Allen Chemistry Handbook for JEE and NEET PDF

HANDBOOK OF CHEMISTRY S. Page CONTENTS No. No. PHYSICAL CHEMISTRY 1. Mole Concept...

HANDBOOK OF CHEMISTRY S. Page CONTENTS No. No. PHYSICAL CHEMISTRY 1. Mole Concept 01 2. Thermodynaics 07 3. Thermochemistry 18 4. Chemical Equilibrium 20 5. Ionic Equilibrium 23 6. Redox 28 7. Electrochemistry 32 8. Chemical kinetics 39 9. Radioactivity 43 10. Liquid Solution 45 11. Solid State 49 12. Gaseous State (Ideal Gas) 53 13. Atomic Structure 55 14. Surface Chemistry 58 INORGANIC CHEMISTRY 1. Some Important Increasing order 61 Periodic Properties 62 2. Chemical Bonding 65 3. s-Block elements 81 4. p-Block elements 86 5. Coordination Chemistry 104 6. d-Block (Transition Elements) 110 7. Metallurgy 114 8. Salt Analysis 118 9. Environment Pollution 124 ORGANIC CHEMISTRY 1. Table for IUPAC Nomenclature 127 2. Isomerism 129 3. Reaction Mechanism 133 4. Practical Organic Chemistry 135 5. Distinction b/w pair of compound 137 6. Hydrocarbons 142 7. Haloalkanes & Grignard Reagents 147 8. Alcohol, Ether and Phenol 151 9. Carboxylic Acid 159 10. Amines 165 11. Benzene diazonium chloride 167 12. Organic Reagents 169 13. Organic Name Reactions 174 14. Polymers 178 E 15. Carbohydrates 180  CHAP TER Chemistry HandBook ALLEN MOLE CONCEPT SOME USEFUL ATOMIC MASS OR RELATIVE ATOMIC MASS OR CONVERSION FACTORS RELATIVE MOLECULAR MASS MOLECULAR MASS Mass of one atom or molecule w.r.t. 1 Å = 10–10 m, 1 nm =10–9 m Mass of one atom or 1/12th of 12C atom 1 pm = 10–12 m molecule in a.m.u. C® 12 1 litre = 10–3 m3 = 1 dm3 C ® 12 amu H2O ® 18 1 atm = 760 mm or torr It is unitless H2O ® 18 amu = 101325 Pa or Nm–2 GRAMS ATOMIC MASS OR 1 bar = 105 Nm–2 = 105 Pa ACTUAL MASS GRAM MOLECULAR MASS 1 calorie = 4.184 J mass of one atom or Mass of one mole of atom or 1 electron volt(eV)=1.6022×10–19 J molecule in grams molecule (1 J = 107 ergs) C ® 12 × 1.6 × 10–24 g C® 12 g H2O®18 ×1.6 × 10–24 g H2O ® 18 g (1 cal > 1 J > 1 erg > 1 eV) It is also called molar mass DEFINITION OF MOLE One mole is a collection of that many entities as there are number of atoms exactly in 12 gm of C-12 isotope. The number of atoms present in exactly 12 gm of C-12 isotope is called Avogadro’s number [NA = 6.022 × 1023] 1g 1u = 1 amu = (1/12)th of mass of 1 atom of C12 = N = 1.66 × 10-24 g A For elements 1 g atom = 1 mole of atoms = NA atoms g atomic mass (GAM) = mass of NA atoms in g. For molecule Mass ( g ) Mole of atoms = For ionic GAM or molar mass compounds 1 g molecule = 1 mole of molecule = NA molecule 1 g formula unit = 1 mole of formula unit = NA formula unit. g molecular mass (GMM) = mass of NA molecule in g. g formula mass (GFM) = mass of NA formula unit in Mass ( g ) g. node06\B0AI-B0\Kota\JEE(Advanced)\Leader\Che\Sheet\Handbook(E+L)\1_PC\Eng\01-Mole concept.p65 Mole of molecule = Mass ( g ) GMM or molar mass Mole of formula unit = GFM or molar mass 1 mole of substance Average or mean molar mass 23 The average molar mass of the different substance Contains 6.022 × 10 particles M1n1 + M2n2 +.... Weighs as much as molecular mass / present in the container Mavg = n1 + n2 +.... atomic mass/ionic mass in grams Here M1, M2 are molar mass of substances and n1, If it is a gas, one mole occupies a n2 are mole of substances present in the volume of 22.4 L at 1 atm & 273 K or container. 22.7 L at STP E 1 Chemistry HandBook ALLEN DENSITIES VAPOUR DENSITY Mass Ratio of density of vapour to the density of Density = volume hydrogen at similar pressure and temperature. Density of any substance Molar mass Relative Density = Density of reference substance Vapour density = 2 Molarity (M) V(L) V(L) ¸ molar mass ¸ 22.7 L mol—1 Number of Volume of gas (in L) mass in (g) at STP × molar mass Moles (n) × 22.7 L mol —1 NA NA Number of Particles STOICHIOMETRY BASED CONCEPT aA + bB ®cC + dD a,b,c,d, represents the ratios of moles, volumes [for gaseous] molecules in which the reactants react or products formed. a,b,c,d, does not represent the ratio of masses. The stoichiometric amount of components may be related as Molesof A reacted Molesof Breacted Moles of Creacted Moles of Dreacted = = = a b c d Amounts may also be related using POAC method. Moles of reactants and products may be related directly using conservation of suitable atoms. r [Concept of limiting reagent] If data of more than one reactant is given then first convert all the data into moles then divide the moles of reactants with their respective stoichiometric coefficient. The reactant having minimum node06\B0AI-B0\Kota\JEE(Advanced)\Leader\Che\Sheet\Handbook(E+L)\1_PC\Eng\01-Mole concept.p65 ratio will be L.R. then find the moles of product formed or excess reagent left by comparing it with L.R. through stoichiometric concept. p PERCENTAGE YIELD : In general, when a reaction is carried out in the laboratory we do not obtain actually the theoretical amount of the product. The amount of the product that is actually obtained is called the actual yield. Knowing the actual yield and theorectical yield the percentage yield can be calculate as : Actual yield % yield = ×100 Theoretical yield The percentage yield of any product is always equal to the percentage extent of that reaction. 2 E  Chemistry HandBook ALLEN p DEGREE OF DISSOCIATION, (a ) : It represents the mole of substance dissociated per mole of the substance taken. M -M A ® n particles; a = (n -o 1).M where, n = number of product particles per particle of reactant M o = Molar mass of 'A' M = Molar mass of final mixture Dissociation decreases the average molar mass of system while association increases it. p PERCENTAGE PURITY : The percentage of a specified compound or element in an impure sample may be given as Actual mass of compound % purity = ´ 100 Total mass of sample If impurity is unknown, it is always considered as inert (unreactive) material. PERCENTAGE DETERMINATION OF ELEMENTS IN COMPOUNDS Mass % of an element in a compound atomicity of an element × atomic mass of an element = ´ 100 molecular mass of compound Methods for organic compounds : (a) Liebig's method : (For Carbon and hydrogen) D ( w ) Organic Compound ¾ ¾® ( w1 ) CO2 + H 2O ( w 2 ) CuO w1 12 w 2 % of C = ´ ´100 ; % of H = 2 ´ ´100 44 w 18 w where w1 = wt. of CO2 produced, w2 = wt. of H2O produced, w = wt. of organic compound taken (b) Duma's method : (for nitrogen) D (w) Organic Compound ¾¾® N2 ® (P, V, T given) CuO use PV = nRT to calculate moles of N2, n. n ´ 28 \ % of N = ´ 100 w (c) Kjeldahl's method : (for nitrogen) node06\B0AI-B0\Kota\JEE(Advanced)\Leader\Che\Sheet\Handbook(E+L)\1_PC\Eng\01-Mole concept.p65 ¾¾® NH3 + H2SO4 ®[molarity and volume (V in L) (w)O.C.+H2SO4 ® (NH4)2SO4 ¾NaOH consumed given] MV ´ 2 ´14 Þ % of N = ´100 w where M = molarity of H2SO4. Some N containing compounds do not give the above set of reaction as in Kjeldahl's method. (d) Sulphur : (w) O.C. + HNO3 ® H2SO4 + BaCl2 ® (w1) BaSO4 w1 32 Þ % of S = ´ ´ 100% 233 w where w1 = wt. of BaSO4, w = wt. of organic compound E 3 Chemistry HandBook ALLEN (e) Phosphorus : O.C + HNO3 ® H3PO4 + [NH3 + magnesium ammonium molybdate] ® MgNH4PO4 D Mg2P2O7 ¾¾® w1 2 ´ 31 % of P = ´ ´100 222 w (f) Carius method : (Halogens) O.C. + HNO3 + AgNO3 ® AgX If X is Cl then colour = white If X is Br then colour = dull yellow If X is I then colour = bright yellow Flourine can't be estimated by this w1 1 ´ (At. wt. of X) % of X = ´ ´ 100 (Mol. wt.of AgX) w EMPIRICAL AND MOLECULAR FORMULA Empirical formula : Formula depicting constituent atoms in their simplest ratio. Molecular formula : Formula depicting actual number of atoms in one molecule of the compound. The molecular formula is generally an integral multiple of the empirical formula. i.e. molecular formula = empirical formula × n molecular formula mass where n = empirical formula mass EXPERIMENTAL METHODS TO DETERMINE ATOMIC & MOLECULAR MASSES For determination of atomic mass : Dulong’s & Petit’s law : Atomic weight of metal × specific heat capacity (cal/gm°-C) » 6.4. It should be remembered that this law is an empirical observation and this gives an approximate value of atomic weight. This law gives better result for heavier solid elements, at high temperature conditions. Experimental methods for molecular mass determination. (a) Victor Meyer’s Method : Victor -Mayer’s method is used to determine molecular weight of volatile compound. (b) Silver Salt Method : AgNO Organic acid (HnA) ¾¾¾¾ ® Silver Salt ¾¾¾Ignite 3 ® Ag node06\B0AI-B0\Kota\JEE(Advanced)\Leader\Che\Sheet\Handbook(E+L)\1_PC\Eng\01-Mole concept.p65 ( W gm ) (x gm) 108 ´ nWæ 108W ö Molar mass of acid = - n ´ 108 + n ´ 1 = n ç - 107 ÷ gmol -1 x è x ø (c) Chloroplatinate Salt Method : H2PtCl6 Ignite B ¾¾¾¾ conc.HCl ® B2 (H2 PtCl6 )n ¾¾¾ ® Pt Organic base Platinic chloride salt (x gm) (Acidity = n) (W gm ) 1 Molar mass of base = (Molar mass of salt – n × Molar mass of H2PtCl6) 2 1 æ W ´ 195 ´ n ö n æ w ´ 195 ö = 2ç - n ´ 410 ÷ = ç – 410 ÷ gmol -1 è x ø 2 è x ø 4 E  Chemistry HandBook ALLEN CONCENTRATION TERMS Concentration Mathematical Concept Type Formula æwö Mass of solute × 100 Percentage by mass %ç ÷ = Mass of solute (in gm) present wè ø Mass of solution in 100 gm of solution. ævö Volume of solute × 100 Volume percentage %ç ÷ = Volume of solution Volume of solute (in cm3) vè ø present in 100 cm3 of solution. æwö Mass of solute × 100 Mass-volume % ç ÷ = Volume of solution Mass of solute (ingm)present èvø percentage in 100 cm3 of solution. Mass of solute × 106 Parts per million ppm = Mass of solution Parts by mass of solute per million parts by mass of the solution Mole of A Mole fraction XA= Mole of A + Mole of B + Mole of C +.... Ratio of number of moles of one component Mole of B XB= Mole of A + Mole of B + Mole of C +.... to the total number of moles. Mole of solute Molarity M= Moles of solute Volume of solution (in L) in one liter of solution. Mole of solute Molality m= Moles of solute in one Mass of solvent(Kg) kg of solvent p MIXING OF SOLUTIONS : node06\B0AI-B0\Kota\JEE(Advanced)\Leader\Che\Sheet\Handbook(E+L)\1_PC\Eng\01-Mole concept.p65 It is based on law of conservation of moles. Total moles M V + M2 V2 (i) Two solutions having same solute : Final molarity = = 1 1 Total volume V1 + V2 M1 V1 (ii) Dilution Effect : Final molarity, M2 = V + V 1 2 p VOLUME STRENGTH OF H2O2 SOLUTION : Labelled as 'volume H2O2’ means volume of O2 (in litre) at 1 bar & 273 K that can be obtained from 1 litre of such a sample when it decomposes according to 2H2O2 ® 2H2O + O2 Volume Strength of H2O2 solution = 11.35 × molarity E 5 Chemistry HandBook ALLEN p PERCENTAGE LABELLING OF OLEUM : Labelled as '% oleum' means maximum amount of H2SO4 that can be obtained from 100 gm of such oleum (mixture of H2SO4 and SO3) by adding sufficient water. For example, 109 % oleum sample means, with the addition of sufficient water to 100 gm oleum sample 109 gm H2SO4 is obtained. % labelling of oleum sample = (100 + x)% x = mass of H2O required for the complete conversion of SO3 in H2SO4 æ 40 ö % of free SO3 in oleum = ç ´ x ÷ % è 9 ø EUDIOMETRY Some basic assumptions related with calculations are: 1. Gay-Lussac's law of volume combination holds good. According to this law, the volumes of gaseous reactants reacted and the volumes of gaseous products formed, all measured at the same temperature and pressure, bear a simple ratio. N2 (g) + 3H2 (g) ¾® 2NH3 (g) 1 vol. 3 vol. 2 vol. Problem may be solved directly is terms of volume, in place of mole. The stoichiometric coefficients of a balanced chemical reactions gives the ratio of volumes in which gaseous substances are reacting and products are formed at same temperature and pressure. 2. The volumes of solids or liquids is considered to be negligible in comparison to the volume of gas. It is due to the fact that the volume occupied by any substance in gaseous state is even more than thousand times the volume occupied by the same substance in solid or liquid states. 2H2 (g) + O2 (g) ¾® 2H2O (l) 2 mole 1 mole 2 mole 2 vol. 1 vol. 0 vol. 3. Air is considered as a mixture of oxygen and nitrogen gases only. It is due to the fact that about 99% volume of air is composed of oxygen and nitrogen gases only. 4. Nitrogen gas is considered as an non- reactive gas. 5. The total volume of non-reacting gaseous mixture is equal to sum of partial volumes of the component gases (Amagat's law). V = V1 + V2 +.............. Partial volume of gas in a non-reacting gasesous mixture is its volume when the entire pressure of node06\B0AI-B0\Kota\JEE(Advanced)\Leader\Che\Sheet\Handbook(E+L)\1_PC\Eng\01-Mole concept.p65 the mixture is supposed to be exerted only by that gas. 6. The volume of gases produced is often given by certain solvent which absorb contain gases. Solvent Gases absorb KOH CO2, SO2, Cl2 Ammonical Cu2Cl2 CO Turpentine oil O3 Alkaline pyrogallol O2 water NH3, HCl CuSO4/CaCl2 H2O 6 E  CHAPTER Chemistry HandBook ALLEN THERMODYNAMICS DEFINITION Deals with interaction of one body with another in terms of energy. System : Part of universe under investigation Surrounding : Rest part of universe except system. Boundary : Devide system & surrounding SYSTEM Open Closed isolated Energy and matter Only energy Neither energy can exchange can exchange nor matter State function Path function Properties which depends only Depends on on initial & final state of system path or process. & not on process or path. e.g. U, H etc. e.g. work, heat THERMODYNAMIC PROPERTIES Extensive Intensive Properties which are dependent Properties which are of matter (size & mass) independent of matter present in system (size & mass) present in system. node06\B0AI-B0\Kota\JEE(Advanced)\Leader\Che\Sheet\Handbook(E+L)\1_PC\Eng\02-Thermodynamics.p65 Extensive Properties Intensive Properties Volume Molar volume Number of moles Density Mass Refractive index Free Energy (G) Surface tension Entropy (S) Viscosity Enthalpy (H) Free energy per mole Internal energy (E & U) specific heat Heat capacity Pressure Temperature Boilling point, freezing point etc E 7 Chemistry HandBook ALLEN PROCESSES Isothermal Isochoric Isobaric Adiabatic Cyclic T = const. V = const. P = const. No heat Initial & exchange final state dq = 0 of system are same Reversible process Irreversible process Slow process Fast process At any time system No equilibrium between and surrounding are system and surrounding in equilibrium. Psys = Psurr ±dP Psys = Psurr ±DP HEAT (q) Energy exchange due to temperature difference : q = CDT, q = nCmDT, q = msDT C = heat capacity Cm = molar heat capacity s = specific heat capacity m = Amount of substance General values of CV & CP for an ideal gas can be taken as follows. CV CP g Atomicity ntr nRot nVib Excl. Vib Incl. Vib Excl. Vib Incl. Vib Excl. Vib Incl. Vib 3 3 5 5 5 5 Mono 3 0 0 2R 2R 2R 2R 3 3 node06\B0AI-B0\Kota\JEE(Advanced)\Leader\Che\Sheet\Handbook(E+L)\1_PC\Eng\02-Thermodynamics.p65 5 7 7 9 7 9 2R 2R 2R 2R Di 3 2 1 5 7 5 13 7 15 7 15 Linear R R 2R 2R 3 2 4 2 2 5 13 Non 4 7 3 3 3 3R 6R 4R 7R Linear 3 6 n1C P1 + nC P2.... * gmix = n1C V1 + nC V2.... 8 E  Chemistry HandBook ALLEN WORK (W) Reversible Irreversible V2 ò Wrev = - Pext × dV V1 Wirr = – Pext(V2–V1) SIGN CONVENTION w (+)ve q system w (- )ve q q INTERNAL ENERGY (E & U) Every system having some quantity of matter is associated with a definite amount of energy, called internal energy. U = UKinetics + UPotential + UElectronic + Unuclear +..... U is a state function & is an extensive property. DU = Ufinal – Uinitial For a given closed system U = f(T, V) æ ¶U ö æ ¶U ö dU = ç ÷ dT + ç ÷ dV è ¶T øv è ¶V ø T FIRST LAW OF THERMODYNAMICS (FLOT) node06\B0AI-B0\Kota\JEE(Advanced)\Leader\Che\Sheet\Handbook(E+L)\1_PC\Eng\02-Thermodynamics.p65 Law of conservation of energy DU = q + W q ENTHALPY Chemical reactions are generally carried out at constant pressure (atmospheric pressure) so it has been found useful to define a new state function Enthalpy (H) as. H = U + PV DH = DU + D(PV) at constant pressure DH = DU + P DV combining with first law. DH = qp E 9 Chemistry HandBook ALLEN Hence transfer of heat at constant volume brings about a change in the internal energy of the system whereas that at constant pressure brings about a change in the enthalpy of the system. æ ¶H ö æ ¶H ö * For a given closed system H = f(P, T) dH = ç ÷ dT + ç ÷ dP è ¶T ø P è ¶P ø T q RELATIONSHIP BETWEEN DH & DU The difference between DH & DU becomes significant only when gases are involved (insignificant in solids and liquids) DH = DU + D(PV) If substance is not undergoing chemical reaction or phase change. DH = DU + nRDT In case of chemical reaction DH = DU + (Dng)RT WORK DONE IN VARIOUS PROCESS Isochoric Isobaric Free expansion W =0 W= –Pex(V2–V1) Pext =0 DU=q=nCVDT W=0, DU=0,q=0 ISOTHERMAL dT =0; DU=0 (for ideal gas); q =–W Reversible Isothermal Irreversible Isothermal æV ö é nRT nRT ù Wrev , iso=-nRTln ç 2 ÷ Wirr ,iso = - Pext ê - è V1 ø ë P2 P1 úû æP ö = - nRTln ç 1 ÷ è P2 ø ADIABATIC : P2 V2 - P1V1 q=0 Þ DU =W= nCVDT Þ W = g -1 CP g= node06\B0AI-B0\Kota\JEE(Advanced)\Leader\Che\Sheet\Handbook(E+L)\1_PC\Eng\02-Thermodynamics.p65 CV Þ CP = molar heat capacity at constant P. CP–CV=RÞ CV = molar heat capacity at constant V. For Reversible adiabatic : PVg = constant (Ideal gas) TVg–1 = constant (Ideal gas) For irrversible adiabatic process : DU = –Pext(V2–V1) æT T ö n CVm (T2–T1) = – Pext. nR ç 2 - 1 ÷ è P2 P1 ø 10 E  Chemistry HandBook ALLEN Process Expression Expression DU DH Work on for w for q PV-graph V2 æV ö P1 Reversible w = - nRT ln q = nRTln ç 2 ÷ 0 0 V1 è V1 ø P-(atm) isothermal process P2 P1 æP ö = - nRT ln q = nRT ln ç 1 ÷ P2 è P2 ø V1 V2 P1 Irreversible w = - Pext ( V2 - V1 ) q = Pext ( V2 - V1 ) 0 0 P-(atm) P2 isothermal æ nRT nRT ö = - Pext ç - è P2 P1 ÷ø V1 V2 Isobaric w = - Pext ( V2 - V1 ) q = DH = nC P DT DU = nCV DT DH = nCP DT P-(atm) process = – nRDT V2 V1 P2 P-(atm) Isochoric w=0 q = DU = nC V DT DU = nC V D T DH = nCP DT P1 process V Reversible w = nC V (T2 - T1 ) q=0 DU = nCV DT DH = nCP DT P1 Isotherm P V - P1V1 P-(atm) Adiabatic g adiabatic = 2 2 PV =constant P2 g-1 process TVg–1=constant V1 V2 TP1–g/g=constant Irreversible w = nC V (T2 - T1 ) q=0 DU = nCV DT DH = nCP DT P1 Rev Isotherm P2 V2 - P1V1 P-(atm) Rev Adiabatic adiabatic = nCv(T2–T1) = P2 g-1 æ nRT2 nRT1 ö - Pext ç - ÷ process è P2 P1 ø V1 V2' V2 node06\B0AI-B0\Kota\JEE(Advanced)\Leader\Che\Sheet\Handbook(E+L)\1_PC\Eng\02-Thermodynamics.p65 T2 P V - P1V1 ò CVdT n=0 w= 2 2 q= n=1 DU = nCV DT DH = nCP DT P-(atm) Polytropic n -1 T1 n=¥ n=g R(T2 - T1 ) R(T2 - T1 ) w= w= ( n - 1) ( n - 1) V1 V2 Cyclic Area encolsed in q = –w 0 0 Process PV-diagram P w For clockwise it is –ive V it is +ive E 11 Chemistry HandBook ALLEN q SPONTANEOUS PROCESS : A process which takes place on it's own (without any external assistance). The driving force of a spontaneous process is large or finite. q CARNOT CYCLE P (P1,V1)A q2 T2 B(P2,V2) T1 T2 (P4,V4)D T1 q1 C(P 3,V3) V V2 AB – Isothermal reversible expansion q2 = –wAB = nRT2 ln V1 BC – adiabatic reversible expansion wBC = nCV(T1 – T2) æV ö CD – Isothermal reversible compression q1 = –wCD = nRT1 ln ç 4 ÷ è V3 ø DA – adiabatic reversible compression wDA = nCV(T2 – T1) -w Total q1 + q 2 T2 - T1 carnot efficiency h = = = q2 q2 T2 q1 q 2 + = 0 for rev. cycle T1 T2 Ñò Ñò dS = 0 q rev = T node06\B0AI-B0\Kota\JEE(Advanced)\Leader\Che\Sheet\Handbook(E+L)\1_PC\Eng\02-Thermodynamics.p65 Entropy (denoted by S) is state function q rev DS = ò T The direction of a spontaneous process and that it eventually reaches equilibrium, can be under- stood on the basis of entropy concept introduced through the second law of thermodynamics. 12 E  Chemistry HandBook ALLEN q STATEMENTS OF SECOND LAW OF THERMODYNAMICS (i) No cyclic engine is possible which take heat from one single source and in a cycle completely convert it into work without producing any change in surrounding. Source E w Sink (ii) In an irreversible process entropy of universe increases but it remains constant in a reversible process. DSsyt + DSsur = 0 for rev. process DSsyt + DSsurr > 0 for irrev. process DSsyt + DSsurr ³ 0 (In general) q PHYSICAL SIGNIFICANCE OF ENTROPY One can think entropy as a measure of the degree of randomness or disorder in a system. The greater the disorder in a system, the higher is the entropy. (i) The entropies of substance follow the order, S(g) > S(l) > S(s) (ii) If more no. of gaseous moles are present on product side, DrS will be +ive (since gas is more disordered than solid or liquid). (iii) Entropy rises with increasing mass, other thing being same e.g. atomicity in gas phase. e.g. F2(g) S° = 203 J/K-mole Cl2(g) S° = 223 J/K-mole Br2(g) S° = 245 J/K-mole (iv) Entropy increases with chemical complexity For CuSO4.nH2O n=0 n=1 n=3 n=5 S° = 113 150 225 305 node06\B0AI-B0\Kota\JEE(Advanced)\Leader\Che\Sheet\Handbook(E+L)\1_PC\Eng\02-Thermodynamics.p65 q CALCULATION OF ENTROPY CHANGE FOR AN IDEAL GAS General Expression T2 V2 DS = nCV ln + nR ln T1 V1 T2 P = nCp ln + nR ln 1 T1 P2 * Reversible & irreversible isothermal expansion or contraction of an ideal gas V2 DS = nR ln V1 E 13 Chemistry HandBook ALLEN * Isobaric heating or cooling : æT ö DS = nCP ln ç 2 ÷ è T1 ø * Isochoric heating or cooling : æT ö DS = nCV ln ç 2 ÷ è T1 ø * Adiabatic process : T2 V DS = nCV ln + nR ln 2 for irreversible process T1 V1 DS = 0 for reversible adiabatic compression and expansion. q ENTROPY CALCULATION Process DSSys. DSSurr. V2 Isothermal DSSys. = nRln DSSurr. = – DSSys. V1 reversible V2 -q Sys WSys - Pext (V2 - V1 ) Isothermal DSSys. = nRln DSSurr. = = = V1 T T T irreversible Adiabatic DSSys. = 0 DSSurr. = 0 reversible T2 P Adiabatic DSSys. = nC Pln + nRln 1 DSSurr. = 0 T1 P2 irreversible T2 Isochoric DSSys. = nC Vln DSSurr. = – DSSys. node06\B0AI-B0\Kota\JEE(Advanced)\Leader\Che\Sheet\Handbook(E+L)\1_PC\Eng\02-Thermodynamics.p65 T1 reversible T2 -q sys - nC V DT Isochoric DSSys. = nC Vln DSSurr.= = T1 Tsurr Tsurr irreversible 14 E  Chemistry HandBook ALLEN q THIRD LAW OF THERMODYNAMICS ‘‘At absolute zero, the entropy of a perfectly crytalline substannce is zero’’, which means that at absolute zero every crystalline solid is in a state of perfect order and its entropy should be zero. By virtue of the third law, the absolute value of entropy (unlike absolute value of enthalpy) for any pure substance can be calculated at room temperature. T q rev ST – S0K = ò 0 T Since S0K = 0 T q rev ST = ò 0 T Absolute entropies of various substances have been tabulated and these value are used to calculate entropy changes for the reactions by the formula; DSr = åS (products) – åS (reactants) q VARIATION OF DSr WITH TEMPERATURE & PRESSURE : T2 ( DSr )T - ( DSr )T = ( DC P ) r ln 2 1 T1 p1 ( DSr )p - ( DSr )p = Dn g R ln 2 1 p2 Similarly ( DH r )T - ( DH r )T = ( DC p )r ( T2 - T1 ) 2 1 {Krichoff's equation} ( DU r )T - ( DU r )T = ( DC V ) r ( T2 - T1 ) 2 1 q GIBBS FREE ENERGY (G) AND SPONTANEITY : A new thermodynamic state function G, the Gibbs free energy is defined as : G = H – TS at constant temperature and pressure node06\B0AI-B0\Kota\JEE(Advanced)\Leader\Che\Sheet\Handbook(E+L)\1_PC\Eng\02-Thermodynamics.p65 DG = DH – T DS If (DG)T,P < 0 Process is irreversible (spontaneous) (DG)T,P = 0 Process is reversible (DG)T,P > 0 process is impossible (non spontaneous) The use of Gibbs free energy has the advantage that it refers to the system only (and not surroundings). To summaries, the spontaneity of a chemical reaction is decided by two factors taken together (i) The enthalpy factor (ii) The entropy factor E 15 Chemistry HandBook ALLEN The equation DG = DH – T DS takes both the factors into consideration. (DHr)T,P (DSr)T,P (DGr) Remarks – ve + ve Always –ve Reaction is spontaneous + ve – ve Always +ve Reaction non spontaneous + ve + ve At low temperature, DG = + ve Non spontaneous At high temperature, DG = – ve Spontaneous – ve – ve At low temperature, – ve Spontaneous – ve – ve At high temperature, + ve Non spontaneous q VARIATION OF GIBB'S FUNCTION (G) WITH TEMPERATURE AND PRESSURE : G = H – TS = U + PV – TS dG = dU + P dV - T d S + VdP - SdT dG = VdP – SdT * At constant temperature DG = VdP æ ¶G ö or ç ÷ =V è ¶P ø T * At constant pressure DG = –SdT æ ¶G ö ç ÷ = -S è ¶T ø P Relationship between DG & Wnon–PV dU = q + WPV + Wnon–PV for reversible process at constant T & P dU + pdV – TdS = Wnon–PV dH – TdS = Wnon–PV (dGsystem)T, P = Wnon–PV (dGsystem)T,P = (Wnon–PV)system node06\B0AI-B0\Kota\JEE(Advanced)\Leader\Che\Sheet\Handbook(E+L)\1_PC\Eng\02-Thermodynamics.p65 Non-PV work done by the system = decrease in gibbs free energy q SOME FACTS TO BE REMEMBERED : (a) Standard condition * For gases/solid / liquid P = 1 bar * For ion / substance in solution Concentration = 1M (b) DGr = (DGf)product – (DGf)reactant DHr = (DHf)product – (DHf)reactant DSr = (DSf)product – (DSf)reactant (All above equation will be derived in thermochemistry) 16 E  Chemistry HandBook ALLEN Relationship between , DGº and equilibrium constant DG = DGº + RTlnQ At equilibrium DG = 0 DGº = – RT ln Keq DHº – TDSº = – RT ln Keq -D r H° D r S° ln Keq = + For Exothermic reaction RT R D rS° D r H° ln K1 = - ln K R RT1 D r S° D r H ° ln K2 = - 1/T R RT2 æ K ö DH° æ 1 1 ö ln ç 2 ÷ = ç - ÷ è K1 ø R è T1 T2 ø node06\B0AI-B0\Kota\JEE(Advanced)\Leader\Che\Sheet\Handbook(E+L)\1_PC\Eng\02-Thermodynamics.p65 E 17 Chemistry HandBook CH APTER ALLEN THERMOCHEMISTRY ENTHALPY OF REACTION (DHR) BOND ENTHALPY Amount of heat evolved or absorbed during a reaction at constant pressure. Average amount of enthalpy required to dissociate one mole gaseous bond into separate gaseous ENTHALPY OF FORMATION (DfH) atoms. (May be endothermic or exotherm ic) æ Sum of bond enthalpy ö æ Sum of bond enthalpyö DrH = ç ÷ø - çè of gaseous product ÷ø Change in enthalpy when one mole of a substance è of gaseous reactant is formed from its constituent elements present in standard state. RESONANCE ENERGY * For elements DfHo = 0 (for standard state) DHoresonance = D f H o (experimental) - D f Ho (calculated) Ex. DfHo [O2(g)] =0, DfHo = S8(rhombic) = 0 = D C Ho (calculated) - D C Ho (experimental) DfHo [P4 (white)]=0, DfH0 = C(graphite) = 0 ENTHALPY OF COMBUSTION (DCH) ENTHALPY OF NEUTRALIZATION (DHneut) (always exothermic) (Always exothermic) Change in enthalpy when 1 mole of a substance Change in enthalpy when one gram equivalent of is completely burnt in oxygen. an acid is completely neutralized by one g- 7 equivalent of a base in dilute solution. C2H6 ( g) + O2 ( g ) ® 2CO2 ( g ) + 3H2O( l ) ; D CHëéC2 H6 ( g)ûù 2 SA + SB ® salt + water ; DHoneut H+(aq) + OH–(aq) ® H2O(l) ; D r Ho = åD CH o ( react.) - åD CH o ( prod.) DH =– 13.7 kCal eq–1 = 57.3 kJ eq–1 Kcal DHcomb In case of weak acid / base or both DHoN < 13.7 eq Calorific value = molecular wt and the difference is enthalpy of ionisation of ionisation of weak species except in case of HF ENTHALPY OF TRANSITION when DHN > 13.7 due to hydration of F–. Enthalpy change when one mole of one allotropic ENTHALPY OF ATOMISATION (DHatom) form changes to another. (always endothermic) C( graphite ) ® C( diamond) D tran Ho = 1.9 kJ mol -1 Change in enthalpy when one mole of