Ebook chemistry basis elements part 1

First Edition, 2009 ISBN 978 93 80168 56 © All rights reserved Published by: Global Media 1819, Bhagirath Palace, Chandni Chowk, Delhi-110 006 Email: globalmedia@dkpd.com Table of Contents The Equilibria Chemical Equilibrium Surface Phenomenon Bioinorganic Chemistry Bio-organic Chemistry The Photochemistry Quantum Chemistry Analytical Chemistry Nuclear Chemistry 10 Organic Name Reactions 11 Reagents in Organic Synthesis 12 Kinetics of Reactions 13 The Macromolecules 14 Fast Reactions 15 Conformational Analysis The Equilibria The Equilibria The E°cell of an aluminium-air battery is 2.73 volts and it involves a 12 electron process The free energy change (DG°) of the battery in kJ Calculated : DG = _ nFE° = _ 12 × 96500 × 2.73 J = _ 3161340 J = _3161.34 Id For the following reaction— H2(g) + C12 (g) 2HCl (g) DG° is _ 262 kJ Calculate the equilibrium constant (K) for the reaction at 298 K DG° = _ 262000 J = _2.303 RT log K log K = = 45.918 K = 8.279 × 1045 Hydrogen-oxygen Fuel Cell Two half-cells of hydrogen-oxygen fuel cell under basic conditions can be depicted as OH_/O2 (g)/ Pt and OH_/H2 (g)/ Chemistry : Basic Elements Pt and their standard electrode potentials at 25°C are 0.4009 and _ 0.8279 V respectively Write the half cell reactions and the complete cell reaction Depict the complete cell and the e.m.f of the cell Calculated : _Pt |H2(g)| OH_| O2(g)| Pt + At anode, the reaction is H2 + 2OH_ ® 2H2O + 2e_ At cathode, the reaction is 2e_ + The cell reaction is H2 + O2 + H2O ® 2OH_ O2 ® H2O The e.m.f of the cell is = 0.4009 _ (_0.8279) = 1.2288 V The reduction potentials of Ag+/Ag and Fe+3/Fe+2 are 0.799 and 0.771 V respectively The equilibrium constant of the reaction Ag + Fe+3 Fe+2 + Ag+ Ag | Ag+ || Fe+3 · Fe+2| Pt(+) E°cell = 0.771 _ 0.799 = 0.028 V At (_ ), Ag ® Ag+ + e_ At (+), Fe3+ + e_ ® Fe+2 Cell Reaction : Ag + Fe3+ ® Ag+ + Fe2+, Here n = log K = = = = _ 0.47329 K = 0.33628 The EMF of the cell, Pb | PbCl2 || AgCl | Ag at 300 K is 0.50 V If the temperature coefficient of EMF is _ × 104 volt deg_1, DH and DS for the cell reaction Calculated : Pb + 2AgCl ® PbCl2 + 2Ag Pb | PbCl2| Cl_ | AgCl | Ag + The cell reaction is Pb + 2AgCl ® PbCl2 + 2Ag The Equilibria DG = _ nFE = _ × 96500 × 0.5 J = _ 96,500 J or _ 96.5 kJ DS = = × 96500 (_2 × 10_4) = _ 38.6 JK_1 DG = DH _ TDS, _ 96,500 = DH _ 300 (_38.6) DH = _ 96,500 _ 300 × 38.6 = _ 108,080 J or, _108.08 kJ The potential of pentane/oxygen fuel cell given that the standard free energies of formation (in kJ/mole) at 298 K are _ 8.2, _ 237.2 and _ 394.4 for pentane, H2O (1) and CO2 (g) respectively Calculated : The chemical reaction that takes place in the pentane-oxygen fuel cell is C2H12 (g) + O2 (g) ® CO2 (g) + 6H2O (1) DG° for the reaction = [5(_ 394.4) + (_ 237.2)] _ [_ 8.2] = _ 1972 _ 1423.2 + 8.2 = _ 3387.0 kJ The electrode reactions are C5H12 + 10 H2O ® CO2 + 32 H + + 32 e_ O2 + 32 H+ + 32 e_ ® 16 H2O DG° = _ n FE°, DG° = _ 3387 × 103 J, n = 32 E° = = 1.0968 volts Potential of Hydrogen-electrode Taking the case of hydrogen electrode, consisting of H2 gas Chemistry : Basic Elements in equilibrium with H+ ions, the electrode reaction, written as reduction reaction is H + + e_ ® H2 (n = 1) Applying Nernst equation, the electrode potential of the hydrogen electrode is given by EH+, H2 = E°H+, H2 _ If H2 gas is at atmospheric pressure, aH2 = \ EH+, H2 = E°H+, H2 Replacing the activity of H+ ions by its molar concentration, we have E°H+, H2 = E°H+, H2 + Since the standard electrode potential of hydrogen electrode is taken as zero E°M+, H2 = Hence EH+, H2 = = 0.0591 log [H+] at 25° C The following cell is used to measure the mean activity coefficient (Y+) of HCl Pt | H2 (g) | HCl (aq) | AgCl (s) | Ag (s) (i) The cell reaction, (ii) activities of H+ and Cl_ ions express the emf of the cell, (iii) the activities in terms of molality of HCl and the mean ionic activity coefficient, obtain an expression for In Y+ in terms of the emf The cell is Pt | H2 (g) | HCl (aq) | AgCl (s) | Ag (s) i.e., The Equilibria it consist of hydrogen and silver-silver chloride electrodes in HCl as the electrolyte The cell reaction will be H2(g)+AgCl (s) ® Ag (s) + H+ (aH+) + Cl_ (aCl_) and the EMF of the cell is Ecel = E°Ag|AgCl _ = (1) where a is the activity of HCl as a whole Now as the activity (a) of HCl at any molality m is related to the mean activity coefficiency Y+ by the expression a = m2Y2+ , substituting this value in eq (1) we get Ecel = Ecel = or, Ecell + = (2) All the quantities on the left hand side of equation (2) are known experimentally Hence to calculate Y±, the value of E°Ag|AgCl is required To determine E°Ag|AgCl the quantity is plotted against m and the result is extrapolated to m = 0, when m = 0, Y+ = 1, therefore from equation(2) E°Ag|AgCl = , i.e., equal to the value of the ordinate The mean ionic activity coefficient of 0.1 molar HCl at 25° C given that the EMF of the cell Calculated : Chemistry : Basic Elements H2 (1 atm) | HC1 (a), AgCl (s) | Ag is 0.3524 V at 25° C and that the standard electrode potential of Ag-AgCl is 0.2224 V at 25°C For the given cell = or, = Putting T = 298 K, R = 8.314 JK_1 mol_1 and F = 96500 C, this equation becomes Ecell + 0.1183 log m = E°Ag|AgCl _ 0.1183 log Y+ Substituting Ecell = 0.3524 V E°Ag|AgCl = 0.2224 V and m = 0.l mol kg_1 0.3524 + 0.1183 log 0.1 = 0.2224 _ 0.1183 log Y± or log Y± = = _ 0.0989 or Y+ = 0.796 The standard reduction potentials of the electrodes Fe+3 | Fe and Fe+2 | Fe are _ 0.035, _ 0.440 V respectively It is easy to oxidise Fe to Fe+2 or Fe to Fe+3 Since the standard reduction potentials are given the standard oxidation potentials will be + 0.036 V and + 0.44 V The standard oxidation potential of Fe | Fe+2 is more positive than that of Fe|Fe+3 electrode So it is easy to oxidise Fe to Fe+2 (a) Write sell electrode for the following reaction Cu (OH)2 (s) ® Cu2+ + 2OH_ (b) Write the cell reaction for the following cell Pt | H2 | HCl | Hg2Cl2 | Hg | Pt The Equilibria (a) Half-cell reactions are Cu(OH)2 (s) + 2e- ® Cu + 2OH_ Cu2+ + 2e_ ® Cu (s) and electrodes are Cu2 + | Cu and Pt | Cu(OH)2 | OH_ (b) The electrode reactions are H+ + e_ ® H2(g) Hg2Cl2 + 2e_ ® 2Hg (1) + 2Cl_ to the H+ form for reuse Normally this might be done by washing the column with 3M H2SO4 but 6M HCI is found to be superior M HCl is found to be superior in this case because it forms tetra chlorocuprate (II) (2 ) and trichlorocuprate(II) (1 ) species with copper(II) which will not bind to the cation exchange sites In multiple extraction of solute from aqueous to organic phase, if the volume of the organic solvent used each time remains constant and equal to the volume of aqueous phase, the weight unextracted after n such operations is : and weight extracted in n operations is Suppose we have a solution containing `m' gm of a substance in `V' ml of solution This solution is repeatedly extracted using every time `v' ml of organic solvent which is immiscible with the first Suppose `ml' be the amount of solute that remains unextracted at the end of first operation Then the distribution coefficient may be written as Analytical Chemistry (1) or m1 = m (KV/KV + v) (2) Exactly in the similar manner, amount m2 that remains unextracted at the end of second extraction, (3) or, m2 = m (KV/KV + v)2 (4) Therefore the weight that remains unextracted at the end of n operations will be m2 = m (KV/KV + v)n and the weight extracted in `n' operations will be given by m mn = m [1 (KV/KV + v)2] Note : The efficiency of extraction increases by increasing the number of operations using only a small quantity of the extracting liquid at each time Synergistic Extraction The phenomenon in which two reagents when used together, extracted a metal ion with enhanced efficiency compared to their individual action is called synergism A common form of synergistic extraction is that in which a metal ion, Mn + is extracted by a mixture of an acidic chelating reagent, HR and an uncharged basic reagent, S The joint action of the reagents is especially pronounced in those cases where the coordination capacity of the metal ion is not fully achieved in the MRn chelate, then the extractant S gives a mixed complex, MRn Sx which is extracted with much greater efficiency than the parent chelate An example is the synergistic influence of zinc in the extraction and AAS determination of trace cadmium in water Chemistry : Basic Elements The weight of Fe(III) left unextracted from 100 ml of a solution having 400 mg of Fe3+ in 6M HCl after two extractions with 25 ml of diethyl ether : D = 150, Calculated : Amount left unextracted, where, W = amount of solute, Vw = volume of aqueous solution, D = 150, V0 = volume of organic phase \ In the extraction of cerium(IV) with 2-thenoyl trifluoroacetone in benzene, the distribution ratio was 999 The volume of organic phase was 20 ml and that of aqueous phase 50 ml, what was the percentage extraction : D = 999, V0 = 20 ml, Vw = 50 ml, Assume E = x is percentage of extraction The electronic transitions will have the lowest energy : n ® p* transition will generally have the lowest energy The effect of addition of Co2+aq to the equilibrium reaction is : Co2+(aq) + 4NCS ® [Co(NCS4)]2 (Pale Pink) (Colourless) (Blue) on the % transmittance of the solution As Co2 + is added to the equilibrium mixture, the concentration of the complex ion in the mixture increases The percentage of absorption by the complex increases and transmittance decreases Analytical Chemistry The Matched : Energy changes involved Region in electromagnetic spectrum (1) Nuclear (a) Infrared (2) Inner shell electrons (b) Radio waves (3) Valence electrons (c) Visible (4) Molecular vibrations (d) Gamma rays (5) Spin orientation (e) X-rays (1)(d), (2)(e), (3)(c), (4)(a), (5)(b) By the use of potential energy curves for ground and excited electronic states, the sequence of steps leading to phosphorescence and fluorescence shown : Raman and Rayleigh scatter can be identified when included in a fluorescence emission spectrum When excitation wavelength is varied, Rayleigh Scatter will vary exactly as the excitation, whereas Raman shift is a constant energy shift Any fluorescence peaks vary only in intensity with the wavelength maxima remaining unchanged Chemistry : Basic Elements After insertion or injection of the sample into the electrothermal atomizer, the temperature profile with ramp heating drown : See the Figure Two advantages and a major limitation of each of the following methods-(a) Flame AAS, (b) electrothermal AAS, (c) FES and (d) AFS, stated : Technique Advantages Limitation Flame AAS (i) Relatively inexpensive Analyte must be in (Atomic absorption, solution spectroscopy) (ii) Fast technique Electrothermal AAS (i) Lower detection limits Less precision, slower (ii) Handles solid samples directly FES (Flame emission (i) It complements AAS It is limited to alkali spectroscopy) and alkaline earth metals (ii) It is fast and inexpensive AFS (Atomic fluore- (i) Good for trace analysis No instrumentation scence spectrometry) is made commercially (ii) It is faster than electrothermal AAS Acetylene-nitrous oxide flame is suitable for elements such as Al, Be and rare earths: These elements form stable oxides The temperature of Analytical Chemistry acetylene-nitrous oxide flame is about 3000° C and it has an advantage that its burning velocity is almost same as that of air-acetylene Consequently it can be easily and safely handled in the laboratory Aqueous solution of a drug (0·10 milli molar) shows a percentage transmission of 50 in a cm cell at 250 run The molar absorptivity, calculated : According to Beer's law A = log = ecl where T is transmittance, c is concentration in moles per litre, l is path length in cm, e is molar absorptivity \ = e × × 104 moles/lit × cm 0·3010 = e × × 104 × e = 0·3010/1 × 104 = 3010 The ratio of number of sodium atoms in 3p excited states to the number in the ground state at 2500 K calculated The average wavelength of the two sodium emission lines involving the 3p to 3s transition is 5892 Å : = 1/5892 Å × 108 cm = 1.698 × 104 cm1 Ej = 1·698 × 104 cm1 × 1·986 × 1016 erg cm1 = 3·372 × 10 12 ergs (since h × c = 1·986 × 1016 erg cm1) 3s and 3p levels have and quantum states respectively Pj / P0 = 6/2 = Hence or, =1·7 × 104 Chemistry : Basic Elements The temperature at which plasma sources operate stated The region where plasma source produces a great number of excited emitted atoms is : Between 7000 and 15000 K In the ultraviolet region plasma source produces a great number of excited emitted atoms The difference between emission and excitation fluorescence spectra is : In emission, excitation is carried out at a fixed wavelength and the emission intensity records as a function of wavelength Excitation fluorescence spectra are obtained by measuring the fluorescence intensity at a fixed wavelength while the wavelength of the exciting radiation is varied In the determination of selenium in urine by AAS it was found that Beer's law applied up to the ppm range If the ppm standard was set at 100 and a blank set at zero The selenium concentration of the sample to read at 73 would be: Selenium concentration of urine = The advantages of molecular fluorescence methods compared to the corresponding absorption spectrophotometric methods and how they arise are : Fluorescence is caused by the absorption of radiant energy and re-emission of some of this energy in the form of light The advantages of molecular fluorescence methods over absorption spectrophotometric methods are (1) The sensitivity of fluorescence methods is generally 10 to 103 times more than the sensitivity of absorption methods (2) These methods have greater specificity and selectivity than absorptiometric techniques This is due to the fact that in luminescence both the wavelength of excitation and emission are selected whereas in absorptiometry only the wavelength for absorption is selected Analytical Chemistry The g-value of the methyl radical whose centre of ESR spectrum appears at 329.4 mT in a spectrometer operating at 9·233 G Hz (h = 6·627 × 1034 Js, electron Bohr Magneton = 9·27 × 1024 JT1) is The number of lines in the ESR spectrum of this methyl radical predicated: or, Hence g = 2·0048, The interaction of an unpaired electron with three equivalent protons as in CH3 radical gives four equally spaced lines according to (n + 1) rule The Polarizability The case with which the electron distribution in a molecule or atom may be distorted by the application of an electric field The basis of turbidimetric and nephelometric analysis is: Measurement of the intensity of the transmitted light as a function of the concentration of the dispersed phase is the basis of turbidimetric analysis and measurement of the intensity of the scattered light (at right angles to the direction of incident light) as a function of the concentration of dispersed phase constitute the basis of nephelometric analysis In the determination of manganese by atomic absorption method, a solution containing an aliquot of unknown strength gives a metre reading of 23 Another sample containing this solution with added 50 ppm of manganese solution gives a metre reading of 92·5 Each reading was corrected for the background The concentration of manganese in the original solution calculated : Chemistry : Basic Elements As absorbance is directly proportional to the concentration of manganese \ 23 = mx, where `x' ppm is concentration of Mn 92·5 = m (x + 50) 92·523 = 50 m or, m = 1·39 and x = = 16·5 ppm Amperometric titration is a better method than polarographic method for quantitative analysis Amperometric titration is a quick, accurate and convenient method similar to potentiometric and conductometric titration and at the equivalence point there is sharp change in diffusion current The galvanometer used need not be calibrated The specific characteristic of capillary does not influence the titration No polarising unit is used, suitable half cell can be easily used for the purpose The Match: List I List II Quantity Measured & Variable Controlled Name of method (1) E, i = (a) Linear potential sweep voltammetry (2) E vs volume of titrant, i = (b) Amperometric titrations (3) i vs E, concentration (c) Ion selective potentiometry (4) i vs E, t (d) Conductometric titrations (5) i vs volume of titrant, E (e) Voltammetry (6) 1/R vs volume of titrant (f) Potentiometric titrations (1)(c), (2)(f), (3)(e), (4)(a), (5)(b), (6)(d) The amperometric titration of iodine with sodium thiosulphate using two indicator electrons (a) The amperometric titration curve and correlate the several regions on this curve with the corresponding regions on the current potential curves sketched Analytical Chemistry (b) This titration curve change if excess iodide ion were initially present would : (a) (b) Using equivalent conductance values, the general form of the titration curve in (a) titration of Ba(OH)2 with HCl, (b) titration of NH4Cl with NaOH sketched : Chemistry : Basic Elements Titration of electroactive substance A with another electroactive substance B was carried out at the potential at which both A and B are reducible at cathode The current vs volume of titrant B curve and label the diagram accordingly sketched : In this titration the current will drop to the end point and there will be increase again to give a V-shave titration curve The residual and limiting current is : Residual or condenser current flowing in a polarographic cell is the sum of the Faradic current (if) and the charging current (ic) Thus Residual current (ir) = if + ic Analytical Chemistry The limiting current (il) is the sum of the diffusion current (id) and the residual current (ir) Thus il = id + ir The merits of cyclic voltammetry are : Merits are in the realm of qualitative or diagnostic experiments A cyclic voltammogram is to the electrochemist what the frequency domain spectrum is to the spectroscopist It gives a picture of a substance electrochemical behaviour while varying the voltage (energy) of an indicator electrode Cyclic voltammetry is capable of rapidly generating a new oxidation state during the forward scan and then probing its fate on the reverse scan Peak height is related to both concentration and reversibility of the reaction Thus it yields information about reaction reversibilities and also offers a very rapid means of analysis for suitable systems This method is particularly valuable for the investigation of stepwise reactions and in many cases direct investigation of reactive intermediates is possible By varying the scan rate, systems that exhibit a wild range of rate constants can be studied and transient species with half lives of milliseconds are readily detected Matched : (1) U.V spectroscopy (a) DC Arc (2) FT IR (b) Interoferometer (3) AES (c) Xenon flash lamp (4) AAS (d) Thermal conductivity detector (5) Gas chromatograph (e) Hollow cathode lamp (6) Fluorescence & (f) Deuterium discharge lamp Phosphorescence spectro photometry (1)(f), (2)(b), (3)(a), (4)(e), (5)(d), (6)(c) The full form of the techniques mentioned below and the quantity measured by them given DSC, DTA, TGA, TMA, DMA, EGA Chemistry : Basic Elements (1) DSCDifferential Scanning Calorimetry (Heats and temperatures of transitions and reactions) (2) DTADifferential Thermal Analysis (Temperature of transitions and reactions) (3) TGAThermo gravimetric analysis (weight change) (4) TMAThermomechanical analysis (Dimension and viscosity changes) (5) DMADynamic mechanical analysis (Modulus, damping and viscoelastic behaviour) (6) EGAEvolved gas analysis (Amount of gaseous products of thermally induced reaction) The probable thermal decomposition patterns of CaC2O4·H2O deduced using the thermogram (obtained in air) given below (Atomic weights of Ca, C and O may be assumed to be 40, 12 and 16 respectively) The probable thermal decomposition of CaC2O4·H2O according to above thermogram will be CaC2O4·H2O ® CaC2O4 + H2O CaC2O4 ® CaCO3 + CO CaCO3 ® CaO + CO2 Analytical Chemistry A purity of a compound can be checked by Differential Scannig calorimetry (DSC) this way: In cases where Van't Hoff equation is applicable and material is pure about 97 to 100% a single endotherm is obtained using a very small sample and heating at the rate of about 1·25° C/min The curve so obtained is integrated to the vertex by parts and the temperatures are tabulated with the partial areas upto that temperature The following equation is employed, where Ts is the instantaneous temperature of the sample in K, To·melting point of the infinitely pure sample (solvent) in K, R = gas constant, DH = Heat of fusion of sample, F = fraction of the total sample melted at Ts Now a plot of temperature as a function of the reciprocal of the fraction melting, 1/F to that temperature is made which should give a straight line of slope KT02/DH with an intercept of T0 AH can be determined from the DSC curve The three types of automatic chemical analyzers with respect to the mode of operation, advantages and disadvantages compared : Analyzer Mode of Operation Advantages Disadvantages (1) Discrete Batch, serial proce- Runs selected test on Expensive prepackaged ssing of sample limited number of reagents and cells, samples, quick start up mechanically complex (2) Continuous Continuous pro- Fast, simple equipment, Difficult to vary test flow cessing of large reagent cost minimised doing a series of number of sam- samples, tubing is ples for same critical series of tests (3) Centrifugal Sample mixed Fast, all samples and Loading samples is with reagent by controls measured under time consuming, can centrifugal force the same conditions run only a limited number of tests ... 0 .15 1 V AgI (s) ® Ag+ + I_ The standard emf of the cell is E° = E°R _ E°L = _ 0 .15 1 _ 0.79 91 = _ 0.95 01 V We know that log Ksp = = _ 16 .1 Ksp = 10 _16 .1 = 7.94 × 10 _17 At 25° C, (ảE/ảT)p = _1. 25... = (2 mol) (96,485 C mol _1) ( _1. 25 × 10 _3 VK _1) = _ 241JK _1 We also know that DH° = _ nF DH° = (2 mol) (96,485C mol _1) [ (1. 36 V _ 298 .15 K) (1. 25 × 10 _3 VK _1) ] DH° = _ 19 1 kJ The Equilibria The... reversible T = 11 1+273 = 384 K _ W = PDV = P(VV_ V1) » PVV = RT ( VV>> VI) = (8. 314 JK _1 mol _1) × (384 K) = 319 3 J mol _1 Chemistry : Basic Elements qp = DHvap = (363.3 Jg _1) × (92 g mol _1) = 33423.6