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S MOLE CONCEPT N   1 gram atom = NA atoms = 6.023 × 1023 atoms L...

S MOLE CONCEPT N   1 gram atom = NA atoms = 6.023 × 1023 atoms L = Gram atomic mass A  1 gram molecule = NA molecules = 6.023 × 1023 molecules = Gram molecular mass W  Mass % of an element  Molecular Mass  Average1relative mass of one molecule  mass of C  12 atom 12 Mass of that element in the compound ×100 RN = Molar mass of the compound  Molecular mass = 2 × VD wt. of metal  The value of n can be obtained by the following  Eq. wt. of metal  wt. of H displaced  1.008 relationship 2 A Molecular mass wt. of metal n Empirical formula mass  Eq. wt. of metal  wt. of oxygen combined  8 K  Normality (N) wt. of metal   35.5 wt. of chlorine combined Gram equivalent of the solute W  1000  A = , Volume of the solution in litre GEM  V in mL  Molecular formula = (Empirical formula)n where GEM is gram equivalent mass of solute. SH  1 amu = 1.66 × 10–24 g (amu - atomic mass unit) Atomic mass  Equivalent mass of an element = Valency w  n= M N Molecular mass where w is weight of substance and M is molar  Equivalent mass of an acid = Basicity mass of substance, n is number of moles A given particles Molecular mass  n= where n is number of moles  Equivalent mass of a base = Acidity 6.022  1023 K Given volume  Equivalent mass of a salt  n= 22.4 lit at STP where n is number of moles A Formula mass = Total +ve or  ve charge  Average atomic mass (RA At.mass)1  (RA At.mass)2  Equivalent mass of an oxidising agent = RA(1)  RA(2) Molecular mass where RA is relative abundance. = Total change in oxidation number  Molarity × GMM (solute) = Normality × GEM  No. of gram equivalents (solute), where GMM is gram molecular mass. Weight of the solute(in g)  Normality and molarity equations : = Equivalent weight of the solute N1V1 = N2V2  No. of milliequivalents M1V1 = M2V2 (For dilution) N Weight of the solute(in g) M 1 V1 M 2 V2 = Equivalent weight of solute × 1000  L n1 n2  Strength of a solution A (For reaction where n1 and n2 are no. of moles of the two reactants in a balanced chemical Wt. of the solute (in g) equation) = Vol. of solution (in litres) W M3(V1 + V2) = M1V1 + M2V2  Parts per million (ppm) of substance A (ppm) (Final molarity on mixing two non-reacting RN solutions) Mass of A = × 106 or  Number of millimoles = Molarity × V in mL Mass of solution  Number of equivalents = Normality × V in L = Vol. of A × 106 Vol. of solution  A Number of milliequivalents M = Normality × V in mL  Molality (m) = MM 2 or K   Number of gram atoms or mole of atoms 1000 Mass of element in gram = m A Gram atomic mass Molarity (M) =  mM 2  1   1 mole = mass of 6.023 × 1023 particles (atoms/  1000  SH molecules) where M2 = molecular mass of solute,  1 mole atoms = Gram atomic mass (or 1 g atom)  = density = 6.023 × 1023 atoms n1 N  1 mole molecules = Gram molecular mass (or  M (n 1M 1  n 2 M 2 ) /  1 g molecule) = 6.023 × 1023 molecules A = 22.4 L at STP Here, n1M1 = mass of solute, n2M2 = mass of solvent  1 mole ionic compound = Gram formula mass K = 6.023 × 1023 formula units i.e., n1M1 + n2M2 = mass of solution. A  W 1. ATOMIC STRUCTURE No. of subshells in main shell = n  Nodes (n–1) = total nodes, l = angular nodes, 2. Total no.of orbitals in main shell = (n)2 (n – l –1) = Radial nodes 3. Total no. of orbitals in subshell - 2l + 1  Orbital angular momentum : 2 N 4. Total no. of electrons in main shell - 2n h 5. Total no. of electrons in sub shell = 2(2l + 1) (  1)  (  1)h 2π L 6. No. of radial or spherical nodes = n – l – 1 14. Bohr’s model formulae 7. Nodal plane :  Radius of nth shell A It is a plane passing through nucleus where probability of finding of electrons is zero. n2 n2 rn  0.529  Å r W No. of nodel plane = l Z Z nh  Velocity of nth shell 8. Angular momentum of electron, mvr = 2 RN Z Z vn   2.185  108 cm / s  V  9. Orbital angular momentum of electron. n n h = (  1) , Z2 2  No of revolutions made by nth shell v = 3 s 1 n = (  1) A n2 V 10. Magnetic moment = n ( n  2) B.M. n3 K Where n = no. of unpaired electrons.  No. of wave made by e – in nth shell h 11. Spin angular momentum = S( S  1) 2 n3 n3 Tn  1.5  10 16  s  v A 12. Maximum no. of lines produced when Z2 Z2 n(n  1)  IE  E  E1 electron falls to ground level, = SH 2 IE = –E1 13. W hen el ectron returns f rom n 2 to n1 state, maximum no. of lines produced IE = 0 – E1 = –E1 (n  n )(n  n  1)  IEH-like atom × Z2 = I.EH atom × Z2 = 2 1 2 1 2 N Z2 1 1   KE = 13.6  eV / atom 1 n2  ν   RZ2  2  2  , A λ  n1 n 2  Z2 [R = 1.0968 × 107 m–1];  P.E. = 27.2  eV/atom n2 K hc h E  hv  ,λ  Z2 λ 2m  K.E.  IE and TE = – 13.6 × eV/atom n2 A  No. of spectral lines produced when an electron drops from nth level to ground level KE Ze 2 2r  KE and TE = TE  2r   2  1 n(n  1) Ze = 2 nh  Angular momentum in orbit mvr =  Heisenberg's Uncertainty Principle = (x) (p) 2  h/4  CHEMICAL BONDING  (i) % ionic character  VSEPR theory Actual dipole moment i. (LP-LP) repulsion > (LP-BP) > (BP-BP) = Calculated dipole moment  100 N ii. NH3 Bond angle 106°45' because L (ii) Dipole moment is helpful in predicting (LP-BP) repulsion > (BP-BP) H2O  104°27' geometry and polarity of molecule. because (LP-LP) repulsion > (LP-LB) > (BP-BP) A  Fajan's Rule: VMCA  Hybridisation : H 2 W Following factors are helpful in increasing covalent character in ionic compounds  MO Configuration i. Small cation Case I : RN ii. Big anion 2s–2p mixing occurs (total e–  14) iii. High charge on cation/anion 1s < *1s < 2s < *2s < 2px = 2py < 2pz < *2px = *2py < *2pz A iv. Cation having pseudo inert gas Case II : configuration (ns2p6d10) K 2s–2p mixing do not occurs (total e– > 14 to 20) e.g. Cu+, Ag+, Zn+2, Cd+2 1s < *1s < 2s < *2s < 2pz < 2pz < 2px=  M.O. Theory: 2py < *2px = *2py < *2pz A i. Bond order = 1/2 (Nb–Na)  Application of H-bonding SH ii. Higher the bond order, higher is the bond Physical State (densile nature)  H-bond dissociation energy, greater is the stability, Melting point (mp)  H-bond shorter is the bond length. on an atom in a Lewis Boiling point (bp)  H-bond N = [total number of valence electrons in the free Viscosity  H-bond atoms] – [total number of non-binding (lone  H-bond A Surface Tension 1 pair) electrons] – [total number bonding 2 Volatilty  1/H-bond K (shared) electrons] Vapour Pressure  1/H-bond  Relative bond strength : A sp3d2 > dsp2 > sp3 > sp2 > sp > p-p (Co-axial) > s-p > s-s > p-p (Co-lateral)  W STATES OF MATTER  Boyle’s Law at constant temperature  Graham’s Law of Diffusion / Effusion and amount 1 Rate of diffusion R  P1V1 = P2V2 = Constant N d  Charle’s Law where d is density of gas at constant L V = kT at constant pressure temperature and pressure A k is the proportionality constant depends r1 d2 upon (i) Amount of Gas (ii) Temperature  r2 d1  Gay Lussac’s Law : W r1 M2 P1 P2   at constant volume r2 M1 T1 T2 RN  Avogadro’s Law  Dalton’s of Partial Pressure : Vn (T and P constant) Calculate the total pressure of mixture of non- V = K4n reacting gas and based on the law of A conservation of amount V1/n1 = V2/n2 (Constant T and P) P1 = PT × x1 (where P1 is a partial pressure, PT  Ideal Gas Equation is a Total pressure, x1 is mole fraction) K PV = nRT Total pressure of Gaseous mixture at constant Where R is Proportionality constant is also (P1 V1  P2 V2 ) known as Gas constant it is same for all Gases temperature : PT  A (V1  V2 ) Value of R in different units Aqueous Tension :Pmoist = Pdry gas + Pwater vapours SH Magnitude Unit 0.0821 Litre-atm K–1 mol–1 Mass of water vapour present in certain volume of air RH  Maximum Mass of water vapour present in same 82.1 ML-atm K–1 mol–1 volume of air saturated by water vapour 62.1 Litre-mm-Hg K–1 mol–1  Molecular Speed N 0.083 Litre bar K–1 mol–1 8.314 Pascal m3 K–1 mol–1 2KT 2RT  A Most probable speed = m M 8.314 × 107 erg K–1 mol–1 8.314 Joule K–1 mol–1 K 8RT 8KT Average speed =  1.987 Cal K–1 mol–1 M m Density ; d = PM/RT A 3RT 3KT (d  P) ; (d  1/T) Root mean square =  M m 8 Vmp : Vav : Vrms = 2: : 3   W  Kinetic Energy  Critical Constant of the Gases 3 Tc or critical temp : Tc = 8a / 27Rb Average Kinetic Energy = KT 2 Pc or critical pressure : PC = a/ 27b2 VC or critical volume : VC = 3b 3 Total Kinetic Energy = nRT 2 PC VC 3 N ZC   (For all real gases) RTC 8 Compressibility Factor (Z)  PV PVm  Z    L  nRT RT   Van der Waal’s Equation Real Gas A  an 2  Ideal gases Real gases  P  2  (V  nb)  nRT where a and b are Van  V  zero volume corrected equation W 2 2 der Waal’s constant. zero attractive force (P + an /V )(V – nb) = nRT PV = nRT non-zero volume a Z=1 some intermolecular force  Boyle’s Temperature : Tb  Rb RN (+)ve (–)ve Z>1 Z 103 [Forward reaction is favoured.]  We have replaced K P by K called thermodyanmic equilibrium constant RN  KC < 10–3 [Reverse reaction is favoured.]  10–3 < KC < 103 [Both reactants and products Significance of G° are present in equilibrium] Significance of G° can be explained from the  Free Energy Charge (G) following points If G° < 0, log K > 0,  K > 1 A a) If G = 0 then reversible reaction would be (i) in equilibrium, KC = 1 Hence, reaction is spontaneas in forward direction. K b) If G = (+) ve then equilibrium will be displaced in backward direction; KC < 1 (ii) Hence G° < 0, log K > 0, K < 1 c) If G = (–) ve then equalibrium will shift in Hence, reaction is non-spontaneous or a A forward direction; reaction proceeds in the forward direction to KC > 1 such a small proceeds in the forward direction to such a small extent that a very small amount SH  (a) KC unit (moles/lit)n of the product is formed. (b) KP unit (atm)n (iii) If G°= 0, log K = 0, K = 1  Reaction Quotient and Equilibrium Constant hence, it represents equilibrium. Consider the following reversible reaction (iv) If G° is large negative number, N A+B C+D K > > 1, the forward reaction is nearly complete. A [C][D]   QC  [A][B] (v) If G° is a very small positive number, K Case I : If QC < KC then : K < < 1, then reverse reaction is nearly complete. [Reactants] > [Products] then the system is not  Mole of Representation of Reversible A at equilibrium Case II : If QC = KC then : reaction. The system is at equilibrium. i. N2 + 3H2 2NH3 K C1 Case- III : If QC > KC then : 1 1 [Products] > [Reactants] K C1  K C1  K C2 K 2C3 The system is not at equilibrium.  W ii. 2NH3 N2 + 3H2 K C2 iv. Decrease of pressure (from less moles to more moles) 1 v. For exothermic reaction decrease in temp. K C1  K C2  K C5 K 2C4 1 (Shift forward) vi. For endothermic increase in temp. (Shift 1 3 forward) iii. N + H2 NH3 K C N 2 2 2 3  Effect of Temperature on Equilibrium Constant L 1 1 K C2  K C3  K C2 3 K C4 According to Von’t Hoff Equation, A 1 3 k = Ae H/RT iv. NH3 N2 + H2 K C4 2 2 Where, K = rate constant, Ea = activation energy, W R = gas constant, T = absolute temperature and 1 e = exponential constant. K  K C2  K C2 C44 K C5 4 RN k H  1 1 log k   2.303R  T  T  2 iv. 2N2 + 6H2 4NH3 K C 1  2 1 5  Le-Chatelier's principle where, T2 > T1 i. Increase of reactant conc. (Shift reaction First case : When H = 0; K2 = K1. A forward) Second case :When H = +Ve; K2> K1 ii. Decrease of reactant conc. (Shift reaction Thirds Case :When H = – ve; K1 > K2 K backward) iii. Increase of pressure (from more moles to less log10 K log10K moles) A T a) Endothermic reaction T (Plots of logK versus T) b) Exothermic reaction SH IONIC EQUILIBRIUM N i. Lewis Acid (e– pair acceptor) d.o.d of strong electrolyte > weak electrolyte d.o.d. dielectric constant of solvent. A CO2, BF3, AlCl3, ZnCl2, normal cation. ii. Lewis Base (e– pair donor)  Buffer solution {Henderson equation} : i. Acidic, pH = pKa + log {Salt/Acid} K NH3, ROH, ROR, H2O, RNH2 normal anions. For maximum buffer action pH = pKa  Dissociation of Weak Acid and Weak Base Range of buffer pH = pKa  1 A i. Weak A ci d, K a = C2/(1 – x) or Ka = C2; < < 1 ii. Alkaline  pOH = pKb + log {Salt/Base} for ii. Weak Base, Kb = C2/(1 – ) or Kb = C2; < < 1 max. buffer action pH = 14 – pKb 1 Range pH = 14 – pKb  1 d.o.d.  µ concentration dilution Moles / lit of Acid or Base mixed d.o.d.  Temperature iii. Buffer Capacity = Change in pH  W  Relation between ionisation constant (Ki) and  Solubility Product degree of ionisation (): Classification of salt on the basis of their solubility 2  2C Ki   = (Ostwald's dilution law) (1  )V (1  ) i. Soluble, Solubility > 0.1 M ii. Slightly Soluble, 0.01 M < Solubility < 0.1 M If is applicable to weak electrolystes for which N  < < 1 then AgCl(s) Ag+ + Cl– Applying the law of chemical equilibrium, we L Ki   KiV  or V  C  have C [Ag  ][Cl  ] A KC  or KC × [AgCl(s)] = [Ag+] [Cl–] Common ion effect : [AgCl(s)]  By addition of X mole/L of a common ion, to a [Ag+] [Cl–] = KC × constant = Ksp W weak acid (or weak base)  becomes equal to Ksp = solubility product Ka  K b  Ksp : product of molar concentrations of the   or  [where  = degree of dissociation] X  X  RN ions (formed in the saturated solution at a given temperature) raised to the power equal  i. If solubility product > ions product then the to the number of times each ion occurs in the solution unsaturated and more of the balanced equation for solubility equilibrium. substance can be dissolved in it.  Application of Solubility Product A ii. If ionic product > solubility product the solution is super saturated (principle of 1. Relation Between Ksp and S precipitation). K General form  Salt of weak acid and strong base :   AxBy  xA+y + yB–x pH = 0.5 (pKw + pKa + log c) a 0 0 A Kh K  K w a–S xS yS h= ; h K (h = degree of hydrolysis) c a Ksp = [A+y ]x [B–x]y SH Salt of weak base and strong acid: = [xs]x × [ys]y = xx.sx. yy.sy pH = 0.5 (pKw – pKb – log c) K sp  x x y y S (x  y) Kw 2. Predicting precipitation in reactions : N h Kb  c (a) If Qsp < Ksp, the solution is unsaturated. Salt of weak acid and weak base : A (b) If Qsp > Ksp, the solution is supersaturated and pH = 0.5 (pKw – pKa – pKb) precipitation takes place. (c) If Qsp = Ksp, the solution is just saturated and K Kw no precipitation takes place. h Ka  K b A  W THERMODYNAMICS  First Law of Thermodynamcis : 1. for isothermal process : E = Q + W P1 Expression for pressure volume work ΔS = 2.303nR log 10 or N P2 W = –PV L V2 Maximum work in a reversible expansion : ΔS = 2.303nR log10 V1 A Pl W = – 2.303 n RT log V2  2.303 nRT log P 2. For isobaric process : V1 2 W Wrev  Wirr T  ΔS = 2.303nCP log  2   T1   qv = cvT = U, qp = cpT = H 3. For isochoric process : RN Enthalpy changes during phase transformation T  ΔS = 2.303 nCV log  2  i. Enthalpy of Fusion  T1  ii. Heat of Vapourisation A 4. For adiabatic process iii. Heat of Sublimation.  q=0 so S = 0  Enthalpy : H = E + PV = E + ngRT K Ssys = 0  Kirchoff's equation : Ssurr. = 0 STotal = 0 A ET2  ET1  C V (T2  T1 ) [constant V] HT2  ET1  CP (T2  T1 ) [contant P]  Entropy and Spotaneity SH STotal = Ssystem + Ssurroundings  Entropy(s) : Case I : For a spontaneous process STotal > 0 Meassure of disorder or randomness Case II : For non spontaneous process STotal < 0 S = SP = – SR N Case III : When process is at equilibrium STotal = 0 q rev V P In a reversible adiabatic process, as q = 0, S   2.03 nR log 2  2.303 n R log 1 A T V1 P2 Ssys =Ssurr =STotal = 0 Entropy Changes in an Ideal Gas  Free energy change : K T2 P G = H – TS, G° = nFE°cell – G = ΔS = 2.303nCp log + 2.303nR log 1 W(maximum) – PV, Gsystem = –TStotal A T1 P2 T2 V ΔS = 2.303nCV log + 2.303nR log 2 T1 V1  W H S G Reaction characteristics – + Alwasy negative Reaction is spontaneous at all temperature + – Alwasy positive Reaction is nonspontaneos at all temperature – – Negative at low temp. Spontaneous at high temp. but positive at high temp. L + + Positive at low temp. Non spontaneous at low temp. but negative at high temp & spontaneous at high temp. A  Isothermal process  Characteristics of Internal energy W 1. T = 0 2. U  0 Ideal gas Real gas 3. q  0 4. H  0 U = f(T) only U =f (T, P or V) RN  Adiabatic process When T is constant When T is constant U = 0, H = 0 U  0, H  0 1. T  0 2. U  0 3. q  0 4. H  0  dU   dU    0   0  dV T  dV T  A Graphical representation of thermodynamic processes  Concept of Heat Capacity K q (1) 1) Isobaric process Heat Capacity (s) C = T 2) Isothermal process P A (2) q 3) Adiabatic process Specific Heat Capacity (s) s = mΔT (4) (3) 4) Isochoric process SH V q Molar Heat Capacity (Cm) Cm = nT for expansion: WIsobaric> WIsothermal > WAdiabatic > WIsochoric  Work Done in Adiabatic Process for compression : As q = 0 N WIsobaric> WAdiabatic > WIsothermal > WIsochoric U = W = nCVT  Work for Osothermal Process W = nCV(T2 – T1) A For expansion for compression nR W= (T  T ) 1 2 1 K 1. V2 > V1 1. V1 > V2 2. W = – ve 2. W = +ve  Relation between CP and CV : A 3. W = –Pext. (V2–V1) 3. W = + Pext. (V1–V2) CP – CV = R 4. W = – Pext. V 4. W = + Pext. V 5. Wmax.no. of moles  W SOLID STATE Total No. of atoms Unit Cell Corners Body Face per unit N cell SCC 1/8 × 8 = 1 ---- ---- 1 L BCC 1/8 × 8 = 1 1 ---- 2 FCC/CCP 1/8 × 8 = 1 ---- 6 × 1/2 = 3 4 A End Centred 1/8 × 8 = 1 ---- 2 × 1/2 = 1 2 W Seven Primitive cells their Possible variations as centred unit cells Possible Axial distances Axial angles RN Crystal system variations or edge lengths Cubic Primitive Body-centred a=b=c  = 90° Tetragonal Primitive, a=bc  = 90° Body-centred A Orthorhombic Primitive, a  a c  = 90° Body- centred, K Face-centred, End-centred Rhombohedral Primitive a=b=c   90° A or Trigonal Hexagonal Primitive a = b c = 90°,  = 120° SH Monoclinic Primitive, a  b c = 90°,  120° End-centred Triclinic Primitive a  b c   90° N S.C. B.C.C. F.C.C. H.C.P. No. of atom 1 2 4 6 A P.E. 52.4% 68% 74% 74% Void space 47.6% 32% 26% 26% K C.N. 6 8 12 12 No. of T.V. 0 0 8 12 A No. of O.V. 0 0 4 6 a 3a a a Relationship between r r r r= 2 4 2 2 2 edge length and radius Type of Packing AAA Type ABCABC Type ABAB AB type  W Max No. of O.V. in one body diagonal = 1 ZM d N A  aq a 1st Nearest distn betn two O.V. = d = density z = number of atom in a unit cell. 2 NA = 6.022 × 1023 Distn betn edge center’s O.V. & Body center’s Square Close Packing a N O.V. = 2 The spheres in the adjacent row lie just one over & show a horizontal & vertical alignment L  3 Co-ordn 4 Diamond = = 0.34 6 A Packing fraction = 78.5% Defects in Crystal Structure : Hexagonal Close Packing Imperfection of Solids W The spheres in every second row are seated in 2 Types the depression. Co-ordn 6 Electronic imperfections Atomic imperfections or RN Packing fraction = 90.75% (91%) Point Defects Tetrahedral Void Co-ordn = 4 Stoichiometric Non-Stoichiometric Impurity defect defect defects rvoid A Radius Ratio= r  0.225 sphere Vacancy Interstitial Schottky Frenkel defects defects defects defects density K T.V’s. Location at body diagonal density  density  density  (constant) Max No. of T.V. in one body diagonal = 2 Shown by Shown by 1st Nearest distn betn two T.V. = a/2 non-ionic solids ionic solids A a Metal excess defect Metal deficiency defect 2nd Nearest distn betn two T.V. = 2 density  SH  Radius ratio and co-ordination number (CN) 3a 3nd Nearest distn betn two T.V. = 2 Limiting radius ratio CN Geometry [0.155  0.225] 3 [Plane triangle] 3a [0.255  0.414] 4 [Tetrahedral] N Distn betn Corner atom & T.V. = 4 [0.414  0.732] 6 [Octahedral] Ratio betn T.V. & O.V. = 2 : 1 [0.732  1] 8 [bcc] A Ratio betn T.V. & O.V. at 1 body digaonal=2:1  Relationship between radius of void (r) and the radius of the sphere (R): r (tetrahedral) K 3a Distance Between O.V. & T.V. = 4 = 0.225 R; r (octahedral) = 0.414 R Octahedral Void  Paramagnetic : Presence of unpaired electrons A [attracted by magnetic field] Co-ordn = 6  Ferromagnetic : r Radius Ratio= void  0.414 Permanent magnetism [] rsphere  Antiferromagnetic : O.V’s. Location at body center & as well as edge center Net magnetic moment is zero []  W SOLUTIONS Pressure of the Gas (or) Henry’a Law : number of equivalents  Normality (N) = volume of the solution in litres  The mass of a gas dissolved per unit volume of solvent is proportional to the pressure of N number of moles the gas at constant temperature.  Molairty (M) = volume of the solution in litres  m  p or m = kp where k is Henry’s constant L  The partial pressure of the gas is proportional  Raoult's law A to the mole fraction of the gas (x) in the P = pA + pB = p° AXA + p° BXB solution” and it is expressed as p = KHx Here  Characteristics of an ideal solution : KH is the Henry’ s law constant. W (i) sol V = 0 (ii) sol H = 0 PP  K H Xgas PP  K Hso lub ility  Relative lowering of vapour pressure RN 1000 PA  PA PA  PA Partial pressure of HCl/torr nB = PA ; PA = X B = nA  nB 500  Colligative  Number of particles ions/moles of solute properties A  Depression of freezing point, Tf = Ktm 0 0.010 0.020  Elevation in boiling point with relative K Mole fraction of HCl in its lowering of vapour pressure Solution in cyclohexane 1000K b  p  p  Tb    1 1 M 1  p  (M1 = mol. wt. of solvent) A KH  Solubility  Solubility Temp  Osmotic pressure (P) with depression in Solubility  p SH dRT massGas  p freezing point Tr P = Tr × 1000K f nGas  p  Relation between Osmotic pressure and other colligative properties : N Positive Deviation Negative Deviation a) Acetone + Ethanol a) Acetone + Aniline PA0  PA dRT i.   Relative lowering of vapour A PA0 MB b) Acetone + CS2 b)Acetone + CHCl3 c) Acetone + C6H6 c) CH3OH+CH3COOH pressure K d) H2O + CH3OH d) H2O + HNO3 dRT ii.   Tb  Elevation in boiling point 1000K b e) H2O + C2H5OH e) CHCl3+C2H5O–C2H5 A f) CCl4 + toulene f) H2O + HCl dRT iii.   Tf  Depression in freezing point g) CCl4 + CHCl3 g)CH3COOH+Pyridine 1000K f h) CCl4 + CH3OH h)CHCl3 + Normal molar mass Observed colligative property i  Observed molar mass  Normal colligative property i) + C2H5OH  W n Solute association  Degree association a = (1 – i) n 1 & If a solute is associated in solutions, n i 1 molecules associate and  is the degree of degree of dissociation ()  association, n 1 nA An ⑦I.I. Abnormal Molar mass I Initial moles 1 0  Electrolytes undergo ionisation in aqueous Number of moles after dissociation 1– n solutions as a result number of particle in the L solution increases hence magnitude of 1i colligative properties increases.  Degree of ionisation, 1 1 A n van’t Hoff’s factor (i) = observed colligativ e properties Colligative properties with Van’t Hoff factor : W a) i = calculated colligativ e property (or) Inclusion of van’t Hoff factor modifies the equations for colligative properties as, Calculated molar mass of solute Relative lowering of vapour pressure of RN b) i = experimental molar mass of solute (or) solvent, p  p total number of moles of particles  iX solute after dissociation or association p c) i = number of moles of particles Depression of freezing point, Tf = iKfm before dissociation or association A Elevation of boiling point, Tf = iKbm Normal / actual / Calculate / Original M wt of solute Osmotic pressure of solution,  = iCST K d) i  Abnormal / Observed / Theorical M of solute wt Solute dissociation (or) Ionisation A If a solute is dissociated or ionised in solutions to give ‘n’ ions and ‘  ’ is the degree of SH ionisation, An nA Initial moles 1 0 Number of moles after dissociation N 1  n A i 1 Degree of ionisation, α  n1 K A  W ELECTROCHEMISTRY *  Under standard conditions (Eo)  Faraday’s 1st Law Eocell = EoRP(cathode) – EoRP(anode) (or) The mass of substance deposited at an electrode N is directly proportional to charge pass through Eocell = EoOP(anode) – EoOP(cathode) (or) it. Eocell = EoOP(anode) + EoRP(cathode) L Mass  Charge (Q)  Nernst Equation WQ A E = E  0.0591 log 10 [Pr oducts ] W = ZQ (Z = electrochemical equivalent) n [Re ac tan ts ] W (Q = Charge in Coulombs)  nE  & E Cell  Eright  Eleft & K eq. = antilog  W=ZIt  0.0591  RN I = current in Ampere G = – nFEcell & t = time in seconds G°= – nFE°cell = – 2.303 RT logKc W = Z Q (Q = 1C) Remember  G  then W=Z & Wmax= + nFE° & G = H + T  T  A P 1 F charge deposits  Application of Nernst Equation K 1 g eq of any substance To find the Ecell of concentration cells EIt So W= sec of substance 0.059 C  F Ecell = log10  2  (for concn cell Eocell = 0) n A  C1  = No. of faradays To find the pH of concentration cell  Faraday’s 2nd Law SH For the measurement of Eq. constant (K) If equal electricity is passed through two or more cells connected in series then the mass 0.059 of substance deposited is directly proportional E 0cell = log 10K C n to equivalent mass N  Calculation of pH of an electrolyte by using a W1  E1 calomel electrode : W2  E2 A E cell  0.2415 pH = W1 W2 0.0591  K E1 E2  Thermodynamic efficiency of fuel cells :  eq  G  nFE cell   A  Degree of dissociation :   For H2 – O2 fuel cells it is 95%. 0eq H H  P = KH.X  Kohlraush's law : 0m  x0A  y0B  W Summing-up the Units of Different Quantities S.N. Physical Symbol Expression Commonly used Units SI Units V 1. Resistance Rq R= omh () omh () I 1 2. Conductance G R= omh–1 (–1) seimen(S) R L a 3. Specific resistance  = ohm cm ohm m 1 A l 1 l l 4. Conductivity k kG   ohm–1 cm–1 (–1 cm–1) S m–1 a Ra p W k 5. Equivalent eq  eq  ohm–1 cm2 eq–1(–1 cm2eq–1) S m2 eq–1 normality RN conductivity k 6. Molar conductivitym m  ohm–1 cm2 eq–1(–1 cm2eq–1) S m2 mol–1 molarity A l 7. Cell constant G* G*  cm–1 m–1 a K Effect of dilution on Conductance Increases A m & eq  Increases Conductivity Decreases SH Debye–Huckel–Onsager equation : m = m – b C (or) eq =oeq – b C m = Molar conductance at given concentration N m = Molar conductance at infinite dilution C = concentration in molarity. A b = Constant value depends on type of electrolyte, solvent & temp. With dilution the Degree of dissociation of weak electrolyte increases, so m increases. K A  W  CHEMICAL KINETICS Rate of reaction = Rate of disappearance of A Decrease in concentration of A Increase in concentration of B  = Rate of appearance of B  Time interval Time interval Order Integrated rate equation Unitss of k Straight line t1/2 proportional obtained by to L plotting t vs A 0 k 1 t [A0 ]  [A] mol L–1 s–1 a–x a W 2.303 a 1 k log s–1 log (a – x) independent of a t ax x 1 1 k RN 2 ta(a x) or L mol–1 s–1 (a  x) a 2.303 b(a x) 1 k log L mol–1 s–1 t(a  b) a(b x) a A 1  1 1  1 1 n k   n 1  Ln–1 mol1–n s–1 t(n  1)  (a  x) n 1 a  (a x)n 1 a n 1 K  Some important graphs of diffrent order of reactions are given below : (a) Plots of rate vs concentrations A Zero order 1st order 2nd order 3rd order Rate Rate Rate Rate SH Concentration Concentration (Concentration)2 (Concentration)3 (b) Plots of integrated rate equations N Zero order First Order Second Order Third Order Intercept = [R]0 A Slope = – k k Slope = k Slope = k Slope  2.303 log [R] [R] 1 1 [R] [R]0 K } Intercept = 1/[R] 0 } Intercept = 1/[R] 0 2 Time ‘t’ Time ‘t’ Time ‘t’ Time ‘t’ A (c) Plots of half-lives vs initial concentration Zero order 1st order 2nd order 3rd order t1/2 t1/2 t1/2 t1/2 Initial conc. (a) Initial conc. (a) 1/a 1/a2  W If reaction completion is given in percent, then This is an equation of straight line of the form take initial concentration (a) 100 and if reaction y = mx + c. completion is given in fraction, then take initial If we draw a graph between log k and (1/T), concentration as 1. we get a straight line with slope equal to Questions based on t1/2 can be solved by another E a method also based on following diagram.. 2.303R N  Order of reaction It can be fraction, zero or any whole number. L  Molecularity of reaction is always a whole Ea Slope = number. It is never more than three. It cannot 2.303R A log k be zero.  First Order Reaction : W 1 2.303 a 0.693 k log 10 & t 1 / 12  [A]t  [A]0 e kt T t ( a  x) k RN Ea can be calculated by measuring the slope of a  Zero Order Reaction : x = kt and t1/ 2  the lines. Ea = –slope × 2.303 R 2k If k1 and k2 are rate constants at temperatures The rate of reaction is independent of the T1 and T2 respectively then, concentration of the reacting substance. A  Time of nth fraction of first order process, k2 Ea  T2  T1  log    k 1 2.303R  T1T2  K   2.303  1  t1/ n  log  Ea k  1  1  Factor e  RT in the Arrhenius equation is known  n  as ‘Boltzmann factor’. A  Amount of substance left after 'n' half lives = -- Activation I Energy SH [A]0 2n The minimum amount of energy absorbed by the reactant mole ules so that their energy  Arrhenius gave a mathematical expression to becomes equal to threshold energy is called deduce the relationship between rate constant activation energy. and temperature. N Or, we can say that it is the difference between  Ea /RT k  Ae threshold energy and the average kinetic A energy possessed by reactant molecules. where, A is frequency factor and it is constant Ea is activation energy Activation energy = Threshold energy – K R is gas constant, T is temperature Average kinetic energy of reactant. On taking log on both sides A Ea Arrhenius equation : log k =   log A 2.303RT Ea ln k = ln A – RT This is an equation of straight Ea Arrhenius equation : log k =   log A 2.303RT  W ET Activation complex Threshold energy Activation Ea energy ER Average kinetic Energy evolved energy of reactants H EP Energy Products Progress of reaction L A Rate constant at (T+10°) Temperature coefficient (n) = Rate constantat T W SURFACE CHEMISTRY RN  Emulsion : Colloidal soln. of two immiscible liquids [O/W emulsion, W/O emulsion]  Emulsifier : Long chain hydrocarbons are added to stabilize emulsion.  Lyophilic colloid : Starchy gum, gelatin have greater affinity for solvent.  A Lyophobic colloid : No affinity for solvent, special methods are used to prepare sol. [e.g. A s2S3, Fe(OH)3 sol] K  Preparation of colloidal solution : i. Disperision methods ii. Condensation method  Properties of colloidal solution : A i. Tyndal effect ii. Brownian movement iii. Coagulation iv. Filtrability SH  Positively charged colloid Negatively charged colloid Hydrated metallic oxide Metal Cu, Ag, Au, Sol Al2O3.xH2O, CrO3.xH2O, Fe2O3.xH2O Metallic sulphides As2S3, Sb2S3, CdS sol N Basic dye stuffs methylene blue sol, Acid dy stuff eosin, congo red Haemoglobin (blood) A Oxide TiO2 Sol Sols of starch, gum gelatin, clay  Hardy Schulze Rule- This rule states that the precipitating effect of an ion on dispersed phase of K opposite charge increases with the valency of the ion. The higher the valency of the flocculating ion, the greater is its pricipitating power. Thus for the A precipitation of As2S3sol (–ve) the precipitating power of Al3+, Ba2+, and Na+ ions is in the order Al3+ > Ba2+ > Na+. Similarly for precipitating Fe(OH)3 sol (positive) the precipitating power of [Fe(CN)6]–3, SO42– and Cl– ions is in the order [Fe(CN)6]3– > SO42– > Cl–  W The minimum concentration of an electrolyte i. Freundlish adsorption isotherm : in milli moles required to cause precipitation For physisorption, Freundlish explained the of 1 litre sol in 2 hours is called variation in adsorption due to the change in FLOCCULATION VALUE. The smaller the pressure graphically and mathematically as flocculating value, the higher will be the follows : coagulating power of the ion. Here Adsorbate - Gas and Adsorbent - Solid N 1 Flocculation value  Flocculation power C D L  x  Gold Number   B m A The number of a hydrophilic colloid that will A jsut prevent the precipitation of 10 ml of P standard gold sol on addition of 1 ml of 10% W NaCl solution is known as Gold number of Case-I At low pressure (A B) that protector (Lyophilic colloid). x P The precipitation of the gold sol is indicated m RN by a colour change from red to blue when the Case-II At high pressure (C D) particle size just increases. x  P0 The smaller the gold number of a protective m Lyophilic colloid, greater is its protection Case - III At intermediate pressure (B C) A power. Note : Gelatin and startch have the maximum x  P1/n [Where n = 1 to ] K & minimum protective powers. m 1 x The resultant condition  KP1/n Protection Capacity  Gold number m A At low pressure n = 1  Effect of temperature : At high pressure n =  SH x At intermediate pressure 1 < n <   Extent of Adsorption  m     The value of (1/n) ranges from 0 to 1. x  Mass of adsorbate Here, N m  Mass of adsorbent x  Mass of adsorbate m  Mass of adsorbent A At constant At constant x pressure x pressure p  pressure of adsorbate gas m m K and n  Constants that depends on the K nature of adsorbate and adsorbent. Temperature  Temperature  [A] Physisorption [B] Chemisorption A  Effect of pressure : The variation in extent of adsorption with change in pressure at constant temperature can be explained with the help of some graphs called as adsorption isotherms.  S BLOCK ELEMENTS L A Synopsis W Elements in which the last electron enters the s–orbital are called s–block elements. Thus, elements of group 1 (alkali metals) and group 2 (alkaline earth metals) constitute s–block elements. RN Alkali MetalsMETALS ALKALI General electronic configuration: ns1 The elements of group 1 : Lithium, sodium, potassium, rubidium, caesium and francium besides hydrogen. A These elements are called alkali metals because they form hydroxides on reaction with water which are strongly alkaline in nature. K l  Physical properties : i) Alkali metals are soft with low melting and boiling points due to weak metallic bonding. A ii) Alkali metals have low density which increases down the group from Li to Cs. iii) On exposure to moist air, all alkali metals except lithium get tarnished quickly so they are always kept in kerosene to protect them from air.