Chemical KineticsContents Topic Page No. Theory 01 - 06 Exercise - 1 07 - 24 Exercise - 2 25 - 33 Exercise - 3 34 - 38 Exercise - 4 39 - 41 Answer Key 42 - 43 Syllabus Chemical Kinetics : Rates of chemical reactions; Order of reactions; Rate constant; First order reactions; Temperature dependence of rate constant (Arrhenius equation). Name : ____________________________ Contact No. __________________ ETOOSINDIA.COM India's No.1 Online Coaching for JEE Main & Advanced 3rd Floor, H.No.50 Rajeev Gandhi Nagar, Kota, Rajasthan 324005 HelpDesk : Tel. 092142 33303 CHEMICAL KINETICS Rate/Velocity of chemical reaction : The rate of change of concentration with time of different chemical species taking part in a chemical reaction is known as rate of reaction of that species. c mol / lit. = = mol lit–1 time–1 = mol dm–3 time–1 Rate = t sec Rate is always defined in such a manner so that it is always a positive quantity. Types of Rates of chemical reaction : For a reaction R P Average rate = Total change in concentrat ion c [R] [P] = = – = Total time taken t t t Instantaneous rate : rate of reaction at a particular instant. c dc d [R] d [P] Rinstantaneous = tlim =– = = 0 dt dt dt t Relation between reaction rates of different species involved in a reaction : For the reaction : N2 + 3H2 2NH3 1d [H2 ] d [N2 ] 1 d [NH3 ] = 3 dt = Rate of reaction = dt 2 dt Order of reaction : Let there be a reaction m1A + m2B products. Now, if on the basis of experiment, we find that R [A]P [B]q Where p may or may not be equal to m1 and similarly q may or may not be equal to m2. p is order of reaction with respect to reactant A and q is order of reaction with respect to reactant B and (p + q) is overall order of the reaction. Integrated rate laws : Zero order reactions : For a zero order reaction General rate law is, Rate = k [conc.]º = constant If C0 is the initial concentration of a reactant and Ct is the concentration at time ‘t’ then Rate = k = C0 Ct ' t' or kt = C0 – Ct or Ct = C0 – kt Unit of K is same as that of Rate = mol lit–1 sec–1. First Order Reactions : k= C0 2.303 log C t t Half life time (t1/2) t1/2 = ln2 0.693 2.303 log 2 = = k k k ETOOSINDIA.COM India's No.1 Online Coaching for JEE Main & Advanced 3rd Floor, H.No.50 Rajeev Gandhi Nagar, Kota, Rajasthan 324005 HelpDesk : Tel. 092142 33303 CHEMICAL KINETICS # 1 Second order reaction : 1 1 – Ct C 0 = kt Pseudo first order reaction : A second order (or of higher order) reactions can be converted into a first order reaction if the other reactant is taken in large excess. Such first order reactions are known as pseudo first order reactions. For A + B Products [Rate = K [A]1 [B]1 2.303 b(a x ) k = t (a b) log a (b x ) Table : Characteristics of Zero, First, Second and nth Order Reactions of the Type A Products Zero Order Differential Rate law First-Order Ä[ A ] = k[A]° Ät (Integrated Rate law) [A]t = [A]0 – kt [ A ] = k[A] t In [A]t = –kt + In [A]0 nth order Second-Order [ A ] = k[A]2 t 1 1 = kt + [A]t [ A ]0 Ä[ A ] k [ A ]n Ät 1 (A t ) n1 1 ( A 0 )n 1 (n 1) kt Linear graph [A]t v/s t Half-life t1/2 = In [A] v/s t [ A ]0 2k t1/2 = (depends on [A]0) 0.693 k (independent of [A]0) 1 1 v/s t [A] ( A t )n1 1 t1/2 = k[ A ] 0 t 1/ 2 v/s t 1 ( A 0 )n1 (depends on [A]0) Methods to determine order of a reaction : By comparison of different initial rates of a reaction by varying the concentration of one of the reactants while others are kept constant r = k [A]a [B]b [C]c if [B] = constant [C] = constant then for two different initial concentrations of A we have r01 = k [A0]1a r02 = k [A0]2a r01 r02 [A ] 0 1 [ A 0 ]2 a Method of half lives : The half lives of each order is unique so by comparing half lives we can determine order for nth order reaction t1/2 t1 / 2 t1' / 2 1 [R0 ]n1 (R '0 )n1 (R 0 )n1 ETOOSINDIA.COM India's No.1 Online Coaching for JEE Main & Advanced 3rd Floor, H.No.50 Rajeev Gandhi Nagar, Kota, Rajasthan 324005 HelpDesk : Tel. 092142 33303 CHEMICAL KINETICS # 2 Methods to monitor the progress of the reaction : Pressure measurement : Progress of gaseous reaction can be monitored by measuring total pressure at a fixed volume & temperature. Let there is a 1st order reaction A(g) k= nB(g) P0 (n 1) 2.303 log nP P t 0 t Final total pressure after infinite time = Pf = nP0 Formula is not applicable when n = 1, the value of n can be fractional also. Do not remember the formula but derive it for each question. By titration method : By measuring the volume of titrating agent we can monitor amount of reactant remaining or amount of product formed at any time. It is the titre value . Here the milliequivalent or millimoles are calculated using valence factors. V0 = vol. of titrant used at t = 0 Vt = vol. of titrant used at ‘t’ V = vol. of titrant used at t = k= [this is exclusively for HCl.] V V0 2.303 log V V t t Optical rotation measurement : It is used for optically active sample. It is applicable if there is at least one optically active species involved in chemical reaction. where are r0, rt, r are angle of optical rotation at time K= t = 0, t = t and t = Y Y0 2.303 log t r rt Effect of temperature on rate of reaction : In early days the effect of temperature on reaction rate was expressed in terms of temperature coefficient which was defined as the ratio of rate of reaction at two different temperature differing by 10ºC (usually these temperatures were taken as 25ºC and 35ºC) T.C. = K t 10 2 to 3 ( for most of the reactions) Kt Arrhenius theory of reaction rate : Enthalpy (H) HR Ea1 Ea2 Threshold enthalpy or energy Reactants H = Hp – HR = Ea1 – Ea2 HP HR = Summation of enthalpies of reactants HP = Summation of enthalpies of reactants H = Enthalpy change during the reaction Ea1 = Energy of activation of the forward reaction Ea2 = Energy of activation of the backward reaction Products Progress of reaction (or reaction coordinate) ETOOSINDIA.COM India's No.1 Online Coaching for JEE Main & Advanced 3rd Floor, H.No.50 Rajeev Gandhi Nagar, Kota, Rajasthan 324005 HelpDesk : Tel. 092142 33303 CHEMICAL KINETICS # 3 Ea – E’a = lowering of activation energy by catalyst. Mechanism of a reaction : Reactions can be divided into Elementary / simple / single step Complex / multi-step ELEMENTARY REACTION : These reaction take place in single step without formation of any intermediate T.50 Rajeev Gandhi Nagar. Rajasthan 324005 HelpDesk : Tel.E. the function of the catalyst is to lower down the activation energy. having energy greater Ea rate e E a / RT dependence of rate on temperature is due to dependence of k on temperature. The rate law for the elementary reaction aA + bB products rate = k[A]a[B]b. Molecularity is defined only for the elementary reactions and not for complex reactions. Ep P. Ea E'a HR HP Products Reaction Coordinate Molecularity and Order : The number of molecules that react in an elementary step is the molecularity of the elementary reaction. E’a = Energy of activation in presence of catalyst.S.E. That is.E. Ea = Energy of activation in absence of catalyst. the orders in the rate law equal the coefficients of the reactants.No.COM India's No. k e Ea / RT k Ae Ea / RT [Arrhenius equation] General characteristics of catalyst : P. Er Reaction coordinates For elementary reaction we can define molecularity of the reaction which is equal to no of molecules which make transition state or activated complex because of collisions in proper orientation and with sufficient energy ETOOSINDIA.T2 Fraction of molecule T2 > T1 T1 Ea e E a / RT represents fraction of molecules K. For an elementary reaction. 092142 33303 CHEMICAL KINETICS # 4 .1 Online Coaching for JEE Main & Advanced 3rd Floor. No elementary reactions involving more than three molecules are known. where a + b = 1. because of very low probability of near-simultaneous collision of more than three molecules. H. A catalyst drives the reaction through a low energy path and hence Ea is less. Kota. 2 or 3. of [R]. a a –x a – xeq.1 Online Coaching for JEE Main & Advanced 3rd Floor.COM India's No.S. or having some mechanism.D. MECHANISMS IN WHICH RDS NOT SPECIFIED : STEADY STATE APPROXIMATION : At steady state d [intermediate ] =0 dt COMPLICATIONS IN 1st ORDER REACTION PARALLEL 1st ORDER REACTION OR COMPETING FIRST-ORDER REACTIONS 1 1 1 Teff = T1 + T2 k1 [B] = k [C] 2 Ea (remember) (remember) E a1 k 1 E a 2 k 2 k1 k 2 REVERSIBLE 1ST ORDER REACTION ( both forward and backward ) A t=0 t=t t = teq. [P] or catalyst with the help of same equilibrium step given in mechanism. ETOOSINDIA. CALCULATION OF RATE LAW/ ORDER MECHANISM IN WHICH R. Rajasthan 324005 HelpDesk : Tel. product or catalyst on reactant side rate law of R. 092142 33303 CHEMICAL KINETICS # 5 .S.S.50 Rajeev Gandhi Nagar.D.D.COMPLEX REACTION : Reaction which proceed in more than two steps.S. we have to specify [ intermediate] in expression of rate law in terms of conc. B 0 x xeq. H.No. involves only reactant. = rate law of reaction RDS is having intermediate on reactant side To calculate order. intermediate E Reaction coordinates For complex reaction each step of mechanism will be having its own molecularity but molecularity of net complex reaction will not be defined. Kota. GIVEN If R. ( sequence of elementary reaction in which any complex reaction procceds) T. (remember) SEQUENTIAL 1ST ORDER REACTION OR CONSECUTIVE FIRST-ORDER REACTIONS k t=0 t k 1 2 A B C a 0 0 a–x y z k 1a y = k k { e k1t – e k 2 t } 2 1 all first order equation (remember) k1 1 tmax. Rajasthan 324005 HelpDesk : Tel.50 Rajeev Gandhi Nagar.No.COM India's No. H. 092142 33303 CHEMICAL KINETICS # 6 . 1 n x x t eq. Kota.kfa x = k k f b kf + kb = 1 e ( k f k b ) t (remember) x eq.1 Online Coaching for JEE Main & Advanced 3rd Floor. = k k n k 1 2 2 CASE- (remember) k1 >> k2 k 1 A CASE : B k 2 C k2 >> k1 Conc [C] a [B] t [A] ETOOSINDIA. xA + yB zC.PART .COM India's No.0 x 10—4 mol L–1 s–1 (B) 2. (B) 8 × 10–2 M s–1 (C) 2 × 10–2 M s–1 (D) 7 × 10–3 M s–1 For the reaction : N2 + 3H2 2NH3. In the formation of sulphur trioxide by the contact process. Hence rate of disappearance of O2 is : (A) 50 g min-1 (B) 40 g min–1 (C) 200 g min–1 (D) 20 g min–1 A-6.25 x 10—4 mol L–1 s–1 —4 –1 –1 (C) 3.2 × 103 (C) 2.1 0 20 40 60 80 100 Time/second (A) 4 × 10–3 M s–1 A-7. 3. Rate of formation of SO3 in the following reaction 2SO2 + O2 2SO3 is 100 g min–1. A reaction follows the given concentration (M)–time graph.5 = dt dt dt (B) 3.1 Online Coaching for JEE Main & Advanced 3rd Floor. If (A) 1. 3 (C) 3.3 where –ve sign indicates rate of disappearance of the reactant.50 Rajeev Gandhi Nagar. 2SO2 (g) + O2 (g) 2SO3 (g) d(O 2 ) The rate of reaction is expressed as – = 2. What is the rate of formation of ammonia. 1 A-3. d[C] d[B] d[ A ] = then x.1. 2. 2.7 × 103 (D) 0.3 0. 3 xA yB d[B] log dt = log dt + 0.4 0.I : OBJECTIVE QUESTIONS *Marked Questions are having more than one correct option. The rate for this reaction at 20 seconds will be: 0. x : y is : (A) 1 : 2 (B) 2 : 1 (C) 3 : 1 (D) 3 : 10 A-4.9 × 103 ETOOSINDIA. For a hypothetical reaction A L.8 × 103. (mole per litre per sec.) (A) 1. If the rate of disappearance of hydrogen is 1. 092142 33303 CHEMICAL KINETICS # 7 .No.5 x 104 mol L-1 sec-1 dt The rate of disappearance of (SO2) will be (A) 5. Kota.2 0. the rate expression is dC A rate = – dt (A) Negative sign represent that rate is negative (B) Negative sign indicates to the decrease in the concentrations of reactant (C) Negative sign indicates the attractive forces between reactants (D) None of the above is correct A-2.5 0.75 x 10 mol L s (D) 50. H. 2 In the following reaction : d[ A ] (D) 2.0 x 10—4 mol L–1 s–1 A-5. Rajasthan 324005 HelpDesk : Tel. Thus.8 × 103 (B) 1. Section (A) : Rate of Reaction A-1.y and z are : = 1. is given. The rate constant of reaction changes when : (A) Volume is changed (B) Concentrations of the reactants are changed (C) Temperature is changed (D) Pressure is changed B-5. the units of rate constant and rate of the reaction are same (A) First order reaction (B) Second order reaction (C) Third order reaction (D) Zero order reaction B-6. The order of reaction is given by : (A) 1 (B) 1/2 (C) 2 The rate constant of nth order has units : (A) Litre1–n mol1-n sec–1 (B) Mol1–n litre1-n sec 2 2 (C) Mol1 n litre n sec–1 dA = K [A]½ [B]1/3 dt (D) 13/12 (D) Mole1–n litren-1 sec–1 B-4. ln the reaction. Kota. Rajasthan 324005 HelpDesk : Tel.50 Rajeev Gandhi Nagar.COM India's No. If the volume of the gaseous system is suddenly reduced to 1/3 of initial volume. the rate of appearance of Z is 0. the rate of the reaction would become(A) 2 × 10–2 mol L–1s–1 (B) 1 × 10–2 mol L–1s–1 –2 –1 –1 (C) 4 × 10 mol L s (D) 2 × 10–1 mol L–1 s–1 B-7. What is the order of a reaction whose rate = KCA3/2 CB–1/2 ? (A) 2 (B)1 (C)– 1/2 (D) 3/2 The rate of certain hypothetical reaction A + B + C products. rate is doubled. The rate of disappearance of X will be (A) 0. dx/dt = k [A]a [B]b . aA + bB Product. For a chemical reaction. If concentration of B is made four times.No. by r = – [C]1/4. A + 2B 3C + D. B-2. + 1/2(d[C]/dt) = – 1/3 (d[D]/dt) = + 1/4 (d[A]/dt) = – (d[B]/dt) The reaction is : (A) 4 A + B 2C + 3D (B) B + 3D 4A + 2C (C) 4A + 2B 2C + 3D (D) B + (1/2) D 4A + 3 A-9. rate is four times. The rate would become – (A) 1/9 times (B) 9 times (C) 1/6 times (D) 6 times B-8. For a gaseous reaction the rate equation is v = k[A][B]. What is relation between rate of disappearance of A and that of B ? (A) – {d [A] / dt} = – {d [B] / dt} (B) – {d [A] / dt} = – {4 d [B] / dt} (C) – {4 d [A] / dt} = – {d [B]/ dt} (D) None of these ETOOSINDIA.25 mol L—1 per min Section (B) : Rate Law B-1. H. When 50% of the reactants are converted into products. 092142 33303 CHEMICAL KINETICS # 8 . 2A + 2B C + D. B-3.05 mol L—1 per hour (B) 0. The initial rate of the reaction is 4 × 10–2 mol L–1 s–1. which of the following expression does not describe changes in the concentration of various species as a function of time : (A) {d [C] / dt} = – {3d [A] / dt} (B) {3d [D] / dt} = {d [C] / dt} (C) {3d [B] / dt} = – {2d [C] / dt} (D) {2d [B] / dt} = {d [A] / dt} A-10. For which of the following.05 mol L–1 per min. If concentration of A is doubled.1 mol L per min (D) 0.A-8.05 mol L—1 per min —1 (C) 0. the order of reaction is one with respect to A and one with respect to B.1 Online Coaching for JEE Main & Advanced 3rd Floor. The rate of a reaction is expressed in different ways as follows . For a chemical reaction 2X + Y Z. – [d (BrO3–) /dt] = K [BrO3– ] [Br¯] [H+]2 It means : (A) Rate constant of overall reaction is 4 sec–1 (B) Rate of reaction is independent of the concentration of acid (C) The changes in pH of the solution will not affect the rate (D) Doubling the concentration of H+ ions will increase the reaction rate by 4 times.0 × 10–2 mol L–1 s–1 (C) 2.0 × 10–2 mol L–1 s–1 .t.9% of the reaction complete – (A) 50 Min (B) 46 Min (C) 49 Min (D) 48 Min ETOOSINDIA. When will 99. k1 . In what period of time would the concentration of X be reduced to 10% of original concentration (A) 20 Min. A and second–order w.25 × 10–3 mol L–1 s–1 (B) 1. successive half lives are equal (C) For a first order reaction. 2NO(g) + 2H2(g) N2(g) + 2H2O(g) the rate expression can be written in the following ways : {d [N2] / dt} = k1 [NO][H2] .5 = 16min (D) t0. rate when half reactants have been turned into products is : (A) 1.0 mol each of A and B introduced into a 1. Which of following statement is false (A) A fast reaction has a larger rate constant and short half-life (B) For a first order reaction.COM India's No.5 = 100s. For the irreversible process. If 1. {– d[NO] / dt} = k1 [NO] [H2] . Wrong data for the first order reaction is – (A) t0. H. t0.t. 99% of a first order reaction was completed in 32 min.0 × 10–2 mol L–1 s–1 Section (C) : The integrated rate laws C-1. second and third order respectively.r. the rate is first–order w. B. {–d[H2] / dt} = k1 [NO][H2] The relationship between k. {d[H2O] / dt} = k[NO][H2] . In acidic medium the rate of reaction between (BrO3)¯ and Br¯ ions is given by the expression. A first order reaction follows the expressions (A) Cte k 1t = C0 (B) Ct = C0 e k 1t (C) ln C0 = –k 1t Ct (D) ln Ct = k 1t C0 C-2.5 =100s. B-12. For the reaction. Rajasthan 324005 HelpDesk : Tel. Which of the following is correct : (A) if [A] = 1 then r1 = r2 = r3 (B) if [A] < 1 then r1 > r2 > r3 (C) if [A] > 1 then r3 > r2 > r1 (D) All B-11. A + B products. Half life period of a zero order reaction is – (A) Independent of concentration (B) Directly proportional to initial concentration (C) Inversely proportional to concentration (D) Directly proportional to the square of the concentration C-4.75 = 200s (C) Both the above (B) t0.75 = 150s C-5.No. What is the half life of a radioactive substance if 75% of any given amount of the substance disintegrates in 60 minutes – (A) 2 Hours (B) 30 Minutes (C) 45 Minutes (D) 20 Minutes C-6. For rate constant is numerically the same for three reactions of first. Kota. (B) 33 Min (C) 15 Min (D) 25 Min C-7.B-9. 092142 33303 CHEMICAL KINETICS # 9 .1 Online Coaching for JEE Main & Advanced 3rd Floor. k1 and k1. t0. For an elementary reaction X (g) Y (g) + Z (g) the half life periods is 10 min.0 L vessel.75 = 32 min t0. and the initial rate was 1.50 Rajeev Gandhi Nagar.r. the half-life is independent of concentration (D) The half life of a reaction is half the time required for the reaction to go to completion C-3. is : (A) k = k1 = k1 = k1 (B) k = 2k1 = k1 = k1 (C) k = 2k1 = k1 = k1 (D) k = k1 = k1 = 2 k1 B-10.50 × 10–3 mol L–1 s–1 (D) 2. C-8.50 × 103 min (D) 4.0 × 10–4 min (C) 2.2 to 0. The order of the reaction is: (A) 0 (B) 1 (C) 2 (D) 3 C-15.303) (C) (ln k / 2. The rate constant of reaction 2 A + B C is 2. For a second order reaction. PCI5(g) PCI3(g) + Cl2(g) is 20 min. (B) Second order kinetics (D) Pseudo first order kinetics The t0. For a first order reaction. if the conc. the ratio k1 /k2 if 90% of (A) has been reacted in time 't' while 99% of(B) has been reacted in time 2t is : (A) 1 (B) 2 (C) 1/2 C-20. What fraction of the substance will be left after an hour the reaction has occurred ? (A) 1/6 of initial concentration (B) 1/64 of initial concentration (C) 1/12 of initial concentration (D) 1/32 of initial concentration C-19. (or 34. (B) Product. and 2. C-16.5 for the first order reaction. what would be the time taken for the conc. the plot of log C against ‘t’ (logC vs 't') gives a straight tine with slope equal to : (A) (k / 2. The reaction is of – (A) Zero order (B) First order (C) Second order (D) Third order C-9. The concentration of sucrose was found to reduce from 0. mole–1 sec–1 after 20 sec. How long will it take a 0. the initial concentration of cane sugar was reduced from 0. In a certain reaction.55 × 10–5 It. (D) 15 × 10/ 8 hr. 80% of the sample will decompose in (A) 15 × 0.303) (B) (– k / 2.No.65 × 10–5 It. (C) 15 × (log5 / log2) hr.COM India's No. A radioactive isotope decomposes according to the first order with half life period of 15 hrs.08M to 0.5M solution to be reduced to 0.1 Online Coaching for JEE Main & Advanced 3rd Floor. sucrose gets hydrolysed into glucose and fructose.665 × 102 min (B) 8. 10% of the reactant decomposes in one hour. The rate constant for a second order reaction 8. The time after which both the concentration will be equal is : (Assume that reaction is first order) (A) 5 min (B) 15 min (C) 20 min (D) concentration can never be equal ETOOSINDIA.1 M in total of 2 hours. Rajasthan 324005 HelpDesk : Tel.8 hr. of a reactant decreases from 0. 30% in three hours and so on the dimensions of the rate constant is : (A) hour–1 (B) mole litre–1 sec–1 (C) litre mole–1 sec–1 (D) mole sec–1 C-17.05 molar in 10 hours.25M in reactant – (A) 8. In the presence of an acid. In presence of HCl. C-12.10 molar in 5 hours and from 0. The time in which the conc.01M – (A) 20 minutes (B) 30 minutes (C) 50 minutes (D) 70 minutes C-10. Kota. to decreases to 0.04M in ten minutes. 2.4 M to 0.0 × 10–4 min C-11. 20 % in two hours. of PCI5 reduces to 25% of the initial conc. In a first order reaction the reacting substance has half-life period of ten minutes. 092142 33303 CHEMICAL KINETICS # 10 . mole–1 sec–1 after 30 sec. The order of the reaction is : (A) zero (B) one (C) two (D) None of these C-18.0 × 10–4 litre mol–1min–1. is close to(A) 22 min (B) 40 min (C) 90 min (D) 50 min C-14.2 M in 1 hour and to 0.50 Rajeev Gandhi Nagar. H. (D) None of these Two substances A (t1/2 = 5 min) and B (t1/2 = 15 min) are taken in such a way that initially [A] = 4[B]. Radioactive decay follows (A) Zero order kinetics (C) First order kinetics C-13. K1 K2 In the following first order reactions (A) Product.83) hr. (B) 15 × (log 8) hr.303) (D) – k.20 to 0.57 × 10–5 It mole–1 sec–1 after 10 sec. 10 0. 1 1 ln a t (C) 0.035 0..070 (B) r = k [A]3 (C) r = k [A] [B]4 (D) r = k [A]2 [B]2 .32 (B) 5. The data for the reaction A + B C is Exp...80 0. t1 and t2.No. 092142 33303 CHEMICAL KINETICS # 11 ..1 Online Coaching for JEE Main & Advanced 3rd Floor. 1 (C) 2. mol–1 min–1.024 0. a3C0 .0 × 107 mole litre–1 sec–1. 1/2 (B) 1. Kota.58 (C) 4.where 'a' is constant. a2C0. ETOOSINDIA. H... what is the ratio. t1/2 = (0. The concentration of A at which rate of the reaction is (1/12) × 10–5 M sec–1 is : (B) (1/20) 5 / 3 M (A) 0. What conclusion can you draw from this graph. For a reaction A Products. 2t ..0 × 10–6 moles of H+. If a I-order reaction is completed to the extent of 60% and 20% in time intervals.070 0. When concentration of reactant in reaction A B is increased by 8 times.. t1/2 1/a (D) None of these C-26.C-21. aC0 .10 0.035 0. [A]0 [B]0 initial rate 1 2 3 4 0. What will be the order of reaction and rate constant for a chemical change having log t50% vs log concentration of (A) curves as : (A) 0. after time interval 0.05 mL) contains 3.11 (D) 8. If the rate constant of disappearance of H+ is 1... 1 C-27. (D) reaction is of zero order and K = 1 1 ln t a The rate constant for the reaction A B is 2 × 10–4 It. How long would it take for H+ in drop to disappear : (A) 6 × 10–8 sec (B) 6 × 10–7 sec (C) 6 × 10–9 sec (D) 6 × 10–10 sec Section (D) : Methods to determine The rate law D-1... Then : (A) reaction is of 1st order and K = (1/1) ln a (B) reaction is of 2nd order and K = (1/tC0)(1–a)/a (C) reaction is of 1st order and K = C-23.. Rajasthan 324005 HelpDesk : Tel. (A) n = 1 . 2 (D) 3. A graph plotted between log t50% vs.. the concentration of reactant are C0 . t1 : t2 ? (A) 6.25 M C-24.. log concentration is a straight line.80 0.012 0..5 M (D) None of these Graph between concentration of the product and time of the reaction A B is of the type Hence graph between – d[A]/dt and time will be of the type : (–d[A]/dt) (B) (A) (C) (D) Time C-25.693 / k) (B) n = 2..024 (A) r = k [B]3 0. t1/2 a (C) n = 1 . The order of the reaction would be (A) 2 (B) 1/3 (C) 4 (D) 1/2 D-2.012 0.50 Rajeev Gandhi Nagar.33 C-22. A drop of solution (volume 0.COM India's No. t. the rate increase only 2 times. – Time [A] 0 20 mol 5 min 18 mol (C) dx = k[A]0 [B]2 dt (D) dx = k[B]1/2 dt d [A] = k and at different time interval. 092142 33303 .[k = 1.No.0 /mol sec] (B) 0.[k = 1. Initial conc. [A] values are : dt 10 min 16 mol At 20 minute.180 IV 0. A is in excess and on changing the concentration of B from 0.30 0.lit–1sec 1] I 0.0 mol/ sec] (C) 2. then order is : dt If (A) 4 (B) 2 (C) 1 (D) 0 D-4.080 III 0.COM India's No.l–1sec–1 I 1 × 10–2 2 × 10–2 2 × 10–4 II 2 × 10–2 2 × 10–2 4 × 10–4 III 2 × 10–2 4 × 10–2 8 × 10–4 Which of the following inference(s) can be drawn from the above data – (a) Rate constant of the reaction 10–4 (b) Rate law of the reaction is k[A][B] (c) Rate of reaction increases four times on doubling the concentration of both the reactant. For the reaction 2A + 3B products.1 M to 0. Kota.1 Online Coaching for JEE Main & Advanced 3rd Floor. rate becomes doubled. The esterification takes place according to the reaction: CH3COOH + C2H5OH CH3COOC2H5 + H2O If each solution is diluted by one litre water the rate would become – (A) 4 times (B) 2 times (C) 0.b and c (B) a and b (C) b and c (D) c alone D-7.[k = 2.10 0.020 II 0.25 times D-5.4 M. of B is doubled and that of A and C is kept constant.4 mol/min ETOOSINDIA. dx =k [A]a [B]b dt dx = k.5 /mol sec] (D) 1.[k = 2. In the reaction : A + 2B + C D + 2E Ther rate of reaction remains unchanged if the conc. rate law is : (A) D-9. The following data pertain to reaction between A and B : S.320 Answer is – (A) 2. (A) 0 (B) 1/2 (C) 1 (D) 3 D-6. [A] [B] Rate mol.l–1 mol. dx = k[A]2 [B]2 dt (B) dx = k[A] [B] dt For the reaction A Products.D-3.50 Rajeev Gandhi Nagar.No.20 0. What is the order with respect to B. Initial rate [mol/] [mol. One litre of 2M acetic acid and one litre of 3M ethyl alcohol were mixed. Select the correct answer – codes :– (A) a.l–1 mol. Using the data given below the order and rate constant for the reaction : CH3CHO(g) CH4(g) + CO(g) would be – Experiment no.5 sec–1] D-8. Rajasthan 324005 CHEMICAL KINETICS # 12 HelpDesk : Tel. Thus. H.5 times (D) 0. A + B Product.40 0. rate will be : (A) 12 mol /min (B) 10 mol/min 15 min 14 mol (C) 8 mol/min (D) 0. Consider a reaction A B + C. Rajasthan 324005 HelpDesk : Tel. then the rate constant for 2A B + 3 C is : (A) (1/1200) ln 1. the total pressure doubled in 3 hrs. In the reaction NH4NO2 (aq. If initially only (A) is present and 10 minutes after the start of the reaction the pressure of (A) is 200 mm Hg and that of over all mixture is 300 mm Hg. If the initial concentraton of A was reduced from 4 M to 2 M in 1 hour and from 2 M to 1 M in 0. The increase in pressure from 100 mm to 120 mm is notices in 5 minutes. then the half life period of a reaction of nth order is directly propotional to (A) an (B) an–1 (C) a1–n (D) an+1 D-12.) N2 (g) + 2 H2O (l) the volume of N2 after 20 min and after a long time is 40 ml and 70 ml respectively. The rate of disapearance of A2 in mm min–1 is : (A) 4 (B) 8 (C) 16 (D) 2 E-4.303 /1200) log (7/3) sec–1 –1 (C) (1/20) log (7/3) min (D) (2. The order of the reaction might possibly be (A) zero (B) first (C) second (D) unpredictable from this data E-2.303 /10) log 1.1 Online Coaching for JEE Main & Advanced 3rd Floor.5 atm (D) 3 atm E-3. The value of rate constant is : (A) (1/20) In (7/4) min–1 (B) (2. Starting with pure A initially. The decomposition of a gaseous substance (A) to yield gaseous products (B). The order of the reaction is (A) 0 (B) 1 (C) 2 (D) 3 D-13.25 sec (D) None of these E-5. Consider the reaction 2A(g) 3B(g) + C(g).25 min–1 –1 (C) (2. If log is plotted against log [A]. If NO2 and F2 are present in a closed vessel in ratio 2 :1 maintained at a constant temperature with an initial total pressure of 3 atm. then graph is of the type : dt dt A Product and (A) (B) (C) (D) D-11. Formation of NO2F from NO2 and F2 as per the reaction 2NO2(g) + F2(g) 2NO2F(g) is a second order reaction. In a gaseous state reaction. what will be the total pressure in the vessel after the reaction is complete? (A) 1atm (B) 2 atm (C) 2.25 sec–1 (B) (2.0 moles litre-1 are 200 sec and 100 sec respectively. (C) follows First order kinetics. H. the order of the reaction is(A) One (B) Zero (C) Two (D) Three Section (E) : Methods to monitor the progress of reaction E-1. Hence rate constant is: (A) (2. / Cu –N2Cl –Cl + N2 Half-life is independent of concentration of reactant. Kota.COM India's No. 092142 33303 CHEMICAL KINETICS # 13 . A2 (g) B(g) + (1/2)C (g).5 hour.303 / 20) log (11/7) min–1 E-6.303 /10) log 5 min–1 (B) (2.5 min–1 –1 (C) (1/10) ln 1.No. After 10 minutes volume of N2 gas is 10 L and after complete reaction it is 50 L.50 Rajeev Gandhi Nagar.303 /10) log 4 min–1 ETOOSINDIA.D-10. The half-life period for a reaction at initial concentrations of 0. dx dx = k[A]2 .303 /10) log 1. If a is the initial concentration of reaction.303 /10) log 2 min (D) (2. first order with respect to NO2 and first order with respect to F2.5 and 1. Chemical reaction occurs as a result of collisions between reacting molecules. (C) the half life depends upon the concentration of the reactants.cal (B) –20 k. A large increase in the rate of a reaction for a rise in temperature is due to (A) increase in the number of collisions (B) the increase in the number of activated molecules (C) The shortening of mean free path (D) the lowering of activation energy F-11. H.98 J ETOOSINDIA. The activation energy of the reaction A + B C + D + 38 k. The decomposition of N2O into N2 & O2 in presence of gaseous argon follow second order kinetics with k = (5.Section(F) : Effect of Temperature F-1. The first order rate constant k is related to temp. Which of the following explains the increase of the reaction rate by catalyst (A) Catalyst decreases the rate of backward reaction so that the rate of forward reaction increases (B) Catalyst provides extra energy to reacting molecules so that they may reduce effective collisions (C) Catalyst provides an alternative path of lower activation energy to the reactants (D) Catalyst increases the number of collisions between the reacting molecules.9 × 104 KJ (B) A = 10–15 and E = 40 KJ (C) A = 1015 and E = 40 KJ (D) A = 10–15 and E = 1. What would be the activation energy of the reaction. For a reaction for whi ch the activation energies of forward and reverse reactions are equal (A) H = 0 (B) S = 0 (C) The order is zero (D) There is no catalyst F-5. Rajasthan 324005 HelpDesk : Tel. Therefore. F-9. Kota.0 x 1011 J 41570 K T (B) 41570 J (K stands for Kelvin units). The energy of activation of the reaction is : (C) 5000 J (D) 345612.4 KJ/mol respectively.No.cal is 20 k.cal (C) 18 k.cal (D) 58 k. 092142 33303 CHEMICAL KINETICS # 14 .COM India's No.0 – (106 /T) Which of the following pair of value is correct ? (A) A = 1015 and E = 1. Rate of which reactions increases with temperature : (A) of any (B) of exothermic reactions (C) of endothermic (D) of none. (B) the rate is independent of the concentration of the reactants. (D) the rate constant has the unit mole It–1 sec–1. F-10.cal F-8. The activation energy of reaction is equal to(A) Threshold energy for the reaction (B) Threshold energy + Energy of the reactants (C) Threshold energy – Energy of the reactants (D) Threshold energy + Energy of the products F-6. For a zero order reaction.50 Rajeev Gandhi Nagar. as log k = 15. The reaction is – (A) Exothermic (B) Endothermic (C) Neither exothermic nor endothermic (D) Independent of temperature F-2. The chemical reactions in which reactants require high amount of activation energy are generally (A) Slow (B) Fast (C) Instantaneous (D) Spontaneous F-4.1 Online Coaching for JEE Main & Advanced 3rd Floor.cal. F-12. the reaction rate is given by (A) Total number of collisions occurring in a unit volume per second (B) Fraction of molecules which possess energy less than the threshold energy (C) Total number of effective collisions (D) None of the above F-3. The activation energies of the forward and backward reactions in the case of a chemical reaction are 30.9 × 104 KJ. Which of the following statement is false : (A) the rate is independent of the temperature of the reaction.0 × 1011 L mol1 s1) e (A) 5. C + D A + B (A) 20 k. F-7.5 and 45. No. the activation energy and the frequency factor of a chemical reaction at 25°C are 3. Trimolecular reactions are uncommon because (A) the probability of three molecules colliding at an instant is very low. (C) the slowest elementary step in sequence of the reactions governs the overall rate of formation of product. G-7.mol1. H = 40kJ. A catalyst lowers Eaf to 20 kJ. 3/2 (C) Not defined.mol1. The value of the rate constant as T is : (A) 2. The ratio of energy of activation for reverse reaction before and after addition of catalyst is : (A) 1. (D) a rate law is often derived from a proposed mechanism by imposing the steady state approximation or assuming that there is a pre-equilibrium. For a given reaction. (D) the probability of many molecules colliding at an instant is high. energy of activation for forward reaction (Eaf) is 80kJ. Select the correct statements : (A) the molecularity of an elementary reaction indicates how many reactant molecules take part in the step. (C) the probability of three molecules colliding at an instant is zero. Which of these proposed mechanism can be consistent with the given information about this reaction ? Mechanism : H2(g) + 2Cl(g) 2HCl(g) + 2(g) Slow Mechanism : H2(g) + Cl(g) HCl(g) + H(g) (A) only G-4. dc = k[NO2]2 the number of dt (D) depends on mechanism The reaction of hydrogen. How much faster would a reaction proceed at 25°C than at 0°C if the activation energy is 65 kJ? (A) 2 times (B) 5 times (C) 11 times (D) 16 times F-14. For the reaction H2 (g) + Br2 (g) 2HBr (g) the experiment data suggested that r = k[H2][Br2]1/2 The molecularity and order of the reaction are respectively : (A) 2.5 (C) 1.6 × 1030 s–1 F-15.0 × 10–4 s–1. Kota. 092142 33303 . (B) the rate law of an elementary reaction can be predicted by simply seeing the stoichiometry of reaction. Calculate the 1 C value of C for which kapp has 90% of its limiting value at C tending to infinitely large values. For the reaction NO2 + CO CO2 + NO the experimental rate expression is – molecules of CO involves in the slowest step will be (A) 1 (B) 2 (C) 3 G-3. ETOOSINDIA. X + 2Y + Z N occurs by the following mechanism (i) X+Y M very rapid equilibrium (ii) M+ZP slow (iii) O+YN very fast What is the rate law for this reaction (A) Rate = k[Z] (B) Rate = k[X] [Y]2 [Z] (C) Rate = [N] (D) neither nor (D) Rate = k[X] [Y] [Z] In the Lindemann theory of unimolecular reactions.0 × 1014 s–1 (C) infinite (D) 3. fast HI(g) + Cl(g) HCl(g) + I2(g) (B) only (C) both and The reaction. it is shown that the apparent rate constant for such a reaction is kapp = k1C where C is the concentration of the reactant k1 and are constants. G-5. 3/ 2 (B) 3/2 . H.mol1 for the reaction.2 (D)2.1/2 G-2. Rajasthan 324005 CHEMICAL KINETICS # 15 HelpDesk : Tel. (A) 106 mole/litre (B) 104 mol/litre (C) 105 mole/litre (D) 5 x 105 mol/litre G-6. 104.F-13.0 (B) 0.0 × 1014 s–1 respectively.0 Section (G) : Mechanism of reactions G-1.50 Rajeev Gandhi Nagar. and iodine monochloride is represented by the equation : 2HCl(g) + 2(g) H2(g) + 2Cl(g) This reaction is first–order in H2(g) and also first–order in Cl(g). The rate constant.4 KJ mol–1 and 6.1 Online Coaching for JEE Main & Advanced 3rd Floor.COM India's No.0 × 1018 s–1 (B) 6. 3/2 (D) 1. (B) the probability of three molecules colliding at an instant is high. given = 9 x 105. 0 × 10–2dm3 mol-1 s–1 and 3.0 kJ mol–1 respectively. H. 092142 33303 CHEMICAL KINETICS # 16 . k1 = k2 for the reaction C+D A+B If H-5.2 M [C] 0. hence.0 kJ mol–1 ETOOSINDIA.2 M 0.2 M 0.2 M (D) 0.0 kJ mol–1 (D) 65.1 M 0. If the corresponding energies of activation of the parallel reactions are 60.3 M 0. dx For an elementary reaction.50 Rajeev Gandhi Nagar. (A) d[A] = k1[A] + k4[D] dt (B) d [ C] = k2[B] k3[C] dt (C) d [D] = k4[D] + k3[D] dt (D) Nothing can be said about order of reactions in this problem For an elementary reaction of reversible nature.8 × 10–5 sec–1 The percentage distribution of B and C H-6. 2B + C is the required reaction (D) none is correct Consider the elementary reaction sequence shown in figure.4 M (B) 0.2 M 0.5 kJ mol–1 (C) 100. what is the apparent overall energy of activation ? (A) 130. given reaction is : 1 2 dt (A) 2A + H-4. C dx dt = k1 [A] [B] – k2[C] [D] in which set of the concentration reaction ceases? [A] [B] [C] [D] [A] [B] (A) 0. Which of the following equations are correct ? H-3.25 M 0.5 M [D] (C) 0.No.2 M 0.1 Online Coaching for JEE Main & Advanced 3rd Floor.83% B and 23. select the correct statement : 2 (A) – d[C] d[ A ]2 d [B] = = is the relation among (B) 2A dt dt dt (C) both are correct H-2.2 M 0.26 × 10–4 sec–1 and K2 = 3. The decomposition follows two parallel first order reactions as : K1 = 1.0 kJ mol–1 (B) 67.0 kJ mol–1 and 70. 1 B C (B) 2A + B (C) 2A C + B–1 (D) None of these At a given temperature. Rajasthan 324005 HelpDesk : Tel. net rate is dt = k[A]2 – k'[C] [B]2 then.3 M 0. net rate is : dx = k [A]2 [B]1 – k [C]. Kota.Section (H) : Complications in first order reactions H-1.COM India's No. (A) 80% B and 20% C (B) 76.4 M 0.4 M The substance undergoes first order decomposition.0 × 10–2 dm3 mol–1 s–1.2 M 0.17%C (C) 90% B and 10% C (D) 60% B and 40% C The rate constant for two parallel reactions were found reactions were found to be 1. 1. Assume this decomposition is a first order. At t = t1 pressure of gas B is : (A) 2.5 s (C) 2. After some time t = t1. Value of t1 in seconds is : (A) 2. (C) 5.No. Kota. It is also given at t = tx. For the (Set-1) : (A) Ea1 > Ea2 if T1 > T2 (C) Ea1 = Ea2 6. For the (Set-1) : (A) if T1 > T2. (B) Ea1 < Ea2 if T1 > T2 (D) Ea1 = 0.5 Ea2 Comparing set-I and II : (A) k4 > k3 & k2 > k1 .COM India's No. (B) 1. k1 < k2 (for endothermic reaction) (D) Ea1 Ea2 5.15 s (C) 1 : 3 : 5 (D) 1 : 3 : 5 Ratio of rate constant at t = 0 to t = t1 to t = tx is : (A) 2 : 3 : 4 (B) 1 : 1 : 1 Comprehension # 2 4. if T2 > T1 (exothermic) ETOOSINDIA. final total pressure is 35 bar.5 bar 2.1 Online Coaching for JEE Main & Advanced 3rd Floor.0 bar (B) 1.25 bar (D) data is insufficient Rate constant (k) = (log 64 – log 49) s–1. 092142 33303 . if T2 > T1 (endothermic) (B) k4 < k3 & k2 > k1 . k1 > k2 always (B) if T1 > T2. if T2 < T1 (exothermic) (D) k4 < k3 & k2 < k1.PART . total pressure is half of the final total pressure at t = tx (a very long time).3 s (D) 1. Rajasthan 324005 CHEMICAL KINETICS # 17 HelpDesk : Tel. H. at a constant temperature.15 s 3. At t = 0 mole fraction of gas C is 1/3.II : MISCELLANEOUS QUESTIONS COMPREHENSION Comprehension # 1 A(g) 2B(g) + C(g) Initially at t = 0 gas A was present along with some amount of gas (C). if T2 < T1 (endothermic) (C) k4 > k3 & k2 > k1 . k1 > k2 (for exothermic reaction) (C) if T1 > T2.50 Rajeev Gandhi Nagar. Let us consider a reaction A(g) B(g) + C(g) At t = 0 a 0 0 At time t a–x x x The rate of reaction is given by the expression represented as k = dx = k(a – x) and integrated rate equation for a given reaction is dt 1 a where a = initial concentration and (a – x) = concentration of A after time t.5 litre vessel initially. At any time we also have 9. if reaction is allowed to takes place at constant pressure & at 298K then find the concentration of B after 100 min.30 (A) 190 min (B) 176. or para positions. we have = k1[A] = k1[A]0 e (k1 k 2 )t . We shall consider the simplest case. The rate law is d[A] = – k1[A] – k2[A] = – (k1 + k2) [A] dt [A] = [A]0 e (k1 k 2 )t . Kota.Comprehension # 3 A reaction is said to be first order if it's rate is proportional to the concentration of reactant. toluene can be nitrated at the ortho. integration of dt [D] = k 2 [ A ]0 (1 – e (k 1 k 2 )t ) k1 k 2 The sum of the rate constants k1 + k2 appears in the exponentials for both [C] and [D]. A k1 [C] = k [D] 2 starting initially with only A Which of the following is correct at time t (A) [A]0 = [A]t +[B]t + [C]t (C) [A]0 = [A]t + [C] t [B] t + 3 2 (B) [A]0 = [A]t + 2 [B]t + 3 [C]t (D) [A]0 = 2 [A]t +[B]t + [C]t 3 ETOOSINDIA. (A) 0. meta.36 M (C) 0. Multiplication by dt and integration from time 0 dt k [A] ( k k ) t (where [C]0 = 0) to an arbitary time t gives [C] = 1 0 (1 e 1 2 ) k1 k 2 d [D] gives Similarly.66 min 8. H. that of two competing irreversible first-order reactions : k k2 D A 1 C and A where the stoichiometric coefficients are taken as unity for simplicity.04 M (B) 0. Thermal decomposition of compound X is a first order reaction.No.50 Rajeev Gandhi Nagar. For example.66 min (D) 156. If 75% of X is decomposed in 100 min. 092142 33303 CHEMICAL KINETICS # 18 . d [C] For C. Rajasthan 324005 HelpDesk : Tel. Starting with 2 moles of A in 12. How long will it take for 90% of the compound to decompose? Given : log 2 = 0.COM India's No. Consider a reaction A(g) 3B(g) + 2C(g) with rate constant 1.09 M (D) None of these Comprehension # 4 Competing First-Order Reactions Frequently a species can react in different ways to give a variety of products. ln t ax 7.66 min (C) 166.386 × 10–2 min–1.1 Online Coaching for JEE Main & Advanced 3rd Floor. No. [ X] t starting with only 'X'. 13. (A) Independent of time (B) (C) Depends upon initial concentration of X (D) [A]0 (ekt –1) (e kt 1) At high temperature acetic acid decomposes into CO2 & CH4 and simultaneously into CH2CO (ketene) and H2O 1 k1 3s (i) CH3COOH CH4 + CO2 1 k 2 4s (ii) CH3COOH CH2CO + H2O What is the fraction of acetic acid reacting as per reaction (i) ? (A) 12. [B] t time [A]t time (C) [C]t (D) [B]t Conc.10. 14. [C]t [B]t [A] t [A]t time time ETOOSINDIA. k1 k 2 k1 Conc.50 Rajeev Gandhi Nagar.1 Online Coaching for JEE Main & Advanced 3rd Floor. B & C at any time 't' is given by [A]t = [A]0 e k1t . [B]t [A]t (B) [C]t Conc. ratio [ Y ] [ Z] t t X 1 11.COM India's No. (A) [C]t conc. 3 4 (B) 3 7 (C) 4 7 (D) none of these starting with pure A ratio of rate of production of B to C is For A (A) Independent of time (C) Depends upon initial concentration of A (B) Independent of temperature (D) Independent of mechanism of reaction Comprehension # 5 For the given sequential reaction k k 2 1 A B C the concentration of A. H. Rajasthan 324005 CHEMICAL KINETICS # 19 HelpDesk : Tel. Kota. [B]t = k1[ A ]0 e k1t e k 2 t ( k 2 k1 ) [C]t = [A0] – ( [A]t + [B]t ) The time at which concentration of B is maximum is (A) (B) 1 k ln 1 k 2 k1 k 2 (C) 1 k ln 1 k1 k 2 k 2 (D) k2 k 2 k1 Select the correct option if k1 = 1000 s–1 and k2 = 20 s–1. 092142 33303 . log 2 = 0.MATCH THE COLUMN 15. 092142 33303 .No. half life for both reactants are equal For the reaction of type A(g) 2B(g) Column-I contains four entries and column-II contains four entries. Rajasthan 324005 HelpDesk : Tel. 16.50 Rajeev Gandhi Nagar. log 3 = 0.I Column . (B) The solution is optically inactive at (q) 7.3) Column .I Column . Kota.5° (C) The equimolar mixture of the products (r) 75. The polarimeter readings in an experiment to measure the rate of inversion of cane suger (1st order reaction) were as follows time (min) : 0 30 angle (degree) : 30 20 – 15 Then match the following. (Use : log 35 = 1.54. 17. H.COM India's No.II (A) d[ B] d[A ] vs for first order dt dt (p) (B) [A] vs t for first order (q) (C) [B] vs t for first order (r) (D) [A] vs t for zero order (s) ETOOSINDIA. (D) The angle at half time (s) laevorotatory Match the following : Column .I Column . Entry of column-I are to be matched with ONLY ONE ENTRY of column-II Column .II (A) The half life of the reaction (p) 120 min.II (A) A + B C + D r = k1 [A] [B] (p) Unit of rate constant possess concentration unit (B) A + B C + D r = k2 [A] [B]º (q) Rate constant for the reaction of both the reactants are equal (C) A + B C + D r = k3 [A]º [B]º (r) Rate of consumption of at least one of the reactants is equal to rate of production of at least one of the products (D) 2A + B 2C + 3D r = k3 [A]º [B]º (s) If both reactants are taken in stoichiometric ratio.2 min.48.1 Online Coaching for JEE Main & Advanced CHEMICAL KINETICS # 20 3rd Floor. Statement-2 is True.136 480 0. (C) Statement-1 is True.order and zero order reactions (r) are same. If we start with 20 mol L–1. 20.1 Online Coaching for JEE Main & Advanced 3rd Floor. Statement-2 is False.693 Assume initial concentration to be same for the both.No. Statement-1 : In a reversible endothermic reaction. Match the following : Column-I Column-II (A) If the activation energy is 65 kJ then how much time faster a reaction proceed at 25°C than at 0°C (p) 2 (B) Rate constant of a first .order reaction is 0. (D) and (E) out of which ONLY ONE is correct. (E) Statement-1 and Statement-2 both are False. (B).18. with thereactants by which more number of molecules are able to cross the barrier per unit of time.0693 min–1. (A) Statement-1 is True. 092142 33303 CHEMICAL KINETICS # 21 . Statement-2 : Reactions with order 1 are either too slow or too fast and hence the time of completion can not be determined.lives of first .50 Rajeev Gandhi Nagar. : The rate of reaction having negative order with respect to a reactant decreases with the increase in concentration of the reactant. Eact of forward reaction is higher than that of backward reaction : The threshold energy of forward reaction is more than that of backward reaction Statement-2 24. [A]0 (M) 0. Statement-2 is NOT a correct explanation for Statement-1. Ratio of rates at the start of reaction is how many times of 0. Statement-1 Statement-2 : The overall rate of a reversible reaction may decrease with the increase in temperature.0677 t1/2 (sec) 240 order of the reaction is 30 (s) 0. Statement-1 Statement-2 : A fractional order reaction must be a complex reaction. ETOOSINDIA. Rajasthan 324005 HelpDesk : Tel. Statement-2 is a correct explanation for Statement-1. it is reduced to 2.COM India's No. Statement-2 is True. 19. Statement-1 : The time of completion of reactions of type A product (order <1) may be determined. 21. 11 (D) The half-life periods are given . (B) Statement-1 is True. : Fractional order of RDS equals to overall order of a complex reaction. H. 23. Statement-2 is True. (C). Statement-1 Statement-2 : A catalyst provides an alternative path to the reaction in which conversion ofreactants into products takes place quickly : The catalyst forms an activated complex of lower potential energy. 22.5 mol L–1 in how many minutes (q) Zero (C) Half .272 960 ASSERTION / REASON Directions : Each question has 5 choices (A). : When the activation energy of forward reaction is less than that of backward reaction. Kota. then the increase in the rate of backward reaction is more than that of forward reaction on increasing the temperature. (D) Statement-1 is False. Statement-1 Statement-2 : Temperature coefficient of an one step reaction may be negative. 40. Order of reaction may be change with change in practical conditions 41. 2 + 2S2O32– S4O62– + 2– . Ct 1 . ETOOSINDIA.No. H. 26. Order and molecularity of a single step reaction may or may not be same. 27. the rate of disappearance of thiosulphate ions is twice the rate of disappearance of 2. For a zero order reaction t3/4 is related to t1/2 as t3/4 = 1. After 200 days. rate = k[N2O]0 = k = constant is a zero order reaction. Kota. The rate of an exothermic reaction increases with the rise in temperature. 28.50 Rajeev Gandhi Nagar.5 t1/2 38. where symbols have standard meaning. n 29. Statement-1 : If the activation energy of reaction is zero temperature will have no effect on the rate constant Statement-2 : Lower the activation energy faster is the reaction TRUE / FALSE 31. C0 2 Statement-2 : Statement-1 : Time taken for the completion of 75% of a st order reaction is double than its t½. Molecularity of a reaction is always whole number. In a complex reaction the rate of overall reaction is governed by the slowest step. Statement-2 Au : N2 O(g) N2 (g) 1/ 2O 2 . of B due to its zero order. 32. Partial orders are never negative. 39. Rajasthan 324005 CHEMICAL KINETICS # 22 HelpDesk : Tel.COM India's No. the value of rate constant for the reaction would be 10s–1. 30.25.93 s. 35. The rate of reaction is uniform in zero order reaction. the half life time of reaction is only defined when conc. Statement-1 : For A + 2B C (rate= K[A]1[B]0). of A and not to conc. 34. Statement-1 : In the reaction. If t1/2 for a first order reaction is 6. fraction left undecaye will be 25%. 37. 36. Statement-2 : The rate of disappearance of 2 is twice the rate of disappearance of S2O32– ions. The activation energy of a catalysed reaction is more than the activation energy of the uncatalysed reaction. Statement-2 : Time taken for completion of any fraction of a st order reaction is a constant value. 33. Statement-1 : Half life of a certain radioactive element is 100 days. of A and B are in stoichiometric ratio Statement-2 : For above given order half life of reaction is directly proportional to conc. Statement-1 : Many reactions occurring on solid surface are zero order reactions.1 Online Coaching for JEE Main & Advanced 3rd Floor. 092142 33303 . Order of a reaction can be written from the balanced chemical equation. 55. 092142 33303 . The rate of the reaction A B having the rate law dt k [ A ] [B] when plotted against time will exhibit a d[A] maximum at some time. the rate remains constant. The rate law of the elementary reaction. 52. At lower temperature. Product can form only when the required orientation and energy conditions are met. Kota. 51. ETOOSINDIA. increase in temperature cause more change in the value of k than at higher temperature. 49. 46. 60.1 Online Coaching for JEE Main & Advanced 3rd Floor. the reaction must be elementary. A catalyst in a chemical reaction increases the forward Ea and decreases the backward Ea 58. 47. Rajasthan 324005 CHEMICAL KINETICS # 23 HelpDesk : Tel. A catalyst in a chemical reaction decreases the activation energy of the forward reaction and increases the activation energy of the reverse reaction 57.50 Rajeev Gandhi Nagar. 56. 54. Molecularity of a reaction may includes the number of product molecules taking part in the reaction. Molecularity defined only for RDS. Every species that appears in the rate law of reaction must be a reactant or product in that reaction. If the partial orders differ from the coefficients in the balanced reaction.COM India's No. 44. A catalyst in a chemical reaction decreases both forward and backward Ea 59. the reaction must be complex. The pre-exponential factor A has the same units for all reactions. If the partial orders are equal to corresponding coefficients in the balanced reaction. greater is the effect on the value of k for a given temperature change.42. For a firstorder reaction. All collisions between reactants yield the desired product. H. 45. 48. If concentration of catalyst appear in rate law (rate = k [catalyst]1 [reactant]1) then it may becomes pseudo first order reaction during the reaction.No. As a first order reaction proceeds at a constant temperature. 3A B must be r = k [A]3 43. Larger the value of Ea . 50. the time required to reduce successively the concentration of reactant by a constant fraction is always same. 53. Activated complex is an intermediate product. 69. The ratio t7/8 / t1/2 for a first order reaction would be equal to _________. For a zero order reaction. it proceeds at _______ rate. 78. 75. Rate constant of a reaction __________ with increase in temperature. For a certain reaction. [Eactivated complex – Ereactants] = ____________. For collision to be effective the energy possessed by the colliding molecules should be equal to or greater than the ____________. the minimum value of activation energy can be _______. 77.1 Online Coaching for JEE Main & Advanced 3rd Floor. In a multistep reaction. A catalyst is chemically ____________ at the end of a reaction 82. 79. 092142 33303 CHEMICAL KINETICS # 24 . 73. 67. The rate constant for the zero order reaction has the dimensions___________. In the reaction. the rate of the reaction is equal to the ______ of the reaction. 81. 74.50 Rajeev Gandhi Nagar. The rate equation r = k [A][B]1/2 suggests that order of overall reaction is _______. A plot of [A] vs t for a certain reaction A B with r = k [A]0 will be a straight line with slope equal to ________. the rate of reaction increases by 4 times when the concentration of M is doubled. The rate law is _________. 71. ETOOSINDIA. The rate of a reaction is ________ to the collision frequency. H. 64. Kota. 62. the rate of disappearance of H2 is _______ the rate of appearance of HI. xM yL. Rajasthan 324005 HelpDesk : Tel. For an endothermic process. For a ______ order reaction the half-life (t1/2) is independent of the initial conc. 76. The value of temperature coefficient is generally between _________. the ________ step is rate determining. 68. 65. In the Arrhenius equation k = Aexp (–E/RT). H2 + I2 2HI.FILL IN THE BLANKS 61.COM India's No. If activation energy of reaction is low. A catalyst increases the rate of the reaction by__________ activation energy. 80. of the reactants. 72. A may be termed as the rate constant at __________. 63.No. a graph of log [A] vs t has a slope equal to __________. For a first order reaction A P. 70. The reactions with molecularity more than three are _________. 66. The order of a reaction is the ________ of the _________ of all the concentration terms in the rate equation. 2.5 × 1011 L mol–1 s–1 10 –1 –1 (C) 1. Given : k = 2.15 × 10–2 The rate constant for the forward reaction A (g) + 2B(g) is 1.I : MIXED OBJECTIVE Single Choice Type 1.0125 M (D) None of these ETOOSINDIA. Starting with initial conc. (C) 900 sec. then half life (t1/2 ) is dt (A) 55.0) | | | | | -1– log (a0 – x) 3.5 × 10 L mol s (D) 1. SO2Cl2(g) SO2(g) + Cl2(g) a graph of log (a0 – x) vs t is shown in figure.1 Online Coaching for JEE Main & Advanced 3rd Floor.PART . The reaction A (g) B(g) + 2C (g) is a first order reaction with rate constant 3465 × 10–6 s–1.. Kota. (A) 0. Reaction A + B C + D follows following rate law : rate = k [A] 1 1 2 [ B] 2 . (C) 62.303 × 10–3 sec–1. 20 minutes time is required for completion of 20 % reaction. What is the rate constant (sec–1)? Time (min) 2 4 6 8 10 (0. (A) 300 sec. find the concentration of A after 200 sec. (B) 4.2 4.13 min (D) None of these If decomposition reaction A (g) B (g) follows first order kinetics then the graph of rate of formation (R) of B against time t will be (A) (B) (C) (D) For the first order decomposition of SO2Cl2(g).50 Rajeev Gandhi Nagar.25 M. If 10–5 moles of A and 100 moles of B are present in a 10 litre vessel at equilibrium then rate constant for the backward reaction at this temperature is : (A) 1.05 M (B) 0. 6.44 min.50× 104 L mol–1 s–1 (B) 1. Starting with 0. when the reaction is allowed to take place at constant pressure and temperature. (D) 1200 sec. of 1 M of A and B each. what is the time taken for concentration of A of become 0.5 × 10–3 s–1 at 100 K.6 × 10–1 (C) 7.1 mole of A in 2 litre vessel.025 M (C) 0.7 × 10–3 (D) 1. Rajasthan 324005 HelpDesk : Tel.COM India's No. (B) 600 sec. 092142 33303 CHEMICAL KINETICS # 25 . (B) 50 min -2– -3– (A) 0.5 × 10–11L mol–1 s–1 5. H. Consider the reaction : A B + C Initial concentration of A is 1 M.No. If d[ B] = k[A]. In respect of the equation k = A exp (– Ea / RT). T 10.5 Given that for a reaction of order n.8 Experiment-1 Experiment-2 0. Rajasthan 324005 CHEMICAL KINETICS # 26 HelpDesk : Tel. The rate constant. t1/2. the activation energy and the Arrhenius parameter (A) of a chemical reaction at 25°C are 3. is given by t1/2 = constant and C0 is the initial concentration. carried out at 27 ºC if 3.4 kJ/mole (C) 100 kJ/mole (D) 88.No.0 × 10–4 s–1.85 kJ/mol (B) 55. 1. E represents (A) The fraction of molecules with energy greater than the activation energy of the reaction (B) The energy above which all the colliding molecules will react (C) The energy below which colliding molecules will not react (D) The total energy of the reacting molecules at a temperature. the integrated form of the rate equation is k = 1 1 1 where C0 t(n 1) Cn 1 Cn1 0 and C are the values of the reactant concentration at the start and after time t.14 kJ/mol (C) 11.0 × 1018 s–1 (B) 6. ( 2 2 ) 1/ 2 xC0 where k is k (D) 0.036 M min–1 (C) 0. For a certain reaction of order n. the time for half change. 092142 33303 .13 M min–1 (D) 1 M min–1 8.COM India's No.65 kJ/mol 12.50 Rajeev Gandhi Nagar. 104.The variation of concentration of A with time in two experiments starting with two different initial concentration of A is given in the following graph.1 Online Coaching for JEE Main & Advanced 3rd Floor. which one of the following statements is correct? (A) R is Rydberg's constant (B) k is equilibrium constant (C) A is adsorption factor (D) Ea is the energy of activation 9.2 1 0. What is the relationship between t3/4 and t1/2 where t3/4 is the time required for C to become 1/4 C0? (A) t3/4 = t1/2 [2n1 + 1] (B) t3/4 = t1/2 [2n1 1] n+1 (C) t3/4 = t1/2 [2 1] (D) t3/4 = t1/2 [2n+1 + 1] ETOOSINDIA. For the first order reaction A — B + C.0 × 1014s–1 respectively. Kota. the Ea (activation energy) of the reaction is [log 3.58] (A) 12 kJ/mole (B) 831.08 M min–1 (B) 0. H.) (A) 0.8 = 0.4 kJ mol–1 and 6.6 × 1030 s–1 11.8 × 10–16 % of the reactant molecules exists in the activated state. What is n? (A) 1 (B) 2 (C) 0 14. The energy of activation of the reaction is (A) 43.0 × 1014 s–1 (C) infinity (D) 3.5 1.8 M? Concentration(M) 7. What is the rate of reaction (M/min) when concentration of A in aqueous solution was 1.57 kJ/mole 13. Rate of a reaction can be expressed by Arrhenius equation as : k = Ae–E/RT In this equation.6 5 10 15 20 time(min. The value of the rate constant at T is (A) 2. The reaction is represented as A(aq) B(aq). A first order reaction is 50% completed in 20 minutes at 27°C and in 5 min at 47°C.97 kJ/mol (D) 6. 5 minutes (D) 165. 200 mm 22. The time elapsed of a certain between 33% and 67% completion of a first order reaction is 30 minutes.465 × 103 min1. Kota.20 at the end of 20 minutes. The values of optical density are 0.93 x 103 min1 23.5 ln 2 min–1 . the half life changes to 500 minute of any concentration of sugar .50 Rajeev Gandhi Nagar. However if pH = 6. 1/2 16. involving a single reactant in each case. 0. Starting with pure A. yields a straight line.93 x 102 min1 (D) 3. (A) 1. 300 mm (D) 0.80. are : (A) 0.47 × 104 mol.5 minutes (B) 12. (i) (ii) (iii) What are the possible orders of the reactions (i).COM India's No. At 373 K.51 x 102 min1 (C) 6. If at t as 30 minutes the conc.51 x 103 min1 The inversion of cane sugar proceeds with half life of 50 minute at pH = 5 for any concentration of sugar. a plot of rate of the reaction on the yaxis. if the initial concentration is 0.No. (A) 6.5 min–1.litre1 min1 20. What is the time needed for 25% completion? (A) 150. (ii).300 mm –1 (C) 0. Calculate the rate constant.35 and 0. 1/2 (C) 0.litre1 min1 (B) 3.5 minutes (C) 180. was 176 mm and after a long time when A was completely dissociated. the concentration of the reactant decrease to 6. 2. 200 mm (B) 0. For a certain reaction.166 × 104 mol. ETOOSINDIA. 1. 1.166 × 103 min1.1 Online Coaching for JEE Main & Advanced 3rd Floor. 100 minutes after the start. 40 minutes and infinite time after the start of the reaction which is first order. (iii). In three different reactions. Initial conc.05 In 3 min–1 . 2.2 mole/litre? (A) 2. 3 (B) 2. versus concentration of the reactant on the xaxis.05 In 3 min . it was 270 mm. C0 = initial conc. 3.5 minutes 17. What is the order of the reaction : (A) 3 (B) zero (C) 1 (D) 2 19.05 In 1. The pressure of A at the end of 10 minutes was : (A) 94 mm (B) 47 mm (C) 43 mm (D) 90 mm 21. a gaseous reaction A 2B + C is found to be of first order. The law expression for sugar inversion can be written as : (A) r = K [sugar]2 [H+]0 (B) r = K [sugar]1 [H+]0 (C) r = K [sugar]1 [H+]1 (D) r = K [sugar]0 [H+]0. The reaction [Co (NH3)5Br]2 + H2O [Co(NH3)5(H2O)]3+ + Br is followed by measuring a property of the solution known as the optical density of the solution which may be taken to be linearly related to the concentration of the reactant. a plot of {[C0 – C] / [C]} against the time t.15.667 × 104 mol.25% of its initial value in 80 minutes. 2 (D) 0. 092142 33303 CHEMICAL KINETICS # 27 . 1. (B) 3. of the reactant and C is the reactant concentration after time t. H. 2. The reaction A(s) 2 B(g) + C(g) is first order.465 × 102 min1.litre1 min1 (C) 3. k (t = 0) of A is C0 while B is 2C0. What is (i) the rate constant and (ii) the rate of the reaction. yields three different curves shown below. of C is C0 /4 then rate expression at t = 30 minutes is : (A) R = 7 C03 k/16 (B) R = 27 C03 k / 32 (C) R = 247 C03 k / 64 (D) R = 49 k C03 / 32 18. and after very long time are 150 mm Hg and 225 mm Hg. the total pressure at the end of 10 min. The value of rate constant and pressure after 40 min.litre1 min1 (D) 2.17 × 103 mol. The pressure after 20 min. A reaction 2A + B C + D is first order with respect to A and 2nd order with respect to B. Rajasthan 324005 HelpDesk : Tel.17 × 102 min1. In a I order reaction A products. 2. Ea / 2. 092142 33303 CHEMICAL KINETICS # 28 . Rajasthan 324005 HelpDesk : Tel. For the reaction 3A Products the value of k = 1 × 10–3 / (mol–min) the value of – d[A] / dt in mol/lt-sec when [A] = 2M is : (A) 6.303) K. A more sensitive (B) B in both cases (C) A in both cases (D) A faster. Where t1 is the time period for 25% completion of the first reaction and t2. log A (D) None of these 29. (D) then the reaction must be A(g) 3B(g) and is a first order reaction. By what factor will it change for the temperature change from 17ºC to 27ºC? (A) 1. (C) (D) 2 1 e2 1 C0 after : (Natural life = 1 ) K (B) two-natural life-time (D) four natural life-time If for a reaction in which A(g) converts to B(g) the reaction carried out at const.693 (D) not defined 2k d[ A ] 1 = k[A]. The Arrhenius relationship of two different reactions is shown below. CD KC = 1014 e–1000/T Temperature T K at-which (KA = KC) is : (A) 1000K (B) 2000K (C) (2000 / 2. Calculate the ratio of time intervals t1 : t2.50 Rajeev Gandhi Nagar.273 : 1 (D) 0. B more sensitive 28. (A) then the reaction must be A(g) 3B(g) and is a first order reaction.1 Online Coaching for JEE Main & Advanced 3rd Floor. (B) then the reaction must be A(g) 3B(g) and is a second order reaction. 0. (A) 0. At a time when t = . A B KA = 1015 e–2000/T . C0 e (B) C0e C0 e Concentration of the reactant in first-order is reduced to (A) one natural life-time (C) three natural life-time 34. Ea / (R ln A) (C) 0. When a graph between log K and 1/T is drawn a straight line is obtained.303 R log A (B) . k k The reaction A Products. (A) 2 M (B) ln 2 M (C) 2 log 2 M (D) 2 ln 2 M 25. The half time of the reaction would be : 0. is zero order while the reaction B Products.71 (C) 1. (C) then the reaction must be A(g) 3B(g) and is a zero order reaction. ETOOSINDIA. For the reaction A + 2B products(started with concentrations taken in stoichiometric proportion). is 1st order. 30.303) K (D) (1000 / 2. concentration of the reactant dt k is : (C0 = initial concentration) (A) 33. The rate of a reaction gets doubled when the temperature changes from 7ºC to 17ºC.2 × 10–2 (C) 2 × 10–4 (D) 4 × 10–3 26. Kota.693 k (B) d [A] = k [A] dt 0. Which reaction is faster at a lower temperature and which is more sensitive to changes of temperature ? (A) B faster.COM India's No. H. for 75% completion of the second reaction. rate law is – (C) [B] . the experimentally determined rate law is – (A) 32.420 : 1 (C) 0.76 31.693 1/ k For a reaction 2A + B product.81 (B) 1.67 × 10–3 (B) 1. Two I-order reactions have half-lives in the ratio 3 : 2.24.91 (D) 1. For what initial concentration of A the half lives of the two reactions are equal.311 : 1 (B) 0. The point at which line cuts y -axis and x -axis respectively correspond to the temp : (A) 0.199 : 1 27. V & T results into the following graph.No. 092142 33303 . The following plot is obtained for such a reaction.50 Rajeev Gandhi Nagar. of A in flask 1 becomes 0.4 × 10–2 s–1 38. H.14 × 10–3 Ms–1 (B) 5.25 M.3 × 10–2 s–1 (D) 8.COM India's No. catalyst Consider the following reactions at 300 K. Flask contains 1 L of 1 M solution of A and flask contains 100 ml of 0.7 × 10–3 Ms–1 (C) 4.56 × 10–4 Ms–1 (D) 1.4 hr (B) 2. The rate constant for the reaction is (A) 4. A B (catalyst reaction) –1 The activation energy is lowered by 8. What will be time for cone.14 × 10–2 Ms–1 36. The potential energy diagram is shown in the fig. If initial concentration of [A] is [A]0. of A in flask to become 0.33 = 28) (A) 15 times (B) 38 times (C) 22 times (D) 28 times 37. Which of the following sets of identifications is correct ? (Assume that the temperature and pressure are constant). A substance ‘A' decomposes in solution following the first order kinetics. the rate law expression is (A) The integerated rate expression is k = (B) The graph of d[A ] = k [A]1/2.56 × 10–3 ms–1.35. For the reaction A B.314 KJ mol for the catalysed reaction.1 Online Coaching for JEE Main & Advanced CHEMICAL KINETICS # 29 3rd Floor. EP ES E+S (B) ES Activated complex EP Activated complex (C) EP Activated complex ES Activated complex (D) E+S ES EP E+P More than one choice Type 40.0 hr (D) unpredictable as rate constant is not given 39. The instantaneous rate of disappearance of the MnO4– ion in the following reaction is 4. (1) (2) (3) (4) (A) E + P.3 M ? : (A) 0. the conc. A simple mechanism for enzyme-catalyzed reaction is given by the following set of equations E + (enzyme) S (reactant) ES (intermediate-1) ES (intermediate-1) EP (intermediate-2) EP E + P (intermediate-2) (enzyme) (product) This is known as the Michaelis–Menten mechanism. Kota. Rajasthan 324005 HelpDesk : Tel.6 × 10–2 s–1 (B) 1.6 M solution. then dt 2 1/ 2 ( A A1 / 2 ) t 0 A Vs t will be (C) The half life period t1 / 2 = K 2[A]10/ 2 (D) The time taken for 75% completion of reaction t 3 / 4 = [A]0 k ETOOSINDIA. After 8 hr.No. Then the rate of appearance of 2 is : 2MnO4– + 10 – + 16H+ 2Mn2+ + 52 + 8H2O 2MnO4– + 10 – + 16H+ 2Mn2+ + 52 + 8H2O (A) 1. A B (uncatalysed reaction). The conversion of vinyl allyl ether to pent-4-enol follows first-order kinetics.4 hr (C) 4. How many times the rate of this catalysed reaction greater than that of uncatalysed reaction ? (Given e3.2 × 10–2 s–1 (C) 2. II : SUBJECTIVE QUESTIONS 1. Given E Reaction H Rate constant Energy of activation PA HA kA EA PB HB kB EB Which of the following is(are) true? (A) a = EB (B) b = EA (C) HA = b d (D) HB = c a PART . keeping the concentrations of all other reactants constant. (a) What is rate of formation of H2O. 5 mole of nitric oxide and 2 mol of Cl2 were brought together.No. (D) Decay constant () of radioactive substance is affected by temperature. Select incorrect statement(s): (A) Unit of pre-exponential factor (A) for second order reaction is mol L–1 s–1. Which of the following statement is incorrect ? (A) The order of reaction is the sum of powers of all the concentration terms in the rate equation. E1 2 In a consecutive reaction system A B C when E1 is much greater than E2. (B) For a zero order reaction. 44. (B) A zero order reaction must be a complex reaction. (b) What is rate of disappearance of H2O2. In the following reaction 2H2O2 (aq) 2H2O () + O2 (g) rate of formation of O2 is 36 g min–1 . (C) A zero order reaction is controlled by factors other than concentration of reactants. In a volume of 2 dm3.5 × 10–3 Ms–1 for a particular time interval.COM India's No. (C) Orders of reactions can not be fractional. Rajasthan 324005 HelpDesk : Tel. The oxidation of iodide ion by peroxy disulphate ion is described by the equation : 3– (aq) + S2O82– (aq) 3– (aq) + 2SO42– (aq) (a) If – [S2O82 ] t = 1.41. 45. (D) The order of a reaction can only be determined from the stoichiometric equation for the reaction. H.4 × 10 3 mole dm3 s. 3. Kota. Which of the following is/are correct statement? (A) Stoichiometry of a reaction tells about the order of the elementary reactions.50 Rajeev Gandhi Nagar. (B) The order of reaction with respect to one reactant is the ratio of the change of logarithm of the rate of the reaction to the change in the logarithm of the concentration of the particular reactant. the yield of B increase with (A) increase in temperature (B) decreases in temperature (C) increase in initial concentration of A (D) decrease in initial concentration of A 43. 092142 33303 CHEMICAL KINETICS # 30 . Consider the decay of P to A and B by two parallel first order reactions as shown in Fig. rate and the rate constant are identical. 42. what is the value of – [ ] for the same time t interval ? (b) What is the average rate of formation of SO42– during that time interval ? 2. (D) A zero order reaction is always elementary reaction.1 Online Coaching for JEE Main & Advanced 3rd Floor. What will be the rate when half of the chlorine has reacted ? ETOOSINDIA. (C) Molecularity is defined only for RDS in a complex reaction. and the initial rate was 2. The reaction 2NO(g) + Cl2(g) 2NOCl(g) is second order in NO and first order in Cl2. (i) What is the order of the reaction with respect to A.0 x 10-3 1. 2NH3 (g) N2 (g) + 3H2 (g) (a) What does d [NH 3 ] denote ? dt (b) What does d [N2 ] d [H2 ] and denote? dt dt (c) If the decomposition is zero order then what are the rate of production of N2 and H2 if k = 2. 7.5 Half--life (seconds) 84 84 83 What is the order of the reaction ? ETOOSINDIA. Calculate the order of the reaction. (iii) What is the overall order. 9.4.s1. Decomposition of ammonia on platinum surface follow the change. and the initial rate of the reaction is double that of the first run. the initial concentration of each reactant is double the respective concentrations in the first run.50 Rajeev Gandhi Nagar.COM India's No. the value of rate constant at For the reaction 3BrO(aq) 0 80 C in the rate law for [BrO ]/t was found to be 0.1 Online Coaching for JEE Main & Advanced 3rd Floor. Kota. 6. 092142 33303 CHEMICAL KINETICS # 31 . In another experiment. the initial concentration of A is double that in the first run.5 × 10–4 Ms–1 ? (d) If the rate obeys – k1 [NH3 ] d [NH3 ] = 1 k [NH ] .012 2.0020 4.4 x 10 0.0080 If the reaction rate given by r = [A]a[B]b. (b) What is the rate after 1. (Given e–0. Substance A reacts according to a first order rate law with k = 5 x 105 s1. (v) Calculate the rate constant. 4. Rajasthan 324005 HelpDesk : Tel. 2. an equimolar mixture of gases was reduced to half the value in 102 seconds. How will the rate of reaction change if the volume of reaction vessel is reduced to 1/4th of its original value? 6.0 hour. What is the rate constant when the rate law is written for (a) [BrO3]/t (b) + [Br ] / t ? 5.0 x 10-3 2. (ii) What is the order of the reaction with respect to B. The reaction A + 2B C + 2D is run three times.024 -3 -3 3. (a) The half life period and initial concentration for a reaction are as follows. In the second run. the initial pressure of 340 mm. (iv) Write the rate law equation.mol1. In a kinetic study of the reduction of nitric oxide with hydrogen.057 L.0 x 10-3 0. H. (a) If the initial concentration of A is 1. 11. the initial pressure of 288 mm. and the initial rate is double that of the first run. 6.No. What is the order of the reaction with respect to each reactant? 10. under the same conditions was reduced to half the value in 140 sec. Consider the following data on the hypothetical reaction between the reactants (A) and (B) give products 2A + 2B C + 3D [A] [B] Rate r [mol –1] [mol –1] [mol –1 s–1] 1.0 M.18 = 0. In the third run. BrO3 (aq) + 2Br (aq) in alkaline solution. then. What is order of reaction? Initial concentration 350 540 158 t1/2 425 275 941 (b) The half-life period for the thermal decomposition of PH3 at three different pressures are given below Initial pressure (mm Hg) 707 79 37.4 x 10-3 0.0 x 10 1.84) 8. what will be the order for decomposition of NH3 if (i) [NH3] is very dt 2 3 very less and (ii) [NH3] is very very high K1 and K2 are constant. The rate law of a chemical reaction given below : 2NO (g) + O2 (g) 2NO2 (g) is given as rate = k [NO]2 [O2]. what is initial rate .0 x 10-3 0.0 x 10-3 1. 50 Rajeev Gandhi Nagar. obeys the following mechanism.79 = 0. The kinetic of hydrolysis of methyl acetate in excess dilute HCl at 25°C were followed by withdrawing 2 mL of the reaction mixture at intervals of (t). e1. A B. 16. 8 D B+C A Time 0 t Volume of reagent V1 V2 V3 The reagent reacts with only B.60 32. H. Find the pressure after time '2t'. Rajasthan 324005 HelpDesk : Tel.24 24.88 atm. slow N2O5 NO2 + NO3 Step . Kota. 17.23 42. also calculate the pressure of reaction mixture after 100 second.79 22. After 55 sec at 400 K. The reaction 2NO (g) + Br2 (g) 2NOBr (g). 31 = 0. What will be the percent decomposition in the same time in a 30% solution maintained at 40 0C? Assume that activation energy remains constant in this range of temperature. 14.58 = 4. When a 20% solution of A was kept at 250C for 20 minutes.858] 13 CI2 (g) + CI2O7(g) 7 O (g) 2 2 For a homogeneous gaseous phase reaction 2A 3B + C the initial pressure was P0 while pressure after time 't' was P.No. requires activation energy of 70 kJ mol1. (i) NO (g) + Br2 (g) NOBr2 (g) Slow (ii) NOBr2 (g) + NO (g) 2NOBr (g) Suggest the rate expression. t (in minute) 0 75 119 259 Titre value (in mL) 19. Find k.03 [ln 15.43 9. fast N2O5 + NO3 3NO2 + O2 4NO2 + O2 2N2O5 suggest the rate expression. adding 50 mL water and titrating with baryta water. ln = 0. Determine the velocity constant of hydrolysis. The decomposition of Cl2O7 at 400 K in gaseous phase to Cl2 and O2 is of order reaction. ETOOSINDIA. A first order reaction. ln = 0.2454. Assume first order reaction.8439] 17. Step . 092142 33303 CHEMICAL KINETICS # 32 .12. The thermal decomposition of N2O5 occurs in the following steps. the pressure of reaction mixture increases from 0. 22.62 to 1.COM India's No. [Given : n 13.79 22. C and D. 25% decomposition took place.39. 18. Calculate the rate constant of given reaction.869.20 26.83 15.1 Online Coaching for JEE Main & Advanced 3rd Floor. and [B]eq. What is now the velocity constant of the backward reaction? 22. For a first order reversible reaction A B . When a catalyst is added this velocity constant is increased to 0.002. H.8 at room temperature. ETOOSINDIA. are intermediates in this reaction? (d) Write the rate law for the rate-determining step. k' N2O2 + O2 2 NO2 [slow step]. (e) Write the rate law for this reaction. (rapid equilibration) . the initial concentration of A and B are [A]0 and zero respectively. and the velocity constant of the forward reaction is 0. Kota. The forward reactions rate for the nitric oxide-oxygen reaction 2NO + O2 2NO2 has the rate law . Assume the mechanism: 2NO N2O2 .No. (f) What is overall order of the reaction ? 20. Rajasthan 324005 HelpDesk : Tel.0045. If concentrations at equilibrium are [A]eq. rate = K [NO]2 [O2] . Given the following steps in the mechanism for a chemical reaction : C A + B (fast) D + E B + C slow) A + E D + F fast) At any time [C] is directly proportional to [A]. derive an expression for the time taken by B to attain concentration equal to [B]eq/2 . Show that the rate law can be explained. (a) What is the stoichiometric equation for the reaction ? (b) Which species.1 Online Coaching for JEE Main & Advanced 3rd Floor. are catalysts in this reaction ? (c) Which species. where in K = kk. if any.COM India's No. 21.19. if any.50 Rajeev Gandhi Nagar. 092142 33303 CHEMICAL KINETICS # 33 . Experiment shows that the equilibrium constant of the reaction : CH3COOC2H5 + H2O C2H5OH + CH3COOH is 2. 2 0.2 (C) 0. The rate constant (s–1) of the reaction is : [JEE-2003.I : IIT-JEE PROBLEMS (PREVIOUS YEARS) * Marked Questions are having more than one correct option. H2(g) or NH3(g). If the same reaction is carried out in the presence of a catalyst at the same rate. then the concentration of N2O5 (in mol L–1) is : [JEE-2000.1 0. Find the half life of the reaction. Identify the correct relationship amongst rate expressions : [JEE-2002.50 Rajeev Gandhi Nagar. [JEE-2004.43 × 10–5 (D) 1. The rate of a first order reaction is 0. Rajasthan 324005 HelpDesk : Tel. AB + hv AB * .38 × 10–4 (C) 1. A hydrogenation reaction is carried out at 500 K.2N2O5 4NO2+ O2. the temperature required is 400 K.04 mole litre–1 s–1 at 10 minutes and 0.1 Online Coaching for JEE Main & Advanced 3rd Floor. of N2(g) .1 0. If '' is the intensity of absorbed light and 'C' is the concentration of AB for the photochemical process. is 3 ×10–5 s–1. (A) 3. Calculate the rate constant of the reaction.45 × 10–5 (B) 1. [JEE-2001.73 × 10–5 Initial rates. [JEE-2004.03 mol litre–1 s–1 at 20 minutes after initiation. H.8 2.PART . of the A + B P at different initial concentrations of A and B ([A]0 and [B]0) are given below: –––––––––––––––––––––––––––––––––––––––––––– [A]0 [B]0 r0 (a) (b) (mol L–1) (mol L–1) (mol L–1 s–1) –––––––––––––––––––––––––––––––––––––––––––– 0. 3/84] (A) 3. Calculate the activation energy of the reaction if the catalyst lowers the activation barrier by 20 kJ mol–1. In the biologically-catalysed oxidation of ethanol.0025 (mol/lit) time (min) 0 40 Initial rate of reaction is in mol //min.04 (D) 0. 3/90] (A) Rate = (C) Rate = d [N2 ] 1 d [H2 ] 1 d [NH3 ] dt 3 dt 2 dt d [N2 ] 1 d [H2 ] 1 d [NH3 ] dt 3 dt 2 dt (B) Rate = d [N2 ] d [H2 ] d [NH3 ] 3 2 dt dt dt (D) Rate = d [N2 ] d [H2 ] d [NH3 ] dt dt dt 6.2 0.73 × 10–4 8.COM India's No.43 × 10–4 (B) 1.1 0. The rate constant for the reaction. N2(g) + 3H2(g) 2NH3(g) The rate of this reaction can be expressed in terms of time derivatives of conc.05 –––––––––––––––––––––––––––––––––––––––––––– Write the rate equation. 1/35] (D) C.05 0. 5/100] 5. 3/100] 3.4 (B) 1.No.00 × 10–4 (D) 5. the concentration of ethanol decreases in a first order reaction from 800 mol dm–3 to 50 mol dm–3 in 2 ×104 s. 2/60] ETOOSINDIA. 3/84] (C) 3. 1/35] (A) 1. Consider the chemical reaction.4 × 10–5 mol L–1 s–1. [JEE-2000.01 0. 092142 33303 CHEMICAL KINETICS # 34 .00 × 10–5 7.1 0. 4. Given X product (Taking 1st order reaction) conc 0. r0.10 0. 1. If the rate is 2. the rate of formation of AB * is directly proportional to (A) C (B) (C) 2 [JEE-2001. Kota. The overall order of the reaction is : [JEE-2007. rate constant of the reaction. [JEE-2005.75 0. When concentration of both the reactants G and H is doubled.386 mol dm–3 of a substance becomes half in 40 k1 seconds and 20 seconds through first order and zero order kinetics.9. Consider a reaction aG + bH Products.No.12 0. 3/163] [R](molar ) 1. 3/163] (B) (C) (D) The concentration of R in the reaction R P was measured as a function of time and the following data is obtained : [JEE-2010. Rajasthan 324005 CHEMICAL KINETICS # 35 HelpDesk : Tel. time taken for 75% completion and find the total pressure when partial pressure of X.5 mol–1 dm3 (B) 1.0. Kota. the rate is doubled.0 × 10 s and 9.6 kJ mol (C) 1. respectively. the temperature (T) dependent rate constant (k) was found to follow the equation 1 + 6. 3/80] 6 –1 –1 –1 –1 (A) 1. Px (mm of Hg) (Partial pressure of X) 800 400 200 Find order with respect to X.0 mol–1 dm3 12.2 kJ mol (B) 6. The number of neutrons emitted when 235 92 U undergoes controlled nuclear fission to 142 54 Xe and 90 38 Sr is : [JEE-2010.0 mol dm–3 (C) 1.0 0. The plot that follows Arrhenius equation is : (A) 14.1 Online Coaching for JEE Main & Advanced 3rd Floor. 3/81] (A) 0 (B) 1 (C) 2 (D) 3 11.) 0.10 t(min. For a first order reaction A P.40 0. [JEE-2010. H. 092142 33303 . (A) 0.3 kJ mol–1 13.0 0.5 mol dm–3 [JEE-2008. Plots showing the variation of the rate constant (k) with temperature (T) are given below. 3/82] (D) 2.05 0.COM India's No. The pre-exponential factor A and the activation energy Ea. the rate increases by eight times.0 s and 16.0 × 106 s–1 and 38.18 The order of the reaction is : 15.6 kJ mol–1 (D) 1. 2/60] 10. initial concentration of 1. when concentration of G is doubled keeping the concentration of H fixed. respectively. Px = 700 mm of Hg. For a reaction 2X(g) 3Y(g) + 2Z(g) Time (min) 0 100 200 the following data is obtained. However. Ratio of the rate k0 constant for first order (k1) and zero order (k0) of the reaction is. Under the same reaction conditions.50 Rajeev Gandhi Nagar.0 × 106 s–1 and 16. 3/163] ETOOSINDIA. are : log k = – (2000) T [JEE-2009. 16. proton. Units of rate constant of first and zero order reactions in terms of molarity M unit are respectively [AIEEE. sec–1 (4) M. the rate of reaction will : [AIEEE.COM India's No. H. For the first order reaction 2N2O5(g) 4NO2(g) + O2(g) (A) the concentration of the reactant decreases exponentially with time (B) the half-life of the reaction decreases with increasing temperature (C) the half-life of the reaction depends on the initial concentration of the reactant (D) the reaction proceeds to 99. (i) and (ii) as shown. neutron (B) Neutron.2002] mn (2) (m + n) (3) (n – m) (4) 2(n – m). 092142 33303 CHEMICAL KINETICS # 36 . positron. ETOOSINDIA. proton (D) positron. the ratio of the new rate to the earlier rate of the reaction will be as [AIEEE. An organic compound undergoes first-order decomposition.No. Y and Z respectivey are : [JEE-2011. neutorn 17.3). Products X. For the reaction A + 2B C. For the reaction system: 2NO(g) + O2(g) 2NO2(g). M sec–1 (2) sec–1. volume is suddenly reduced to half its value by increasing the pressure on it.50 Rajeev Gandhi Nagar.2003] (1) diminish to one-fourth of its initial value (2) diminish to one-eighth of its initial value (3) increase to eight times of its initial value (4) increase to four times of its initial value. position (C) proton. Rajasthan 324005 HelpDesk : Tel. dH2 d[I ] d[HI] 2 dt dt dt [AIEEE.2003] 1 (1) 2 5. On doubling the concentration of A and halving the concentration of B. The differential rate law for the reaction H2 + I2 2HI is : (1) (3) 4. [ t1 / 10 ] [IIT-JEE-2012] PART . rate is given by R = [A] [B]2 then the order of the reaction is : [AIEEE. Bombardment of aluminum by -particle leads to its artificial disintegration into two ways. neutron. 3/163] (A) Proton.II : AIEEE PROBLEMS (PREVIOUS YEARS) 1.sec–1. Kota. What is the value of [JEE-2011.6% completion in eight half-life duration 18. M (3) M. The time taken for its decompositionto 1/8 and 1/10 of its initial concentration are t1/8 and t1/10 respectively. 4/163] [ t 1/ 8 ] × 10? (take log10 2 = 0. position.1 Online Coaching for JEE Main & Advanced 3rd Floor. (2) 1 dH2 1 d[I2 ] d[HI] 2 dt 2 dt dt dH2 d[I2 ] 1 d[HI] dt dt 2 dt (4) 2 dH2 d[I ] d[HI] 2 2 dt dt dt The rate law for a reaction between the substances A and B is given by rate = k [A]n [B]m.2002] (1) 3 (2) 6 (3) 5 (4) 7 3. If the reaction is of first order with respect to O2 and second order with respect to NO.2002] (1) sec–1. sec–1 2. the order of the reaction with respect to NO(g) is [AIEEE.1 Online Coaching for JEE Main & Advanced 3rd Floor. The half . H. 9.084 g (3) 3. decreases from 0.5 minutes [AIEEE. with everything else kept the same. Rajasthan 324005 CHEMICAL KINETICS # 37 HelpDesk : Tel.50 Rajeev Gandhi Nagar. (2) 15 minutes (3) 7. In the respect of the equation k = Ae –Ea/RT in chemical kinetics. 3/120] (1) 1 14. In a first order reaction.2004] (4) 60 minutes The rate equation for the reaction 2A + B C is found to be : rate = k[A] [B]. 12.4 M in 15 minutes. The time taken for the concentration to change from 0. NOBr2 (g) + NO (g) 2NOBr (g) (slow step) If the second step is the rate determining step. (1) k is equilibrium constant (2) A is adsorption factor (3) Ea is energy of activation (4) R is Rydberg constant. If the initial mass of the isotope was 200 g.2006] 13.125 g (4) 4.2003] 7. respectively. Consider an endothermic reaction X Y with the activation energies Eb and Ef for the backward and forward reaction. The presence of a catalyst lowers the activation energy of both (forward and reverse) reactions by 100 kJ mol–1. which one of the following statements is correct : [AIEEE. The enthalpy change of the reaction (A2 + B2 2AB) in the presence of catalyst will be (in kJ mol–1).6.COM India's No.025 M is : (1) 30 minutes 8. the mass remaining after 24 hours undecayed is : (1) 1.167 g.2005] A reaction was found to be second order with respect to the concentration of carbon monoxide.2004] (2) 2.2007. Kota. (1) remain unchanged (2) tripled (3) increased by a factor of 4 (4) doubled The following mechanism has been proposed for the reaction of NO with Br2 to form NOBr. NO (g) + Br2 (g) NOBr2 (g) .042 g 10. [AIEEE. (2) 0 (3) 3 (4) 2 The energies of activation for forward and reverse reactions for A2 + B2 2AB are 180 kJ mol–1 and 200 kJ mol–1 respectively.1 M to 0.8 M to 0.2005] (2) H = U (3) H < U (4) H > U A reaction involving two different reactants can never be : (1) unimolecular reaction (2) first order reaction (3) second order reaction (4) bimolecular reaction [AIEEE. 092142 33303 . 3/120] (2) 20 (3) 300 (4) 120 ETOOSINDIA. In general (1) Eb < Ef 11.life of a radioisotope is four hours. [AIEEE.No.2004] (1) unit of k must be sec–1 (2) t1/2 is a constant (3) rate of formation of C is twice the rate of disappearance of A (4) value of k is independent of initial concentrations of A and B. The correct statement in relation to this reaction is that the : [AIEEE. the rate of reaction will be : [AIEEE. (1) 280 [AIEEE. If the concentration of carbon monoxide is doubled.2007. the concentration of the reactant. The rate of reaction when the concentration of A is 0. 3/105] d [ A ] 1 d [B] dt 4 dt (2) d [ A ] d [B] dt dt (3) d[A] d [B] 4 dt dt (4) d [ A ] 1 d [B] dt 2 dt The half life period of a first order chemical reaction is 6.73 × 10–4 M/min ETOOSINDIA.2008. the concentration of A changes from 0.73 × 10 M/min –3 (2) 3. If the temperature is raised by 50° C. 8/144] (2) 0. H.COM India's No. If it is a zero order reaction? (1) 4 h 19. Its half-life period is 30days. how much time does it take for its concentration to come from 0. rate of disappearance of 'A' related to the rate of appearance of 'B' by the 2 expression.15. (2) Both A and B (3) Neither A nor B (4) A only The rate of a chemical reaction doubles for every 10° C rise of temperature. (2) 100 days For a reaction (3) 1000 days (4) 300 days 1 A 2B.47 × 10 M/min –4 (3) 3. [AIEEE .2010.No. is (1) 1. 8/144] (3) 460.93 minutes.6 minutes (4) 230.03 minutes 18. the rate of the reaction increases by about : (1) 10 times 21.50 Rajeev Gandhi Nagar.50 to 0. Time required for the completion of 99% of the chemical reaction will be (log 2 = 0. Cl2 + H2S H+ + Cl– + Cl+ + HS– (slow) Cl+ + HS– H+ + Cl– + S (fast) B.1 Online Coaching for JEE Main & Advanced 3rd Floor.47 × 10 M/min –5 [AIEEE-2012.1 M to 0. 3/120] (1) 10 days 16.5 h (3) 0.2007. H2S H+ + HS– (fast equilibrium) Cl2 + HS– 2Cl– + H+ + S (slow) (1) B only 20.301) : (1) 23. (1) 17.06 minutes [AIEEE .2010. (A) products.2009.01 M. (2) 24 times [AIEEE-2011] (3) 32 times (4) 64 times For a first order reaction. When the initial concentration of the reactant ‘A’. 4/144] Cl2 (aq) + H2S(aq) S(s) + 2H+ (aq) + 2Cl– (aq) The rate equation for this reaction is rate = k [Cl2][H2S] Which of these mechanisms is/are consistent with this rate equation? A.25 mol L–1. is 2. 4/120] (4) 1.3 minutes The time for half life period of a certain reaction A Products is 1 hour.025 M in 40 minutes. A radioactive element gets spilled over the floor of a room. Rajasthan 324005 CHEMICAL KINETICS # 38 HelpDesk : Tel. after how many days will it be safe to enter the room : [AIEEE.0 mol L–1 . [AIEEE. 092142 33303 .25 h (4) 1 h Consider the reaction [AIEEE . (2) 46. If the initial activity is ten times the permissible value. Kota. What is the effect of temperature on the rate constant of a reaction? How can this effect of temperature on rate constant be represented quantitatively? 8.50 Rajeev Gandhi Nagar. Calculate the rate of reaction after [A] is reduced to 0. How is the rate of reaction affected if the concentration of the reactant is (i) doubled (ii) reduced to half ? 7.0 × 10 –6 mol –2 L 2 s –1. From the rate expression for the following reactions. [B] = 0. In a pseudo first order hydrolysis of ester in water. then what are the units of rate and rate constants? 5. H. the following results were obtained : (i) Calculate the average rate of reaction between the time interval 30 to 60 seconds. Kota.5 × 10–4 mol–1 L s–1? 4. (ii) How is the rate affected on increasing the concentration of B three times? (iii) How is the rate affected when the concentrations of both A and B are doubled? 10. Mention the factors that affect the rate of a chemical reaction.1 Online Coaching for JEE Main & Advanced 3rd Floor. Rate = k (PCH3OCH3)3/2 If the pressure is measured in bar and time in minutes.2 mol L–1. i. (i) 3NO(g) N2O (g) Rate = k[NO]2 – + – (ii) H2O2 (aq) + 3I (aq) + 2H 2H2O (l) + I3 Rate = k[H2O2] [I–] (iii) CH3CHO (g) CH4 (g) + CO(g) Rate = k[CH3CHO]3/2 (iv) C2H5Cl (g) C2H4 (g) + HCl (g) Rate = k[C2H5Cl] 2.PART-NCERT QUESTIONS 1. What are the rates of production of N2 and H2 if k = 2. so the rate can also be expressed in terms of the partial pressure of dimethyl ether. In a reaction between A and B.06 mol L–1. A reaction is first order in A and second order in B. 3. 092142 33303 CHEMICAL KINETICS # 39 .e. The decomposition of NH3 on platinum surface is zero order reaction. (i) Write the differential rate equation. Calculate the initial rate of the reaction when [A] = 0.1 mol L–1. For the reaction : 2A + B A2B the rate = k[A][B] 2 with k = 2.No. Rajasthan 324005 HelpDesk : Tel. The decomposition of dimethyl ether leads to the formation of CH4. the initial rate of reaction (r0) was measured for different initial concentrations of A and B as given below : What is the order of the reaction with respect to A and B? ETOOSINDIA. 9. determine their order of reaction and the dimensions of the rate constants.COM India's No. H2 and CO and the reaction rate is given by Rate = k [CH3OCH3]3/2 The rate of reaction is followed by increase in pressure in a closed vessel.. (ii) Calculate the pseudo first order rate constant for the hydrolysis of ester. A reaction is second order with respect to a reactant. 6. The reaction between A and B is first order with respect to A and zero order with respect to B. 12. The half-life for radioactive decay of 14C is 5730 years. (ii) Find the half-life period for the reaction. ETOOSINDIA.11. Kota. one of the products is 90Sr with half-life of 28. 15. How much time will it take to reduce the initial concentration of the reactant to its 1/16th value? 17. (iii) Draw a graph between log[N2O5] and t.1 years. Fill in the blanks in the following table: 13. The rate constant for a first order reaction is 60 s–1. Rajasthan 324005 HelpDesk : Tel. (iv) What is the rate law ? (v) Calculate the rate constant. 092142 33303 CHEMICAL KINETICS # 40 . During nuclear explosion. An archaeological artifact containing wood had only 80% of the 14C found in a living tree. (vi) Calculate the half-life period from k and compare it with (ii). Estimate the age of the sample. how much of it will remain after 10 years and 60 years if it is not lost metabolically. The experimental data for decomposition of N2O5 [2N2O5 4NO2 + O2] in gas phase at 318K are given below : (i) Plot [N2O5] against t. 16.No.1 Online Coaching for JEE Main & Advanced 3rd Floor.COM India's No. H. If 1µg of 90Sr was absorbed in the bones of a newly born baby instead of calcium. Calculate the half-life of a first order reaction from their rate constants given below : (i) 200 s–1 (ii) 2 min–1 (iii) 4 years–1 14.50 Rajeev Gandhi Nagar. The following results have been obtained during the kinetic studies of the reaction : 2A + B C + D Determine the rate law and the rate constant for the reaction. show that time required for 99% completion is twice the time required for the completion of 90% of reaction.5 × 104 s–1? 29. If the energy of activation is 179. The rate of a reaction quadruples when the temperature changes from 293 K to 313 K. The decomposition of A into product has value of k as 4.0 mol L–1.50 Rajeev Gandhi Nagar. 24. the following data are obtained. Sucrose decomposes in acid solution into glucose and fructose according to the first order rate law. Calculate the concentration of A remaining after 100 s if the initial concentration of A is 1. Calculate the rate constant. The rate constant for the first order decomposition of H2O2 is given by the following equation : log k = 14. Calculate k at 318 K and Ea. Calculate the energy of activation of the reaction assuming that it does not change with temperature.25 × 104K/T Calculate Ea for this reaction and at what temperature will its half-period be 256 minutes? 28. A first order reaction takes 40 min for 30% decomposition. Predict the rate constant at 30° and 50°C. 21.418 × 10–5 s–1 at 546 K. Consider a certain reaction A Products with k = 2. 23. The decomposition of hydrocarbon follows the equation k = (4.No.0 × 10–2 s–1. What fraction of sample of sucrose remains after 8 hours ? 26. The rate constant for the decomposition of N2O5 at various temperatures is given below : Draw a graph between ln k and 1/T and calculate the values of A and Ea. The time required for 10% completion of a first order reaction at 298K is equal to that required for its 25% completion at 308K.COM India's No.5 × 103 s–1 at 10°C and energy of activation 60 kJ mol–1. For a first order reaction.1 Online Coaching for JEE Main & Advanced 3rd Floor. Rajasthan 324005 HelpDesk : Tel.00 hours. 20. Calculate t1/2. Kota.18. 19.9 kJ/mol. If the value of A is 4 × 1010 s–1.5 × 1011s–1) e–28000K/T 27. SO2 Cl2 (g) SO2 (g) + Cl2 (g) Calculate the rate of the reaction when total pressure is 0. k = (4.5 × 1011s–1) e–28000K/T. H. For the decomposition of azoisopropane to hexane and nitrogen at 543 K. what will be the value of pre-exponential factor. At what temperature would k be 1. Calculate Ea.34 – 1. ETOOSINDIA. The rate constant for the decomposition of hydrocarbons is 2. with t1/2 = 3.65 atm. 25. 30. The following data were obtained during the first order thermal decomposition of SO2Cl2 at a constant volume. 22. 092142 33303 CHEMICAL KINETICS # 41 . (C) p. rate constant 73. (B) s . 23. C-26. 24. (ABCD) H-1. 16. (B) (B) (C) (B) (D) (D) (B) (C) (D) (C) (A) (C) (D) (D) (B) A-5. T 60. C-9. C-19. (C) (D) (B) (C) (D) (D) (ABC) 75. 29. s (A) r . (a) 4. 14.5 × 10–4 M sec–1 . s . 31. F-8. r. 70. T 52. (D) q (D) 22. (D) q 16. F-15. (B) 31. C-1. D-6. T 42. H. B-10. (A) (A) (D) (B) (D) (AC) SUBJECTIVE 1. C-5.No. (B) 28.COM India's No. 26. (B) r. F-14. (D*) 3. (A) C-20. 15. T 59. (C) p. (D) p. (D) (D) (B) (D) (C) (B) (C) (A) (C) (B) (D) (B) (A) (B) A-6. 36. 25. 47. H-6. (D) B-11. (A) 29. (D) q 18. A-8. T 41. 2 and 3 76. C-23. 20.* (D) A-7. 37. (C) F-13.019 M1 s1 . (B) C-6.2 × 105 M/s 2. 45. 39. (D) (B) (D) (C) (A) (D) 5. (C*) 6. T 39. F-5. 7. (a) 40.5 gm min–1 (b) 76. F-10. T 55. (A) 30. 17. r. F-1. T 43. B-6. H-4. C-4. C-8. 18. 34. D-10. B-2. H-3. F-11. D-2. –k 77. (C) p . 11. 27. 6. F 32. r. (c) 2. (B) C-3. G-3. (C) (D) (B) (A) (B) (B) (C) (B) (D) (B) (A) (A) (B) (B) 1. B-7. Rajasthan 324005 HelpDesk : Tel. (A*) 2. D-12. (C) 20. F 50. C-16. (C*) 14. PART # II (B*) 4. (A) C-27. D-9. B-5. E-2.0 × 10–3 M sec–1 4. (B) C-10. 35. (B) B-8. B-9. C-22. s . 10. T 56. threshold energy 62. 30. F 34. F-7. 15. E-3. infinte temp.5 gm min–1 (a) 0. 40. (C) G-5. (b) 0.5 × 10–4 M sec–1 (a) 5 × 105 M/s (b) 4. (C*) 9. 26. F 46. (A*) 13. T 35. C-7. G-4. q. 3. 9. D-1. rare increases 71. (D) D-11. 1 1 2 PART # I A-4. first 80. k 2. (C) s. 43. 65. (C) (C) (B) (C) (C) (C) (CD) 3. Activation energy for forward reaction 79. C-14. B-3. 54. 33. (A) F-9. T 51. F-3. exponents 1. slowest 72. T 53.1 Online Coaching for JEE Main & Advanced 3rd Floor. (A) p. C-21. C-11.5 × 10–3 M sec–1 (b) 3. G-2. 092142 33303 CHEMICAL KINETICS # 42 . (A) F-2. (A) 27. (C) D-4. 41. A-3. (C) (B) (D) (D) (C) (C) (ABC) 4. Kota. (A*) 5. (A) B-1. rate = k[M]2 78. D-3. E-6. fast 69. (B) p . (C) 24. 19. 5. 33. D-5. D-13.50 Rajeev Gandhi Nagar. q. 3 74. 32. 28. s . A-9. (C*) 10. half 63. (D) G-1. (A) r . equal to H 64. (B) E-1. E-4. F 44. 7. (B*) 11. G-6. mol L–1s–1 66. (B) q. (C) C-13. 12. C-2.32 × 10–4 M sec–1 64 (a) Rate of decomposition of NH3. 8. (B) (B) (D) (A) (D) (C) (C) (B) (D) (B) (A) (B) (D) (B) A-2. F 57. 17. H-2. 4. C-12. (C) E-5. directly proportional 67. (C) C-17. sum. 38. (A*) 7. 13. (A) s. (A) 25. 61. 19. unchanged 6. C-15. T 38. or for zero Ea 82.038 M1 s1 (b) Rate of formation of N2 and rate of formation of H2 (d) (i) 1 . 40. (ii) 0 ETOOSINDIA. F 36. lowering 68.EXERCISE # 1 A-1. r. G-7. C-18. D-8. F 37. (C) H-5. 21. (A) 23. T 48. 42. (D) B-4. A-10. (C) 21. B-12. T 49. F 45. 22. (B) (C) (B) (A) (B) (AD) 7. (B) (D) (B) (A) (C) (ACD) (B) (C) (D) (B) (B) (B) (C) (A) (A) (C) (A) (C) (B) (C*) (C*) (C) T F T F F EXERCISE # 2 2. 44. 8. F-12. (B) C-24. (C) D-7. (B) F-6. C-25. (B*) 12. F 58. F-4.303 81. (d) rate = k [B][C]. D) 1.13 2.10 2. Kota.1 (D) 1.925 × 10–4 s–1 8.10% 18.93 × 10–3 min–1.33 atm. (i) 2nd order.21 n2 = 6. Rate of reaction = rate of diappearance of A = 0.17 (A.50 Rajeev Gandhi Nagar.4 2.001607 22.15 (1) (4) (2) 2. 092142 33303 . (a) n = 2.9 2X(g) 3Y(g) + 2Z(g) t=0 800 – – t=0 800 – 2x 3x 2x = (800 + 3x) from given data in time 100 min the partial pressure of X decreases from 800 to 400 so t1/2 100 min. 0.005 mol litre–1min–1 6.4 t1/2 = 24 min.693 K f Kb EXERCISE # 3 PART-I 1. (iii) Third order.15 3 (3) (3) (4) 2.6 2. k [N2O5] 0. 3.2 2. (iv) r = K[A]2[B] (v) k = 3.9 2. 3. 13.5 (A) 1. Rate = k [NO]2 [Br2] 19. (c) C. t1/ 2 Time taken for 75% completion = 2 × t1/2 = 200 min.1 Online Coaching for JEE Main & Advanced 3rd Floor. H. order with respect to B = 0 3 11.897 kJ mol–1 ETOOSINDIA.3 (B) 1. 1. 1.66 × 10–6 Ms–1 Order of the reaction is 2. 4. Rate constant K = 1. (e) rate = k'[A][B] (f) 2. (a) 2 B+F 2E. (ii) 1st order.10 1.8 (a) R0 = k[A0].14 2.18 (4) (1) (3) 2. B.20 (3) (4) (3) 2. 2P0 15.7 2.11 2. Since half left is independent of initial concentration so reaction must Ist order with respect to X.16 1.8.12 (D) 1.No.13 (A) (A) 1.D. Ea = 52. 10. 1. t= (2 P0 P)2 P0 14. (b) First Order 12. Now when Px = 700 = 800 – 2x so x = 50 mm of Hg so total pressure = 800 + 3x = 950 mm of Hg (D) 1.471 × 10–19 2.12 2.5 2. ( V3 V1) 1 ln ( V V ) t 3 2 16.27 × 10–3 min–1 17. 5.8 2. 9.19 (1) (2) (2) EXERCISE # 4 1.11 (A) 1. % decomposition = 67.3 2.58 × 10–2.2 EA = 100 KJ/mol 1.17 (4) (1) (2) 2. (b) A.14 0 1.COM India's No.7 (A) 1.18 9 PART-II 2. (b) 0. 1. Rajasthan 324005 CHEMICAL KINETICS # 43 HelpDesk : Tel.6 (B) 1. 21.5 XY Rate = k[X]2 The rate will increase 9 times t = 444 s 9. 1.1 2.16 (1) (3) (1) 2.5 sec–1. rav = 6. Also in next 100 min Px decreases from 400 to 200 to again t1/2 = 100 min. 2.33 × 105 mole–2 +2 sec–1 order with respect to A = 1.
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