CHEMISTRY XI CHAPTERS 8-1 BKLET-
Contents: Page No. Chapter 8 Chemial Equilibrium 181-199 Chapter 9 Redox Reations 00-19 Chapter 10 s & p Blok Elements part 1 0-49 Chapter 11 s & p Blok Elements part 50-77 Chapter 1 s & p Blok Elements part 78-98
Chemial Equilibrium Whenever we hear the word Equilibrium immediately a piture arises in our mind an objet under the influene of two opposing fores. For hemial reations also this is true. A reation also an exist in a state of equilibrium balaning forward and bakward reations. Equilibrium and its dynami nature. (1) Definition : Equilibrium is the state at whih the onentration of reatants and produts do not hange with time. i.e. onentrations of reatants and produts beome onstant. () Charateristis : Following are the important harateristis of equilibrium state, (i) Equilibrium state an be reognised by the onstany of ertain measurable properties suh as pressure, density, olour, onentration et. by hanging these onditions of the system, we an ontrol the extent to whih a reation proeeds. Conentration Rate of reation Time Equilibrium state Time Produts Reatants Constant level utlet Equilibrium state (ii) Equilibrium state an only be ahieved in lose vessel, but if the proess is arried out in an open vessel equilibrium state annot be attained beause in an open vessel, the reverse proess will not take plae. (iii) Equilibrium state is reversible in nature. (iv) Equilibrium state is also dynami in nature. Dynami means moving and at a mirosopi level, the system is in motion. The dynami state of equilibrium an be ompared to water tank having an inlet and outlet. Water in tank an remain at the same level if the rate of flow of water from inlet (ompared to rate of forward reation) is made equal to the rate of flow of water from outlet (ompared to rate of bakward reation). Thus, the water level in the tank remains onstant, though both the inlet and outlet of water are working all the time. (v) At equilibrium state, Rate of forward reation = Rate of bakward reation (vi) At equilibrium state, G = 0, so that H = T S. () Types : Equilibrium in a system implies the existene of the following types of equilibria simultaneously, (i) Thermal equilibrium : There is no flow of heat from one part to another i.e. T = onstant. (ii) Mehanial equilibrium : There is no flow of matter from one part to another i.e. P = onstant. (iii) Physial equilibrium : There is the substane exist in three states: solid, liquid and gaseous. (iv) Chemial equilibrium : There is no hange in omposition of any part of the system with time. Inlet G= 0 Physial equilibrium. The physial equilibrium is a state of equilibrium between the same hemial speies in different phases (solid, liquid and gaseous). The various equilibria whih an exist in any physial system are, Chapter 8 181
Chemial Equilibrium Solid Liquid Liquid Vapour Solid Gas(vapour) Solid Saturated solution of solid in a liquid Gas(vapour ) Saturated solution of gas in a liquid (1) Solid-liquid equilibrium Rate of transfer of moleules from ie to water = Rate of transfer of moleules from water to ie Rate of melting of ie = Rate of freezing of water Y Rate Vaporisation Free energy hange and solid-liquid equilibrium in water : For ie-water system free energy hange ( G), at 7 K, and one atmosphere pressure is zero i.e., G = 0; Ie Water; H ( ) H ( ) (i) At temperature higher than 7 K, and 1 atm pressure, G < 0. Thus, the proess in the forward diretion would beome favourable and ie will melt to give more water. (ii) At temperature less than 7 K, and 1 atm pressure, G > 0. Thus, the reverse reation will beome favourable, and more ie will be formed from liquid water. () Liquid-vapour equilibrium : A liquid plaed in an open ontainer disappears ompletely. After some time vapours of the liquid held in an open ontainer an esape out to the atmosphere. Thus, when vapour of liquid exists in equilibrium with the liquid, then Rate of vaporisation = Rate of ondensation, H ( ) H ( ) l v Conditions neessary for a liquid-vapour equilibrium (i) The system must be a losed system i.e., the amount of matter in the system must remain onstant. (ii) The system must be at a onstant temperature. (iii) The visible properties of the system should not hange with time. s Condensation () Solid-vapour equilibrium : Certain solid substanes on heating get onverted diretly into vapour without passing through the liquid phase. This proess is alled sublimation. The vapour when ooled, gives bak the solid, it is alled disposition. Solid Vapour Time l X The substanes whih undergo sublimation are amphor, iodine, ammonium hloride et. For exampie, Ammonium hloride when heated sublimes. NH Cl ( ) NH Cl ( ) 4 s heat ool 4 v (4) Equilibrium between a solid and its solution : When a saturated solution is in ontat with the solid solute, there exists a dynami equilibrium between the solid and the solution phase. Sugar moleules Saturated solution of sugar Solid sugar Chapter 8 18
Chemial Equilibrium Solid substane Solution of the substane Example : Sugar and sugar solution. In a saturated solution, a dynami equilibrium is established between dissolved sugar and solid sugar. Sugar (s) Sugar (aq) At the equilibrium state, the number of sugar moleules going into the solution from the solid sugar is equal to the number of moleules preipitating out from the solution, i.e., at equilibrium, Rate of dissolution of solid sugar = Rate of preipitation of sugar from the solution. (5) Equilibrium between a gas and its solution in a liquid : Gases dissolve in liquids. The solubility of a gas in any liquid depends upon the, (i) Nature of the gas and liquid. (ii) Temperature of the liquid. A hilled soda water bottle fizzes out when opened beause, soda water is a solution of arbon (iii) Pressure of the gas over the surfae of the solution. Henry s law : At a ertain temperature, the mass of a gas whih dissolves in a definite volume of a liquid is proportional to the pressure of the gas in equilibrium with the solution. m P or m = KP ; (where K is the proportionality onstant) Thus, at a onstant temperature, the ratio of the molar onentration of the gas in the solution and into the atmosphere is onstant. Limitations of Henry s law Henry s law is appliable to ideal gases only. Henry s law should be applied only at low pressures beause real gases behave like ideal gases at low pressures. Henry s law is not appliable to gases whih reat hemially with the solvent. Henry s law will not apply to the solution of gases like ammonia ( NH ) and hydrogen hloride (HCl) in water beause these gases reat hemially with water. Note : Chemial equilibrium. dioxide gas, C ( ) in water at high pressure. As soon as the bottle is opened under normal g atmospheri onditions, the dissolved gas esapes out to reah a new equilibrium state, so that the pressure of the gas inside the bottle beomes equal to the atmospheri pressure. At low pressure, the solubility of the gas in water dereases. The equilibrium between different hemial speies present in the same or different phases is alled hemial equilibrium. There are two types of hemial equilibrium. (1) Homogeneous equilibrium : The equilibrium reations in whih all the reatants and the produts are in the same phase are alled homogeneous equilibrium reations. Example : (i) C H H l) + CH CH ( ) CH + 5 ( l (ii) N g) + H ( ) NH g ( g (iii) S ( g) + ( ) g ( ) gas phase S ( g) gas phase CC H 5 ( l) H liquid ( l) phase () Heterogeneous equilibrium : The equilibrium reations in whih the reatants and the produts are present in different phases are alled heterogeneous equilibrium reations. Chapter 8 18
Chemial Equilibrium Example : (i) CaC ( ) Ca ( s) + C ( g) Note : s (ii) NaHC ( s ) Na C s ) + C ( g ) + H ( ) (iii) H ( ) H ( ) l g ( g + (iv) Ca H) ( s) + H ( ) Ca ( aq) + H ( aq) ( l The equilibrium expression for heterogeneous reations does not inlude the onentrations of pure solids beause their onentrations remain onstant. Reversible and irreversible reations. A hemial reation is said to have taken plae when the onentration of reatants dereases, and the onentration of the produts inreases with time. The hemial reations are lassified on the basis of the extent to whih they proeed, into the following two lasses; (1) Reversible reations : Reations in whih only a part of the total amount of reatants is onverted into produts are termed as reversible reations. (i) Charateristis of reversible reations (a) These reations an be started from either side, (b) These reations are never omplete, () These reations have a tendeny to attain a state of equilibrium, (d) This sign ( ) represents the reversibility of the reation, (e) Free energy hange in a reversible reation is zero ( G = 0), (ii) Examples of reversible reations (a) Neutralisation between an aid and a base either of whih or both are weak e.g., CH CH + Na H CH CNa + H (b) Salt hydrolysis, e.g., Fe Cl + H Fe ( H) + 6HCl () Thermal deomposition, e.g., PCl 5( g) PCl ( g) + Cl ( g) Q Ca + C ; HI ( g) H + I ( g) CaC( s) ( s) ( g) (d) Esterifiation, e.g., CH CH + CH5H CH CCH5 + H (e) Evaporation of water in a losed vessel, e.g., H ( l) H ( g) Q (f) ther reations, e.g., N ( g) + H ( g) NH ( g) ( g) + Q ; S ( g) + ( g) S ( g) + Q () Irreversible reations : Reations in whih the entire amounts of the reatants are onverted into produts are termed as irreversible reations. (i) Charateristis of irreversible reations (a) These reations proeed only in one diretion (forward diretion), (b) These reations an proeed to ompletion, () The arrow ( ) is plaed between reatants and produts, (d) In an irreversible reation, G < 0, (ii) Examples of irreversible reations Chapter 8 184
Chemial Equilibrium (a) Neutralisation between strong aid and strong base e.g. NaH + HCl NaCl + H + 1.7 kal (b) Double deomposition reations or preipitation reations e.g. BaCl + H S BaS + HCl ; AgN ( aq) + NaCl( aq) AgCl( g) + NaN( aq) ( aq) 4 ( aq) 4 ( s) ( aq) () Thermal deomposition, e.g. Mn, ( s ) ( s) KCl KCl + Where [A] and [B] are the molar onentrations of the reatants A and B respetively, k is a onstant of heat 4 heat 4 Pb ( N ) Pb + N + ; NH N N + H (d) Redox reations, e.g., SnCl ( aq) + FeCl ( aq) SnCl 4 ( aq) + FeCl ( aq) (e) ther reations, e.g., Law of mass ation. NaCl + H S4 NaHS4 + HCl n the basis of observations of many equilibrium reations, two Norwegian hemists Guldberg and Waage suggested (1864) a quantitative relationship between the rates of reations and the onentration of the reating substanes. This relationship is known as law of mass ation. It states that The rate of a hemial reation is diretly proportional to the produt of the molar onentrations of the reatants at a onstant temperature at any given time. The molar onentration i.e. number of moles per litre is also alled ative mass. It is expressed by enlosing the symbols of formulae of the substane in square brakets. For example, molar onentration of A is expressed as [A]. Let us onsider a simple reation between the speies A and B : A + B Produts Aording to law of mass ation, rate of reation, r [A] [B] = k [A] [B] proportionality for the forward reation and is known as rate onstant. The rate onstant is also alled veloity onstant. Now, if the onentration of eah of the reatants involved in the reation is unity, i.e., [A] = [B] = 1, then, rate of reation, r = k 1 1 or r = k Thus, the rate onstant of a reation at a given temperature may be defined as the rate of the reation when the onentration of eah of the reatants is unity. For a general reation, aa + bb + C Produts The law of mass ation may be written as : Rate of reation, a r = k[ A] [ B] [ C] b Thus, the law of mass ation may be restated as, The rate of a hemial reation at any partiular temperature is proportional to the produt of the molar onentrations of reatants with eah onentration term raised to the power equal to the number of moleules of the respetive reatants taking part in the reation. The number of moleules of a reatant taking part in a reation is also alled its stoihiometri oeffiient. For example, a, b and. in the above equation are alled stoihiometri oeffiients of A, B and C. respetively. Chapter 8 185
Chemial Equilibrium Equilibrium onstant. (1) Equilibrium onstant in terms of law of mass ation : The law of mass ation may be applied to a reversible reation to derive a mathematial expression for equilibrium onstant known as law of hemial equilibrium. k / is alled equilibrium onstant and has a onstant value Let us onsider a simple reversible reation, A + B X + Y in whih an equilibrium exists between the reatants (A and B) and the produts (X and Y). The forward reation is, A + B X + Y Aording to law of mass ation, Rate of forward reation [ A][ B] = k [ A][ B] Where k f is the rate onstant for the forward reation and [A] and [B] are molar onentrations of reatants A and B respetively. Similarly, the bakward reation is ; X + Y A + B Rate of bakward reation [ X][ Y] = k [ X][ Y] Where k b is the rate onstant for the bakward reation and [X] and [Y] are molar onentrations of produts X and Y respetively. At equilibrium, the rates of two opposing reations beome equal. Therefore, at equilibrium, Rate of forward reation = Rate of bakward reation k [ A][ B] k [ X][ Y] k k f b f = [ X][ Y] = [ A][ B] The ombined onstant K, whih is equal to or f b [ X][ Y] K = [ A][ B] for a reation at a given temperature. The above equation is known as law of hemial equilibrium. For a general reation of the type : aa + bb C + dd The equilibrium onstant may be represented as : k b [ C] [ D] K = a [ A] [ B] where the exponents a, b, and d have the same values as those in the balaned hemial equation. Thus, the equilibrium onstant may be defined as, The ratio between the produts of molar onentrations of the produts to that of the molar onentrations of the reatants with eah onentration term raised to a power equal to its stoihiometri oeffiient in the balaned hemial equation at a onstant temperature. () Charateristis of equilibrium onstant (i) The value of equilibrium onstant is independent of the original onentration of reatants. For example, the equilibrium onstant for the reation, b f d b Fe + + ( aq) + SCN ( aq) = FeSCN ( aq) ; + [ FeSCN ] 1 = 18.0 L + K = mol (at 98 K) [ Fe ][ SCN ] Whatever may be the initial onentrations of the reatants, to be 18.0 L mol 1 at 98 K. Chapter 8 186 + Fe and SCN ions, the value of K omes out
Chemial Equilibrium (ii) The equilibrium onstant has a definite value for every reation at a partiular temperature. However, it varies with hange in temperature. For example, the equilibrium onstant for the reation between hydrogen and iodine to form hydrogen iodide is 48 at 717 K. [ HI] H ( g) + I ( g) = HI( g) ; K = = 48 [ H ][ I ] For example, if equilibrium onstant, K, for the reation of ombination between hydrogen and iodine at 717 K is 48 For this reation, the value of K is fixed as long as the temperature remains onstant. (iii) For a reversible reation, the equilibrium onstant for the forward reation is inverse of the equilibrium onstant for the bakward reation. [ HI] H ( g) + I ( g) HI (g); K = = 48 [ H ][ I ] Then, the equilibrium onstant for the deomposition of hydrogen iodide is the inverse of the above equilibrium onstant. In general, [ H ][ I ] 1 1 HI(g) H ( g) + I ( g) ; K = = = = 0. 0 [ HI] K 48 K forward reation = K 1 bakward reation (iv) The value of an equilibrium onstant tells the extent to whih a reation proeeds in the forward or reverse diretion. If value of K is large, the reation proeeds to a greater extent in the forward diretion and if it is small, the reverse reation proeeds to a large extent and the progress in the forward diretion is small. (v) The equilibrium onstant is independent of the presene of atalyst. This is so beause the atalyst affets the rates of forward and bakward reations equally. (vi) The value of equilibrium onstant hanges with the hange of temperature. Thermodynamially, it an be shown that if K 1 and K be the equilibrium onstants of a reation at absolute temperatures T 1 and T. If H is the heat of reation at onstant volume, then H 1 1 log K log K1 = (Van t Hoff equation).0 R T T1 The effet of temperature an be studied in the following three ases (a) When H = 0 i.e., neither heat is evolved nor absorbed log K log K1 = 0 or log K = log K1 or K = K1 Thus, equilibrium onstant remains the same at all temperatures. (b) When H = +ve i.e., heat is absorbed, the reation is endothermi. The temperature T is higher than T 1. log K log K1 = + ve or log K > log K1 or K > K1 The value of equilibrium onstant is higher at higher temperature in ase of endothermi reations. () When H = ve, i.e., heat is evolved, the reation is exothermi. The temperature T is higher than T 1. log K log K1 = ve or log K 1 > log K or K 1 > K The value of equilibrium onstant is lower at higher temperature in the ase of exothermi reations. (vii) The value of the equilibrium onstant depends upon the stoihiometry of the hemial equation. Chapter 8 187
Chemial Equilibrium Examples : (a) If the equation (having equilibrium onstant K) is divided by, then the equilibrium onstant for the new equation is the square root of K i.e. K. For example, the thermal dissoiation of S an be represented in two ways as follows, S ( g ) S ( g ) + ( ) and S ( ) S g ) + 1/ ( ) g (b) Similarly, if a partiular equation is multiplied by, the equilibrium onstant for the new reation (K ) will ] [ ] [ S ] K = and [ S g [ S][ ] K = ; [ S ] 1 / K = ( g K or ( K) be the square of the equilibrium onstant (K) for the original reation i.e., 1 / K = K () If the hemial equation for a partiular reation is written in two steps having equilibrium onstants K 1 and K, then the equilibrium onstants are related as K = K 1 K For example, the reation N g) + ( ) N ( g ) with equilibrium onstant (K) an be written in two steps : ( g N ( g) + ( g) N (g) ; (Equilibrium onstant = K 1 ) N ( g ) + ( g ) N ( g) ; (Equilibrium onstant = K ) Now, Therefore, [ N] K 1 = and [ N ][ ] [ N] K = [ N] [ ] [ N] [ N] [ N] K 1 K = = = K [ N ][ ] [ N] [ ] [ N ][ ] () Types of equilibrium onstant : Generally two types of equilibrium onstants are used, (i) K It is used when the various speies are generally expressed in terms of moles/litre or in terms of molar onentrations. (ii) K p It is used when in gaseous reations, the onentration of gases expressed in terms of their partial pressures. K p is not always equal to K. K p and K are related by the following expression, where, R = Gas onstant = 0.081 bar dm 1 1 mol k ; T = Temperature in Kelvin K p = K ( RT) n = number of moles of gaseous produts number of moles of gaseous reatants in hemial equation (4) Unit of equilibrium onstant : Equilibrium onstant K has no units i.e., dimensionless if the total number of moles of the produts is exatly equal to the total number of moles of reatants. n the other hand if the number of mioles of produts and reatants are not equal, K has speifi units. Units of K p and K and the value of n Value of n Relation between K p and K Units of K p Units of K 0 K p = K No unit No unit >0 K p > K (atm) n (mole l 1 ) n <0 K p < K (atm) n (mole l 1 ) n n Chapter 8 188
Chemial Equilibrium (5) Appliations of equilibrium onstant : Knowing the value of the equilibrium onstant for a hemial reation is important in many ways. For example, it judge the extent of the reation and predit the diretion of the reation. (i) Judging the extent of reation We an make the following generalisations onerning the omposition of equilibrium mixture. (a) If K > 10, produts predominate over reatants. If K is very large, the reation proeeds almost all the way to ompletion. (b) If K < 10, reatants predominate over produts. If K is very small, the reation proeeds hardly at all. () If K is in the range This is illustrated as follows, 10 to Reation proeeds hardly at all 10, appreaiable onentration of both reatants and produts are present. (ii) Reation quotient and prediting the diretion of reation : The onentration ratio, i.e., ratio of the produt of onentrations of produts to that of reatants is also known as onentration quotient and is denoted by Q. [ X][ Y] Conentration quotient, Q =. [ A][ B] It may be noted that Q beomes equal to equilibrium onstant (K) when the reation is at the equilibrium state. At equilibrium, Q = K = K = K. Thus, p (a) If Q > K, the reation will proeed in the diretion of reatants (reverse reation). (b) If Q < K, the reation will proeed in the diretion of the produts (forward reation). () If Q = K, the reation mixture is already at equilibrium. Thus, a reation has a tendeny to form produts if Q < K and to form reatants if Q > K. This has also been shown in figure, 10 K Both reatants and produts are present at equilibrium 10 Reation proeeds to ompletion Q K Q K Q K Reatants Produts Equilibrium Reatants Produts Chapter 8 189
Chemial Equilibrium (6) Calulation of equilibrium onstant : We have studied in the harateristis of the equilibrium onstant that its value does not depend upon the original onentrations of the reatants and produts involved in the reation. However, its value depends upon their onentrations at the equilibrium point. Thus, if the equilibrium onentrations of the speies taking part in the reation be known, then the value of the equilibrium onstant and vie versa an be alulated. Homogeneous equilibria and equations for equilibrium onstant (Equilibrium pressure is P atm in a V L flask) n = 0 ; K p = K n < 0 ; K p < K n > 0 ; K p > K H ( g) + I ( g) g HI ( ) N ( + H g) ( g) NH ( g) S + ( g) ( g) S ( g) PCl 5 ( g) PCl + Cl Initial mole 1 1 0 1 0 1 0 1 0 0 Mole at (1 x) (1 x) (1 x) ( x) x ( x) (1 x) x (1 x) x x Equilibrium x Total mole at equilibrium (4 x) ( x) (1 + x) Ative masses Mole fration Partial pressure K K p 1 V x 1 x 1 x p 1 V x 1 x x V x 1 x x p p 4x 1 V x 1 x ( x) 1 x x V V 1 x x x ( x) V x x x 1 V x 1 x x x V x x 1 V x 1 x 1 + x 1 x (1 x) Px x P P ( x)_ P 1 x x 1 x P P P ( x) ( x) x x x 1 + x 4x V ( 1 x) 7 ( 1 x) 4 ( 1 x) 4x 16x ( x) x ( x) ( 1 x) 4 7( 1 x) P P ( 1 x) ( g) 1 x V x + x x P 1 + x ( g) x V x ( 1 x)v Px ( 1 x ) Heterogeneous equilibria and equation for equilibrium onstant (Equilibrium pressure is P atm) 1 x V x + x x P 1 + x NH HS( ) NH ( ) + H S( ) C ( s) + C( g) C ( g) NH C NH ) NH ( g) + C ( ) 4 s g g 4 ( s g Initial mole 1 0 1 1 0 1 0 0 0 Mole at equilibrium (1 x) x (1 x) (1 x) (1 x) x x x x Total moles at equilibrium (solid not inluded) x (1+x) x Mole fration x 1 x = 1 1 x 1 1 + x x 1 + x Partial pressure P P 1 x P P 1 + x x P P 1 + x K p P 4 4P x 4P (1 x ) 7 Chapter 8 190
Chemial Equilibrium (7) Equilibrium onstant and standard free energy hange : Standard free energy hange of a reation and its equilibrium onstant are related to eah other at temperature T by the following relation, when, G o G o =.0 RT log K = ve, the value of equilibrium onstant will be large positive quantity and when, G o = + ve, the value of equilibrium onstant is less than 1 i.e., low onentration of produts at equilibrium state. Example : 1 In the reversible reation A + B C + D, the onentration of eah C and D at equilibrium was 0.8 Solution: (d) Example : mole/litre, then the equilibrium onstant K will be [MP PET 1986] (a) 6.4 (b) 0.64 () 1.6 (d) 16.0 Suppose 1 mole of A and B eah taken then 0.8 mole/litre of C and D eah formed remaining onentration of A and B will be (1 0.8) = 0. mole/litre eah. K [ C][ D] 0.8 0.8 = = = 16.0 [ A][ B] 0. 0. For the system A(g) + B(g) C(g), the equilibrium onentrations are (A) 0.06 mole/litre (B) 0.1 mole/litre (C) 0.16 mole/litre. The K for the reation is [CPMT 198] eq (a) 50 (b) 416 () 4 10 (d) 15 [ C] 0.16 Solution: (a) For reation A + B C; K eq = = = 50 [ A][ B] 0.06 0.1 0.1 Example : Molar onentration of is 96 gm, it ontained in litre vessel, ative mass will be (a) 16 mole/litre (b) 1.5 mole/litre () 4 mole/litre (d) 4 mole/litre weight weight 96 Solution: (b) Ative mass = M.wt. = = = = 1.5 mol / litre Volume M.wt. Volume Example : 4 Solution: (a) moles of PCl 5 were heated in a losed vessel of litre apaity. At equilibrium, 40% of PCl 5 is dissoiated into PCl and Cl. The value of equilibrium onstant is [MP PMT 1989] (a) 0.66 (b) 0.5 ().66 (d) 5. At start, PCl 5 PCl + Cl At equilibrium. 60 100 0 40 100 0 40 100 Volume of antainer = litre 40 40 K 100 100 = = 0.66 60 100 Example : 5 A mixture of 0. mole of H and 0. mole of I is allowed to reat in a 10 litre evauated flask at The reation is H + I HI the K is found to be 64. The amount of unreated I at equilibrium is (a) 0.15 mole (b) 0.06 mole () 0.0 mole (d) 0. mole 500 o C. [KCET 1990] Chapter 8 191
Chemial Equilibrium Solution: (b) [ HI] K = ; [ H ][ I ] 64 = x 0.0 0.0 4 x = 64 9 10 ; x = 8 10 = 0. 4 x is the amount of HI at equilibrium. Amount of I at equilibrium will be 0.0 0.4 = 0.06 mole Example : 6 The rate onstant for forward and bakward reations of hydrolysis of ester are minute respetively. Equilibrium onstant for the reation, 1.1 10 and 1.5 10 per Solution: (d) CH CC H 5 + H CH CH + C H 5H is [AIIMS 1999] (a) 4. (b) 5. () 6. (d) 7. K f = 1.1 10, K 1.1 10 f K b = 1.5 10 ; K = K b = 1.5 10 = 7. Example : 7 For the reation PCl ( g) + Cl ( g) PCl 5 at 50 o C, the value of K is 6, then the value of K p on the same temperature will be [MNR 1990; MP PET 001] (a) 0.61 (b) 0.57 () 0.8 (d) 0.46 Solution: (a) n g = 1 = 1 K = K ) p n ( RT ; K p = K ( RT) 1 Sine R = 0.081 litre atm k 1 mol 1, T = 50 o C = 50 + 7 = 5 K = 6(0.081 5) 1 = 0. 61 Example : 8 If K p for reation A ) + B( ) C ( g) + D( g) is 0.05 atm at 1000K its K in term of R will be [CBSE PMT 1989] Solution: () Example : 9 Solution: (d) (a) K p 5 10 R 4 = K ( RT) ( g g (b) 5 R () n 5 10 R 1 5 10 5 10 = K ( R 1000) K = R 5 5 p (d) None of these If the equilibrium onstant of the reation HI H + I is 0.5, then the equilibrium onstant of the reation H + I HI would be [MP PMT 1989, 95] (a) 1.0 (b).0 ().0 (d) 4.0 K for the IInd reation is reverse of Ist for reation 1 1 HI will be K = = = 4. K 0.5 HI H + I is 0.5K for reation, H + I Example : 10 If equilibrium onstant for reation AB A + B, is 49, then the equilibrium onstant for reation Solution: (a) AB A + B 1 1 AB A + B, will be [MP PMT 00; EAMCET 1998] (a) 7 (b) 0 () 49 (d) 1 K [ A][ B = [ AB] ] = 49 Chapter 8 19
Chemial Equilibrium 1 1 For reation AB A + B K = 1/ 1/ [ A] [ B] [ AB] K = K = 49 = 7 Example : 11 For the reation N ( g) N ( g) + ( g) Solution: (b) 6 K = 1.8 10 at o C 185. At 185 o C, the value of K for the 1 reation N ( g) + ( g) N ( g) is [UPSEAT 000] (a) 6 0.9 10 (b) Reation is reversed and halved. 1 K = ; K 1 K = = 6 1.8 10 7.5 10 () 7.5 10 1.95 (d) 10 1.95 10 Example : 1 In an equilibrium reation for whih G 0 = 0 the equilibrium onstant K p should be [BHU 1987] (a) 0 (b) 1 () (d) 10 Solution: (b) If 0 0 G = 0 and G =.0RT log K p Example : 1 Solution: (d) log K p = 0, K = 1 p what is the equilibrium onstant at 5 o C for HI ( g) H ( g) + I ( g) 0 G for HI( g ) + 1.7kJ / mole (a) 4.0 (b).9 ().0 (d) 0.5 G 0 =.0RT log K p =.0 8.14 10 98 log K 1.7 =.0 8.14 10 98 log K p K p = 0.5 p [KCET 199] Example : 14 It is found that the equilibrium onstant inreases by a fator of four when the temperature is inreased from o 5 C to o C 40. The value of o H is (a) 5.46 1 kj mol (b) 171.67 1 kj mol () 89.4 1 kj mol (d) 71.67 kj mol 1 Solution: (d) Using the equation, ( K log ( K ) ) p o 5 C H 1 =.0R T1 T p o 40 C 1, we get log 4 = H 1 1.0 8.14 7 + 5 7 + 40 H = 71.67 kj mol 1 Example : 15 K for the reation A ( g) + B( g) C(g) is.0 at 400K. In an experiment a mol of A is mixed with mol of B in a 1-L vessel. At equilibrium mol of C is formed. The value of a will be, (a) 4.5 mol (b) 9.5 mol ().5 mol (d).5 mol Chapter 8 19
Chemial Equilibrium Solution: (d) A( g) + B a a x ( g) x C g ( ) 0 x From the equation, x = x = 1.5 K 4 = x ( a x)( b x) ; 4 (1.5) = ; ( a 1.5)( 1.5) 4.5 = a =.5 ( a 1.5)(1.5) Example : 16 For the reation (g) Solution: (d) (a) ( g) 1 1 / / A +, AB is % dissoiated at a total pressure of P. Then AB ( g) B( g) P = K p (b) P 4K p AB ( n ) K p A g + B g ) ( ) ( 1 0 + / 1 0 + / 1 / 1 / 1 1 + + eq / m = = = P A P P AB P = 8K p B = 4 1 / 1 / P P 4 / 4 / / P 4 / = = () P = K p (d) p = 8K p 1 8 P Example : 17 The total pressure observed at equilibrium in the dissoiation of solid ammonium arbamate at a ertain temperature is.0 atm. The equilibrium onstant K p is (a) 4.185 (b) 1.185 ().76 (d) 1.07 Solution: (b) NH CNH ) K p 4 ( s NH + g C g ( ) 1 ( ) P P 1 4 = PNH P = C = P P P 7 4 = () 7 = 1.185 Example : 18 At the equilibrium of the reation N 4 ( g) N ( g ), the observed molar mass of N 4 is 77.70 g. The perentage dissoiation of N 4 is (a) 8.4 (b) 46.7 ().4 (d) 18.4 Solution: (d) α = M M Th obs M obs ( n 1) ; Molar mass of N 4 9 = g mol 1 9.00 77.70 Here, n = ; α = = 0.184 = 18.4% 77.70( 1) 1 Example : 19 In the equilibrium H ( g) H ( g) + ( g), the extent of dissoiation of water when p = 1 atm and K =.08 10 is approximately (a) % (b) 0.% () 0% (d) 1% Chapter 8 194
Chemial Equilibrium 1 Solution: (a) For the equilibrium H ( g) H ( g) + ( g) / 1 / P K p = α P = 1atm α = (. K p ) / = 0.005 % Le-Chatelier and Braun (1884), Frenh hemists, made ertain generalizations to explain the effet of Fators whih Change the State of Equilibrium : Le-Chatelier's Priniple. hanges in onentration, temperature or pressure on the state of system in equilibrium. When a system is subjeted to a hange in one of these fators, the equilibrium gets disturbed and the system readjusts itself until it returns to equilibrium. The generalization is known as Le-Chatelier's priniple. It may stated as : Change in any of the fators that determine the equilibrium onditions of a system will shift the equilibrium in suh a manner to redue or to ounterat the effet of the hange. The priniple is very helpful in prediting qualitatively the effet of hange in onentration, pressure or temperature on a system in equilibrium. This is appliable to all physial and hemial equilibria. (1) Effet of hange of onentration : Aording to Le-Chatelier's priniple, If onentration of one or all the reatant speies is inreased, the equilibrium shifts in the forward diretion and more of the produts are formed. Alternatively, if the onentration of one or all the produt speies is inreased, the equilibrium shifts in the bakward diretion forming more reatants. Thus, Inrease in onentration of any of the reatants Inrease in onentration of any of the produts the Shifts equilibrium to Forward diretion the Shifts equilibrium to Bakward diretion () Effet of hange of temperature : Aording to Le-Chatelier's priniple, If the temperature of the system at equilibrium is inreased (heat is supplied), the equilibrium will shift in the diretion in whih the added heat is absorbed. In other words, the equilibrium will shift in the diretion of endothermi reation with inrease in temperature. Alternatively, the derease in temperature will shift the equilibrium towards the diretion in whih heat is produed and, therefore, will favour exothermi reation. Thus, Inrease in temperature Shifts the equilibrium in the diretion of Endothermi reation Derease in temperature Shifts the equilibrium Exothermi reation in the diretion of () Effet of hange of pressure : Pressure has hardly effet on the reations arried in solids and liquids. However, it does influene the equilibrium state of the reations that are arried in the gases. The effet of pressure depends upon the number of moles of the reatants and produts involved in a partiular reation. Aording to Le- Chatelier's priniple, Inrease in pressure shifts the equilibrium in the diretion of dereasing gaseous moles. Chapter 8 195
Chemial Equilibrium Alternatively, derease in pressure shifts the equilibrium in the diretion of inreasing gaseous moles and pressure has no effet if the gaseous reatants and produts have equal moles. Thus, Inrease in pressure Derease in pressure Shifts the equilibrium Dereasing gaseous moles in the diretion of Shifts the equilibrium Inreasing gaseous moles in the diretion of (5) Effet of atalyst : Catalyst has no effet on equilibrium. This is beause, atalyst favours the rate of (4) Effet of volume hange : We know that inrease in pressure means derease in volume, so the effet of hange of volume will be exatly reverse to that of pressure. Thus, dereasing the volume of a mixture of gases at equilibrium shifts the equilibrium in the diretion of dereasing gaseous moles while inreasing the volume shifts the equilibrium in the diretion of inreasing gaseous moles. Thus, Inrease in volume Derease in volume Shifts the equilibrium in the diretion of Inreasing gaseous moles Shifts the equilibrium in the diretion of Dereasing gaseous moles forward and bakward reations equally. Therefore, the ratio of the forward to reverse rates remains same and no net hange ours in the relative amount of reatants and produts present at equilibrium. Thus, a atalyst does not affet the position of the equilibrium. It simply helps to ahieve the equilibrium quikly. It may also be noted that a atalyst has no effet on the equilibrium omposition of a reation mixture. Thus, Catalyst does not shift the equilibrium in any diretion (6) Effet of addition of inert gas : The addition of an inert gas (like helium, neon, et.) has the following effets on the equilibrium depending upon the onditions : (i) Addition of an inert gas at onstant volume : When an inert gas is added to the equilibrium system at onstant volume, then the total pressure will inrease. But the onentrations of the reatants and produts (ratio of their moles to the volume of the ontainer) will not hange. Hene, under these onditions, there will be no effet on the equilibrium. (ii) Addition of an inert gas at onstant pressure : When an inert gas is added to the equilibrium system at onstant pressure, then the volume will inrease. As a result, the number of moles per unit volume of various reatants and produts will derease. Hene, the equilibrium will shift in a diretion in whih there is inrease in number of moles of gases. Thus, Addition of an inert gas V ons tan t No effet on the equilibrium. = Addition of an inert gas P ons t Shifts the equilibrium in the diretion of = tan Inreasing gaseous moles. Chapter 8 196
Chemial Equilibrium Appliations of Le-Chatelier's Priniple. The Le-Chateliers priniple has a great signifiane for the hemial, physial systems and in every day life in a state of equilibrium. Let us disuss in brief a few appliations. (1) Appliations to the hemial equilibrium : With the help of this priniple, most favourable onditions for a partiular reation an be predited. (i) Synthesis of ammonia (Haber s proess): + NH + kal (exothermi) N H 1 vol vol vol (a) High pressure ( n < 0) (b) Low temperature () Exess of N and forward reation. diretion. (ii) Formation of sulphur trioxide : S + S + 45 kal (exothermi) vol 1 vol vol H (d) Removal of NH favours (a) High pressure ( n < 0) (b) Low temperature () Exess of S and, favours the reation in forward (iii) Synthesis of nitri oxide : N + N 4. kal 1 vol 1 vol vol (endothermi ) (a) High temperature (b) Exess of N and () Sine reation takes plae without hange in volume i.e., n = 0, pressure has no effet on equilibrium. (iv) Formation of nitrogen dioxide : N + N 7.8 Kal vol 1 vol + vol (a) High pressure (b) Low temperature () Exess of N and favours the reation in forward diretion. (v) Dissoiation of phosphours pentahloride : PCl PCl Cl kal 5 + 15 (a) Low pressure or high volume of the ontainer, n > 0 (b) High temperature () Exess of PCl 5. () Appliations to the physial equilibrium : Le-Chatelier's priniple is appliable to the physial equilibrium in the following manner; (i) Melting of ie (Ie water system) : (Greater Volume) 1 vol 1 vol Ie Water x kal (Lesser Volume) (In this reation volume is dereased from 1.09.. to 1.01.. per gm.) (a) At high temperature more water is formed as it absorbs heat. (b) At high pressure more water is formed as it is aompanied by derease in volume.() At higher pressure, melting point of ie is lowered, while boiling point of water is inreased. (ii) Melting of sulphur : S(s) S ( l) x kal (This reation aompanies with inrease in volume.) (a) At high temperature, more liquid sulphur is formed. (b) At higher pressure, less sulphur will melt as melting inreases volume.() At higher pressure, melting point of sulphur is inreased. (iii) Boiling of water (water- water vapour system) : Water Water Vapours x kal (Low volume) 1 vol (Higher volume) (It is aompanied by absorption of heat and inrease in volume.) (a) At high temperature more vapours are formed.(b) At higher pressure, vapours will be onverted to liquid as it dereases volume.() At higher pressure, boiling point of water is inreased (priniple of pressure ooker). Chapter 8 197
Chemial Equilibrium (iv) Solubility of salts : If solubility of a salt is aompanied by absorption of heat, its solubility inreases with rise in temperature; e.g., NH 4 Cl, K S4, KN et. KN ( aq KN x kal e.g., ( s) + ) ( aq) n the other hand if it is aompanied by evolution of heat, solubility dereases with inrease in temperature; CaCl, Ca( H), NaH, KH et. Ca( H) ( aq) Ca( H x kal ( s ) + ) ( aq) + (iii) Carriage of oxygen by haemoglobin in blood : The haemoglobin (Hb) in red orpusles of our blood () Appliation in every day life : We have studied the appliation of the Le-Chatelier's priniple to some equilibria involved in the physial and hemial systems. In addition to these, the priniple is also useful to explain ertain observations whih we ome aross in every day life. A few out of them are disussed below, (i) Clothes dry quiker in a windy day : When wet lothes are spread on a stand, the water evaporates and the surrounding air tends to get saturated thus hampering the proess of drying. n a windy day when breeze blows, the nearby wet air is replaed by dry air whih helps the proess of evaporation further. Thus, lothes dry quiker when there is a breeze. (ii) We sweat more on a humid day : The explanation is the same as given above. In a humid day, the air is already saturated with water vapours. This means that the water that omes out of the pores of the body as sweat does not vaporise. This will result in greater sweating in a humid day. arries oxygen to the tissues. This involves the equilibrium, Hb ( s) + ( g) Hb ( ) The blood that is in equilibrium with the oxygen of the air in the lungs finds a situation in the tissues where the partial pressure of oxygen is low. Aording to Le-Chatelier's priniple, the equilibrium shifts towards the left so that some of the oxyhaemoglobin hanges to haemoglobin giving up the oxygen. When the blood returns to the lungs, the partial pressure of the oxygen is higher and the equilibrium favours the formation of more oxyhaemoglobin. (iv) Removal of arbon dioxide from the tissues by blood : Blood removes C from the tissues. The equilibrium is, C ( g) + H ( l) H C ( aq) H + ( aq ) HC + ( aq ) Carbon dioxide dissolves in the blood in the tissues sine the partial pressure of C is high. However in the lungs, where the partial pressure of C is low, it is released from the blood. (v) Sweet substanes ause tooth deay : Tooth enamel onsists of an insoluble substane alled hydroxyapatite, Ca 5( P4 ) H. The dissolution of this substane from the teeth is alled demineralization and its formation is alled remineralization. Even with the healthy teeth, there is an equilibrium in the mouth as Ca 5( P4 s ) H( ) exothermi endothermi 5Ca + ( aq) + P 4 s ( aq) + H When sugar substanes are taken, sugar is absorbed on teeth and gets fermented to give H + ions. The H + ions produed distrub the equilibrium by ombining with H to form water and with P4 to form HP 4. Removal of produts ause the equilibrium to shift towards right and therefore, Ca 5 ( P4 ) H dissolves ausing tooth deay. Relation between vapour density and degree of dissoiation. ( aq) In the following reversible hemial equation. A yb Initial mol 1 0 At equilibrium (1 x) yx x = degree of dissoiation Number of moles of A and B at equilibrium = 1 x + yx = 1 + x( y 1) If initial volume of 1 mole of A is V, then volume of equilibrium mixture of A and B is, Molar density before dissoiation, moleular weight m D = = volume V = [ 1 + x ( y 1)] V Chapter 8 198
Chemial Equilibrium Molar density after dissoiation, m d = [ 1 + x( y 1)] V ; D = [ 1 + x( y 1 d )] ; x = D d d( y 1) y is the number of moles of produts from one mole of reatant. d D is also alled Van t Hoff fator. In terms of moleular mass, M m x = ; Where M = Initial moleular mass, m = moleular mass at equilibrium ( y 1) m Thus for the equilibria (I) PCl 5( g) PCl ( g ) + Cl ( g), y = (II) N 4( g) N, ( g ) y = (III) N N D d ( d D) x = (for I and II) and x = (for III) d d Also D = Moleular weight (theoretial value) d = Moleular weight (abnormal value) of the mixture ***, y 4 = 1 Chapter 8 199
Redox Reations Chemial reations involve transfer of eletrons from one hemial substane to another. These eletron transfer reations are termed as oxidation-redution or redox-reations. Redox reations play an important role in our daily life. These reations are aompanied by energy hanges in the form of heat, light, eletriity et. Generation of eletriity in batteries and many industrial proesses suh as prodution of austi soda, KMn 4, extration of metals like sodium, iron and aluminium are ommon examples of redox reations. Moleular and Ioni equations. (1) Moleular equations : When the reatants and produts involved in a hemial hange are written in moleular forms in the hemial equation, it is termed as moleular equation. Examples : (i) Mn + 4 HCl MnCl + H + Cl (ii) FeCl + SnCl FeCl + SnCl4 In above examples, the reatants and produts have been written in moleular forms, thus the equation is termed as moleular equation. () Ioni equations : When the reatants and produts involved in a hemial hange are ioni ompounds, these will be present in the form of ions in the solution. The hemial hange is written in ioni forms in hemial equation, it is termed as ioni equation. Examples : (i) + + Mn + 4 H + 4 Cl Mn + Cl + H + Cl + + + 4+ (ii) Fe + 6Cl + Sn + Cl Fe + 4Cl + Sn + 4Cl In above examples, the reatants and produts have been written in ioni forms, thus the equation is termed as ioni equation. () Spetator ions : In ioni equations, the ions whih do not undergo any hange and equal in number in both reatants and produts are termed spetator ions and are not inluded in the final balaned equations. Example : + + Zn + H + Cl Zn + H + Cl (Ioni equation) + + Zn + H Zn + H (Final ioni equation) In above example, the Cl ions are the spetator ions and hene are not inluded in the final ioni balaned equation. (4) Rules for writing ioni equations (i) All soluble ioni ompounds involved in a hemial hange are expressed in ioni symbols and ovalent substanes are written in moleular form. H, NH, N, N, S, C, C, et., are expressed in moleular form. (ii) The ioni ompound whih is highly insoluble is expressed in moleular form. (iii) The ions whih are ommon and equal in number on both sides, i.e., spetator ions, are anelled. (iv) Besides the atoms, the ioni harges must also be balaned on both the sides. The rules an be explained by following examples, Example : Write the ioni equation for the reation of sodium biarbonate with sulphuri aid, The moleular equation for the hemial hange is, NaHC + H S 4 Na S 4 + H + C NaHC, H S 4 and Na S 4 are ioni ompounds, so these are written in ioni forms. + + Na + HC + H + S 4 Na + S4 + H + C + Chapter 9 00
Redox Reations + Na and S 4 ions are spetator ions; hene these shall not appear in the final equation. HC + H + H + C To make equal harges on both sides, HC + H + H + C HC should have a oeffiient. Example : Write the following ioni equation in the moleular form if the reatants are hlorides. In order to balane the hydrogen and arbon on both sides, the moleules of H and C should have a oeffiient respetively. HC + H + H + C This is the balaned ioni equation. = or HC + H + = H + C Conversion of ioni equation in moleular form an be explained by following example, Fe + +Sn + Fe + +Sn 4+ For writing the reatants in moleular forms, the requisite number of hloride ions are added. Similarly 8 Fe + + + 6Cl + Sn + Cl or FeCl + SnCl Cl ions are added on R.H.S. to neutralise the harges. + 4+ Fe + 4Cl + Sn + 4Cl or FeCl + SnCl 4 Thus, the balaned moleular equation is, FeCl +SnCl = FeCl + SnCl 4 xidation-redution and Redox reations. (1) xidation : xidation is a proess whih involves; addition of oxygen, removal of hydrogen, addition of non-metal, removal of metal, Inrease in +ve valeny, loss of eletrons and inrease in oxidation number. (i) Addition of oxygen (a) Mg + Mg (xidation of magnesium) (b) S + S (xidation of sulphur) () C + C (xidation of arbon monoxide) (d) Na S +H Na S 4 +H (ii) Removal of hydrogen (a) H S + Cl HCl + S (b) 4HI + H + I o Cu / 00 C () C H 5 H CH CH + H (d) 4HCl + Mn MnCl + H + Cl (xidation of sodium sulphite) (xidation of hydrogen sulphide) (xidation of hydrogen iodide) (xidation of ethanol) (xidation of hydrogen hloride) (iii) Addition of an eletronegative element or addition of Non-metal (a) Fe + S FeS (xidation of iron) (b) SnCl + Cl SnCl 4 () Fe + F FeF (xidation of stannous hloride) (xidation of iron) (iv) Removal of an eletropositive element or removal of metal (a) KI + H KH + I (xidation of potassium iodide) Chapter 9 01
Redox Reations (b) K Mn 4 + Cl KCl + KMn 4 (xidation of potassium manganate) () KI + Cl KCl + I (xidation of potassium iodide) (v) Inrease in +ve valeny and Derease in ve valeny Inrease in +ve valeny 4,,, 1, 0, +1, +, +, +4 Derease in ve valeny (a) (b) Fe + + Fe + 4+ + e Sn Sn + e (+ve valeny inreases) (+ve valeny inreases) 4 () [ Fe( CN) ] [ Fe( CN) + e ( ve valeny dereases) (d) 6 6 ] 4 Mn + e ( ve valeny dereases) Mn 4 (vi) Loss of eletrons (also known as de-eletronation) (a) (b) () (d) (e) (f) H 0 H 4 M + + e M M 1 M 0 M +1 M + M + M e e e e e e e e Loss of eletrons +4 M (Formation of proton) 0 + H H + e (De-eletronation of hydrogen) Fe + Fe + + + + e Mg Mg e (De-eletronation of + Fe ) (De-eletronation of Magnesium) 4 Mn + e (De-eletronation of Mn 4 Mn 4 ) Cl Cl + e (De-eletronation of hloride ion) (g) 0 + + Fe Fe 6e (De-eletronation of iron) (vii) Inrease in oxidation number (a) Mg 0 Mg + (From 0 to +) + 4 + (b) [ ] [ ] () Fe ( CN) Fe ( CN) (From + to +) 6 6 0 (From 1 to 0) Cl Cl () Redution : Redution is just reverse of oxidation. Redution is a proess whih involves; removal of oxygen, addition of hydrogen, removal of non-metal, addition of metal, derease in +ve valeny, gain of eletrons and derease in oxidation number. (i) Removal of oxygen (a) Cu + C Cu + C (Redution of upri oxide) (b) H C C + H Coke Steam Water gas + (Redution of water) Chapter 9 0
Redox Reations () (d) Fe 4 + 4 H Fe + 4H (Redution of Fe 4 ) C 6 H 5H + Zn C6 H 6 + Zn (Redution of phenol) (ii) Addition of hydrogen (a) (b) Cl + H HCl (Redution of hlorine) S + H H S (Redution of sulphur) () CH 4 + H CH 6 (Hydrogenation of ethene) (iii) Removal of an eletronegative element or removal of Non-metal (a) HgCl + SnCl Hg Cl + SnCl4 (Redution of meruri hloride) (b) () + FeCl + H FeCl HCl (Redution of ferri hloride) FeCl + H S FeCl + HCl + S (Redution of ferri hloride) (iv) Addition of an eletropositive element or addition of metal (a) HgCl + Hg Hg Cl (Redution of meruri hloride) (b) CuCl + Cu CuCl (Redution of upri hloride) (v) Derease in +ve valeny and Inrease in ve valeny (a) (b) Derease in +ve valeny 4,,, 1, 0, +1, +, +, +4 Inrease in ve valeny Fe + Fe + (+ve valeny dereases) Sn 4+ Sn + (+ve veleny dereases) 4 () [ Fe ( CN) ] [ Fe( CN) ( ve valeny inreases) (d) 6 6 ] Mn ( ve valeny inreases) 4 Mn 4 (vi) Gain of eletrons (also known as eletronation) 4 M M M 1 M 0 M +1 M + M + M +4 M +e +e +e +e +e +e +e +e Gain of eletrons + (a) Zn ( aq) + e Zn( S) (Eletronation of + Zn ) (b) () (d) (e) Pb + + e Pb 0 (Eletronation of 7+ + Mn + 5e Mn (Eletronation of + + Fe + e Fe (Eletronation of 4+ + Sn + e Sn (Eletronation of + Pb ) 7+ Mn ) + Fe ) 4+ Sn ) Chapter 9 0
Redox Reations (f) Cl + e Cl (Formation of hloride ion) 4 (g) [ Fe ( CN) ] + e [ Fe( CN) (Eetronation of [ Fe ( CN) 6 ] ) 6 6 ] (vii) Derease in oxidation number (a) + 0 (From + to 0) Mg Mg 4 (b) [ ] [ ] () 6 Fe( ) 6 Fe ( CN) CN (From + to +) Cl 0 Cl (From 0 to 1) () Redox-reations (i) An overall reation in whih oxidation and redution takes plae simultaneously is alled redox or oxidation-redution reation. These reations involve transfer of eletrons from one atom to another. Thus every redox reation is made up of two half reations; ne half reation represents the oxidation and the other half reation represents the redution. (ii) The redox reations are of following types (a) Diret redox reation : The reations in whih oxidation and redution takes plae in the same vessel are alled diret redox reations. (b) Indiret redox reation : The reations in whih oxidation and redution takes plae in different vessels are alled indiret redox reations. Indiret redox reations are the basis of eletro-hemial ells. () Intermoleular redox reations : In whih one substane is oxidised while the other is redued. For example, Al + Fe Al Fe + Here, Al is oxidised to Al while Fe is redued to Fe. (d) Intramoleular redox reations : In whih one element of a ompound is oxidised while the other is redued. For example, Here, KCl + KCl +5 Cl in KCl is redued to 1 Cl in KCl while in KCl is oxidised to (iii) To see whether the given hemial reation is a redox reation or not, the moleular reation is written in the form of ioni reation and now it is observed whether there is any hange in the valeny of atoms or ions. If there is a hange in valeny, the hemial reation will be a redox reation otherwise not. For example, (a) Ba + H S4 BaS4 + H (b) CuS4 + 4NH [ Cu( NH ) 4 ] S4 In above examples there is no hange in the valeny of any ion or atom, thus these are not redox reations. 0. (iv) Some examples of redox reations are, e (a) xidation HgCl + SnCl Hg Cl + SnCl + e Redution 4 Here meruri ion is redued to merurous ion and stannous ion is oxidised to stanni ion, i.e., meruri ion ats as an oxidising agent while stannous ion ats as a reduing agent. Chapter 9 04