C1-Allens Made Chemistry Theory {PART-1}

3 .Dipole moment and percentage ionic characterThe measured dipole moment of a substance may be used to calculate the percentage ionic character of acovalent bond in simple molecules.1 unit charge = Magnitude of electronic charge = 4.8 × 10–10 e.s.u.1 D = 1 × 10–18 e.s.u × cm.% ionic character = Observed dipole momentTheoritical dipole moment × 100Theoritical dipole moment is confined when we assume that the bond is 100% ionic and it is broken into ionswhile observed dipole moment is with respect to fractional charges on the atoms of the bond.ELECTROVALENT OR IONIC BOND(a)The chemical bond formed between two or more atoms as a result of the complete transfer of one ormore electons from one atom to another is called Ionic or electrovalent bond.(b)Electro possitive atom loses electron (group IA to IIIA)(c)Electro negative atom gains electron (group VA to VIIA)(d)Electrostatic force of attraction between cation and anion is called ionic bond or electrovalent bond.(e)Total number of electron lose or gained is called electrovalency.Ex.(i) Mg O2, 8, 2 2, 62e–+ Mg + O+2–2electrovalency of Mg = 2electrovalency of O = 2(ii)CaCl2,8,8,22,8,7One eCl2,8,7–One e–+ Ca + 2Cl+2–electrovalency of Ca = 2electrovalency of Cl = 1(iii) Ca + O2e– Ca + O+2–2electrovalency of Ca = 2electrovalency of O = 2(f)The force of attraction is equal in all direction so ionic bond is non-directional.(g)A definite three dimensional structure is formed called crystal lattice.(h)Ionic compound do not have molecular formula. It has only empirical formula.Ex.NaCl is empirical formula of sodium chloride

FORMATION OF IONIC COMPOUND - BORN HABER CYCLE(a)When elements react to form compounds, G (the free energy of formation) is negative. For a reactionto proceed spontaneously, the free energy of the products must to lower than that of the reactants.(b)Usually the energy changes are measured as enthalpy values H, and G is related to H by the equationG = H – T SIn many cases ethalpy values are used instead of free energy values, and the two are almost the same ifthe term T S is small. At room temperature T is almost 300 K, so G and H are similar when thechange in entropy S is very small. Entropy changes are large when three is a change in physical state,Ex. solid to liquid, or liquid to gas, but otherwise entropy changes are usually small.(c)A whole series of energy changes is involved when one starts from the elements and finishes with an ioniccrystal. These changes are shown in the Born-Haber cycle.Formation of NaCl (s) involvesNa + Cl NaCl (s)2  (g)(s)12HfSD2Na(g)Na+(g)Cl(g)Cl–(g)I.P. –e–– EGE+ e–– UthusS + I.P. + D2 – EGE – U = Hfhere S = heat of sublimation of Na(s)I.P. = ionisation potential of Na(g)D = bond dissociation energy of Cl2EGE = electron gain enthalpy of Cl(g)U = lattice energyH = enthalpy heat of formation of NaCl(s)fIf lattice is MgX (s) then2S + (IP + IP ) + D – 2EGE – U = H12fwhere (IP + IP ) = total ionisation energy to form Mg122+(g) from Mg .(g)FACTORS FAVOURING IONIC BONDING(a)Ionisation energy :Amount of energy required to remove an electron from the outermost orbit of an isolated gaseous atomto form the +ve ion or cation. (energy absorbed)Lesser Ionization energy Greater tendency to form cation.Ex.Na > Mg > Al++2+3Cation formation tendency Cs > Rb > K > Na > Li+++++

(b)Electron affinity :Amount of energy released when an electron is added to an isolated gaseous atom to form–ve ion (anion) energy released.Higher electron affinity Greater tendency to form anionEx.Cl > F > Br > I––––F > O > N ––2–3(c)Lattice energy - (Energy released)The energy released in the formation of 1g mole electrovalent compound from isolated gaseous ions is calledlattice energy (U) of that compound.Higher lattice energy Greater will be the stability or strength of ionic compound.Factors affecting lattice energy :(i)Magnitude of charge U z z (Ionic charge) +– Lattice energy Magnitude of chargeNaClMaCl2AlCl3Na+Mg+2Al+3– Lattice energy increases– Size of cation decreases.(ii)Size of Cation :- Lattice energy 1rrLiClNaCl KClRbClCsCl– Size of cation increasing– Size of anion is constant– Lattice energy decreases.PROPERTIES OF IONIC COMPOUND(a)Physical state – Ionic compounds are hard, crystalline and brittle due to strong electrostatic force of attraction.Brittleness {Same charged ions comes nearer. So they repell each other}++++++++++++++++++++AttractionRepulsion(b)Isomorphism : Different ionic compounds, having same configuration/geometry of ions are isomorphsof each other and phenomenon is known as isomorphismEx.NaF, MgO, ZnSO · 7H O, FeSO · 7H O. All alums M 'SO · M \"(SO ) · 24H O.4242224 32(c)Boiling point and melting point –Ionic compounds have high boiling point and melting point due to strong electrostatic force of attractionamong oppositely charged ions.(d)Conductivity – It depends on ionic mobility.(i)In solid state - No free ions - Bad conductor of electricity.(ii)In fused state or aqueous solution Due to free ions - Good conductor of electricity.Conductivity order Solid state < fused state < Aqueous solution

(e)Solubility – Highly soluble in water (Polar solvents) Ex. NaCl in water(i)The Na ions get associates with - vely charged 'O' of water+(ii)And Cl ions associates with +vely charged 'H' of water.–Na +OHH + +– –=Cl–(iii)Thus charge on Na and Cl decreases and electrostatic force of attraction also decreases which leads+–to produce free ion.(iv)The energy released due to interaction between solvent and solute is called solvation energy. If water isused as solvent it is called hydration energy.(v)For an ionic compound to be soluble in water – Hydration energy > Lattice energyLattice energy 1So lub ilityHydration energy Solubility.Hydration energy (H) 11rr { r & r are radius of cation and anion}+–(vi)Hydration energy mainly depends on the cation radius because the value of 1r is negligible in comparision to 1r(vii)Down the group both the lattice energy & hydration energy decreases, if decrease in lattice energy isgreater than hydration energy, solubility increases down the group and vice versa.FACTOR AFFECTING SOLUBILITY :(a)Dielectric constant of solvent –The capacity of solvent to neutralise the charge of ionic compounds is called Dielectric constant. It isrepresented by (i)Water has maximum dielectric constant ( = 80)(CH OH = 35)3,(Acetone = 21)(C H OH = 27)25,(Ether = 4.1)(Benzene = 2.3)H O > CH OH > CH CH OH > CH COCH > CH OCH > C H2332333366(ii)Ionic compounds are more soluble in the solvents, having high dielectric constant.(iii)H SO and H O have high dielectric constant but these are not a good solvent due to oxidising nature.2422(b)Size of ion :(i)Keeping size of cation constant, the lattice energy decreases with the increase of anionic radius.Hence order of solubility of LiX in water is LiF < LiCl < LiBr < LiIAs solubility1lattice energy

(ii)In LiI covalent nature is more according to Fajan's rule but HE > LE therefore LiI is more soluble in water.(iii)Keeping size of anion constant, the hydration energy decreases with the increase of cationic radius.Hence order of solubility of MSO will be – BeSO > MgSO > CaSO > SrSO > BaSO ( Exception of444444Fajan's rule)(iv)If size of cation and anion is very large, solubility decreases from top to bottom.(v)Solubility decreases in a period (as ionic nature decreases and covalent nature increases)NaCl > MgCl > AlCl23TRANSITION FROM IONIC TO COVALENT BOND – FAJANS' RULE(a)Just as a covalent bond may have partial ionic character an ionic bond may also show a certain degree ofcovalent character. When two oppositely charged ions aproach each other closely, the cation wouldattract the electrons in the outer shell of the anion and simultaneously repel its nucles. This producesdistortion or polarization of the anion, which is accompanied by some sharing of electrons between theions, i.e., the bond acquires a certain covalent character. The formation of a covalent bond between twoions may be illustrated with reference to formation of Agl.Ag+Ag+Ag ...+I –I –I –separate ionsdistorted ioncovalent character in ionic bond(b)FACTORS INFLUENCING ION – DEFORMATION OR INCREASING COVALENT CHARACTER(i)Large charge on the ions:The greater the charge on the cation, the more strongly will it attract the electrons of the anion. Forexample, Al can distort Cl ion more than Na ion. So aluminium chloride is a covalent compound3+–+whereas NaCl, AlF , AgF are ionic.3(ii)Small cation and large anion:For a small cation, the electrostatic force with which its nucleus will attract the anion will be more.Moreover a large anion cannot hold the electrons in its outermost shell, especially when they are attractedby a neighbouring cation. Hence there will be increased covalent charcter with a small cation and a largeanion, as in AgI.(iii)Cation with a pseudo-inert gas type of electronic configuration:A cation with a 18 electron in outermost shell such as Ag ([Kr] 4d ) polarizes anions more strongly than+10a cation with a 8 electron arrangement as in K . The 'd' electrons in Ag do not screen the nuclear charge++as effectively as the 's' and 'p' electron shell in K . Thus Ag is more covalent than K , although Ag and+II+K ions are nearly of the same size. Cuprous and mercurous salts are covalent.+The above statements regarding the factors, which influence covalent character, are called Fajans' rules.It can thus be seen easily that there is nothing like a purely ionic compound or a purely covalent compound.Polarisation power of a cation is usually called ionic potential or charge density.Ionic potential (phi) = Ch arge on cationSize of cation

APPLICATION OF THE CONCEPT OF POLARISATION :(a)To determine covalent and ionic character of molecule : -Covalent characterIonic characterFrom left (larger size) to right (smaller size) in a period increases so covalent character increases.Na , Mg+++Al+++ Si++++-charge increases-size decreasesincreases-Covalent character increases with particular anionFrom top to bottom in a group decreases so covalent, character decreases.Li+Na+-Size increases (charge is fix)K +-decreasesRb+-Hence covalent character decreases with particular anionCs+(b)To determine nature of oxide : - < 2.2(Basic oxides) = 2.2 to 3.2(Amphoteric oxides)Neutral oxides doesn't react with acid & base eg. H O, CO etc.2Amphoteric oxides (Al O etc.) reacts with acid & base23 > 3.2(Acidic oxide)Ex.Li O, Na O, K O, Rb O, Cs O22222– decreases–Basic character increases(c)To determine conductivity of metal halides (MX)If  < 2.2 MX - ionic natureIf  > 2.2 MX - covalent nature

(d)Formation of complex compounds :-Smaller the cation, more will be the tendency of forming complex compounds.High value of shows tendency of forming complex compounds.If is low No or less tendency of forming complex compounds.s-block metals (larger size) doesn't have the tendency to form complex compounds.Exceptions - Li, Be, Mg (small size)d-block metals have the tendency to form complex compounds (small in size, high charge).(e)To determine thermal stability of metal carbonates : stability   1 (Covalent nature)M O — C MO + CO+22–OO ––More smaller the size of metal cation, its polarisation capacity increases - strength of M—O bond increasesand C—O bond decreases. So thermal stability of carbonates decreases.CO3–2O + CO–22From top to bottom thermal stability of carbonates increases (as size of cation increases)(Ionic character increases or covalent character decreases)BeCO3CaCO3 – decreases (covalent character decreases)MgCO3 – size of cation increasesSrCO3 – Thermal stability increasesBaCO3(f)To explain colour of compounds : -More the covalent character, more will be the colour intensity.Colour density (Covalent character)Ex.AgClAgBrAgI(White) (Light yellow (Dark yellow)SnO2SnS2(White)(Yellow)(g)To explain diagonal relationship :-Since the value of for Be is almost of the same order as that of diagonally situated Al . Hence have2+3+many similar properties. Similarly value of for Li is equal to Mg++2 , so have diagonal relationship.

MOLECULAR ORBITAL THEORY (MOT):Given by hund & Mulican(a)Two atomic orbital come nearer & then overlap each other to form two molecular orbitals (MO)(b)Combination of two atomic orbital (AO) forms two molecular, orbital (MO):AO + AO ABMO (antibonding molecular orbital)BMO (bonding molecular orbital)(c)Both orbitals can be filled by electrons according to Auffbau principle.(d)Energy of BMO < Energy of ABMO.(e)Order of energy in molecular orbitals of di-atomic molecules.(i)From H to N :221s < *1s < 2s < *2s < 2p = 2p < zy2p < *2p = *2p < *2pxzyxFor N molecule2  2 p2 p  px 2s2s2s  2s 1s1s1s  1s^Increasing energy iN2mpz  py  pz

(ii)For O & F :221s < *1s ; < 2s < *2s < 2p < 2p = 2p < *2p = *2p < *2pxzyzyxFor O molecule2  pz  p x  py2 p2 pp x  py pz 2s2s2s  2s 1s1s1s^Bond order = (8–4) = 2½  1s* Having two unpairedelectrons so paramagneticIe  = antibonding molecular orbital= bonding molecular orbitalBond order = baNN2 N = No. of electron in bonding MO'sbN = No. of electron in antibonding MO'sa(g)If bond order = 0, it means species does not exist.(h)Bond order of 1, 2 & 3 corresponds to a single bond, double & triple bond respectively.(i)Bond order stability of molecule bond length (j)If the molecule has one or more unpaired electron, it will be paramagnetic, while if all the electrons arepaired it will be diamagnetic.Ex.H = Configuration : 2(1s) 2 * (1s)0Bond order = baNN2  = 202  = 1,Hence H – H (dimagnetic)

MO Electronic Configuration of Some MoleculesMolecule Total no.MO configurationBond orderMagneticor ionof electrons behaviourH 22 ( s) 21DiamagneticH 2+1 ( 1s)10.5ParamagneticH 23( 1s) , ( * 1s)210.5ParamagneticHe24 ( 1s) , ( * 1s)220DiamagneticN 214 KK ( 2s) , ( * 2s) ,22 3Diamagnetic ( 2p ) =( 2p ) , ( 2p )y2z2 x2O 216KK ( 2s) , ( * 2s) ( 2p ) ,22 x22Diamagnetic ( 2p ) =( 2p ) ( * 2p ) ( * 2p )y2z2 y1z1O 2+15Remove one electron from2.5Paramagnetic * 2p from Oz2O 22+14Same as that of N23DiamagneticO 217KK ( 2s) , ( * 2s) ( 2p ) ,22 x21.5Paramagnetic ( 2p ) ,( 2p )y2z2 ( *2p ) ,( *2p )y2z1O 22–18KK ( 2s) , ( * 2s) ( 2p )22 x21Diamagnetic ( 2p ) ,( 2p )y2z2 ( *2p ) ,( *2p )y2z2F218Same as above1DiamagneticNe22018 as above and ( * 2p )x20DiamagneticCO14Same as in N23DiamagneticNO15Same as in O2 +2.5ParamagneticNO+14Same as in N23DiamagneticNO2+13KK( 2s) ( * 2s) ( 2p )22 x22.5Paramagnetic ( 2p ) ,( 2p )y1z1NO–16Same as in O22ParamagneticCN13Same as in NO2+2.5ParamagneticCN–14Same as in N23DiamagneticHYDROGEN BOND(a)An atom of hydrogen linked covalently to a strongly electronegative atom can estabilish an extra weakattachment to another electronegative atom in the same or different molecules. This attachment is calleda hydrogen bond.(b)To distinguish from a normal covalent bond, a hydrogen bond is represented by a broken line,e.g. X – H ... Y where X and Y are two electronegative atoms. The strength of hydrogen bond is quite lowabout 2-10 kcalmol or 8.4-42 kJmol .–1–1

(c)Conditions of hydrogen bonding:(i)The molecule must contain a highly electronegative atom linked to H-atom. (If E.N. polarity of bond )(ii)The size of the electronegative atom should be small.(size electrostatic atraction )Ex.H N · ·· H N ····· H N ··· H — N—· ·——· ·HHHHHHHHH O ··· H O · ·· H — O,—· ·—· · + + + + + + + – – –HH HH + + + – –H F ····· H F——(d)Strength of H-bond :H .............. Fbond dissociation energy = 41.8 kJ mol–1H .............. Obond dissociation energy = 29.3 kJ mol–1H .............. Nbond dissociation energy = 12.6 kJ mol–1(e)Effect of H-bond(i)It causes the association of many molecules.(ii)Due to hydrogen bond molecules are associated and show high moleculer weight.(iii)M.P. & B.P. of the molecules increases(iv)Viscosity & surface tension of the molecules increases.(v)The compounds which can form H-bond with the covalent molecules are soluble in suchsolvents.Examples : Alcohol & Ammonia are water soluble.Type of H-bonding :(a)Intermolecular(b) Intramolecular( a )Intermolecular H-bond:(i)This type of H-bonding takes place between two moleules. Ex. ROH, H O, R - OH & H O22R|O – H ... O – H,R|H|O – H ... O – H,H|R|O – H ... O – H ... O – HH|R|(ii)In such compounds molecular wt., M.P, & B.P. are high.(iii)Extent of H-bonding viscosity & density .(b)Intramolecular Bond:(i)It is the H-bonding in the same molecule of a compound. The bonding also known as chelation.(ii)Solubility in water, M.P. & B.P. of these compounds decreases.(iii)The value of acid strength of acid depends on the relative stability of the acid and its canjugated baseif the interamolecular H-bonding stablises the conjugate base then, the acid strength increase and ifthe conjugate acid is stabilised in this way then the acid strength decreased.

H-bond has serious consequences on the physical properties. These are –(i)H O is liquid at room temperature whereas H S is gas at room temperature although H S has greater222molecular weight.(ii)Ice is lighter than water, although it solid because in ice H O molecules are extending their H-bonds to full2length and six water molecules produce a cage like structure which is porous in nature leading to smallermass and larger volume, thus lowering the density.(iii)Alcohols have higher boling points than corresponding alkanes and there of the same moleculear mass.This can be explained by the intermolecular H-bonds existing among the R-OH molecules whereas ethersand hydrocarbon are not capable to have such strong intermolecular forces. thus they have lower bolingpoints.H O > R-OH > R-OH 2ExceptionalH S < R-SH < R-S-R 2NormalH-bond increasesHO > R-OH > R-O-R2B.P. increasesR-O-R  no H-bondH O 2 H-bondsR-OH  H-bond.(iv)Alcohol and water are miscible in any proportion due to formation of H-bonds.(v)Ammonia is excessively soluble in water whereas other gases are partially soluble. This is due to thetremendous capacity of NH molecule to generate 4-H-bonds.3NH > RNH > R NH322(vi)Solubility order of different amines and ammonia in water follows the order NH > RNH > R NH. This322can be explained by the capacity of these elements to form H-bonds.(vii)KHF exists whereas KHCl , KHBr , KHI do not because KHF is constituted by 2 ions22222i.e.K & HF +2–[F ... H – F – – – +][Cl ... H – Cl]– existdoes not existKHF is a red orange coloured solid.2(viii)o-nitro phenol is steam volatile and less H O soluble than its coresponding para isomer because in2o-nitro phenol, intra molecular H-bond exists which reduces the capacity of the molecule to produceintermolecular H-bonds with other molecules like water. This brings interamolecular forces among the o-nitrophenol molecules resulting into high volatility.(ix)Acetic acid has a molecular weight of 60 in benzene and it shows a very high boiling points. This is due todimerization of molecule.CH – C3O H O–. . .O ... H OC CH3

(x)Salicylic acid has very high acidity due to chelation of salicylate ion through the H-bondO ··· HCO–(xi)Chelation means to grab atoms in vicinity as much as possible.(xii)Maleic acid has greater acidity than furmaric acid. This can also be explained by chelation of monomaleateanion which has a H-bond between carboxylate anion adn the unionized carboxylic acid.H — CH — COCCOO –OH(xiii)Normally when 2 hydroxyl groups are present on the same carbon atom i.e. gem diols are unstable, but\"chloralhydrate\" is a stable molecule due to formation of H-bond.Cl ..... H — OCl C — C — HCl ..... H — O(xiv)CH – C – CH – C – CH332OOIn the above compound, (acetyl acetone) enol form is stable because of the intramolecular H-bond.CH 3CCHCH 3OOHCVANDER WALL'S FORCES(a)These are the weakest type of inter molecular forces that exist among the molecules which being significantchange in physical properties.(b)These are non-directional, non-valence cohesive forces. These attractive forces being played betweenthe two molecules are independent of the presence of other molecules.(c)Solid, liquid or gaseous states of many molecules are explained on the basis of inter molecular forcesother than covalent, ionic or metallic bonds. Although inert gases do not form any type of bond but mayexist in liquid and solid states. This shows that the atoms of inert gases are attracted by each otherthrough some type of inter molecular forces. These intermolecular forces are called Vander Walls forcesand may be of the following types:

(d)Ion-dipole attraction : Polar molecules are atracted by ions. The negative pole is attracted by cationand positive pole attracted by the anion. This type of attraction is called ion dipole attraction, ion-dipoleattraction is abserved generally in the process of solvation when sodium chloride (Na Cl ) is dissolved in+–water because negative poles of water aggregate around Na ions and positive poles around Cl ions.+–Na ······ O+HH(e)Dipole-dipole attraction : The force of attraction between the oppositely charged poles of two polarmolecules (for example : H O, H–F, NH etc.) is called dipole-dipole attraction. This type of attraction is23weaker than the ion-dipole attraction.HH O······ H–ClHHHHOO······(f)Ion-induced dipole attraction : When non polar molecules come in contact with ions, its electron cloudgets polarised and the oppositely charged end of it is attracted by the ion. For example attractionbetween Na and Cl molecule.+2Na + Cl – Cl + Na ........ +Cl – Cl + –(g)Dipole-induced dipole attraction : This type of cohesive forces occurs in a mixture of polar and nonpolar molecules. The former induce polarity in non polar molecules by disturbing their electron system.for example force of attraction between Cl and H O.2 2HH O + Cl – Cl HHO ........ Cl – Cl  + –(h)Induced-Induced dipole : The weak intermolecular forces operating in similar non polar gaseousmolecules are called London forces. These forces are very weak in nature and exists only at lowtemperature. For example weak intermolecular forces in F , Cl , N , molecules and in nobal gasses.222 + + – –++

INTRODUCTIONChemical reaction : Symbolic representation of any chemical change in terms of reactants and products iscalled chemical reaction.Ex.N + 3H 22 2NH3TYPES OF CHEMICAL REACTION :On the basis of physical stateHomogeneous reactionHeterogeneous reactionAll reactants and products are in sameReactants and products are in more than onephasephaseN (g) + 3H (g) 22 2NH (g)3Zn(s) + CO (g) 2 ZnO(s) + CO(g)On the basis of directionReversible reactionIrreversible reaction(i)Chemical reaction in which products(i) Chemical reaction in which products cannotcan be converted back into reactantsbe converted back into reactants.N + 3H 22 2NH3AgNO + NaCl 3 AgCl + NaNO33Fe + 4H O 2 Fe O + 4H342NaCl + H SO 24 NaHSO + HCl4H + I 222HIZn + H SO 24 ZnSO + H422KClO 3 2KCl + 3O2(ii)Proceed in forward as well as(ii) Proceed only in one direction (forward).backward direction.(iii)To obtain reverisible reactions, if anyone(iii) Generally possible in open container.of the reactant or product is in gaseousstate, then the reaction should be carriedout in closed vessel.CaCO (s) 3 CaO (s) + CO (g) 2(iv)These attain equilibrium.(iv) These do not attain equilibrium.(v)Reactants are never completely(v) Reactants are completely convertedconverted into products.into products.(vi)Generally thermal decomposition in(vi) Generally thermal decomposition inclosed vessel.open vessel.PCl 5(g) PCl3(g) + Cl (g)2PCl 5(g) PCl3(g) + Cl (g)2On the basis of speed.Fast reactionsSlow reactions(i)Generally these reactions are ionic(i) Generally these reactions are molecularreactions.reactions.HCl + NaOH  NaCl + H O2H + I 22 2HIAcid Base Salt WaterCHEMICAL EQUILIBRIUM

On the basis of heatExothermic reactionEndothermic reaction(i)Heat is evolved in these chemical reaction(i) Heat is absorbed in these chemical reactionR  P + x kcalR + x kcal  Por R  P – x kcal(ii)Change in heat energy(ii) Change in heat energyQ = (+) veQ = (–) ve(iii)Change in enthalpy(iii) Change in enthalpyH = (–) veH = (+) veHQ Eg. : Formation reactionEg. : Dissociation reactionException N + O22 2NO/N O/NO22 O + F 22O F /OF2 22Active mass : The term active mass means the concentration of the reactants & products expressed inmoles per litre (molar concentration). Active mass is usually expressed by enclosing the symbol of thereactant in square bracket [ ]Active mass= molesVolume in litres= wgrams(w)mol.wt.(M ) Volume in litres(V) = ww 1000MV(mL) The active mass of solids and pure liquids is a constant quantity (unity) and solvent (excess) is considered asone. Because there is no change in activity with the change in quantity or volume of vessel.Molar concentration= wlit.wMV = w M  ( = density in g/lit)=density of the substancemolecular mass of the substanceas density of pure solids and liquids is constant and molecular mass is also constant.But this is not applicable to the substance in aqueous solution or gaseous state because their amount in agiven volume can vary.Following other names of active mass can also be use :(i)mole/lit.(ii)gram mole/lit.(iii)gram molecules/lit.(iv)molarity(v)Concentration(vi)Effective concentration(vii)active quantity(viii)n/v(ix)C(x)M(xi)[ ]Examples :(a)25.4 g of iodine is present in 2 litres of solutionthen225.4I254 2   =0.05 mole/litre(b)8.5 g ammonia is present in a vessel of 0.5 litre capacity then38.5NH1mole / litre17 0.5 (c)Active mass of C (s) or S(s) or Zn(s) is equal to 1.

RATE OF REACTION :The change in concentration of reactants or products per mole in unit time is known as rate of the reaction.Rate of reaction= (–) change in concentration of rectantstimedcdt reactants.Here negative sign indicate that concentration of reactants decrease with time.Rate of reaction= +change in concentration of productstime= + dcdt   productsHere possitive sign indicate that concentration of products increase with time.Note :The concentration change may be positive or negative but the rate of reaction is always positive.Unit of rate of reaction = mole/lit.sec = molelit.sec = mole lit sec–1–1For example A  BFor reactant  d[A]dt[concentration decreases with time]For product  d[B]dt[concentration increases with time]Ex.2A + 3B  C + 4DRate of dissappearance of A = d[A]dt Rate of dissappearance of B = d[B]dt Rate of appearance of C = +d[C]dtRate of appearance of D = +d[D]dtRate of reaction (ROR) = 1 d[A]2 dt = 1 d[B]3dt = +d[C]dt = +1 d[D]4 dtNote :Rate of reaction is always for per mole.aA + bB  cC + dDRate of reaction = 1 d[A]1 d[B]1 d[C]1 [D]adtbdtc dtd dt  Ex.For the reaction2SO + O 222SO3rate of reaction is 2.5 × 10 moles/lit.sec. then find out the rate of dissappearance of SO .–42Ans.ROR = 221 d[SO ]d[O ]2dtdt  = 2.5 × 10 mole/lit.sec.–4– 2 d[SO ]dt = 5 × 10 mole/lit.sec.–4

FACTORS AFFECTING RATES OF REACTIONS :( a )State of matter : The decreasing order of rate of reaction in gas, liquid and solid state are-gs ( b )Temperature : Rate of reaction temperature( c )Concentration : Rate of reaction concentration.( d )Catalyst : Positive catalyst increases the rate of reaction.CHEMICAL EQUILIBRIUM :The state of the reversible chemical reaction at which rate of forward reaction becomes equal to rate of backward reaction. RfA + B C + DR = forward rate of reactionf RbR = backward rate of reactionbi.e.R = RfborThe state of the reversible chemical reaction at whichthe measurable properties of the system like temperature,concentration, colour, density etc. don't undergo any changewith time at the given set of conditions is said to be chemicalequilibrium conditions. Rate of forward reaction decreases asthe concentration of products increases, rate of backwardreaction also starts increasing.At a certain stage, rate of forward reaction becomes equal to rate of backward reaction called equilibrium state.AT EQUILIBRIUM STATE :Rate of forward reaction = Rate of backward reactionAt this state of equilibrium forward and backward reactions proceeds with same speed.The stage of the reversible reaction at which the concentrations of the reactants and products do notchange with time is called the equilibrium state.The equilibrium state is dynamic in nature.The reaction does not stop, but both the opposing reactions are going on continously with same speeds.CHARACTERISTICS OF EQUILIBRIUM :( a )Chemical equilibrium is dynamic in nature means the reaction, although appears to be stopped, butactually takes place in both the directions with the same speed.( b )To obtain equilibrium, if anyone of the reactant or product is in gaseous state then the reaction should becarried out in closed vessel.( c )At a given temperature and pressure of equilibrium the properties like concentration, colour, densityremains constant.( d )In a reversible chemical reaction the equilibrium state can be attained in lesser time by the use of positivecatalyst.A catalyst doesn't change the equilibrium state becuase it increases the rate of both forward and back-ward reaction simaltaneously by changing the path of reaction and it helps in attaining equilibriumrapidly.( e )In order to prevent escape of products, equilibrium is reached in only in closed vessels in reversiblereactions.( f )Homogeneous equilibrium is the equilibrium in which the reactants and products are in the same phase.3252CH COO C H ( ) H O( )    CH COOH( ) C H OH( )325 ( g )Heterogeneous equilibrium is the equilibrium in which the reactants and products are in two or more phases.Zn(s) + CO (g)2ZnO(s) + CO(g)Rtime taken to attainequilibriumRate of backwardreactionRate of forwardreactiontimeeffect ofcatalystEquilibriumstate (R = R )fb

Note :(i)Whenever question doesn't ask about direction, then we take forward direction only.(ii)In a reversible reaction if forward reaction is exothermic then the backward reaction will be endothermicand vice versa.TYPES OF EQUILIBRIA :There is two types of equilibria :1 .Physical equilibria :If in a system only physical state (phase) is changed and then equilibrium is established, (i.e. there is no chemicalchange) the equilibrium is called physical equilibrium.e.g. Fusion of ice, evaporation of water, dissolution of salts and absorption of gases in liquid etc.Following are the types of common physical equilibria :-( i )Liquid-Vapour equilibria : In a closed vessel, the vapours above the liquid are in equilibrium atgiven temperature.Ex.H O ( ) 2 H O (g)2(ii)Solid-Liquid equilibria : This equilibrium can be established only at melting point of solid. At thisstage solid and liquid phases exist simultaneously in equilibrium.Ex.H O (s) 2 H O ( ) at melting point2(iii)(Solute-Solvent), Saturated solution equilibria : If the rate of dissolution of solids in liquid isequal to the rate of crystallization of solid from solution i.e. solution is saturated with respect to solidthen saturated solution equilibria established, provided temperature is constant.Ex.NaI (s) 2H O Na (aq.) + I (aq.)+– Note:Above example is of solubility of sparingly soluble salt, which only depends on temperature.(i v)(Gas + Solvent), Saturated solution equilibria : In such equilibriums, solvents is saturated withrespect to gas i.e. rate of entering of gas molecules in solvent is equal to rate of escaping of gas moleculesfrom solvents. Above phenomenon can be observed in closed container at definite temperature.Ex.Dissolved CO in cold drinks. Dissolved O in water etc.22Note :(i)The solubilities of gases in liquid is a function of pressure of gas over liquid.(ii)Henry's law can be applied on such system, that states, the mass of gas dissolved in a given mass ofsolvent at any temperature is proportional to the pressure of the gas above the solvent.C gP or C = k Pgggwherek = Henry's constantC = Solubility of gas in the solution (mol L )g–1P = Pressure of the gasg(iii)One should not compare it with liquid vapour equilibria.2 .Chemical equilibria :When chemical change occur in a reversible reaction i.e. reactants convert into products and products alsoconvert into reactants under similar conditions of pressure and temperature, the reaction is said to be inchemical equilibria.(i)H (g) + I (g) 22 2HI (g) (formation of HI)(ii)SO (g) + Cl (g)2 2 SO Cl (g) (formation of SO Cl )2222(iii)PCl (g) + Cl (g)32 PCl (g) (formation of PCl )55(iv)2NH (g)3  N (g) + 3H (g) (Decomposition of NH )223

EXAMPLES :Ex.Chemical equilibrium is a condition :(A) where all species have same concentration(B) where all species have constant concentration with respect to time.(C) where all species have concentration = 1(D) all of aboveSol.( B )Chemical equilibrium defined as when all species have constant concentration with respect to time.Ex.Example of physical equilibria, is :(A) H (g) + I (g) 22 2HI (g)(B) CaCO (s)3  CaO (s)+ CO (g)2(C) H O (s) 2 H O ( )2(D) PCl (g) 5 PCl (g) + Cl (g)32Sol.( C )Physical equilibria does not include any chemical change.Ex.At equilibrium :(A) the energy of system is minimum(B) the entropy of system is maximum(C) the energy of system is maximum(D) the entropy of system is minimumSol.(A,B)It is the compromising stage of minimum energy and maximum entropy.LAW OF MASS ACTION OR LAW OF CHEMICAL EQUILIBRIUM :The law of mass action is given by Guldberg and Waage.According to them at a given temperature rate of reaction is proportional to product of active masses ofreactants at that instant raised to the powers which are numerically equal to the number of their respectivemolecule in the stoichiometric equation describing the reaction.Derivation of equilibrium constant :Consider a reversible homogeneous reaction which has attained equilibrium state at particular temperature :A + B C + DLet the active masses of A, B, C and D be [A] [B] [C] & [D] are respectively.According to law of mass action :rate of forward reaction [A] [B]rate of backward reaction [C] [D]R = K [A] [B]ffR = K [C] [D]bbWhere K and K are forward and backward rate constants respectively.fbAt equilibriumR = RfbK [A] [B] = K [C] [D]fbfbCDKKAB            cCDKAB           fcbKKK K is known as equilibrium constant K has a definite value for every chemical reaction at particular tempera ture.cc

qFor a general reactionm A + m B + m C 123 n M + n N + n O123312mmmf r[A] [B] [C]r = K ff312mmm[A] [B] [C][K = forward rate (velocity) constant]f312nnnb r[M] [N] [O]r = K bb312nnn[M] [N] [O][K = backward rate (velocity) constant]bAt equilibriumr = rfbK f312mmm[A] [B] [C] = Kb 312nnn[M] [N] [O]123123nnnfcmmmbMNOKKKABC                uThe equilibrium constant, at a given temperature, is the ratio of the rate constants of forward and backward reactions.Ex.Write down the equilibrium constant for the following reactions.(a)N + 3H 222NH3(b)PCl 5 PCl + Cl32(c)3A + 2B C + 4D(d)CaCO (s) 3 CaO (s) + CO (g)2(e)2KClO (s) 3 2KCl(s) + 3O (g)2(f)CH COOH ( ) + C H OH ( ) 325  CH COOC H ( ) + H O ( )325 2(g)NH (aq) + H O 32 NH (aq) + OH (aq)4+–(h)H O ( ) 2  H O (g)2Ans. (a)2 3322NHKNH  (b)325PClClKPCl  (c)432CDKAB             (d)K = [CO ] (Active mass of solid is 1)2(e)K =[O ]23(f)3252325CH COOC HH OKCH COOHC H OH     (here H O is not in excess)2(g)43NHOHKNH   (here H O is in excess (solvent) so its concentration doesn't change.)2(h)K = [H O]2(g)qPossible value of K[0 < K < ]When K = 1[Product] = [Reactant]When K > 1[Product] > [Reactant]When K < 1[Product] < [Reactant]As K stability of products stability of reactant time to attain equilibrium t 1/KqStability of reactants and products2XO (g)  X (g) + O (g) ;22K = 1 × 101242XO (g) 2 X (g) + 2O (g) ;22K = 2.5 × 10210K > K12So the stability of XO > XO2For reactants, stability increases when value of K decreases.For products, stability increases when value of K increases.(more is the value of equilibrium constant, more is the formation of product means more is the stability ofproduct.)qTime taken to attain equilibrium increases when value of K decreases.

qFORMS OF EQUILIBRIUM CONSTANT :There are two forms.(i) Concentration form (K )C(ii) Partial pressure form (K )PPartial pressure : The individual pressure exerted by the gases substance of the total pressure is called partialpressure of the gases substance.Gaseous molesClosed vesselP totalAn mol1 Cn mol3 Bn mol2 Dn mol4 Partial pressure=Moles of substanceTotal moles PtotalMoles of subs tanceMole fractionTotal moles= mole fraction × Ptotallet n + n + n + n = N12341AtnPP ,N 2BtnPP ,N 3CtnPP ,N 4DtnPPN P + P + P + P = PABC DtotalQ.A vessel contains 5 mole of A & 10 moles of B. If total pressure is 18 atm. Find out partial pressure of gases.Ans. P = A515 × 18 = 6 atmP = B1015× 18 = 12 atmuWhen the reactants and products are in gaseous state then partial pressure can be used instead of concentra-tion. At a definite temperature, as the partial pressure of a substance is proportional to its concentration in thegas phase.m A + m B 12n C + n D12If partial pressure of A, B, C and D at equilibrium are P , P , P and P respectively, thenABCD    1212nncDPmmABPPKPPRELATION BETWEEN K AND K :PCThis relation can be established for reaction not involving liquids because kp is not defined for liquids.Consider a reversible reactionm A + m B 12n C + n D121212nnCmmCDKAB           1212nnCDPmmAB(P ) (P )K(P ) (P ) For an ideal gasPV = nRTn PRTV = active mass RTn = number of mole and V = Volume in litreSonV = molar concentration or active massP = [ ] RTat constant temperature P [ ]

P = [A] RT,AP = [B] RT,BP = [C] RT,CP = [D] RTDSo K11221212nnnnpmmmmC(RT)D(RT)A(RT)B(RT)        K p 12121212nnnnmmmmCDRTRTAB             1212nnmmPCKKRTn = (n + n ) (m + m )1212= total number of gaseous molecules of products – total number of gaseous molecules of reactants. n gPCKKRTEx.Find the values of K for each of the following equilibria from the value of K .cp(a) 2NOCl (g)  2NO (g) + Cl (g)2K = 1.8 × 10 at 600 Kp–2(b) CaCO (g) 3 CaO (s) + CO (g)2K = 167 at 1173 KpSol.(a) 2NOCl (g)  2NO (g) + Cl (g)2K = 1.8 × 10p–2n = 3 – 2 = 1gK = K (RT)pcn g K =cp KRT=21.8 100.0821 600= 3.65 × 10–4(b) K = 167pn = 1gK = K (RT)pcn g = K × (RT)c K =cp KRT=1670.0821 1173= 1.734Ex.At 540 K, 0.10 moles of PCl are heated in 8 litre flask. The pressure of the equilibrium mixture is found to5be 1.0 atm. Calculate K and K for the reaction.pcSol.PCl (g) 5 PCl (g) + Cl (g)32 0.1 0 0(0.1–x) x xK = c325[PCl ][Cl ][PCl ] = xx880.1x8  = 2x8(0.1x) ........(i)From gas lawPV = nRT 1 × 8 = (0.1 + x) × 0.082 × 540 x = 0.08........(ii)From eqs. (i) and (ii)K =c0.08 0.088 (0.1 – 0.08)= 4 × 10 mol L–2–1K = K (RT)pcn g( n = +1) gR=gas constant=0.0821 lit. atm. K mol–1–1=8.314 J K mol–1–1 =1.98 2 Cal K mol~–1–1

= 4 × 10 × (0.082 × 540) = 1.77 atm–2 Ex.At a given temperature and a total pressure of 1.0 atm for the homogeneous gaseous reaction, N O242NO ,2the partial pressure of NO is 0.5 atm.2(a) Calculate the value of K .p(b) If the volume of the vessel is decreased to half of its original volume, at constant temperature, what arethe partial pressures of the components of the equilibrium mixture ?Sol.For equilibrium system, N O 24 2NO , the total pressure is 1.0 atm2The total pressure = 242N ONOPP = 124 N O P= 0.5 atm and 2 NO P= 0.5 atm(i)K =p2242NON OPP=2(0.5)0.5= 0.5 atm(ii) As volume is decreased to half its original volume, equilibrium is disturbed and the new initial conditionsfor the re-establishment of new equilibrium are24 N O P = 1.0 atm and 2 NO P = 1.0 atmAccording to Le Chatelier's principal, when volume is decreased, the system moves in that directionwhere there is decrease in number of moles. Hence, the system (here will moves in reverse direction, asthere is a decrease in mole ( n = 2 – 1 = 1), i.e. NO will be converted into N O .224Let, the decrease in pressure of NO be x atm.2 N O24 2NO2 Initial pressure (atm) 1.01.0At equilibrium1+x/21–xK =p2(1x)(1x / 2)= 0.5  4x – 9x + 2 = 02x = 2 or 0.25 (x 2 as initial pressure = 1.0)  x = 0.2524 N OxP12   = 1.125 atm and 2 NO P= 1 – x = 0.75 atmEx.At temperature T, a compound AB (g) dissociates according to the reaction, 2AB (g) 22 2AB (g) + B (g)2with degree of dissociation , which is small compared to unity. Deduce the expression for in terms of theequilibrium constant K and the total pressure P.pSol.2AB (g) 2  2AB (g) + Br (g)2 Initial (mole) 1 0 0At eq. (mole)1 – /2 Total moles at equilibrium = (1 + /2)At eq. (p.p)P11/ 2    P.1/ 2 P. / 2(1/ 2) K =p222ABBr2ABPPPK = p222P.P. / 21/ 21/ 21P1/ 2         K =p32P2(1) (1/ 2)  But 1 >>   K =p 3P2 

 =1 / 3p 2KP uThe K is expressed by the units (mole lit ) and K by (atm) .C  1nP nIn terms of mole fraction, equilibrium constant is denoted by K .XFor general reaction aA + bB  mC + nDK =X mnCDabAB(X ) (X )(X ) (X )K = K (RT)px n g............(i)n = (m + n) – (a + b)gWhen n = 0, K = K = K gpCXSome General Equilibrium Expressions :(a)H (g)2+I (g)22HI (g)Initially a b 0At equilibrium(a–x)(b–x) 2xK = c222[HI][H ][I ]=2(2x)(ax)(bx)=24x(ax)(bx)K = p222HIHI(p )pp=222(2x)P(ab)axbx.P.Pabab   =24x(ax)(bx)So K = Kc p( n = 0)(b)2NO (g) N (g) + O (g)22Initially a 0 0At equilibrium(a–x) x/2 x/2K = c222[N ][O ][NO] = 2x / 2 x / 2(ax) =2P2xK4(ax)( n = 0)(c)CH COOH + C H OH 3( ) 25( )  CH COOC H + H O325( )2( ) Initially a b 0 0At equilibrium (a–x) (b–x) x xK =c3252325[CH COOC H ][H O][CH COOH][C H OH]=2x(ax)(bx)K should not be given for this reactionp(d)PCl (g)5 PCl (g) + Cl (g)32 Initially a 0 0 At equilibrium(a–x) x xActive mass(ax)vxvxvK =c325[PCl ][Cl ][PCl ]=xxvv(ax)v=2x(ax)v 

K =p325PClClPClppp=xx.P.PaxaxaxPax        =2x P(ax)(ax)=222x Pax (e)N (g)2+3H (g) 2 2NH (g)3 Initially a b 0At equilibrium (a–x) (b–3x) 2xActive mass(ax)vb3xv 2xvK =c23322[NH ][N ][H ]=232xvaxb3xvv  =2234x V(ax)(b 3x)K =p3222NH3NH(p)p(p )=232xPab 2x(ax)P(b3x)P(ab 2x)(ab 2x)       =22324x (ab 2x)(ax)(b 3x) P Ex.In the reaction, H (g) + I (g) 22 2HI(g) the concentration of H , I and HI at equilibrium are 10.0, 6.0 and2228 moles per litre respectively. What will be the equilibrium constant?(A) 30.61(B) 13.066(C) 29.40(D) 20.90Sol.( B )H (g) + I (g) 22 2HI(g)Applying law of mass action,K =c222[HI][H ][I ]Given[H ] = 10 mol L2–1[I ] = 6.0 mol L2–1[HI] = 28.0 mol L–1So,K = c2(28.0)(10) (6.0) = 13.066Ex.For a gas phase reaction at equilibrium,3H (g) + N (g) 22 2NH (g), the partial pressures of H and N are 0.4 and 0.8 atmosphere respectively.322The total pressure of the entire system is 2.4 atmosphere. What will be the value of K if all the pressures arePgiven in atmosphere ?(A) 32 atm–2(B) 20 atm–2(C) 28.125 atm–2(D) 80 atm–2Sol.( C )N (g) + 3H (g) 22 2NH (g),3Partial pressures at equilibrium0.8 0.4 [2.4 – (0.8 + 0.4) = 1.2]

Applying law of mass action,K = P3222NH3NH[P][P ][P ]=1.2 1.20.8 0.4 0.4 0.4K = 28.125 atmP–2Ex.When ethanol and acetic acid were mixed together in equilimolecular proportion 66.6% are converted intoethyl acetate. Calculate K . Also calculate quantity of ester produced if one mole of acetic acid is treated withc0.5 mole and 4 mole of alcohol respectively.(A) 4, 0.93, 0.43(B) 0.93, 4, 0.43(C) 0.43, 0.93, 4(D) 4, 0.43, 0.93Sol.( D ) CH COOH + C H OH 325 CH COOC H + H O325 2 1 1 0 0 1–0.666 1–0.666 0.666 0.666K = c3252325[CH COOC H ][H O][CH COOH][C H OH]=[0.666][0.666][0.333][0.333] = 4(a) Let x moles of ester is formed from 1 mole of acid and 0.5 mole of alcohol, then K =c2x(1x)(0.5x)2x4 (1x)(0.5x) x = 0.43(b) K =c2x(1x)(4x) or 4 =2x(1x)(4x) x = 0.93Ex.Starting with 3 : 1 mixture of H and N at 450°C, the equilibrium mixture is found to be 9.6% NH ; 22.6%223N and 67.8 % H by volume. The total pressure is 50 atm. What will be the value of K . The reaction is -22PN + 3H 22 2NH3(A) 3.25 × 10 atm–5–2(B) 5.23 × 10 atm (C) 6.23 × 10 atm–5–2 –5–2(D) 8 × 10 atm–5–2Sol.( B )The ratio of number of moles will be the same as the ratio of volume. According to Dalton's law, the partialpressure of a gas in a mixture is given by the product of its volume fraction and the total pressure. Therefore,the equilibrium pressure of each gas is,3 NH9.6P50 atm = 4.8 atm100 2 N22.6P50 atm = 11.3 atm100 2 H67.6P50 atm = 33.9 atm100  Total pressure = 50 atmK =P3222NH3NH[P][P ][P ] ; Substituting the values of partial pressures,K =P23(4.80 atm)(11.3 atm)(33.9 atm)= 5.23 × 10 atm–5–2Ex.K for the reaction A(g) + 2B(g) P 3C(g) + D(g) ; is 0.01 atm. What will be its K at 1000 K in terms of R ?c(A) 51.0 10R(B) 5R5 10(C) 5 × 10 R–5 (D) none of theseSol.( A )We know thatKP = K (RT) or K =c ncPnK(RT)Here n = 4 – 3 = 1

T = 1000 K, K = 0.01PK =c10.01(R 1000) = 51.0 10REx.0.5 mole of H and 0.5 mol of I react in 200 L flask at 448° C. The equilibrium constant K is 50 for22cH + I 22 2HI,(a) What is the K ?P(b) Calculate mol of I at equilibrium.2Sol.H + I 22 2HI0.5 0.5 0 Initial(0.5–x) (0.5–x) 2x at equili.(a) Since n = 0  K = KPc(b) 50 =224x(0.5x)  or 2x500.5x  x = 0.39 mol of I = 0.5 – 0.39 = 0.11 mol2uThree cases may arise :(a)When n = 0K = K (RT) = KPC0CFor example :N + O 222NOH + I 222HIuK and K are unit less in this case.CP(b)When n = +veK > KPCFor example :PCl5PCl + Cl321CPKmole litKatm ( n = 1)2NH3N + 3H2222C2PKmole litKatm ( n = 2)(c)When n = veK < KPCN + 3H 222NH322C2PKmolelitKatm ( n = –2)uFactors affecting equilibrium constant :( a )Mode of representation of the reaction :A + B C + DThe equilibrium constant for the reactionCCDKAB            If the reaction is reversedC + D A + Bthen,1cABKCD             

The two equilibrium constant related as c1c1KK Ex.For N + 3H 22 2NH if K = 5 then find K ' for reverse reaction.3CCAns. K ' = 1/5 = 0.2C(b)Stoichiometry of the reaction :When a reversible reaction can be written with the help of two or more stoichiometric equation, the value ofequilibrium constant will be numerically different.For reaction2NO 2N + 2O22222C2 2NOKNO   For reaction NO2221NO212 221c2NOKNO   The two constants are related as 1cCKK ( c )Temperature : The value of equilibrium constant changes with the change of temperature.If K and K be the equilibrium constants of a reaction at absolute temperatures T and T and H is the heat1212of reaction at constant volume, then :2d( nk)HdTRT 221121KH11loglog Klog KK2.303R TT (According to Vant Hoff equation)(i)H = 0 (neither heat is absorbed or evolved)log K – log K = 021log K = log K12K = K12Thus, equilibrium constant remains the same at all temperaturesIf temp. T is higher than T2121110 , log K – log K = TT 21ve H2.303R (ii)When H = +ve (endothermic reaction)log K log K > 021orlog K > log K21K > K21The value of equilibrium constant is higher at higher temperature in case of endothermic reactions.K TC(iii)When H = ve (exothermic reaction)log K log K < 021log K < log K21 K < K21The value of equilibrium constant is lower at higher temperature in the case of exothermic reactions.K 1/TCuThe value of equilibrium constant is independent of the following factors :-(a) Initial concentrations of reactants.(b) The presence of a catalyst.

(c) The direction from which the equilibrium has been attained.(d) Presence of inert materials.Ex.For the reaction,A  B, H for the reaction is –33.0 kJ/mol.Calculate :(i)Equilibrium constant K for the reaction at 300 Kc(ii)If E (f) and E (r) in the ratio of 20 : 31, calculate E (f) and E (r) at 300 K.aaaaAssuming pre-exponential factor same for forward and reverse reaction.Sol.(i)H = E – E = – 33 kJa(f)a(r)Progress of reactionEHEa (r)Ea(f)k = fEa(f)/RTAek = bEa( r)/RTAek = cfbEE/ RTa(r )a(f )kek  333 108.314 300e = 5.572 × 10 at 300 K5 (ii)a(f)a(r )EE = 2031E – E = – 33kJa(f)a(r) E – a(r)3120 × E = – 33kJa(f) E = a(f)33 2011  = 60kJ E = + 93 kJa(r)Ex.The equilibrium constant for the reaction H (g) + S (s) 2 H S(g) ; is 18.5 at 925 K and 9.25 at 1000 K2respectively. The enthalpy of the reaction will be :(A) – 68000.05 J mol–1(B) –71080.57 J mol (C) – 80071.75 J mol–1–1(D) 57080.75 J mol–1Sol.( B )Using the relation,log 21KK = 2112HTT2.303RT Tlog 9.2518.5=H2.303 8.314× 75925 1000–0.301 = H 752.303 8.314 925 1000  H = –71080.57 J mol .–1Ex.The reaction CuSO .3H O(s) 42 CuSO .H O(s)+ 2H O (g) ; the dissociation pressure is 7 × 10 atm at422–325°C and H° = 2750 cal. What will be the dissociation pressure at 127°C ?Sol.For given reactionK = pp  22H Oso K (25°C) = (7 × 10 ) atmp –3 2 2 = 4.9 × 10 atm–5 2Since H° = 2750 cal, so using Vant Hoff eq.

logPPK (127 C)K (25 C) =H2.303 R 400298400 298logP5K (127 C)4.9 10=27502.303 2102119200 K (127°C) = 3.2426 × 4.9 × 10 = 1.58 × 10p–5 –4so 2H O p at 127° = pK (127 C) = 41.58 10 = 1.26 × 10 atm.–2Law of Mass Action as Applied to Heterogeneous Equilibrium :In such cases the active mass of pure solid and pure liquid is taken as unity and the value of equilibrium constantis determined by the gaseous substances only.For example : The dissociation of CaCO in closed vessel.3CaCO (s) 3 CaO(s) + CO (g)2K = [CO ],C2K = PpCO2uPCl (s) 5 PCl ( ) + Cl (g)32K = [Cl ],C2K = Pp Cl2u2H O( ) 2 2H (g) + O (g)22K = [H ] [O ], K = (p ) (p )C222P H 22 O 2u3Fe(s) + 4H O(g) 2 Fe O (s) + 4H (g)3422244H2CP442H O(p)[H ]K, K[H O](p)Ex.One mole of ammonium carbamate dissociate as shown below at 500 K.NH COONH (s) 24 2NH (g) + CO (g)32If the pressure exerted by the released gases is 6.0 atm, the value of K is -P(A) 7 atm(B) 3 atm(C) 32 atm(D) 8 atmSol.( C )Applying the law of chemical equilibrium, we getK = (PpNH3) (P 2 CO2)Since total pressure is 6 atm, the partial pressures of NH (g) and CO (g) are32(PNH3) = 6 × 23 = 4 atm(PCO2) = 6 × 13 = 2 atmK = [4.0] [2.0] = 32.0 atmp2 Ex.For the reaction.CaCO (s) 3 CaO (s) + CO (g) ; K = 1.16 atm. at 800°C. If 40 g of CaCO was put into a 20 L2p3container and heated to 800°C, what percent of CaCO would remain unreacted at equilibrium.3Sol.K = p2 CO P = 1.16 atmn(CO ) = PV/RT = 21.16 200.0821 1073 = 0.26335 molmoles of CaCO initially present3 = 40/100 = 0.4 mol

So % decomposition of CaCO =30.26335100 = 65.83 % decomposed0.4Hence 34.17 % remain unreacted.Ex.For the reaction :SnO (s) + 2H (g) 22 2H O (g) + Sn (s)2Calculate K at 900 K where the equilibrium steam hydrogen mixture was 35 % H by volume.p 2Sol.K =p222H O2H(P)(P )given H is 35% by volume at constant temperature in closed vessel (P V)2so 2H O P= 0.65 atm and 2 H P= 0.35 atmK =p 20.650.35 = 3.448LE-CHATELIER'S PRINCIPLE :According to this principle. If a system at equilibrium is subjected to a change of concentration, pressure ortemperature, the equilibrium is shifted in such a way as to nullify the effect of change.( a )Change in concentration : In an equilibrium increasing the concentrations of reactants results in shifting theequilibrium in favour of products while increasing concentrations of the products results in shifting theequilibrium in favour of the reactants.( b )Change of pressure : When the pressure on the system is increased, the volume decreases proportionately.The total number of moles per unit volume increases. According to Le-Chatelier's principle, the equilibriumshift in the direction in which there is decrease in number of moles.If there is no change in number of moles of gases in a reaction, a pressure change does not affect theequilibrium.( c )Change in temperature :- If the temperature at equilibrium is increased reaction will proceed in the directionin which heat can be used. Thus increase in temperature will favour the forward reaction for endothermicreaction.Similarly, increase in temperature will favour the backward reaction in exothermic reactions.APPLICATION OF LE-CHATELIER'S PRINCIPLE :( A )Chemical equilibria :( a ) Formation of HIH (g) + I (g) 22 2HI(g) + 3000 Cals( i )Effect of concentration : When concentration of H or I is increased at equilibrium, the system moves in22a direction in which decreases the concentration. Thus the rate of forward reaction increases thereby increas-ing the concentration of HI.(ii)Effect of pressure :- In formation of HI, there is no change in the number of moles of reactants and products( n = 0). Thus it is not affected by the change in pressure or volume.(iii)Effect of temperature :- The formation of HI is exothermic reaction. Thus the backward reaction movesfaster when temperature is increased. i.e. formation of HI is less.In short favourable conditions for greater yield of HI :High concentration of H and I .22Low temperature.No effect of pressure( b ) Formation of NO :N + O 22 2NO – 43200 cals.( i )Effect of concentration : When concentration of N or O is increased, the system moves in a direction in22

which N or O is used up or rate of forward increases.22(ii)Effect of pressure : The formation of NO is not affected by change in pressure.( n = 0).(iii)Temperature : The formation of NO is endothermic. Thus increase in temperature favours to forwardreaction.Favourable conditions for greater yield of NO :High concentration of N and O .22High temperature.No effect of pressure( c ) Dissociation of PCl :5PCl (g) 5PCl (g) + Cl (g) 15000 cals.32( i )Effect of concentration : When concentration of PCl is increased at equilibrium, the rate of forward5reaction increases as to decrease the added concentration. Thus dissociation of PCl increases.5(ii)Effect of pressure : The volume increases in the dissociation of PCl when pressure is increased, the system5moves in the direction in which there is decrease in volume. Thus high pressure does not favour dissociation ofPCl .5(iii)Effect of temperature :- The dissociation of PCl is an endothermic reaction. Thus increase of temperature5favours the dissociation.Favourable conditions for dissociation of PCl are :-5High concentration of PCl .5Low pressure.High temperature.( d ) Synthesis of ammonia :N (g) + 3H (g) 222NH (g) + 22400 Cals.3The favourable conditions for greater yield of NH are :-3High concentration of N and H .22High pressure.Low temperature.( e ) Formation of SO :32SO (g) + O (g) 22 2SO + 45200 Cals.3The favourable conditions for greater yield of SO are :-3High concentration of SO and O .22High pressure.Low temperature.Ex.In reaction,CO(g) + 2H (g) 2 CH OH (g)3H° = – 92 kJ/mol–1concentrations of hydrogen, carbon monoxide and methanol become constant at equilibrium. What willhappen if :(A) volume of the reaction vessel in which reactants and products are contained is suddenly reduced to half ?

(B) partial pressure of hydrogen is suddenly doubled?(C) an inert gas is added to the system at constant pressure?(D) the temperature is increased ?Sol.For the equilibrium,CO (g) + 2H (g) 2 CH OH (g)3K = c322[CH OH][CO][H ] K = p32CH OH2COHPPP(A) When the volume of the vessel is suddenly reduced to half, the partial pressures of various species getsdoubled. Therefore, Q =p32CH OHp2COH2P1K42P2PSince Q is less than K , the equilibrium shift in the forward direction producing more CH OH.pp3(B) When partial pressure of hydrogen is suddenly doubled, Q changes and is no longer equal to K .p pQ =p32CH OHp2COHP1K4P2PEquilibrium will shift from left to right.(C) When an inert gas is added to the system at constant pressure, equilibrium shifts from lower number ofmoles to higher number of moles (in backward direction).(D) By increasing the temperature, K will decrease and equilibrium will shift from right to left.pSPECIAL POINTS :(1)Irreversible reaction proceeds in one direction and completed with time while reversible reaction proceeds inboth direction and are never completed.(2)Equilibrium is defined as the point at which the rate of forward reaction is equal to the rate of backwardreaction.(3)Chemical equilibrium is dynamic in nature and equilibrium state can be approached from both sides.(4)Active mass is molar concentration of the substance. Active mass of solid and pure liquid is taken as unity.(5)Equilibrium constant has definite value for every chemical reaction at a given temperature. It is independent ofconcentration and catalyst.(6)If a reaction can be expressed as the sum of two or more reactions then overall K will be equal to the productCof the equilibrium constant of individual reaction.Example :SO (g) + 212 O (g) 2 SO (g) 3K1NO (g) 2 NO(g) + 12 O (g) 2 K2thenSO (g) + NO (g) 22 SO (g) + NO(g) 3 KSoK = K K12(7)Change in temperature, pressure or concentration favours one of the reactions and thus shift the equilibriumpoint in one direction.(8)A catalyst ables the system to reach a state of equilibrium more quickly.(9)Pressure and volume has no effect on the reaction in which there is no change in the number of moles.(10)If the concentration of reactants is increased and product is removed, the reaction will take place in forwarddirection.

(11)Free energy changeG = G° + 2.303 RT log Q At equilibrium G = 0, (T is in Kelvin), Q = K soG° = – 2.303 RT log K, where K is equilibrium constant.Ex.G° for ½ N + 3/2 H 22 NH is –16.5 kJ mol . Find out K for the reaction at 25°C. Also report K3–1PPand G° for N + 3H 22 2NH at 25°C.3Sol.log K = – P G2.303 RT  = 316.5 102.303 8.314 298 = 2.8917K = antilog (2.8917) = 779.41PNow given reaction N + 3H 22 2NH can be obtained by multiplying eq. 1/2 N + 3/2 H 322 NH3by 2.so K ' = (779.41) = 6.07 × 10p 2 5and G° = –2.303 RT log K' = – 2.303 × 8.314 × 298 log (6.07 × 10 ) Jp5G° = –32.998 kJ mol .–1Ex.For the gaseous reaction CO + H O 2CO + H the following thermodynamics data are given.2 2H°300 K = – 41.16 kJ mol ; S°–1 300 K = –0.0424 kJ mol .–1H°1200 K = – 32.93 kJ mol ; S°–1 1200 K = –0.0296 kJ mol .–1Assuming partial pressure of each component at 1 atm. determine the direction of spontaneous reaction at(i) 300 K(ii) 1200 K. Also calculate K for the reaction at each temperature.P Sol.Using G° = H° – T S°G°300 K = –41.16 – 300 (–0.0424)= –28.44 kJ mol–1so reaction is spontaneous in given direction since G° is negativeCO + H O 2 CO + H22at 1200 KG°1200 K = –32.93 – 1200 (–0.0296)= 2.56 kJ mol–1so reaction will not be spontaneous in given direction, but reverse reaction spontaneous i.e.CO + H 2 CO + H O2We know G° = – 2.303 RT log KPso K (300 K)P = antilog 328.44 102.303 8.314 300  = 8.8 × 104

K = (1200 K)P= antilog 32.59 102.303 8.314 1200 = 0.77REACTION QUOTIENT AND EQUILIBRIUM CONSTANT :Consider the follwong reversible reactionA + B  C + DThe reaction quotient (Q ) is ratio of the product of active masses of the products and product of active massesCof the reactants, at any given time.Q = C [C][D][A][B]The concentration are not necessarily equilibrium concentration.[At equilibrium Q = K ]CCCase I :If Q < K then : [Reactants] > [Products]CC then the system is not at equilibriumThe value of [Product][Reactant] is smallFor establishment of equilibrium the reaction will go in forward direction. [Reactants  Products]Case II :If Q = K then : The system is at equilibrium and the concentration of the species C,D,B,A are at equilibrium.CC Case III :If Q > K then : [Product] > [Reactants]CCThe system is not at equilibrium.The value of [Product][Reactant] is largeFor establishment of equilibrium the reaction will go in backward direction. [Products  Reactants]Ex.A mixture of 4.2 moles of N , 2.0 moles of H and 10.0 moles of NH is introduced into a 10.0 L reaction vessel at223500 K. At this temperature, equilibrium constant K is 1.7 × 10 , for the reaction N (g) + 3H (g)c2222NH (g)3 (i)is the reaction mixture at equilibrium ?(ii) if not, what is the direction of the reaction?Sol.[N ] = 24.210 = 0.42 M[H ] = 22.010 = 0.2 M[NH ] = 31010 = 0.1 MFor these concentration, reaction quotient (Q) for the reactionN (g) + 3H (g) 22 2NH (g) is3

Q = 23322[NH ][N ][H ]= 23(0.1)(0.42) (0.2) = 2.976But K = 1.7 × 10c2(i)Since Q K , hence reaction is not at equilibrium.c(ii)Also Q < K , the reaction will proceed from left to right.cCALCULATION OF DEGREE OF DISSOCIATION FROM VAPOUR DENSITY MEASUREMENT :Degree of dissociation :Degree of dissociation of a substance at a particular temperature is defined as the fraction of total number ofmoles dissociated into simpler molecules at that particular temperature.Degree of dissociation ( ) = No. of moles dissociatedTotal no. of moles takenDegree of dissociation can be calculated from vapour density measurements for those substance which areaccompanied by change in the number of moles.e.g.PCl 5 PCl + Cl32Initial moles 1 0 0No. of moles after(1– ) dissociationTotal number of moles = 1 – + + = (1 + )Let, volume occupied by the vapour per mole = V litres.Initial vapour density = DVapour density after dissociation = dAs, vapour density 1V D 1Vd 1(1)V D1 / V1d(1)V = (1 + )  = D(Dd)1 dd As, mol. wt. = 2 × vapour densityAlso, = tooMMM M = theoretical molecular masstM = observed (experimental) molecular masso M can be calculated from the mass of definite volume of the vapour at particular temperature.o Also, PV = nRT = oWRTMM = oWRTRTvPP where = density of the vapour.

In general, if one mole dissociates to give n moles of products, thenAnB10(1 – )n Total no. of moles = 1 – + n = 1 + (n – 1)D1 (n 1)d    = (Dd)(n 1) dAlso, = tooMM(n 1) MLet us, consider the reaction, 2NH (g) 3 N (g) + 3H (g)22If the initial moles of NH (g) be 'a' and x moles of NH dissociated at equilibrium.332NH (g) 3 N (g) + 3H (g)22Initial moles a 0 0At equilibriuma – x x23x2Degree of dissociation ( ) of NH is defined as the number of moles of NH dissociated per mole of NH .333 If x moles dissociate from 'a' moles of NH then, the degree of dissociation of NH would be 33xa.We can also look at the reaction in the following manner.2NH (g) 3 N (g) + 3H (g)22Initial moles a 0 0At equilibriuma(1 – ) a23a2ora – 2x' x' 3 x'where = 2x 'aHere, total number of moles at equilibrium is a – 2x' + x' + 3x' = a + 2x'Mole fraction of NH = 3a 2x 'a 2x 'Mole fraction of N = 2x 'a 2x 'Mole fraction of H = 23x 'a 2x 'The expression of K ispK = p33TT22Tx '3x 'PPa2x 'a2x 'a 2x 'Pa2x '   = 42T2227x'P(a 2x')(a2x')In this way, you should find the basic equation. So, it is advisable to follow the below mentioned steps whilesolving the problems. Write the balanced chemical reaction (mostly it will be given). Under each component write the initial number of moles. Do the same for equilibrium condition. Then derive the expression for K and K accordingly.pc

Ex.Vapour density of the equilibrium mixture NO and N O is found to be 42 for the reaction,224N O 24 2NO . Find2(a) Abnormal molecular weight(b) Degree of dissociation(c) Percentage of NO in the mixture2Sol.(a) For the reaction, N O 24 2NO2 Observed value of vapour density = 42 Abnormal molecular weight = 42 × 2(d = 42)(b) Theoretical molecular weight = 92 2 × D = 92 D = 92462  =Dd4642d42= 0.095(c) N O 24 2NO2 1 0 (1 – ) 2 0.905 0.19Total moles at equilibrium = 1 + = 1 + 0.095 % of NO = 2 2100(1)  = 0.19100 = 17.35 %1.095Ex.The equilibrium constant of the reaction A (g) + B (g) 22 2AB (g) at 50°C is 50. If one litre flask containingone mole of A is connected to a two litre flask containing two moles of B , how many moles of AB will be22formed at 323 K.Sol.A (g) + B (g) 22 2AB (g) ; K = 50cInitial mole 1 2 0At eq. mole 1 – x 2 – x 2xAt eq. conc.1 x3 2x3 2x3K = c22x3501 x2x33   23x – 75x + 50 = 02x = 0.934 or 2.326Only 0.934 values is permissibleSo, moles of AB = 1.868Ex.Calculate the % age dissociation of H S (g) if 0.1 mole of H S is kept in a 0.4 L vessel at 900 K. The value22of K for the reaction, 2H S (g) c 2 2H (g) + S (g), is 1.0 × 10 .224Sol.2H S 2 2H + S22Volume of vessel = V = 0.4 LLet, x be the degree of dissociation

Moles2H S 22H2 +S 2Initially0.100At equilibrium0.1 – 0.1x0.1x0.1x/2K = c 22222[H ] [S ][H S]= 220.01x0.01v2v0.01 0.01xv    = 10 4 x = 0.02 or 2% dissociation of H S2Ex.The degree of dissociation of HI at a particular temperature is 0.8. Find the volume of 1.5 M sodiumthiosulphate solution required to react completely with the iodine present at equilibrium in acidic condition,when 0.135 mol each of H and I are heated at 440 K in a closed vessel of capacity 2.0 L.22Sol.First find the value of K for dissociation of HI from its degree of dissociationc2HI  H + I (degree of dissociation is 0.8)22Concentrations2HI H2 +I2Initially1.000At new equilibrium1.0–0.80.40.4K =c222[H ][I ][HI]=2(0.4)(0.4)4(0.2)Now, we have to start with 0.135 mol each of H and I and the following equilibrium will be established.22H + I 22 2HI with K = ¼cConcentrations H +2 I2 2HIInitially0.1350.135 0At new equilibrium0.135–x0.135–x 2x K =c222[HI][H ][I ]=2(2x)1(0.135x)(0.135x)4  x = 0. 135/5 = 0.027 molesNow, find the moles of I left unreacted at equilibrium.2n = 0.135 – 0.027 = 0.108 molesI2I reacts with sodium thiosulphate (Na S O ) as follows :22 232Na S O + I 2 232 Na S O + 2NaI2 46 Applying mole concept, we have 2 moles of Na S O2 23  1 mole of I2 0.108 moles of I 2 × 0.108 = 0.216 moles of Na S O are used up22 33 Moles = MV (M = Molarity, V = volume in litres)nn 0.216 = 1.5 V V = 0.144 lt = 144 mL.

SR nRvvUU HNbKCKPKCKPKP C1H2+2 0KP= C(0KP= CNNN–(t22 2+20KP= C(0KP= CNNN+(t3P5 3+2+ 1KP= C(1KP> C( (p1o t4N2O4 2+ 1KP= C(1KP> C(+(p–o t523 2+2+ 2KP= C(2KP> C(+(p2o t6N2+2 3-KP= C(–KP< C(–(p–o t7P3+2 5-KP= C(–KP< C(–(p–o t822+ 223-KP= C(–KP< C(–(p–o tRbad(pgm1l–gm1l–gm2l–gm–l2gm–l+gm–l+CopLNHHNHHLHHLHHLHLHHLHHLHH

MEMORY TIPS1 .Law of mass action. It was put forward by Guldberg and Waage. It states that the rate at which a substancereacts is directly proportional to its active mass and hence the rate at which substances react together is directlyproportional to the product of their active masses. Active mass means molar concentration.2 .Law of chemical equilibrium. For the reaction aA + bB  xX + yY, xyab[X] [Y]K , called equilibrium[A] [B]constant which is constant for a reaction at constant temperature.3 .Equilibrium constant in terms of concentrations (K ) is K =ccxyab[X] [Y][A] [B].It has units = (mol L )–1(x + y) – (a + b)Equilibrium constant in terms of pressures is K = p xyXY abABP PP PIt has units = (atm)(x + y) – (a + b)Expressed in terms of activities (in place of molar concentration), equilibrium constant is dimensionless.4 .Relation between K and K . K and K are related to each other as K = K (RT)pcpcpc n gWhere n = (n – n ) gaseous gpr5 .Concentration Quotient condition or Reaction Quotient (Q). For the reaction aA + bB  xX + yY,at any other than the stage of equilibrium, the expression xyab[X] [Y][A] [B]= Q is called concentration quotient orreaction quotient.(i)If Q = K, the reaction is in equilibrium.(ii)If Q < K, Q will tend to increase till it becomes equal to K. Hence, reaction proceeds in the forwarddirection.(iii)If Q > K, Q will tend to decrease. As a result, the reaction will proceed in the backward direction.6 .Effect of temperature on K. For aA + bB fbkk C + D, K = fbkk.For exothermic reaction, k decreases with increase of temperature, so K decreases.fFor endothermic reaction, k increases with increase of temperature, so K increases.f7 .Effect of adding inert gas at equilibrium.(i)For reactions in which n = n , there is no effect of adding an inert gas at constant volume or atprconstant pressure on the equilibrium.(ii)For reaction in which n > n (e.g. PCl pr5 PCl + Cl ), there is no effect of adding inert gas on the32equilibrium at constant volume but at constant pressure, equilibrium shifts in the forward direction.8 .Le Chatelier's principle states that \"if a system in equilibrium is subjected to a change of concentration,temperature or pressure, the equilibrium shifts in a direction so as to undo the effect of the change imposed.\"9 .Van't Hoff reaction isotherm. It is an equation which gives the relationship between standard free energychange ( G°) of a reaction and its equilibrium constant (K ),pi.e. G° = – RT lnKpThis equation helps to calculate G° of a reaction at temperature T if its equilibrium constant at this temperatureis known or vice-versa.1 0 .Van't Hoff equation. This equation gives the variation of equilibrium constant of a reaction with temperature.The equation is

p2d ln KHdTRT The integrated form of this equation is221112KHTTlogK2.303RT T where H° = enthalpy change of the reaction (assumed to the constant in the temperature range T to T ).121 1 .Units of equilibrium const. = (mol L )–1(x + y) – (a + b) or (atm)(x + y) – (a + b)1 2 .Degree of dissociation of PCl or N O is given by =524Ddd=tooMMM where D = theoretical vapour density and d = vapour density after dissociation (observed V.D.), M = theoreticalt(calculated) molecular mass and M = observed molecular mass.o1 3 .From integrated form of van't Hoff equation, viz.log 21KHK2.303R  2112TTT TWe may conclude that(i)If H° = 0, i.e. no heat is evolved or absorbed in the reaction.log (K /K ) = 0, i.e. K /K = 1 or K = K .2121 21So, equilibrium constant does not change with temperature.(ii)If H° = +ve, i.e. heat is absorbed in the reaction, thenlog (K /K ) = +ve or log K > log K or K > K .2121 21So, equilibrium constant increases with increase in temperature.(iii)If H° = –ve, i.e. heat is evolved in the reaction, thenlog (K /K ) = –ve, i.e. log K < log K or K < K .2121 21So, equilibrium constant decreases with increase in temperature.

SOLVED PROBLEMS (SUBJECTIVE)Ex .1(i)Consider the heterogeneous equilibriumCaCO (s) 3 CaO (s) + CO (s)2K = 4 × 10 atmp–2..........(i)C (s) + CO (g) 2 2CO (g)K ' = 4.0 atmp......... (ii)Calculate the partial pressure of CO (g) when CaCO and C are mixed and allowed to attain equilibrium at3the temperature for which the above two equilibria have been studied.(ii)Calculate the pressure of CO gas at 700 K in the heterogeneous equilibrium reaction.2CaCO (s) 3 CaO (s) + CO (g)2If G° for this reaction is 120.2 kJ/mol.Sol.(i)For Eq. (i), K = p2 CO PFrom Eq. (ii), K ' = Pp22COCo / P K × K ' = (P ) = 4 × 10 × 4 = 16 × 10 atmppCO2 –2–22 P = CO2216 10 atm = 0.4 atm(ii)G° = –2.303 RT log Kp log K = –p G2.303RT–3111120.2 10 Jmol2.303 (8.314JK mol ) (700 K) K = 1.00 × 10 atm = Pp–9 CO2Ex .2For the dissociation reaction N O (g) 24 2NO (g)2derive the expression for the degree of dissociation in terms of K and total pressure P.pSol. N O (g) 24 2NO (g)2Let initial no. of moles 1 0Moles at equilibrium (1– ) 2K =p g2242nNON OnPnn=       2(2 1)2P(1) (12 ) =     12 4P(1) (1)=   224P(1)P4PK =  22(1)=   21121p4P 1K pp4PKK 2 =  ppK4PK=  ppK4PK

Ex .3The value of K is 1 × 10 atm at 25°C for the reaction, 2NO + Cl p–3–12 2NOCl. A flask containsNO at 0.02 atm and 25°C. Calculate the moles of Cl that must be added if 1% of NO is to be converted2to NOCl at equilibrium. The volume of the flask is such that 0.2 moles of the gas produce 1 atm pressureat 25°C (Ignore the probable association of NO to N O ).22Sol.Let, initial pressure of added Cl is p atm.2 2NO + Cl 22NOCl Initial 0.02atm p atm 0At equilibrium  0.020.021000.01 p100 0.02100 = 2 × 10 (100 – 1) = p–10 atm = 2 × 10 atm–4–4–4 = 198 × 10 atm–4K = p22NOCl2NOClPPP 10 = –34 24 24(2 10 )(198 10 )(p 10 ) p – 10 = –4234(198)10= 0.102 p = 0.102 + 0.0001 = 0.1021 atmVolume of the vessel can be calculated as follows,PV = nRTor V = nRTP = 0.2 0.082 298L = 4.887 L1Again applying, (PV = nRT) we can calculate the number of moles of Cl2n =Cl2PVRT=0.1021 4.8870.082 298=0.0204 mol.Ex .4When 1-pentyne (A) is treated with 4 N alcoholic KOH at 175°C, it is converted slowly into an equilibriummixture of 1.3% 1-pentyne (A), 95.2% 2-pentyne (B) and 3.5% of 1,2-pentadiene (C). The equilibriumwas maintained at 175°C. Calculate G° for the following equilibria.B  A1 G?B  C2 G?From the calculated value of 1 G  and 2 G  indicate the order of stability of (A), (B) and (C). Write areasonable reaction mechanism showing all intermediates leading to (A), (B) and (C).Sol.Pentyne KOH 2-Pentyne + 1,2-Pentadiene (A) (B)(C)At eqm.% 1.3 95.53.5K = c[B][C][A]= 95.2 3.51.3=256.31..........(i)From eqm. B  AK = 1[A][B]From Eqs. (i) and (ii), K = 1c[C]K=3.5256.31=0.013..........(ii)G°= –2.303 RT log K101= – 2.303 × 8.314 × 448 log 0.01310= 16178.4= 16.1784 kJ

Stability order for A and B is B > ASimilarly, B  CK = 2[C][B] = c2K[A][B] = 256.31 3.195.2 95.2 = 0.0876  G° = –2.303 RT log K2 102= –2.303 × 8.314 × 448 log 0.087610= 9068.06 J = 9.068 kJThus, stability order for B and C is B > CTotal order of stability is B > C > A.Ex .5The density of an equilibrium mixture of N O and NO at 1 atm is 3.62 g/L at 288 K and 1.84 g/L at242348 K. Calculate the entropy change during the reaction at 348 K.Sol.N O 24 2NO2Case (i)PV = nRT = mixwRT m mmix = wRTdRTVPP = 3.62 × 0.082 × 288 = 85.6Let, a mole of N O and (1–a) mole of NO exist at equilibrium242 a × 92 + (1 – a) × 46 = 85.6 a = 0.8624 N O n= 0.86 mol,2 NO n= 0.14 molK =p10.14 0.1410.861   = 0.0228 atm at 288 K.Case (ii)m =mixdRTP= 1.84 × 0.0821 × 348 = 52.57Let, a' mol of N O and (1 – a') mol of NO exist at equilibrium242 a' × 92 + (1 – a') × 46 = 52.57 a' = 0.1424 N O n= 0.14 mol,2 NO n= 0.86 mol K = p10.86 0.86 10.141    = 5.283 atm at 348 Klog1021ppKK =H2.303R2112TTT T log 105.2830.0228=H2 2.303348288348 288H = 18195.6 cal = 18.196 KcalG = –2.303 RT log Kp= – 2.303 × 2 × 348 × log 5.283= –1158.7 cal.S = HGT   = 18195.6 1158.7348 = 55.62 cal

Ex .6For the reaction, [Ag(CN) ]2–  Ag + 2CN , the equilibrium constant, K at 27°C is 4.0 × 10+–c–19.To find the silver ion concentration in a solution which is originally 0.10 M in KCN and 0.03 M inAgNO .3Sol.Ag + 2CN +– [Ag(CN) ]2–K ' = c22[Ag(CN) ][Ag ][CN ]=c1K= 2.5 × 1020........(i)Very high value of K ' show that complex forming equilibrium is spontaneous and almost all the Ag ionc+would have reacted leaving xM in solution : Ag + 2CN +–2 [Ag(CN) ]initial0.03M 0.1M 0At eqm. xM (0.1 – 0.03 × 2x)M 0.03 MK ' = 2.5 × 10 =c2020.03x(0.1 0.03 2x) x = [Ag ] = 7.5 × 10 M+–18Ex .7In an experiment, 5 moles of HI were enclosed in a 10 litre container. At 817 K equilibrium constant forthe gaseous reaction, 2HI (g)  H (g) + I (g), is 0.025. Calculate the equilibrium concentrations of22HI, H and I . What is the fraction of HI that decomposes?22Sol.Let, 2n be the number of moles of HI which is decomposed, the number of moles of H and I produced22will be n mole each. Then molar concentrations of various species at equilibrium are[HI] = (5 2n)10  mol/L,[H ] = 2n10 mol/L, and [I ] = 2n10 mol/LAlso, K = c222[H ][I ][HI]=2nn10 105 2n10 0.025 =22n(52n)Solving for n, we get n = 0.6 [HI] = 52 0.610 = 3.810= 0.38 mol/L [H ] = 20.610 = 0.06 mol/L [I ] = 20.610 = 0.06 mol/LFraction of HI decomposed = 2 0.65  = 0.24 or 24%

Ex .80.5 moles of N and 3 moles of PCl are placed in a 100 litre container heated to 227°C. The equilibrium25pressure is 2.05 atm. Assuming ideal behaviour, calculate the degree of dissociation of PCl and value of5K for its dissociation.pSol.Dissociation of PCl is written as5PCl (g) 5 PCl (g) + Cl (g)32 Let, x be the no. of moles of PCl decomposed at equilibrium5PCl (g)5 PCl (g) + Cl (g)32Initial moles3 0 0Moles at eqm.3–x x xNow, total gaseous moles in the container = nTn = moles of (PCl + PCl + Cl ) + moles of NT5 3 22n = 3 – x + x + x + 0.5 = 3.5 + xTThe mixture behaves ideally, hence PV = n RTTLet us, calculate no. of moles by using gas equation n = TPVRT=2.05 1000.0821 400 n = 5TNow, equating the two values of n , we haveT3.5 + x = 5  x = 1.5 degree of dissociation = 1.5/3 = 1/2 = 0.5Now, K =p325PClClPClPPP5 PCl P = 3xP = 3.5x1.52.055 = 0.615 atm3 PCl P =1.55× 2.05 = 0.615 atm2 Cl1.5P5 × 2.05 = 0.615 atmK = p325PClClPClPPPatm  K =p 0.615 0.6150.615 = 0.615 atmK = 0.615 atmpNote :The inert gases like N or noble gases (He, Ne etc.) though do not take part in the reaction, but still they2affect the degree of dissociation and equilibrium concentrations for the reactions in which n 0. They add to the total pressure of the equilibrium mixture (p n).Ex .9For the reaction, CaCO (s) 3 CaO(s) + CO (g) ; K = 0.059 atm at 1000 K. 1 g of CaCO is placed2 3in a 10 litre container at 1000 K to reach the equilibrium. Calculate the mass of CaCO left at equilibrium.3Sol. CaCO (s) 3 CaO(s) + CO (g)2 At equilibrium a – xx xHere, a = initial moles of CaCO3K = PpCO2 = 0.059nCO2 = 2 CO PVRT=0.059 100.082 1000 = 7.2 × 10 moles–3Moles of CaCO left = 0.01 – 0.0072 = 0.00283Mass of CaCO left = 0.28 g3

Ex. 10The value of K for the reaction, 2H O (g) + 2Cl (g) p22 4HCl (g) + O (g) is 0.035 atm at 500°C,2when the partial pressures are expressed in atmosphere. Calculate K for the reaction,c21O (g) + 2HCl (g) 2 Cl (g) + H O (g)2 2Sol.K = K (RT)pc nn = moles of products – moles of reactants = 5 – 4 = 1R = 0.0821 L atm/mol/K, T = 500 + 273 = 773 K 0.035 = K (0.0821 × 773)cK = 5.515 × 10 mol Lc–4–1 K ' for the reverse reaction would be cc1K K ' =c415.515 10=1813.24 (mol L )–1 –1When a reaction is multiplied by any number n (integer or a fraction) then K ' or K ' becomes (K ) or (K )cpcn pnof the original reaction. K for c12O (g) + 2HCl (g) 2  Cl (g) + H O (g)22is 1813.24 = 42.58 (mol.L )–1 –½Ex. 11K for the reaction N O (g) p242NO (g) is 0.66 at 46°C. Calculate the percent dissociation of N O at22446°C and a total pressure of 0.5 atm. Also calculate the partial pressure of N O and NO at equilibrium.242Sol.This problem can be solved by two methods.Method 1 : Let, the number of moles of N O initially be 1 and is the degree of dissociation of N O .2424N O 24 2NO2Initial moles 1 0Moles at equilibrium 1– 2Total moles at equilibrium = 1 – + 2 = 1 + 24 N O1p1    × PT2 NOT2pP1   K = p2242NON Opp=2T4P(1)(1)  =2240.51   = 0.5, i.e. 50% dissociationHence, partial pressure of N O = 0.167 atm.24and partial pressure of NO = 0.333 atm.2Method 2 : Let, the partial pressure of NO at equilibrium be p atm, then the partial pressure of N O at224equilibrium will be (0.5 – p) atm. K = p 2p0.66(0.5p)p + 0.66 p – 0.33 = 02 On solving, p = 0.333 atm.2 NO p= 0.333 atm and 24 N O p= 0.167 atm.

Ex. 12Ammonium hydrogen sulphide dissociated according to the equation,NH HS (s) 4NH (g) + H S (g). If the observed pressure of the mixture is 2.24 atm at 106°C, what is32the equilibrium constant K of the reaction ?pSol.The reaction is NH HS (s) 4 NH (g) + H S (g).32If is the degree of dissociation of equilibrium,Total moles of NH and H S at equilibrium = 2 .32Partial pressure = Moles of substanceTotal pressureTotal no.of moles3 NH p2× P = 0.5 P ; p2H S2× P = 0.5 PK = p32NHH Spp = 0.5 P × 0.5 P = 0.25P2Substituting the value of P = 2.24 atm.K = 0.25 × 2.24 × 2.24 = 1.2544 atmp2Alternatively :At equilibrium 3 NH p + 2H S p = 2.24 atmAs 32NHH Spp3 NH p=2.242= 1.12 atm K = 1.12 × 1.12 = 1.2544 atmp2Ex. 13In a mixture of N and H , initially they are in a mole ratio of 1 : 3 at 30 atm and 300°C, the percentage22of ammonia by volume under the equilibrium is 17.8%. Calculate the equilibrium constant (K ) of thepmixture, for the reaction, N (g) + 3H (g) 22 2NH (g).3Sol.Let, the initial moles N and H be 1 and 3 respectively (this assumption is valid as K will not depend on22pthe exact number of moles of N and H . One can even start with x and 3x)22N (g) + 3H (g) 22 2NH (g)3 Initially1 3 0At equilibrium1–x 3–3x 2xSince % by volume of a gas is same as % by mole,2x42x = 0.178 x = 4 0.178(2 2 0.178)  = 0.302 Mole fraction of H at equilibrium = 233x42x = 0.6165Mole fraction of N at equilibrium = 1 – 0.6165 – 0.178 = 0.20552 K = p3222NHT3NTHT(XP )(XP )(XP )=23(0.178 30)(0.2055 30)(0.6165 30)K = 7.31 × 10 atm .p–4–2Ex. 14Given below are the values of H° and S° for the reaction at 27°C,SO (g) + 212O (g) 2 SO (g)3H° = 98.32 kJ/mol S° = –95 J/mol. Calculate the value of K for the reaction.pSol.log K = 10pH2.303 RT  +S2.303 R  log K = 10p983202.303 8.314 300 – 952.303 8.314 K = 1.44 × 10 atmp 1212 