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Chemical Equlibrium Contents Topic Page No Theory 01 - 05 Exercise - 06 - 20 Exercise - 20 - 34 Exercise - 35 - 38 Exercise - 38 - 41 Answer Key 42 - 43 Syllabus Chemical Equlibrium : Types of Equlibrium, Relation between Kp and Kc, Homogeneous and Hetrogeneous Equlibrium, Relation Between Degree of Dissociation and Vapour Density, Reaction Quotient, Factors Affecting Equlibrium Constant, Le-Chateliers Principle, Simultaneous Equlibrium Name : Contact No ETOOS ACADEMY Pvt Ltd F-106, Road No.2 Indraprastha Industrial Area, End of Evergreen Motor, BSNL Lane, Jhalawar Road, Kota, Rajasthan (324005) Tel : +91-744-242-5022, 92-14-233303 CHEMICAL EQUILIBRIUM Types of chemical reactions Types of chemical reactions Irreversible reaction Reversible reaction The reaction which proceeds in one direction (forward The reaction which proceed in both the direction under the same direction) only set of experimental conditions Reactants are almost completely converted into products Reactants form products and products also react to form reactants Products not react to form reactants again in backward direction These are possible in closed vessels (a) Do not attain equilibrium state Such reactions are represented by single arrow {} Examples – Precipitation reactions e.g NaCl(aq) + AgNO3(aq) NaNO3(aq) + AgCl (b) Neutralization reactions e.g HCl(aq) + NaOH(aq) NaCl(aq) + H2O 2KCl(s) + 3O2(g) (c) 2KClO3 (s) (a) Attain the equilibrium state and never go to completion ) ( Represented by double arrow ( ) or Examples :– Homogeneous reactions- only one phase is present (i) Gaseous phase– H2(g) + I2(g) 2HI(g) 2NO(g) [Birkland eyde process (HNO3)] N2(g) + O2(g) 2NH3(g) (Haber’s process) N2(g) + 3H2(g) (ii) Liquid phase (d) Reactions in open vessel :– CH3COOC2H5(l)+ H2O(l) CH3 COOH(l) + C2H5OH(l) Even a reversible reaction will become irreversible if it is Heterogeneous reactions– More than one phases are present carried out in open vessel Ex CaCO3(s) CaO(s) + CO2(g) CaCO3(s) CaO(s) + CO2(g) Open NH3(g) + H2S(g) NH4HS(s) Closed NH3(g) + H2S(g) (b) NH4HS(s) vessel vessel STATE OF CHEMICAL EQUILIBRIUM : At equilibrium : (i) Rate of forward reaction (rf) = rate of backward reaction (rb) (dynamic nature) (ii) All measurable parameters become constant with respect to time ETOOS ACADEMY Pvt Ltd F-106, Road No.2 Indraprastha Industrial Area, End of Evergreen Motor, BSNL Lane, Jhalawar Road, Kota, Rajasthan (324005) Tel : +91-744-242-5022, 92-14-233303 CHEMICAL EQUILIBRIUM_ADVANCED# Types of equilibria on the basis of process Types of equilibria on basis of physical state Homogeneous equilibrium When all reactants and products are in same phase H2(g) + Cl2(g) 2HCl(g) Heterogeneous equilibrium When more than one phase are present 3Fe(s) + 4H2O(g) Fe3O4(s) + 4H2(g) SO2(g) + NO2(g) 2Na2O2(s) + 2H2O() SO3(g) + NO(g) 4NaOH + O2(g) CHARACTERISTICS OF CHEMICAL EQUILIBRIUM : The nature and the properties of the equilibrium state are the same regardless of the direction from which it is achieved It can be achieved in both directions Equilibrium is dynamic in nature It means that at microscopic level reaction has not stopped It appears that no change is occuring but But both the opposing reactions are proceeding at the same rate So there is no net change.Thus equilibrium is not static in nature A catalyst can alter the rate of approach of equilibrium but does not change the state of equilibrium By using catalyst, the equilibrium can be achieved in different (more/less) time, but the relative concentrations of reactants and products are same irrespective of the presence or absence of a catalyst Equilibrium can be observed by constancy of some observable properties like colour, pressure, concentration, density, temperature, refractive index etc.which may be suitable in a given reaction At equilibrium, free energy changeG = Equilibrium state can be affected by altering factors like pressure, volume, concentration and temperature etc.(Le chateliers Principle) System moves toward an equilibrium state spontaneously even if it is disturbed It will return to original state LAW OF MASS ACTION : To illustrate the law of mass action, consider the following general reaction at constant temperature, aA(g) + bB(g) cC(g) + dD(g) or Rf = k f [A]a[B]b …(i) Where kf = rate constant of forward reaction or Rr = kr[C]c [D]d …(ii) where kr = rate constant of reverse reaction ETOOS ACADEMY Pvt Ltd F-106, Road No.2 Indraprastha Industrial Area, End of Evergreen Motor, BSNL Lane, Jhalawar Road, Kota, Rajasthan (324005) Tel : +91-744-242-5022, 92-14-233303 CHEMICAL EQUILIBRIUM_ADVANCED# At equilibrium, Rate of forward reaction = Rate of reverse reaction i.e., Rf = Rr So, from equations (i) and (ii), we get kf[A]a [B]b = k r[C]c[D]d or, k f [C]c [D]d k r [A]a [B]b or kf [C]c [D]d Kc kr [A]a [B]b Where, Kc is the equilibrium constant in terms of molar concentration Equilibrium Constant (Kp) in terms of Partial Pressures: Kp PCc PDd PAa PBb …(2) RELATIONSHIP BETWEEN KP AND KC : K p K c (RT) n Where, n = (c + d) – (a + b) Applications of Equilibrium constant : Predicting the direction of the reaction Reaction Quotient (Q) At each point in a reaction, we can write a ratio of concentration terms having the same form as the equilibrium constant expression This ratio is called the reaction quotient denoted by symbol Q It helps in predicting the direction of a reaction [C]c [D] d The expression Q = [ A ]a [B]b at any time during reaction is called reaction quotient.The concentrations [C], [D] , [A], [B] are not necessarily at equilibrium Predicting the extent of the reaction : [Pr oduct ] eq K = [Re ac tan t ] eq Case-I If K is large (K > 103) then product concentration is very very larger than the reactant ([Product] >>[Reactant]) Hence concentration of reactant can be neglected with respect to the product In this case, the reaction is product favourable and equilibrium will be more in forward direction than in backward direction Case-II If K is very small (K < 10–3) [Product] T1 then K2 > K1 provided H = +ve (endothermic reaction) K2 < K1 if H = –ve (exothermic reaction) In the above equation, the unit of R and H/T should be same Relation between equilibrium constant & standard free energy change Gº = – 2.303 RT log K Where G° = standard free energy change T = Absolute temperature, R = universal gas constant Homogeneous liquid system : Formation of ethyl acetate : The reaction between alcohol and acid to form ester is an example of homogeneous equilibrium in liquid system CH3COOH(l ) + C2H5OH(l ) CH3COOC2H5(l ) + H2O(l ) [CH3 COOC 2H5 ][H2O] KC = [CH COOH][C H OH] Observed molecular weight and Observed Vapour Density of the mixture molecular weight of A n (g) Observed molecular weight of An(g) = total no of moles at equilibriu m Mobs = a.Mth a(1 (n 1) ) = Mth [1 (n 1)] where Mth = theoritical molecular weight (n = atomicity) Mmixture = MA n [1 (n 1)] , MA n = Molar mass of gas An Vapour density (V.D) : Density of the gas divided by density of hydrogen under same temp & pressure is called vapour density MA n D = vapour density without dissociation = Mmix d = vapour density of mixture = observed v.d = D = + (n – 1) d M Mo Dd T (n 1) d (n 1)M0 where MT = Theoritical molecular wt M0 = observed molecular wt or molecular wt of the mixture at equilibrium EXTERNAL FACTORS AFFECTING EQUILIBRIUM : LE-CHATELIER’S PRINCIPLE When a system (reaction) in equilibrium is subjected to change in any of the factors that determine the equilibrium condition, the equilibrium shifts in such a manner so as to reduce or counteract the effect of that change Consider the following reaction in equilibrium 2NH3(g) N2(g) + 3H2(g) ETOOS ACADEMY Pvt Ltd F-106, Road No.2 Indraprastha Industrial Area, End of Evergreen Motor, BSNL Lane, Jhalawar Road, Kota, Rajasthan (324005) Tel : +91-744-242-5022, 92-14-233303 CHEMICAL EQUILIBRIUM_ADVANCED# (I) Effect of change of concentration : Increase in concentration of reactant propel the reaction in forward direction and increase in concentration of product propel the reaction in backward direction Effect of change of pressure : (a) When n = ; no effect (II) (III) (IV) (1) (a) (b) When n ; If pressure increases the reaction shifts in the direction of lesser number of moles If pressure decreases the reaction shifts in the direction of greater number of moles Addition of inert gas : (a) At constant volume – no effect (b) At constant pressure – When n = ; no effect When n ;Reaction shifts in the direction of greater number of moles Effect of temperature : (a) If H = +ve (endothermic) an increase in temperature, shifts the reaction in forward direction and vice– versa (b) If H = –ve (exothermic) an increase in temperature, shifts the reaction in backward direction and vice– versa Application of Le-Chatelier Principle to Physical Equilibrium : Effect of Pressure on Melting Points : When volume of solid decrease on melting i.e., in liquid state volume is lesser, then by increasing pressure on such solids the melting point will go down, because at high pressure melting is facilitated eg Ice, diamond, Carborundum (SiC) etc Solid (b) (2) Liquid When volume increase on melting then by increasing pressure, the above equilibrium will shift in the reverse direction, hence process of melting will be lowered and more heat is required, thus melting point will rise eq Iron, Copper, NH4Cl, NaCl etc Effect of Pressure on Boiling point : When pressure is raised then condensation of vapour into liquid take place, thus vapour pressure will decrease Now more heat is required to equate vapour pressure with atmospheric pressure, hence boiling point will increase FREE ENERGY AND CHEMICAL EQUILIBRIUM : Standard Free Energy and Equilibrium Constant : The change in free energy for a reaction taking place between gaseous reactants and products represented by the general equation aA + bB cC + dD According to Van’t Hoff reaction isotherm G G RT ln pcC pdD = G0 + RTlnQp p aA p Bb the condition for a system to be at equilibrium is that G = Thus at equilibrium pc pdD G RT ln Ca G RT ln K 0p b pA pB Whence G0 = – RTlnK0p Hence ln K p G RT ETOOS ACADEMY Pvt Ltd F-106, Road No.2 Indraprastha Industrial Area, End of Evergreen Motor, BSNL Lane, Jhalawar Road, Kota, Rajasthan (324005) Tel : +91-744-242-5022, 92-14-233303 CHEMICAL EQUILIBRIUM_ADVANCED# PART - I : OBJECTIVE QUESTIONS * Marked Questions are having more than one correct option Section (A) : Equilibrium, Equilibrium constant ,law of mass action, Reaction Quotient A-1 A chemical reaction, A B, is said to be in equilibrium when : (A) rate of forward reaction is equal to rate of backward reaction (B) conversion of A to B is only 50% complete (C) complete conversion of A to B has taken place (D) only 25% conversion of A to B taken place A-2 The equilibrium concentration of x, y and yx are 4, and respectively for the equilibrium 2x + y yx2 The value of equilibrium constant, KC is (A) 0.625 (B) 6.25 (C) 0.0625 (D) 62.5 A-3 mole of A are mixed with mole of B and mole of C are formed at equilibrium, according to the reaction, A+B C+D the equilibrium constant is : (A) (B) (C) (D) A-4 For the following reaction at 250°C, PCl3(g) + Cl2(g) PCl5(g) the value of KC is 26 then the value of Kp at same temperature will be (A) 0.57 (B) 0.61 (C) 0.83 (D) 0.91 A-5 For a reversible reaction, the rate constants for the forward and backward reactions are 2.38 ×10–4 and 8.15 × 10–5 respectively The equilibrium constant for the reaction is – (A) 0.342 (B) 2.92 (C) 0.292 (D) 3.42 A-6 If different quantities of ethanol and acetic acid were used in the following reversible reaction, CH3COOH() + C2H5OH() CH3COOC2H5() + H2O() the equilibrium constant will have values which will be ? (A) different in all cases (B) same in all cases (C) higher in cases where higher concentration of ethanol is used (D) higher in case where higher concentration of acetic acid is used A-7 Chemical equilibrium is dynamic in nature because – (A) The equilibrium in maintained quickly (B) Conc of reactants and products become same at equilibrium (C) Conc of reactants and products are constant but different (D) Both forward and backward reactions occur at all times with same speed A-8 Which of the following statements is false in case of equilibrium state – (A) There is no apparent change in properties with time (B) It is dynamic in nature (C) It can be attained from either side of the reaction (D) It can be attained from the side of the reactants only A-9 Starting with the reactants , At any moment before a reversible reaction attains equilibrium it is found that – (A) The rate of the forward reaction is increasing and that of backward reaction is decreasing (B) The rate of the forward reaction is decreasing and that of backward reaction is increasing (C) The rate of both forward and backward reactions is increasing (D) The rate of both forward and backward reactions is decreasing ETOOS ACADEMY Pvt Ltd F-106, Road No.2 Indraprastha Industrial Area, End of Evergreen Motor, BSNL Lane, Jhalawar Road, Kota, Rajasthan (324005) Tel : +91-744-242-5022, 92-14-233303 CHEMICAL EQUILIBRIUM_ADVANCED# A-10 A chemical reaction A B is said to be in equilibrium when (A) Complete conversion of A to B has taken place (B) Conversion of A to B is only 50% complete (C) Only 10% conversion of A to B has taken place (D) The rate of transformation of A to B is just equal to rate of transformation of B to A in the system A-11 In a chemical reaction Kp is greater than Kc when (A) the number of molecules entering into the chemical reaction is more than the number of molecules produced (B) the number of molecules entering into the chemical reaction is the same as the number of molecules produced (C) the number of molecules entering into the chemical reaction is less than the number of molecules produced (D) the total number of moles of reactants is less than the number of moles of products A-12 Which of the following statements is not correct about the equilibrium constant ? (A) Its value does not depend upon the initial conc of the reactants (B) Its value does not depend upon the initial conc of the products (C) Its value does not depend upon temperature (D) Its value does not depend upon presence of catalyst A-13 In a chemical equilibrium, the equilibrium constant is found to be 2.5 If the rate constant of backward reaction is 3.2 × 10–2, the rate constant of forward reaction is (A) 8.0 × 10–2 (B) 4.0 × 10–2 (C) 3.5 × 10–2 (D) 7.6 × 10–3 A-14 K1 and K2 are the rate constants of forward and backward reactions The equilibrium constant K of the reaction is (A) K1 × K2 A-15 (B) K1 – K2 (B) 40 (D) K1 K K1 – K 2HI(g) is 50 What is the value of KC (C) 50 (D) 70 In which of the following reaction, the value of KP will be equal to KC – (A) N2(g) + O2(g) (C) 2NH3 (g) A-17 K1 K2 The value of KP for the reaction H2(g) + I2(g) (A) 30 A-16 (C) 2NO (g) (B) PCl5 (g) N2(g) + 3H2(g) PCl3 (g) + Cl2(g) (D) 2SO2 (g) + O2(g) 2SO3 (g) Select the correct expression regarding the relation between KP and KC for the reaction aX(g) + bY(g) (A) KP = KC(RT)a+b bZ(g) + aW (g) (B) KP = KC (a b) (C) KP = KC RT (D) KP = KC A-18 The equilibrium constant KC for the decomposition of PCl5 is 0.625 mole / lit at 300ºC Then the value of KP is (A) 2.936 atm (B) 0.0625 atm (C) 6.25 atm (D) 0.00625 atm A-19 The reaction A(g) + B(g) C(g) + D(g) proceeds to right hand side upto 99.9% when starting bwith equal moles of A and B The equilibrium constant K for the reaction will be (A) 104 (B) 105 (C) 106 (D) 108 A-20 For the reaction, 2NO2 (g) 2NO (g) + O2(g), KC = 1.8 × 10–6 at 185ºC At 185ºC, the value of KC for the reaction NO(g) + O (g) NO2(g) is 2 (A) 0.9 × 10 (B) 7.5 × 102 (C) 1.95 × 10–3 (D) 1.95 × 103 ETOOS ACADEMY Pvt Ltd F-106, Road No.2 Indraprastha Industrial Area, End of Evergreen Motor, BSNL Lane, Jhalawar Road, Kota, Rajasthan (324005) Tel : +91-744-242-5022, 92-14-233303 CHEMICAL EQUILIBRIUM_ADVANCED# A-21 If in the reaction N2O4(g) 2NO2(g), is the part of N2O4 which dissociates per mole, then the number of moles at equilibrium will be (A) (B) (C) (1–)2 (D) (1 + ) A-22 The equilibrium concentration of X, Y and YX2 are 4, and moles respectively for the equilibrium 2X(g) + Y(g) YX2(g).The value of KC is (A) 0.625 (B) 0.0625 (C) 6.25 (D) 0.00625 A-23 At 444º C, the equilibrium constant K for the reaction 2AB(g) A2(g) + B2(g) is degree of dissociation of AB will be (A) 10% (B) 20 % (D) 50% A-24 For the reaction A(g) + B(g) (A) A-25 A-26 A-27 K (B) 2P For N2(g) + 3H2(g) (A) KP = KC 2NH3(g) (B) KP = KC RT (D) K 1 K–1 (C) P (D) (C) KP = KC (R T)–2 P (D) KP = KC (R T)–1 In the reaction C2H6(g), the equilibrium constant can be expressed in units of (B) mol2 litre–2 (C) litre mol–1 (D) mol litre–1 Equilibrium concentration of HI, I2 and H2 is 0.7, 0.1 and 0.1 moles/litre Calculate the equilibrium constant for the reaction : 2HI(g) – (B) 36 (C) 49 The equilibrium constant for the reaction Zn (s)+CO2(g) (A) A-30 (C) NO(g) + NO2(g) ; total pressure = P, degree of dissociation = 50% (A) 3P I2(g) + H2(g) (A) 0.36 A-29 K +1 For the reaction : N2O3(g) Then Kp would be – C2H4(g) + H2(g) (A) litre–1 mol–1 A-28 C(g) + D(g), the degree of dissociation would be – (B) K 1 (C) 30% The 64 Pco Pco2 (B) [ ZnO] [ Zn] (C) (D) 0.49 ZnO (s) + CO (g) is - PZnO PCO (D) PZnPCO2 PZ nPCO PZnO PCO For the reaction C (s) + CO2 (g) 2CO (g) the partial pressure of CO2 and CO are 2.0 and 4.0 atm respectively at equilibrium The Kp for the reaction is – (A) 0.5 (B) 4.0 (C) 8.0 (D) 32.0 A-31 A-32 The unit of equilibrium constant for the reaction H2 + I2 (A) Mole–1 litre (B) Mole–2 litre (C) Mole litre 2HI is – (D) None KP for the gaseous reaction – KC (a) A + B 2C (b) A 4B (c) A + B + 2C would be respectively (A) (RT)–3 , (RT)2, (RT)º (C) (RT)–3 , (RT)2, (RT) 4D (B) (RT)–3 , (RT)–2, (RT)–1 (D) None of the above ETOOS ACADEMY Pvt Ltd F-106, Road No.2 Indraprastha Industrial Area, End of Evergreen Motor, BSNL Lane, Jhalawar Road, Kota, Rajasthan (324005) Tel : +91-744-242-5022, 92-14-233303 CHEMICAL EQUILIBRIUM_ADVANCED# A-33 In Which of the following equilibria, the value of KP is less than KC (A) H2 + I2 2Hl (C) N2 + O2 A-34 (B) N2 + 3H2 2NO 2NH3 (D) CO + H2O In a reversible reaction A CO2 + H2 B, the initial concentration of A and B are a and b in moles per litre and the equilibrium concentrations are ( a x) and (b + x) respectively; express x in terms of k1, k2, a and b (A) A-35 k 1a k 2b k k (B) k 1a k 2b k1 k (C) k 1a k 2b k1k The equilibrium constant of the reaction SO2(g) + ½O2(g) (D) k 1a k 2b k 1k SO3(g) is × 10–3 atm–1/2 The equilibrium constant of the reaction 2SO3(g) 2SO2(g) + O2(g) would be : (A) 250 atm (B) × 10 atm (C) 0.25 × 104 atm (D) 6.25 × 104 atm Section (B) : Degree of Dissociation, Vapour Density , Average Molar Mass B-1 For which of the following reactions, the degree of dissociation cannot be calculated from the vapour density data I 2H(g) H2(g) + I2(g) II 2NH3(g) N2(g) + 3H2(g) III 2NO N + O IV PCl PCl + Cl (g) (A) I and III 2(g) 2(g) 5(g) (B) II and IV 3(g) (C) I and II 2(g) (D) III and IV B-2 The degree of dissociation of SO3 is at equilibrium pressure P0 Kp for 2SO3(g) 2SO2(g) + O2(g) is 3 (A) [(P0 )/2(1 – ) ] (B) [(P03)/(2+)(1 – )2] (C) [(P02)/2(1 – )2] (D) None of these B-3 In a container equilibrium N2O4 (g) 2NO2 (g) is attained at 25°C The total equilibrium pressure in container is 380 torr If equilibrium constant of above equilibrium is 0.667 atm, then degree of dissociation of N2O4 at this temperature will be : (A) B-4 (B) (C) (D) D d In the dissociation of N2O4 into NO2, (1 + ) values with the vapour densities ratio is as given by : [-degree of dissociation, D-vapour density before dissociation, d-vapour density after dissociation] (A) B-5 (B) In the above question, varies with (A) (B) (C) (D) D according to : d (C) ETOOS ACADEMY Pvt Ltd F-106, Road No.2 Indraprastha Industrial Area, End of Evergreen Motor, BSNL Lane, Jhalawar Road, Kota, Rajasthan (324005) Tel : +91-744-242-5022, 92-14-233303 (D) CHEMICAL EQUILIBRIUM_ADVANCED# 81 An exothermic reaction is represented by the graph : (A) (B) (C) (D) Multiple Choice Correct : 82 In which of the following reactions is Kp < Kc ? (A) CO(g) + Cl2(g) (C) 2BrCl(g) 83 COCl2(g) Cl2(g) + Br2(g) (B) CO(g) + 3H2(g) (D) I2(g) CH4(g) + H2O(g) 2I(g) The dissociation of phosgene, which occurs according to the reaction COCl2(g) CO(g) + Cl2(g) is an endothermic process Which of the following will increase the degree of dissociation of COCl2? (A) Adding Cl2 to the system (B) Adding helium to the system at constant pressure (C) Decreasing the temperature of the system (D) Reducing the total temperature of the system 84 The equilibrium of which of the following reactions will not be disturbed by the addition of an inert gas at constant volume ? (A) H2(g) + I2(g) (C) CO2(g) + 2H2(g) 2HI(g) CH3OH(g) (B) N2O4(g) 2NO2(g) (D) C(s) + H2O(g) CO(g) + H2(g) 85 A box contains CO(g), Cl2(g) and COCl2(g) in equilibrium at 1000 K The removal of CO(g) will (A) decrease the concentration of COCl2 (B) increase the concentration of Cl2 (C) increase the concentration of COCl2 (D) reduce the concentration of CO as well as Cl2 86 An industrial fuel, ‘water gas’, which consists of a mixture of H2 and CO can be made by passing steam over red-hot carbon The reaction is C(s) + H2O(g) CO(g) + H2(g) , H = +131 kJ The yield of CO and H2 at equilibrium would be shifted to the product side by (A) raising the relative pressure of the steam (B) adding hot carbon (C) raising the temperature (D) reducing the volume of the system 87 For the equilibrium 2SO2(g) + O2(g) 2SO3(g), H = -198 kJ, the equilibrium concentration of SO3 will be affected by (A) doubling the volume of the reaction vessel (B) increasing the temperature at constant volume (C) adding more oxygen to the reaction vessel (D) adding helium to the reaction vessel at constant volume 88 The following reaction attains equilibrium at high temperature N2(g) + 2H2O(g) + heat 2NO(g) + 2H2(g) The yield of NO is affected by (A) increasing the nitrogen concentration (B) decreasing the hydrogen concentration (C) compressing the reaction mixture (D) none of these 89 N2(g) + 3H2(g) 2NH3 + heat In this reaction, the direction of equilibrium will be shifted to the right by (A) increasing the concentration of nitrogen (B) compressing the reaciton mixture (C) removing the catalyst (D) decreasing the concentration of ammonia N2(g) + 3H2(g) 2NH3 + heat 90 The dissociation of ammonia carbonate may be represented by the equation NH4CO2NH2(s) 2NH3(g) + CO2(g) The equilibrium will shift from right to left if there is (A) a decrease in pressure (B) an increase in temperature (C) an increase in the concentration of ammonia (D) an increase in the concentration of carbon dioxide ETOOS ACADEMY Pvt Ltd F-106, Road No.2 Indraprastha Industrial Area, End of Evergreen Motor, BSNL Lane, Jhalawar Road, Kota, Rajasthan (324005) Tel : +91-744-242-5022, 92-14-233303 CHEMICAL EQUILIBRIUM_ADVANCED# 29 91 Which of the following statements about the reaction quotient Q are correct? (A) the reaction quotient, Q and the equilibrium constant always have the same numerical value (B) Q may be >, < , = Keq (C) Q(numerical value) varies as reaction proceeds (D) Q = at equilibrium 92 Which of the following factors will increase solubility of NH3(g) in H2O? NH3(g) + H2O(aq) (A) increase in pressure (C) liquefaction of NH3 93 94 NH4OH(aq) (B) addition of water (D) decrease in pressure Which of the following factors will affect solubility of CaO in H2O? (A) pressure (B) temperature (C) addition of water (D) volume For the following equilibrium NH4HS(s) NH3(g) + H2S(g) partial pressure of NH3 will increase (A) if NH3 is added after equilibrium is established (B) if H2S is added after equilibrium is established (C) temperature is increased (D) volume of the flask is decreased 95 For the gas phase reaction carried out in a vessel, C2 H6 C2 H4 + H2 the equilibrium concentration of C2H4 can be increased by (A) increasing the temperature (B) decreasing the pressure (C) removing some H2 (D) adding some C2H6 96 For the reaction 2A(g) + B(g) 2C(g) ; H = +13.6 kJ which of the following will increase the extent of the reaction at equilibrium (A) increasing the temperature (B) increasing the pressure (C) addition of catalyst (D) removing C 97 The reactions in which the yield of the products cannot be increased by the application of high pressure (A) 2SO2(g) + O2(g) 2SO3(g) (B) NH4HS(s) NH3(g) + H2S(g) (C) N2O4(g) 98 2NO2(g) (D) N2(g) + 3H2(g) (i) N2(g) + O2(g) 2NO(g), (ii) N (g) + O (g) 2 2 2 (iii) 2NO(g) (iv) NO(g) 2NH3(g) K1 NO(g) ; K2 N2(g) + O2(g) ; K3 1 1 N (g) + O (g) ; 2 2 K4 Correct relation between K1, K2 and K4 is/are : (A) K1 × K3 = 99 (B) K1 K = (C) K3 K2 = (D) None When NaNO3 is heated in a closed vessel, oxygen is liberated and NaNO2 is left behind At equilibrium, NaNO3 (s) NaNO2 (s) + O (g) 2 (A) addition of NaNO2 favours reverse reaction (B) addition of NaNO3 favours forwards reaction (C) increasing temperature favours forward reaction (D) increasing pressure favours reverse reaction ETOOS ACADEMY Pvt Ltd F-106, Road No.2 Indraprastha Industrial Area, End of Evergreen Motor, BSNL Lane, Jhalawar Road, Kota, Rajasthan (324005) Tel : +91-744-242-5022, 92-14-233303 CHEMICAL EQUILIBRIUM_ADVANCED# 30 100 The dissociation of ammonium carbamate may be represented by the equation NH4CO2NH2(s) 2NH3(g) + CO2(g) H0 for the forward reaction is negative The equilibrium will shift from right to left if there is (A) a decrease in pressure (B) an increase in temperature (C) an increase in the concentration of ammonia (D) an increase in the concentration of carbon dioxide 101 For the reaction PCl5(g) PCl3(g) + Cl2(g), the forward reaction at constant temperature is favoured by (A) introducing an inert gas at constant volume (B) introducing chlorine gas at constant volume (C) introducing an inert gas at constant pressure (D) introducing PCl5 at constant volume 102 When NaNO3 is heated in a closed vessel, oxygen is liberated and NaNO2 is left behind At equilibrium (A) addition of NaNO2 favours reverse reaction (B) addition of NaNO3 favours forward reaction (C) increasing temperature favours forward reaction (D) increasing pressure favours reverse reaction 103 For the gas phase reaction, C2H4 + H2 C2H6 (H = – 32.7 kcal), carried out in a closed vessel, the equilibrium moles of C2H4 can be increased by (A) increasing the temperature (B) decreasing the pressure (C) removing some H2 (D) adding some C2H6 104 For the gas phase exothermic reaction, A2 + B2 C2, carried out in a closed vessel, the equilibrium moles of A2 can be increased by (A) increasing the temperature (B) decreasing the pressure (C) adding inert gas at constant pressure (D) removing some C2 105 Consider the equilibrium HgO(s) + 4I– (aq) + H2O (l) HgI42– (aq) + 2OH– (aq), which changes will decrease the equilibrium concentration of HgI42– (A) Addition of 0.1 M HI (aq) (B) Addition of HgO (s) (C) Addition of H2O (l) (D) Addition of KOH (aq) 106 Decrease in the pressure for the following equilibria : H2O (s) H2O(l) result in the : (A) formation of more H2O (s) (B) formation of more H2O(l) (C) increase in melting point of H2O(s) (D) decrease in melting point of H2O(s) PART - II : SUBJECTIVE QUESTIONS Br2 (g) KP = 0.15 atm at 90°C If NOBr, NO and Br2 are mixed at this temperature having partial pressures 0.5 atm, 0.4 atm and 0.2 atm respectively, will Br2 be consumed or formed ? For the reaction NOBr (g) NO(g) + The KP values for three reactions are 10–5 , 20 and 300 then what will be the correct order of the percentage composition of the products mole of N2 and moles of H2 are placed in 1L vessel Find the concentration of NH3 at equilibrium, if equilibrium pressure is atm and the equilibrium constant at 400K is The value of Kc for the reaction, N2(g) + 2O2(g) value of equilibrium constant for (i) 2NO2(g) N2(g) + O2(g) ; 27 2NO2(g) at a certain temperature is 400 calcualte the (ii) 1/2 N2(g) + O2(g) ETOOS ACADEMY Pvt Ltd F-106, Road No.2 Indraprastha Industrial Area, End of Evergreen Motor, BSNL Lane, Jhalawar Road, Kota, Rajasthan (324005) Tel : +91-744-242-5022, 92-14-233303 NO2(g) CHEMICAL EQUILIBRIUM_ADVANCED# 31 From the following data : (i) H2(g) + CO2(g) H2O(g) + CO(g) K2000K = 4.4 (ii) 2H2O(g) 2H2(g) + O2(g) K2000K = 5.31 x 10–10 (iii) 2CO(g) + O2(g) 2CO2(g) K1000K = 2.24 x 1022 State whether the reaction (iii) is exothermic or endothermic? In an experiment starting with mole of ethyl alcohol, mole of acetic acid and mole of water at T0C, the equilibrium mixture on analysis shows that 54.3% of the acid is esterfied Calculate the equilibrium constant of this reaction Derive an expression for Kc and Kp for the reaction N2(g) + 3H2(g) 2NH3(g) Assuming that in a container of volume V, initially mole of N2 and moles of H2 were taken and at equilibrium 2x moles of NH3 is formed The degree of dissociation at a certain or given temperature of PCl5 at atm is found to be 0.4 At what pressure, the degree of dissociation of PCl5 will be 0.6 at the same temperature? Also calculate the equilibrium constant for the reverse reaction (i) Calculate the percentage dissociation of H2S(g), if 0.1 mole of H2S is kept in 0.4 litre vessel at 1000K For the reaction, 2H2S(g) 2H2(g) + S2(g), the value of Kc = 1.6 × 10–6 (ii) A sample of HI was found to be 22% dissociated when equilibrium was reached What will be the degree of dissociation if hydrogen is added in the proportion of mole for every mole of HI originally present, the temperature and volume of the system being kept constant? 10 Calculate the degree of dissociation and Kp for the following reaction PCl5(g) PCl3(g) + Cl2(g) t=0 a 0 t=t a –x x x Since for a mole, x moles are dissociated 11 The vapour density of PCl5 at 250ºC is found to be 57.9 Calculate percentage dissociation at this temperature 12 The vapour density of a mixture containing NO2 and N2O4 is 38.3 at 33°C calculate the no of moles of NO2 if 100g of N2O4 were taken initially 13 What is the effect on the following equilibrium if each of the indicated stresses is applied? N2(g) +O2(g) NO 2(g) + heat (a) Increase in N2 concentration (b) Decrease in temperature (c) Increase in volume of vessel (d) addition of a catalyst ETOOS ACADEMY Pvt Ltd F-106, Road No.2 Indraprastha Industrial Area, End of Evergreen Motor, BSNL Lane, Jhalawar Road, Kota, Rajasthan (324005) Tel : +91-744-242-5022, 92-14-233303 CHEMICAL EQUILIBRIUM_ADVANCED# 32 14 Kc for the reaction N2O4 2NO2 in chloroform at 291 K is 1.14 Calculate the free energy change of the reaction when the concentration of the two gases are 0.5 mol dm–3 each at the same temperature (R = 0.082 lit atm K–1 mol–1.) 15 Calculate the pressure of CO gas at 700K in the heterogeneous equilibrium reaction CaCO3(s) CaO(s) + CO2(g) if 16 G0 for this reaction is 130.2 kJ mol–1 For the equilibrium NiO(s) + CO(g) Ni(s) + CO2(g), G0 (J mol–1) = – 20,700 – 11.97 T Calculate the temperature at which the product gases at equilibrium at atm will contain 400 ppm (parts per million) of carbon monoxide 17 (i) Calculate the partial pressure of HCl gas above solid a sample of NH4Cl(s) as a result of its decomposition according to the reaction: NH4Cl(s) NH3(g) + HCl(g) Kp = 1.04 × 10–16 (ii) Calculate the equilibrium constant of a reaction at 300 K if reaction is 29.29 kJ mol–1 18 G0 at this temperature for the For the formation of ammonia the equilibrium constant data at 673K and 773K respectively are 1.64 × 10–4 and 1.44 × 10–5 respectively Calculate heat of the reaction Given R=8.314 JK–1mol–1 19 The equilibrium constant KP for the reaction N2(g) + 3H2(g) 2NH3(g) is 1.6 10-4 atm at 400oC What will be the equilibrium constant at 500oC if heat of the reaction in this temperature range is 25.14 k cal? 20 The equilibrium constant for the reaction H2 + I2 2HI; is found to be 64 at 450°C If mole of hydrogen are mixed with mole of iodine in a litre vessel at this temperature; what will be the concentration of each of the three components, when equilibrium is attained ? If the volume of reaction vessel is reduced to half; then what will be the effect on equilibrium? 21 gm of PCl5 were completely vaporized at 250°C in a vessel of 1.9 litre capacity The mixture at equilibrium exerted a pressure of one atmosphere Calculate the degree of dissociation KC and Kp for this reaction 22 If a mixture of N2 and H2 in the ratio : at 50 atmosphere and 650°C is allowed to react till equilibrium is reached Ammonia present at equilibrium was at 25 atm pressure Calculate the equilibrium constant for the reaction N2(g) + 3H2(g) 2NH3(g) 23 0.96 g of HI were heated to attain equilibrium 2HI(g) H2 (g) + I2(g) The reaction mixture on titration requires 15.7 mL of N/10 hypo solution Calculate degree of dissociation of HI 24 Would 1% CO2 in air be sufficient to prevent any loss in weight when M2CO3 is heated at 120oC ? M2CO3(s) M2O(s) + CO2(s) Kp = 0.0095 atm at 120oC How long would the partial pressure of CO2 have to be to promote this reaction at 120oC ? 25 In a container of constant volume at a particular temparature N2 and H2 are mixed in the molar ratio of 9:13 The following two equilibria are found to be coexisting in the container N2 (g) + 3H2 (g) 2NH3 (g) N2H4 (g) N2 (g) + 2H2 (g) The total equilibrium pressure is found to be 3.5 atm while partial pressure of NH3 (g) and H2(g) are 0.5 atm and atm respectively Calculate of equilibrium constants of the two reactions given above ETOOS ACADEMY Pvt Ltd F-106, Road No.2 Indraprastha Industrial Area, End of Evergreen Motor, BSNL Lane, Jhalawar Road, Kota, Rajasthan (324005) Tel : +91-744-242-5022, 92-14-233303 CHEMICAL EQUILIBRIUM_ADVANCED# 33 26 A saturated solution of iodine in water contains 0.33g of I2 per litre of solution More than this can dissolve in KI solution because of the following equilibrium I2(aq) + I– I3– A 0.1 M KI solution actually dissolves 12.5g of I2/litre, most of which is converted to I3– Assuming that the concentration of I2 in all saturated solutions is the same, calculate the equilibrium constant for the above reaction 27 For the reaction Ag(CN)2– Ag+ + 2CN–, the KC at 25°C is 10–19 Calculate [Ag+] in solution which was originally 0.1 M in KCN and 0.03 M in AgNO3 28 When baking soda is heated in a sealed tube, following equilibrium exits: 2NaHCO3(s) Na2CO3(s) + CO2(g) + H2O(g) If the equilibrium pressure is 1.04 atm at 398 K, calculate the equilibrium constant for the reaction at 398 K 29 For the reaction N2(g) + H2(g) 2 NH3(g) Show that degree of dissociation of NH3 is given as : 3 P 1 K p 1 / where ‘P’ is equilibrium pressure If Kp of the above reaction is 78.1 atm at 400°C, calculate KC 30 (i) The vapour density of a mixture consisting of NO2 and N2O4 is 38.3 at 26.7°C Calculate the number of moles of NO2 in 100 gm mixture (ii) Establish a relationship between Kc and Kp for the following reactions 31 (a) N2(g) + O2(g) 2NO(g) (b) (NH (NH )2CO CO3(s) 2NH + CO + CO OH 4)2 3(s) 2NH 3(g)3(g) 2(g) +H 2(g)2+ (g) O(g) Calculate the Kc and Kp for the following reactions and also deduce the relationship between Kc and Kp (i) 2SO2(g) + O2(g) 2SO3(g) 32 (ii) N2(g) + H2(g) NH3(g) 2 At 400K for the gaseous reaction 2A + 3B 3C + 4D the value of Kp is 0.05 Calculate the value of Kc (R = 0.082 dm3 atm K–1 mol–1) ETOOS ACADEMY Pvt Ltd F-106, Road No.2 Indraprastha Industrial Area, End of Evergreen Motor, BSNL Lane, Jhalawar Road, Kota, Rajasthan (324005) Tel : +91-744-242-5022, 92-14-233303 CHEMICAL EQUILIBRIUM_ADVANCED# 34 PART - I : IIT-JEE PROBLEMS (PREVIOUS YEARS) * Marked Questions are having more than one correct option When 3.06 g of solid NH4HS is introduced into a two litre evacuated flask at 27ºC,30% of the solid decomposes into gaseous ammonia and hydrogen sulphide [JEE-1999, 7/120] (i) Calculate kC & kP for the reaction at 27ºC (ii) What would happen to the equilibrium when more solid NH4HS is introduced into the flask For a chemical reaction 3X(g) + Y(g) X3 Y(g), the amount of X3 Y at equilibrium is affected by [JEE-1999, 2/80] (A) temperature and pressure (C) pressure only (B) temperature only (D) temperature, pressure and catalyst For the reversible reaction, N (g) + 3H 2(g) 2NH at 500°C, the value of K P is 1.44 × 10 –5 when partial pressure is measured in atmospheres The corresponding value of K C, with concentration in mole litre –1, is [JEE 2000, 1/35] (A) 1.44 10 5 (B) (0.082 500) 2 1.44 10 5 (8.314 773) 2 (C) 1.44 10 5 (0.082 773)2 (D) 1.44 10 5 (0.082 773) 2 When two reactants, A & B are mixed to give products C & D, the reaction quotient Q, at the initial stages of the reaction [JEE-2000, 1/35] (A) is zero (B) decrease with time (C) is independent of time (D) increases with time At constant temperature, the equilibrium constant (KP) for the decomposition reaction N2O4 2NO2 is expressed by KP = ( x P) , where P = pressure, x = extent of decomposition Which one of the following (1 x ) statements is true ? [JEE 2001, 1/35] (A) KP is increases witn increase of P (B) KP is increases witn increase of x (C) KP is increases witn decrease of x (D) KP remains constant with change in P and x Consider the following equilibrium in a closed container [JEE 2002, 3/90] N2 O4 (g) 2NO2 (g) 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 (KP) and degree of dissociation ()? (A) neither KP nor changes (B) both KP and change (C) KP changes, but does not change (D) KP does not change but changes N2 + 3H2 NH3 K = × 106 at 298 K = 41 at 400 K [JEE 2006, 3/184] Which statements is correct? (A) If N2 is added at equlibrium condition, the equilibrium will shift to the forward direction because according to IInd law of thermodynamics the entropy must increases in the direction of spontaneous reaction (B) The condition for equilibrium is 2GNH3 = 3GH2 + GN2 where G is Gibbs free energy per mole of the gaseous species measured at that partial pressure (C) Addition of catalyst does not change Kp but changes H (D) At 400 K addition of catalyst will increase forward reaction by times while reverse reaction rate will be changed by 1.7 times ETOOS ACADEMY Pvt Ltd F-106, Road No.2 Indraprastha Industrial Area, End of Evergreen Motor, BSNL Lane, Jhalawar Road, Kota, Rajasthan (324005) Tel : +91-744-242-5022, 92-14-233303 CHEMICAL EQUILIBRIUM_ADVANCED# 35 The value of log10K for a reaction A B is : (Given : r H298K = –54.07 kJ mol–1, r S 298K = 10 JK–1 mol–1 and R = 8.314 JK–1 mol–1 ; 2.303 x 8.314 x 298 = 5705) (A) [JEE 2007, 3/162] (B) 10 (C) 95 (D) 100 STATEMENT-1 : For every chemical reaction at equilibrium, standard Gibbs energy of reaction is zero STATEMENT-2 : At constant temperature and pressure, chemical reactions are spontaneous in the direction of decreasing Gibbs energy [JEE 2008, 3/163] (A) Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1 (B) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1 (C) Statement-1 is True, Statement-2 is False (D) Statement-1 is False, Statement-2 is True (E) Statement-1 and Statement-2 both are False 10.* The thermal dissociation equilibrium of CaCO3(s) is studied under different conditions : CaCO3(s) CaO(s) + CO2(g) For this equilibrium, the correct statement(s) is(are) (A) H is dependent on T (B) K is independent of the initial amount of CaCO3 (C) K is dependent on the pressure of CO2 at a given T (D) H is independent of the catalyst, if any 11 [JEE_2013] The Ksp of Ag2CrO4 is 1.1 × 10–12 at 298 K The solubility (in mol/L) of Ag2CrO4 in a 0.1 M AgNO3 solution is: [JEE_2013] –11 –10 –12 –9 (A) 1.1 × 10 (B) 1.1 × 10 (C) 1.1 × 10 (D) 1.1 × 10 PART - II : AIEEE PROBLEMS (PREVIOUS YEARS) Change in volume of the system does not alter the number of moles in which of the following equilibriums : [AIEEE 2002] (1) N2(g) + O2(g) 2NO(g) (3) N2(g) + 3H2(g) 2NH3(g) (2) PCl5(g) (4) SO2Cl2(g) PCl3(g) + Cl2(g) SO2(g) + Cl2(g) In which of the following reactions, increase in the volume at constant temperature don’t effect the number of moles of at equilibrium : [AIEEE 2002] (1) 2NH3 N2 + 3H2 (3) H2 (g) + O2 (g) (2) C (g) + (1/2) O2 (g) H2O2 (g) CO (g) (4) none of these Consider the reaction equilibrium 2SO2 (g) + O2 (g) 2SO3 (g) ; H° = – 198 kJ On the basis of Le Chatelier’s principle, the condition favourable for the forward reaction is [AIEEE 2003] (1) lowering of temperature as well as pressure (2) increasing temperature as well as pressure (3) lowering the temperature and increasing the pressure (4) any value of temperature and pressure For the reaction equilibrium, N2O4 (g) 2NO2 (g) the concentrations of N2O4 and NO2 at equilibrium are 4.8 × 10–2 and 1.2 × 10–2 mol L–1 respectively The value of Kc for the reaction is [AIEEE 2003] –1 –1 –1 –3 –1 –1 (1) 3.3 × 10 mol L (2) × 10 mol L (3) × 10 mol L (4) × 10 mol L ETOOS ACADEMY Pvt Ltd F-106, Road No.2 Indraprastha Industrial Area, End of Evergreen Motor, BSNL Lane, Jhalawar Road, Kota, Rajasthan (324005) Tel : +91-744-242-5022, 92-14-233303 CHEMICAL EQUILIBRIUM_ADVANCED# 36 What is the equilibrium constant expression for the reaction P4 (s) + 5O2 (g) P4 O10 (s) ? P4 (s) + 5O2 (g) P4 O10 (s) ? (1) KC = [P4O10]/[P4] [O2]5 (2) KC = 1/[O2]5 (3) KC = [O2] (4) KC = [P4O10] / 5[P4] [O2] [AIEEE 2004] For the reaction, CO(g) + Cl2 (g) [AIEEE 2004] (1) 1/RT COCl2 (g) then Kp / Kc is equal to : (2) 1.0 (3) The equilibrium constant for the reaction, N2 (g) + O2 (g) of Kc for the reaction, NO(g) NO(g) (1) 2.5 × 102 (4) RT RT 2NO(g) at temperature T is × 10–4 The value 1 N2 (g) + O (g) at the same temperature is 2 1 N (g) + O (g) 2 2 (2) 0.02 (3) × 10–4 [AIEEE 2004] (4) 50 For the reaction, 2NO2 (g) NO(g) + O2 (g), [AIEEE 2005] (KC = 1.8 × 10–6 at 184°C) (R = 0.0831 kJ/(mol.K)) When Kp and Kc are compared at 184°C it is found that (1) Whether Kp is greater than, less than or equal to Kc depends upon the total gas pressure (2) Kp = Kc (3) Kp is less than Kc (4) Kp is greater than Kc The exothermic formation of ClF3 is represented by the equation : [AIEEE 2005] Cl2 (g) + 3F2 (g) 2ClF3 (g) ; rH = – 329 J which of the following will increase the quantity of ClF3 in an equilibrium mixture of Cl2 , F2 and ClF3 (1) Adding F2 (2) Increasing the volume of container (3) Removing Cl2 (4) Increasing the temperature 10 An amount of solid NH4HS is placed in a flask already containing ammonia gas at a certain temperature at 0.50 atm pressure Ammonium hydrogen sulphide decomposes to yield NH3 and H2S gases in the flask When the decomposition reaction reaches equilibrium, the total pressure in the flask rises to 0.84 atm? The equilibrium constant for NH4HS decomposition at this temperature is : [AIEEE 2005] (1) 0.11 (2) 0.17 (3) 0.18 (4) 0.30 11 Phosphorus pentachloride dissociates as follows in a closed reaction vessel PCl5(g) PCl3(g) + Cl2(g) If total pressure at equilibrium of the reaction mixture is P and degree of dissociation of PCl5 is x, the partial pressure of PCl3 will be : [AIEEE 2006] 2x P (2) 1– x x P (1) x 1 12 The equilibrium constant for the reaction, SO3(g) x P (4) 1– x x P (3) x 1 SO2(g) + O (g) 2 is KC = 4.9 × 10–2 The value of KC for the reaction 2SO2(g) + O2(g) (1) 416 (2) 2.40 × 10–3 (3) 9.8 × 10–2 13 2SO3(g) will be (4) 4.9 × 10–2 For the following three reactions a, b and c, equilibrium constants are given: (A) CO(g) + H2O(g) CO2(g) + H2(g); K1 (B) CH4(g) + H2O(g) CO(g) + 3H2(g); K2 (C) CH4(g) + 2H2O(g) CO2(g) + 4H2(g); K3 Which of the following relations is correct ? (1) K2 K3 = K1 (2) K3 = K1K2 (3) K3 K23 = K12 ETOOS ACADEMY Pvt Ltd F-106, Road No.2 Indraprastha Industrial Area, End of Evergreen Motor, BSNL Lane, Jhalawar Road, Kota, Rajasthan (324005) Tel : +91-744-242-5022, 92-14-233303 [AIEEE 2006] [AIEEE 2008, 3/105] (4) K K K CHEMICAL EQUILIBRIUM_ADVANCED# 37 14 The equilibrium constants K p and K p for the reactions X 2Y and Z P + Q, respectively are in the ratio of : If the degree of dissociation of X and Z be equal then the ratio of total pressures at these equilibria is [AIEEE 2008, 3/105] (1) : (2) : (3) : (4) : 36 15 If 10–4 dm3 of water is introduced into a 1.0 dm3 flask at 300 K, how many moles of water are in the vapour phase when equilibrium is established ? [AIEEE 2010, 4/144] (Given : Vapour pressure of H2O at 300 K is 3170 Pa ; R= 8.314 J K–1 mol–1) (1) 5.56 × 10–3 mol (2) 1.53 × 10–2 mol (3) 4.46 × 10–2 mol (4) 1.27 × 10–3 mol 16 A vessel at 1000 K contains CO2 with a pressure of 0.5 atm Some of the CO2 is converted into CO on the addition of graphite If the total pressure at equilibrium is 0.8 atm, the value of K is : [AIEEE 2010, 4/144] (1) 1.8 atm (2) atm (3) 0.3 atm (4) 0.18 atm NCERT QUESTIONS A liquid is in equilibrium with its vapour in a sealed container at a fixed temperature The volume of the container is suddenly increased (a) What is the initial effect of the change on vapour pressure ? (b) How rates of evaporation and condensation change initially ? (c) What happens when equilibrium is restored finally and what will be the final vapour pressure ? What is Kc for the following equilibrium when the equilibrium concentration of each substance is: [SO2] = 0.60M, [O2] = 0.82M and [SO3] = 1.90M ? 2SO2(g) + O2(g) 2SO3(g) At a certain temperature and total pressure of 105Pa, iodine vapour contains 40% by volume of I atoms I2 (g) 2I (g) Calculate Kp for the equilibrium Write the expression for the equilibrium constant, Kc for each of the following reactions: (i) 2NOCl (g) 2NO (g) + Cl2 (g) (ii) 2Cu(NO3)2 (s) 2CuO (s) + 4NO2 (g) + O2 (g) (iii) CH3COOC2H5 (aq) + H2O(l) CH3COOH (aq) + C2H5OH (aq) – (iv) Fe + (aq) + 3OH (aq) Fe(OH)3(s) (v) I2 (s) + 5F2 2IF5 Find out the value of Kc for each of the following equilibria from the value of Kp : (i) 2NOCl (g) 2NO (g) + Cl2 (g); Kp = 1.8 × 10–2 at 500 K (ii) CaCO3 (s) CaO(s) + CO2(g); Kp = 167 at 1073 K For the following equilibrium, Kc = 6.3 × 1014 at 1000 K NO2 (g) + O2 (g) NO (g) + O3 (g) Both the forward and reverse reactions in the equilibrium are elementary bimolecular reactions What is Kc, for the reverse reaction ? Explain why pure liquids and solids can be ignored while writing the equilibrium constant expression? Reaction between N2 and O2– takes place as follows : 2N2 (g) + O2 (g) 2N2O (g) If a mixture of 0.482 mol N2 and 0.933 mol of O2 is placed in a 10 L reaction vessel and allowed to form N2O at a temperature for which Kc = 2.0 10–37, determine the composition of equilibrium mixture ETOOS ACADEMY Pvt Ltd F-106, Road No.2 Indraprastha Industrial Area, End of Evergreen Motor, BSNL Lane, Jhalawar Road, Kota, Rajasthan (324005) Tel : +91-744-242-5022, 92-14-233303 CHEMICAL EQUILIBRIUM_ADVANCED# 38 Nitric oxide reacts with Br2 and give nitrosyl bromide as per reaction given below : 2NO (g) + Br2 (g) 2NOBr (g) When 0.087 mol of NO and 0.0437 mol of Br2 are mixed in a closed container at constant temperature, 0.0518 mol of NOBr is obtained at equilibrium Calculate equilibrium amount of NO and Br2 10 At 450K, Kp = 2.0 × 1010 bar for the given reaction at equilibrium 2SO2(g) + O2(g) 2SO3 (g) What is Kc at this temperature ? 11 A sample of HI(g) is placed in flask at a pressure of 0.2 atm At equilibrium the partial pressure of HI(g) is 0.04 atm What is Kp for the given equilibrium ? 2HI (g) H2 (g) + I2 (g) 12 A mixture of 1.57 mol of N2, 1.92 mol of H2 and 8.13 mol of NH3 is introduced into a 20 L reaction vessel at 500 K At this temperature, the equilibrium constant, Kc for the reaction N2 (g) + 3H2 (g) 2NH3 (g) is 1.7 10 Is the reaction mixture at equilibrium ? If not, what is the direction of the net reaction ? 13 The equilibrium constant expression for a gas reaction is, kc [NH3 ]4 [O2 ]5 [NO]4 [H2 O]6 Write the balanced chemical equation corresponding to this expression 14 One mole of H2O and one mole of CO are taken in 10 L vessel and heated to 725 K At equilibrium 40% of water (by mass) reacts with CO according to the equation, H2O (g) + CO (g) H2 (g) + CO2 (g) Calculate the equilibrium constant for the reaction 15 At 700 K, equilibrium constant for the reaction: H2 (g) + I2 (g) 2HI (g) –1 is 54.8 If 0.5 mol L of HI(g) is present at equilibrium at 700 K, what are the concentration of H2(g) and I2(g) assuming that we initially started with H2(g) HI(g) allowed it to reach equilibrium at 700K ? 16 What is the equilibrium concentration of each of the substances in the equilibrium when the initial concentration of ICl was 0.78 M ? 2ICl (g) I2 (g) + Cl2 (g); Kc = 0.14 17 Kp = 0.04 atm at 899 K for the equilibrium shown below What is the equilibrium concentration of C2H6 when it is placed in a flask at 4.0 atm pressure and allowed to come to equilibrium ? C2H6 (g) C2H4 (g) + H2 (g) 18 Ethyl acetate is formed by the reaction between ethanol and acetic acid and the equilibrium is represented as : CH3COOH (l) + C2H5OH (l) CH3COOC2H5 (l) + H2O (l) (i) Write the concentration ratio (reaction quotient), Qc, for this reaction (note: water is not in excess and is not a solvent in this reaction) (ii) At 293 K, if one starts with 1.00 mol of acetic acid and 0.18 mol of ethanol, there is 0.171 mol of ethyl acetate in the final equilibrium mixture Calculate the equilibrium constant (iii) Starting with 0.5 mol of ethanol and 1.0 mol of acetic acid and maintaining it at 293 K, 0.214 mol of ethyl acetate is found after sometime Has equilibrium been reached ? 19 A sample of pure PCl5 was introduced into an evacuated vessel at 473 K After equilibrium was attained, concentration of PCl5 was found to be 0.5 × 10–1 mol L–1 If value of Kc is 8.3 × 10–3, what are the concentrations of PCl3 and Cl2 at equilibrium ? PCl3 (g) + Cl2 (g) PCl5 (g) ETOOS ACADEMY Pvt Ltd F-106, Road No.2 Indraprastha Industrial Area, End of Evergreen Motor, BSNL Lane, Jhalawar Road, Kota, Rajasthan (324005) Tel : +91-744-242-5022, 92-14-233303 CHEMICAL EQUILIBRIUM_ADVANCED# 39 20 One of the reaction that takes place in producing steel from iron ore is the reduction of iron(II) oxide by carbon monoxide to give iron metal and CO2 FeO (s) + CO (g) Fe (s) + CO2 (g); Kp = 0.265 atm at 1050K What are the equilibrium partial pressures of CO and CO2 at 1050 K if the initial partial pressures are : pCO= 1.4 atm and pCO = 0.80 atm ? 21 Equilibrium constant, Kc for the reaction N2 (g) + 3H2 (g) 2NH3 (g) at 500 K is 0.061 At a particular time, the analysis shows that composition of the reaction mixture is 3.0 mol L–1 N2, 2.0 mol L–1 H2 and 0.5 mol L–1 NH3 Is the reaction at equilibrium ? If not in which direction does the reaction tend to proceed to reach equilibrium ? 22 Bromine monochloride, BrCl decomposes into bromine and chlorine and reaches the equilibrium : 2BrCl (g) Br2 (g) + Cl2 (g) for which Kc = 32 at 500 K If initially pure BrCl is present at a concentration of 3.3 × 10–3 mol L–1, what is its molar concentration in the mixture at equilibrium ? 23 At 1127 K and atm pressure, a gaseous mixture of CO and CO2 in equilibrium with soild carbon has 90.55% CO by mass C (s) + CO2 (g) 2CO (g) Calculate Kc for this reaction at the above temperature 24 Calculate a) f G and b) the equilibrium constant for the formation of NO2 from NO and O2 at 298K NO (g) + 1/2O2 (g) NO2 (g) f G (NO2) = 52.0 kJ/mol f G (NO) = 87.0 kJ/mol f G (O2) = kJ/mol 25 Does the number of moles of reaction products increase, decrease or remain same when each of the following equilibria is subjected to a decrease in pressure by increasing the volume ? (a) PCl5 (g) PCl3 (g) + Cl2 (g) (b) CaO (s) + CO2 (g) CaCO3 (s) (c) 3Fe (s) + 4H2O (g) Fe3O4 (s) + 4H2 (g) 26 Which of the following reactions will get affected by increasing the pressure ? Also, mention whether change will cause the reaction to go into forward or backward direction (i) COCl2 (g) CO (g) + Cl2 (g) (ii) CH4 (g) + 2S2 (g) CS2 (g) + 2H2S (g) (iii) CO2 (g) + C (s) 2CO (g) (iv) 2H2 (g) + CO (g) CH3OH (g) (v) CaCO3 (s) CaO (s) + CO2 (g) (vi) NH3 (g) + 5O2 (g) 4NO (g) + 6H2O(g) 27 The equilibrium constant for the following reaction is 1.6 × 105 at 1024K H2(g) + Br2(g) 2HBr(g) Find the equilibrium pressure of all gases if 10.0 bar of HBr is introduced into a sealed container at 1024K 28 Dihydrogen gas is obtained from natural gas by partial oxidation with steam as per following endothermic reaction : CH4 (g) + H2 O (g) CO (g) + 3H2 (g) (a) Write as expression for Kp for the above reaction (b) How will the values of Kp and composition of equilibrium mixture be affected by (i) increasing the pressure (ii) increasing the temperature (iii) using a catalyst ? ETOOS ACADEMY Pvt Ltd F-106, Road No.2 Indraprastha Industrial Area, End of Evergreen Motor, BSNL Lane, Jhalawar Road, Kota, Rajasthan (324005) Tel : +91-744-242-5022, 92-14-233303 CHEMICAL EQUILIBRIUM_ADVANCED# 40 29 Describe the effect of : (a) addition of H2 (b) addition of CH3OH (c) removal of CO (d) removal of CH3OH on the equilibrium of the reaction: 2H2(g) + CO (g) CH3OH (g) 30 At 473 K, equilibrium constant Kc for decomposition of phosphorus pentachloride, PCl5 is 8.3 × 10-3 If decomposition is depicted as, PCl5 (g) PCl3 (g) + Cl2 (g) r H = 124.0 kJ mol–1 (a) write an expression for Kc for the reaction (b) what is the value of Kc for the reverse reaction at the same temperature ? (c) what would be the effect on Kc if (i) more PCl5 is added (ii) pressure is increased (iii) the temperature is increased ? 31 Dihydrogen gas used in Haber’s process is produced by reacting methane from natural gas with high temperature steam The first stage of two stage reaction involves the formation of CO and H2 In second stage, CO formed in first stage is reacted with more steam in water gas shift reaction, CO (g) + H2O (g) CO2 (g) + H2 (g) If a reaction vessel at 400 °C is charged with an equimolar mixture of CO and steam such that pCO = pH O = 4.0 bar, what will be the partial pressure of H2 at equilibrium ? Kp = 10.1 at 400°C 32 Predict which of the following reaction will have appreciable concentration of reactants and products: (a) Cl2 (g) 2Cl (g) Kc = × 10–39 (b) Cl2 (g) + 2NO (g) 2NOCl (g) Kc = 3.7 × 108 (c) Cl2 (g) + 2NO2 (g) 2NO2Cl (g) Kc = 1.8 33 The value of Kc for the reaction 3O2 (g) 2O3 (g) is 2.0 × 10–50 at 25°C If the equilibrium concentration of –2 O2 in air at 25°C is 1.6 × 10 , what is the concentration of O3 ? 34 The reaction, CO (g) + 3H2 (g) CH4 (g) + H2O (g) is at equilibrium at 1300 K in a 1L flask It also contain 0.30 mol of CO, 0.10 mol of H2 and 0.02 mol of H2O and an unknown amount of CH4 in the flask Determine the concentration of CH4 in the mixture The equilibrium constant, Kc for the reaction at the given temperature is 3.90 ETOOS ACADEMY Pvt Ltd F-106, Road No.2 Indraprastha Industrial Area, End of Evergreen Motor, BSNL Lane, Jhalawar Road, Kota, Rajasthan (324005) Tel : +91-744-242-5022, 92-14-233303 CHEMICAL EQUILIBRIUM_ADVANCED# 41 EXERCISE # A-1 A-8 (A) (D) A-2 A-9 (C) (B) A-3 A-10 (C) (D) PART # I A-4 (B) A-11 (C) A-5 A-12 (B) (C) A-6 A-13 (B) (A) A-7 A-14 (D) (C) A-15 (C) A-16 (A) A-17 (D) A-18 (A) A-19 (C) A-20 (B) A-21 (D) A-22 A-29 B-1 B-8 C-4 C-11 C-18 C-25 C-32 E-2 (B) (A) (A) (A) (C) (C) (A) (C) (B) (ABD) A-23 A-30 B-2 B-9 C-5 C-12 C-19 C-26 C-33 E-3 (B) (C) (B) (A) (B) (C) (D) (B) (C) (B) A-24 A-31 B-3 B-10 C-6 C-13 C-20 C-27 C-34 E-4 (A) (D) (B) (C) (D) (D) (C) (A) (A) (B) A-25 A-32 B-4 B-11 C-7 C-14 C-21 C-28.* D-1 E-5.* (C) (A) (A) (C) (B) (C) (D) (ACD) (D) (AD) A-26 A-33 B-5 C-1 C-8 C-15 C-22.* C-29 D-2 E-6 (C) (B) (B) (C) (C) (B) (BCD) (A) (A) (D) A-27 A-34 B-6 C-2 C-9 C-16 C-23 C-30 D-3.* E-7 (C) (A) (C) (A) (C) (A) (B) (D) (A) (A) A-28 A-35 B-7 C-3 C-10 C-17 C-24 C-31 E-1 (C) (D) (B) (D) (B) (A) (B) (D) (A) 14 (A) (B) 22 29 36 43 C B T T 14 21 28 35 42 49 56 63 70 77 (A) (B) (A) (A) (C) (A) (A) (B) (A) (D) (C) 90 97 104 (BCD) (BC) (ABC) PART # II 15 19 23 30 37 (D) (B) (A) (A) q, s C A T (C) (A) 16 (C) (B) p (C) p (D) r 24 D 31 A 38 F 10 17 20 25 32 39 (AB) (B) (B) (B) (C) 11 (C) 12 (B) 13 (D) (A) 18 (A) P,R,S (B) P,Q,R,S (C) P,Q,R,S (D) Q (A – r) ; (B – r) ; (C – q) ; (D – p) 21 C A 26 B 27 A 28 A A 33 D 34 T 35 F T 40 T 41 F 42 F 44 F 45 46 F F 47 EXERCISE # 15 22 29 36 43 50 57 64 71 78 84 91 98 105 (C) (B) (D) (A) (B) (D) (B) (B) (A) (A) (C) (C) (ABCD) (BC) (ABC) (CD) 16 23 30 37 44 51 58 65 72 79 85 92 99 106 (B) (B) (C) (C) (B) (B) (B) (B) (C) (B) (A) (A) (AB) (AB) (CD) (AC) 10 17 24 31 38 45 52 59 66 73 80 86 93 100 (A) (C) (B) (B) (A) (D) (B) (D) (A) (B) (B) (D) (AC) (B*C) (BCD) 11 18 25 32 39 46 53 60 67 74 81 87 94 101 (C) (B) (B) (B) (A) (A) (B) (B) (B) (A) (D) (A) (ABC) (ACD) (CD) 12 19 26 33 40 47 54 61 68 75 82 88 95 102 (B) (A) (B) (A) (B) (A) (A) (B) (B) (B) (A) (AB) (ABC) (ABCD) (CD) ETOOS ACADEMY Pvt Ltd F-106, Road No.2 Indraprastha Industrial Area, End of Evergreen Motor, BSNL Lane, Jhalawar Road, Kota, Rajasthan (324005) Tel : +91-744-242-5022, 92-14-233303 13 20 27 34 41 48 55 62 69 76 83 89 96 103 (B) (B) (B) (C) (B) (B) (C) (D) (D) (D) (A) (BD) (ABD) (BCD) (ABCD) CHEMICAL EQUILIBRIUM_ADVANCED# 42 SUBJECTIVE Kp = 300 reaction is exothermic (i) 2.34% (ii) 0.29 13 14 (a) Right (b) Right – 19.67 lit atm 17 (i) 1.02 0.76 (i) 0.0025 mole litre–1 (ii) 20 litre–1/2 mole–1/2 4.0 16 x ( x ) 27(1 x ) P 0.677 atm 12 0.435 10 KP = 2 P (c) Left (d) No shift will occur 15 1.94 × 10–10 atm 16 (ii) 7.91 × 10 19 × 10 atm 1.43 × 10–5 23 27 0.209 7.5 10–18 24 28 – 0.0005 0.2704 atm 30 (i) 0.435 (ii) (a) Kp = Kc (b) Kp = Kc (RT)4 31 (i) K p –8 Kc RT –6 20 (ii) K p 11 1 2 0.16 21 25 29 Kc RT 32 80% T = 399K 18 – 105.216kJ 22 1.677 10–3 0.6 atm–2 26 –1 1.413 moles litre 708 4.648 × 10–5 EXERCISE # PART # I (i) 8.1 10 5 mol/, 0.049 atm2 (ii) No change (A) (D) 11 (D) (B) (D) 10.* (ABD) 12 (2) (1) 13 (1) (2) (D) (B) (B) (D) PART # II 15 (1) (4) (4) 16 (4) (1) (1) 10 (3) (1) 11 (3) (1) 14 (4) (4) EXERCISE # 12 15 17 18 19 21 24 30 12.229 2.67 × 104 (i) 4.33 × 10–4 (ii) 1.90 1.59 × 10–15 [N2] = 0.0482 mol L–1, [O2] = 0.0933 mol L–1, [N2O] = 6.6 × 10–21 mol L–1 0.0352 mol of NO and 0.0178 mol of Br2 10 7.47 × 1011 M–1 11 4.0 Qc = 2.379 × 103 No, reaction is not at equilibrium 14 0.44 –1 0.068 mol L eact of H2 and I2 16 [I2] = [Cl2] = 0.167 M, [ICI] = 0.446 M [C2H6]eq = 3.62 atm (i) [CH3COOC2H5] [H2O] / [CH3COOH] [C2H6OH] (ii) 3.92 (iii) Value of Qc is less than Kc therefore equilibrium is not attained 0.02 mol L–1 for both 20 [PCO] = 1.739 atm, [PCO2] = 0.461 atm No, the reaction proceeds to form more products 22 × 10–4 mol L–1 23 0.149 –2 (a) – 35.0( kJ, (b) 1.365 × 10 27 [PH ]eq = [PH ]eq = 2.5 × 10 bar, [PHBr] = 10.0 bar 2 (b) 120.48 31 [H2]eq = 0.96 bar 33 2.86 × 10–28 M 34 5.85 × 10–2 ETOOS ACADEMY Pvt Ltd F-106, Road No.2 Indraprastha Industrial Area, End of Evergreen Motor, BSNL Lane, Jhalawar Road, Kota, Rajasthan (324005) Tel : +91-744-242-5022, 92-14-233303 CHEMICAL EQUILIBRIUM_ADVANCED# 43