y . Sol. : Given, z = 16x3 ‒ 10x2y2 + 7y4 ‒ 8 ∂z = 16∙3x2 ‒ 10∙2x∙y2 + 0 ‒ 0 ∂x = 48x2 ‒ 20xy2 Again ∂∂zx = 0 ‒ 10x2∙2y + 7∙4y3 ‒ 0 = ‒ 2x2y + 28y3 Ans. Q. 18. Determine the maximum value of r for the curve 22 ab [c4 =+](#bookmark162) [r2 sin2θcos](#bookmark163) [c4 2 2 22](#bookmark164) — = a cosec 0 + b sec 0 Sol. : r = a2 + a2 cot2 0 + b2 + b2 tan2 0 = (a2 + b2 + 2ab) + (a2 cot2 0 + b2 tan2 0 - 2ab) = (a + b)2 + (a cot 0 - b tan 0)2 the value of c2 will be minimum, if r2 ( a cot 0 - b tan 0) = 0 or 2a tan2 0 = — b c2 In this position minimum value of —= = ( a + b ) r r2 1 Hence, the maximum value of ~ = “ 7“? ’ c4 ( a + b )2 c2 Ans. or maximum value for = ( a + b ) Q. 19. Draw a graph of function x2 ‒ 4x + 5 = 0. Sol. : (i) There y = x2 ‒ 4x + 5 (ii) Tabu for the different values of x corresponding to y x ‒4 ‒3 ‒2 ‒1 0 1 2 3 4 5 6 7 y 37 26 17 10 5 2 1 2 5 10 17 26 When x corresponds to œ, y also corresponds to œ. Similarly, When x corresponds to -œ, y corresponds to œ. - (iii) Differentiating given equation dy = 2x ‒ 4 = 2 (x ‒ 2) dx Y – X ![](AU B. Sc. I Chemistry-III(OK)_files/AU20B.20Sc.20I20Chemistry-III(OK)-3.png)– 10 – 15 – 20 dy when x > 2, dx = +ve quantity dy when x < 2, dx = ‒ve quantity Hence, curve does not cut x-axis. Because on placing y = 0, the values of x comes out to be imaginary. - (iv) By plotting the various points of tabu, following graph is obtained – Y which is a parabola. Q. 20. Evaluate, - (i) I tan2x dx - (ii) sec x dx sec x + tan x - Sol. (i) I tan2 x dx = | (sec2 x - 1) dx = I sec2x dx -1 dx = tan x ‒ x +c Ans. - (ii) I secx dx = secx ∙secx‒tanx∙dx sec x + tan x J sec x + tan x sec x - tan x 2 = sec2 x‒secxtanx dx sec2 x‒tan2 x 2 sec x‒secxtanx J 1 = | sec2 x dx -1 sec x tan x dx = tan x ‒ sec x + c Ans. Q. 21. Write a short note on Isotropy and Anisotropy. Ans. Isotropy and Anisotropy : Amorphous substances are said to be isotropic because they exhibit the same value of any property in all directions. Thus refractive index, thermal and electrical conductivities, coefficient of thermal expansion in amorphous solids are independent of the direction along which they are measured. Crystalline substances, on the other hand, are anisotropic and the magnitude of a physical property varies with directions. For example, in a crystal of silver iodide, the coefficient of thermal expansions is positive in one direction and negative in the other. Similarly, velocity of light in a crystal may vary with direction in which it is measured. Thus, a ray of light passing through a Nicol prism splits up into two components, each travelling with different velocity (double refraction). Explanation of Isotropy and Anisotropy In amorphous substance, as in liquids, the arrangement of particles is random and disordered. Therefore, all direction are equivalen and properties are independent of direction. On the other hand, the particles in a crystal are arranged and well ordered. Thus, the arrangement of particles may be different in different direction. This is illustrated in Fig. in which a simple two dimensional arrangement of two different types of particles has been shown. When a property is measured along AB and DC, the value along CD will be different from that along AB. This is so because there is only one type of particles along AB while there are two types of particles in alternate positions along CD. This explains why crystalline substances are anisotropic. B C A Q. 22. Write short notes on symmetry elements. Ans. Symmetry elements : A symmetry element is a geometrical entity such as a line (or axis), a plane or a point with respect to which one or more symmetry operations may be carried out. The various types of elements are as follows : - (i) Plane of symmetry : A crystal is said to possess a plane of symmetry when an imaginary plane passing through the centre of crystal can divided into two parts such that one is the exact mirror image of the other. The standard rotation for a plane of symmetry is indicated by σ. - (ii) Axis of symmetry : It is a line about which the crystal may be rotated so that it represents the same appearance more than one during a complete revolution. If the equivalent configuration occurs twice, thrice, four and six time, i.e., after rotation of 180°, 120º, 90º and 60º, the axes of rotation are known as two-fold (diad), three fold (triad), four-fold (tetrad) and six-fold (heaxad), axes of symmetry respectively. The axis of symmetry is indicated by C. If it is two-fold, it is indicated by C2. Similarly, three-fold end four-fold axes of symmetry are indicated by C3 and C4 respectively. - (iii) Centre of symmetry : It is a point that any line drawn through it will meet the surface of the crystal at equal distances on either side. It is important to mention here that a crystal may possess a number of planes or axis of symmetry but it can only have one centre of symmetry. Now we will illustrate the various elements of symmetry as shown in Fig. ![](AU B. Sc. I Chemistry-III(OK)_files/AU20B.20Sc.20I20Chemistry-III(OK)-4.jpg)represent the three-fold axis, four-fold and two-fold aixs of symmetry respectively, Figs. (g) and (h) represent the centre of symmetry. ![](AU B. Sc. I Chemistry-III(OK)_files/AU20B.20Sc.20I20Chemistry-III(OK)-5.png)(e) (f) (h) ![](AU B. Sc. I Chemistry-III(OK)_files/AU20B.20Sc.20I20Chemistry-III(OK)-6.png) A cube has thirteen axis of symmetry (three four-fold, four three-fold, six two-fold), nine planes of symmetry and one centre of symmetry, i.e., 23 elements of symmetry altogether. Q. 23. Write short notes on the following : - (a) Law of constancy of symmetry - (b) Law of rational intercepts. Ans. (a) Law of constancy of symmetry : According to this law, “All crystal of one and the same substance have the same symmetry.” This law of crystallography may also be stated as : “All crystal of the same substance possess the same elements of symmetry.” The total number of planes, lines and centres of symmetries possessed by a crystal is termed as the elements of symmetry of the crystal. - (b) Law of rational intercepts : The law of rationality of indices or intercepts, proposed in 1784 by Hauy states that “It is possible to choose along the three co-ordinate axes unit distances (a, b, c) not necessarily of the same length, such that the ratio of the three intercepts of any plane in the crystal is given by ma : nb : pc, when m, n and p are either inegral whole numbers, including infinity, or fractions of whole numbers.” Explanation : For describing the geometry of a crystal usually three noncoplanar co-ordinate axes are chosen. These are known as crystallographic axes generally selected arbitrarily. These three axes may coincide or be parallel to the edges between the principal phases of the crystal. Generally these axes may be at right angles to one another or may be at some desired angles. It is important to remark that any face of the crystal may cut one or more axes; the face can be extended if it is found necessary. Now select one of the faces which interset the three axes. The distances of the points where the standard face cuts the three axes from the origin are called intercepts. Suppose the intercepts made by this face on the three axes are a, b and c (Fig.). Now if any other face cuts the three axes with intercepts x, y and z, that it can be written as x : y : z = ha : kb : lc .....(1) Where h, k and l are simple small integers (1, 2, 3...........) From Eq. (1) it follows that all faces cut a given axis at distances from origin which bear a simple ratio to one another. This is the law of rational intercepts. Q. 24. Write short note on Tyndall Effect. Ans. Tyndall Effect : When a strong beam of light is concentrated on a colloidal solution the path of the beam is illuminated by a bluish light, The phenomenon is known as Tyndall effect. The cause of this phenomenon is the scattering of light by the colloidal particles. The scattering of light cannot be due to simple reflection because the size of the particles is smaller than the wavelength of visible light but it is the particles themselves become self luminous due to absorption of light energy, which is then scattered as a light of shorter waves, Its importance lies counting the number and size of colloidal particles, to demonstrate the heterogenity of sols and to detect the solid suspended impurities in solution. MICROSCOPE ![](AU B. Sc. I Chemistry-III(OK)_files/AU20B.20Sc.20I20Chemistry-III(OK)-7.png) BRIGHT CONE DARKNESS ![](AU B. Sc. I Chemistry-III(OK)_files/AU20B.20Sc.20I20Chemistry-III(OK)-8.png) ![](AU B. Sc. I Chemistry-III(OK)_files/AU20B.20Sc.20I20Chemistry-III(OK)-9.png) VIEW UNDER VIEW FROM SIDE COLLOIDAL ULTRAMICROSCOPE SOLUTION Q. 25. Write a short note on electrophoresis. ![](AU B. Sc. I Chemistry-III(OK)_files/AU20B.20Sc.20I20Chemistry-III(OK)-10.png) Ans. Cataphoresis or Electrophoresis : The colloidal particles carries either positive or negative charge. On passing the electric current, they move towards opposite electrodes. This movement of colloidal particles in electric fields is known as electrophoresis and the movement of positively charged colloid, like ferric hydroxide solution is termed as cataphoresis. Its importance lies removal of carbon (which is in colloidal form, as negative colloid) from smoke, in removal of dirt from sewage water, in electro deposition of rubber and in fractionation of proteins, polysaccharide etc. Q. 26. Give the application of liquid crystals. Ans. Applications of Liquid Crystals Uses : (i) Liquid crystals are becoming important substances because of their electrical and optical properties and use in electronic display instruments. For example, these have been use in gas liquid chromatography and in digital displays like pocket calculators, digital wrist watches, etc. - (ii) Liquid crystals are also used as commercial lubricants. - (iii) These are most suited to biological functions. Since the colour of liquid crystal depends sensitively on temperatures, these have been used to measure skin temperature. - (iv) These are used for detecting tumours in the body by employing method called thermography. - (v) These are employed as solvents for studying the structure of an isotropic molecules spetroscopically. - (vi) Due to their electrical and mechanical properties lying between crystalline solids and isotropic liquids these are used in gas liquid chromatrography. Q. 27. Write a short note on Associated Colloids. Ans. Micelleas or Associated Colloids : Substance whose molecules aggregate spontaneously in given solvent to form particles of colloidal dimensions are called associated colloids.The molecules of soaps and detergents are usually smaller than the colloidal particles. However, in concentrated solutions, these molecules associated and form aggregate of colloidal size. These aggregates of soaps or detergents molecules are called micellis. Soaps and detergents are strong electrolytes and when dissolved in water they furnish ions. C17H35COONa → C17H35COO‒ + Na+ Na+ Na+ ![](AU B. Sc. I Chemistry-III(OK)_files/AU20B.20Sc.20I20Chemistry-III(OK)-11.png)\+ Na Na+ Na+ Na Fig. Micell The negative ion aggregate to form a micelle of colloidal size. The negative ion has a long hydrocarbon chain and a polar group (‒ COO‒) at one end. In micelle formation the long hydrocarbon chain (tail) which is insoluble in water is directed towards the centre while the soluble polar head is on the surface in contact with water. The charge on the micelle is responsible for the stability of this sytem. The cleaning action of soap is due to these micelles. The grease stain is absorbed into the interior of the micelle and gets detached from the fabric. The dust particles sticking on the stain are also removed. Q. 28. Write short notes on the following : - (a) Source of electrical charge on colloidal particles - (b) Protection and gold number. Ans. (a) Source of electrical charge on colloidal particles : According to modern views the charge on colloidal particles is due to preferential absorption of ions on their surface. An absorbed ions followed by oppositely charged ions, present in the medium forming a diffuse layer i.e., electrical double layer. A colloidal solution of Fe(OH)3 formed when a fresh precipitate of Fe(OH)3 is treated with dilute FeCl3 solution, is positively charged due to absorption of Fe+++ on its surface. The diffuse layer of Cl‒ is formed by the dissociation, of FeCl3 and can be expressed as : \[Fe(OH)3Fe+++\]3Cl‒ or \[Fe(OH)3\]Fe+++3Cl‒ +++e+ ++ +e+ ++ + \+ ++ e+ +e+ +e+ ++e ++ COLLOID PARTICLES ++ e++ \+ + ++ e+ + ++ + ++ + 3 e y+ ++ (Electrical double layer of ferric ions (Fe+++) absorbed on sol particle of Fe(OH)3 with Cl ions forming diffuse layer) - (b) Protection and gold number : The number of milligrams of protective colloid which must be added to 10 c.c. of given gold soll so that it is just prevented from coagulation by 1 c.c. of 10% NaCl solution is known as the Gold number of protective colloid. The smaller the gold number the higher is the protection power of hydrophilic colloid. The gold number of some of the protective colloids is given below : Substance Gold Number Gelatin Albumin Gum Arabic Starch 0∙005―0∙05 0∙8―0∙08 0∙15―0∙10 1∙4—25 The water or solvent having colloids can be prevented from being coagulated by addition of electrolyte by previous addition of more stable water loving colloid. The protective action differs from substance to substance and is measured in terms of Gold Number and can be defined as above. Q. 29. What do you mean by liquid crystals ? Discuss its types. Ans. Liquid Crystals : When a solid melts, the vibrations of its constituent particles acquire such magnitude that they lapse into translatory motions. However, in some cases molecular or ionic clusters do not lose their identity beyond the melting point of the solid. In such solids the breakdown of clusters in some direction is very slow and they yield very viscous cloudy liquids at a more or less sharp characteristics temperature, known as the transition temperature. If the temperature is increased beyond transition temperature, the cloudiness disappears again sharply at a new temperature, known as melting point of a solid. Beyond the melting point, the substance usually behaves as an ordinary liquid. Between the temperature. i.e., the transition and melting point the cloudy liquid shows double refraction, a behaviour usually exhibited by crystalline solids. This account the cloudy liquids state is some time called mesomorphic state or regarded to be made up of liquid crystal. Liquid crystal state is intermediate between the liquid state and the solid state. Solid State ![](AU B. Sc. I Chemistry-III(OK)_files/AU20B.20Sc.20I20Chemistry-III(OK)-12.png) One such substance that forms liquid crystal is ![](AU B. Sc. I Chemistry-III(OK)_files/AU20B.20Sc.20I20Chemistry-III(OK)-13.png) The term ‘liquid crystal’ is made up of liquid and crystal. The word liquid is there because these tend to take the shape of the container and the work crystals is there because they still contain one or two dimensional arrays. The term ‘liquid crystals’ is not satisfactory because the substance in this state do not have properties of crystalline state. Actually they are more like liquids in having properties such as mobility, surface tension, viscosity, etc. To understand the liquid crystal, let us take an example. The compound 5-chloro-6-n-heptyloxy-2-naphthoic acid, exists in ordinary crystalline form. On heating to 438.5 K, the crystal fuses sharply yielding turbid liquid. If this is spread on a flat plate, one may see steps or ridges; it is birefringent, showing coloured areas under polarized light and having different optical properties parallel and perpendicular to the plate. Thus, it shows anisotropic behaviours (they have different physical properties in different directions). The turbid liquid is termed as liquids crystal. Cl ![](AU B. Sc. I Chemistry-III(OK)_files/AU20B.20Sc.20I20Chemistry-III(OK)-14.png)—n—C7H15 Q. 30. Give the kinetic molecular description of liquid state. Ans. Kinetic molecular description of liquid state : The liquid state lies between the gaseous and the solid state in the sense that their is neither the ordered arrangement of constituents nor the complete disorder as in gases. It is generally observed that some of the properties of liquids closely resemble thos of gase while some of the properties approach those of gases. In the light of kinetic molecular theory, liquid may be regarded as a continuation of gases into the region of small volumes and high intermolecular attraction. The cohesive forces in a liquid have been stronger than those in gases even at high pressures, and are sufficiently high to keep the molecules cofined to a definite volume. The positions of the molecules in liquids are not rigidly fixed. These forces are not strong enough to entirely eliminate the movements of the molecules in the liquid. Hence the molecules in the liquid state have much shorter mean free path as compared to gas molecules. Properties of liquids can be explained on the basis of their following characteristics : - (a) Appreciable forces of attractions exist between the molecules of liquid. These are about 106 times as strong as in gases. The force amongst the constituents of a liquid disallow them from separating spontaneously from each other, but they are not strong enough to hold the molecules in fixed positions. - (b) The molecules in a liquid are in a state of random motion although the extent of randomness in appreciable much smaller in comparison to gases. The molecules are relatively close together. Most of the space in the liquid gets occupied by its molecules and only a small fraction of the space is available to them for their free movement. This explains the high density, in compressibility and slow diffusion of liquids in comparison to gases. - (c) The average kinetic energy of the molecules in a liquid is proportional to the absolute temperature. Increase in temperature would increase the proportion of the energized molecules, lowers the attractive forces between the molecules and consiquently increases the vapour pressure of liquid. From the above arguments, it is evident that most of the characteristics properties of the liquid arises due to the nature and magnitude of the intermolecular forces between the molecules. Q. 31. Write short notes on : - (a) Thermography, - (b) Seven segment cell. Ans. (a) Thermography : Cholestric type liquid crystals are used in the detection of tumour in body. This method is termed as thermography. Like the solid crystal, liquid crystal can diffract light. Only one of the wavelength of white light is reflected by the crystal which appears coloured. As the temperature changes, the distance between the layres of molecules also changes. Therefore, the colour of the reflected light changes correspondingly. The cholestric liquid crystals undergoes a series of colour changes with temperature. Becasue the temperature of healthy body is different from that part which is effected by tumour, so the wavelength or colour of rays diffracted by these parts will be different. On this basis tumour is detected in body by its help. These crystals are used in indicator types to monitor body temperature or to spot areas of over heating in mechanical systems. - (b) Seven segment cell : Another important application of liquid crystals is the formation of seven segment cell. It is a device which is commonly used in digital systems for the demonstration of digits. Seven segment cells are used in pocket calculators and digital watches for the demonstration of digits from 0 to 9. a f e ![](AU B. Sc. I Chemistry-III(OK)_files/AU20B.20Sc.20I20Chemistry-III(OK)-15.png)d In this cell there are seven fragments from a to g as shown in fig. In it digits are b converted into binary codes and then decoding is performed by which appropriated numbers shine which are formed on screen. c In it when a thin layer of nematic liquid crystal is placed between two electrodes and an electric field is applied, the polarmolecules are pulled out of alignment. This causes the crystal to be opaque. Transparency returns when electric signal is removed. Due to this property it is used in the number displays of digital watches, electronic calculators and other instruments. Q. 32. What do you mean by coagulation ? Discuss Hardy-Schulze’s law. Ans. Coagulation : The colloidal particles carry similar type of charge and hence remain far apart from each other. On removing or neutralizing this charge, they will come nearer to each other forming large particle and thus precipitates out. This is known as coagulation or floculation. For example, on addition of a little sodium phosphate (Na3PO4) to a colloidal solution of ferric hydroxide (FeOH2), the later precipitates out. It is due to the fact the negatively charged PO4 ions are attracted by positively charged colloidal particles of Fe(OH)3 with the result that the charge of the latter gets neutralized. The neutral particles get together to from bigger particles and finally settle down as precipitate. Hardy-Schulze’s Law : The colloidal solutions are coagulated by addition of electrolytes. The coagulating power of different electrolytes depends upon the nature of oppositely charged ions. Hardy and Schulze gave a law governing this phenomenon. “The greater the valency of oppositely charged ion (i.e., the ion whose charge is opposite to that of colloidal particle) the greater is the coagulating power of ion.” For example in coagulation of negative sol like that of As2S3 the trivalent Al+++ ions are more effective than bivalent Ba++ ions or monovalent K+. Q. 33. Prove that, - 9 1 RTc [a = RTcVc andb =](#bookmark169) - [88 P](#bookmark170) - 91 RT Sol. : Derivation of a = RTcVc and b = - [88 P](#bookmark171) Since Vc for a gas is not known as accurately as Pc and Tc it is always advantageous to express the two constants ‘a’ and ‘b’ in terms of Pc and Tc. From Van der Waals equation, the equation of critical state can be calculated as shown, RTc (i) 3V = Tc + b c Pc and 1 b = 3 Vc (ii) On equating the value of ‘b’ from eqn. (ii) in eqn. (i), we get RTc 1 3V c = p + 3Vc (ui) Pc or V c 3 RTc = 8 Pc On squaring both sides, Vc2 9 R2Tc2 " 64 ' P? (lv) According to the equation of critical state, 3Vc2 = a 1a Pc or Vc = 3 Pc On putting the value of Vc2 in equation (iv), we get 1 a 9 R2Tc2 or 3 Pc 64 Pc2 27 R2Tc2 ........(v) a 64 Pc2 Now, substituting the value of ‘b’ from eqn. (ii) in eqn. (i), we get RT 3 x 3b = ---c + b Pc RTc Pc or b 1 RTc = 8' IT ........(") F 3 RTcI On putting the value of Pc = l^’“ I in eqn. (v), we get a 27 8 V = 77R2Tc2 x-x — 64 3 RTc a 9 = RTcV ........(vii) 8 Q. 34. Write short notes on collision frequency. or 8b = Ans. Collision Frequency : The mean free path (L), multiplied by the number of collision per second gives the root mean square velocity. Consequently, the mean free path is known as also the molecular velocity the number of collisions per second also known as collision frequency can be calculated as v Number of collision per second = L At N. T. P., v for hydrogen is 1,83,000 cm sec“1 and L is 1-78 x io-5 cm. No. of collision per second 183000 1-78x10-5 1,02,865x10-5 Q. 35. What are the limitations of Van der Waal’s equations ? Ans. Limitations of Van der Waals equations : Van der Waals equation explains satisfactory the general behaviour of real gas. It is valid over a wide range of pressure and temperature. However, it fails to give exect agreement with experimental data at very high pressure and low temperature. Dieterici proposed a modified Van der Waals equation. This is known as Dieterici equation. For one mole of gas it may be stated as P( v - b)=RTe - a/VRT Hence the term (a) and (b) hare the same significance as in Van der Waals equation. Q. 36. Write explanatory note on law of corresponding states. Ans. Law of corresponding states : Van der Waals equation contains three constants a, b and R which may be expressed in terms of critical constants as : Vc b - -3-, a = 3PcVc2 8 PcVc R =-- 3 Tc On substituting these values in Van der Waals equations 3 2I p + G v ###### V2 K H Vc 3 8 PcVc \---T 3 Tc (i) Divide the above equation both sides by PcVc P + 3PcVc2 V 2 V - c Vc 3 Vc 8 PcVc T 3 Tc PcVc (ii) or P FV I2 \+ 3 ###### Pc HG V KJ Let 3V -1 Vc P — = n Pc P= 8T Tc 1=♦ X=0 , Vc , Tc (iii) Where, n, 0 and 0 are called reduced pressure, reduced volume and reduced temperature respectively. On substituting these values in equation (iii) it becomes, ##### FI n + . b30- 1g = 80 (iv) ##### H 0 K This is called reduced Van der Waal’s equation. It involves neither R nor the Van der Waal’s constants a and b. Hence, it is a general equations applicable to all substance liquids and gases i.e., all fluids. It also follows from this equation that if two substances have the same reduced temperature and the same reduced pressure they will have the same reduced volume. These two substances are said to be in corresponding states and this generalisation is called the Law of Corresponding states. Q. 37. State and explain law of equipartition energy. Ans. Principle of equipartition energy : Since the velocity of molecules in space (u) is related to the component velocities (ux, uy, uz) by, u2 = u 2 + u 2 + u 2 xyz Multiply to both the sides of the above equation by 21 m we get, 4 mu2 = I mux 2 +1 muv 2 +1 mu.2 2 2x2y2z Since molecules move in the container in random manner therefore they do not have any preferential direction. Thus, the velocities along three axis are equally probable. i.e., u = u = u xy z or 1 mu 2 = 1 mu 2 = 2x 2y 1 2 2 muz2 or (KE)x = (KE)y = From kinetic theory of gases ‒ (KE)z KE = 3/2KT Therefore, (KE)x = (KE)y = (KE)z = 21 kT Thus, the energy term in particular component contribute equally to the total energy. This is called Law of equipartition of energy. This can be defined as : “If the energy of the individual molecule can be written in the form of a sum of terms each of which is proportional to the square of either velocity component or as a co-ordinate then each of these squares terms contribute 21 KT to the average energy.” Q. 38. What do you mean by mean free path ? Ans. Mean Free Path : The mean distance transversed by a gas molecule without coming into collision with another is called its mean free path. It is denoted by the symbol L. If ‘v’ is the average velocity of a molecule and ‘z’ is the no. of collision suffered by a single molecule per second, then the average distance covered by a molecule between two successive collision namely, the mean free path, is given by vv 1 L z 21 nvc2N 21 no2N (i) For an ideal gas, containing N molecules per c.c. (i.e., V = 1), p x 1 = nRT = N/N0RT where N0 is Avogadro’s number. PN0 = P RT = kT .(ii) where k (= R/N0) is the Boltzmann constant. Substituting the above value of N in equation (i), we get : kT L = J2P n (iii) It is clear from the eqn. (iii) that for a gas, the mean free path is directly proportional to temperature and inversely proportional to pressure. Q. 39. Explain homogeneous and heterogeneous catalysis with suitable example. Ans. Homogeneous Catalysis : When the catalyst is present in same phase as the phase of reactants, the phenomenon is known as homogeneous catalysis. For example : - (i) Oxidation of SO2 to SO3 in presence of nitric oxide (NO) 2SO2 + O2 jNiW^ 2SO3 (gas) (gas) catalysis (gas) - (ii) Hydrolysis of ethyl acetate in the presence of an aqueous solution of an acid CH3COOC2H5 + H2O HCl (liquid) > CH3COOH + C2H5OH (liquid) (liquid) (catalyst) (liquid) (liquid) - (iii) Hydrolysis of sugar is catalysed by H+ ions fernished by H2SO4 c.^o.. + Ho H2SO4 (liquid) CH o + CHA 12 22 11 2 > 6 12 6 6 12 6 (liquid) (liquid) glucose fructose The reactant and catalyst both are in gaseous state in the first case and both are in liquid state in the second and third case. Heterogeneous Catalysis : When the catalyst is in a different phase than that of reactant, the phenomenon is known as heterogeneous catalysis. For example : - (i) Manufacture of ammonia by Haber’s process, in presence of Fe an Mo. N2 + 3H2 + \[Fe\]> 2NH3 + \[Fe\] (gas) (gas) (catalyst) (gas) (solid) - (ii) Combination of SO2 and O2 in the presence of Pt to form SO3. SO2 + O2 + \[Pt\]\_\_\_\_> 2SO3 + \[Pt\] (gas) (gas) (catalyst) (gas) (solid) - (iii) Hydrogenation of vegetable oils in the presence of the finally divided nickel as catalyst. Vegetable Oil + H2 Ni (solid) ) Vegetable ghee (liquid) (gas) In the above reaction one of the reactant is in liquid state and the other is gaseous state while the catalyst is in solid state. - (iv) Oxidation of ammonia into nitric oxide in the presence of Pt gauze as a catalyst in Ostwald’s process. 4NH3 + 5O2 Pt (solid) . 4NO + 6H2O (gas) (gas) (gas) (liquid) The reactants are in gaseous state while the catalyst is in solid state. Q. 40. Optical rotation of a carbohydrate in 0∙5N‒HCl at different intervals of time as Time (minute) 0 10 20 30 40 x Optical rotation +32∙4 +28∙8 +25∙5 +22∙4 +19∙6 ‒11∙1 Prove that the reaction is of first order. Sol. : Rate constant for first order reaction is, 2 • 303 r0 - r x k = log10 0 - t r 1 — r x where, r0 = optical rotation at the start of the reaction. - r1 = optical rotation at certain interval of time. - r2= optical rotation at the end of the reaction. Hence, r0 ‒ r1 = +32∙4 ‒ (‒11∙1) = + 43∙5 - t rr ‒ r0 [2•303.43](#bookmark182) 10 28∙8 ‒ (‒11∙1) = 39∙9 k = log10 = 0∙008625 10 39 - 9 [2 - 303.43](#bookmark183) - 12 25∙5 ‒ (‒11∙1) = 36∙6 k = log10 = 0.008625 20 36 - 6 [2 - 303,43](#bookmark184) 30 22∙4 ‒ (‒11∙1) = 33∙5 k = log10 = 0.008649 20 33 - 5 2■303. 43-5 40 19∙6 ‒ (‒ 11.1) = 30∙7 k = log10 = 0∙0086 30 30■7 Since the rate constant k, is approximately the same, hence the reaction is of first order. Q. 41. Write short note on : - (i) Auto-Catalyst, - (ii) Induced Catalyst, - (iii) Catalytic Poisons. Ans. (i) Auto-Catalyst : When one of the product formed in the reaction acts as a catalyst, the substance is known as auto-catalyst and the phenomenon is known as auto-catalysis. For example : (a) Hydrolysis of ethyl acetate by water is an autocatalytic reaction, since acetic acid liberated in this reaction acts as a catalyst. CH3COOC2H5 + H2O---> CH3COOH + C2H5OH (auto-catalyst) - (b) The oxidation of oxalic acid by acidic KMnO4 is catalysed by the presence of Mn2+ ions formed in the solution. In the beginning the colour of KMnO4 disappears slowly but as Mn2+ is formed in the solution the colour discharges rapidly. So, Mn2+ ions acts as auto-catalyst. 2MnO4" + 5C2O.2 + 16H+----> Mn2+ + 10CO2 + 8H2O (colourless) - (ii) Induced Catalyst : When one reaction influence the rate of other reaction which does not occur under ordinary conditions, the phenomenon is known as induced catalysis. For example, (a) Sodium sulphide solution is readily oxidised in air but sodium arsenite solution is not oxidised by passing a current of air through it. If air is passed through a mixture of sodium sulphite and arsenite solution. - (b) The reduction of HgCl2 with oxalic acid is very slow, but KMnO4 is reduced readily with oxalic acid. If, however oxalic acid is added to a mixture of KMnO4, thus induces the reduction of HgCl2. - (iii) Catalytic Poisons : Substances which destroy, the activity of the catalyst by their presence are known as catalytic poison. Some of the examples are : - (a) The presence of traces of As2O3 in the reacting gases reduce the activity of platinized asbestos which is used as catalyst in contact process for the manufacture of H2SO4. - (b) The activity of iron catalyst is destroyed by the presence of H2S or CO in the synthesis of ammonia by Haber’s process. - (c) The platinum catalyst used in the oxidation of hydrogen is poisoned by CO. The poisoning of the catalyst is due to the preferential absorption of poision on the surface of the catalyst, thus reducing the space available for adsorption of reacting molecules. Q. 42. Discuss in detail the mechanism of enzyme catalysed reaction. Ans. Mechanism of Enzyme Catalysis : There are a number of cavities present on the surface of colloidal particles of enzymes. These cavities are of characteristic shape and possess active groups such as —NH2, —COOH, — SH, —OH etc. These are actually the active centres on the surface of enzyme particles. The molecules of the reactant (substrate), which have complementary shape, fit into these activities just like a key fits into a lock. On account of the presence of active groups, an activated complex is formed which then decomposes to yield the products. Michaelis and Menten suggested the following mechanism for enzyme catalysis : Step 1 : Binding of enzyme to substrate to form an activated complex E + S ES Step 2 : Product formation in the activated complex. ES ----> EP Step 3 : Decomposition of EP into products and enzyme again, EP ----> P + E This mechanism accounts for the high specificity of enzymatic reactions. Q. 43. Write short notes on Pseudo-molecular reaction. Ans. Pseudo-Unimolecular reactions. In inversion of cane-sugar and hydrolysis of methyl acetate in presence of dilute acids apparently two molecules of reactant take part, but actually belong to first order reaction as the concentration of water in dilute solution does not appreciably alter and the kinetic of the reactions are determined by the concentration of cane-sugar or ester. Such reactions are also called Pseudo-Unimolecular Reactions i.e., Bimolecular reactions become unimolecular. Q. 44. 50% of a gas decomposed in second order way in 40 minutes. What would be the time taken for 75% of the gas to decompose? Sol. : For second order reaction, 1x K = X ,----V t a b a - x g Given t = t1/2 = 40 minutes a – x = a/2 x = a/2 On putting these values in the equation. 1 K = tx 1 or = ~X t 1 or = 40 a/2 aba/2g 1 a 1 - X — a Substituting the value of K in equation K 1 40a or t x or a x ba - xg :. t 1x = -X—---7 t a b a - x g x1 = —---X a (a - x) t = a - x = 0∙75, = 0∙25 a 0 ■ 75 =---=3 0 ■ 25 = 40 × 3 = 120 minutes. Thus, the time taken for 75% of the gas to decompose is 120 minutes. Q. 45. Explain the rate of a reaction. Discuss the factors affecting the rate of a reaction. Ans. Rate of reaction. The rate at which the concentration of reacting substance changes with time is known as Rate of Reaction or Velocity of Reaction. Suppose a very small amount of a substance, dx, changes into a very small time dt, then the rate of reaction is expressed by, dx dt Factors affecting the reaction rate - (i) Nature of reactants : Reaction between polar or ionic molecules such as double displacement and neutralisation reaction are very fast occur almost instantaneously. On the other hand, reactions in which reacting molecules are covalent take longer time than ionic reactions. - (ii) Effect of concentration on reaction rates : On changing the concentration of the reaction the number of collisions changes and, hence the reaction rate is changed. Increase in concentration of the reactant increase the reaction rate and vice-versa. - (iii) Temperature effect : On raising the temperature the rate of all single reactions increases e.g., a change for 25°C to 35°C double the rate of most of reactions. The change in rate takes place because of increase with the specific reaction rate R, with increase of temperature. The temperature T1 and T2 K are related to the K1 and K2 respectively. l K2 = Ea 10K1 2⋅303R 11 - T1 T2 R is gas constant (1∙987 cal/mole/degree). Ea is Arrhenius activation energy or Arrhenius constant. - (iv) State of sub-division and catalyst : The state of sub-division of the substance and catalyst considerable enhances the rate of reaction by providing an alternative course for the chemical reaction. Q. 46. 10 ml of ethyl acetate were added to a flask containing 100 ml of 0∙1 N-HCl placed in a thermostat maintained at 25°C. 5 ml of the reaction mixture were withdrawn at different intervals of time and after chilling titrated against standard alkali, the following data were obtained : Time (min) 0 75 119 183 ∞ Alkali used (ml) 9∙6 12∙10 13∙30 14∙75 21∙05 From the above data find out the order of reaction. Ans. Hydrolysis of ethyl acetate in acidic medium is as follows : H+ CH3COOC2H5 + HOH → CH3COOH + C2H5OH It will be first order reaction if the values of ‘k’ obtained by this given equation comes to the constant. or k = 2⋅303 t log10 2⋅303 t a ba - xg bV∞ -V0g log10bV∞-Vtg Where V0, Vt and V∞ are the volume of alkali used at the beginning of the reaction, after ‘t’ internal of time and at the end of the reaction, respectively. a = V∞ ‒ V0 = 21∙05 ‒ 9∙62 = 11∙43 The volume of ‘k’ at different interval of time (t) will be as follows : Time V∞ ‒ Vt 2⋅303 t V∞ log10V∞ - - V0 Vt = k 75 (21∙05 ‒ 12∙10) = 8∙95 2-3031 11-43 1 75 log10 8⋅95 = 0∙003259 min‒1 2-3031 11-43 1 119 (21-05- 13-30) = 7^5 2( 303 log|o ‘^ T4? = 0-003264 min"1 2-303i 11-43 1 183 (21^05 - 1435) = 630 183^10 630 = 0^003254 min-1 Since the values of ‘k’ come out to be constant, the acidic hydrolysis of ethyl acetate is first order reaction. Q. 47. 50% of H2O2 decomposes in 10 minutes in an experiment. Find out its velocity coefficient if it is a reaction of first order (log102=0∙3010) Sol. : The expression for the first order reaction is, 2•303, a k = log10 t b a — x g Given, t = 10 minutes a = 100 x = 50 On substituting these values in the rate constant expression of first order, 2•303, K = 10 l 2 • 303 100 og10(100 - 50) 10 2 • 303 = 10 = 0∙06932 x log2 x0•3010 Thus the velocity coefficient is 0∙06932 Q. 48. The density of hydrogen at 0°C and 760 mm pressure is 0∙00009 gm/cc. Determine its RMS velocity. Density of mercury is 13∙6 gm/cc. Sol. : v= rms 3PV M 3P (M/V) Where D is the density of gas. v rms 3 x 76 x 13-6 x 981 {0•00009 18900 cm/sec. Q. 49. Explain the molecularity and order of a reaction. Describe the difference between them. Ans. Molecularity and order of reaction. Molecularity of a reaction is the number of molecules involved in the reaction or the number of molecules required for the reaction to take place. Thus, when only one molecule is involved in the reaction, it is known as unimolecular reaction. For example : - (i) Decomposition of hydrogen peroxide H2O2 ^ H2O x jO2 - (ii) Dissociation of bromine moleculed into atoms Br2 ^ 2Br In those cases where the number of molecules required for the reaction to take place are two, are called bimolecular reaction and where three molecules are necessary are called as trimolecular and so on. Examples of bimolecular reaction are. CH3COOC2H5 + NaOH ^ CH3COONa + C2H5OH H2 + I2 ^ 2HI Order of reaction is defined as “the number of reacting molecules whose concentration alter as a result of chemical change.” For example the decomposition of H2O2 is a unimolecular reaction and as the concentration of only one reactant changes hence it is also a first order reaction. Similarly, the hydrolysis of ethyl acetate by NaOH is of second order. CH3COOC2H5 + NaOH ^ CH3COONa + C2H5OH Because here the concentration of both reactions changes. Difference between molecularity and order of reaction : In fact till recently molecularity and order of reaction were regarded as synonymus. But it has now been found that there are certain cases in which these are not identical. For example in hydrolysis of ethyl acetate in presence of dilute acid (as a catalyst) or inversion of canesugar into glucose and fructose apparently two moleucles take part but actually the concentration of only one of reactants changes i.e., the concentration of water in dilute solution does not appreciably alters. Hence the reaction though bimolecular but is of first order. Such reactions are called Pseudo-Unimolecular Reactions. H+ CH3COOC2H5 + H2O ^ CH3COOH + C2H5OH H+ C13H22O12 + H2O ^ C6H!2O6 + C6H!2O6 Canesugar Glucose Fructose Q. 50. The specific rate of reaction of a chemical reaction at 300Å is 0∙001 minute‒1 which becomes 0∙002 minute‒1 at 310Å. Calculate the energy of activation of the reaction. 66 Vikas, 2010 (A. U.) Sol. : We know, that the expression for the energy of activation is ki = AE L ± - 1 O k 2 2 • 303R N T1 T2 q or Given that, On putting AE = 2 • 303R T2 = 310, k1 = 0∙001, T2 x Ti T2 - Ti logk2 k1 these values in the above expression, T1 = 300 k2 = 0∙002, R = 1∙987 AE = 2 • 303 x 1.987 x x log 310 - 300 0 • 002 0 • 001 93000 = 2 ■ 303 x 1-987 x------x 0 ■ 3010 10 = 12,800 cals/gram mole Thus, the energy of activation of the reaction is 12,800 calories per gram mole. Section ‘C’ Essay Type Questions Q. 1. What do you mean by operating system ? Explain in brief. Ans. Definition and Function : An operating system (OS in short) is a master control program which runs the computer and acts as a scheduler. It controls the flow of signals from the CPU to various parts of the computer. It is the first program loaded (copies into the computer’s memory after the computer is switched on. The operating system is an important component of the computer system, because it sets the standards for the application programs that run in it. All programs must be written to ‘talk to’ the operating system. Popular operating systems include MS-DOS, OS/2, Window 95 and UNIX. The macintosh uses Finder and Multifinder. Digital system uses the VMS and Ultrix operating systems. The operating system performs the following function : - 1\. Processor management, that is, assignment of processors to different tasks being performed by the computer system. - 2\. Memory management, that is, allocation of main memory and other storage areas to the system programs as well as user programs and data. - 3\. Input/Output management, that is coordination and assignment of the different input and output devices while one or more programs are being executed. - 4\. File management, that is, the storage of files on various storage devices and the transfer of these files from one storage device to another. It also allows all files to be easily changed and modified through the use of text editors or some other file manipulation routines. - 5\. Interpretation of commands and instruction. - 6\. Coordination and assignment of compilers, assemblers, utility programs, and other software to the various users of the computer system. - 7\. Establishment of data security and integrity. That is, it keeps different programs and data in such a manner that they do not interfere with each other. Moreover, it also protects itself from being destroyed by any user. - 8\. Maintenance of internal time clock and log of system usage for all users. - 9\. Facilitates easy communicaiton between the computer system and the computer (human) operator. An operating system perform a wide variety of jobs. Each of these jobs are performed by one or more computer programs and all these programs are jointly known as an operating system. Out of the complete operating system, normally one control program resides in the main memory of the computer system. This control program is known as the resident program or the resident routine. The other programs are stored on the disk and are called transient programs or transient routines. These program include utility programs, compilers, assemblers, etc. The control program transfers these program into the main memory and executes them as and when they are needed. It may be recalled here that the capacity of the main memory of any computer system is very small as compared to its secondary storage devices like disks. This is because main memory is very expensive as compared to secondary storage device. This is the reason why only the control program is stored in main memory and the rest of the operating system is stored on disks. In effect, besides the hardware, each computer system consists of an operating system that enables a user to effectively use the system. Thus, as shown in fig the OS tends to isolate the hardware from the user. The user communicated with the OS, supplies application programs and input date, and receives output results. The efficiency of an operating system and the overall performance of a computer installation is judged by a combination of two main factors. They are : Hardware System Software Application Software User Fig. The in-between software layers isolate the hardware of a computer system from its users. - 1\. Throughout : It is the total volume of work performed by the system over a given period of time. - 2\. Turnaround Time : It is also known as response time and is defined as the interval between the time a user submits his job to the ystem for processing and the time he receives results. Response time is especially important where many different users share the use of the system and the overall progress of their work depends upon their receiving prompt results from the system. - Q. 2. Integrate the following : - (i) x cos x dx (ii) dx 1+sin x Sol. : (i) J x cos xdx Taking x as first function and cos x as second function J x cosxdx = x J cos xdx - J ^ J dxx. J cos xdx \] dx = x sin x - Jl.sinxdx = x sin x + cos x (ii) dx J 1 + sin x - 1 sin x -—dx\* (1 + sin x)(1 - sin x) \[Multiplying denominator and numerator by (1 ‒ sin x)\] = f 1 - sin2x dx (1 - sin2 x) = 1 dx ‒ sinx dx cos2 x cos2 x ∫ sec2 xdx ‒ ∫secxtanxdx = tan x ‒ sec x Ans. Q. 3. What do you mean by computer languages ? Explain merits and demertis of Machine Languages. Ans. Analogy with Natural Languages A language is a system of communication. With our natural language such as English, we communicate to one another our idea and emotion. Similarly, a computer language is a mean of communication used to communicate between people and the computer. With the help of a computer language a programmer tells a computer what he wants it to do. All natural languages (English, French, German etc.) use a standard set of symbols for the purpose of communication. These symbols are understood by everyone using that language. We normally call this set of symbols the vocabulary of that particular language. For example, the words we use in English are the symbols of English language that make up its vocabulary. Each word has definite meaning which can be looked up in a dictionary. In a similar manner, all computer languages have a vocabulary of their own. Each symbol of the vocabulary has definite unambiguous meaning which can be looked up in the manual mean for that language. Hence, each symbol of a computer language is used to tell the computer to do a particular job. Each natural language has a systematic method of using symbols of that language. In English, this method is given by the rules of grammar. These rules tell us which words to use and how to use them. Similarly, the symbols of a particular computer language must also be used as per set rules which we known as the syntax rules of the language. In case of a natural language, people can use poor or incorrect vocabulary and grammar and still make themselves understood. However, computer, being a machine, are receptive only to exact vocabulary used correctly as per syntax rules of the language being used. Thus, in case of a computer language, we must stick by the exact rules of the language if we want to be understood by the computer. Unless a programmer adheres exactly to the syntax rules of a programming language, even down to the correct punctuation marks, his commands will not be understood by the computer. Programming languages have improved throughout the years, just as computer hardware has improved. They have progressed from machine-oriented language that use strings of binary 1s and 0s to problem oriented languages that use common mathematical and/or English terms. However, all computers languages can be classified in the following three broad categories : - (a) Machine language - (b) Assembly language - (c) High-level language We shall now examine the evolution and nature of each type of language. Machine Language Although computers can be programmed to understand many different computer languages, there is only one language understood by the computer without using a translation programme. This language is called the machine language or the machin code of the computer. Machine code is the fundamental language of a computer and is normally written as strings of binary 1s and 0s. The circuitry of a computer is wired in such a way that it immediately recognized the machine language and converts it into the electrical signals needed to run the computer. An instruction prepared in any machine language has a two-part format, as shown in fig. The first part is the command or operation, and it tells the computer what function to perform. Every computer has an operation code or opcode for each of its functions. The second part of the instruction is the operand, and it tells the computer where to find or store the data or other instructions that are to be manipulated. Thus, each instruction tells the control unit of the CPU what to do and the length and location of the date fields that are involved in the operation. Typical operations involve reading, adding, subtracting, writing and so on. OPCODE (Operation code) OPERAND (Address/Location) Fig. Instruction Format We already known that all computer use binary digits (0s and 1s) for performing internal operation. Hence, most computer’s machine language consists of strings of binary numbers and is the only one the CPU directly undrstands. When stored inside the computer, the symbols which make up the machine language program are made up of 1s and 0s. For example, a typical program instruction to print out a number on the printer might be 101100111111010011101100 The program to add two numbers in memory and print the result might look something like the following : 001000000000001100111001 001100000000010000100001 011000000000011100101110 101000111111011100101110 000000000000000000000000 This is obviosuly not a very easy language to learn, partly because it is difficult to read and understand and partly because it is written in a number system with which we are not familiar. But it will be surprising to note that some of the first programmers, who worked with the first few computers, actually wrote their programs in binary form as above. Since human programmers are more familiar with the decimal number system, most of them preferred to write the computer instructions in decimal, and leave the input device to convert these to binary. In fact, without too much effort, a computer can be wired so that instead of using long string of 1s and 0s we can use the more familiar decimal numbers. With this change, the preceding program appears as follows : 10001471 14002041 30003456 50773456 00000000 The set of instructions codes, whether in binary or decimal, which can be directly understood by the CPU of a computer without the help of a translating program, is called a machine code or machine language. Thus, a machine language program need not necessarily be coded as strings of binary digit (1s and 0s). It can also be written using decimal digits if the circuitry of the computer being used permits this. Advantages and Limitations of Machine Language Programs written in machine language can be executed very fast by the computer. This is mainly because machine instructions are directly understood by the CPU and no translation of the program is required. However, writing a program in machine language has several disadvantages which are discussed below : - 1\. Machine dependent : Because the internal design of every type of computer is different from every other type of computer and needs different electrical signals to operate, the machine language also is different from computer to computer. It is determined by the actual design or construction of the ALU, the control unit, and the size as well as the word length of the memory unit. Hence, suppose after becoming proficient in the machine code of a particular computer, a company decides to change to another computer, the programmer may be required to learn a new machine language and would have to rewrite all the existing programs. - 2\. Difficult to program : Although easily used by the computer, machine language is difficult to program. It is necessary for the programmer either to memorize the dozens of code numbers for the commands in the machine’s instruction set or to constanlty refer to a reference card. Moreover, a machine language programmer must be an expert who knows about the hardware structure of the computer. - 3\. Error prone : For writing programs in machine language since a programmer has to remember the opcodes and he must also keep track of the storage location of data and instruction, it becomes very difficult for him to concentrate fully on the logic of the problem. This frequently results in program errors. Hence, it is easy to make errors while using machine code. - 4\. Difficult to modify : It is difficult to correct or modify machine language programs. Checking machine instructions to locate errors is about as tedious as writing them initially. - Q. 4. One card is drawn from a pack of 52 cards, each of the 52 cards being equally likely to be drawn. Find the probability that : - (i) the card drawn is red. - (ii) the card drawn is a king. - (iii) the card drawn is red and a king. - (iv) the card drawn is either red or a king. Solution. Let S draw the sample space. The n(s) = 52 - (i) Since E1 be the event of drawing a red card. Since the numbers of red cards is 26, we have n(E1) = 26 [n(E1) 261](#bookmark207) ., P(a red card)=p(Ei) = — = 52 = ^ - (ii) Let E2 be the event of drawing a king Since the number of kings is 4, we have n(E2) = 4 [n(E ) 41](#bookmark208) ... p(a king)=P(E2) = — = 52 = - - (iii) Let E3 be the event of drawing a red card which is a king Since the number of red kings is 2, we have n(E3) = 2 [n(E ) 21](#bookmark209) P(a red king)=P(E3) = = 52 = 26 - (iv) Let E4 be the event of drawing a red card and a king. Clearly, there are 26 red cards (including kings) and there are two more kings. Thus, n(E4) = 26 + 2 = 28 . P (a red card or a king) = P(E4) n (E4) = 28 = 7 Ans. nS = 52 = 13 Q. 5. Differentiate the following : ex (i) ex (ii) (iv) sin x3 2x+3 x2 ‒5 (iii) (v) sin3x ex 1+sin x (vi) esinx Solution. (i) Using the quotient rule, we have d dx ![](AU B. Sc. I Chemistry-III(OK)_files/AU20B.20Sc.20I20Chemistry-III(OK)-16.png) xd ( ex ) dx xd e (x) dx x 2 xex - ex 1 ex ( x -1) (ii) ![](AU B. Sc. I Chemistry-III(OK)_files/AU20B.20Sc.20I20Chemistry-III(OK)-17.png) 2 x + 3 ^ x2 - 5 ) (x2 - 5) d (2x + 3) - (2x + 3) d (x2 - 5) dx dx (x2 - 5)2 (x2 - 5) x 2 - (2x + 3)-2x = (x2 - 5)2 2 x2 -10 - 4 x2 - 6 x = (x2 - 5)2 -2 x2 - 6 x -10 = (x2 - 5)2 -2( x2 + 3 x + 5) = (x2 - 5)2 (iii) d ( \_e\_ dx I 1 + sin x (1 + sinx)dex -ex d(1 + sinx) dx dx (1 + sin x)2 (1 + sin x)ex - ex cos x (1 + sin x )2 (1 + sin x - cos x)ex (1 + sin x )2 (iv) Let y = sin x3 Put x3 = t, then y = sin t and t = x3 dy dt ■ — = cos t and — = 3x dt dx 2 So, dy dy dt \---=---•--- dx dt dx = 3x2 cos t = 3x2 cos x3 (v t = x3) Hence, — (sin x3) = 3 x2 cos x3 dx - (v) Let y = sin3x = (sin x)3 dt Put sin x = t, So that y = ex and — - cos x dy dy dt . — = —•— dt dt dx = et∙cosx = esinx∙cos x - (vi) Let y = esin x = (sin x)3 . dy = esin x — (sin x) dt dx sin x = e ∙cos x sin x = cos x∙e Ans. Q. 6. What do you mean by binary number system ? How decimal fraction is written in binary fractions ? Write 0.700 in binary system. Ans. Binary Number System : The binary number system uses only two digits, 0 and 1. In short, a binary digit, is called a bit. The storing or computing electronic elements of a computer have only two stable states. The output of such electronic circuits at any time is either HIGH or LOW. These states can be represented by 1 and 0 respectively, that is HIGH is represented by 1 and LOW by 0. Due to this characteristic of electronic circuits a computer can understand information composed of only 0 s and 1s. So all computers perform their internal manipulations on binary digits only. But programmers use data and other information in the form of decimal digits, alphabets and special symbols. This information is converted to binary codes within the computer because a digital computer can operate on binary bits only. Thus, knowledge of the binary number system is needed to understand the working principle of a digital computer. Conversion from decimal to binary system : For this we have to successively divide the decimal number by two unitll it is reduced to zero. When on division by two there is a remainder of one this becomes a binary digit 1. And if there is no remainder it becomes a binary 0. The binary operation is built up from right to left. Conversion of fractions of decimal system to binary system : The following steps are used for converting fractions of decimal system to binary system. Step 1 : Multiply the fractions of the decimal system by the redix (R) of the required system (redix is 2 for binary system) Step 2 : Record the resulting integer as left most digit of the new base number. Let f1 be the fractional past. Step 3 : Multiply ‘f1’ by redix R (2 for binary system) again Step 4 : Record integer part as the next digit to the right side of the new number. Lef f2be the new fraction. Step 5 : Repeat the above steps till the fractional past becomes totally zero. In some cases the fractional past never becomes zero. In such cases stop the calculation after repeatation of the sequence. Solution of the example 0∙700 x 2 f1^1 • 400 x 2 f2 ^ 0 • 800 x 2 f3 ^1 • 600 x 2 f4 ^ 1 • 200 x 2 f5 ^ 0 • 400 x 2 f; ^ 0 • 800 x 2 f7 ^1 • 600 x 2 f4 ^ 1 • 200 or (0∙700)10 = (0. f1 . f2 . f3 ........f5) = (0∙10110011)2 Q. 7. What is software ? Give the relationship between hardware and software. Discuss its types. Ans. What is Software ?. It is important to note that a computer cannot do anything n its own. It must be instructed to do a desired job. Hence it is necessary to specify a sequence of instructions that a computer must perform to solve a problem. Such a sequence of instructions written in a language that can be understood by a computer is called a computer program. Software consists of programs that drive your computer to perform a specific job such as drafting letters and reports. Without software, a computer will not be able to do any useful work for you. This is similar to the case of a cassette player. Without inserting a recorded cassette, it cannot play songs. Relationship between Hardware and Software. In order for a computer to produce useful output, the hardware and software must work together. Thus, there is a special relationship between hardware and software. Both are complementary to each other. Nothing useful can be done with the computer hardware on its own and software cannot be utilized without supporting hardware. To take an analogy, a cassette player and the casettes purchase from market are hardware. However, the songs recorded on the cassettes are its software. To listen to a particular song, first of all that song has to be recorded on one of the cassette which should then be mounted on the cassette player and played. Similarly, to get a particular job done by a computer, the relevant software should be loaded in the hardware before, processing starts. It is immediately evidence that hardware is necessary but software is vital. Type of Software. Computer software is normally classified into two broad categories : application software, also known as an application package is a set of one or more programs designed to carry out operation for a specified application. For example, a payroll package produces pay slips as the major output and an application package for processing examination results produces mark-sheets as the major output alongwith some other statistical reports. Similarly, a program written by a scientist to solve his particular research problem is also an application software. The programs that constitute an application package are know as application programs and the person who prepares application programs is known as an application programmer. Another software type is systems software, also known as a system package, is a set of one or more programs, designed to control the operation of a computer system. These problems do not solve specific problems. They are general programs written to assist humans in the use of the computer system by performing tasks, such as controlling all of the operations, required to move data into and out of a computer and all the steps in executing an application program. In general, system packages support the running of other software; communication with peripheral devices (printers, card reader, disk and tape device etc.); support the development of other types of software; and monitor the use of various hardware resources (memory, peripheral, CPU etc.) Thus, systems software makes the operation of the computer system more effective and efficient. The programs and the person who prepares systems software is referred to as a systems programmer. Without systems software, application packages could not be run on the computer system. However, the production of sytems software is a complex taks that requires extensive knowledge and considerable specialized training in computer science. A computer without some kind of systems software would be very ineffective and most likely impossible to operate. The relationship between hardware, software and the user of the computer system is shown in fig. At the centre of any computer system is hardware and surrounding the hardware is system software. Finally, the user interacts with application software. Hardware System Software Application Software User Fig. Relationship between Hardware, Software and the User of a Computer System Q. 8. What do you mean by assembly language ? Discuss its demerits and advantages. Ans. Assembly Language One of the first steps in improving the program preparation process was to substitute letter symbols menemonics for the numeric operation codes of machine language. A menemonic (or memory aid) is any kind of mental trick we use to help us remember. Menemonics come in various shpaes and sizes, all of them useful in their own way. Fox example, a computer may be designed to interpret the machine code of 1111 (binary) or 15 (decimal) as the operation ‘subtract’, but is is easire for a human being to remember it as SUB. All computers have the power of handling letters as well as numbers. Hence, a computer can be taught to recognize certain combination of letters or numbers. It can be taught (by means of a program) to substitute the number 14 every time it sees the symbol ADD, substitute the number 15 every time it sees the symbol SUB, and so forth. In this way, the computer can be trained to translate a program written with symbols instead of numbers into the computer’s own machine language. Then we can write program for the computer using symbols instead of numbers and have the computer do its own translating. This makes it easier for the programmer, because he can letters, symbols and menemonics instead of numbers for writing his programs. For example, the preceding program that was written in machine language for adding two numbers and printing out the result could be written in the following way : CLA A ADD B STA C TYP C HLT Which would mean “take A, add B, store the result in C, type C, and halt.” The computer, by means of a translating program, would translate each line of this program into the corresponding machine language program. The language which substitutes letters and symbols for the numbers in the machine language program is called an assembly language or symbolic language. A program written in symbolic language that uses symbols instead of numbers is called an assembly code or a symbolic program. The translator program that translates an assembly code into the computer’s machine code is called as assembler. The assembler is a system program which is supplied by the computer manufacturer. It is written by system programmers with great care. It is so called because in addition to translating the assembly code into machine code, it also, ‘assembles’ the machine code in the main memory of the computer and makes it ready for execution. A symbolic program written by a programmer in assembly language is called a source program. After the source program has been converted into machine language by an assembler, it is reffered to as an object program. As shown in fig. the input to an assembler is a source program written in assembly language and its output is an object program which is in machine language. Assembly Language Program INPUT ------\* ASSEMBLER OUTPUT Machine \---------► Language Program One-to-one (Source program) ---------------» (Object program) Correspondence Advantages of Assembly Languages over Machine Languages : Assembly languages have the following advantages over machine languages : - 1\. Easier to understand and use : Assembly languages are easier to understand and use because menemonics are used instead of numeric opcodes and suitable names are used for data. The use of menemonics means that comments are usually not needed; the program itself is understandable. Symbolic programming also saves a lot a time and effort of the programmer because it is easier to write a compared to machine language program. - 2\. Easy to locate and correct errors : While writing programs in assembly language, fewer errors are made, and those that are made are easier to find and correct because of the use of menemonics and symbolic field names. - 3\. Easier to modify : Assembly language programs are easier for people to modify than machine language programs. - 4\. No worry about addresses : One of the greatest advantage of assembly language is that it eliminates worry about address for instructions and data. Limitations of Assembly Language : The following disadvantages of machine language are not solved by using assembly language : - 1\. Machine dependent : Because each instruction in the symbolic language is translated into exactly one machine language instruction, assembly languages are machine dependent. - 2\. Knowledge of hardware required : Since assembly language are machine dependent, so the programmer must be aware of a particular machine’s characteristics and requirements as the program is written. - 3\. Machine level coding : In case of an assembly language, instructions are still written at the machine-code level that is , one assembler instruction is substituted for one machine-code instruction. Machine and assembly languages being machine dependent are referred to as low-level languages. Q. 9. What is high level language ? Describing compiler and interpreter give the advantages of high level language. Ans. High Level Language We have already seen that writing of programs in machine language or assembly language requires a deep knowledge of the internal structure of the computer. While writing program in any of these languages, a programmer has to remember all the operation codes (numeric, or menemonic) of the computer and know in detail what each code does and how it affects the various registers of the computer. In order to facilitate the programmers to use computers without the need to know in detail the internal structure of the computer, high-level languages were developed. High-level languages, instead of being machine based, are oriented more towards the problem to be solved. These languages enable the programmer to write instructions using English words and familiar mathematical symbols. So it becomes easier for him to concentrate on the logic of his problem rather than getting involved in programming details. High-level languages are basically symbolic languages that use English words and/or mathematical symbols rather than menemonic codes. In other words, a high-level language is a symbolic language with nothing but macroinstruction. Every instruction which the programmer writes in a high-level language is translated into many translation and not one-to-one as in the case of assembly language. It is due to this reason that high-level language are so called. High-level languages are also known as problem-oriented languages because the micro instructions are especially picked to be useful for solving particular types of problems. Compilers Since a computer hardware is capable of understanding only machine level instructions, so it is necessary to convert the instructions of a program written in high-level language to machine instructions of a program written in high-level language to machine instructions before the program can be executed by the computer. In case of a high-level language, this job is carried out by a compiler. Thus, a compiler is a translating program that translates the High-Level Language Program INPUT ------\* COMPILER OUTPUT ---------► Machine Language Program One-to-many (Source program) ■+ (Object program) Translation instructions of a high-levle language into machine language. A compiler is so called because it compiles a set of machine language instructions for every program insturction of a high-levle language is called a source program. After this source program has been converted into machine language by a compiler, it is referred to as an object program. As shown in fig. the intup to a compiler (program) is a source program written in a high-level language and its output is an object program which consists of machine language instructions. Interpreters An interpreter is another type of translator used for translating high level languages into machine code. It takes one statement of a high-level language and translates it into a machine instruction which is immediately executed. Translation and execution alternate for each statement encountered in the high-level language program. In other words, an interpreter translates one instruction, and the control unit executes the resulting machine code, the next instructions is translated, and the control unit executes the machine code insturction, and so on. This differs from a compiler which merely translates the entire source program into an object program and is not involved in its execution. In case of a compiler the whole source program is translated into and equivalent machine language program. The object code, thus obtained, is permanently saved for future use and is used every time the program is to be executed. So repeated compilation is not necessary for repeated execution of a program. However, in case of an interpreter, no object code is saved for future use because the translation and the execution processes alternate. The next time an instruction is used, it must once again be interpreted and translated into machine language. Interpreter are often employed with micro computers (small computers). The advantages of an interpreter over a compiler is fast response to changes in the source program. Moreover, a compiler is a complex program comapred to an interpreter. Interpreters are easy to write and they do not require large memory space in the computer. The interpreter, however, is a time consuming translation method because each statement must be translated every time it is executed form the source program. Thus, a compiled machine language program runs much faster than an interpreter program. Assemblers, compilers, and interpreters are systems software that translates a source program written by the user to an object program which is meaningful to the hardware of the computer. These translators are also referred to as language processsors since they are used for processing a particular language. Advantages of High-level Languages : High level languages enjoy the following advantages over assembly and machine languages : - 1\. Machine independence : High-level languages are machine independent. This is a very valuable advantage because it means that a company changing computer—even to one from a different manufacturer—will not be required to rewrite all the programs that it is currently using. - 2\. Easy to learn and use : These languages are very similar to the languages normally used by us in our day-to-day life. Hence they are easy to learn and use. The programmer need not learn anything about the computer he is going to use. He need not worry about how to store his numbers in the computer, where to store them, what to do with them, etc. - 3\. Lower program preparation cost : Writing programs in high-level languages requires less time and effort which ultimately leads to lower program preparation cost. - 4\. Easier to maintain : Programs written in high-level languages are easier to maintain than assembly language or machin language programs. This is mainly because they are easier to understand and hence it is easier to locate, correct and modify instructions as and when desired. Limitations of High-level Languages : Two disadvantages of high-level languages are : - 1\. Lower efficiency : Generally, a program written in assembly language or machine languages is more efficient than one written in high-level language. That is, the programs written in high-levle languages take more time to run and require more main storage. - 2\. Lack of flexibility : Because the automatic feature of high-level languages always occur and are not under the control of the programmer, they are less flexible than assembly languages. An assembly language provide programmers access to all the special feature of the machine they are using certain types of operations which are easily programmed using the machine’s assembly languages, are impractical to attempt using a high-level language. This lack of flexibility means that some task a cannot be done in a high-level language, or can be done only with great difficulty. The credit for the development of the first high-level language is usually given to Dr. Greece Hopper who described the idea of a compiler and its language as early as 1952. Two languages were developed under Dr. Hopper’s directions : FLOWMATIC was a commercial and business language which could easily be put together from the contents of a flow chart, whereas MATHEMATIC was a mathematical language. These two languages were an early example of the development of high-level language in different directions—business and commercial, and science and mathematics. Since, then, many other high-level languages have been produced. Today there are over 200 high-level languages. However, most of these are for very special purposes or are designed to solve problems in a very specific applications area. Some of the most common high-level languages are : FORTRAN, which stands for Formula Translations; COBOL is an acronym for Common Business Oriented Languages; BASIC (Beginners All-purposes Symbolic Instructions Code); PASCAL; PL/I which stands for programming language one. Q. 10. What are Weiss and Miller’s indices ? Explain in brief. ![](AU B. Sc. I Chemistry-III(OK)_files/AU20B.20Sc.20I20Chemistry-III(OK)-18.png) Ans. Weiss Indices : Let us consider another face which makes intercepts x´, y´ and z´ on three axes where the standard intercepts are a, b and c such that x´ : y´ : z´ = 2a : b : 2c ...(1) From Eq. (1) if follows that the ratio of in-Ytercepts in terms of the standard is 2 : 1 : 2. These ratios characterize and represent any plane of the crystal. These co-efficients are generally known as Weiss indices of the plane. In some other words, Weiss indices may be defined as follows : “The ratio of the distances from the origin at which a face intersects the crystallographic axes are called the Weiss’s parameters of the crystal face.” Miller’s Indices. Miller indices of a plane are the reciprocals of the distances from the origin at which a given face intersects the three axes. Let us consider a plane which is parallel to Z-axis. Then the plane will cut the Z-axis at infinity. Immediately one would suggest that the ratio of the intercepts would be ...(2) x : y : z = 3 a : b : <x> c In the above case the use of Weiss indices is awkard and has been universally replaced by Miller’s indices. Miller’s indices are generally obtained by taking the reciprocals of the coefficients of a, b and c and when it is found necessary, the ratio is multiplied by the least common multiple to get integral 11 values. This Miller indices of Eq. (3) will be - : 1 : —, i.e., 1 : 2 : 0. This plane or face is simple indicated as (120). Let us consider another plane which is perpendicular to one axis and parallel to the other axis, then the Weiss indices will be 1 : — : — and the Miller’s indices will be 1 1 1 — 1 : or 1 : 0 : 0 — This face is indicated by (100) plane, If the plane makes an intercept on the negative side, say, a : -b : »c, the Miller’s indices for the plane would be 110 : the bar above would indicate intersection of the plane on the negative side of the axis. The symbol 110 means minus unity. X ![](AU B. Sc. I Chemistry-III(OK)_files/AU20B.20Sc.20I20Chemistry-III(OK)-19.png)Q. 11. What do you mean by space lattice and unit cell ? Explain also simple cubic lattice, face-centered cubic space lattice and body-centered cubic space lattice with suitable example. Ans. (a) How to represent Crystal lattice and Unit cell ? The crystal lattice of a substance is depicted by showing the position of particle (structural units) in space. These positions are represented by bold dots (or circles) and are referred to as lattice points or lattice sites. The overall shape and structure of a crystal system is governed by that of the units cell of which it is composed. A unit cell has one atom or ion at each corner of the lattice. Also, there may be atoms or ions in faces and interior of the cell. A cell with an interior point is called the body centered cell. A cell which does not contain any interior points is known as the primitive cell. That is, a primitive cell is a regular three dimensional unit cell with atoms or ions located at its corners only. Parameters of the unit cells. ![](AU B. Sc. I Chemistry-III(OK)_files/AU20B.20Sc.20I20Chemistry-III(OK)-20.png) Fig. IIIustration of parameters of a unit cell. In 1850, August Bravis, a French mathematician observed that the crystal lattice of substances may be categorised into seven types. These are called Bravis lattices and the corresponding unit cells are referred to as Bravis unit cells. The unit cells may be characterised by the following parameters : - (a) relative lengths of the edges along the three axes (a, b, c). - (b) the three angles between the edges (a, p, y). The parameters of a unit cells can be illustrated as in Fig. Parameters of the seven bravis unit cells are listed in Table. CUBIC UNIT CELLS These are the simplest unit cells. These unit cells are particularly impor- ![](AU B. Sc. I Chemistry-III(OK)_files/AU20B.20Sc.20I20Chemistry-III(OK)-21.png) ![](AU B. Sc. I Chemistry-III(OK)_files/AU20B.20Sc.20I20Chemistry-III(OK)-22.png) ![](AU B. Sc. I Chemistry-III(OK)_files/AU20B.20Sc.20I20Chemistry-III(OK)-23.png) Fig. The primitive unit cells for the seven crystal systems. Where two or more of the axes are equal. the same letter is shown in each. Right angles (90°) are shown as 7. The heavy line indicates the hexagonal unit cell. tant for two reasons. First, a number of ionic solids and metals have crystal lattices comprising cubic unit cells. Second, it is relatively easy to make calculations with these cells because in them all the sides are equal and the cell angles are all 90º. Table. The seven unit cells Crystal system Relative axis length Angles Examples Cubic (isometric) a = b = c a = p = y = 90° Na+Cl‒, Cs+Cl‒, Ca2+(F)2, Ca2+ O2‒ Tetragonal a = b \* c a = p = y = 90° (K+)2 PtCl62‒, Pb2+ WO42‒, NH4+Br‒ Orthorhombic a \* b \* c a = p = y = 90° (K+)2SO42‒, K+NO3‒, Ba2+ SO42‒, Ca2+ CO32‒ (apagonite) Rhombohedral (trigonal) a = b = c a = p = y\* 90° Cu2+CO32‒ (calcite), Na+NO3‒ Hexagonal a = b ^ c a=p=90°, y= 120° Agl, SiC, HgS Monoclinic a \* b \* c a=p=90°, y \*90° Ca2+SO42‒, 2H2O, K+ClO3, (K+)4 Fe(CN)64‒ Triclinic a \* b \* c a / p / y\* 90° Cu2+ SO42‒ . 5H2O (K+)2Cr2O7‒2 ![](AU B. Sc. I Chemistry-III(OK)_files/AU20B.20Sc.20I20Chemistry-III(OK)-24.png)SIMPLE CUBE BODY-CENTERED CUBE FACE-CENTERED CUBE Fig. Three cubic unit cells. Three types of Cubic unit cells There are three types of cubic unit cells : - (1) Simple cubic unit cell; (2) Body-centred cubic unit cell; (3) Face-centred cubic unit cell. A simple cubic unit cell is one in which the atoms or ions are occupying only the corners of the cube. A body-centred cubic unit cell has one particle at the centre of the cube in addition to the particles at the corners. A face-centred cubic unit cell has one particle at each of the six faces of the cube apart from the particles at the corners. Q. 12. What do you mean by X-ray Crystallography ? Derive Bragg’s equation nλ : 2d sin θ for the reflection of X-ray by crystals. How internal structure of a crystal is determined by its application ? Describe rotating crystal method for the crystal analysis. Ans. X-RAY CRYSTALLOGRAPHY A crystal lattice is considered to be made up of regular layers or planes of atoms equal distance apart. Since the wavelength of x-ray is comparable to the ![](AU B. Sc. I Chemistry-III(OK)_files/AU20B.20Sc.20I20Chemistry-III(OK)-25.png)CANCELLATION (DARK AREA) Fig. Diffraction patterns produced by crystals RE-INFORCEMENT (BRIGHTY DPOT) ![](AU B. Sc. I Chemistry-III(OK)_files/AU20B.20Sc.20I20Chemistry-III(OK)-26.png) interatomic distances, Laue (1912) suggested that crystal can act as grating to X-ray. Thus when a beam of X-ray is allowed to fall on a crystal, a large number of image of different intensities are formed. If the diffracted waves are in the same phase, they reinforce each other and a series of bright spots are produced on a photographic plate placed in their path. On the other hand, if the diffracted waves are out of phase, dark spots are caused on the photographic plate. From the overall diffraction patterns produced by a crystal, we can arrive at the detailed information regarding the position of particles in the crystal. The study of crystal structure with the help of X-rays is called X-ray crystallography. Bragg’s Equation In 1913, the father-and-son, W.L. Bragg and W.H. Bragg worked out mathematical relation to determine interatomic distances from X-ray diffraction patterns. This relation is called the Bragg equation. They showed that : - (1) the X-ray diffracted from atoms in crystal planes obey the laws of reflection. - (2) the two rays reflected by successive planes will be in phase if the extra distance travelled by the second ray is an integral number of wavelengths. Derivation of Bragg Equation Fig. given here shows a beam of X-rays falling on the crystal surface. Two successive atomic planes of the crystal are shown separated by a distance d. Let the X-rays of wavelength λ strike the first plane at an angle θ. Some of the rays will be reflected at the same angle. Some of the rays will penetrate and get second plane. These rays will reinforce those reflected from the first plane if the extra distance travelled by them (CB + BD) is equal to integral number, n, of wavelengths. That is, ![](AU B. Sc. I Chemistry-III(OK)_files/AU20B.20Sc.20I20Chemistry-III(OK)-27.png)Fig. Reflection of X-rays from two different planes of a crystal. nλ = CB + BD ...(i) Geometry shows that CB = BC = AB sin θ ...(ii) From (i) and (ii) it follows that - nλ = 2AB sin θ or nλ = 2d sin θ This is known as the Bragg equation. The reflection corresponding to n = 1 (for a given series of planes) is called the first order reflection. The reflection corresponding to n = 1 is the second order reflection and so on. Bragg equation is used chiefly for determination of the spacing between the crystal planes. For X-rays of specific wavelength, the angle θ can be measured with the help of Bragg X-ray spectrometer. The interplanar distance can then be calculated with the help of Bragg equation. Determination of crystal structure by rotating crystal method : The apparatus used by Bragg is shown in Fig. A beam of X-rays of known wave- X-RAY ![](AU B. Sc. I Chemistry-III(OK)_files/AU20B.20Sc.20I20Chemistry-III(OK)-28.png)ROTATION OF CRYSTAL length falls on a face of the crystal mounted on a graduated turn table. The diffracted rays pass into the ionisation chamber of the recorder. Here they ionise the air and a current flows between the chamber wall and an electrode inserted in it which is connected to an electrometer. The electrometer reading is proportional to the intensity of X-rays. As the recorder along with the crystal is rotated, the angles of maximum intensity are noted on the scale. Thus, values of θ for n = 1, 2, 3. etc. are used to calculate the distance d between the lattice planes parallel to the face of the crystal. Q. 13. Represent Crystal lattice of NaCl and KCl. Ans. Structure of Rock Salt, NaCl The structure of sodium chloride (Fig. b) was studied by the Bragg’s X-ray spectrometer. The intensity of the ionisation current was measured for different angles of rotation of the crystal. Then, a curve was plotted between the current intensity and glancing angles as shown in Fig. (a) for (100) and (110) and (111) faces. From the graph it is evident that, - (i) The intensities of current fall of regularly for the faces (100) and (110), and - (ii) There is a difference for the (111) face. In case of (111) face, the first order spectrum is abnormally weak, the second as abnormally strong, the third as abnormally weak and fourth as abnormally strong. (100) 1 (110) 2 3 4 5 (111)1 √2 1 2√2 1 1 3√2 3√2 √3 3√3/2 2√3 ![](AU B. Sc. I Chemistry-III(OK)_files/AU20B.20Sc.20I20Chemistry-III(OK)-29.png) It was proved that the (100) face of a sodium chloride crystal yielded reflection maxima at glancing angles of 5∙9º, 11∙85º, and 18∙15º. If we say that these are representing first, second and third order reflections, respectively, then their since must be in the ratio of 1 : 2 : 3. These are in good agreement with our expectations. Let us now consider the Bragg’s equation - 2 d sin 0 = n X or d = (n X/2) sin 0 ...(1) Therefore, for a particular order of reflection, d 1 ....(2) « sin 0 The first order reflections are found to have the glancing angles as 5∙9º, 8∙4º and 5∙2º for the (100), (111) planes of sodium chloride. Applying equation (2), we have 111 d(100) ■ d(n0) ■ d(in) = sin 5.9° :sin8.4° :sin5.2° 111 = 0-103'0-146'0-091 = 1 : 0∙704 : 1∙155 As we know that for a face-centred cubic lattice d100 : d110 : d111 = 1 : 0∙704 : 1∙154 These rations are same as obtained for the unit cells of a face-centred cubic lattice. Thus, the structure of NaCl is face-centred lattice. Now, the problem is to decide about the structural units of the crystals. These units may be molecules, atoms or ions. This problem is solved by considering the plane (111). In planes (111), the first order spectrum is abnormally weak, the second abnormally strong, the third abnormally weak and fourth abonormally strong. The alternation of these intensities are due to alternate layers or planes of chlorine atoms separated by layers of sodium atoms. Also, it is evident from the alternations in the intensities that the real unit of matter in crystal lattice is the atoms and not the molecules and not the ions. Bragg gave the structure of NaCl as shown in Fig. (b). From the structure, it follows that : - (i) the rock salt crystals have face-centred cubic lattices with sodium and chlorine atoms arranged alternately in all directions parallel to the three rectangular axes and - (ii) each unit cell of sodium chloride consists of 14 sodium atoms and 13 chlorine atoms, and - (iii) each chlorine atom is surrounded by six sodium atoms and each sodium atom is surrounded by six chlorine atoms. Structure of KCl The study of structure of Sylvine was done by Bragg’s X-ray spectrometer. The intensities of ionisation current ware determined for glancing angles. By plotting the curve between the current intensity and glancing angles, two types of the curves are observed as shown in Fig. (9a) for (100) (110) and (111) faces. The first order spectrum from the (100), (110) and (111) planes of KCl was observed at the glancing angles 5∙38º, 7∙61º and 9∙38º respectively. Therefore, 111 :: d(100) ■ d(i10) ■ d(111) sin5.38° sin7.6r> s^.38° 111 = 0-0938:0-1326:0-1326 = 1 : 0∙704 : 0∙575 ![](AU B. Sc. I Chemistry-III(OK)_files/AU20B.20Sc.20I20Chemistry-III(OK)-30.png)But we know for a simple cubic lattice, d100 : d110 : d111 = 1 : 0∙707 : 0∙575. It, therefore, follows that KCl crystal has a simple cubic lattice whereas NaCl has a face-centred cubic lattice. Expected similarity between NaCl and KCl : In almost every property, KCl resembles NaCl, suggesting that the fundamental lattice should be the same, i.e., the face-centred lattice. The expected behaviour is supported by the following observations : The first order reflection from (100) planes of the rock salt and KCl occurred at 5∙9º and 5∙3º respectively. As we know that the value of d is inversely proportional to sin 0, it means diooKCl sin5 - 9° \-----------=---------= 1-11 d100NaCl sin5- 3° Since the volume of a small cubic lattice is (d111)3, it follows that if the potassium and sodium chlorides have the same crystal lattice structure, the quantity (1∙11)3 = 1∙37 should give the ratio of molecular volumes of the two salts but the experimental value is 1∙38 which suggests that the two must have same space lattice. i.e., face-centred cubic lattice. But the Bragg’s study indicated that KCl should have a simple cubic lattice. Thus, there exists some anomaly. How to solve the above anomaly ? The above anomaly can be explained on the basis that, “the X-ray scattering factor for an atom is equal to the number of extraplanetary electrons, (viz., atomic number) at small grancing angles.” The atomic numbers of potassium and chlorine are 19 and 17 respectively. As these numbers are not very much different, the X-rays are unable to detect any difference between the two kinds of atoms. If we assume that all the atoms are identical, it means that the face-centred arrangement of NaCl becomes a simple cubic arrangement for KCl. In (111) face of KCl, the atomic numbers of K and Cl are so identical that the reflected intensities are nearly same and the odd orders are completely cut off. Due to the above mentioned facts, KCl appears to be a simple cubic lattice. On the other hand, in case of NaCl there is difference between the atomic number of sodium (11) and chlorine (17). It means that their scattering factors are different and hence, give rise to a face-centered cubic lattice. From the above discussion, we can safely claim that “Both the rock salt (NaCl) and (KCl) have the same structure.” Although the structures of NaCl and KCl are same, there is an important difference which may be observed by applying the expression (1) to both the crystals. Therefore, d100(NaCl) sin5-3° \_ 1 d100(KCl) = sin5 ■ 9° = 1-11 From the expression, it is evident that the fundamental units are more widely spaced in the crystal of KCl than in the crystal of NaCl. Q. 14. Give the types of liquid crystals ? Also give their properties and uses. Ans. Liquid crystals are of two types : - (a) Nematic liquid crystals - (b) Cholesteryl liquid crystals. - 1\. Nematic Liquid Crystals : These are closer to true an isotropic liquids than are the nematic phases. These liquid crystal have a low viscosity and flow readily. In polarised light they appear to have mobile thread-like structure. The molecules are aggregated together in groups or swarms with their axis parallel to one another. These molecules move sideways or up and down along their length. Each molecule can rotate around its axis. Fig. shows the arrangement of molecules in the nematic type liquid crystal. In an electric or a magnetic field the swarms of nematic liquid crystals get oriented in the same direction. The turbidity in nematic phase is due to the scattering of light as we find in finely powdered glass. With rise in temperature swarms diminish in size and at the m.p., these are too small to scatter light. The liquid becomes clear. Properties : (i) Because of the translational order, nematic crystals flow like a normal liquid. - (ii) Because of the orientation of the molecules about a preferred axis (known as the director axis) nematic state gives rise to an isotropy in several properties, e.g. optical, diamagnetic, viscosity and elasticity. - (iii) Degree of orientational order has been found to decrease with increase of temperature and nematic-isotropic transition occurs with a weak latent heat of transition (= 1 kilo Joule per mole) and a small volume increase (= 0.5%) - (iv) The centres of gravity of the molecules have no long range order and therefore there is no Bragg peak in the X-ray diffraction pattern. The positions between the centre of gravity of neighbouring molecules are correlated similarly as those of conventional liquid. - (v) The molecules tend to be parallel to some common axis, labelled by a unit vector, a. This is supported by all macroscopic tensor properties. The difference between refractive index measured with polarizaitons parallel or normal to a is quite large. - (vi) The direction of a is arbitrary in space. In practice it is impsed by the guiding effect of wall of the container. - (vii) The direction of a and ‒a are indistinguishable. If the individual molecules carry a permanent dipole there are many dipoles up and many dipoles down. - (viii) These phases are found only with material which do not distinguish between right and left. Each constituent molecule must be identical to its mirror image or if it is not, it should be a racemic mixture. The substances which show nematic behavious are given in Table 1 - Table 1 Compound yielding nematic type liquid crystals Compounds Transition Point Melting Point (K) (K) p-azoxy phenetole 410 440 Anisaldazine 438 453 p-Methoxy cinnamic acid 443 459 p-azoxyanisole 389 408 Dibenzal benzidine 507 533 - 2\. Cholesteryl Liquid Crystals : Some liquid crystals, such as cholesteryl esters, show in addition to their nematic behaviour some colour effect under polarised light. These have been given the special name cholesteryl phase. Because of the optically acitve molecules, the cholesteric state has a spontaneous twist about an axis normal to the director. The helical arrangement is right or left handed. Cholesteryl phase possesses a very high optical rotaroy power. The plane of polarisaiton of light is rotated by a few thousand degree per mm which is thousand times greater than the rotatory power of a solid crystal like quartz. The pitch of the hlix decreases with increase of temperature and the wavelength of reflection follows out. In this type of liquid crystal the molecular axis are aligned, and the molecules are arranged in layer in which the orientation of axis shifts in a regular way in going on layer to the next. These are named as cholesteric crystal because the skeleton of the substance passing through this state is similar to that of the cholesterol. These liquid crystal are characterised by their very high optical rotation. The pitch of the spiral and the reflected colour depends sensitively on the temperature. The first substance in which mesomorphism was detected i.e., cholesteryl benzoate, is of the cholesteric type; its transition temperature is 146°C and melting point 178∙5°C. Uses : (i) Liquid crystals are becoming important substances becuse of their electrical and optical properties and use in electronic display instruments. For example, these have been use in gas liquid chromatography and in digital displays like pocket calculators, digital wrist watches, etc. - (ii) Liquid crystals are also used as commercial lubricants. - (iii) These are most suited to biological functions. Since the colour of liquid crystal depends sensitively on temperatures, these have been used to measure skin temperature. - (iv) These are used for detecting tumours in the body by employing method called thermography. - (v) These are employed as solvents for studying the structure of an isotropic molecules spetroscopically. - (vi) Due to their electrical and mechanical properties lying between crystalline solids and isotropic liquids these are used in gas liquid chromatrography. Q. 15. Throw light on the structure of liquids. Ans. Structure of liquids : In a perfect crystal a complete ordered arrangements of the atoms, ions or molecules constituting the crystal exists. The intermolecular forces are strong enough to hold them together. The constituents vibrate about their mean positions but can not execute translational motion. Hence, both the short range as well as the long range orders exist in crystals. The molecules in a gas have complete random motion and the intermolecular forces between the molecules are small and are effective only at short distances. There is no possibility of any kindly of ordered structure in gases. In liquids, on the other hand, the situation somewhat lies in between the two. The cohesive forces in liquids are stronger than those in gases. These forces, however, are not strong enough to disallow considerably the transaltional motion of the individual molecules. In terms of the arrangement of the constituents a liquid is having short range order but lacks long range order. JOOOCK oooooc jOOOOOOk '»OOOOOO«' 900000 ^OOOOr Solid [o o o oo](#bookmark238) [o o o oo](#bookmark239) [o o o oo](#bookmark240) [o o o oo](#bookmark241) [o o o oo](#bookmark242) [o o o oo](#bookmark243) [o o o oo](#bookmark244) [o o o oo](#bookmark245) Liquid o o o o o o o o o o o o Gases When a crystal melts there is, in general an increase in volume of about 10% of about 3% intermolecular spacings. The molecules in the liquid state still remain in the vicinity of the other molecules surrounding them. The intermolecular forces tend to decrease as the distance between the molecules increases. In other words, the short range forces exist while the long range forces are negligible. The thermal motion tends to introduce a disorder in the structure but the motion is not as random as in a gas. Some ordered arrangement still perisist. The ordered arrangement in the crystal is not completely destroyed on melting. As the temperature get further increases, the thermal motion of the molecules increase the kinetic energy, decrease the order unit at the boiling point is completely vanishes. The sharpness of the melting point could be explained by two dimensional model for solid, liquids and gases as shown in fig. Introduction of a small region of disorder into a crystal bring about the disturbance of the long range order and destroys the crystalline arrangement. This explains the abrupt variation in the properties between solids and liquids.J. D. Bernal postulated that an atom is surrounded by 5 atoms insted of normal six atoms as is the requirement for a closest packed structure for solid. The remaining atoms (circles in the figure) are drawn in the utmost possible ordered arrangement. One point of abnormal co-ordination number is sufficient to bring about a long range disorder. That is, when the normal motion causes a disorder in one region, it spread in all direciton destorying the entire regular structure. Liquids, therefore, may be regarded to have structure like that of a crystal with a difference that the ordered arrangement extends over a short region instead of over the whole mass. This is termed as the short range order or long range disorder. It is to be noted that the short range order in liquid structure is continuously changing because of the thermal motions. Q. 16. Give reasons for the following : - (i) Bleeding is stopped by adding alum. - (ii) Where river and sea water meet a delta is formed. - (iii) Alum is added to drinking water supply. - (iv) Colloidal particles move under and applied electric field. - (v) When As2S3 and Fe(OH)3 sols are mixed precipitation is occured. - (vi) Sols cannot filtered by ordinary filter paper. - (vii) Colloidal particles make zig-zag motion. Ans. (i) Blood is a colloidal solution of an albuminoid substance. The styptic action of alum and ferric chloride solution is due to coagulation of blood forming a clot which stops further bleeding. - (ii) River water is colloidal solution of clay. Sea water contains a number of electrolytes. When river water meats the sea water, the electrolytes present in sea water coagulate the colloidal solution of clay which get deposited with the formation of delta. - (iii) They water obtained from natural sources often contains bacteria and suspended impurities. Alum is added to such water so as to destroy the bacteria as well as to coagulate the suspended impurities and make water fit for drinking purpose. - (iv) Colloidal particles always carry an electric charge. This charge is of the same type on all the particles in a given colloidal solution and may be either positive or negative, therefore, when an electric current is passed through a colloidal solution the particles moves towards either of the oppositely charged electrodes. For example, if the colloidal solution of As2S3 is joined with electric circuit, then due to the negative charge on As2S3 particles, these move towards anode. - (v) It is due to the presence of positive charge on Fe(OH)3 sol and negative charge on As2S3 sol. When equivalent amount of As2S3 and Fe(OH)3 are mixed, they both get coagulated (natural coagulation) and give the precipitate. - (vi) Colloidal particles passes through ordinary filter paper. However, these particles do not pass through parchment and other fine membranes. This is due the fact that size of the colloidal particles is lesser than the diameter of the holes of ordinary filter paper. Hence, colloids cannot be filtered by ordinary filter paper. - (vii) The colloidal particles make zig-zag motion. It is due to the unequal bombardment of colloidal particles by the molecules of dispersion medium. With the increase in the size of the particles, the probability of the unequal bombardment decreases and the Brownian movement also disappears. Q. 17. What do you mean by colloidal state or sol ? Describe various types of colloidal dispersion. Ans. Colloidal state or sol. A substance is said to be in colloidal state if the size of its particles is in between 10‒4 to 10‒7 cms while the size of particle of true solution and suspension lies between 10‒3 cm and onward and 10‒2 to 10‒8 respectively. 10‒2 to 10‒4 cm 10‒4 to 10‒7 10‒7 to 10‒8 True solution colloidal state suspension The particle of the colloidal substance constituste the dispersed phase while liquid medium in which they are present is known as dispersion medium. The whole system consisting of dispersed phase any dispersion medium is called Colloidal solution or sol. ![](AU B. Sc. I Chemistry-III(OK)_files/AU20B.20Sc.20I20Chemistry-III(OK)-31.jpg)COLLOIDAL MILL Mechanical Dispersion. The substance and the dispersion medium are introduced in a colloidal mill consisting of two metal discs rotating in opposite direction at a speed of 70,000 revolutions per minute. The space between them is 10‒4 to 10‒7 cm. It is used for preparing dye stuffs, printing inks, paints, dental cream etc. Electro-dispersion. (Bredig’s Arc, method, 1881). The rods of a metal whose colloidal solution is to be prepared are made electrodes and the container is kept in cold freezing mixture. On passing an electric current, an are is produced which vapourises the metal and these vapours are immediately condensed to the particles of colloidal state and are stabilized by alkali. Freezing Mixture ![](AU B. Sc. I Chemistry-III(OK)_files/AU20B.20Sc.20I20Chemistry-III(OK)-32.png)Peptisation. A process of transferring back a freshly prepared precipitate into a colloidal state is called peptisation. For example, freshly prepared precipitate of ferric hydroxide readily passes into colloidal state on treatment with dil. FeCl3 solution (peptising agent). ![](AU B. Sc. I Chemistry-III(OK)_files/AU20B.20Sc.20I20Chemistry-III(OK)-33.png)Freshly Precipitated Fe(OH)3 Colloidal Particles of Fe(OH)3 It is due to the preferential absorption of one of the ions of electrolyte and preferable those which are common of them. The ferric ions, Fe3+ are absorbed on particles of Fe(OH)3 constituting the precipitate where by the + ve charges comes on their surface. Due to similar charges on each particles, they get separated yielding colloidal particles. Q. 18. Discuss the Maxwell distribution law of molecular velocities. What is the effect of temperature on velocity distribution ? Ans. Maxwell’s Distribution Law of Molecular Velocity : Gas molecules have different velocities due to their frequent colliding with each other and the walls of the container. The change in velocities is due to redistribution of the energy and velocity of the colliding molecules. A particular fraction of the molecules at any time have same velocity. Maxwell and Boltzman, using theory of probability, have shown that distribution of molecular velocities depends on the temperature and the molecular weight of a gas and is given by : dn0 4π n' 3/2 α U -Mc/RT c2 dc 2πRT VW e This equation is known as the Maxwell Boltzmann distribution law, where dn0 /n' the fraction of total number of molecules n’, having velocities between c and dc, M is the molecular weight of the gas and T the absolute temperature. Dividing the equation by dc, we get : 3/2 1 ⋅ dn0 = 4πFG M IJ e-Mc/2RT⋅ c2 n' dc H 2πRT K The left hand side of this equation represents the probability P, of finding molecules having a particular velocity, c. By knowing the molecular weight of a gas, velocity of a fraction of molecules at any temperature can be determined with the help of the above equation. Effect of temperature on velocity distribution : The most probable velocity increases with rise in temperature as shown in the figure. The entire distribution curve in fact, shifts to the right with rise in temperature, as shown. The rise in temperature, therefore considerably increase the fraction of the molecules having high velocity (most probable velocities). This can readily be understood form the presence of the factor. ![](AU B. Sc. I Chemistry-III(OK)_files/AU20B.20Sc.20I20Chemistry-III(OK)-34.png)![](AU B. Sc. I Chemistry-III(OK)_files/AU20B.20Sc.20I20Chemistry-III(OK)-35.png)-Mc2/2RT in equation (ii). The exponent has a negative sign and the temperature T is in the denominator. The factor, therefore, increases markedly with increase in temperature. This factor is known as Boltzmann factor. Further knowing that 12 mc2 is the K. E. of one molecule of the gas having velocity c, the factor - -Mc2/RT = -E/RT ee where E gives the K. E. per molecule of the gas. Evidently, the greather is the remperature the greater is the value of E. Q. 19. (a) Define gram molecular specific heat. Starting with kinetic equation of gas calculate the ratio of Cp and Cv. How does this vary with molecular complexity of the gas ? - (b) The Specific heat of a mono-atomic gas is 0∙075 at constant volume. Determine its molecular weight. Ans. (a) Specific Heat : Specific heat of a substance is defined as the amount of heat required to raise the temperature of 1 gm of substance through 1ºC. Gram Molecular specific heat or molar heat is equal to heat required to raise the temperature of 1 gm mole (i.e., mol. wt. expressed in grams) of substance through 1ºC. Molar heat = Sp. heat x mol. wt. Evidently molar specific heat is equal to specific heat multiplied by its molecular weight. The value of specific heat depends on the conditions under which a substance is heated. Heating can be conducted either, at constant pressure or at constant volume. The two values differ appreciately from each other in gases because their coefficient of expansion are quite large, whereas in case of liquids and solids the two values are nearly the same. Specific heat at constant volume (Cv) It is defined as the amount of heat required to raise the temperature of 1 gm. of gas through 1°C, while the volume is kept constant and the pressure is allowed to increase. It is possible to calculate its value by making use of the kinetic theory. Considering one gram molecule of a gas at the temperature T. Its kinetic energy is 12 mv2 . From the kinetic equation PV = 13 mnv2 = -3 x 2 mnv2 = RT or 12 mnv2 = K. E. 23 RT It is the temperature raised by 1°C to (T + 1) Å The kinetic energy becomes 23 R(T + 1) :. Increase in kinetic energy = | R(T + 1) - -3 RT = 23 R When a gas is heated at constant volume, the heat supplied is utilized for: - (i) Increasing the kinetic energy of the gas molecules, which is equal to 23 R. - (ii) Increasing the intra-molecular energy i.e., vibrational and rotational. Consequently in such case the molecular heat will be greater than 23 R by a factor x. C = 3 R + x v2 The value of x varies from gas to gas and is zero for monoatomic. Specific Heat at Constant Pressure (Cp) It may be defined as the amount of heat required to raise the temperature of one gm. of gas through 1°C, the pressure remaining constant while the volume is allowed to increase. When a gas is heated at constant pressure the heat supplied is utilized for : - (i) Increasing the kinetic energy of the gas molecules, - (ii) Increasing the intra-molecular energy, - (iii) Performing external work for expanding gas. - :. Cp = (Increase in K. E. per degree rise of temp. \+ increase in intra-molecular energy \+ external work done for expanding the gas) If during heating, the gas expands from V at T A to AV at (T + 1) A at constant pressure P. PV= RT at T° P(V + AV) = R (T + 1) at (T + 1°) External work done = P(V + AV) - PV = R(T + 1) - RT = R Hence, R cals. must be added to the value of 3/2 cals in order to get the thermal equivalent of the energy supplied to one gm. molecule of the gas in the form of heat when the temperature is raised by 1°C C = 3/2R + R = 5/2R Cpp = 5/2R + x The value of x varies from gas to gas and is zero for monoatomic gases. Ratio of specific or molar heats, Cp 2R + x CV = |R + x In case of mono-atomic gas no increase in intra-molecular energy takes place i.e., x = 0. Cp 52R Y = C = Ar = 1-66 Cv 2 R Variation of Cp/Cv with molecular Complexity of the gas In case of di and tri atomic gases, however, the ratio does not remain the same. In such cases the heat supplied is utilized in not only increasing the kinetic energy of the molecules, but also their rotational and vibrational energies. This amount thus varies with the atomicity of gases and is found to be greater than 3/2R by a factor x. For mono-atomic gases, like helium and argon, x =0; for di-atomic gases like hydrogen and oxygen x = R while for poly atomic gases like CO2, SO2 etc., x = 3/2 R. Thus, Cp 2R + x For any gas Y = — Cv For mono-atomic gases |r+x Cp Y ~ v For di-atomic gases x = R |R + 0 5 = |R + 0= 3 = ^ Cp Y=c“ Cv (b) Solution : |R + R 7 = 2 = = 1∙40 |R + R 5 We know that, Gram Molecular Heat = Specific heat x Mole. wt. It is known to us that gram molecular heat at constant volume (Cv) is 3 calories. Thus on keeping the values of gram molecular heat and Cv in the above expressions, we get : Gram Molecular heat = Cv x Mol. wt. 3 = 0^075 x M 3 0 ■ 075 M = = 40 Hence molecular weight of the gas is 40. Q. 20. Explaining the isotherm of CO2 gas, define the following : - (i) Critical temperature, - (ii) Critical pressure, - (iii) Critical volume. Ans. Isotherms of Carbon Dioxide : The curve representing the variations of volume and pressure at constant temperature are called isotherms. For an ideal gas the product PV is constant and hence the isotherms would be rectangular hyperbolas. Andrews in 1869 studied the isotherms of CO2 at different temperatures. It can be seen from the diagram that the isotherms of CO2 at 50°C is nearly rectangular hyperbola (Fig) similar to that of an ideal gas, the isotherm (Fig) at 81°C remains horizontal for a short while at the point F, thus showing a great decrease of volume for a small change of pressure. This deformity in the curve is coincident with the appearance of liquid CO2. At still lower temperatures the horizontal portions of the curves are much more pronounced. Now on studying the isotherm at 21°C, the portion AB indicates the compression of CO2 vapour. At B liquefaction starts and the curve BC remains horizontal while the gas is changing to liquid at constant pressure. At C the liquefaction is complete and the curve CD rises almost vertically, indicating a small decrease of volume with increase of pressure, which is the characteristic of a liquid. It is clear from these considerations that any point within the parabolic area indicated by the dotted line, both vapour and liquid are present, while outside this area either liquid or vapour only is present. I ![](AU B. Sc. I Chemistry-III(OK)_files/AU20B.20Sc.20I20Chemistry-III(OK)-36.png)VOLUME ----► The isotherm EFG marks the boundary between gas and vapour. At the point F, the distinction between liquid and vapour disappears and CO2 exists in a state called the critical state. the point F is called the critical point, the isotherm passing through this point is known as the critical isotherm and the temperature corresponding to this isotherm (31°C) is called the critical temperature. ![](AU B. Sc. I Chemistry-III(OK)_files/AU20B.20Sc.20I20Chemistry-III(OK)-37.png)- (i) Critical Temperature (Tc) : It can be defined as the temperature Tc, above which a gas cannot be liquefied how so ever pressure is applied. - (ii) Critical Pressure (Pc) : The minimum pressure required to liquefy a gas at the critical temperature is known as critical pressure. - (iii) Critical Volume (Vc) : The volume occupied by one mole of a gas at a critical state is called critical volume. (b) Critical constant : Van der Waal’s equation is F P+a I b V- b g = RT a ab or PV + —- Pb - —2 = RT Multiplying throughout by V2, we get PV3 + aV ‒ PV2b ‒ ab = RTV2 or PV3 + aV ‒ PbV2 ‒ ab ‒ RTV2 = 0 Dividing by P and arranging descending powers of V, we get .....(i) ....(ii) F RT I aV ab V - V21 + b +---=0 HP K P P ....(iii) This is the general gas equation of Van der Waal. It is a cubic equation which when solved would give three values of V. Under critical conditions the distinction between liquid and the vapour disappears and the three values of V become identical. This volume is called critical volume. i.e., or or or V V ‒ V c = Vc (Vc = critical volume) = 0 (V‒Vc)3 = 0 V3 ‒ 3V2Vc + 3Vc2V ‒ Vc3 = 0 .(iv) This is the equation of critical state. To change equation (iii) into (iv) an equation of critical condition, we put : T= Tc (critical temp); P = Pc (critical pressure) Equation (iii) becomes V3 - RT A RT aV ab V2 —c- + b +---= o P P V Z’ Z’ Pc K cc (v) Hence (iv) and (v) are alike. Equating the power of V in eqn. (iv) and (v), we get RT 3V = + b c Pc a 3Vc2 = P c ab - Vc3 = Pc Dividing eqn. (viii) by (vii), we get - Vc = 3b ....(vi) ....(vii) ....(viii) ....(ix) Substituting the value of Vc in (viii), we get a Pc 27b2 Substituting the values of V and P in (vi), we get 8a c c T = c 27Rb ....(x) ....(xi) Q. 21. Discuss in brief the other equation of states for explaining the behaviours of real gases. Ans. Although Van der Waals equation has achieved success in explaining the behaviours of real gases but an appreciable deviation have also been observed at too low temperatures and too high pressure of the gases. A large number of other equations have been suggested to represent P = V‒ T relations of the gases more satisfactorily. Some of them are empirical and some are theoretical. Some important equations of state are given on the next page. - 1\. Dieterici Equation : This equation makes an allowance for the forces of attraction on molecules and its effect on the pressure. The equation is : P(V - b) = RTe-a/RT Where e is the base of natural log. This eqn. agrees with the Van der Waals eqn. over a small range of pressure. At high pressure, the two equations differ considerably and Dieterici eqn. gives resutls which are more in agreement with the exp. results. - 2\. Claussius Equation : Claussius showed that the factor a in the Van der a Waals eqn. is not independent of temperature and used the term t(v + c) for a . The proposed eqn. is : FI a ##### Hib v+c g2 K b V - b g = RT Where c is some constant. This equation explains P ‒ V ‒ T relationship fairly well for some gases and not for all gases. - 3\. Berthelot Equation : In this equation, Berthelot empirically determined the value of the constant ‘a’ and ‘b’ in terms of critical temperatue (Tc) and pressure (Pc). The equation is FI PV = nRT 1+----- H 128PTK #### F 3I 1--T #### H TK The eqn. is found to be very accurate only for pure gases. - 4\. Kammerlinigh Onnes Equation : This is an empirical equation and gives PV as a power series of the pressure at any given temp. This eqn. is written as : PVm = A + BP + CP2 + DP3 + ................ P is the pressure and Vm is the molar volume. The coefficients A, B, C, D etc., are known respectively as the 1st, 2nd, 3rd , 4th etc. virial coefficients (Greek, virial = force). When the pressure is very low, only the 1st coefficient A, is equal to RT is significant and others are neglected. When the pressure is high, the other coefficients are also significant and are taken into consideration. The order of significance of those coefficients is their order in the eqn. The values of all those coefficients although constant at a given temperature, vary with temperature. This eqn. can be made to explain the experimental result with fair accuracy. Q. 22. (a) Derive Van der Waals equation. Give units of van der Waals contants use in it. Ans. Van der Waals Equation : The gas equation PV = RT implies that at constant temperature PV is constant, i.e., if PV is plotted against P at constant temperature, the curve obtained should be straight line. The gases which obey the gas equation are known as perfect gas or ideal gas. Actually no gas is ideal (or perfect) at different temperatures and pressures. ![](AU B. Sc. I Chemistry-III(OK)_files/AU20B.20Sc.20I20Chemistry-III(OK)-38.png)PV (Atmosphere) -------► (Curve between P and PV of H2, N2 and CO2 at constant Temp.) Regnault and Amagat studies the effect of change of pressure on PV of - H2, N2 and CO2 and different temperature and notes as follows : - (1) At Low Pressure : PV for all gases, expect H2, has lower value than expected for an ideal gas. - (2) At Higher Pressure : PV for all gases has higher value than expected for an ideal gas. - (3) At Low Temperature : The deviations are much more pronounced than at high temperature. - (4) Exceptional behaviour of H2 : PV for H2 has greater value than for an ideal gas. Reasons for Deviations Van der Waal attributed that the reasons for deviations of real gases from gas equations is due to the following wrong assumptions in the kinetic theory : - (i) The actual volume of the molecules is negligible as compared with the total volume of the gas. ![](AU B. Sc. I Chemistry-III(OK)_files/AU20B.20Sc.20I20Chemistry-III(OK)-39.png) (Effect of Temp. change on Nitrogen) - (ii) There are no attractive forces among the molecules. Both these assumptions are not true, particularly at low temperature and high pressure. Improvements : Hence Van der Waal (1837) made the following two corrections so that all the gases may obey the general gas equations. - (i) Volume corrections. The volume of a gas is considered equal to the vacant space of the container. This holds good at low pressure when the volume of the molecules is negligible but it is by no means negligible at higher ![](AU B. Sc. I Chemistry-III(OK)_files/AU20B.20Sc.20I20Chemistry-III(OK)-40.png)Fig I Fig II Fig III pressure when the molecules come closer as shown in figure II & III. Thus, if the molecules do occupy some volume, than the vacant space, available for them to move about in volume V is not equal to V but less than this, say, it is equal to (V ‒ b), where b is volume occupied total number of moleucles present in 1 gm. mole of the gas and is known as the ‘limiting volume’. It can be calculated as follows : If the radius of the molecule (which is considered as non-compressible spherical particle) is ‘r’, the closest approach of the molecule can be represented as shown in the fig. The lower molecule excludes the centre of the upper molecule to be present in a space equal to the volume of the dotted sphere y n(2r)3 or 8 x | nr3. In the same manner, the upper molecule also excludes the vol of a 1 3 molecule =3 nr ![](AU B. Sc. I Chemistry-III(OK)_files/AU20B.20Sc.20I20Chemistry-III(OK)-41.png)Excluded Volume ^ (2r3) centre of the lower molecule from a space 8x|nr3. Thus, in the closest approach the two molecule mutually exclude a volume equal to 8 x | nr3. The volume excluded for one molecules is 4 x 4- nr3 or four times the molecular volume. Hence for 1 gm mole ofmolecule (Avogadro’s number) N = 6^02 x 1022, the excluded volume (limiting volume) denoted by ‘b’ becomes equal to 4Ne 3 nr2 j. Thus the correct volume Vi = (V-b). - (ii) Pressure correction : A molecule lying in the middle of the vessel (C), fig. is being uniformly attracted on all sides by the neighbouring molecules, thus neutralizing each other. However, a molecule which approaches the wall of the vessel (W) is unequally attracted by the molecules which are present on one side of it. Hence it will strike the wall with lower velocity and will exert a lower pressure, say p. Thus the corrected or ideal pressure, should be (P + p). ![](AU B. Sc. I Chemistry-III(OK)_files/AU20B.20Sc.20I20Chemistry-III(OK)-42.png) Now, the force of attraction exerted on a single molecule about to strike the wall depends upon the number of molecules per unit volume i.e., directly upon the density of the gas. Further, the number of molecules striking the wall at any given instant also depend directly upon the density of the gas. Thus the total inward attractive pull ‘p’ is proportional to the squares of the density of the gas. i.e., or p « d2 1 p ^ V2 (V = volume of 1 gm. mole) or a p = V2 (a = constant) Thus, the gas equation for one mole of a real gas become This is known as Van der Waals equation. The constants ‘a’ and ‘b’ are known as Van der Waal’s constants. Units of ‘a’ and ‘b’ In Vand der Waals equation units of ‘a’ and ‘b’ depend upon the units of P and V. We know then force of attraction if molecules due to pressure p « d2 but, for n mole gas, whole volum V and molecular weight, nM d=-- V Hence, P « nM IJ2 V KJ but M for a specific gas is permanent. Therefore, or P « ny V2 P = an2 V2 PV2 a = 2 n Hence unit of ‘a’ Pressure x (Volume) (mole)2 or atom litre2 mole2. The unit of ‘b’ will be the unit of volume i.e, litre/mole or cc/mole. (b) Calculate the values ‘a’ and ‘b’ for CO2 if Tc = 31°C, Pc = 72∙8 atoms and R = 0∙082 lit atm. per deg. per mole. Sol. : We know that, T = 8a/27Rb ......(i) P = a/27b2 .....(ii) By dividing eqn. (i) by (ii) Tc 8b or — = — Pc R b Tc x R .....(iii) or = Pc x 8 Given that, T c = 304Å, P = 72∙8, R = 0∙082 By putting the above values in eqn. (iii) 304x0■083 b= 72 ■ 8 x 8 = 0∙0428 litre Mole‒1 By putting the value of b in the following equation, 8a Tc 27Rb On solving the above equation, 27RbTc a = 8 .....(iv) \_ 27 x 0 ■ 082 x 0 ■ 0428 x 304 =8 = 3∙6 Atm. litre2 Mole‒2 Q. 23. (a) Explain enzyme catalysis with suitable examples. - (b) What are the general characteristics of enzyme catalysed reactions ? Ans. Enzyme Catalysis : Enzymes are complex nitrogenous organic compouds. They are produced in the living cells of plants and animals. These are actually high molecular weight protein molecules, Enzymes form colloidal solutions in water and are very effective catalysts. They catalyse numerous reactions, especially those connected with natural processess. Numerous reaction occur in the bodies of animals and plants to maintain the life process. These reactions are catalysed by enzymes. The enzymes are thus termed as biochemical catalyst and the phenomenon itself is knows as biochemical catalysis. Many enzymes have been obtained in pure crystalline state from living cells. However, the first enzyme was synthesized in the laboratory in 1969. The following are some examples of enzyme catalysis. - (i) Inversion of canesugar C12H12O11+ II O -■ - e >C6H12O6 + C6H12O6 (cane sugar) (glucose) (fructose) - (ii) Conversion of glucose into ethyl alcohol C6H12O6 zymase > 2C2H5OH + 2CO2 (glucose) (ethyl alcohol) - (iii) Conversion of starch into maltose - 2 (CfHioOs) n + n H2O > n C„H2!OU (starch) (maltose) - (iv) Conversion of maltose into glucose C12H22O11 + H2O maltose > 2C6H12O6 (maltose) (glucose) - (v) Decomposition of urea into ammonia and carbon dioxide : The enzyme ureas catalysis this decomposition. NH2CONH2 + H2O —urease 2NH3 + CO2 - (vi) Conversion of ethyl alcohol into acetic acid : The mycoderma aceti enzyme converts dilute solutions of alcohol into acetic acid and water. C2H5OH + O2 mycodermaaceti > CH3COOH + H2O - (vii) Conversion of milk into curd : It is an enzymatic reaction brought about by lactic bacilli enzyme present in curd. - (viii) In stomach, the pepsin enzyme converts proteins into peptieds while in intestine, the pancreas hysin converts protein into amino acids on hydrolysis. - (b) Characteristics of Enzyme Catalysts : (1) Enzymes form a colloidal solution in water and hence they are very active catalysts. - (2) Like inorganic catalysts they cannot disturb the final state of equilibrium of a reversible reaction. - (3) They are highly specific in nature, i.e., one catalyst cannot catalyse more than one reaction. - (4) They are highly specific to temperature. The optimum temperature of their activity is 35°C to 40°C. They are deactivated at 70°C. - (5) Their activity is increased in the presence of certain substance, known as co-enzymes. - (6) A small quantity of enzyme catalyst is sufficient for large change. - (7) They are destroyed by U. V. light. - (8) Their efficiency is decreased in presence of electrolytes. Q. 24. (a) How does the period of half charge depends upon the initial concentrations of the first and second order. - Sol. : t1/2 (half life period) and initial concentration : First order reaction and t1/2. For first order reaction we know, k = 2 • 303 t log10 a b a - x g The order of reaction is taken known from the following consideration : - (i) If half period is independent of concentration, it is first order reaction. - (ii) If half period is inversely proportional to initial concentration, it is second order reaction. 1 t x — a - (iii) It half period is inversely proportional to square of inital concentration, it is third order reaction. t x —2 a Where ‘a’ is the initial concentration. a 2\. when half of the original amount ‘a’ has decomposed, then x = Substituting it in equation for first order reaction, or k= t1/2 2 • 303 t1/2 log10 2 • 303 t1/2 log102 2 • 303 k log102 a 2 Since this equation does not involve the concentration of the reactant, hence t1/2 or “Half Life Period” of reaction is independent of the intial concentration. Second order reaction and t1/2. For second order reaction, 1a \---X—:----- 11/2 aba - xg At half change, x = a/2, then substituting it in the equation 1 k= t1/2 a/2 a/2 ###### F aI a a G a —ax — ###### H 2K 2 or or 11 = —x- t1/2 a 11 1 t 1/2 - 4 a or t 1/2 \* a Thus, the time required for half change in second order reaction inversely proportional to inital concentration of the reactants. (b) To study the rate of decomposition of H2O2, a same amount of reaction mixture was titrated against standard KMnO4 solution. The different data obtained are as follows : Time (minute) 0 10 20 30 Vol. of KMnO4 used (ml) 46∙1 29∙8 19∙6 12∙3 Prove that the decompostion of H2O2 is a first order reaction. Sol. : The rate constant of first order reaction is, k = 2 • 303 t log10 a b a - x g The volume of KMnO4 used, evidently, corresponds to the undecomposed H2O2. Hence the volume of KMnO4 used at zero times corresponds to the intial concentration ‘a’ and the volume used after time ‘t’, correspond (a–x) at that time. On keeping these values in the above equation, we get, [2•303,46](#bookmark300) t = 10 minutes; k = log10 = 0∙04385‒1 [1029](#bookmark301) [2•303,46](#bookmark302) t = 20 minutes; k = log10 = 0∙04295‒1 20 19 • 6 [2 • 303,46](#bookmark303) t = 30 minutes; k = log10 = 0∙04405‒1 [3012](#bookmark304) Since the value of k is approximately the same, it shows that the decomposition of H2O2 in aqueous solution is a first order reaction. Q. 25. (a) Derive the rate law constant for second order reaction. (b) In an experiment ethyl acetate is saponified by sodium hydroxide. The progress of the reaction is performed by titrating 25 c.c. of the reaction mixture with standard acid at different intervals. By taking equal concentrations of ester and alkali, the following data was obtained in the experiment : Reading 1 2 3 4 5 Time (minute) 0 5 15 25 55 Vol. of Acid 16 10∙24 6∙13 4∙32 2∙31 (a – x) c.c. Prove that the order of reaction is two. How much amount of ethyl acetate would have been decomposed after 35 minutes ? Ans. (a) In a second order reaction, the minimum number of molecules required for the reaction to proceed is two i.e., 2A ^ Product or A + A ^ Product Suppose ‘a’ gm. mole be the initial concentration of each reactant and (a – x) gm. mole be their concentration after any time ‘t’. Then we have or dx — — k ( a—x ) ( a — x ) dt = k (a – x)2 (Law of mass action) dx ? = kdt (a-x)2 (1) In order to get the value of k, on integrating this expression (i), we have or i — \[kdt J (a - x)2 J or —1— = kt + C (C = integration const.) ( a - x ) When t = 0 and x = 0. ......(2) Substituting these values in the above equation (2), we get 1 = C a ....(3) On substituting the value of C in eqn. (2), we have 1 1 \-------= kt + ( a - x ) a or 11 \-----— — — kt (a - x) a ( a - a + x ) , x , - \---------i = kt or --------= kt a(a – x) a(a – x) or 1x \- x-------= k t a( a – x) ...(4) The equation (4) is known as the equation for a second order. - (b) The reaction being bimolecular the equation by using equivalent concentration of the reactant is, 1x k =--- t a(a – x) In this case the volume of acid used corresponding to amount of unused NaOH. Thus, the volume of acid used at 0 (zero) minute i.e., at the start of the reaction corresponds to initial concentration a and the volume used after time t corresponds to a – x at the time. Therefore here a = 16 - (i) After 5 minutes the value of (a – x) is 10∙24 c.c. and the value on x is 16 ‒ 10∙24 = 5∙76 c.c. on substituting these values in the equation, we have 1 5-76 k = — x--------= 0 ■ 00703 15 16 x 10 ■ 24 - (ii) After 15 minutes, the value of (a – x) is 6∙13 c.c. and the value of x is 16 ‒ 6∙13 = 9∙87 c.c. On substituting these values in the equation, we have 1 9-87 k = —x-------= 0 - 00673 15 16 x 6-13 - (iii) After 25 minutes the value of (a –x) is 4∙32 c.c. and the value of x is - 16 ‒ 4∙32 = 11∙68 c.c. On substituting these values in the equation, we have r 1 11 - 68 k = — x--------= 0 - 00675 25 16 x 4 - 32 - (iv) After 55 minutes, the value of (a – x) is 2∙31 c.c. then value of x is - 16 ‒ 2∙31 = 13∙69 c.c. On substituting these values in the equation, we have, k = \_L x 13'69 = 0■00673 55 16x2■31 Similar values of k in the above calculations are family constant, the reaction is of t second order. The average value of k + 0∙00681. The amount of ethyl acetate decomposed after 35 minutes is calculated as follows : 1x k = - x------ t a( a – x) 0 • 00681 = — X---x--- 35 16(16 – x) or 0 • 00681 = 35 x 16------- (16 – x ) or 3∙8136 (16 ‒ x) = x or x = 12∙63 c. c. After 35 minutes the amount of ethyl acetate decomposed is 12∙63 c.c. Q. 26. (a) Discuss an expression for the first order rate constant of a reaction. - (b) Give Integration and Equifraction methods employed for the determination of order of reaction. Ans. (a) Consider a general unimolecular (first order) reaction. A--> Product Suppose ‘a’ is the initial concentration of A of which ‘x’ reacted in time so that (a ‒ x) gm/moles remain unreacted. A--> Product a 0 ...before reaction a - x a ...after time ‘t’ Applying the law of mass action, the reaction rate, dx t x( a -x) or dx — = k ( a - x ) dt or dx (a – x) = k dt ...(1) Integrating the above equation (1) f d' = fkd( J ( a — x ) J or ‒loge (a – x) = kt + c ...(2) (where ‘c’ is integration const.) When t = 0, x = 0, the eqn. (2) becomes ‒loge a = c Substituting the value of ‘c’ in eqn. (2) we have ‒loge (a – x) = kt ‒ loge a or loge a ‒ loge (a – x) = kt or log e -a- = ktx a–x or 1a ...(3) k = -log e----- t a–x , 2•303, a k = —-—logic--- t a–x \[loge x = 2∙303 x log10 x\] This is known as the equation for first order. - (b) Determination of order of reaction : - (1) Integration Method. In this method known quantities or reactions are mixed and the progress of the reaction is determined by analysing the reaction mixture after regular intervals. The experimental values are substituted in the velocity coefficient equation of first, second and third orders. The order of reaction is then known by that equation which gives constant value of K. 2 • 303i - (1) k = —t— logic a a–x 1a - (11) k = - x----- t a–x (Ist order reaction) (IInd order reaction) (iii) k = — x 2t x(2a – x) 22 a ( a – x ) (IIIrd order reaction) - (2) Equifraction method or half-change method. In this method the time at which one half of the original substance has disappeared is determined and the experiment is repeated with different initial concentration of the reactants. Q. 27. (a) What do you mean by temperature coefficient and activation energy of a reaction ? Derive an expression for determining the activation energy of a reaction. - (b) The value of rate constant for this reaction is 3∙46×10‒5 at 25ºC. N,O. ^ NO. + 10, 25 24 2 2 and 4^87 x 10-3 at 65°C. Calculate the activation energy for the reaction. Ans. (a) Temperature coefficient. The rate of a reaction increases considerably with the increase in the temperature. The temperature coefficient of a chemical reaction is defined as the ratio of the velocity constant of a reaction at two temperatures separated by 10 degrees centigrade, usually 25 degree and 35 degrees centigrade. Temperature coefficient ‒ K35º / K25º If the logarithm of reaction velocity be plotted against the reciprocal of absolute temperature, the various points obtained are all on a straight line for most of the chemical reactions. This variation, therefore, can be represented by the equation, a - b log = T Where K is the specific reaction rate, T is the absolute temperature and ‘a’ and ‘b’ are the positive empirical constants, the values of which can be obtained from the intercept and slope respectively of the liner plot. The exponential form of the above equation known as Arrhenius equation and is, k = Ae‒Ea/RT ...(i) Where, A is called the Arrhenius factor and Ea is the Energy of action, which represents the minimum energy required for the collision between molecules to be effective and determines the influence of temperature on reaction velocity. On taking logarithms. equation (ii) may be written as : -E„ ..(ii) log k = or + lOgeA RT Differentiating with respect to temperature, we get, or d log10 k dT d loge k dT Ea RT2 Ea 2 ■ 303RT2 ....(iii) Intergrating the above equation between the limits K = K1 at T = T1 and K = K2 at T = T2 we have, - k, Ea F T - T log 2 a 2 1 ....(iv) - 10 k1 2 ■ 303R H T1T2 It is clear from equation (iv) that if the two values of K are known at tow different temperatures, Ea can be calculated. The energy of activation can also be calculated by plotting a graph between log K and 1/T, the graph plotted in this way is simple line whose slope EE a or 2 ■ 303 4■576. After knowing slope, E can be calculated - (b) Solution : It is given that k 1 = 3-46 x 10 5 T1 = 273 + 25 = 298 K k2 = 4^87 x 10"3 T2 = 273 + 65 = 338 K On putting the above values in the following equations : k log 2 = k1 E a x F T2-T11 2 ■ 303 R H T1 x T2 J =\_\_\_\_\_\_\_\_\_Ea\_\_\_\_\_\_\_\_\_G 338K-298K J (2 ■ 303) (8 ■ 314JK-1 mol-1 H298K x 338K = 103584 J Mol‒1 = 103∙584 kJ Mol‒1 Q. 28. Explain the modern Adsorption theory of Catalysis. Ans. Adsorption Theory : This theory explains the mechanism of heterogeneous catalysis. The old point of view was that when the catalyst is in solid state. The reactants are in gaseous state or in solution, the molecules of the reactants are absorbed on the surface of the catalyst. The increase in concentration of the reactants on the surface influences the rate of reaction (law of mass action). Adsorption being an exothermic process, the heat of adsorption is taken up by the surface of the catalyst which is utilized in enhencing the chemical activity of the reacting molecules. The view does not explain the specificity of a catalyst. Adsorption is broadly ot two types : physical and chemical. The chemical adsorption is specific and involves chemical combination on the surface of the catalyst. The modern adsorption theory is the combination of intermediate compound formation and the old adsorption theory. The catalytic activity is locaslised on the surface of the catalyst. The mechanism involves five steps : - (i) Diffusion of the reactant to the surfact of the catalyst. - (ii) Some form of association between the catalyst surface and the reactants ocurs. This is assumeb to be adsorption. —O—O—O— A —O—O—O—A \+ B —O—O—O— —O—O—O—B [I II](#bookmark305) Catalyst surface having Adsorption of reacting free valencies molecucles [I](#bookmark306) —O—O—O— —O—O—O—A \+ AB —O—O—O— —O—O—O—B [I](#bookmark307) Product Activated Complex Fig. Adsorption of Reacting Molecules, Formation of a Activated Complex and Desorption of Products - (iii) Occurrence of chemcial reaction on the catalyst surface. - (iv) Desorption of reaction products from the catalyst surface. - (v) Diffusion of reaction products away from the catalyst surface. The catalyst surface is a seat of chemical forec of attraction. There are free valencies on the surface of the catalyst. When a gs comes in contact with such a surface,its molecules are held up there due to loose chemcial combination. It different molecules are adsorbed, side by side, they may react and new molecules so farmed may evarporate leaving the way for the forest reactant molecules. In case free valencies are responsible for the catalytic activity, it follows that with the increase of these valencies on the surface of a catalyst, the catalytic acitivity will be greatly enhanced. The free valencies can be increased in the following two ways : - (i) By Sub-division of the Catalyst : Number of free valencies increases on disintegration. Finally powdered or colloidal catalyst particles having large surface area are very rich in free valencies. ...Ni—Ni—Ni... ...Ni ...Ni... ...Ni... Ill I I I ...Ni—Ni—Ni... ------► ...Ni... + ...Ni... + ...Ni... III I I I ...Ni—Ni—Ni... ...Ni... ...Ni... ...Ni... 12 free valencies 24 free valencies Fig. Increase in No. of Free Valencies Actually it is observed that finally divided nickel and colloidal platinum act as efficient catalysts. - (ii) By Rough Surface of the Catalyst : There are a number of active spots in the form of edges, corners, crackes and peaks on the rough surface. They give rise to an increase in number of free valencies. These active spots enhance the adsorption and thereby increases the catalytic efficiency of the catalyst. The adsorption theory explains the following facts of heterogeneous catalysis : - 1\. The surface of the catalyst is used again and agian due to alternate adsorption and desorption. Thus, a small quantity of catalyst can catalyst large amounts of reactants. - 2\. Chemcial adsorption depends on the nature of the adsorbent and adsorbate. Hence, catalysts are specific in action. - 3\. The energy of adsorption compensates the activation energy of the reacting molecules to some extent. Thus, reaction occurs in faster rate. - 4\. Greater efficiency of the catalyst in finally divided state and rough surface. - 5\. It adequately explains the poisoning of catalysts. The poisons are preferentially adsorbed at the active centres of the catalyst. The effect reduces the free valencies for the reacting molecules and thus, the catalytic activity decreases. - 6\. Promoters are responsible for increasing the roughness of the surface of the catalyst. This effect increase the free valencies for the reacting molecule and thus, the catalytic activity of the catalyst in increased. B. Sc. I, Chemistry III 121