- Chemical Kinetics Topic | Rate of a Chemical Reaction and Factors Influencing Rate of a Reaction Previous Years’ Examinati on Questions 1+ Mark Questions 1. Define order of a reaction, [Foreign 2012; 2007; All Indin 2011, 2010, 2009, 2008; Delhi 2011C, 2010, 2008C] 2. Identify the reaction ofdér fom the following rate constant, k = 2.3 x 10~ 3 mol" s7! woe [AIL India 20136) 3. Why does the rate of a reaction not remain constant through out the reaction process? 4. Define rate of a reaction. (Delhi 2010; All India 2010) For the reaction, Cly{g) + 2NO (g) + 2NOCIg) ‘The rate law is expressed as. =kICLJ INO}? * P Practice the Real Questions What is the overall order of this reaction? (Delhi 2007} Expris the rate of the following reaction in terns of disappesirance of hydragen in the reaction. 3H,[g) + Naig) > 2NHy(g) TALI india 2007] ‘ihe reaction, A +38 —> C dbeys the rate equation, rate & & (A)? [B"* What is the order of reaction? [AH India 2007) For the reaction, A——>8, the rate of reaction becomes twenty. seven. times when the concentration of A is increased three “times. What is the” order of the. reaction? [Delhi 2008}a 10. Chapterwise CBSE Solved Papers Chemistry ‘The reaction; A +B C has zero order, What is the rate equation? (Delhi 2006) What is meant by elementary step ia reaction? TAIT Indio 2008, Marks Questions ih, 12 13. wld 15. 16. Ie 18. 1. tks {For a reaction A+B, the sate law is given by, r= KAJ? [87 What is the order of this reaction? Ai)_A first osder reaction is found to have a rate constant k =5.5 x 107" s”, Find ‘the’ half-life of the reaction, {All India 2013} A reaction is of second arder with respect to'areactant. Now is its rate affected if the concentration of the reactant is (i). doubled (ii) reduced to halt . TAll Zndia 2042; Dethi 2009; “= Foreign 2008, 2008] What do you understand by the rate jaw and rate constant of a reaction? Identify the order of a reaction if the units of its rate constants are (i) Ut mols" Gi}. Lmott st [Welhi 2012; All India 20121 Distinguish’ between: rate” expression and rate constant of a reaction. [Dethi 2011} A reaction is of second order with respect toa reactant. How is thé ‘rate of reaction affected if the concentration of the reactant. is reduced to half? What is the unit of rate constant for such a reaction? fA India 2011) Express clearly what do you understand by rate. expression’ and rate ‘constant of a reaction? [Foreign 2011) Define (i order of reaction (ii) elementary step in a reaction, [Foreign 2011] tivo itiain differences between order of ‘@ reaction and molecularity of a reaction. (etki 2013¢} Identify-giving teasans, the réaction order from each of the following rate constants: 3x10? Lott st K=3.05107% st Temi 20110) 20, 21 22, 23 24, 25. 26. “on, 28. 29, Explain the terms (i) Rate determining step of a reaction (i) Motecutarity of a reaction, TAM India 2011€, 2010) A reaction is of first order in reactant A ancl of second order in reactant 8, Flow is the: rate of this reaction affected when (i) the concentration of B as alone is increased! to three times (i) the concentrations of A as well as B are doubled? (Dell 2010), Discuss the effect of any fourfactors, which affect the rate of a chemical reaction, {Delhi 2016C; All India 2008} Explain the difference between the average rate and instantaneous rate of a chemical reaction. TAH India 20100] Distinguish between molecularity and order. of a reaction. [Al India 2010; Dethi 20080] Define the following {i} Elementary step in a reaction (i Rate of a reacion, [All India 2009; Foreign 2009] Explain the term order of reaction. Derive the unit for first order rate constant Wethi 20091 What is mednt by rate constant, & of a reaction? if the concentration be expressed in molt'units and time in seconds, what would be the units for & (i) fora zero order reaction and (i) for a first order reaction? [Dethi 2008; Foreign 2008] The decomposition of NH, on platinum surface is a zero order reaction. What are the rales of production of -N, and H, if, k22.5x107 moll's (Dethi 2008, 2007; AH India 2008; Foreign 2008] List the factors Ga’ which the: rate of a ~ chemical reaction depends: 30. (Delhi 2008; All India 2008; Foreign 2008} For'a chemical reaction, what ié the effect of a catalyst on the following? (i) Activation enetgy of the reaction. (i) Rate constant of the reaction: [Ali India 2008C)Chemical Kinetics BL. Nitric oxide reacts with hydrogen to’ give nitrogen and water 3NO + 2H, +N; +2H,0 The kinetics of this reaction is explained by the following steps (2 NO +Hy Nj + H,0, (slow) tii) Hj, +H, —» 2H,0 (fast) What is the predicted rate law? (Delhi 2007] 82, Define rate law. Give example. [All Indi 2006) 83, (i) What is rate constant? . (if) On what factors it’ depends? . [Delhi 2006C] 84, At 300° the thermal dissociation of HI is found to be 20%. What will be the. equilibrium concentrations of H, abd |; ia the system Hy +1,7==*2HI at this temperature if“ the ~ equilibrium concentration of Hl in it be 0.96 mol 7"? (Delhi 2006C; All Indin 2006C} 85. Define zero order reaction. Give its unit. » {AH India 2006C} 36. Write units of rate constant k for zero order reaction, first order, second order and nth order-reaction: [All india 2006C] 3 Marks Questions BT. The-reaction, No{g)'+ Oy (g) = 2NO(g) contributes to air pollution wherever a fuel is burnt in-air-at a high: temperature. At 1500 K; equilibrium: constant K for it is 1.01075. Suppose in a case IN3] =0.80 mott"! and [0,] =0:20 molt before any reaction occurs. Calculate the equilibrium concentrations of the reactants . and the product after the mixture has been heated to 1500 K. __ [All Indin 2012] 38. For the reaction, 2NO ig) + Cl (g) + 2NOCI (gh The following data: were collected, All the measurements were takeri at 263 K: (@) Write the expression for rate law. (i) Calculate the value of rate constant and specify its units. (ii) What is the initial rate of disappearance of Cl, in experiment 4? _ (Belhi 20127 39. Consider the reaction, 24.4 B——» C+D Following result Were. “obtained in experiments designed to study the rate of reaction 30107 i ox 197 {i Write the rate faw for the reaction. (ii) Calculate the vatue of rate constant for the reaction, (ii) Which of the: following’ possible reaction mechanisms is consistent with the rate law found in Lo ASB C4E | (slows AtE—9D (fast H B39 C +E (slow) - AFE SF (fast) A+F—+D (fast). [Foreign 2012) 40... The following results have been obtained tic studies of the reaction. 03M | 02M 179% 10° mint 03M . joan | 288 10" rin? 4 OAM | OA | 240% 107M mint Determine rate law and the rate constant for the reaction. {Delhi 201001Chapterwise CBSE Solved Papers Chemistry. . 41. The data given below is forthe reaction, NO; (g) “> NO} ig) + Og) OB4x 107 F 25x10 oan ra? = tex 0 Determine for this reaction, {i) Order of reaction.. Rate faw Rate constant: TAI India 2008C) 5 Marks Question 42, (i) A reaction is second orderin A and first order in B. fa) Write the differentia rate equation. (b) How. is the rate affected on increasing the concentration of A three times? {c) How is the rate affected when the conceritrations ofboth A and Bare doubled? (ii) A first order seaction takes 40 min for 30% decomposition. Calculate ty» for this reaction. (Given, log 1.428 =0,1548). (ethi 20131 ‘Step-by-Step Solutions 1. Order of a reaction . The sum of the powers of the concentration of the reactants of a chemical reaction in the rate law expression is called the “order of that chetnical reaction. a Rate = [AJ* (6) ‘Osler of reaction x It'is the seconet orcler reaction: a IC is because,.the concentration of reactants goes an decreasing with time. a 4, Rate of a reaction Change in concentration of reactant OF product per unit time is known as rate of a réaction. a 5. Overall order of the reaction =142=3. mM 4 ety 1 e 532 a The reaction is of second order. 8. Let ekIAIs ‘According to question; 27r=k (3A}" therefore, MN G3 3°, Order of reaction, a =3 co KAT’. a OF Rate = KTAI BI? =: @ 10. Eich: step: of a. complex: reaction’ is called » "elementary step of the reaction, © 1 ASB —9P Givens =k (A! (6F Order ive A et Order wre B=2 2 Overallorier=242=2 a 2 2 GD k=5.5 x10 st For first order reaction, ty, 0993 wn = 0693. ina 55x10" =0.426 «10's =1.26 1085 (iy normat 12 QW wit me fate! tawe for * concentration of reactant and for the: conditions when. concentration is doubled or reduced to halt. (il) Now: compare the normat concentration condition’ with changed ‘concentration to find the effect on rate. LetRis the initial concentration of reactant and is the initial rate. So, the rate law for the second onter reaction, 3 Rate, eR WWChemical Kinetics 48, When the concentration of the reactant is doubled, ie, Re=2R The new, Take,’ ccf]? = [2K]? = 4 IRI? oi) From Eq. (and (ii AR Nope ae aire 4 “je, the rate becomes four times on doubling “13. the concentration is reactant, a (i) When the concentration is reduced to half, pak 2 New rate, (7 a(R} -f From Eq. (i) and Gi), eR wet ght That is the rate becomes one fourth of the initial rate when concentration is réduced to half, (1) Rate’ faw. It is an’ expression which gives relationship between rate of reaction and concentration of reactants.. For -a. general teaction, aA + bB + eC +dD. fay tal” ang” a » The above equation is known as rate law or 14. rate expression; In the’ rate ‘equation. &° is: known as “rate constant. It is equal: to the. rate of reaction, when Concentration of reactants is equal to 1. (i Zero onder «lay {ii) Second order (2) Rate expression’ is'a way of expressing rate of reaction in terms of concentration of reactants, @.g:, for a general reaction’ aA +bB=rcC +d Rate =k EA)" [6] a Rate. constant (4) is equal to the rate of reaction when molar concentration of resictants is unity. Its units depends upon the order of reaction. (1) 22, 15. The rate of reaction will become one fourth. a The units of rate constant is L maf 5"! (second. . order reaction) a 16. Refer ans. 14. | @ IZ. i) Refer ans. 1. . o (ii) Each step of a complex réaction is called the elementary step of the reaction. - (1) 18. Order of a reaction is defined as the sum of the exponents of the concentration. terms . in experimental rate law of reaction: 0 It can be zero, 1,2,3 or fractional. Malecularity is the number of reacting particlés, which collide simultaneously to bring about the chemical change, it is a theoretical concept. ft is always a whole number. 4) 19. (i Units of k indicates that the reaction is of second order reaction. cy (ii) Units of & indicates that it #8 a’ fist order reaction. a, 20. (i) The slowest step. in_ the action. mechanism isthe’ rate determining step of a reaction. aM Gi) Molecularity of a reaction refer ans: 18. (1) 21. Since, the reaction i of first order'wrt A and second order wet B, the rate law’ can be given as Rate =k [A] [6]? (i) When the concentration of & is increased to three times (38), the rate Would be Rate, =k [Al (38)? Rate, = 9k [Al [8] 2g x Rate ++ Rate is increased by 9 times, aw Gi)’ Rate, =k [2.4] [28)7 Rate = 2x 2x 2xk (ATI)? =8 x. Rate. Rate js increased by 8 times. 2 ay Factors influencing the ‘ate oF a chemigal: reaction are {) Nature of reactants Different feactants require different amount of energies. for breaking the: ‘old bond “and for. theChapterwise CBSE Solved Papers Chemisiry + formation -of new borids. Hence, the (ii) Rate of reaction is defined as the change reactivity of a substance is related to the in concentration of reactants or products: ., ease with which the specific bonds are per unit time. Its units mots! 4p broken or formed. e.g., amy 2NO + O,——» 2NO, (fast) 28, (i) Order of a reaction Refer ans. 18. (1) Unit for first order rate constant: a 2CO + O,—» 2CO, tslow} of reactants: Rate of reaction. is directly proportionat to the concentration of the reactants. Gry mot L! Son i wom =k [rol v Gi) Temperature Rate of reaction, increases 8 with increase in temperature, (ey tiv) Catalyst Catalyst incréases the rate of ¥ + eae 27. The rate constant, & is equal to’ the rate of a reaction by lowering down the activation reaction when the concentration of the energy. I provides an alternative path to reactants is unity. tt depends upon the order of the reaction. (vay reaction and temperature, Units of & @ 2., Average. rate of reaction is defined as the 4) For a zero order reaction, rate=k [R]? change in the concentration of any one of the . reactants or product per unit time. a ‘Average rate of reaction 4 aay Change in ‘concentration fea mol | 7 Time interval ) For first order reaction, Rate =KiRl = (1/2) ZAIRE es Rate_mol Os at “RL Imei “Instantaneous rateof reaction at any instant of. ~ time is the rate of change of concentration of 28. 2NH, (g) == N,(g) + 3Hy(g) any one. of the. reactants. of, products at that Rate of reaction. Particular instant of time, Fora reaction, RP : “a : aie, A) aN) 1 ath) im dt dt AL ETAL r= very small interval of time) For zero order reaction, 24. = rate of reaction =k= 2.5x°10"* mol U's AIND 2 Bot Sra) 1 isthe ruber fee gis the sum atthe Bt Te Or moles ay molesiles: undergoing | powers of the simultaneous’ © | concentration AMO 32.5 104 collision in the elementary | teres in the sate At rection [ow expression. 2758104 mol Cts ay fii) [has Only: whofe’ number |. can have zet0, pals uiholeramber of” 28." Factors affecting rate ofa chemical reaction ~ frdetional values, (i) Nature of the reactants i cies theoretical concept. [it is deterined ii) Concentration of the reactants I _pevipentaly Gi) Surface area of the reaictants 2) {iv) Temperatire “at “which” reaction ‘takes 25.._() Elementary step in a reaction place. Refer ans. 17 tiit a ) Presence of a catalyst. @Chemical Kinetics 30. 31. 32. 33. 34. 35. {HA catalyst deczeaises the activation energy. It provides. an. alternative path to the reaction. @ (ii) A catalyst increases the rate of 2 reaction therefore, rate constant of the reaction increases. ao Slowest step in the rédctiori mechanism is the rate determining step. Therefore, according «10 rate=k ENOPIH]. rate law, @ Rate law or rate equation ft is the expression which ‘relates: the rate. of. reaction with concentration of the reactants, ¢.g., a CHCh + Ch——> CCl, + HCL t Rate = KICHChI EChI? w Rate aww for any reaction cannot be predicted by mefely looking at the balanced chemical equation, i.e, theoretically but it is doterniined experimentally. Rafer ans. 27: @. PHI = Hye by 2mole <1 molt mol Equilibrium concentration of Hi =0.96 molL"* Equilibrium concentration of Hi= 80% ofx= 0.96 molt" {because 20% HI is dissociated. 0.96% 101 mol Lt 9.96% 10024 2 mat C are mel a Concentration) of. - HE’: dissociated — at equilibrium = 20% of 1:2 mol L™ oe 20. B= * E2= 0.24 mol 700 2 O.12molt- | (1 The ‘reactions in’ Which’ rate’ of reaction is indeperiient of. the concentration of the reactants is called zero ordet reaction: ay Rate [R)°=4; where, & is rate constant, “I = Molt Thus, its unit ismol Ut s 5 m 36. Order of reaction Units of k Zoro order reaction mot Os 37. First order reaction st Second otder reaction © mol" Ls“* nth order reaction (mol eee ‘ \ 1 re aa2t \2 } Na (gh. +) Oz 4g) 4 INO Initially 080 0.20 0 Atequil.(0B0~ 8 (0.20~ 2x “Ino? “ax? * (WKA[O3] 0.80 —9) 10.20 ~ x) =1.0% 10% a ax? 51.0 x 10° (0.16 ~0.80 x =0.20 x + x°). A107 x? 2016 =x et or 4x 10% x! x? + x-016 50, 4x10? 24 x-016=0.. Using quadratic equation, * Abt yb? =4ae 2a and where, a= 4% 10°, ay? =016 4x 4x10°x 2% 4x105 +6.3%107; x==6.3x10 @ (Negative value is not taken as concentration can never be negative) INO} 46.3% 10 21.26% 10 mol/L (Nal =0.80 ~(63x 10) =0.799 mol/L. [0,] =0.20- (6.3 x 107%) =0.199 mol/L. a338. Chapterwise CBSE Solved Papers Chemistry. () Rate taw may be written as Rate =k [NOP [Cl]? ‘The initial fate becomes Rate}y =kINOP {Ch}? Corinparing experiment’1 and 2,” Rate =k (015)? (015) =0.60 Rate, =k (015) (030)" =1.20 Dividing Eq. (i) by Rates 5)? (0.30)" (Rate, (045)? (045)" 0.60 or. 2g! ast +-Order with respect to Cl, Comparing experiment 1 and 3, Rate, =k (0.15? 0215)? 2060 (Rated, =k 0.30)" (0.15)? = 2.40 Dividing Eq. (i) by @ . ‘ © .300 0.15)" Ki045) 045)". 060 Rated, or : par ‘Thus, order with respect to NO is 2. Rate law =k INOF [Cl] a (ii) The rate law for the reaction. Rate =kINOF {Ch © Rate constant. can be “calculated by substituting: the. value of rate, INO} and “Teh fr any expe 0.003375 2177 77 mol? t? mit NOP ICL 2177.77 x 0.25)? (0.25) =2.78 Mimin om 39. {i} Rate law may be written as Rate =k {APP {87 Comparing experiment (2) and (3), we get 0.2? (O27 =3x 107 (0.2)? @.4" =6x10°7 Dividing Eq. (ii) by (i), we get (Rated, 10.2 ©." - 6 x10° Rated, 10.2) 0.2 3x 107 2952 qi ~ Comparing experiment (1) and (2): (Rate, =k OW? 0.0" 1.5 «107% (0.2)? 0.2)" =3.0<10"> iv) Dividing Eq. (iv) by (i 3x10" _ K (0.2 0.2) 15x10? Ki0-1° 0.08 2 =(2)" (2)* qet 122? where, 2 pe This, the raté law is rate =k EAI® (8 =: 6 ae (ii) Rate =k {8} Rate. 3x107) 0D, 215x103 mint ay (iB —S C+E slow) 15 reaction, which is consistent with the rate law which is rate =k [6h ‘ (ty 40. Let rate law in terms of rate of formation of D be aka oP oar Baie: Gi) 7.2% 107 =k (0.3) (0.2) (ii) 2.88% 1077 =0.3)* (0.4) tiv) 2.40% 107 =k0.4)" (1) Divide Eq. (iv) by (i, 424" cast a). the: possibleChernical Kinetics Divide Eq. (iii) by ci, Order with respect to. =1 Order with respect to B = 2 Rate law = al =k IAL! DD On pulting the value of ‘a arid ‘b into any” equation, say () . 6.0% 107? M min”! =k (0.1. M) (0.1M)? . k= 6M? min a 41. Let rate law in terms of rate of disappearance of NsO be rate =k (N;Ogl" {34x10 =k [1.13x 1077]" (i) 25 x 10°5 =k [0.84x 107)" Divide Eg. (i) by ti), we get 34x10 _k[1.13x 1079" 25x10 k[0.84x 10)" 1.36= (1.35]" coast, a (i) Ortier of reaction =1 «1i2) {ii) Rate =k [NOs] (Rate faww expression (1/2) (iif} Rate constant, __ Rate ~ 34x 1073 ino 1.13%: 107 = 30.09 «10°? min” a 42, (i) A reattion is second order in A and first order in 8. 5 (a) ilferentia ate equation = Rate” =i care. at SUES ay tb) When the concentration of A is increased: three times i.e, 3A thea Rate =k [3AVI6) = OMAP 1B] =9[Rate} This show that rate will increase 9 times the initial rate. a (©) When concentration of both A and & is doubled then, Rate =k [2APRBE = BfRatG This show that rate will increase 8 times the initial rate. a iy AP t=0a 0 tet fax oe For 30% decomposition, it take 40 niin, which means after 40 min, reactant left 70% of its initial concentration. ‘Thus, aay or tog 4. 8a aay, 2.303, = 23 og 1.428: a0 8 (ay. = k 2000891 mintTopic ? Integrated Rate Equation, Pseudo First Order Reaction Previous Years’ Exami nation Questions: Practice the Real Questions 1 Mark Question . (i) What fraction of the reactant will be * ft after an hour of the reaction has 4: A reaction is 50% complete in 2 hand 75% i i complete in 4h. What is the order of the occurred? Welhi 20116) reaction? TAM tndia 20061 3. Whe re pcudo fs onde estonst Cie . cone example of such reactions. 2 Marks Questions ° {All Fadia 2012) 2. A reactant has a half-life of 10 min. 4. Ina first order reaction, the concentzation of, 6). Cafcufate the rate constant for the first the reactant is reduced from 0.6 mol L"! 19 order reaction.Chemical Kinetics 6. a 8. 7 0.2 moi Lin 5 min, Calculate the. rate constant of the reaction. [AM India 20110} The rate constant for a zero order reaction in A i80,0030 mol Ls. Haw long will it take for the initial concentration of A to fall from 0.10 M10 0.075 Mz [Delhi 2010; Foreign 2010) Show that for a first order reaction, the time required for half the change half-life period} is independent of intial concentration, [Delhi 20096 ‘A first order ‘cecomposition reaction takes 40 min far 30% decomposition. Calculate itStj;, value, (Dethi 2008; All India 2008] Derive the general form. of expression, for the halflife of frst order reaction. (Delhi 2008; AH India 2008] Calculate the altslife of a first order reaction whose rate constant is 200 s"'. [Welhi 20071 3 Marks Questions 10. 11. Nitrogen pentoxide decomposes according to the equation ; 2NLO; tg)» 4NO} (g)+ Onfe) “This first order reaction was allowed to proceed at 49°C and the data below were collected : (i). Calculate the, rate’ constant.” Include units with your answer. Gi), What’ Will be ‘the’ concentration - of NO, after 100 mint (iii) Calculate the initial rate of reaction. (Delhi 2013) ‘The thermal decomposition of HCOOH is a first order reaction with a rate constant of 24x 10 st at a. certain. temperature. Calculate how jong will it take for three. fourth of initial qtaritity of HCOOH to decompose? (log 0.25 = 0.6021) TAM India 20111 42, 13. a 15. Nitrogen pentoxide decomposes according to equation : > 2N,05(g) +9 4NO3(g)+ Osg) This first order reaction was-allowed te proceed at 40°C and the data below were {Foreign 2011) collected : (i) Calculate the rate constant. Include units with your answer. i) Calculate initial rate of reaction. After how many minutes witl [N,O,] be equal to 0.350 M? A first order reaction has a rate constant value of 0.00510 min”! f we begin with 0.10 M concentration of the reactant, how much of the reactant will remain aiter 20h? {Foreign 2014, 2009, Delhi 2009; AN India 2009}. The décomposition of phosphine; PH; pro- ceeds according to the following equation. APH,(g)—> Pytgi+ GH3{e) It is found that the feaction follows the following rate equation Rate =K{PH\} The half-life of PH, is 37.9 sat120°C, {How much time is fequited for 2th of * "PE to decompose? (ii), What fraction of the original sampie of PH, remains behind after 1 min? {AH India 2010; Foreign 2009} The decomposition of a compound “is found to follow a first order rate law, If it takes 15 min for 20% of original material to react, calculate 4) the rate constant: Gi the time at which 10% of the oxiginal. material rémains unreacted. (Dethi 2010C}Chapterwise CBSE Solved Papers Chemistry 16. Ina pseuicia first ofdar hydrolysis of ester in water, the following results are obtained : tinsecondss 0.30 60 90 (Ester) M O55 0.31 O17 0.85 {i Calcutate the average’ fate of reaction between the time interval 30 to 60 s, (i) Calculate the pseudo first order rate constant for the hydrolysis of ester. TAIL India 20100) 17, The halflife for a first order reaction is 5x10%s. What percentage of the initial reactant will react in 2 hours? Calculate. [All Indin 2009C] 18. A first order reaction is 20% complete in 5 min. Calculate the time taken for the reaction to be 60% complete. [Al India 2009C} 19,- Hydrogen peroxide, HO, (aq) decomposes toH,0 () and 2 (g}in a reaction that is of first order in HO, and has a rate constant, 21.06% 10°) min GY How long will it take 15% of a sample ° of HJO} to decompose? 4). How [ong will take 85% ofa sample of H,O,to decompose? [Foreign 2009) * 20. The halflife for decay of radioactive '¥C is 5730 yr, An archaeological anifact containing wood had only 80% of the *C found in a » living tree, Estimate the age of the sample. (Delhi 2008; Al India 2008; Foreign 2008). 21. first order reaction is 15% complete in 20 min, How tong will it take to be 60% coniplete? (Delhi 2007, 20060) 22, The rate constant fora first order reaction is 60s. How much time will it take to reduce the concentration of the reactant to ib thof its initial value?” [Atl tadin 20071 23. Show that in case of first order reaction, the lime required for 99.9% of the reaction to complete is 10 times that required for half of the reaction to take place. (log 2= 0.3010) [Foreign 20061 24, Prove that’ the- time’ required for: the completion of 3 of the frst order is wwice the time required for the completion of half of the reaction. [AI India 2006) Step-by-Step Solutions 1. First order reaction, SINCE tos, = 2sor a 0.693"_0.693.. 2, katt OO = 9 o Oba GN=No iy =No(3 a * Qwhere; Ng = initial amount of reactant and N = ‘amount of reactant left after time, 0 3. Pseudo first. order reaction The’ reaction which is bimolecular but has’ order one, is, called pseudo first order reaction. e.g., acidic hydrolysis of ester. . a CHCOOC,H; + H,0 == CH,COOH + CHO (1) 4, Rate constant, 2.303 [Rlo, SO lo Given, Rly =0.6' mol L! [9 =0.2 mol Ct =5 min p= 2303 1500 =r los 5°85 aayChemical Kinetics: 22393 tag's (log 320.4771) k 22303. G:a7710.2197 mia ‘ « = 5. For zero order reaction, rate constant; ke Bhat =i Given, [R}p =0. son [R}=0.075 Mand (1) k= 0.0030 mol 1" s:! (Alp = 1RY o When, tty then [i = 8] Re 303 |, te on fa o 2 for’ a first: order. reaction, halflife: period's constant: It" is") independent’ of initial © concefitration of thé reacting speciés:.“-< ( TQ). 60 Totind ts frst calculate & by using the 2800 FAB i Ha, ferme S00 ay Gi) Caleblate. toa by Using. the formulas ese GB : Giver; t= 40min 30% decomposition means [Rjp = 100 Mand [R]=70M_. (1/2) 2 Ro (RE 03 jog 100 70 logt.42a 303x 0.1548 40 = 0.0089:min (v2) 0.693 = 5. t= = 7.865 mit 2 Bagge 277-865 min (1) 8: Refer ans. 6, - : (2) an 10s a 200 loglSa oR 400M .289 M, t= 20 min 03, 0.400. log ——— 0.289 01625 min = 40min, IRT=0.209M" 2303 5, 10.400] 30. °8 0,209} [0.209], Similarly, whent =60min, (RI =0.151 2303, {04001 60. 0. 454]. 2303 0.4337 60 = 0.01623 mi .01625 + 0.01623 + 0.01623 3 =O. 016236min"! E ao. ‘fi k= = 2303 jog (le t 8 TR 1.6236 « 10?Chapterwise CBSE Solved Papers Chemistry 0.4 _1, 6236 i 2.303 04 <> =antilog 0.705 (Rl = 0.705 =0.07889M > (1) Initial rate = k [N,Os} 0.016236 x04: =0.00649 molt" st a LL, Given, three. fourths: of- initial quantity of HCOOH is decomposed means that initial concentration ={Rjgy °° Concentration ater time t=, ae st Thus, for the first ordeét reaction, rate constants" k 2 to Se ra) nit a 2a 1073571 = 2303 jg Ro Ce yd, 2,303 tent «logs 0 ae ios 8 o or = 208 0.6021 “Zax 10 . 77% 1075 wy 12, (i) Referans. 104." a (ii) Refer ans. 10 (ii), ay 2,303 |. lo’ Gin k= 22S Jog Ro eR o 18. Given, k= 0.00510 min”! [Ry =0.40M and t=3h=3%60 min, 3 jlo Jog flo cc) 3x60 = 2303. jog 10.10} wo 000510 i) 0.1] _+80x 0.00510: k = 180 0.00510. 9. oa 2 | 2.303 3986 [33] antitog 0.3986 a =0.0399M ay 14, Given, tyy'=37:95, Initial concentfation =[Rlg Concentration after tine, t = fe (because 3h of the PF; is decomposed) o ex 0,893) 0.6934 fyg = 379) 2.303 45 Rol ke, “E logs 2.303 ‘0593 379 : 2.303% 37.9% 0.6020 0.693 am (i t= and = 0.01835" 37.9. 12303 5g lo k ia) 60 = 2303, (Alo “00183. Rl Flo. 60 « 0.0183. or 8 et “3.308 (2) a (Blo - antilog 0.4768 We = 2.997 os)Chemical Kinetics Let the initial amount is 100, (R)=33 x 3% @ 15. (i) 20% decomposition means RyI =100, IR] #100 -20 =80 2.303). Ro k= log 2 t IR} 2,303), 100 = fos 15 80 9148 mint (ty fii) Time. at which: 10% “of the’ original material remains unreacted. {Ro =100, [R} #10 303 jog la RE 100 — 10=1) tog logto=1) 2155.6 mini 03) z 16. (i) Average.rate of reaction between the time interval 30 t0 60 s is = 4.67%.107 st cu) iy 22:303- IR Soll fog Gok =5 tos ae 32303 og 1.774 30. 21.91% 10% Alt = 60's [R] =0.17 M =1.96% 107 "1. Att =90 5, R]=0,085'M 2303+), 0.55 90: 0.085 2,303 = «0.81 3g * 0.8109 Average kee LOX 107 + 1.96.10" + 2.07% 1077 3 = 1.98« 107 @ 17. Given, ty) =5 x10" s 5260x605. 0.693 0.693." ta 3x10" 2303 455 [Aly t (a) 0.693 | _2:303 15, (Aly 5x10* 2% 60« 60...” [A 0.693 x 2x 60x 60 5x10" x 2,303 0433 = antilog 0.0433 wo and k o [A] 1xi00., . aan = 90.497% 4 (Aly © 1.705 os 0 48. When a first order reaction is'20% complete in 5 min, [Aly =100 M, t=Smin [Al=100 ~20=80M | and 2:303 jg, le 1/2) ‘ Al ‘Chapterwise CBSE Solved Papers Chemistry. ° 2.303 tog 100 5.27 BO 2.303 Sa tog 1.25 zs 2:303 0.0969 5 (iy ‘= 0.0446 min" 2) Time taken when the reaction is 60% complete [Alp =100 M and [A] =100 -60 = 40M, k = 0.0446 min"! 2303. (Alo 2.303 jog lo eal 303 ‘jpg 100, t 0.0446 2303 “pose 82> _2:303x'0.3979 0.0446 1 aC) 19. (i) When 15% ‘ofa sample. of H,O, is decomposed, For a first order reaction, 2,303 |, [Rlo ka fogt i Given, k=1.06x 107 min’, [Rp =100M, : TRiafter time, 0 =100 15 285M" (22308. jog 100 160x107" 85 238 tog 1.176 1.06% 107 : 2.303% 0.0704 459.9 min (3) 1,06x 10" os 2 (i) When "85%. ‘of a“sariple’“of HO, is decomposed, [R] =100.-85 =15 M “2.303 | 7,100 2.303 jog 100. Tox 107 Ps = 7303 og 6.667 1.06% 107 2.303 x 0.8240 21790.25 min (4) 2 20, Given, ty, =5730 yi, (Rls =100 and [R]=80._- 0.693 0.693 b tant k o23 12 decay constant tae 5730 yer (2) 2.303 Il. 2303 jog [lo eR 23035730, 100 yj) 0.693 30 2303x5730 og a5 (wy 0.693 2,303 #5730 , 9 969 0693 =184518 yr o 21. Refer ans. 18. 2112.7 min t (3) 22. Given, k=60 57, IR] 27M. i and {R] =. M and R= 55 2.303 |, [Rlo = 22% tog Ba a k (Al ws 1=2:303 jog t0 = 0.03845 Q oo 2.303, - (A 23, k=“ log 2 ay fl < (Reaction is 99.9% completed, it’ means that {Al =100 M and [R}=100 -99-9 30.1M) 2.303, 100 22333 fog 22 to Od 1,2 2.303 2228 logo". aay T / : o) For hatf-life of the reaction. 0.693: ta a 24. Refer ans. 23. Q).Temperature Dependence of the Rate of a Reaction, Collision of Chemical Reaction Previous Years’ Examination Questions 41 Mark Question 1 Define activation energy. TAI! India 2012 2081, 3010, 2000, 2006; Delhi 2007] 2 Marks Questions 2. The rate of most tedctions become double when their temperature is raised from 298 K to 308 K. Calculate their activation energy. (Given, R= 8.314) mol K™) (Delhi 20110; Foreign 2007] 8, The rate ofa reaction becomes four times when the temperature changes from 300 K to 320K. Caleillate the energy’ of; activation’ of the reaction, assuming that it does not change with temperature. - [AH India 2010; Foreign 2008] (R= 8,314 JK" mol) 4. What is the effect of temperature on the rate constant of a reaction? How can this tempera- ture effect on’ rate constant: be: expressed quantitatively? (Dethi 20106 ; All India 2008C] 5. The first ‘ordér. rate'constant. for. the. decomposition of ethy! iodide by the reaction, CyHyi @)—> CH, (e)+ Ag) at 600K “is 1.60x 10°%s:', Its energy of activation is > 209 ki/mal. Calcifaté the rate’ constant of the reaction at 700 K. (all India 2007) 3 Marks Questions 6. The rate of a reaction becomes four times when the temperature changes from 293K to 313K. Calculate the energy of the activation (E,) of the reaction assuming that it does not change with temperature. [Welhi 2010} (R=8314 JK" mol, fog 40.6021] 7. The activation energy for the reaction 2HI(g) —> H3.(g) + Ip(g) is 209.5 k) mol at Practice the Real Questions 581K, Calculate the fraction of molecules having energy equal to or greater than activation energy. [R= 8.31 JK“ mol} FAL Indin 2610}, 8. Fora decomposition reaction the vatues of rate constant k at two different temperatures are given below + ky = 2.15% 107% mol"! slat 650 K ky = 2.39% 1077 Linol”! sat 700 K Calculate the value of activation energy for this reaction: (R= 8,314 JK-' mol”). {All India 2009; Foreign 2009} 9. The . - decomposition. . of. phosphine, PH, (g}—> Py(g)-+ 634g) has the rate law, rate =k [PH] The rate constant is 6.0% 10~ s-* at 300 K and activation..energy - is 3.05% 10° J mol. Calculate the value of rate constant at 340 K . [Given, R=8.34JK"' mol] eth 2008C} 10. The fate of a particular reaction triples when temperature change. from: 30°C to 100°C. Calculate the activation energy of the reaction. (Given, log 3=¢ 0:4771; R= 8.314 |! mol"). ©: [Delhi 2008) Ll; A certain reaction is 50% complete in 20 min at 300 K and the same reaction is again 50% complete in’: mint at 330K: Calculate the activation energy if it is a first order reaction.“ | [R= 8.314 JK*! mol"; log 4=0.6021 TAIL India 2006]. 12. (i) For a first order reaction, show that time required for 99% completion is twice the time required for the completion of 90% of reaction.Chemical Kinetics i) Rate constant ‘k’ of a reaction varies with temperature 'T" according to the equation E, 1 logk = logs =£__{ 1). opt = fog sacral) Where F, is the activation energy, When a graph is plotted for log k vs - a straight fine with a slope of ~ 4250 K is obtained, Calculate ‘£,' for the reaction (R =8.314} K7! mol”). Step-by-Step Solutions 1. Itis extra energy which must be supplied to feaclants so that their energy becomes equal to threshold value ‘and these can change into products, It is denoted by E,.: mM 2. Given, T, =298 K, Ty =308 K and ky/k, =2 and R=8.344) mol“ K7" pfien| Bah oa T30IR| Thy on? = 2 | “3303x8314 | 298% 30 6 10 303% 8314 298x308 E, =52897.78 Jmol! 8. Refer ans: 2: Ans.’ is55327.581 Jmol? gay Hint Be 4,7, =300K, Ty =320K 1 . 4. Most of the chemical réactions are‘accelerated by increase in temperature. it has been found that: fora:chemical «reaction with rise in temperature by 10%, the rate constant is nearly doubled. a Temperature coefficients) Rate constaht (+10) K Rate constant at TK 5. Given, ¥, =600 K; T)=1700k, £, =209 kJ mof! = 209000 Jmat"t and ky = 1. : ow log = 209000 x 100. ky 2,303 x 8314 600 x 700 = 2.5989 { _ antilog2.598 ky =3,971K 10? 971% 107. ky 56.35 1079 5 w 6 Given, 7, 293K, T2313K) 7k, R,=8314. Vk" mol", fog 4=0.6021: ky By [R21 log $2 = na | ot a) Bi, Sal a o Eo 313 ~29: log d Fa) 313-293 °8 satan | ° 20 1.6021 = mat | 20 oo till i E. 0602119147 «91709 205. =5286294) mol“? w 7, Given, €, =209.5k} mol" = 209500 J mol" R=8.314 JK" mol! T =581K Fraction of motecules,= ~é° 2,303 RT 7209500 © 2.303% 8.314 x 581 a). logx= log xChapterwise CBSE Solved Papers Chemistry log x =~18.8323 a x= antitog 19.677 xeta7ixio” ow 8. Given, & =2.15 x10 Linot's"!, 7, =650 K 2.39 x 1077 Lot's, T, = 700K, 314 Kt mol? 10. Refer ans. 2. a 1. CP) 0 First find & and & at temperatures * 300 K and 350 K respectively by using =} log 2.39%1077, Ey At 300 K, ty) =20 min, 2asnio® 2.303% 8314" 120-993 inet 700 - 650 50x 700 At 350 K, ty, #5 min, : 0.693. fog tie Oy pa min o 2303%6,314 455000 i) thet £, x50 oe 2) (3 o 1.0457 =-—__ + 457 5503x0314» 455000 Ky) 23038) Te Te 1.0457 «2.303% 8.314% 458000 yy log 22 wf [ 359-300 50 3 "2.303% 8.314 | 300% 350 =182201.71} mol" log 4= E,x50 9, Given, » T)=300K, T, =310K, “3.303% 8.314% 300 350 £, =3.05 «10% mot 06020 x 2303x8314 300 x 350 ee 5 30 2 otk S By60 «100 Ska =F £, =24205.8 Jmol" = 24.206 kJ mol k is ts): aS 2 Bu), 12. (i) For a first order reaction: ky} 2303R[ 7, Te w A Pp te) 3,05 «10% [2ig=200) boo ke J 2,303 x 8.314 | 300 x 310. a} Suppose. ‘t’. time iaatied for 99% i) Sas ei0% x10 completion of reaction then, i = = ee 2303x8.314% 300" 310 x=99% of a : (a-9=1% ofa 21.7126 a 1 peas 100. 2303, 2 k Fane 2303 jog 2* 100 a a2 gto? t pp) @ a wChemical Kinetics (6) Suppose ‘t’ time is needed for the 90% ®s, completion of the reaction then, x=90% of a dX 510% of a 1022 100. 10 Hence, the time required for 99% - Completion of first order reaction is twice the time required for 90% completion. (i) We know, Rl logk=tog A ~==4_( 1 2303 (7 a Where, £, = Activation energy Temperature (in K Gas constant This equation is analogous to straight line equation y=mx +c 1 250K uw On plotting y ws x, we get straight line iwith slope ‘m’ and intercept ‘c’. Similary, when we plot logit: v/s 1/7, we get Slope: 4250 Ks . SOK aa ay = E, = 4250 x 2303x8314 =8137536 J mol7!