2 Chemical Equilibrium Introduction Whenever we hear the word Equilibrium Immediately a picture arises in our mind an object under the influence of two opposing forces. For chemical reactions also this is true. A reaction also can exist in a state of equilibrium balancing forward and backward reactions, ‘Symbolic representation of any chemical change in terms of reactants and prod reaction, 1 Types of chemical reaction 1s is called chemical (@ | On the basis of physical state Homogeneous reactions All reactants and products are in same phase N, (g) + 3H(g) === 2NH.g) Heterogeneous reaction Reactants and produ more phase 2n(s) + CO, G) ——=Zn0(s) + CO) are in two or (©) | On the basis of direction Reversible reaction () Chemical reaction in which products can be converted back into reactants Hoth === 2HI reversible reaction Chemical reaction in which products cannot be convert back into reactants Zn + H,SO.—> ZnSO. + (i) Proceed in forward as well as backward Proceed only in forward direction (ii) These attain equilibrium These do not attain equilibrium (iv) Reactant are converted Into products never completely Reactants are nearly completely converted into product (@ Generally thermal dissociations are held in closed vessel Pcl, Cl, @) +Cl,@) Generally thermal decompositions are held in open vessel 2KCIOs(s) —> 2KCK(s) + 309(9) (© | On the basis of speed Fast rea ns () Generally these reactions are ionic in nature HCI + NaOH —> NaCl + Ho Acid Base salt Water Slow reactions Generally these reactions are molecular In nature He + b— 2H (@) | On the basis of heat Exothermic reaction () Heat is evolved in these type of chemical reactions R—>P +x kcal Endothermie reaction Heat is absorbed in these type of chemical reactions R—>P-xkcal Chemical Equilibrium 99100 Itis an experimental fact that most of the process including chemical reactions, when carried out in a closed vessel, do not go to completion, Under these conditions, a process starts by itself oF by initiation, continues for some time at diminishing rate and ultimately appears to stop. The reactants may still be present but they do not appear to change into products any more Aquarium) * Definition: “Equilibrium is the state at which the concentration of reactants and products donor change with time. ie. concentrations of reactants and products become constant * Characteristics: Following are the important characteristics of equilibrium state, 1 Products § € g Reactants 8 Time —> () Equilibrium state can be recognised by the constancy of certain measurable properties such as pressure., density, colour, concentration ete. by changing these conditions of the system, we can control the extent to which a reaction proceeds. (i) Equilibrium state can on! y be achieved in dose vessel, but if the process is carried out in an open vessel equilibrium state cannot be attained because in an open vessel, the reverse process may not take place (i) Equilibrium state is reversible in nature. (iv) Equilibrium state Is also dynamic in nature. Dynamic means moving and at a microscopic level, the system Is in motion. The dynamic state of equilibrium can be compared to water tank having an inlet and outlet. Water in tank can remain at the same level if the rate of flow of water from inlet (compared to rate of forward reaction) is made equal to the rate of flow of, water from outlet (compared to rate of backward reaction). Thus, the water level in the tank remains constant, though both the inlet and outlet of water are working all the time. Chemical Equilibrium() At equilibrium state, Rate of forward reaction = Rate of backward reaction Equilibrium state Rate of reaction > Time —> Types: Equilibrium in a system implies the existence of the following types of equilibrium simultaneously, () Physical equilibrium: It Is a state of equilibrium between the same chemical species in different phases (Solid, Liquid and Gas) (i) Chemical equilibrium: There is no change in composition of any pan of the system with time. 2.1 Physical equilibrium The various equilibrium which can exist in any physical system are, Solic — liquic Liquid = Vapour Solic —= Vapour Solic = Gas( vapour) Solic — Saturated solution of solid in a liquid Gas(Vapour) ——= Saturated solution of gas in a liquid () Solid-liquid equilibrium 1,06) =H, Otliquia) Rate of transfer of molecules from ice to water = Rate of transferor molecules from water to ice Rate of melting of ice = Rate offreez.ing of water (I) Liquid-vapour equilibrium: When vapour of a liquid exists in equilibrium with the liquid, then Rate of vaporisation _—=_Rate of condensation H,O(liquie) ——H,O(vapour) Chemical Equilibrium 101102 (ul) (Iv) Conditions necessary for a liquid!-vapour equilibrium are () The system must be a closed system ie., the amount of matter in the system must remain constant (i) The system must be at a constant temperature Vaporisation Eandensation Rate—> < Time—> x (lil) The visible properties of the system should not change with time. Solld vapour equilibrium: Certain solid substances on heating get converted directly into vapour without passing through the liquid phase. This process is called sublimation. The vapour when cooled, glves back the solid, it is called disposition, Solid == Vapour The substances which undergo sublimation are camphor, iodine, ammonium chloride etc. For example, Ammonium chloride when heated sublimes NH,CI(Solid) =S—=NH,Cl vapour) Equilibrium between a solid and Its solution: When a saturated solution Is in contact with the solid solute, there exists a dynamic equilibrium between the solid and the solution phase. Solid substance —— Solution of the substance Example Sugar and sugar solution. In a saturated solution, a dynamic equilibrium is established between dissolved sugar and solid sugar Sugar(solid) ——* Sugar (aqueous) At the equilibrium state, the number of sugar molecules going into the solution from the solid sugar is equal to the number of molecules precipitating out from the solution, ie. at equilibrium, Rate of dissolution of solid sugar = Rate of precipitation of sugar from the solution. Chemical Equilibrium? (V)__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, () Nature of the gas and liquia. (i) Temperature of the liquicl (ill) Pressure of the gas aver the surface of the solution. 2.2 Chemical Equilibrium Characteristics of chemical equilibrium (@) It is @ dynamic equilibrium ie. at this stage, reaction takes place in both the directions with same speed so, there is no net change. (b) At equilibrium the reaction proceeds both the side, equally (©) At equilibrium, both reactants and products are present and their concentration do not change with respect to time. (#) The state of equilibrium is not effected by the presence of catalyst: It only helps to attain the equilibrium state in less or more time, {e) Change in pressure, temperature or concentration favours one of the reactions and thus shifts the equilibrium point in one direction. Rate of Reaction In a reaction , there is change in concentration of reactant or product per mole in unit time, it is known as rate of the reaction. (\changeinconcentrationofreactant (« Here negative sign indicate that concentration of reactants decrease with time. (ae (at changeinconcentrationof products time Rate of reaction product Here positive sign Indicate that concentration of products Increase with time. Note: The concentration change maybe positive or negative but the rate of reaction is always positive. mole/lit, mole sec, lit sec Unit of rate of reaction = = mole lit"! sec! Foe example A > 8 For reactant > 441 [concentration decrease with time] For reactant -» +461 (concentration increase with time] [d {A), 6 {8} are change in concentration of A & B in time alt] ‘At equilibrium, since there is no net change in concentration of reactant or product, So rate of reaction is zero. (At equilibrium) Chemical Equilibrium 103104 Law of Mass Action @ (b) This law was given by Guldberg and Waage. At agiven temperature, the product of the concentration of products each raised to the corresponding stoichiometric coefficients in the balanced chemical equation divided by the product of the concentrations of the reactants raised to the corresponding stoichiometric coefficients has a constant value. A+yBo==C1D Rate of chemical reaction r « [A] [B] r= KIA] [8] Mathematical Expression @ Ww For unitary stoichiometries coefficients At the constant temperature, let us consider the following reversible reaction A+B=—=C+D According to law of mass action - Rate of forward reaction rex [A [B] or tr = Ke [A] [BI where K-is the rate constant of the forward reaction Rate of backward reaction Tex (C}ID] or —fe= Ke (C] (D] where Ky is the rate constant of the backward reaction. At equilibrium: Rate of reaction = Rate of forward reaction ~ Rate of backward reaction = 0 K- [A] [B] ~ Ke [C] [D] = 0 _ (€i10) KAI) (c1b) Keg ~ (C0) “ TAI] Here, k Is equilibrium constant of given reversible reaction For non-unitary stoichiometric coefficient nA + n.B === mC + m0 reste {er (or fay (By Note: [A], (Bl), [C], [D] are molar concentration of reactants and products, for dilute solution Chemical Equilibrium? Equilibrium Constants, Ke. Ke. Kee & Kx There are various metiiods for measuring equilibrium constant in terms of concentration, pressure, mole fraction () Equilibrium constant in term of concentration Consider an equilibrium reaction as XQ) +Y¥Q=—=Z@ for this reactions, which is in equilibrium, there exist an equilibrium constant (Kee) represented as - ll “~ BM for the given equilibrium, irrespective of the reacting species (ie. either X + Y or ZorX+ZorY +ZorX +¥+2Z) and their amount we start with, the ratio, ral is always constant at a given temperature. The given expression involves all variable terms(variable term means the concentration of the involved species changes from the start of the reaction to the stage when equilibrium is reaches) (2) XM - DIN] Thus, for the given equilibrium, it seems that Kee and Ke are same but in actual practice for some other equilibrium, they are not same. so the ratio can also referred as Ke (i) Equilibrium constant in terms of pressure ‘Assuming that the gases, X, Y and Z behave ideally PV= nRT p= 2 Rt=crT v P Cnt Pot Be _ [xX] Rr aa Rt and [Z] ar P, RT) _ pert TRYTR) PR RT) (RT, Pe RTP, xP, The LHS of the above expression is a constant since Ke, R and T, all are constant This implies that RHS is also a constant, which is represented by Ke Chemical Equilibrium 105106 0) Pp, PxP, Thus, expression of K, involves partial pressures of all the involved species and represents the ratio of partial pressures of product to reactants of an equilibrium reaction, If the phase of reactant X from gaseous to pure solid. Then the equilibrium reaction can be shown as X(s) + Yq) == 2) Its equilibrium constant (Kea) would be @ “BIN ‘The concentration of X is the number of moles of X per unit volume of solic, As we known, the concentration of all pure solids (and pure Liquids) is a constant as It Is represented by d/M (where d and M represents its density and molar mass). This ratio of d/M will be a constant whether X is present initially or at equilibrium, Keg 1X) = 2) 2 My) ™) Thus expression of Ke involves only those species whose concentration changes during the reaction. The distinction between Kea and Ke is that the expression of Keg Involves all the species (whether they are pure solids, pure liquids, gases, solvents or solutions) while the Ke expression involves only those species whase concentration is a variable (like gases and solutions). Thus, expression of Ke is devoid of pure components (like pure solids and pure liquids) and solvents. R 2 constant, represented by Ke. Equilibrium constant in terms of both concentration & pressure Consider the following equilibrium X(s0l) + ¥@) === Z@) + AO) _ IAI IM If concentration of X, ¥, Z and A Is expressed in terms of partial pressures Chemical EquilibriumPP bar, Thus, Ke, can exist only for that equilibrium which satisfies these two conditions (@) At least one of the reactant or product should be In gaseous phase and (©) No component of the equilibrium should be in solution phase (Because When solution is present, the equilibrium constant would be called Kx) (iv) Equilibrium constant in terms of moles fraction Kec A+ B=—==c:D Mole fraction at equilibrium Xa+ Xe Xe Xo Kye Bet Xo Characteristics of Equilibrium Constant () The expression for equilibrium constant, K is applicable only when concentrations of the reactants and products have attained their equilibrium values and do not change with time. (i) The value of equilibrium constant is independent of initial concentration of the reactants and product (li) Equilibrium constant has one unique value for a particular reaction represented by a balanced equation at a given temperature, (iv) The equilibrium constant for the reverse reaction is equal to the inverse of the equilibrium constant for the forward reaction 1,@ +1,.@ == 2HI9) () The equilibrium constant K, for a reaction is related to the equilibrium constant of the corresponding reaction whose equation is obtained by multiplying or dividing the equation. 2) + h(g) == 2HI(g) Fe a ke Chemical Equilibrium 1071 1 7 h@+ Fh@ == HI@ . Pal °° FERT (vi) Ifreaction is performed in multiple steps A == B: overall reaction Step-1: Ag) == C(@) + Df) ke Step-2 Cg) ——= F(a) Kr Step-3: Dg) + E@) —=—= BQ) Ks Ag) == 8@) Kn then Kr = Kor. Kez . Kes Factors which do not influence equilibrium constant fa) (o) (©) @ Concentration of reactants and products. Pressure and volume. Presence of catalyst. Addition of the inert gas at constant Pressure and volume, Factors which influence the equilibrium constant: A B c 108 Made of representation of chemical reaction. Stoichiometry of reaction, ‘Temperature. Mode or representation or reaction - Ifwe take reaction 2HI == Hy + bb ‘Then, we write the value of equilibrium constant Ke, for the above reaction as following. _ Hh) Kore Tay ° Now, if we take reaction Hy +, = 2H Then. we write the value of equilibrium constant K, for above reaction as following THI 1 (H,0,] Ail) Chemical Equilibrium? B. Stoichiometry of the reaction Method of writing the equation of the reversible reaction is called as stoichiometry of the reaction. Now, we consider the formation of HI(g) by the combination of H.(9) and 12(9) H+ bh === 2HI The expression of its equilibrium constant is (HIP {HJTL) If the equation of above reaction Is written by following method K, 1 1 7 Mag) + 5 kg) —= HI@) The expression for the equilibrium constant is Ho HI, on the basis of comparing both the equilibrium constant equation. a a a Note: When we divide @ reaction by a factor ‘n’ in the equation, the value of new equilibrium constant is equal to the root of n of the previous equilibrium constant, For Example: Suppose, the equilibrium constant for the following reaction. Ke, A+ B=—=C+D Is Ky then) for the reaction No.1 texto non the value of the equilibrium constant K; is equal to nyk, or (K;)!" Ka = KY (C) Temperature Increase in temperature favours the endothermic reaction and decrease in temperature favours the exothermic reaction for the forward reaction so for exothermic reactions, the value of Ke and Ke decrease will rise in temperature while for endothermic reaction the value of Ke and IS, increases with rise in temperature, This type of variation in equilibrium constant with temperature given by Van't Hoff equation as follows - aH ft log Ks okie 32 3| aH fad 2303 [T,_T, Chemical Equilibrium 109 orm0 Where, K: « equilibrium constant at temperature T2 K; = equilibrium constant at temperature Tr AH = Energy of reaction of constant temperature R = Molar gas constant According to the temperature, reaction are of three types. (@) _Non-thermic reaction means AH = 0 log Ks = log K; log Kz = log Ki There is no effect of temperature on this type of reaction. (b) — Endothermic reaction = AH = (+) ve log K; = log Ks = (+) ve means Ks > Ky On increasing of temperature. equilibrium constat type of reaction (©) Exothermic reaction = AH log Ke = fog K; = (-) ve, means Ki < Ki On the increase of temperature equilibrium constant will decreases exothermic reaction, Units of Ke and Kp will also increase for this oF The concentration is expressed in me term of moles per litre, Therefore, units of Ke will be {moles litre"). In the same way, partial pressure are measured by the unit of atmospheres and therefore units of Kp will be (Atmospheres) Value of An _| Unit of Ke Unit of Ke o No uni No unit 0 (Moles F™ my ° (Moles Fy" aim Relation between Keand K. Let us consider the following reaction nA+nB——=mc+mp The value of Ke for the reaction is, _ (emo) Tare According to gas law PV =n RT Chemical Equilibriumfn _ numberofmoles Here 7 = = [] = Active mass. i PIP) PIP, on putting the value of'p’ in the formula of Kp by the equation (I) (ce1Rt)” (ORT) (lAIRT) (BIRT) x - NOMA TAP By RT Ke = KART) Kp = K(RT)" {an = (rm +m.) = (rt ne) ny = number of moles of gaseous products ~ number of moles of gaseous reactants T = Absolute temperature Example 1 At 700 K, the equilibrium constant Ke, for the reaction 2803(g) 280x(9) + O.(g) is 1.8 x 10% kPa, What Is the numerical value of Ke for this reaction at the same temperature (A) 3.09 x 1077 mole litre (B) 9.03 x 10°? mole litre’ (C) 5.05 x 10-? mole litre? (0) 5.05 x 10-* mole litre! Ans. (A) Solution: We know the relationship Ko = Ko(RT) Here Ke = 1.80 « 107 Ky = 182107 101.3 = 1.78 x 10° atm R = 0.0821 litre atm K-! mot! an=3-261 T= 700K _ Ke 17810 (RN © 00821700 = 3.09 x 10 mole titre” Ko Chemical Equilibrium m= Applications of Equ um Constant Consider some applications of equilibrium constant and use it to answer question like () predicting the extent of a reaction on the basis of its magnitude (li) predicting the direction of the reaction. (Predicting tile extent of a reaction ‘The magnitude of equilibrium constant is very useful especially in reactions of industrial importance. An equilibrium constant tells us whether we can expect a reaction mixture to contain a high or low concentration of product(s) at equilibrium. (It is important to note that an equilibrium constant tells us nothing bout the rate at which equilibrium is reached), In the expression of Ke of Kp, product of the concentrations of products is written in numerator and the product of the concentrations of reactants is written in denominator. High value of equilibrium constant indicates that product(s) concentration is high and its low value indicates that concentration of the product(s) in equilibrium mixture is low, Hig) + Brel) === 2HBr(g) (Pa) eee BR) The large value of equilibrium constant indicates that concentration of the product, HBr k= is very high and reaction goes nearly to completion Similarly, equilibrium constant for H.(g) + Cl(g) —— 2HCI(9) Is very high and reaction goes virtually to completion = Leet] [afer] Thus, large value of K> or Kc (larger than about 10"), favour the products strongly 4.0 x10" For intermediated values of K (approximately in the range of 10% to 10%), the concentrations of reactants and products are comparable. Small values of equilibrium constant (smaller than 10”), favour the reactants strongly. At 298 K for reaction, Na(g) + 02(g) === 2NO(g) Noy INTO] The very small value of Kc implies that reactants Np and Q; will be the predominant species in the reaction mixture at equilibrium. 12 Chemical Equilibrium? The equilibrium constant Is also used to find In which direction the reaction will proceed Predicting the direction of tire reaction for a given concentration of reactants and products. For this purpose, we calculate the Reaction Quotlent(Q). The reaction quotient Is defined in the same way as the equilibrium constant (with molar concentrations to give Oc, or with partial pressure to give Op) at any stage of reaction. For a general reaction DAs nB SS mC + md ¢ - (ermor cartel Then, if Qc > Ke, the reaction will proceed in the backward direction if Qe < Ke, the reaction will move in the forward direction If Qc= Ke, the reaction mixture Is already at equilibrium, In the reaction, Hp(g) + b(g) <= 2HI(Q). if the molar concentrations of H>, I» and HI are 0.1 mol L" respectively at 783 K, then reaction quotient at this stage of the reaction is (HIP. (o4y [HL] (0.10.2) Kc for this reaction at 783 K is 46 and we find that Qc < The reaction, therefore, will move to right i.e, home H.(g) and L(g) will react to form, more Hi(g) and their concentration will decrease till Qs = Ke Degree of Dissociation Degree of dissociation Is the fraction of a mole oftlle reactant that underwent dissociation It Is represented by ‘a’ number ofmolesofreactant dissociated umber ofmolesofreactant present initially For example, Let the equilibrium reaction Is the dissociation equilibrium of NHs into No and Ho 1 3 NH, (@) =, 9) + 3H, (9) 5Ni@ +5) Moles initially a o oO Moles at equilibrium (1a) = 3a Here, a represented the degree of dissociation. Chemical Equilibrium 113Example 2: 2.56 gm of sulphur $(5) is taken which Is in equilibrium with its vapour according to reaction, Ss(s) —= Si) if vapours occupies 960 m/ at 1 atm & 273 K then the degree of dissociation of Se(s)will be [Given: R = 0.08] (A) 05 (8) 055 (oa (0) 0.44 Ans, (8) Solution 256 _ 554 8x32 Sals) 4 8s@ 0.01 (1-0) 8x001 xa PV = nRT 1x 322 _ 01.8 x a) x 0.08 x 273 7000 0.55 Calculation of Kp & Ke {@) Homogeneous equilibrium in gaseous phase (6) Homogeneous equilibrium in solution phase (©) Equilibrium constant for various heterogeneous equilibrium (@) Homogeneous equilibrium in gaseous phase Formation of Nitric Oxide: (An = 0) A Calculation of Ke ‘Suppose the initial concentration of N, and O, is a and b respectively x is the degree of dissociation. N+ Op == 2NO Initial moles a b ° moles at tquilibrlum (@») o%) 2x (=x) (b=) 2x ive me 1 ca Active mass (mol 7 T 7 Here, V is the volume of container in litre According to the law of mass action [NOP IN,1[0,] m4 Chemical EquilibriumSubstituting the values in the aboOve equation Ke = i x) bd) Ke for this reaction is independent of V ot reaction container B. Calculation of Ky All the things being same as above, except pressure. Let P atmosphere Is the pressure at equilibrium, M+ Oo == 2No Initial moles a b oO moles at equilibrium (a-x) (b-”) 2x Total no. of moles = (a-x) + (b-x) + 2x = (a + b) The partial pressure of the above three species can be calculated as below _ (a-nP * ~@ee) p, (oP Gb) (2x)P ™ (a+b) According to the law of mass action [rol lle substituting the value of Pyo, P,P, in the above equation of Ke - (2xp]’ eles °* T@=w)P]] xP =r (rb) ae acc Chemical Equilibrium 5Thermal Dissociation of Phosphorus pentachloride - (An > 0) A. Calculation of Ke - Suppose one mole of PCIk is take in a closed container of V litre Further at equilibrium x mol of PCIs dissociated PC == PCh + Ch Initial moles 1 0 0 moles at equilibrium ce) x x Concentration (mol /) x x vo According to law of mass action PCI, “= Pe] The formula of Ke has V in the denominator, hence the equilibrium will be affected by V of the reaction container for the given reaction, i xed then 1= x =1 So. Kea & v = Kev way xo WW If we increase the volume, the dissociation x Is also increased. 8 Calculation of ke Pls — Pcl, ch initial motes 1 0 ° moles at equilibrium 1x x x Total no. of moles at equilibrium, (=) +x = x= (+) moles According to law of mass action Pres, *Pa, ko Ne Chemical EquilibriumAt equilibrium P,,, = 22 TW) xP 9) fy = HHP 6 ae) Substituting the values in the above equation of Ke - (22/22) ko The equation of Ke is not independent of pressure. suppose, xecl then Ko = eP kK, P The degree of dissociation of PCls is inversely proportional to the square root of pressure So, decrease of pressure increases dissociation of PCls Formation of Ammonia = (An < 0) A alculation of Ke No + 3H2 ——— 2NHs Initial moles 1 3 0 moles at equilibrium (1x) (3%) x \ wd According to law of mass action (NH,)" IN IH] Substituting the values in the above equation- Chemical Equilibrium 7ev (1-x)(8-3x) acy! 27(1-x)* Ke ‘The formula of Ke bas V in the numerator, hence the equilibrium will be affected by V of the reaction container Dependence If, x <« 1 then, (I= x) 4xv" °° aT 2. 2K x aw a. 1 co xed ve If we increase the volume of the container the degree of dissociation x is decreased B. Calculation of Kp Ne 3h, == NH Initial concentration 1 3 0 moles at equilibrium (-9 (- 30) 2x Total number of moles at equilibrium (1 = x) + (3 = 3x) + 2K = (4 = 2x) According to the law of mass action (Pos) (@.)=(,) _ OxP ss" (@=2x) At equilibrium P “4 p - G=30P a2 Substituting the values in the above equation of Ke, 4x2(4 — 2x)? =x)(3— 39)? jx? (2— x)? Ty 18 Chemical EquilibriumThe equation of Kp is not independent of pressure Suppose, x << 1 then, (=I and (2x) =4 axe ae a P? xaP If we Increase the pressure the above degree of dissociation x is also increased Example 3 At a certain temperature (7), the equilibrium constant (K.) Is 1 for the reaction No(g) + 3H)(g) ——= 2NHA(g) If 2 moles of Nz, 4 moles of Ha, 6 moles of NH; & 3 moles of inert gas are introduced into a two litre rigid vessel at constant temperature T. It has been found that equllibrium concentration Of H:, & NHs are equal then what is the equilibrium concentration of Nz (In M)? Solution j = 18 Qe > Ke So reaction will proceed in backward direction, Nog) + 34.9) == 2NHs at equilibrium:2 + x 4+ 3x 6-2x At equilibrium [H,(g)] = [NH,(9)] 442K _ 6-2x 2 2 [N.@)] at equilibrium (b) Homogeneous equilibrium in solution phase Formation of ethyl acetate Equilibrium is represented as CaHs(OH)() + CHsCOOH() === CH.COOC.Hs()) + H,0() Initial moles 1 1 0 ° Moles at equilibrium x x x x Active mass(mol = = ( ) Vv Vv Chemical Equilibrium ng1 [c1,c000,4 Jh.0] x O-*)0=x) Determine the amount of ester present under equilibrium when 3 moles of ethyl alcohol react with | mole of acetic acid, when equilibrium constant of the reaction is 4 Solution: 120 © CH.COOH + CHsH = == * CHsCOOCHs + H.0 Bax x x v v v 3x? — 16x +12 = 0 x= 0.903 Amount of ester at equilibrium = 0.903 mole Equilibrium constant for various heterogeneous equilibrium Heterogenous equilibrium results from a reversible reaction involving reactants and product that are in different phases. the law of mass action is applicable to a homogeneous equilibrium and is also applicable to a heterogeneous system. (@) Decomposition of solid CaCO; into solid CaO and gaseous COz Let ‘a! moles of CaCO; are taken in a vessels of volumer ‘V' litre at temperature TK Caco\(s) === Cad{s) + CO.(9) Moles initially a 0 0 Moles at equilibrium a-x x x cao] [co. a= (eoleo.] [caco,] As CaCO, and Ca0(s) are pure solids, so their concentration is unity x + o Assuming COz gas to behave ideally at the temperature & pressure of the reaction. the molar concentration of CO, can be written using ideal equation as Feo, RT Chemical Equilibrium? Since Ke, R and T are constant, their product will also be a constant referred as Ke, Koa Po, = AF (2) Example 4 At 87°C, the following equilibrium is established He(@) + S(s) ——=* H,S(Q), Ke = 8 x 107 1 0.3 mole hydrogen and 2 mole sulphur are heated to 87°C In a 2L vessel, what will be the partial pressure of H.S approximately at equilibrium, [Use R= 0.08 atm, L/mol. K] (A) 032atm (@) 0.43atm (©) 0.62atm (0) 40atm Ans. (A) Solution k= BS@) a. to2 - [H@ 03 Ore bore 0.024 = 1.08 x -007 fgg = QO2HOOBH S60, Lg apaten " 2 Multiple Equilibrium In multiple equilibrium the product molecules (s) in one equilibrium system are involved in a second equilibrium process +8 —=cl@ + _ (2118) A@) +8) —= c@ + 9@) «, = el + £@ <= F@) + (FG) c@) + E@) —= F@ +6@ ue Overall reaction: A(g) + B(@) + E(@) ——=* DG) + Fig) + Gig) Sie In this case, one of the product molecule, C(g) of the first equilibrium reaction combines with E(@) to give F(g) and G(Q) in another equilibrium reaction. so in the overall, C(g) will not appear on either sce The equlibrium constant (K,) of the overall reaction can be obtained If we take the procuct of the expression of (K,.) and! (K.,) kK xk, < (C10), (ENG) _ {DIFIIG) oe Tale] {CHE} TAIIBILET Kee Ke, = Ke, Chemical Equilibrium 121122 If a equilibrium reaction can be expressed as the sum of two or more equilibrium reactions, the equilibrium constant for the overall reaction Is given by the product of the equilibrium constant of the individual reactions. Simultaneous Equilibrium In simultaneous equilibrium more than one equilibrium are established in a vessel at the same. time and anyone of the reactant or product Is common in more than one equilibrium. then the equilibrium concentration of the common species in all the equilibrium would be same For example, if we take CaCOs(s) and C(s) together in a vessel of capacity V' litre and heat it at temperature 'T' K, then CaCO; decomposes to CaO(s) and CO,(g). Further, evolved CO, combines with the C(s) to give carbon monoxide. Let the moles of CaCO; and carbon taken initially be 'a' and ‘b’ respectively CaCOs(s) Ca0{s) + CO.(9) Moles at equilibrium ax x (cy) CO.(g) + C(s) ——* 2C0(g) Moles at equilibrium Oy) (b-y) ay Thus, as CO, Is common in both the equilibrium so Its concentration is same in both the equilibrium constant expression. Equilibrium constant for first equilibrium, «,, = [CO.] [cor _ @ytv | ay? [co,] V’&-y) Vy) Equilibrium constant for second equilibrium, K., = Equilibrium Constant as Per Kinetics According to the kinetic theary of gases, in any gaseous system, different gas molecules travel with different speeds. The molecular collision with low energy can never cause bond cleavage and hence can not result the product formation. Only those molecular collision result the fomlation of product in which the molecules collides with a certain minimum energy Threshold energy - The minimum amount of energy, which the colliding molecules must posses in order to make the chemical reaction to occur, is known as Threshold energy, & Activation energy - ‘The minimum amount of energy required to make active participated of almost all molecules in a reaction is called Activation energy, E,. The activation energy is equal 10 E, ~ Ep, where Es is the average energy level of reactant molecules. Activation energy for forward reaction = Threshold energy ~ Potential energy of reactants ‘The activation energy of reaction depends on the nature of reactant and temperature. It decreases with increase in temperature but the decrease is so small that it is normally considered temperature independent Chemical EquilibriumE, = Threshold energy E, = Activation energy of forward reaction Ey = activation energy of backward reaction Potential energy of reactants =Potential energy of products Energy —> Reactants¥=— Product Pe Reaction co-ordinates —> Ag) ==) +C@) ala] at 7 Kr IAI~ Ke (8) [Cc] At equilibrium =A K. _ (BIC) KO According to Arrhenius equation Where k= A. e% A; pre-exponential factor 7 E,: activation energy [at temp 1, K = Kil Ink = Ina- At [at temp Ts, K = Ko Chemical Equilibrium 123Calculation of Degree of Dissociation by Vapour Density 124 Measurement Reactions in which there is a change in the number of moles after dissociation, the extent of dissociation can be determined by vapour density measurement Consider the following reaction - Initially 1 o 0 molesat (1 - a) a oa equilibrium, (‘a’ is the degree of dissociation) Total number of moles at equilibrium = (1- a) +a+ a= (1+ a) Vis the volume occupied by 1 mol of PCIs(s) which have vapour density is ‘D’ before dissociation and after dissociation is ‘a’, Under the same conditions, the volume occupied by (1 + @) moles at equilibrium would be (1 + a) V litre, Density © pet coy 4 or Sv (+a) or Vea) 2 x Vapour density where M, = calculate molecular mass Mo = observed molecular mass Note: When one mole of reactant on dissociation gives 'n moles of gaseous products. A—— 1B Initial moles 1 ° Moles after dissociation 1-a na Total moles of equilibrium 1-a+na=1+a(n-1) D D D-d) 2 iin da, 5 (=a or @ mp 4 Chemical Equilibrium? Example § Ans. N.O; on decomposition gives NO and NO;, they are found to be in equilibrium at 300 K. If the vapour density of such an equilibrium mixture is 23.75, calculate percentage by mass of N:Os in ‘the equilibrium mixture? (A) 80% (8) 60% (C) 40% (0) 20% © Solution: NO; D-d _ 38-2372 at 23.75(2-1) WHOFNO, 145 04x76 Totalwt 06x30106%46+04xT6 1 =06 100 Mass % of N.O, in the equilibrium mixture = 40% Gibb’s Free Energy and Equilibrium Constant Gibb’s free energy(G) of a system is defined as the thermodynamic quantity of the system, the decrease In whose value during a process Is equal to useful work done by the system Standard free energy change Is defined as the free energy change for a process at 298 K and | atm pressure in which the reactants in their standard state are converted to products in ‘their standard state. It is denoted as AG* Note: Standard free energy change ( AG") is not the energy change at equilibrium. AG" is related to K (equilibrium constant) by the relation AG’ = -RT ink AG" = -2.303 RT log kK. K may elther be Ke oF Ke. The units of AG” depends only on RT. T is always in Kelvin, and if R is in Joules, AG* will be in joules and if R is calories then AG* will be in calories. Example 6: NO and Br, at initial partial pressures of 98.4 and 41.3torr, respectively, were allowed to react at 300K. At equilibrium the total pressure was 110.5torr. Calculate the value of the equilibrium constant and the standard free energy change at 300 K for the reaction 2NO(g) + Bra(g) 2 NOBrig) Solution: 2NO@) = Br{g) === 2NOBr(g) Initial pressure 98.4 a3 ° At equilibrium 98.4-x 4a 3-3 x Total pressure at equilibrium is 110.5 torr Chemical Equilibrium 128126 984—K4413-F +x= 105 X= 58.4 torr Now 1 atm = 760 torr = 7.68 «107 atm Pyoer = 7.68 * 10-4 atm; Pro = 98.4 — 40 torr = 6.26 x 10% atm Py, 2 M3 = E9121 toor = 159 « 10.21 Pron | (r68x102)) Ke 34 atm [PINO)} [Pa,.] (5.26% 107) (159%10") AG® = =2,303 RT log K = = 2.303 (1.99) x 10°? (300) (log 134) = 292k eal = 12.2 ks [If R is used as 1.99 cal/mol K, then AG* will be in cal. If R is used as 8.314 J/mol K, then AG* will be in joules, But Ke must be in (atm)**] Le Chatellier Principle Chemical equilibrium represents a balance between forward and reverse reactions. In most cases, this balance is quite delicate. Changes in concentration, pressure, volume and temperature may disturb the balance and shift the equilibrium position so that more or less of the desired product is formed. There Is a general rule(named Le Chaterlier principle) that helps Us to predict the direction in which an equilibrium reaction will move when a change in concentration, pressure, volume or temperature accurs. Le Chatelier's principle state that if an external stress Is applied to a system at equilibrium, the system adjusts in such a way that the stress is partially offset. The word “stress” here implies a change in concentration, pressure. volume, addition of an inert gas or temperature that removes a system from the equilbrium state. Le Chatelier principle can be explained using the following equilibrium reaction PCI(g) === PCha(g) + Ch(g) Let the moles of PCls, PCI; and Cl, at equilibrium be a, b and ¢ respectively, Also let the volume of the container in which equilibrium is established be 'V' litre and the total pressure of the system at equilibrium be Pr atm Pal (pBeve] arbse Ko = bexP, °* Jlaebeo) ° ‘The total pressure of the system (P.) can be given as(assuming all gases at equilibrium behave ideally under the given conditions) Chemical Equilibrium(a+bsc)Rr v Port jb Vv Pr P. Inserting the value of ——* in equation (i), we get (a+b+c) xRT (2 axV Let us examine the effect of change of certain parameters like moles of reactant, moles of product, volume, temperature, addition of inert gas and addition of catalyst on the given equilibrium, (@) Change in number of moles of reactant If we add ‘d' moles of PCls to the equilibrium mixture, the equilibrium would be disturbed bexRT id thi si and the expression 2 jon moves in the becomes Qp. As Qs < Kp, so the net reai forward direction till Q> becomes equal to Ko, Thus for any equilibrium, when more reactant Is added to (or some product is removed from) an equilibrium mixture, the net reaction moves in the forward direction (as Q < K) to establish a new equilibrium state (b) Change In number of moles of product Let ‘d’ moles of PCl; (or Cl.) are added to the equilibrium. The equilibrium would be unc srs and thus the expression PHDERT you pocome Or, Snes Oe > Ke, SO ax the net reaction moves in the reverse direction till Q» becomes same as Kp. Thus for any equilorium, when product Is added to (or some reactant Is removed from) an equilibrium mixture, the net reaction moves in the reverse (backward) direction (as Q > K) to establish a new equilibrium state (©) Change in volume Let the volume of the container be Increased from V to V' litre. The equilibrium would bexRT becomes Op. The value of Op is less than Kp, be disturbed and the expression so the net reaction moves In the forward direction to establish new equilibrium. But when the volume of the container is decreased, the reaction moves in the backward direction to again attain the equilibrium state Thus for any equilibrium, on increasing the volume of the container, the net reaction shifts in the direction of more moles of the gases while on decreasing the volume of the vessel, the reaction goes in the direction of fewer moles of the gases Chemical Equilibrium 127128 (d) Addition or an inert gas The effect of addition of an Inert gas can be studied under two conditions, (), at constant volume (ii) at constant pressure. 0 i) At constant volume Let ‘d! moles of an inert gas are added to the equilibrium mixture at constant volume. The total number of moles of the system increases so is the pressure of the system but the partial pressure of all the species would still be same. Let Py RT the total pressure becomes Py then ——t—— = “Las R, T and V are fatb+ctd) VV bexRT ax constant, so the expression would still be equal to Ks. then net reaction does not move at all Thus for any equilibrium when an inert gas is added at constant volume. the equilibrium remains unaffected whether the equilibrium reactions have An equal to zero or non-zero. At constant pressure Now, let % moles of an inert gas are added to the equilibrium mixture at constant pressure to keep the pressure constant, volume of the vessel should increases. Let the volume of the vessel increases from V to V' litre, So the expression be xRT Tr becomes Qo As the value of Qe < Ke, so the net reaction moves in the ax forward direction to establish new equilibrium state ‘Thus, addition of an inert gas at constant pressure has the same effect as produced by the Increased volume of the container. Thus, for equilibrium having An = 0, when an inert gas is added at constant pressure, the equilibrium remains unaffected (since V does not appear in the expression of Ks) while for equilibrium having An + 0, the addition of an inert gas at constant pressure causes reaction to move in the direction of more moles of the gases. Addition of a catalyst A catalyst enhances the rate of a reaction by lowering the reactions’ activation energy Actually a catalyst lowers the activation energy of the forward reaction and the reverse reaction to the same extent, so the presence of a catalyst does not alter the equilibrium constant not does it shift the position of an equilibrium system. Adding a catalyst to a reaction mixture that is not at equilibrium will simply cause the mixture to reach equilibrium faster Chemical Equilibrium? If Kp Increases, the net reaction moves forward while If Ke decreases, the net reaction (Change in temperature moves backward The variation of K» with temperature Is given by Van't Hoff equation as. k, aH [-4| where T.> Th kK, ~ 23038|7, 7, All reactions are either endothermic or exothermic in nature. For an endothermic reactions, AH is positive and with an increase in lemperature of the system to Ty K from TK, the RHS of the expression becomes positive. Thus. equilibrium constant at higher tomporature (Kk (k) But for an exothermic reaction, AH Is negative and on Increasing the temperature of the ) would be more than the equillrlum constant at lower temperature system from T; K to To K. the RHS of the expression becomes negative. So the equilibrium constant at higher temperature would be less than equilibrium constant at lower temperature. The give equilibrium, PCIs(g) === PCL(@) + Cl(Q) is endothermic in nature. So, with the increase of temperature fram 7; K to Tz K. Ke and Qp both increases. Therefor equilibrium shifts in the forward direction, Thus, for an endothermic reaction (AH = positive), with the increase of temperature, net reaction moves in the forward direction and the decreases in temperature favours backward reaction while for an exothermic reactlon (AH = negative), net reaction moves In the backward direction with the increase of temperature and in forward direction with the decrease temperature In general, with the increases of temperature, net reaction moves in that direction where the heat is absorbed and the effect of increasing temperature is nullified, (@) Change in more than one parameter For the given equilibrium, if the number of moles of PCI; is increased four folds and the volume of the vessel is doubled, then the equilibrium would be disturbed. The expression RXGART would becomes Op. Since Qs > ke! so the net reaction moves in reverse ax direction till Qe becomes equal to Ke. Thus, when two or more parameters are simultaneously changed for any equilibrium, find the changed value of © and K and compare them. If Q = K, there will be no effect on the reaction, if @ > K. the net reaction moves in the backward direction, While if Q < K., net reaction moves in the forward direction Chemical Equilibrium 129)Application of Le cbatelier principle on physical equilibrium A. Melting of Ice H.0(s) —— H,0() Ice water more volume. less volume If we increase the pressure the equilibrium will in the direction of less volume. Hence: the rise of pressure, more ice will melt Into water Le. melting point of Ice Is decreased by rise of pressure, B. —Vaporization of liquid H,0() — —-H.0@) water vapour less volume more volume Vaporization of a liquid is endothemlic pro sin the nature Le. the evaporation of a liquid into its vapour Is completed by absorption of heat. so the rise of temperature will favour vaporization. On the other hand in this process, on increase of pressure the equilibrium will shift in the direction of less volume means water cannot be convened into vapour and boiling point increases. © Melting of Sulphur Sulphur(s) ——= — Sulphur() less volume more volume On increase In pressure. the equilibrium will shift towards less volume means solid is not converted into liquid and thus, melting point of sulphur increases Example 7 Following two equilibrium is simultaneously established in a container PCI(g) —— PCh(g) + Cig) COW + Clg) == coct(g) IF some Ni(s) is introduced in the container forming Ni (CO), (g) then at new equilibrium (A) PCI; concentration will increase (8) PCI, concentration will decrease (C) Cl, concentration will remain same (D) CO concentration will remain same ans. (8) 130 Chemical EquilibriumKP ak for different reactions S| Reaction] An | Relation Values] Values | Unit. | Unit |AH | Condition No between of Ke of Ke of for for Kp and Ke Ke | Ke obtaining more product 1] Hove =2Hi [0 an 4x2 [None | None |-ve tow Een] Hoe (exother | temperature mic) | No. pressure High concentratio n of reactant 2 | 2H Hoek | 0 KART? x x None | None | +e High a= | 40=x) (endoth | temperature ermic) | No pressure High concentratio nof reactant 3 Pc OS | KART xp Melk [atm | we High Plt W-av (endoth | temperature ermic) | Low pressure High concentratio 1 of reactant | Nae = || Kea kART | Ke | axe Molt [atm | we High 2No, T-0v (endoth | temperature ermic) | Low pressure High concentratio n of reactant 5. |2NH = [2 KARTE [> Ke 27Ke Zixp | MoFE | atm? | we High Not3He (endoth | temperature ermic) | Low pressure High concentratio nof reactant 6 [Nah = ex [e [atm [ve [tow 2NHs PHI» | TG—x)*p"] mor (endoth | temperature ermic) | Low pressure High concentratio Keke n of reactant Chemical Equilibrium 1317 | Pchich = eK] ow xaos) umol [am* [ve iow Pcls = wp (endoth | temperature ermic) | Low pressure High concentratio n of reactant 8 [250,02 Ke= KART? xv Xx) [emol [atm [ve | low =2503 to (yp {endoth | temperature ermic) | Low pressure High concentratio n of reactant 1 For the reaction PCIe(g) ——= PCL(g) + Ch@) The moles of each component PCls, Cl: and Cl: at equilibrium were found to be 2. If the total pressure Is 3 atm. The Ke will be (A) atm. (8) 2atm (©) 3am (0) 15 atm. Ans. (A) Solution Total Moles = 2+2+2=6 Q.2 For the reaction —— Hl The value of equilibrium constant is 9.0. The degree of dissociation of HI will be- 2 (@) 2/5 sia (0) v2 Ans. (8) Solution Equilibrium constant of the reaction Heth === 2H So the equilibrium constant for the dissociation of Hil ie, 2H === H, + b will be 1/9. 2H Ht lh 1 o o x x x 2 z 132 Chemical Equilibriumor 2-2x = 3x ox=2 x= 2/5 For the reaction No = 2NH;, Ni: He were taken in the ratio of 1: 3. Up to the point of a3 equilibrium 50% each reactant has been reacted. If total pressure at equilibrium is P. The partial pressure of ammonia would be- (A) P/3 (8) P/6 (Prva (wo) PB Ans. (A) Solution Ne + 3H) === 2NH, Initially 1 3 0 At equilibrium 105 3-15 2x05 Total moles = (1-05) + (3-15) +1=3 Fs Q4 — Ina reaction vessel of 2 litre capacity 3 moles of Nz reacts with 2 moles of O, to produce 1 mole of NO. What is the molar concentration of Nz at equilibrium? (A) 1.28 (8) 150 (0.75 (0) 20 Ans. (A) Solution: M+ Op === 2NO 302 0 3-x 2-x 2x. 2k 1: x 08 3-05 [Nal = 1.25 0.5 Hi was heated in a sealed tube at 440°C till the equilibrium was established. The dissociation of HI was found to be 22%, The equilibrium constant for dissociation is - (A) 0.282 (8) 0.0786 (©) 0.0199 (0199 Ans, (©) Chemical Equilibrium 133Solution: The equilibrium of the dissociation of Hoo H+ lL 1 0 ° 2222 2) 400 100x2 j00%2 0.78 om on Ke = OXON on99 078x078 0.6 —AL87°C, tile following equilibrium Is established H,(g) + S(s) === HS) Ke = 7 x 107 If 0.50 mole of hydrogen and 10 mole of sulphur are heated to 87°C In 1.0 L vessel, what will be the partial pressureofH2S at equilibrium? (A) 0.966atm (8) 138atm (©) 0.0327atm (0) 9.66atm, Ans, (A) Solution: Hg) + Ss) === 4H,S(g) [Hs] aeree I a / (3 RT 00327 «0.0871 «360 0.966 atm P, Q.7 At some temperature N,Ox Is dissociated to 40% & 50% into NO; at total pressure P; & P, atm respectively, then the ratio of Py & Pa is we @t ost (0) None of these 5 a 7 Ans. (B) Solution: N09) = == -2N0,(g) at Pi (t= Eq) 1-04 2(0.4) =06 =08 at P) (t = Fa) 1-05 2(0.5) =05 temperature is same, Ko is same. 134 Chemical Equilibrium14x06 (ay x15x05 Q.8 AG" for the dissociation of the dimer (Ay = 2A) in benzene solution at 27°C Is 6.909 kcal/mol. if 8 moles of A is dissolved in 10dm? of benzene at 27°C. What Is the ratio of equilibrium concentration of monomer to dimer ([AI/IA |)? Given: R=2 Cal/ mol. K (A) |: 200 @) I 100 (©) 200:1 (©) 800:1 Ans. (A) Solution AG? == RT In Kee 6.909 x 1000 - = 2 x 300 x 2.303 log Ke =5 = log Ke or Ke = 10-* KL = Ke = 10° 2a =A oe-mx x K. Is very high S02 =08 >x=04 0.8 - 2x =y ror = 24 y= (04 x10)" = 2x107 w IAL _y | 2x10” ial x04 ee 7000 ~ 700 9 For the equilibrium at 27°C. LICLANHs(s) ——=* LICLNH.(6) + 2NH(Q): Ke = 9 atm? A 24.63 litre flask contain 1 mol LiCI.NH:. How many moles of NH; should be added to flask at this temperature to drive backward reaction for completion. Solution Ke = 9 atm? pi, = 9 atm Py, = 3atm 3 x 24.63 = Ny, xRx300 ny, = 3 (at equillbrium) LICL.3NHs(s) == LICI.NHs(s) + 2NHs(g). 1 n ° m2 n-2-3 n= moles Chemical Equilibrium 135Ina system, the equilibrium reaction: 2NHs(g) === No(g) + 3H2(g) was studied starting with NH; and Ne(inert gas). It Is found that at 10 atm and 700 K, the equilibrium gaseous mixture contains 10 mole % each of NH(@) and Ne(g). Calculate Ke (in atm?) Solution X +X, = 1-01- 01-08 1 fx08 302 x, = 3x08 306 4 Py, = 0.110 = 1 atm Py, = 0.2 x 10 = 2 atm P, = 0.6 «10 =6atm Ko = 2x 216 Ke = 432 (atm)? Ans. O.11 Ata certain temperature the equilibrium constant Ke is 0.25 for the reaction AQ) + B:lg) = Cx(Q) + D.(Q) If we take 1 mole of each of the four gases in a 10 litre container, what would be equilibrium concentration of A(a) Solution Ix a s| u Q > Ke so reaction will proceed In backward direction Ax(g) + Bog) == C9) + alg) 1O+x 1O+x 10-x 10-x concentration at equilibrium 0 10 10 Ce) os - LOF 305 305+05 Tex 70 15x=05 0.333 [Aig = 2X = LX = 0.13 Ans. 7 10 136 Chemical Equilibrium? Q.12 Ata certain temperature, equilibrium constant Ke = 4 x 10° for the reation N.{g) + 02g) ——= 2NO() If we take 15 mole of NO and 2 mole each of Nz & O; in § litre vessel, what would be the equilibrium concentration of NO (in mole/litre)? Solution: (1.5/5)? rey reaction will proceed in backward direction Ng) + 0) == 2NO() => 0.5625 1 O> Ke Initial moles 2 2 18 moles at eq? 2+ x 2ex 15 ~ 2x com ateqr 2#%* EK) 5 5 5 __ Nowy IN@)[2.0) 18 ~2x = yoo4 202 Zax 1 = 2x =0.4 + 0.2x x=08 1.5-2x Equillbrium concentration of NO = 201M Chemical Equilibrium 137Single Correct Type Question Properties of equilibrium and equilibrium constant 1 138 ‘A chemical reaction is at equilibrium when (A) Reactants are completely transformed into product: (8) The rates of forwa (C) Formation of products is minimised (0) Equal amounts of reactants and products are present and backward reactions are equal Which of the following statement is Incorrect (A) At equilibrium, vapour pressure of solution and refractive index of equilibrium mixture becomes constant. (B) Equilibrium can be attained in both homogenous and heterogeneous reaction (C) At equilibrium concentration of reactants and products becomes constant (©) Equilibrium is dynamic in nature and equal Concentrations of reactants and products at equilibrium can be like: “ VS ‘\ (0) All of these time Rate of reaction curve for equilibrium can be like: [ry = rate of forward, ry = rate of backward] rate] ae a” @) rn time Tne time ‘rate Initially the reactions in the container a & b are at equilibrium when the products & reactants are put together in a container c then at the equilibrium the total number of different chemical compounds are — No He NA, 5 (7 (Ce (8 Chemical Equilibrium10, " Y In a chemical equilibrium, the rate constant for the backward reaction is 2 x 10 and the equilibrium constant Is 15. The rate constant for the forward reaction Is: (A) 2x10" (8) 5 x10 (0) 3x 104 (0) 9.0 x10 The equilibrium constant for the reaction N,(Q) + 0,{g) <= 2NO(g) at temperature T Is 4 x 10-*, The value of equillbrium constant for the 1 1 reaction NO (@) =* > Nxi@) + 5 O2(@) at the same temperature is: ~ (a) 0.02 (@) 50 (©) 4104 (0) 25 102 The equilibrium constant for the given reaction 1 SOs(g) = SOQ) + 5 Ox(Q) | K = 5 x 107 ‘The value of K. for the reaction 250,(g) + O2(g) = 2S0s(g), will be (A) 400 (8) 2.40 x 1077 (0) 98 x 107 (0) 4.9 «107 For the following three reactions 1, 2 and 3, equilibrium constants are given () CO) + H.0() = COxQ) + Hila) k (2) CH(g) + 109) = COW) + 3HX9) Ke (3) CHa(g) + 2H20(g) = CO2(g) + 4H.(9) Ks Which of the following relations is correct 7 WK =k (8) Kiks = ki (C) Ks = Kike (D) Ks Kk? Equilibrium constant for following reactions respectively Ki, Ka and Ks No+ 3H, = 2NHs Ki No+¥ OQ, 2NO Kk 1 Het 50. 2 HO Kk bie k 5 2NHs + 502 <2 2NO + 3H.O Ke Which of the following relation is incorrect. Kyxk, Ky (B) Ke = Ky x Ka/({Ka)? (C) Ko = Sulfide ion in alkaline solution reacts with solid sulfur to form polysulfide ions having formulas S#, $s, Se and so on. The equilibrium constant for the formation of S$, Is 12 ( Ki) & for the formation of Sv is 132 (K,), both from $ and S*What is the equilibrium constant for the formation of $.2- from S.- and S? an @ 2 (cy 132 {D) None of these Chemical Equilibrium 13913 14 16 16. W 18. 19. 20, 140 For a reaction N; + 3H; <2 2NHs, the value of Ke does not depends upon: {@) Initial concentration of the reactants (b) Pressure (©) Temperature @ Catalyst (A) Only © abc (abd Wabead For any reversible reaction if concentration of reactants increases then effect on equilibrium, constant: (A) Depends on amount of concentration (8) Unchange {C) Decrease (0) Increase If some He gas is introduced into the equilibrium PClig <= PCliw + Clap at constant pressure and temperature then equilibrium constant of reaction: (A) Increase (8) Decrease (©) Unchange {D) Nothing can be sald The equilibrium constant for the reaction: Nx(g) + 02g) ——*2NO(q) at 2000 K Is 4 x 10" In presence of a catalyst the equilibrium is established ten times faster at the same temperature What Is the value of equilibrium constant in presence of catalyst (A) 40 x 10 (8) 4x10" () 4 x 108 (0) 25x 10+ The equilibrium constant (Ke)for the reaction 2X(g) + 2¥(@) == 2Z(@) Is given as: txil2¥) pay) (zp (zp (A) (8) —— (C) D) “Ra © © Reyer © dim For the reaction CUSOL5H:Oy = CUSO,3H:O+ 24:0 Which one is correct representation:— A) Kr Pyoop (@) K = [hor ()Ky=Ke(RT? DAIL For hypothetical equilibrium, 4A(g) + 5B(g) == 4X(g) + 6Y(g) The unit of Ke will be: {A) litre mole" (B) mole litre? (C) litre mole-2 (D) mole? litre? What is the unit of Ke for the reaction 7 CS; (g) + 4H: g) = CH, (g) + 22S (@) (A) atm (B) atm? (C) atm? (D) atm The equilibrium concentration of B[(B).Jfor the reversible reaction A = B can be evaluated by the expression:— k, (A) Ke [ay @ cb [Ale (D) keke [AJ ©) ky Pa © Chemical Equilibrium2 22. 23 24. For which reaction is Kp = Ke— (A) 2NOCI = 2NO(g) + CL(g) (8) No(@) + 3Ha(g) = 2NHs(g) (C) H(g) + bo(g) = 2H! (g) (0) 250,(g) + O.(g) = 250,(9) K, log Z + log RT = O is true relationship for the following reaction: c (AVC = PCL + Cl (8) 280; + 0, = 250; (C) No + 3Hz == 2NHs (0) (8) and (C) both For which reaction at 298 K, the value of 2 will be maximum and minimum respectively (@) N.0x(g) —=2NO,(9) (©) 280:(9) + Ox(9) == 280s(9) ©) X@ + YQ) == 42) (d) A@) + 38) ——=7C@) Adc @)db ©) cb Oda For the following gases equilbrium, NOs (g) <= 2NO: (g) Kp is found to be equal to Ke. This is attained when temperature is orc (8) 273k (1k (2) 12.19 k Application of equilibrium constant (Homogeneous equilibrium) 25 26. When Ke >> 1 for a chemical reaction, (A) the equilibrium would be achieved rapidly {B) the equilibrium would be achleved slowly {C) product concentrations would be much greater than reactant concentrations at equilibrium {D) reactant concentrations would be much greater then product concentrations at equilibrium, Equilibrium constant of some reaction are given as under @x—y K = 10" by =z K=2% 107 (PrP —a K=3x 104 @R—=s k=2% 108 Initial concentration of the reactants for each reaction was taken be equal: Review the above reaction and indicate the reactions in which the reactants and products respectively were of highest concentration Adc @ca (ad (be Chemical Equilibrium 14128. 29. 30, a 32. 142 What should be the value of Ke for the reaction 28021) + Ox) <2 2SOsy, if the amount are SOs = 48 g, SO,= 12.8 g and O; = 9.6 g at equilibrium and the volume of the container is one litre? (A) 64 (8) 30 (cy 42 as A certain quantity of PCls was heated in a 10 litre vessel at 250°C. At equilibrium the vessel contains 0.1 mole of PCI; 0.20mole of PCI; and 0.2 mole of Cl; The equilibrium constant of the reaction PCIs(g) ——* PCLg) + Cig) is. (A) 0.02 (8) 008 (©) 0.04 (0) 0.025 PCIs(g) == PChs(g) + Chala) In above reaction, at equilibrium condition mole fraction of PCI; is 0.4 and mole fraction of Cl. is 0.3. Then find out mole fraction of PCIs PCIs(g) === PChi(g) + Cla(g) (A) 03 (@)07 joa (06 In the reaction 2P(g) + Q(g) = 3R(Q) + S(g). If 2 moles each of P and Q taken initially in a1 litre flask. At equilibrium which Is true: A) [PI < [0] (8) [P] = [0] (C) [0] = IRI (0) None of these For the reaction A + 28 = 2C = D, initial concentration of A Is a and that of B is 1.5 times that of A. Concentration of A and D are same at equilibrium. What should be the concentration of B at equilibrium? m3 a) % o 0) All of the ab WG @5 OF (0) Allof the above The figure show the change in concentration of species A and B as a functional of time. The equilibrium constant Ke for the reaction A(g) = 2B (9) is: (A) Ke>t @k«1 (©) kK (0) data insufficient Ke = 9 for the reaction, A + B = C + D, If one mole of each A and B are taken, then amount of € in equilibrium is: wi (0.25 (075 (a) None of these Chemical Equilibrium24. 35 36. 37. 38. 29. 40, 4 ? ‘The equilibrium N, (g) + O7(g) = 2NO (q) is established in a reaction vessel of 2.5 L capacity, ‘The amounts of Nz and O, taken at the start were respectively 2 moles and 4 moles, Half a mole of nitrogen has been used up at equilibrium. The molar concentration of nitric oxide Is: (a)o2 (@)o4 06 (01 For the reaction 3 A (g) + 8 (g) = 2 C (g) at a given temperature, Ke = 9.0 , What must be the volume of the flask, if a mixture of 2.0 mol each of A, B and C exist in equilibrium 7 ol @ ou (C36 {D) None of these In the reaction Aig) + 28() == 2C(g) , if 2 mole of A, 3.0 moles of B and 2.0 moles of C are placed in a 2L flask. If equilibrium concentration of is 0.5 mole/L, The value of equilibrium constant (Ke) will be: (a) 0.073 (@) 0147 (©) 0.05 (0) 0.026 The equilibrium constant for the reaction AQ) + 28) = Cig) is 0.25 dm*mor. In a volume of 5 dm', what amount of A must be mixed with 4 mol of B to yleld 1 mol of C at equilibrium, (A) 3 moles (8) 24 moles (C) 26 moles (0) None of these A 20.0 litre vessel initially contains 0.50 mole each of H2 and Iz gases. These substances react and finally reach an equilibrium condition. Calculate the equilibrium concentration of HI if Kea = 49 for the reaction Ho + Iz = 2HI (A) 0.78M (8) 0.039 M (€) 0.033 M (0) 0.021 M Equimolar concentrations of H, and |, are heated to equilibrium in a 2 litre flask. At equilibrium, ‘the forward and the backward rate constants are found to be equal, what percentage of initial concentration of H has reacted at equilibrium for the reaction Hs +k == 2HI (A) 33% (8) 66 % (©) 50% (0) 40% (©) 20% When alcohol (C:H:OH ()) and acetic acid (CH:COOH (/)) are mixed together in equimolar ratio at 27°C, 33% of each Is converted into ester. Then the Ke for the equilibrium CiHsOH() + CH:COOH () == CHaCOOC He (I) + H.0(0 is a eu «9 wv In the reaction, PCls = PCI; + Cl, the amount of each PCls, PCI and Chis 2 mole at equilibrium and total pressure is 3 atmosphere. The value of Ky will be (A) 1.0 atrn. (@) 3.0 atm (©) 29 atm (0) 6.0 atm Chemical Equilibrium 14343 44, 45. 46. 47 48, 144 PC = PCl: + Cl. in the reversible reaction the moles of PCls, PCls and Cl. are a, b and c respectively and total pressure is P then value of Ke is: be b ber c “ a ar ® apo ° © a(a+b+c) © fa+bec) The equilibrium constant, Ke for the reaction 2S0,(g) + Ox(g) = 2S0(g) is 4.0 atm~ at 1000 K, What would be the partial pressure of O; if at equilibrium the amount of SO, and SO; is the same? (A) 16.0atm, (8) 0.25 atm (c) atm (0) 0.75 atm For the reaction Adlg) + 2Br =+ 2C.(g) ‘the partial pressure of A,, 8: at equilibrium are 0,80 atm and 0.40 atm respectively. The pressure of the system is 2.80 atm. The equilibrium constant Ks will be (A) 20 (8) 50 (©) 0.02 (0) 0.2 The equilibrium constant Ke for the reaction, AQ) + 28() = 3C(Q) is 2 « 107 What would be the equilibrium partial pressure of gas C if initial pressure of gas A & B are 1&2 atm respectively. (A) 0.0625 atm (8) 01875 atm (©) 0.21 atm {D) None of these At 675 K, Ha(g) and CO» (g) react to form CO(g) and H:0 (9), Ke for the reaction is 0.16. Ifa mixture of 0.25 mole of H,(g) and 0,25 mal of CO, is heated at 675 K, mole % of CO(g) in equilibrium, mixture is wm (8) 14.28 (©) 28.57 (D) 33.33 (NH? The reaction quotient Q for Nz{g) + 3H.(g) = 2NHa(g) is given by Q man The reaction will proceed In backward direction, when (A) O= Ke, ()O Ke (0) Q=10 N,Ox(@) == 2NO,(Q), Ke = 4. This reversible reaction Is studied graphically as shown in figure. Select the correct statement {A) Reaction quotient has maximum value at point A (8) Reaction proceeds left to right at a point when [N04] = [NO.] = 0.1M {C) Ke = @ when point D or F is reached: {D) Both (8) and (C) are correct Chemical Equilibrium? Degree of dissociation (a) 49. Ag) = 3A) In the above reaction, the initlal concentration of A; Is "a" moles/lit.Ifx Is degree of dissociation Of As, The total number of moles at equilibrium will be: ax a a-ax Wa- > (8) 5- ax © ( 3 ) (D) a + 2ax 50. For the reaction: P = Q + R. Initially 2 moles of P was taken. Up to equilibrium 0.5 moles of P was dissociated. What would be the degree of dissociation (A) 05 (B)1 (C) 0.25 (0) 42 51. A moles of PCls are heated at constant temperature in closed container. If degree of dissociation for PCls is 0.5 calculate total number of moles at equilibrium: (A) 45 (B)6 (3 (o)4 82. The degree of dissociation of SO; is « at equilibrium pressure p°. Ky for 280x(9) <= 280,49) + O2(9) we @ Pe eet (0) None of these 2(1-a) (2+a)(1-ay 2(1-a) 53. For the reaction: 2HI (g) = H.(9) + L(g), the degree of dissociated (a) of Hi(g) is related to equilibrium constant Kr by the expression [14 2K, 2k, (8) 2 © (0) Y (rae Tale 84. Two sample of HI, each of 5 gm. were taken separately into different vessels of volume § and 10 litres respectively at 27°C . The extent of dissociation of HI will be: (A) More in 6 litre vessel (8) More in 10 litre vessel (©) Equal in both vessel (0) None of these 55. The vapour density of N.O« at a certain temperature is 30, What is the % dissociation of N,O, at this temperature? (A) 53.3% (8) 106.6% (C) 26.7% (0) None 56. Vapour density of the equilibrium mixture of the reaction 2NH: (9) = Nz (@) + 3He (g) Is 6.0 Percent dissociation of ammonia gas Is: (A) 13.88 (8) 58.82 (©) 41.66 {D) None of these Chemical Equilibrium 145The degree of dissociation is 0.5 at 800 K and 2 atm for the gaseous reaction PC = PCh + Ch Assuming ideal behaviour of all the gases. Calculate the density of equilibrium mixture at 800 K and 2 atm: (A) 4.232 g/L (8) 64 g/t. (C)84 g/t (0) 2.2 g/t Application of equilibrium constant (Heterogenous equilibrium) 58. 60, 61 In the reaction C(s) + CO.(g) = 2CO{@), the equilibrium pressure Is 12 atm. If 50% of CO2 reacts ‘then Kp will be: (A) 12 alm (8) 16 atm (©) 20 atm (0) 24 atm Solid ammonium carbamate dissociate to give ammonia and carbon dioxide as follows NH.COONH,(s) = 2NHA(g)_ + CO.(g) Which of the following graph correctly represents the equilibrium. For NH Choose the correct options: Pow tl es + | Ate (a) Pressure} [eae (8) Pressure Ip, Time(s) —> nes t t A Den (©) Pressure © aut — [a Aes time(s) —> time(s) —> Chemical Equilibrium? 74. A liquid is in equilibrium with its vapour at its boiling point . On the average the molecules in Physical Equilibrium the two phases have equal: {A) inter molecular forces (8) potential energy {C) kinetic energy (0) none of these 78. Au(s) = Au() Above equilibrium is favoured at: (A) High pressure low temperature (8) High pressure high temperature {C) Low pressure, high temperature (0) Low pressure, low temperature 76, Agas*X' when dissolved in water heat Is evolved, Then solubility of 'X' will Increase: (A) Low pressure, high temperature (8) Low pressure, low temperature {C) high pressure, high temperature (0) high pressure, low temperature 77. When the pressure is applied over system ice =: water what will happen (A) More water will form (8) More ice will form {C) There will be no effect over equilibrium (0) Water will decompose in H. and Q, 78. Some quantity of water is contained in @ container as shown in figure. As neon is added to this system at constant pressure, the amount of liquid water in the vessel vapour —| water —| (A) increases (8) decreases (©) remains same (0) changes unprediictably Simultaneous equilibria 79. The two equilibria, ABlag) = A'(aq) + BM(eq) and ABleq) + B-(aq) = ABM(aq) are simultaneously maintained In a solution with equilibrium constants, K; and K, respectively. The ratio of concentration of A* to AB," in the solution is: (A) directly proportional to the concentration of B- (aq) (8) inversely proportional to the concentration of B (aq) {C) directly proportional to the square of the concentration of B- (aq) {D) inversely proportional to the square of the concentration of B- (aq) Chemical Equilibrium 149The reactions PCl(g) = PCh(g) + Ch(g) and COCI.(g) = CO(g) + Cl{g) are simultaneously in equilibrium at constant volume. A few males of CO(g) are introduced into the vessel. After some time, the new equilibrium concentration of (A) PCls will remain unchanged (8) Ch. will be greater {C) PCIs will become less (0) PCI: will become greater ANSWER KEY 1 8 2 © 3 © 4 WM & © 6 © 7 & a Wo © 0 @ 7 AW 2 (| 2B B® Ww © bs © 6 © 7 © 8 @ 0 & 20 | 2 © 2 @ 2 @ 4 © 2% © 2% & 227 B& 2 © 2. W 30 @ 3. @ 32 A 88) 84) 3% ©) 32 © 3% @ 39 A 4 & 4 A 42 © 43 (8) 44 «(450 @) 46) 8, 50, () 51.) 5%) 83.) ) 85H) e758) 5) 6) 64. A) 65, (0) 6A), Be), 7.) Mm A 72 © 7 @ 7% © 7% ©) 7% O 7 wW 7 (8) 79 (0) Ba) 150 Chemical EquilibriumMultiple Correct Type Question 1 Following two equlliorlum is simultaneously established In a container PCL(g) a2 PCL(g) + Cl(9) COI) + CL(@) = COC) If some Ni(s) is introduced in the container forming Ni (CO), (@) then t new equilibrium (A) Pel, concentration will increase (8) PCh concentration will decrease (€) Cl. concentration will remain same (0) CO concentration will remain same. For the reaction Pt Is(g) <= PCh(g) + Ch(g). the forward reaction at constant temperature Is favoured by (A) introducing an inert gas at constant volume (8) introducing chlorine gas at constant volume (©) introducing an inert gas at constant pressure (0) introducing PC! -onstant volume. When NaNO: is heated in a closed vessel, oxygen is liberated and NaNO; Is left At equilibrium (A) addition of NaNO; favours reverse reaction (B) addition of NaNO; favours forward reaction (©) increasing temperature favours forward reaction (0) increasing pressure favours reverse reaction For the gas phase reaction, C:Hs + Hy <= C:He (AH = -32.7 kcal), carried out in a closed vessel ‘the equilibrium moles of C;Hs can be increased by (A) increasing the temperature (8) decreasing the pressure (C) removing some Hz (0) adding some CH. Phase dlagram of CO; is shown as following 395 217 1298304 «) Based on above find the carrect statement ) (A) 298K Is the normal boiling point of liquid CO. (8) At 1 atm & 190 K CO, will exist as gas, (C) COx{s) will sublime above 195K under normal atmospheric pressure (0) Melting point & boiling point of CO, will increase on increasing pressure Chemical Equilibrium 151The equilibrium between, gaseous isomers A, B and C can be represented as Reaction Equilibrium constant Ag) =B@) Ke B@=c@ K c@ =A Ks = 0.6 If one mole of A is taken in a closed vessel of volume 1 litre, then (A) [A] + [8] + [C] = 1 Mat any time of the reactions (8) Concentration of ¢ is 4.1 M at the attainment equilibrium in all the reactions 1 The value of K is (©) The value of Kis 55 (0) Isomer [A] is least stable as per thermodynamics. For the gas phase exothermic reaction, A: + By = Cz, carried out in @ closed vessel, the equilibrium moles of Az can be increased by (A) increasing the temperature (8) decreasing the pressure {C) adding inert gas at constant pressure (0) removing some C2 Consider the equilibrium Hg0(s) + 41> (aq) + H.0 (0. Hgle? (aq) + 20H> (aq), which changes will decrease the equilibrium concentration of Hgli?= (A) Addition of 0.1 M HI (aq) (8) Addition of Hg0 (s) (©) Addition of H.0 (1) (0) Addition of KOH (2a) Decrease in the pressure for the following equilibria: H»O (s) <= H,O(¢) result in the (A) formation of more H.0 (s) (B) formation of more H,0(4) {C) increase in melting point of H.0(s) {D) decrease in melting point of H,0(5) Assertion Reason 10, 152 Statement=1: Ammonia at a pressure of 10 atm and CO, at a pressure of 20 atm are introduced into an evacuated chamber. If Ky for the reaction NH.COONHs (5) = 2NHs (g) + CO, (g) is 2020 atm’, the total pressure after a long time is less than 30 atm Statement=2: Equilibrium can be attained from both directions. (A) Statement-1 is true, statement-2 is true and statement-2 is correct explanation for statement-1 {B) Statement-1 Is true, statement-2 is true and statement-2 Is NOT the correct explanation for statement-1 (C) Statement-1 is true, statement-2 is false. {D) Statement-1is false, statement-2 is true. Chemical EquilibriumComprehension Type Question 1 12. 13 14, 18 16. W Paragraph for Question Nos. 11 to 14 In 2 7.0 L evacuated chamber, 0.50 mol H, and 0.50 mol |, react at 427°C H,{g) + L(g) = 2HI(g). At the given temperature, Ke = 49 for the reaction. What is the value of Kp 7. a7 @)49 (©) 245 (©) None What Is the total pressure (atm) in the chamber? (a) 83.14 (8) 8314 a2 (0) None How many moles of the iodine remain unreacted at equilibrium? (A) 0.388, (one (©) 0.25 () 0.128 What is the partial pressure (atm) of HI in the equilibrium mixture? (A) 6.385 (8) 12.77 (©) 40.768 (0) 646.58 Paragraph for Question Nos. 15 to 17 Influence of pressure, temperature, concentration and addition of Inert gas on a reversible chemical reaction In equilibrium can be explained by formulating the expr constant Ke or Kp for the equilibrium. On the other hand Le Chatelier principle can be theoretically used to explain the effect of P, T or concentration on the physical or chemical equilibrium both, For the reaction: PCls = PCI; + Cl. increase of pressure shows: (A) An Increase in degree of dissociation and a decrease In Ke {B) A decrease in degree of dissociation and a decrease in Ke (C) An increase in degree of dissociation but Ke remains constant (0) A decrease in degree of dissociation but Ke remains constant ion for equilibrium For the reaction: 250, + 0, =: 280s; AH = -ve, An increase in temperature shows: {A) More dissociation of SO; and a decrease in Ke (B) Less dissociation of SO; and an increase in Ke {©) More dissociation of SO; and an increase in Ke (D) Less dissociation of SO; and an decrease in Ke For the reaction: Fevhigy (24) + SCN-(aq). = [Fe(NCS)]” (aq): in equilibrium if tittle more Red aqueous solution of FeCl, Is added, than: (A) Forward reaction becomes more and red colours is deepened (B) Backward reaction becomes more and red colour faintens {C) Solution becomes colourless (D) None of these Chemical Equilibrium 153Match the column 18. 19. Column | (A) Kee Ke (8) Introduction of inert gas at constant pressure will decrease the concentration of reactants (©) K*ris dimensionless {D) Temperature increase will shift Column-1 (Reactions) (A) Oxidation of nitrogen (P) Na(g) + O2(g) + 180.6 kJ = 2NO@) (8) Dissociation of N,O«(g) (Q) N,O.(g) + 57.2 KJ = 2NO,(g) (C) Oxidation of NH.(g) (R) ANHs(g) + 50.(9) = 4NO() + 6H,0(9) + 905.6 kJ (D) Formation of NO.(g) (Ss) NO(g) + Os(g) ....NO2(g) + O,(g) + 200 kJ Matching list type 20, Code: 15 19. 154 Column Il (P) No + 3H) <2 2NHy (Q) PCs (g) = PCI 9) + Ch @) ® 2NO; (g) = N.0s (9) (S)__NHs (g) + HI (g) = NH« (s) Column-tI (Favourable conditions) Addition of inert gas at constant pressure Decrease in pressure Decrease in temperature Increase In temperature Column-Hl (if ais negligible wert. 1, V =1 litre) an2x an3x ie a= (2K) (cd) 6 (ACD) 7 (asc) © 3 © 4 A (A)-P.R'S; (B)-PO.RS: (C)-P.O.RS; (D)-0: column-I (Reaction) ©) 2X@) = Y@ +Z@) “ (2) X@) = Y@ + Z@) @) ®) 3X) = YO +2) @ ()— 2X(@) = Ya) + 2219) (4) Pp Qa oR s Wo4 1 3 2 (B) 2 4 1 3 (c) 1 4 2 3 (D) 2 3 1 4 ANSWER KEY (8) 2 (cD) 3. (cD) 4. (ABCD) 5. (cD) 9 (AC) 10. (co) 1 (8) 12. © 6 W 17 #18 (A)-S; (B)-P,Q,S; (C)-P,Q,R; (D)-R- 20. (© Chemical EquilibriumApplication of equilibrium constant (Homogeneous equilibrium) 1. Consider the gas-phase hydration of hexafluoroacetone, (CF:):CO (CF.)» CO @) + H.0 (9) == (CF), CLOH), (@) At 76°C, the forward and reverse rate constants are ky = 0.18 M*' Sand ky = 6 x 10s", What is the value of the equillorium constant Ke? 2. The progress of the reaction A = NB with time, is presented in figure. Determine Sen os 03 ae 01 / M 3. 5 OT — Time (min.) (i) the value of n, (ll) the equilibrium constant K. 3. IF Ke = 0.5 « 10° at 1000 K for the reaction Nz (g) + O2 (g) = 2NO (g), what is Ke at 1000 K for 1e reaction 2NO (g) —= Ny (g) + 02 (9)? 4. The vapour pressure of water at 27°C is 0.2463 atm. Calculate the values of Ky and Ke at 27°C for the equilibrium H.0 (D) = H.0 (9) 5. 2 moles of A & 3 moles of B are mixed in 1 litre vessel and the reaction is carried at 400°C according to the equation: A + B <= 2C. The equilibrium constant of the reaction is 4. Find the number of moles of C at equilibrlum 6. AL 400 K, Ke = 25 x 10° far the reaction CH, (q) + 2H:8 = CS,(@) + 4H)(Q). A 10.0 L reaction vessel at 1400 K contains 2.0 mol of CH,, 3.0 mol of C52, 3.0 mol of Hy and 4.0 mol of H)S. Is the reaction mixture at equilibrium? IF not, in which direction does the reaction proceed to reach equilibrium? Chemical Equilibrium 155Degree of dissociation (a) 7 10, When 36.89 N:O, (g) Is introduced inte a 1.0-litre flask at 27°C . The following equilibrium reaction occurs N2O« (g) = 2NO> (g) : Ks fa) Calculate Ke of the equilibrium reaction. 0.1642 atm {b) What are the number of moles of N,O. and NO; at equilibrium? (©) What is the total gas pressure in the flask at equilibrium? {q) What is the percent dissociation of N.0. At some temperature and under a pressure of 4 atm, PCle Is 10% dissociated . Calculate the pressure at which PCl; will be 20% dissociated, temperature remaining same. PCI. dissociates according to the reaction PCls = PCIs(Q) + Cle(g). At 523 K, Kp = 1.78 atm. Find ‘the density of the equilibrium mixture at a total pressure of 1 atm The system NOs = 2.NO, maintained in a closed vessel at 60° C & a pressure of 5 atm has an average (Le. observed) molecular weight of 69, calculate Kp. At what pressure at the same temperature would the observed molecular weight be (230/3) ? Application of equilibrium constant (Heterogenuous equilibrium) 1 12 13 14 156 At 90°C, the following equilibrium Is established: Halg) + S(6) = HiS(Q) Ke = 68 x 107 IF 0.2 mol of hydrogen and 1.0 mol of sulphur are heated to 90°C in a 1.0 litre vessel, what will be the partial pressure of H.S at equilibrium? A sample of CaCOs(s) is introduced into sealed container of volume 0.821 litre & heated to 100K until equilibrium is reached, The equilibrium constant for the reaction CaCO,(s) = CaO(s) + CO%(g) is 4 x 10 atm at this temperature. Calculate the mass of CaO present at equilibrium, 20.0 grams of CaCO,(s) were placed In a closed vessel, heated & maintained at 727° C under equilibrium CaCOx(s) = CaO{s) + CO2(g) and it is found that 75% of CaCOs was decomposed What Is the value of K,? The volume of the container was 15 litres For the equilibrium CuSO, 5H,0(s) = CuSOs.3H,0(s) + 2H,0@) Kp = 2.25 x 10~* atm? and vapour pressure of water is 22.8 Torr at 298 K. CuSO. 5H,0(5) Is efflorescent (.e., loses water) when relative humidity Is Chemical EquilibriumLe-Chatelier’s principle 5 16. W 18, Using Le Chatelier's principle, predict the effect of () decreasing the temperature and (W) increasing the pressure on each of the following equilibria: (A) Ni@) + 3H) <= 2NHs(g) + Heat (8) NxG@) + O2{g) = 2NO() + Heat (C) H.0@) + Heat = Hs @) +O: () (0) 2CO () + O2 (g) = 2CO2 g) + Heat Suggest four ways in which the concentration of hydrazine, Not, could be increased in an equilibrium described by the equation No (g) + 2He (g) = NoHs (g) AH = 95 kJ {a) Methanol, a liquid fuel that could possibly replace gasoline, can be prepared from water gas and additional hydrogen at high temperature and pressure in the presence of a suitable catalyst, Write the expression for the equilibrium constant for the reversible reaction, 2H, (g) + CO (9) <2 CHLOH (g) AH = - 90.2 kJ {b) Assume that equilibrium has been established and predict how the concentration of H:, CO and CHsOH will differ at a new equilibrium if (1) more Hy Is added (2) CO is removed. (3) CHOH Is added. (4) the pressure on the system Is increased (5) the temperature of the system is increased. (6) more catalyst is added. Fram the following data () Ha(g) + COx(g)_ H;0(g) + COW) Kooooe (ii) 2H20(g)2H.(g) + O2(9) Koooox = 5.31 x 10-1 (il) 2CO() + 04(g) 2C0,(g) Kyoooe = 2.24 x 10? State whether the reaction (iil) is exothermic or endothermic? Simultaneous equilibria 19. 20. For given simultaneous reaction X(s) = A@) + Bs) + Cig) K, = 500 atm Y(s) = DG) + Aig) + Els) 2000 atm Find total pressure at equilibrium. When NO & NO; are mixed, the following equilibria readily obtained: 2NO, = NiO« Kp= 68 atm” NO + NO: <2 NiO» Ke=? In an experiment when NO & NO; are mixed in the ratio of 1: 2, the total final pressure was §.05 atm & the partial pressure of NO, was 1.7 atm. Calculate {a) the equilibrium partial pressure of NO. (b) Ke for NO + NO, N:Os Chemical Equilibrium 18716. "7. 18. 19. 20, 158 ANSWER KEY 250 Wn-2;()K-12M 2x10" Kp = 0.2463, Ke = 0.01M 2.4 mole The reaction is not an equilibrium because Q. > Ke. The reaction will proceed from right to left to reach equilibrium (a) 6.667 «10° mol L (b) n (N,0:)=0.374 mol: n (NO,)=0.052 mol : (c) 10.49 atm (a) 6.44% 0.97 atm 2Ig iit Kp= 2.5 atm, P= 1.5 atm oon atm 224mg 0.821 atm Less than 50 % () When decreasing temperature (@) Forward (©) Forward (©) Backward (d) Forward (li) Increasing the pressure (a) Forward (b) No change (©) Backward (d) Forward add Nz, add Ha, Increase the pressure, heat the reaction (a) K = [CH,OHI/[H2}°(CO} (b) 1. [Ha] increase, [CO] decrease, [CHOH] increase ; 2. [Ha] increase, [CO] decrease, [CH:OH] decrease ; 3. [H.] Increase, (CO) increase, [CH:OH] increase ; 4. (Hs increase, [CO] inerease,[CH,OH] increase ; 5. [Hz] increase, [CO] increase, [CHOH] decrease : 6. no change Exothermic. 100 fa) 1.08 atm, (b) 3.43 atm Chemical Equilibrium2 1 One mole of N,O; (g) at 300 K is left in a closed container under one atm . It is heated to 600 K when 20 % by mass of NsO, (g) decomposes to NO» (g) . Calculate resultant pressure. 2, Solid Ammonium carbamate dissociates as: NH; COONH, (s) <= 2NH\(g) + CO2(9) Ina closed vessel solid ammonium carbamate is in equiliirium with its dissociation products At equilibrium, ammonia is added such that the partial pressure of NH, at new equilibrium now equals the original total pressure. Calculate the ratio of total pressure at new equilibrium to ‘that of original total pressure. 3 For the reaction A(g) + 28g) = C(g) + Dig) : K. IF the initial moles of A,B,C and D are 0.5, 1, 0.8 and 3.5 moles respectively in a one litter vessel What is the equilibrium concentration of B? 4. Two solids X and Y dissociate into gaseous products at a certain temperature as follows: X(5) = A(@) + C(g), and ¥(s) = 8(g) + C(g). At a given temperature, pressure over excess solid X is 40 mm and total pressure over solid Y is 60 mm. When they are preset in separate containers. Calculate {@) the values of Kp for two reactions (in mm) (6) the ratio of moles of A and B in the vapour state over a mixture of X and Y (6) the total pressure of gases over a mixture of X and ¥ 5. When 1 mole of A(g) Is introduced in a closed rigid 1 litre vessel maintained at constant temperature the following equilibria are established Ag = Bg) + c@) K cg = D@ + Big) Ke, The pressure at equilibrium Is twice the initial pressure. Calculate the value of ANSWER KEY 1 (2.4 atm) (31/27) 3. (2* 104) 4. (2) 400mm?, 900mm*: (b) 4: 9; (c) 72.15 mm Hg 5. (a) Chemical Equilibrium 159160 Previous Year In reaction A + 2B =: 2C + ©, initial concentration of 8 was 15 times of |Al, but at equilibrium. the concentrations of A and B became equal. The equilibrium constant for the reaction Is: ‘JEE(Main)-2013] Or Qe @12 (4) 8 Ne(g) + 3H2(g) = 2NHA(Q), Ki a [JEE(Main)-2013] Ne(g) + O2(g) = 2NO(Q), Ke (2) 1 Ho(g) + 7 Ox(g) = H20(Q), Kr (3) The equation for the equilibrium constant of the reaction 2 NH(g) + 5 02(g) = 2NO (g) + 3H,0(9), ( in terms of Ky, Ke and K, is: KK KK, K 2 @ (4) KK: Ke 1 For the reaction SOng) + 7 Or <2 SOse:. If Kp = Ke (RT)" where the symbols have usual meaning hen the value of x is: (assuming ideality) £(Main)-2014] 1 ; 5 @)1 @- M-5 For the decomposition of the compound, represented as (JEE(Main)-2014] NH,COONHs(s) <= 2NH3(g) + CO,(g) the Ky=2.9 x 10-* atm’. IF the reaction is started with 1 mol of the compound, the total pressure at equilibrium would be (1) 38.8 «10 atm (2)1.94 «107 atm (3) 5.82 « 107 atm (4) 7.66 « 10 atm The equilibrium constants at 298 K for a reaction A + B = C + D Is 100. If the Initial concentration of all the four species were 1M each, then equilibrium concentration of D (in mol >) will be: JEE(Main)-2016] () 1182 (2) 0.182 (3) 0.818 (4) 1.818 Chemical Equilibrium10. ? Which of the following lines correctly show the temperature dependence of equilibrium constant, K, for an exothermic reaction 7 [JEE(Main)2018] Ink ( Aand oD (2) Aand B (3) Band c (4) C and D Consider the following reversible chemical reactions: [JEE(Main) Jan-2019] Ad(g) + 8{g) Kis 208(Q) ...(1) 6ABG@) Ks 3A.) + 38.) (2) (1) Kiky = 3 Ke= kr (3) Ke= Ke (4) Kiko = : The value of Ky/Ke for the following reactions at 300 K are, respectively {At 300 K, RT = 24,62 dm? atm mol") (JEE(Main) Jan-2019] Naig) + O29) = 2NO(g) N:04(g) <= 2NO.(9) Na(g) + 3H.(g) = 2NH,(g) (1) 24.62 dm* atm mol", 606.0 dm* atm? mol, 1.65 x 10 dmé atm? mol? (2)1, 4.1. 10% dem atm! mol, 606.0 dm’ atm? mol? (3) 1, 24.62 dm? atm mol, 606.0 dmé atm? mol? (4) 1, 24.62 dm? atm mol", 1.65 x 10% dmr* atm? mol? 5.1g NH4SH is introduced in 3.0 L evacuated flask at 327°C, 30% of the solid NH«SH decomposed to NH and HS as gases. The Ky of the reaction at 327°C Is (R= 0.082 Latm — mol~'K", Molar mass of S = 32 g mol/~!, molar mass of N = 14 g mol") (JEE(Main) Jan-2019] (1) 0.242 x 10-4 atm? (2) 0.242 atm? (1x10 atm (4) 4.9. 10-4 atm? Consider the reaction (JEE(Main) Jan-2019] N, (g) + 3H, (Q) == 2NH, (9) The equilibrium constant of the above reaction is Ks. If pure ammonia is left to dissociate, the partial pressure of ammonia at equilibrium is given by (Assume that Py,» @ ® pe Ki a e Chemical Equilibrium 16112. 13 “4 6 16. 162 ‘Two solids dissociate as follows (JEE(Main) Jan-2019] A(s) == Big) + Clg) : K,, =x atm? Dis) = Cig) + Ele); K, The total pressure when both the solids dissociate simultaneously is ery @) 2(feFyJam Ks atm (8) Yeryatm yal Ina chemical reaction, A + 28 Ss 2€ + D, the Initial concentration of 8 was 1.5 times of the concentration of A, but the equilibrium concentrations of A and B were found to be equal. The equilibrium constant(K) for the aforesaid chemical reaction is: [JEE(Main) Jan-2019] 1 OF Qa 16 yt For the following reactions, equilibrium constants are given: [JEE(Main) Apr.-2019] S(6) + Ox(g) = S0,(g): Ki = 10" 28(s) + 302(q) = 2S04(g): K2 = 10° The equilibrium constant for the reaction, 250,(g) + O.(g) = 250,(9) is: () 10 (2) 10 (10 (4) 10” For the reaction, 250,(g) + 0.(g) = 2S0,(g), AH = -57.2 kJ mol" and Ke = 1.7x10'* Which of the following statement is INCORRECT 7 [JEE(Main) Apr.-2019] (1) The equilibrium constant Is large suggestive of reaction going to completion and so no catalyst is required (2) The addition of inert gas at constant volume will not affect the equilibrium constant (3) The equilibrium constant decreases as the temperature increases. (4) The equilibrium will shift in forward direction as the pressure increases In which one of the following equilibria, Ke #* Ke 7 [JEE(Main) Apr.-2019] (1) 2HI(g) = H2(g) + 12(9) (2) NO.(g) + SO,(g) = NO(g) + SOs(g) (3) 2C{s) + On(g) = 2C0() (4) 2NO@) = Nol) + O2(9) Reactant A represented by square is in equilibrium with product 8 represented by circles. Then value of equilibrium constant Is [JEE(Main)-2020] ot (2 3 (aya Chemical EquilibriumW 18, 9. 20, 2 ? 8. the variation of the rate of the forward (@) and reverse (b) For the equilibrium a; reaction with time is given by (JEE(Main)-2020] of<— al] 3 7 é ——S Time @ : | 4) | : Consider the following reaction: N,O.(g) = 2NOA(q): AH? = +58K) For each of the following cases (a, b). the direction In which the equilibrium shifts Is {@) Temperature is decreased [JEE(Main)-2020] (b) Pressure is increased by adding N 2 at constant T (1) (@) Towards product, (b) towards reactant (2) (@) Towards reactant, (b) no change (3) (@) Towards reactant, (b) towards product (4) (@) Towards product, (b) no change For a reaction X + ¥ = 22, 1.0 mol of X, 15 mol of ¥ and 0.5mol of Z were taken in a1 L vessel and allowed to react. At equilibrium, the concentration of Z was 1.0 mol L-!. The equlliorium x constant of the reaction Is ==. The value of x is [JEE(Main)-2020] For the reaction Fe:N(s) + 3 H.(g) === 2Fe(s) + NHs(g) [JEE(Main)-2020] (1) Ke=Ke(RT! (2) Ke = KelT) (3) Ko= Kp (RTP? (4) Ke (RT? The value of Ke Is 64 at 800K for the reaction No @) + Sib @) == 2NIb @) ‘The value of Ke for the following reaction is, NHs(g) —=4n(@ + FH [JEE(Main)-2020] ot ad oe mt 3 a 7 Chemical Equilibrium 16322 23 24 25 26. 164 In the Figure shown below reactant A (represented by square) is an equilibrium with product B oO oO oo) oO oo oO ° oO Oo oO oo (represented by circle). The equilibrium constant is: (JEE(Main)-2020] (4 (2)8 Ql (4) 2 The stepwise formation of [Cu(NHs]* is given below: JEE(Main)-2021] Cu + Ny = [Cu(NHs)al* [CU(NHy)2}" + NHy = [Cu(NHL)aI [Cu(NH)." + NH; == [Cu(NH) [CUu(NH)sP' + NH === [Cu(NHs)«)* The value of stability constants ki, Ko, Ky and Ks are 10*, 1.58 x 102, 5 x 10? and 10? respectively. The overall equilibrium constants for dissociation of [Cu(NH))sI Is x * 10%, The value of x is (Rounded off to the nearest integer) At 1990 K and 1 atm pressure, there are equal number of Cl, molecules and Cl atoms in the reaction mixture. The value of Ke for the reaction Cla = 2Cliy under the above conditions is x 10°. The value of is (Rounded off to the nearest integer) ‘JEE(Main)-2021] ‘A homogeneous ideal gaseous reaction AByjy === A(g) ~ 2Bq is carried out in a 25 litre flask at 27°C, The initial amount of AB, was 1 mole and the equilibrium pressure was 1.9 atm. The value of Ke is x x 10°. The value of is _[R = 0.08206dm’ K" mo!"] JEE (Main) -2021] Consider the reaction N,O,(g) === 2NO,(g). The temperature at which Ke = 20.4 and Ks = 600.1, is____K. (Round off the Nearest Integer) [Assume all gases are Ideal and R = 0.0831 L bar K"' and mol] (JEE(Main)-2021] Chemical Equilibrium21 28. 29. 30, 6 22. 29. ? For a reaction at equilibrium 1 A@ = BO) + 7c) the relation between dissociation constant (K), degree of dissociation (a) and equilibrium pressure (p) is given by: (JEE(Main)-2022) a a (ap)? (a) x=? (-36f 0-0) (1+a)0-4 PCk dissociates as PCI (g) = PCs (Q) + Ch @) 5 moles of PCI; are placed in a 200 litre vessel which contains 2 moles of Nz and is maintained at 600 K. The equilibrium pressure Is 2.46 atm. The equilibrium constant Kp for the dissociation of PCleis__« 10. (nearest Integer) (Given: R = 0.082 L atm K~ mol: Assume ideal gas behaviour) [JEE(Main)-2022] 2NOCIIg) = 2NO(g) + Cla(Q) In an experiment, 2.0 moles of NOCI was placed in a one-Iitre flask and the concentration of NO after equilibrium established, was found to be 0.4 mol/L. The equilibrium constant at 30°C is___x 10 [JEE(Main)-2022] 4.0 moles of argon and 5.0 moles of PCls are introduced into an evacuated flask of 100 litre capacity at 610 K. The system Is allowed to equilibrate. At equilibrium, the total pressure of mixture was found to be 6.0 atm. The Ke for the reaction Is [Given R= 0.0821 atm K-! mol- [JEE(Main)-2022] () 2.25 (2) 6.24 (3) 1213 (4) 15.24 ANSWER KEY oO 2 @ 8 MW 4 @ 5 @ 6 BM 7 ®™ 8 @ 02 @@ W @ 2 @ 2B M 4 WH GB) 16 (217. 2) 18. 2) 19 16 20. i) 2 a) (23) 2.) 25,74) 26, (BBA) 27.) 28. (107) (125) 30. (i) Chemical Equilibrium 165166 eaeresicy The degree of dissociation is 0.4 at 400K & 10 atm for the gaseous reaction PC. = PCI + Ch(g). Assuming ideal behaviour of all gases. Calculate the density of equilibrium mixture at 400K & 1.0 atm pressure [EE-1999] For the reversible reaction: (JEE-2000) Na(Q) + 3H4(9) 2 2NH(@) at 500°C. The value of Ke is 1.44 x 10-5, when partial pressure is measured in tmospheres. The corresponding value of Ke with concentration in mol L-* is: (A) 1.44 « 10° (0.082 x 500)" (8) 1.44 x 10 (8.314 x 773)? (C) 1.44 « 10° (0.082 x 500)* (0) 1.44 x 10 (0.082 « 773) When two reactants A and B are mixed to give products C and D, the reaction quotient 0, at the initial stages of the reaction: 2000} (A) is zero (8) decrease with time (C) independent of time (0) increases with time When 3.069 of solid NHsHS is introduced into a two litre evacuated flask at 27°C, 30% of the solid decomposes into gaseous ammonia and hydrogen sulphide. LEE 2000} (i) Calculate Ke & Ke for the reaction at 27°C. (i) What would happen to the equilibrium when more solid NH«HS is introduced into the flask? At constant temperature, the equilibrium constant (Ke) for the decomposition reaction. N,Os<2 2NO; is expressed by Ke = 4x°P/(I — x°) where P is pressure, x is extent of decomposition Which of the following statement is true 7 [JEE-2001] (A) kr increases with increase of P (8) Ke increases with increase of x (C) Ke increases with decrease of x (0) Ke remains constant with change in P or x Consider the following equilibrium in a closed container: N,O.(g) = 2NOx(9) At a fixed temperature, the volume of the reaction container is halved. For this change, which of the following statements holds true regarding the equilibrium constant (K:) and degree of dissociation (a): (JEE-2002] (A) Nelther Ks nor a changes (8) Both K> and a change (C) Ke changes, but a does not change (0) Ke does not change, but a changes IF Ag’ + NHs = [Ag(NHS)}* > Ki = 1.6 x 10! and JEE-2006] TAg(NH:)]° + NHy == [Ag(NHi}a} ; Ko = 6.8 x 107 The formation constant of [Ag(NHs)2]° is: (A) 6.08 x 10% (8) 68x 10% (16 «10° (0) 1.088 « 107 Chemical Equilibrium? The thermal dissociation equilibrium of CaCO%(s) is studied under different conditions. CaCOs(s) = Cad(s) + COXg) [JEE-2013] For this equilibrium, the correct statement(s) istare) (A) AH Is dependent on T {B) K Is independent of the initial amount of CacOs {C) Kis dependent on the pressure of CO, at a given T {D) AH Is independent of the catalyst, if any For the following reaction, the equilibrium constant K- at 298 K Is 1.6 x 10", Fe (aq) +S*(aq) ——= Fe (5) [EE (Advanced) -2019] When caual volumes of 0.06 M Fe* (aq) and 0.2 M $* (aq) solutions are mixed, the equilibrium concentration of Fe“(aq) Is found to be Y « 10-7 M. The value of Y is, ANSWER KEY 454gdm" 2 © 3 o (i) Ke= 8.1 x 10° mol L?; Ke = 4.91 x 10°? atm? (li) No effect 0) 6 O) 7 0 (ABD) 9. 8.92 or 893 Chemical Equilibrium 167