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. Rate = = = mol lit–1 time–1 = mol dm–3 time–1 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 Total change in concentrat ion c [R] [P] Average rate = Total time taken = = – = t t t Instantaneous rate : rate of reaction at a particular instant. c dc d [R] d [P] Rinstantaneous = tlim 0 = =– = t dt dt dt Relation between reaction rates of different species involved in a reaction : For the reaction : N2 + 3H2 2NH3 d [N2 ] 1d [H2 ] 1 d [NH3 ] Rate of reaction = = 3 dt = 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 C0 Ct Rate = k = or kt = C0 – Ct or Ct = C0 – kt ' t' Unit of K is same as that of Rate = mol lit–1 sec–1. First Order Reactions : 2.303 C0 k= log C t t Half life time (t1/2) 2.303 log 2 ln2 0.693 t1/2 = = = 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 CHEMICAL KINETICS # 1 HelpDesk : Tel. 092142 33303 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 First-Order Second-Order nth order Ä[ A ] [ A ] [ A ] Ä[ A ] k [ A ]n Differential Rate law = k[A]° = k[A] = k[A]2 Ät Ät t t 1 1 1 1 (Integrated Rate law) [A]t = [A]0 – kt In [A]t = –kt + In [A]0 = kt + n1 [A]t [ A ]0 (A t ) ( A 0 )n 1 (n 1) kt 1 1 Linear graph [A]t v/s t In [A] v/s t [A] v/s t ( A t )n1 v/s t [ A ]0 0.693 1 1 Half-life t1/2 = t1/2 = t1/2 = k[ A ] t 1/ 2 2k k 0 ( A 0 )n1 (depends on [A]0) (independent of [A]0) (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 a r01 [A ] 0 1 r02 [ A 0 ]2 Method of half lives : The half lives of each order is unique so by comparing half lives we can determine order 1 for nth order reaction t1/2 [R0 ]n1 t1 / 2 (R '0 )n1 t1' / 2 (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 CHEMICAL KINETICS # 2 HelpDesk : Tel. 092142 33303 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) nB(g) 2.303 P0 (n 1) k= 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 [this is exclusively for HCl.] Vt = vol. of titrant used at ‘t’ V = vol. of titrant used at t = 2.303 V V0 k= 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 t = 0, t = t and t = 2.303 Y Y0 K= 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) K t 10 T.C. = Kt 2 to 3 ( for most of the reactions) Arrhenius theory of reaction rate : HR = Summation of enthalpies of reactants Threshold enthalpy Enthalpy (H) Ea1 Ea2 or energy HP = Summation of enthalpies of reactants H = Enthalpy change during the reaction HR Reactants Ea1 = Energy of activation of the forward reaction H = Hp – HR = Ea1 – Ea2 Ea2 = Energy of activation of the backward reaction HP Products Progress of reaction (or reaction coordinate) ETOOSINDIA.COM India's No.1 Online Coaching for JEE Main & Advanced CHEMICAL KINETICS # 3 3rd Floor, H.No.50 Rajeev Gandhi Nagar, Kota, Rajasthan 324005 HelpDesk : Tel. 092142 33303 Rajasthan 324005 CHEMICAL KINETICS # 4 HelpDesk : Tel. Molecularity is defined only for the elementary reactions and not for complex reactions. the orders in the rate law equal the coefficients of the reactants. the function of the catalyst is to E'a lower down the activation energy. Fraction of T2 molecule T1 T2 > T1 Ea e E a / RT represents fraction of molecules K. H. Products Ea – E’a = lowering of activation energy by catalyst. Kota.COM India's No. No elementary reactions involving more than three molecules are known. P.E.E. k e Ea / RT k Ae Ea / RT [Arrhenius equation] General characteristics of catalyst : P.E. 2 or 3. because of very low probability of near-simultaneous collision of more than three molecules.50 Rajeev Gandhi Nagar.S. That is. where a + b = 1.No. 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. HR Ea = Energy of activation in absence of catalyst. For an elementary reaction.1 Online Coaching for JEE Main & Advanced 3rd Floor. Reaction Coordinate Molecularity and Order : The number of molecules that react in an elementary step is the molecularity of the elementary reaction. HP E’a = Energy of activation in presence of catalyst. having energy greater Ea rate e E a / RT dependence of rate on temperature is due to dependence of k on temperature. Ep 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. Ea A catalyst drives the reaction through a low energy path and hence Ea is less. 092142 33303 . The rate law for the elementary reaction aA + bB products rate = k[A]a[B]b. GIVEN If R. involves only reactant. [P] or catalyst with the help of same equilibrium step given in mechanism.1 Online Coaching for JEE Main & Advanced CHEMICAL KINETICS # 5 3rd Floor. E intermediate 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. Rajasthan 324005 HelpDesk : Tel. of [R]. ETOOSINDIA.D. 092142 33303 . Kota.S.S.COM India's No.50 Rajeev Gandhi Nagar. a – xeq. xeq. MECHANISMS IN WHICH RDS NOT SPECIFIED : STEADY STATE APPROXIMATION : d [intermediate ] At steady state =0 dt COMPLICATIONS IN 1st ORDER REACTION PARALLEL 1st ORDER REACTION OR COMPETING FIRST-ORDER REACTIONS 1 1 1 Teff = T1 + T2 (remember) [B] k1 = k (remember) [C] 2 E a1 k 1 E a 2 k 2 Ea k1 k 2 REVERSIBLE 1ST ORDER REACTION ( both forward and backward ) A B t=0 a 0 t=t a –x x t = teq.S. product or catalyst on reactant side rate law of R. we have to specify [ intermediate] in expression of rate law in terms of conc. = rate law of reaction RDS is having intermediate on reactant side To calculate order.D. ( sequence of elementary reaction in which any complex reaction procceds) T. or having some mechanism.No.COMPLEX REACTION : Reaction which proceed in more than two steps. CALCULATION OF RATE LAW/ ORDER MECHANISM IN WHICH R.S. H.D. 1 Online Coaching for JEE Main & Advanced CHEMICAL KINETICS # 6 3rd Floor.COM India's No. Kota.No. kfa ( k f k b ) t x = k k f b 1 e (remember) 1 x eq. kf + kb = n x x (remember) t eq. H.50 Rajeev Gandhi Nagar. Rajasthan 324005 HelpDesk : Tel. SEQUENTIAL 1ST ORDER REACTION OR CONSECUTIVE FIRST-ORDER REACTIONS 1 k 2 k A B C all first order equation t=0 a 0 0 t a–x y z k 1a y = k k { e k1t – e k 2 t } (remember) 2 1 1 k1 tmax. = k k n k (remember) 1 2 2 CASE- k1 >> k2 1 k 2k A B C CASE : k2 >> k1 Conc a [C] [B] [A] t ETOOSINDIA. 092142 33303 . 4 0.5 0. 1 (B) 3.50 Rajeev Gandhi Nagar.5 = then x.3 0. Rajasthan 324005 CHEMICAL KINETICS # 7 HelpDesk : Tel. Section (A) : Rate of Reaction A-1. x : y is : (A) 1 : 2 (B) 2 : 1 (C) 3 : 1 (D) 3 : 10 A-4. 2. If the rate of disappearance of hydrogen is 1.5 x 104 mol L-1 sec-1 dt The rate of disappearance of (SO2) will be - (A) 5. In the following reaction : xA yB d[ A ] d[B] log dt = log dt + 0. A reaction follows the given concentration (M)–time graph.25 x 10—4 mol L–1 s–1 (C) 3. 2SO2 (g) + O2 (g) 2SO3 (g) d(O 2 ) The rate of reaction is expressed as – = 2.COM India's No. What is the rate of formation of ammonia. 3.0 x 10—4 mol L–1 s–1 (B) 2.75 x 10 —4 mol L s–1 –1 (D) 50. H.7 × 103 (D) 0. 092142 33303 . Thus. 2.8 × 103 (B) 1. In the formation of sulphur trioxide by the contact process.I : OBJECTIVE QUESTIONS *Marked Questions are having more than one correct option.) (A) 1.3 where –ve sign indicates rate of disappearance of the reactant.1 0 20 40 60 80 100 Time/second (A) 4 × 10–3 M s–1 (B) 8 × 10–2 M s–1 (C) 2 × 10–2 M s–1 (D) 7 × 10–3 M s–1 A-7.1. For the reaction : N2 + 3H2 2NH3.9 × 103 ETOOSINDIA. 2 (D) 2.2 × 103 (C) 2.No. xA + yB zC.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 d[ A ] d[B] d[C] A-2. Kota. Rate of formation of SO3 in the following reaction 2SO2 + O2 2SO3 is 100 g min–1. For a hypothetical reaction A L. 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-6.0 x 10—4 mol L–1 s–1 A-5. (mole per litre per sec. 3 A-3. The rate for this reaction at 20 seconds will be: 0. If = = 1.1 Online Coaching for JEE Main & Advanced 3rd Floor.8 × 103. 3 (C) 3.y and z are : dt dt dt (A) 1. PART . The rate of a reaction is expressed in different ways as follows . Kota. The initial rate of the reaction is 4 × 10–2 mol L–1 s–1. 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. 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 order of reaction is given by : (A) 1 (B) 1/2 (C) 2 (D) 13/12 B-3. the order of reaction is one with respect to A and one with respect to B. is given. dx/dt = k [A]a [B]b .1 Online Coaching for JEE Main & Advanced 3rd Floor. 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 (D) Mole1–n litren-1 sec–1 B-4. If concentration of A is doubled. 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. 092142 33303 . For a gaseous reaction the rate equation is v = k[A][B].25 mol L—1 per min Section (B) : Rate Law B-1. For a chemical reaction 2X + Y Z. For a chemical reaction.A-8. The rate of certain hypothetical reaction A + B + C products. the rate of appearance of Z is 0.05 mol L–1 per min.05 mol L—1 per hour (B) 0.No. What is the order of a reaction whose rate = KCA3/2 CB–1/2 ? (A) 2 (B)1 (C)– 1/2 (D) 3/2 dA 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. A + 2B 3C + D. H. If the volume of the gaseous system is suddenly reduced to 1/3 of initial volume. Rajasthan 324005 CHEMICAL KINETICS # 8 HelpDesk : Tel. rate is four times.COM India's No.05 mol L—1 per min (C) 0. ln the reaction. the rate of the reaction would become- (A) 2 × 10–2 mol L–1s–1 (B) 1 × 10–2 mol L–1s–1 (C) 4 × 10 mol L s –2 –1 –1 (D) 2 × 10–1 mol L–1 s–1 B-7.1 mol L —1 per min (D) 0. 2A + 2B C + D. 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. The rate would become – (A) 1/9 times (B) 9 times (C) 1/6 times (D) 6 times B-8.50 Rajeev Gandhi Nagar. by r = – = K [A]½ [B]1/3 dt [C]1/4. If concentration of B is made four times. When 50% of the reactants are converted into products. aA + bB Product. rate is doubled. The rate of disappearance of X will be - (A) 0. {– d[NO] / dt} = k1 [NO] [H2] . 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.5 =100s.B-9. second and third order respectively. rate when half reactants have been turned into products is : (A) 1.t. For an elementary reaction X (g) Y (g) + Z (g) the half life periods is 10 min. Wrong data for the first order reaction is – (A) t0.0 L vessel.COM India's No. {d[H2O] / dt} = k[NO][H2] . B. 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-1.25 × 10–3 mol L–1 s–1 (B) 1.t.5 = 16min (C) Both the above (D) t0.5 = 100s. H. 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-11.75 = 200s (B) t0. 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. – [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. In what period of time would the concentration of X be reduced to 10% of original concentration - (A) 20 Min. 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. For the irreversible process. k1 .50 Rajeev Gandhi Nagar. successive half lives are equal (C) For a first order reaction. {–d[H2] / dt} = k1 [NO][H2] The relationship between k.50 × 10–3 mol L–1 s–1 (D) 2.0 × 10–2 mol L–1 s–1 (C) 2.No. (B) 33 Min (C) 15 Min (D) 25 Min C-7. A first order reaction follows the expressions k 1t C0 Ct (A) Cte = C0 (B) Ct = C0 e k 1t (C) ln = –k 1t (D) ln = k 1t Ct C0 C-2. B-12. 092142 33303 . 2NO(g) + 2H2(g) N2(g) + 2H2O(g) the rate expression can be written in the following ways : {d [N2] / dt} = k1 [NO][H2] .75 = 150s C-5. For rate constant is numerically the same for three reactions of first. is : (A) k = k1 = k1 = k1 (B) k = 2k1 = k1 = k1 (C) k = 2k1 = k1 = k1 (D) k = k1 = k1 = 2 k1 B-10. t0. When will 99. A and second–order w. and the initial rate was 1. For the reaction. Rajasthan 324005 CHEMICAL KINETICS # 9 HelpDesk : Tel. Kota. k1 and k1.0 mol each of A and B introduced into a 1.0 × 10–2 mol L–1 s–1 . 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. t0.r. If 1.1 Online Coaching for JEE Main & Advanced 3rd Floor.75 = 32 min t0.9% of the reaction complete – (A) 50 Min (B) 46 Min (C) 49 Min (D) 48 Min ETOOSINDIA. the rate is first–order w. In acidic medium the rate of reaction between (BrO3)¯ and Br¯ ions is given by the expression.r. A + B products. 5 for the first order reaction. The time in which the conc. mole–1 sec–1 after 30 sec. Kota. Two substances A (t1/2 = 5 min) and B (t1/2 = 15 min) are taken in such a way that initially [A] = 4[B]. 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. In presence of HCl. 80% of the sample will decompose in - (A) 15 × 0. (C) 15 × (log5 / log2) hr.0 × 10–4 litre mol–1min–1.C-8.10 molar in 5 hours and from 0. (B) Product. 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 (D) None of these C-20. 20 % in two hours. 10% of the reactant decomposes in one hour. 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. (B) 15 × (log 8) hr. For a first order reaction. the initial concentration of cane sugar was reduced from 0. The order of the reaction is: (A) 0 (B) 1 (C) 2 (D) 3 C-15.303) (B) (– k / 2.No.65 × 10–5 It. In a first order reaction the reacting substance has half-life period of ten minutes. the plot of log C against ‘t’ (logC vs 't') gives a straight tine with slope equal to : (A) (k / 2.COM India's No.50 Rajeev Gandhi Nagar. 2.25M in reactant – (A) 8. The rate constant of reaction 2 A + B C is 2. (D) 15 × 10/ 8 hr. to decreases to 0.08M to 0. The t0. The reaction is of – (A) Zero order (B) First order (C) Second order (D) Third order C-9. For a second order reaction.83) hr. mole–1 sec–1 after 20 sec. 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 K1 K2 C-19. The rate constant for a second order reaction 8. of PCI5 reduces to 25% of the initial conc.0 × 10–4 min (C) 2.8 hr. C-16.0 × 10–4 min C-11. H. How long will it take a 0. Rajasthan 324005 CHEMICAL KINETICS # 10 HelpDesk : Tel.4 M to 0. is close to- (A) 22 min (B) 40 min (C) 90 min (D) 50 min C-14. Radioactive decay follows - (A) Zero order kinetics (B) Second order kinetics (C) First order kinetics (D) Pseudo first order kinetics C-13.303) (D) – k. In a certain reaction.2 M in 1 hour and to 0. C-12.01M – (A) 20 minutes (B) 30 minutes (C) 50 minutes (D) 70 minutes C-10. The order of the reaction is : (A) zero (B) one (C) two (D) None of these C-18. In the presence of an acid. sucrose gets hydrolysed into glucose and fructose. and 2. what would be the time taken for the conc.5M solution to be reduced to 0. PCI5(g) PCI3(g) + Cl2(g) is 20 min. 092142 33303 .20 to 0.55 × 10–5 It.2 to 0.04M in ten minutes.50 × 103 min (D) 4.303) (C) (ln k / 2.665 × 102 min (B) 8. of a reactant decreases from 0. A radioactive isotope decomposes according to the first order with half life period of 15 hrs.1 Online Coaching for JEE Main & Advanced 3rd Floor.57 × 10–5 It mole–1 sec–1 after 10 sec. (or 34. The concentration of sucrose was found to reduce from 0. if the conc. In the following first order reactions (A) Product.1 M in total of 2 hours. 012 0. 1/2 (B) 1.070 0.10 2 0. Kota. The concentration of A at which rate of the reaction is (1/12) × 10–5 M sec–1 is : (A) 0. the rate increase only 2 times. The rate constant for the reaction A B is 2 × 10–4 It. t1/2 1/a (C) n = 1 . H. When concentration of reactant in reaction A B is increased by 8 times.035 0. 1 C-27.50 Rajeev Gandhi Nagar... log concentration is a straight line.. [A]0 [B]0 initial rate 1 0. 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-1.80 (A) r = k [B]3 (B) r = k [A]3 (C) r = k [A] [B]4 (D) r = k [A]2 [B]2 . t1 : t2 ? (A) 6. a2C0. ETOOSINDIA.05 mL) contains 3.012 0. t1 and t2.. 2t .10 4 0. 1 (C) 2. t. A graph plotted between log t50% vs..33 C-22. 2 (D) 3. Rajasthan 324005 CHEMICAL KINETICS # 11 HelpDesk : Tel.COM India's No.024 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) 0.25 M (B) (1/20) 5 / 3 M (C) 0. mol–1 min–1. If a I-order reaction is completed to the extent of 60% and 20% in time intervals.58 (C) 4. aC0 .11 (D) 8. The data for the reaction A + B C is Exp. 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) (A) (B) (C) (D) Time C-25. what is the ratio.80 3 0.693 / k) (D) None of these C-26. a3C0 .. t1/2 = (0.5 M (D) None of these C-24. the concentration of reactant are C0 .. For a reaction A Products..No.. The order of the reaction would be - (A) 2 (B) 1/3 (C) 4 (D) 1/2 D-2....C-21. 092142 33303 ..0 × 10–6 moles of H+. What conclusion can you draw from this graph. If the rate constant of disappearance of H+ is 1... A drop of solution (volume 0. after time interval 0.1 Online Coaching for JEE Main & Advanced 3rd Floor.070 0.035 0.024 0.. t1/2 a (B) n = 2. (A) n = 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 1 1 1 1 (C) reaction is of 1st order and K = ln a (D) reaction is of zero order and K = ln t t a C-23..0 × 107 mole litre–1 sec–1.32 (B) 5.where 'a' is constant. 5 times (D) 0. then order is : dt (A) 4 (B) 2 (C) 1 (D) 0 D-4.5 /mol sec] (D) 1.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. – = k and at different time interval.5 sec–1] D-8.0 /mol sec] (B) 0.0 mol/ sec] (C) 2. Select the correct answer – codes :– (A) a.180 IV 0. A + B Product. What is the order with respect to B.40 0. H.l–1 mol.COM India's No. Initial conc.b and c (B) a and b (C) b and c (D) c alone D-7. of B is doubled and that of A and C is kept constant.1 Online Coaching for JEE Main & Advanced 3rd Floor. Using the data given below the order and rate constant for the reaction : CH3CHO(g) CH4(g) + CO(g) would be – Experiment no. [A] [B] Rate mol. [A] values are : dt Time 0 5 min 10 min 15 min [A] 20 mol 18 mol 16 mol 14 mol At 20 minute.25 times D-5. Rajasthan 324005 CHEMICAL KINETICS # 12 HelpDesk : Tel.No. For the reaction A Products.50 Rajeev Gandhi Nagar. For the reaction 2A + 3B products. Thus.[k = 1.[k = 2. rate law is : dx dx dx dx (A) = k[A]2 [B]2 (B) = k[A] [B] (C) = k[A]0 [B]2 (D) = k[B]1/2 dt dt dt dt d [A] D-9. 092142 33303 . =k [A]a [B]b dt dx If = k.20 0. In the reaction : A + 2B + C D + 2E Ther rate of reaction remains unchanged if the conc. dx D-3.l–1 mol.020 II 0.080 III 0. Kota. One litre of 2M acetic acid and one litre of 3M ethyl alcohol were mixed.1 M to 0.10 0.No. rate will be : (A) 12 mol /min (B) 10 mol/min (C) 8 mol/min (D) 0. rate becomes doubled. (A) 0 (B) 1/2 (C) 1 (D) 3 D-6.[k = 1. A is in excess and on changing the concentration of B from 0.30 0. The following data pertain to reaction between A and B : S.320 Answer is – (A) 2.lit–1sec 1] I 0.[k = 2. 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) 0.4 M.4 mol/min ETOOSINDIA. Initial rate [mol/] [mol. Starting with pure A initially. After 10 minutes volume of N2 gas is 10 L and after complete reaction it is 50 L. Consider the reaction 2A(g) 3B(g) + C(g). then graph is of the type : dt dt (A) (B) (C) (D) D-11.) N2 (g) + 2 H2O (l) the volume of N2 after 20 min and after a long time is 40 ml and 70 ml respectively. then the rate constant for 2A B + 3 C is : (A) (1/1200) ln 1. In a gaseous state reaction. The rate of disapearance of A2 in mm min–1 is : (A) 4 (B) 8 (C) 16 (D) 2 E-4.25 sec–1 (B) (2. H.303 /10) log 5 min–1 (B) (2. In the reaction NH4NO2 (aq.1 Online Coaching for JEE Main & Advanced 3rd Floor. Kota. what will be the total pressure in the vessel after the reaction is complete? (A) 1atm (B) 2 atm (C) 2. A2 (g) B(g) + (1/2)C (g). If a is the initial concentration of reaction. 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-12.303 /1200) log (7/3) sec–1 (C) (1/20) log (7/3) min–1 (D) (2. The half-life period for a reaction at initial concentrations of 0. Hence rate constant is: (A) (2.25 min–1 (C) (2. first order with respect to NO2 and first order with respect to F2.5 atm (D) 3 atm E-3. If log is plotted against log [A]. Formation of NO2F from NO2 and F2 as per the reaction 2NO2(g) + F2(g) 2NO2F(g) is a second order reaction. (C) follows First order kinetics. dx dx D-10.25 sec –1 (D) None of these E-5. The order of the reaction might possibly be (A) zero (B) first (C) second (D) unpredictable from this data E-2.303 /10) log 4 min–1 ETOOSINDIA.5 and 1. Rajasthan 324005 CHEMICAL KINETICS # 13 HelpDesk : Tel. Consider a reaction A B + C.303 / 20) log (11/7) min–1 E-6.5 min–1 (C) (1/10) ln 1. The decomposition of a gaseous substance (A) to yield gaseous products (B). The increase in pressure from 100 mm to 120 mm is notices in 5 minutes.No.303 /10) log 2 min–1 (D) (2. 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.5 hour. The value of rate constant is : (A) (1/20) In (7/4) min–1 (B) (2. 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. 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.COM India's No.50 Rajeev Gandhi Nagar. The order of the reaction is - (A) 0 (B) 1 (C) 2 (D) 3 D-13.0 moles litre-1 are 200 sec and 100 sec respectively. the total pressure doubled in 3 hrs. 092142 33303 . 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.303 /10) log 1. / Cu –N2Cl –Cl + N2 Half-life is independent of concentration of reactant.303 /10) log 1. A Product and = k[A]2 . (D) the rate constant has the unit mole It–1 sec–1.0 – (106 /T) Which of the following pair of value is correct ? (A) A = 1015 and E = 1. as log k = 15.0 x 1011 J (B) 41570 J (C) 5000 J (D) 345612.No.COM India's No. Rate of which reactions increases with temperature : (A) of any (B) of exothermic reactions (C) of endothermic (D) of none. (C) the half life depends upon the concentration of the reactants. The chemical reactions in which reactants require high amount of activation energy are generally - (A) Slow (B) Fast (C) Instantaneous (D) Spontaneous F-4. 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 energy of activation of the reaction is : (A) 5. The decomposition of N2O into N2 & O2 in presence of gaseous argon follow second order kinetics with 41570 K k = (5. For a zero order reaction.9 × 104 KJ.cal. The reaction is – (A) Exothermic (B) Endothermic (C) Neither exothermic nor endothermic (D) Independent of temperature F-2.50 Rajeev Gandhi Nagar. The activation energy of the reaction A + B C + D + 38 k.cal is 20 k.98 J ETOOSINDIA.0 × 1011 L mol1 s1) e T (K stands for Kelvin units). 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-6.9 × 104 KJ (B) A = 10–15 and E = 40 KJ (C) A = 1015 and E = 40 KJ (D) A = 10–15 and E = 1. (B) the rate is independent of the concentration of the reactants. F-7. F-9. F-10.1 Online Coaching for JEE Main & Advanced 3rd Floor.Section(F) : Effect of Temperature F-1.5 and 45.4 KJ/mol respectively. 092142 33303 .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.cal F-8. H. The activation energies of the forward and backward reactions in the case of a chemical reaction are 30. What would be the activation energy of the reaction. 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. C + D A + B (A) 20 k. 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. The first order rate constant k is related to temp.cal (C) 18 k. Chemical reaction occurs as a result of collisions between reacting molecules. F-12. Rajasthan 324005 CHEMICAL KINETICS # 14 HelpDesk : Tel. Kota.cal (D) 58 k. Which of the following statement is false : (A) the rate is independent of the temperature of the reaction. Therefore. 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) 2.50 Rajeev Gandhi Nagar. 104. The reaction. H.6 × 1030 s–1 F-15. G-7. Trimolecular reactions are uncommon because (A) the probability of three molecules colliding at an instant is very low.0 (B) 0. H = 40kJ.COM India's No. The ratio of energy of activation for reverse reaction before and after addition of catalyst is : (A) 1. The rate constant. 3/ 2 (B) 3/2 . (C) the probability of three molecules colliding at an instant is zero.1 Online Coaching for JEE Main & Advanced 3rd Floor. For the reaction NO2 + CO CO2 + NO the experimental rate expression is – = k[NO2]2 the number of dt molecules of CO involves in the slowest step will be - (A) 1 (B) 2 (C) 3 (D) depends on mechanism G-3.mol1. given = 9 x 105.0 × 1014 s–1 respectively. The reaction of hydrogen. the activation energy and the frequency factor of a chemical reaction at 25°C are 3. In the Lindemann theory of unimolecular reactions.0 Section (G) : Mechanism of reactions G-1. (B) the probability of three molecules colliding at an instant is high. 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-14.4 KJ mol–1 and 6. 3/2 (D) 1. Select the correct statements : (A) the molecularity of an elementary reaction indicates how many reactant molecules take part in the step. ETOOSINDIA. (A) 106 mole/litre (B) 104 mol/litre (C) 105 mole/litre (D) 5 x 105 mol/litre G-6. it is shown that the apparent rate constant for such a k1C reaction is kapp = where C is the concentration of the reactant k1 and are constants. 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) Rate = k[X] [Y] [Z] G-5. and iodine monochloride is represented by the equation : H2(g) + 2Cl(g) 2HCl(g) + 2(g) This reaction is first–order in H2(g) and also first–order in Cl(g). The value of the rate constant as T is : (A) 2.1/2 dc G-2.No. energy of activation for forward reaction (Eaf) is 80kJ. 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) fast HI(g) + Cl(g) HCl(g) + I2(g) (A) only (B) only (C) both and (D) neither nor G-4.2 (D)2. (B) the rate law of an elementary reaction can be predicted by simply seeing the stoichiometry of reaction. 3/2 (C) Not defined.mol1. For a given reaction. Calculate the 1 C value of C for which kapp has 90% of its limiting value at C tending to infinitely large values.0 × 10–4 s–1. Kota. 092142 33303 . A catalyst lowers Eaf to 20 kJ.5 (C) 1.F-13. (C) the slowest elementary step in sequence of the reactions governs the overall rate of formation of product. (D) the probability of many molecules colliding at an instant is high.0 × 1014 s–1 (C) infinite (D) 3. (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.mol1 for the reaction. Rajasthan 324005 CHEMICAL KINETICS # 15 HelpDesk : Tel.0 × 1018 s–1 (B) 6. 2 M 0.25 M 0.2 M H-5. The rate constant for two parallel reactions were found reactions were found to be 1.COM India's No.2 M 0.0 kJ mol–1 and 70.83% B and 23. At a given temperature. net rate is : dx = k [A]2 [B]1 – k [C].0 kJ mol–1 (D) 65. The substance undergoes first order decomposition. For an elementary reaction of reversible nature.0 kJ mol–1 respectively.5 kJ mol–1 (C) 100. If the corresponding energies of activation of the parallel reactions are 60.8 × 10–5 sec–1 The percentage distribution of B and C (A) 80% B and 20% C (B) 76. hence. what is the apparent overall energy of activation ? (A) 130.2 M (D) 0. 092142 33303 . given reaction is : dt 1 2 1 (A) 2A + C (B) 2A + B C (C) 2A C + B–1 (D) None of these B H-4. Kota.26 × 10–4 sec–1 and K2 = 3.3 M 0. For an elementary reaction.0 kJ mol–1 ETOOSINDIA.4 M (B) 0. H. Which of the following equations are correct ? d[A] d [ C] (A) = k1[A] + k4[D] (B) = k2[B] k3[C] dt dt d [D] (C) = k4[D] + k3[D] (D) Nothing can be said about order of reactions in this problem dt H-3.0 kJ mol–1 (B) 67.1 M 0.1 Online Coaching for JEE Main & Advanced 3rd Floor.2 M 0.No.3 M 0.5 M (C) 0.4 M 0.0 × 10–2dm3 mol-1 s–1 and 3.50 Rajeev Gandhi Nagar. Consider the elementary reaction sequence shown in figure. select the correct statement : 2 d[ A ]2 d [B] d[C] (A) – = = is the relation among (B) 2A 2B + C is the required reaction dt dt dt (C) both are correct (D) none is correct H-2.2 M 0. Rajasthan 324005 CHEMICAL KINETICS # 16 HelpDesk : Tel. net rate is dt = k[A]2 – k'[C] [B]2 then.Section (H) : Complications in first order reactions dx H-1.0 × 10–2 dm3 mol–1 s–1. The decomposition follows two parallel first order reactions as : K1 = 1.17%C (C) 90% B and 10% C (D) 60% B and 40% C H-6.4 M 0. k1 = k2 for the reaction A+B C+D dx If dt = k1 [A] [B] – k2[C] [D] in which set of the concentration reaction ceases? [A] [B] [C] [D] [A] [B] [C] [D] (A) 0.2 M 0.2 M 0. k1 < k2 (for endothermic reaction) (D) Ea1 Ea2 5.5 bar (B) 1. Rajasthan 324005 CHEMICAL KINETICS # 17 HelpDesk : Tel.5 Ea2 6.1 Online Coaching for JEE Main & Advanced 3rd Floor.15 s 3. Kota. H. Comparing set-I and II : (A) k4 > k3 & k2 > k1 . At t = t1 pressure of gas B is : (A) 2.3 s (D) 1.No.15 s (B) 1.50 Rajeev Gandhi Nagar. Value of t1 in seconds is : (A) 2. For the (Set-1) : (A) Ea1 > Ea2 if T1 > T2 (B) Ea1 < Ea2 if T1 > T2 (C) Ea1 = Ea2 (D) Ea1 = 0. 1.0 bar (D) data is insufficient 2. Rate constant (k) = (log 64 – log 49) s–1. For the (Set-1) : (A) if T1 > T2. total pressure is half of the final total pressure at t = tx (a very long time). k1 > k2 (for exothermic reaction) (C) if T1 > T2. It is also given at t = tx.25 bar (C) 5. at a constant temperature. if T2 < T1 (exothermic) (D) k4 < k3 & k2 < k1. 092142 33303 . if T2 < T1 (endothermic) (C) k4 > k3 & k2 > k1 . final total pressure is 35 bar. After some time t = t1. if T2 > T1 (exothermic) ETOOSINDIA.5 s (C) 2. Ratio of rate constant at t = 0 to t = t1 to t = tx is : (A) 2 : 3 : 4 (B) 1 : 1 : 1 (C) 1 : 3 : 5 (D) 1 : 3 : 5 Comprehension # 2 4. k1 > k2 always (B) if T1 > T2. At t = 0 mole fraction of gas C is 1/3. Assume this decomposition is a first order. if T2 > T1 (endothermic) (B) k4 < k3 & k2 > k1 .COM India's No. PART .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). Let us consider a reaction A(g) B(g) + C(g) At t = 0 a 0 0 At time t a–x x x dx The rate of reaction is given by the expression = k(a – x) and integrated rate equation for a given reaction is dt 1 a represented as k = ln where a = initial concentration and (a – x) = concentration of A after time t. For example. t ax 7. [C] k1 At any time we also have = k [D] 2 9. Kota.66 min (C) 166. meta. Consider a reaction A(g) 3B(g) + 2C(g) with rate constant 1. we have = k1[A] = k1[A]0 e (k1 k 2 )t .No. 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.COM India's No. (A) 0.1 Online Coaching for JEE Main & Advanced 3rd Floor. We shall consider the simplest case. 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. H. A starting initially with only A Which of the following is correct at time t (A) [A]0 = [A]t +[B]t + [C]t (B) [A]0 = [A]t + 2 [B]t + 3 [C]t [B] t [C] t 2 (C) [A]0 = [A]t + + (D) [A]0 = [A]t +[B]t + [C]t 2 3 3 ETOOSINDIA. Starting with 2 moles of A in 12.66 min (D) 156.386 × 10–2 min–1.04 M (B) 0.Comprehension # 3 A reaction is said to be first order if it's rate is proportional to the concentration of reactant. 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] k 2 [ A ]0 Similarly.36 M (C) 0.66 min 8.09 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. How long will it take for 90% of the compound to decompose? Given : log 2 = 0. Thermal decomposition of compound X is a first order reaction.50 Rajeev Gandhi Nagar. 092142 33303 . integration of gives [D] = (1 – e (k 1 k 2 )t ) dt k1 k 2 The sum of the rate constants k1 + k2 appears in the exponentials for both [C] and [D]. or para positions. toluene can be nitrated at the ortho.5 litre vessel initially. The rate law is d[A] = – k1[A] – k2[A] = – (k1 + k2) [A] [A] = [A]0 e (k1 k 2 )t . If 75% of X is decomposed in 100 min. Rajasthan 324005 CHEMICAL KINETICS # 18 HelpDesk : Tel. dt d [C] For C. H.1 Online Coaching for JEE Main & Advanced 3rd Floor. 092142 33303 . Rajasthan 324005 CHEMICAL KINETICS # 19 HelpDesk : Tel. The time at which concentration of B is maximum is k1 1 k 1 k k2 (A) (B) ln 1 (C) ln 1 (D) k 2 k1 k 2 k1 k 2 k1 k 2 k 2 k 2 k1 14. [C]t [C]t conc. For A starting with pure A ratio of rate of production of B to C is (A) Independent of time (B) Independent of temperature (C) Depends upon initial concentration of A (D) Independent of mechanism of reaction Comprehension # 5 For the given sequential reaction 1 k 2 k A B C the concentration of A.50 Rajeev Gandhi Nagar. X starting with only 'X'. Conc. ratio [ Y ] [ Z] t t 1 (A) Independent of time (B) (e kt 1) (C) Depends upon initial concentration of X (D) [A]0 (ekt –1) 11. B & C at any time 't' is given by k1[ A ]0 [A]t = [A]0 e k1t .No. Kota. [ X] t 10. At high temperature acetic acid decomposes into CO2 & CH4 and simultaneously into CH2CO (ketene) and H2O 1 1 (i) CH3COOH CH4 + CO2 k1 3s (ii) CH3COOH CH2CO + H2O k 2 4s What is the fraction of acetic acid reacting as per reaction (i) ? 3 3 4 (A) (B) (C) (D) none of these 4 7 7 12. [B]t = e k1t e k 2 t ( k 2 k1 ) [C]t = [A0] – ( [A]t + [B]t ) 13. (A) (B) [B]t [A]t [B] t [A]t time time [C]t [C]t Conc. Conc.COM India's No. Select the correct option if k1 = 1000 s–1 and k2 = 20 s–1. (C) (D) [B]t [B]t [A]t [A] t time time ETOOSINDIA. 54. log 2 = 0.I Column . Rajasthan 324005 HelpDesk : Tel.1 Online Coaching for JEE Main & Advanced CHEMICAL KINETICS # 20 3rd Floor.5° (C) The equimolar mixture of the products (r) 75. Kota.2 min. 092142 33303 .I Column . (Use : log 35 = 1.I Column .II (A) A + B C + D (p) Unit of rate constant possess concentration r = k1 [A] [B] unit (B) A + B C + D (q) Rate constant for the reaction of both r = k2 [A] [B]º the reactants are equal (C) A + B C + D (r) Rate of consumption of at least one of the r = k3 [A]º [B]º reactants is equal to rate of production of at least one of the products (D) 2A + B 2C + 3D (s) If both reactants are taken in stoichiometric r = k3 [A]º [B]º ratio.50 Rajeev Gandhi Nagar. Match the following : Column . log 3 = 0.48.3) Column . For the reaction of type A(g) 2B(g) Column-I contains four entries and column-II contains four entries. Entry of column-I are to be matched with ONLY ONE ENTRY of column-II Column . H.II d[ B] d[A ] (A) vs for first order (p) dt dt (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. (B) The solution is optically inactive at (q) 7.II (A) The half life of the reaction (p) 120 min.No. 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. half life for both reactants are equal 17. (D) The angle at half time (s) laevorotatory 16.COM India's No.MATCH THE COLUMN 15. (B) Statement-1 is True. Statement-2 is NOT a correct explanation for Statement-1. (q) Zero If we start with 20 mol L–1. Ratio of rates at the start of reaction is how many times of 0. Statement-2 is a correct explanation for Statement-1. Statement-1 : The time of completion of reactions of type A product (order <1) may be determined.18. (D) Statement-1 is False. Statement-2 : Fractional order of RDS equals to overall order of a complex reaction. 23.order reaction is 0. Statement-2 : When the activation energy of forward reaction is less than that of backward reaction. with thereactants by which more number of molecules are able to cross the barrier per unit of time.5 mol L–1 in how many minutes (C) Half . (E) Statement-1 and Statement-2 both are False. it is reduced to 2. Statement-2 : The rate of reaction having negative order with respect to a reactant decreases with the increase in concentration of the reactant. Statement-1 : Temperature coefficient of an one step reaction may be negative.1 Online Coaching for JEE Main & Advanced 3rd Floor. Statement-1 : The overall rate of a reversible reaction may decrease with the increase in temperature.50 Rajeev Gandhi Nagar. H. 21.No. Statement-1 : In a reversible endothermic reaction. Statement-2 is True. 22. (A) Statement-1 is True. then the increase in the rate of backward reaction is more than that of forward reaction on increasing the temperature. 092142 33303 . (s) 30 [A]0 (M) 0.lives of first .order and zero order reactions (r) 11 are same. (C) Statement-1 is True.0693 min–1.0677 0. Eact of forward reaction is higher than that of backward reaction Statement-2 : The threshold energy of forward reaction is more than that of backward reaction 24. Statement-1 : A catalyst provides an alternative path to the reaction in which conversion ofreactants into products takes place quickly Statement-2 : The catalyst forms an activated complex of lower potential energy. (D) and (E) out of which ONLY ONE is correct. (B). Match the following : Column-I Column-II (A) If the activation energy is 65 kJ then how much time (p) 2 faster a reaction proceed at 25°C than at 0°C (B) Rate constant of a first . (D) The half-life periods are given . ETOOSINDIA. Statement-2 is True.272 t1/2 (sec) 240 480 960 order of the reaction is ASSERTION / REASON Directions : Each question has 5 choices (A). Statement-2 : Reactions with order 1 are either too slow or too fast and hence the time of completion can not be determined. Statement-2 is True. (C). Statement-1 : A fractional order reaction must be a complex reaction. Kota. 20.COM India's No.693 Assume initial concentration to be same for the both. Rajasthan 324005 CHEMICAL KINETICS # 21 HelpDesk : Tel. Statement-2 is False.136 0. 19. Order of reaction may be change with change in practical conditions 41.93 s. 39. 34. 30. 33. For a zero order reaction t3/4 is related to t1/2 as t3/4 = 1. The rate of reaction is uniform in zero order reaction. Order of a reaction can be written from the balanced chemical equation. Au Statement-2 N2 (g) 1/ 2O 2 .5 t1/2 38. Statement-1 : Half life of a certain radioactive element is 100 days. 36. 37. Statement-1 : In the reaction. Statement-1 : For A + 2B C (rate= K[A]1[B]0).50 Rajeev Gandhi Nagar. ETOOSINDIA. the value of rate constant for the reaction would be 10s–1. the rate of disappearance of thiosulphate ions is twice the rate of disappearance of 2. of B due to its zero order. H. Statement-2 : Time taken for completion of any fraction of a st order reaction is a constant value. After 200 days. 35.25. 32. Molecularity of a reaction is always whole number. 26. Statement-1 : Time taken for the completion of 75% of a st order reaction is double than its t½. 092142 33303 . If t1/2 for a first order reaction is 6. Order and molecularity of a single step reaction may or may not be same. rate = k[N2O]0 = k = constant is a zero order reaction. 2 + 2S2O32– S4O62– + 2– . Partial orders are never negative. The activation energy of a catalysed reaction is more than the activation energy of the uncatalysed reaction. 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. fraction left undecaye will be 25%.1 Online Coaching for JEE Main & Advanced 3rd Floor. In a complex reaction the rate of overall reaction is governed by the slowest step.No. 40. The rate of an exothermic reaction increases with the rise in temperature. Statement-2 : The rate of disappearance of 2 is twice the rate of disappearance of S2O32– ions. Kota.COM India's No. the half life time of reaction is only defined when conc. where symbols have standard meaning. : N2 O(g) 28. of A and not to conc. Statement-1 : Many reactions occurring on solid surface are zero order reactions. of A and B are in stoichiometric ratio Statement-2 : For above given order half life of reaction is directly proportional to conc. C0 2 29. 27. n Ct 1 Statement-2 : . Rajasthan 324005 CHEMICAL KINETICS # 22 HelpDesk : Tel. As a first order reaction proceeds at a constant temperature. the rate remains constant. H. 49. increase in temperature cause more change in the value of k than at higher temperature. 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. 56.No. 45.42. For a firstorder reaction. 48. Every species that appears in the rate law of reaction must be a reactant or product in that reaction. Molecularity defined only for RDS. the reaction must be elementary. All collisions between reactants yield the desired product. Product can form only when the required orientation and energy conditions are met. 54. A catalyst in a chemical reaction increases the forward Ea and decreases the backward Ea 58. 3A B must be r = k [A]3 43. 46. Larger the value of Ea . Kota. the reaction must be complex. greater is the effect on the value of k for a given temperature change. 47. The rate law of the elementary reaction. 50. 53.50 Rajeev Gandhi Nagar. If the partial orders differ from the coefficients in the balanced reaction. ETOOSINDIA. d[A] 51. At lower temperature. The pre-exponential factor A has the same units for all reactions. 092142 33303 . 44. The rate of the reaction A B having the rate law dt k [ A ] [B] when plotted against time will exhibit a maximum at some time. 55. A catalyst in a chemical reaction decreases the activation energy of the forward reaction and increases the activation energy of the reverse reaction 57.COM India's No. Molecularity of a reaction may includes the number of product molecules taking part in the reaction. If the partial orders are equal to corresponding coefficients in the balanced reaction. the time required to reduce successively the concentration of reactant by a constant fraction is always same.1 Online Coaching for JEE Main & Advanced 3rd Floor. Activated complex is an intermediate product. Rajasthan 324005 CHEMICAL KINETICS # 23 HelpDesk : Tel. 52. 60. A catalyst in a chemical reaction decreases both forward and backward Ea 59. 66. 80. Rajasthan 324005 CHEMICAL KINETICS # 24 HelpDesk : Tel. the ________ step is rate determining. For a certain reaction. 68. H. 76.1 Online Coaching for JEE Main & Advanced 3rd Floor. 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 ________. Rate constant of a reaction __________ with increase in temperature. The order of a reaction is the ________ of the _________ of all the concentration terms in the rate equation. the minimum value of activation energy can be _______. 63. [Eactivated complex – Ereactants] = ____________. A catalyst is chemically ____________ at the end of a reaction 82. If activation energy of reaction is low. 62. For an endothermic process. the rate of disappearance of H2 is _______ the rate of appearance of HI. The rate law is _________. In the Arrhenius equation k = Aexp (–E/RT). For a ______ order reaction the half-life (t1/2) is independent of the initial conc. H2 + I2 2HI.No. xM yL. 73.50 Rajeev Gandhi Nagar. 81. 71. ETOOSINDIA. 67. For a zero order reaction. 72.COM India's No. In the reaction. The reactions with molecularity more than three are _________. the rate of the reaction is equal to the ______ of the reaction.FILL IN THE BLANKS 61. In a multistep reaction. Kota. The rate of a reaction is ________ to the collision frequency. of the reactants. The ratio t7/8 / t1/2 for a first order reaction would be equal to _________. For a first order reaction A P. 79. the rate of reaction increases by 4 times when the concentration of M is doubled. The rate constant for the zero order reaction has the dimensions___________. For collision to be effective the energy possessed by the colliding molecules should be equal to or greater than the ____________. 77. 78. The value of temperature coefficient is generally between _________. The rate equation r = k [A][B]1/2 suggests that order of overall reaction is _______. 092142 33303 . 74. 75. 69. a graph of log [A] vs t has a slope equal to __________. A may be termed as the rate constant at __________. 70. 64. 65. A catalyst increases the rate of the reaction by__________ activation energy. it proceeds at _______ rate. What is the rate constant (sec–1)? Time (min) (0. 20 minutes time is required for completion of 20 % reaction.5 × 10 L mol s–1 –1 (D) 1.0) 2 4 6 8 10 | | | | | log (a0 – x) -1– -2– -3– (A) 0. PART .303 × 10–3 sec–1.05 M (B) 0.1 Online Coaching for JEE Main & Advanced 3rd Floor. then half life (t1/2 ) is dt (A) 55. Reaction A + B C + D follows following rate law : rate = k [A] 2 [ B] 2 .I : MIXED OBJECTIVE Single Choice Type 1.44 min.15 × 10–2 4. The rate constant for the forward reaction A (g) + 2B(g) is 1. Rajasthan 324005 CHEMICAL KINETICS # 25 HelpDesk : Tel. (D) 1200 sec.6 × 10–1 (C) 7.25 M.No.7 × 10–3 (D) 1. (B) 50 min (C) 62.0125 M (D) None of these ETOOSINDIA. what is the time taken for concentration of A of become 0.50 Rajeev Gandhi Nagar. Given : k = 2. Starting with initial conc. (A) 300 sec.5 × 10–3 s–1 at 100 K.025 M (C) 0. 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) 3. 6. H. The reaction A (g) B(g) + 2C (g) is a first order reaction with rate constant 3465 × 10–6 s–1.13 min (D) None of these 2.5 × 10–11L mol–1 s–1 1 1 5. Consider the reaction : A B + C Initial concentration of A is 1 M. when the reaction is allowed to take place at constant pressure and temperature.. d[ B] If = k[A].2 (B) 4. Kota. (B) 600 sec. For the first order decomposition of SO2Cl2(g).50× 104 L mol–1 s–1 (B) 1. SO2Cl2(g) SO2(g) + Cl2(g) a graph of log (a0 – x) vs t is shown in figure.COM India's No. 092142 33303 . Starting with 0. 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. (A) 0. (C) 900 sec. of 1 M of A and B each. find the concentration of A after 200 sec.1 mole of A in 2 litre vessel.5 × 1011 L mol–1 s–1 10 (C) 1. 8 M? Concentration(M) 1. For a certain reaction of order n.97 kJ/mol (D) 6. the time for half change.8 = 0.85 kJ/mol (B) 55.COM India's No. 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. the integrated form of the rate equation is k = 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. t1/2. The rate constant.036 M min–1 (C) 0.0 × 1014 s–1 (C) infinity (D) 3.58] (A) 12 kJ/mole (B) 831. the Ea (activation energy) of the reaction is [log 3. is given by t1/2 = xC0 where k is k constant and C0 is the initial concentration.4 kJ mol–1 and 6.8 Experiment-1 Experiment-2 0. 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.1 Online Coaching for JEE Main & Advanced 3rd Floor. 092142 33303 . Rajasthan 324005 CHEMICAL KINETICS # 26 HelpDesk : Tel.0 × 10–4 s–1. In respect of the equation k = A exp (– Ea / RT).No.14 kJ/mol (C) 11. The energy of activation of the reaction is (A) 43.6 5 10 15 20 time(min. The value of the rate constant at T is (A) 2.) (A) 0.08 M min–1 (B) 0.0 × 1014s–1 respectively. The reaction is represented as A(aq) B(aq).57 kJ/mole ( 2 2 ) 1/ 2 13. Kota. T 10.4 kJ/mole (C) 100 kJ/mole (D) 88. 104.5 1. Given that for a reaction of order n. 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. H.7. Rate of a reaction can be expressed by Arrhenius equation as : k = Ae–E/RT In this equation.8 × 10–16 % of the reactant molecules exists in the activated state.2 1 0. What is n? (A) 1 (B) 2 (C) 0 (D) 0.13 M min–1 (D) 1 M min–1 8.6 × 1030 s–1 11. 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. the activation energy and the Arrhenius parameter (A) of a chemical reaction at 25°C are 3.65 kJ/mol 12. What is the rate of reaction (M/min) when concentration of A in aqueous solution was 1. For the first order reaction A — B + C.50 Rajeev Gandhi Nagar. carried out at 27 ºC if 3. A first order reaction is 50% completed in 20 minutes at 27°C and in 5 min at 47°C.5 1 1 1 14.0 × 1018 s–1 (B) 6. 05 In 1. H. If at t as 30 minutes the conc.No. Rajasthan 324005 CHEMICAL KINETICS # 27 HelpDesk : Tel. was 176 mm and after a long time when A was completely dissociated. However if pH = 6. In three different reactions.300 mm (C) 0. versus concentration of the reactant on the xaxis. 3 (B) 2. 1/2 16. the concentration of the reactant decrease to 6. The pressure after 20 min. 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 rate of the reaction on the yaxis. 0. are : (A) 0.05 In 3 min–1 . yields a straight line. (iii). of the reactant and C is the reactant concentration after time t.17 × 102 min1.15. The values of optical density are 0. For a certain reaction. 1. and after very long time are 150 mm Hg and 225 mm Hg. yields three different curves shown below. Initial conc. 2 (D) 0. The time elapsed of a certain between 33% and 67% completion of a first order reaction is 30 minutes.465 × 103 min1. 092142 33303 . 2.17 × 103 mol.litre1 min1 (C) 3. the half life changes to 500 minute of any concentration of sugar .80. if the initial concentration is 0.50 Rajeev Gandhi Nagar. What is (i) the rate constant and (ii) the rate of the reaction. 2.465 × 102 min1.5 minutes (C) 180.166 × 103 min1. 2. Kota.667 × 104 mol.1 Online Coaching for JEE Main & Advanced 3rd Floor. A reaction 2A + B C + D is first order with respect to A and 2nd order with respect to B. 1.51 x 103 min1 23. ETOOSINDIA. (A) 1.5 ln 2 min–1 . In a I order reaction A products.93 x 103 min1 (B) 3. 40 minutes and infinite time after the start of the reaction which is first order. 300 mm –1 (D) 0. 200 mm 22. 1. 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 18.47 × 104 mol.20 at the end of 20 minutes.05 In 3 min . 100 minutes after the start.COM India's No. (ii). 1/2 (C) 0. Calculate the rate constant. (t = 0) of A is C0 while B is 2C0.93 x 102 min1 (D) 3. What is the order of the reaction : (A) 3 (B) zero (C) 1 (D) 2 19. The reaction A(s) 2 B(g) + C(g) is first order. The pressure of A at the end of 10 minutes was : (A) 94 mm (B) 47 mm (C) 43 mm (D) 90 mm 21. At 373 K. (A) 6.25% of its initial value in 80 minutes.2 mole/litre? (A) 2.35 and 0. Starting with pure A. (i) (ii) (iii) What are the possible orders of the reactions (i).5 min–1. 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. a plot of {[C0 – C] / [C]} against the time t.litre1 min1 20.litre1 min1 (D) 2. What is the time needed for 25% completion? (A) 150. C0 = initial conc. 2.5 minutes (D) 165.166 × 104 mol.5 minutes (B) 12. The inversion of cane sugar proceeds with half life of 50 minute at pH = 5 for any concentration of sugar. it was 270 mm. involving a single reactant in each case.51 x 102 min1 (C) 6. a gaseous reaction A 2B + C is found to be of first order.5 minutes k 17.litre1 min1 (B) 3. 200 mm (B) 0. the total pressure at the end of 10 min. The value of rate constant and pressure after 40 min. 3. (A) then the reaction must be A(g) 3B(g) and is a first order reaction. log A (D) None of these 29.311 : 1 (B) 0.693 0.COM India's No.1 Online Coaching for JEE Main & Advanced 3rd Floor. H.81 (B) 1. 092142 33303 . Two I-order reactions have half-lives in the ratio 3 : 2.273 : 1 (D) 0. is zero order while the reaction B Products. concentration of the reactant dt k is : (C0 = initial concentration) C0 C0 1 (A) (B) C0e (C) 2 (D) e e C0 1 1 33. When a graph between log K and 1/T is drawn a straight line is obtained. For what initial concentration of A the half lives of the two reactions are equal. (A) 2 M (B) ln 2 M (C) 2 log 2 M (D) 2 ln 2 M 25.303 R log A (B) .303) K. Calculate the ratio of time intervals t1 : t2.76 31. 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) 6.50 Rajeev Gandhi Nagar. The half time of the reaction would be : dt 0. Concentration of the reactant in first-order is reduced to after : (Natural life = ) e2 K (A) one natural life-time (B) two-natural life-time (C) three natural life-time (D) four natural life-time 34. Ea / (R ln A) (C) 0. For a reaction 2A + B product. For the reaction A + 2B products(started with concentrations taken in stoichiometric proportion). (C) then the reaction must be A(g) 3B(g) and is a zero order reaction. B more sensitive 28. If for a reaction in which A(g) converts to B(g) the reaction carried out at const. (A) 0. CD KC = 1014 e–1000/T Temperature T K at-which (KA = KC) is : (A) 1000K (B) 2000K (C) (2000 / 2.71 (C) 1. is 1st order. The Arrhenius relationship of two different reactions is shown below. The point at which line cuts y -axis and x -axis respectively correspond to the temp : (A) 0. 30. At a time when t = . rate law is – = k[A].420 : 1 (C) 0.67 × 10–3 (B) 1. ETOOSINDIA. for 75% completion of the second reaction.199 : 1 27. Ea / 2. k k 24. Rajasthan 324005 CHEMICAL KINETICS # 28 HelpDesk : Tel. Where t1 is the time period for 25% completion of the first reaction and t2. The reaction A Products. A more sensitive (B) B in both cases (C) A in both cases (D) A faster.693 0.693 (A) (B) (C) (D) not defined k 1/ k 2k d[ A ] 1 32. Which reaction is faster at a lower temperature and which is more sensitive to changes of temperature ? (A) B faster.303) K (D) (1000 / 2.2 × 10–2 (C) 2 × 10–4 (D) 4 × 10–3 26.91 (D) 1. (B) then the reaction must be A(g) 3B(g) and is a second order reaction. Kota. By what factor will it change for the temperature change from 17ºC to 27ºC? (A) 1. A B KA = 1015 e–2000/T .No. V & T results into the following graph. (D) then the reaction must be A(g) 3B(g) and is a first order reaction. the d [A] experimentally determined rate law is – = k [A] [B] . The rate of a reaction gets doubled when the temperature changes from 7ºC to 17ºC. A B (catalyst reaction) The activation energy is lowered by 8.0 hr (D) unpredictable as rate constant is not given 39. What will be time for cone. 092142 33303 . After 8 hr.314 KJ mol for the catalysed reaction. 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 d[A ] 40.33 = 28) (A) 15 times (B) 38 times (C) 22 times (D) 28 times 37.3 × 10–2 s–1 (D) 8. Rajasthan 324005 HelpDesk : Tel.14 × 10–3 Ms–1 (B) 5.4 hr (C) 4.4 hr (B) 2.7 × 10–3 Ms–1 (C) 4. The conversion of vinyl allyl ether to pent-4-enol follows first-order kinetics. The potential energy diagram is shown in the fig. Flask contains 1 L of 1 M solution of A and flask contains 100 ml of 0. Which of the following sets of identifications is correct ? (Assume that the temperature and pressure are constant). How many times the rate of this –1 catalysed reaction greater than that of uncatalysed reaction ? (Given e3. For the reaction A B.6 M solution. the conc.50 Rajeev Gandhi Nagar.35. Kota.14 × 10–2 Ms–1 catalyst 36.6 × 10–2 s–1 (B) 1.COM India's No. A B (uncatalysed reaction). then dt 2 1/ 2 (A) The integerated rate expression is k = ( A A1 / 2 ) t 0 (B) The graph of A Vs t will be K (C) The half life period t1 / 2 = 2[A]10/ 2 [A]0 (D) The time taken for 75% completion of reaction t 3 / 4 = k ETOOSINDIA. The instantaneous rate of disappearance of the MnO4– ion in the following reaction is 4. H. The following plot is obtained for such a reaction.25 M. A simple mechanism for enzyme-catalyzed reaction is given by the following set of equations E + S ES (enzyme) (reactant) (intermediate-1) ES EP (intermediate-1) (intermediate-2) EP E + P (intermediate-2) (enzyme) (product) This is known as the Michaelis–Menten mechanism. (1) (2) (3) (4) (A) E + P. the rate law expression is = k [A]1/2. of A in flask 1 becomes 0.2 × 10–2 s–1 (C) 2. The rate constant for the reaction is (A) 4.4 × 10–2 s–1 38. Then the rate of appearance of 2 is : 2MnO4– + 10 – + 16H+ 2Mn2+ + 52 + 8H2O 2MnO4– + 10 – + 16H+ 2Mn2+ + 52 + 8H2O (A) 1. If initial concentration of [A] is [A]0.1 Online Coaching for JEE Main & Advanced CHEMICAL KINETICS # 29 3rd Floor.3 M ? : (A) 0.56 × 10–4 Ms–1 (D) 1. Consider the following reactions at 300 K. of A in flask to become 0.56 × 10–3 ms–1. A substance ‘A' decomposes in solution following the first order kinetics.No. (B) For a zero order reaction.No. Kota. Select incorrect statement(s): (A) Unit of pre-exponential factor (A) for second order reaction is mol L–1 s–1. keeping the concentrations of all other reactants constant.COM India's No.41. Given 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 . The reaction 2NO(g) + Cl2(g) 2NOCl(g) is second order in NO and first order in Cl2. Consider the decay of P to A and B by two parallel first order reactions as shown in Fig. 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.5 × 10–3 Ms–1 for a particular time interval. (D) Decay constant () of radioactive substance is affected by temperature. (D) A zero order reaction is always elementary reaction. (D) The order of a reaction can only be determined from the stoichiometric equation for the reaction. 44. 3. What will be the rate when half of the chlorine has reacted ? ETOOSINDIA. (B) A zero order reaction must be a complex reaction. H. 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. In the following reaction 2H2O2 (aq) 2H2O () + O2 (g) rate of formation of O2 is 36 g min–1 . and the initial rate was 2. (C) A zero order reaction is controlled by factors other than concentration of reactants. (C) Molecularity is defined only for RDS in a complex reaction. 092142 33303 . (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. Rajasthan 324005 HelpDesk : Tel.1 Online Coaching for JEE Main & Advanced CHEMICAL KINETICS # 30 3rd Floor. Which of the following is/are correct statement? (A) Stoichiometry of a reaction tells about the order of the elementary reactions. 45. 5 mole of nitric oxide and 2 mol of Cl2 were brought together. rate and the rate constant are identical. In a consecutive reaction system A B C when E1 is much greater than E2. what is the value of – for the same time t t interval ? (b) What is the average rate of formation of SO42– during that time interval ? 2.II : SUBJECTIVE QUESTIONS 1. (b) What is rate of disappearance of H2O2.4 × 10 3 mole dm3 s. E1 2 E 42. In a volume of 2 dm3. The oxidation of iodide ion by peroxy disulphate ion is described by the equation : 3– (aq) + S2O82– (aq) 3– (aq) + 2SO42– (aq) [S2O82 ] [ ] (a) If – = 1. (a) What is rate of formation of H2O. (C) Orders of reactions can not be fractional.50 Rajeev Gandhi Nagar. 7.012 2.4 x 10 -3 0.5 Half--life (seconds) 84 84 83 What is the order of the reaction ? ETOOSINDIA.0 M. and the initial rate of the reaction is double that of the first run. 6.0 hour.0 x 10-3 0. (i) What is the order of the reaction with respect to A. 092142 33303 .0 x 10 -3 1.s1. What is the order of the reaction with respect to each reactant? 10.0020 4. In a kinetic study of the reduction of nitric oxide with hydrogen. 11.84) 8. The reaction A + 2B C + 2D is run three times.0 x 10-3 1. an equimolar mixture of gases was reduced to half the value in 102 seconds.COM India's No. and the initial rate is double that of the first run.4. 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. under the same conditions was reduced to half the value in 140 sec. (iv) Write the rate law equation. In the third run. In the second run. What is the rate constant when the rate law is written for (a) [BrO3]/t (b) + [Br ] / t ? 5. Calculate the order of the reaction. H. 2. The rate law of a chemical reaction given below : 2NO (g) + O2 (g) 2NO2 (g) is given as rate = k [NO]2 [O2].1 Online Coaching for JEE Main & Advanced 3rd Floor. (v) Calculate the rate constant. (b) What is the rate after 1. Kota. (Given e–0. For the reaction 3BrO(aq) BrO3 (aq) + 2Br (aq) in alkaline solution.0 x 10-3 0. then. the initial pressure of 288 mm.50 Rajeev Gandhi Nagar. Substance A reacts according to a first order rate law with k = 5 x 105 s1. Decomposition of ammonia on platinum surface follow the change. Rajasthan 324005 CHEMICAL KINETICS # 31 HelpDesk : Tel. the initial concentration of A is double that in the first run. (iii) What is the overall order.024 3. (a) If the initial concentration of A is 1.No. the initial concentration of each reactant is double the respective concentrations in the first run.mol1.057 L. 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. How will the rate of reaction change if the volume of reaction vessel is reduced to 1/4th of its original value? 6. In another experiment. the initial pressure of 340 mm. (a) The half life period and initial concentration for a reaction are as follows.5 × 10–4 Ms–1 ? d [NH3 ] k1 [NH3 ] (d) If the rate obeys – = 1 k [NH ] . (ii) What is the order of the reaction with respect to B. the value of rate constant at 0 80 C in the rate law for [BrO ]/t was found to be 0. 6.0 x 10-3 2. what is initial rate .4 x 10-3 0.18 = 0. 4.0 x 10-3 1. 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] 1.0080 If the reaction rate given by r = [A]a[B]b. 2NH3 (g) N2 (g) + 3H2 (g) (a) What does d [NH 3 ] denote ? dt d [N2 ] d [H2 ] (b) What does 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. 9. The reaction 2NO (g) + Br2 (g) 2NOBr (g). Calculate the rate constant of given reaction. 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). 17. The decomposition of Cl2O7 at 400 K in gaseous phase to Cl2 and O2 is of order reaction. C and D.43 9. H. D A B+C Time 0 t Volume of reagent V1 V2 V3 The reagent reacts with only B.23 42.869. Find the pressure after time '2t'.03 22. 18. e1. 8 15.COM India's No. 31 7 [Given : n = 0.79 [ln = 0. N2O5 + NO3 3NO2 + O2 2N2O5 4NO2 + O2 suggest the rate expression.79 22.50 Rajeev Gandhi Nagar. ln = 0. Find k. ETOOSINDIA. adding 50 mL water and titrating with baryta water. Rajasthan 324005 CHEMICAL KINETICS # 32 HelpDesk : Tel. 25% decomposition took place. (i) NO (g) + Br2 (g) NOBr2 (g) Slow (ii) NOBr2 (g) + NO (g) 2NOBr (g) Suggest the rate expression. the pressure of reaction mixture increases from 0.88 atm.39. 14. ln = 0.8439] 17. Assume first order reaction. requires activation energy of 70 kJ mol1. When a 20% solution of A was kept at 250C for 20 minutes. Kota. slow Step . 16. A first order reaction.60 32.12. Determine the velocity constant of hydrolysis. N2O5 NO2 + NO3 fast Step .No. also calculate the pressure of reaction mixture after 100 second. A B.2454.83 15.1 Online Coaching for JEE Main & Advanced 3rd Floor. After 55 sec at 400 K. 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.79 22.58 = 4. obeys the following mechanism. The thermal decomposition of N2O5 occurs in the following steps. For a homogeneous gaseous phase reaction 2A 3B + C the initial pressure was P0 while pressure after time 't' was P.62 to 1.24 24.20 26.858] CI2O7(g) CI2 (g) + O (g) 13 2 2 13. 092142 33303 . t (in minute) 0 75 119 259 Titre value (in mL) 19. Assume the mechanism: 2NO N2O2 .0045. are intermediates in this reaction? (d) Write the rate law for the rate-determining step. ETOOSINDIA. are catalysts in this reaction ? (c) Which species.19. Experiment shows that the equilibrium constant of the reaction : C2H5OH + CH3COOH CH3COOC2H5 + H2O is 2. derive an expression for the time taken by B to attain concentration equal to [B]eq/2 . When a catalyst is added this velocity constant is increased to 0. Given the following steps in the mechanism for a chemical reaction : A + B C (fast) B + C D + E slow) D + F A + E fast) At any time [C] is directly proportional to [A]. rate = K [NO]2 [O2] . if any. If concentrations at equilibrium are [A]eq. The forward reactions rate for the nitric oxide-oxygen reaction 2NO + O2 2NO2 has the rate law . 092142 33303 .COM India's No. (e) Write the rate law for this reaction.1 Online Coaching for JEE Main & Advanced 3rd Floor.002. (a) What is the stoichiometric equation for the reaction ? (b) Which species. (rapid equilibration) . What is now the velocity constant of the backward reaction? 22. (f) What is overall order of the reaction ? 20. and the velocity constant of the forward reaction is 0. if any. and [B]eq.No.50 Rajeev Gandhi Nagar. k' N2O2 + O2 2 NO2 [slow step]. Kota. 21. the initial concentration of A and B are [A]0 and zero respectively. where in K = kk. Show that the rate law can be explained.8 at room temperature. Rajasthan 324005 CHEMICAL KINETICS # 33 HelpDesk : Tel. For a first order reversible reaction A B . H. 05 –––––––––––––––––––––––––––––––––––––––––––– (a) Write the rate equation. Rajasthan 324005 HelpDesk : Tel. If the rate is 2. If '' is the intensity of absorbed light and 'C' is the concentration of AB for the photochemical process. Kota. PART .4 × 10–5 mol L–1 s–1.1 Online Coaching for JEE Main & Advanced CHEMICAL KINETICS # 34 3rd Floor. [JEE-2001.8 2. The rate constant for the reaction.43 × 10–5 (D) 1. 092142 33303 . Find the half life of the reaction.2 0.43 × 10–4 (B) 1.2 0. [JEE-2004.00 × 10–4 (D) 5.No. 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 (mol L–1) (mol L–1) (mol L–1 s–1) –––––––––––––––––––––––––––––––––––––––––––– 0. In the biologically-catalysed oxidation of ethanol. Consider the chemical reaction.05 0. the temperature required is 400 K.73 × 10–4 (C) 3.I : IIT-JEE PROBLEMS (PREVIOUS YEARS) * Marked Questions are having more than one correct option. N2(g) + 3H2(g) 2NH3(g) The rate of this reaction can be expressed in terms of time derivatives of conc. is 3 ×10–5 s–1. 3/84] (A) 3.1 0. AB + hv AB * . If the same reaction is carried out in the presence of a catalyst at the same rate. H.0025 (mol/lit) time (min) 0 40 Initial rate of reaction is in mol //min. H2(g) or NH3(g).38 × 10–4 (C) 1. 1. of N2(g) . the rate of formation of AB * is directly proportional to [JEE-2001. 1/35] (A) C (B) (C) 2 (D) C. The rate of a first order reaction is 0.03 mol litre–1 s–1 at 20 minutes after initiation. Calculate the activation energy of the reaction if the catalyst lowers the activation barrier by 20 kJ mol–1. Initial rates.45 × 10–5 (B) 1.10 0.50 Rajeev Gandhi Nagar. 2/60] ETOOSINDIA. Identify the correct relationship amongst rate expressions : [JEE-2002. 1/35] (A) 1.1 0.1 0. then the concentration of N2O5 (in mol L–1) is : [JEE-2000.73 × 10–5 8. r0. Given X product (Taking 1st order reaction) conc 0. 3/84] (A) 3.00 × 10–5 7. the concentration of ethanol decreases in a first order reaction from 800 mol dm–3 to 50 mol dm–3 in 2 ×104 s. A hydrogenation reaction is carried out at 500 K. [JEE-2000. The rate constant (s–1) of the reaction is : [JEE-2003. [JEE-2004. 3/100] 3. 5/100] 5.4 (B) 1. 4.04 (D) 0.2 (C) 0. (b) Calculate the rate constant of the reaction. 3/90] d [N2 ] 1 d [H2 ] 1 d [NH3 ] d [N2 ] d [H2 ] d [NH3 ] (A) Rate = (B) Rate = 3 2 dt 3 dt 2 dt dt dt dt d [N2 ] 1 d [H2 ] 1 d [NH3 ] d [N2 ] d [H2 ] d [NH3 ] (C) Rate = (D) Rate = dt 3 dt 2 dt dt dt dt 6.2N2O5 4NO2+ O2.1 0.01 0.COM India's No.04 mole litre–1 s–1 at 10 minutes and 0. time taken for 75% completion and find the total pressure when partial pressure of X. the rate increases by eight times.75 0.0. 3/81] (A) 0 (B) 1 (C) 2 (D) 3 11.0 × 106 s–1 and 38. Px = 700 mm of Hg. Rajasthan 324005 CHEMICAL KINETICS # 35 HelpDesk : Tel.COM India's No. when concentration of G is doubled keeping the concentration of H fixed.3 kJ mol–1 13. Kota.2 kJ mol –1 (B) 6. For a first order reaction A P. the rate is doubled. The pre-exponential factor A and the activation energy Ea. rate constant of the reaction.5 mol–1 dm3 (B) 1. 3/82] (A) 0.No.18 The order of the reaction is : 235 142 90 15.1 Online Coaching for JEE Main & Advanced 3rd Floor. Under the same reaction conditions. 3/80] 6 –1 (A) 1. The concentration of R in the reaction R P was measured as a function of time and the following data is obtained : [JEE-2010.12 0. 3/163] (A) (B) (C) (D) 14. respectively.10 t(min.9. the temperature (T) dependent rate constant (k) was found to follow the equation 1 log k = – (2000) + 6. Consider a reaction aG + bH Products.0 0. 092142 33303 . 3/163] [R](molar ) 1. 3/163] ETOOSINDIA. The overall order of the reaction is : [JEE-2007. respectively. [JEE-2005. H.0 × 10 s and 9.6 kJ mol –1 –1 (C) 1. 2/60] 10.40 0. [JEE-2008.0 × 106 s–1 and 16. The plot that follows Arrhenius equation is : [JEE-2010. are : T [JEE-2009.5 mol dm–3 (D) 2. For a reaction 2X(g) 3Y(g) + 2Z(g) the following data is obtained. However. The number of neutrons emitted when 92 U undergoes controlled nuclear fission to 54 Xe and 38 Sr is : [JEE-2010. initial concentration of 1. Px (mm of Hg) Time (min) (Partial pressure of X) 0 800 100 400 200 200 Find order with respect to X.50 Rajeev Gandhi Nagar.05 0. Plots showing the variation of the rate constant (k) with temperature (T) are given below.) 0.0 mol–1 dm3 12.0 s and 16. When concentration of both the reactants G and H is doubled.0 0.386 mol dm–3 of a substance becomes half in 40 k1 seconds and 20 seconds through first order and zero order kinetics.0 mol dm–3 (C) 1. Ratio of the rate k0 constant for first order (k1) and zero order (k0) of the reaction is.6 kJ mol–1 (D) 1. 1 Online Coaching for JEE Main & Advanced 3rd Floor.16. For the reaction system: 2NO(g) + O2(g) 2NO2(g).2003] 1 (1) (2) (m + n) (3) (n – m) (4) 2(n – m). M (3) M. (i) and (ii) as shown.2002] (1) 3 (2) 6 (3) 5 (4) 7 3. Products X. M sec–1 (2) sec–1. What is the value of × 10? (take log10 2 = 0. volume is suddenly reduced to half its value by increasing the pressure on it.2002] dH2 d[I ] d[HI] dH2 d[I2 ] 1 d[HI] (1) 2 (2) dt dt dt dt dt 2 dt 1 dH2 1 d[I2 ] d[HI] dH2 d[I ] d[HI] (3) (4) 2 2 2 2 dt 2 dt dt dt dt dt 4.50 Rajeev Gandhi Nagar. Y and Z respectivey are : [JEE-2011. mn 2 5. The differential rate law for the reaction H2 + I2 2HI is : [AIEEE. For the reaction A + 2B C.COM India's No. positron. H. rate is given by R = [A] [B]2 then the order of the reaction is : [AIEEE. 092142 33303 .II : AIEEE PROBLEMS (PREVIOUS YEARS) 1. If the reaction is of first order with respect to O2 and second order with respect to NO. proton (C) proton.2002] (1) sec–1. the rate of reaction will : [AIEEE. Rajasthan 324005 CHEMICAL KINETICS # 36 HelpDesk : Tel. proton. An organic compound undergoes first-order decomposition. sec–1 (4) M. [ t1 / 10 ] [IIT-JEE-2012] PART . neutron (D) positron. neutron. For the first order reaction [JEE-2011. On doubling the concentration of A and halving the concentration of B.6% completion in eight half-life duration 18. sec–1 2.No. position (B) Neutron. Bombardment of aluminum by -particle leads to its artificial disintegration into two ways. 3/163] (A) Proton. ETOOSINDIA.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. neutorn 17.sec–1. The rate law for a reaction between the substances A and B is given by rate = k [A]n [B]m.3). Units of rate constant of first and zero order reactions in terms of molarity M unit are respectively [AIEEE. position. 4/163] 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. Kota. the ratio of the new rate to the earlier rate of the reaction will be as [AIEEE. The time taken for its decompositionto 1/8 and 1/10 [ t 1/ 8 ] of its initial concentration are t1/8 and t1/10 respectively. the rate of reaction will be : [AIEEE. H. with everything else kept the same. If the concentration of carbon monoxide is doubled. A reaction was found to be second order with respect to the concentration of carbon monoxide.8 M to 0.2006] (1) remain unchanged (2) tripled (3) increased by a factor of 4 (4) doubled 13. The time taken for the concentration to change from 0. If the initial mass of the isotope was 200 g.COM India's No.2004] (1) 30 minutes (2) 15 minutes (3) 7. The correct statement in relation to this reaction is that the : [AIEEE.5 minutes (4) 60 minutes 8. The rate equation for the reaction 2A + B C is found to be : rate = k[A] [B]. The enthalpy change of the reaction (A2 + B2 2AB) in the presence of catalyst will be (in kJ mol–1).life of a radioisotope is four hours. which one of the following statements is correct : [AIEEE. 3/120] (1) 1 (2) 0 (3) 3 (4) 2 14.2004] (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.2005] (1) Eb < Ef (2) H = U (3) H < U (4) H > U 11. In a first order reaction. 7. decreases from 0. NO (g) + Br2 (g) NOBr2 (g) .125 g (4) 4. [AIEEE. 10.4 M in 15 minutes. In the respect of the equation k = Ae –Ea/RT in chemical kinetics. The energies of activation for forward and reverse reactions for A2 + B2 2AB are 180 kJ mol–1 and 200 kJ mol–1 respectively.50 Rajeev Gandhi Nagar. Consider an endothermic reaction X Y with the activation energies Eb and Ef for the backward and forward reaction.1 M to 0. the order of the reaction with respect to NO(g) is [AIEEE.2004] (1) 1. 092142 33303 . In general [AIEEE. The presence of a catalyst lowers the activation energy of both (forward and reverse) reactions by 100 kJ mol–1. respectively.No. the mass remaining after 24 hours undecayed is : [AIEEE.2005] (1) unimolecular reaction (2) first order reaction (3) second order reaction (4) bimolecular reaction 12. Kota.6.2003] (1) k is equilibrium constant (2) A is adsorption factor (3) Ea is energy of activation (4) R is Rydberg constant. NOBr2 (g) + NO (g) 2NOBr (g) (slow step) If the second step is the rate determining step.2007. 9. the concentration of the reactant.2007. The half .084 g (3) 3.042 g (2) 2. Rajasthan 324005 CHEMICAL KINETICS # 37 HelpDesk : Tel. A reaction involving two different reactants can never be : [AIEEE.1 Online Coaching for JEE Main & Advanced 3rd Floor.025 M is : [AIEEE. 3/120] (1) 280 (2) 20 (3) 300 (4) 120 ETOOSINDIA.167 g. The following mechanism has been proposed for the reaction of NO with Br2 to form NOBr. 50 Rajeev Gandhi Nagar. For a reaction A 2B.2010. Time required for the completion of 99% of the chemical reaction will be (log 2 = 0. Its half-life period is 30days. Kota. 092142 33303 .25 mol L–1. When the initial concentration of the reactant ‘A’.5 h (3) 0.06 minutes (3) 460.3 minutes 18.73 × 10 M/min –3 (2) 3.1 Online Coaching for JEE Main & Advanced 3rd Floor. H.6 minutes (4) 230.2009.01 M.2008. (A) products. Rajasthan 324005 CHEMICAL KINETICS # 38 HelpDesk : Tel.47 × 10 M/min –4 (3) 3. The half life period of a first order chemical reaction is 6.0 mol L–1 . Consider the reaction [AIEEE .2007.47 × 10 M/min –5 (4) 1. [AIEEE. A radioactive element gets spilled over the floor of a room.50 to 0. rate of disappearance of 'A' related to the rate of appearance of 'B' by the 2 expression. Cl2 + H2S H+ + Cl– + Cl+ + HS– (slow) Cl+ + HS– H+ + Cl– + S (fast) B. If the temperature is raised by 50° C. 3/105] d [ A ] 1 d [B] d [ A ] d [B] d[A] d [B] d [ A ] 1 d [B] (1) (2) (3) 4 (4) dt 4 dt dt dt dt dt dt 2 dt 17.No.301) : [AIEEE . 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. If it is a zero order reaction? [AIEEE .25 h (4) 1 h 19. the rate of the reaction increases by about : [AIEEE-2011] (1) 10 times (2) 24 times (3) 32 times (4) 64 times 21. The rate of a chemical reaction doubles for every 10° C rise of temperature.93 minutes. 4/120] (1) 1. 8/144] (1) 4 h (2) 0. H2S H+ + HS– (fast equilibrium) Cl2 + HS– 2Cl– + H+ + S (slow) (1) B only (2) Both A and B (3) Neither A nor B (4) A only 20.025 M in 40 minutes.03 minutes (2) 46. the concentration of A changes from 0. The rate of reaction when the concentration of A is 0. For a first order reaction.COM India's No.73 × 10–4 M/min ETOOSINDIA.15. 3/120] (1) 10 days (2) 100 days (3) 1000 days (4) 300 days 1 16. is 2. The time for half life period of a certain reaction A Products is 1 hour. If the initial activity is ten times the permissible value. how much time does it take for its concentration to come from 0.2010. is [AIEEE-2012. after how many days will it be safe to enter the room : [AIEEE. 8/144] (1) 23.1 M to 0. 3. (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. determine their order of reaction and the dimensions of the rate constants. In a reaction between A and B. In a pseudo first order hydrolysis of ester in water. 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.1 Online Coaching for JEE Main & Advanced 3rd Floor. PART-NCERT QUESTIONS 1. 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. then what are the units of rate and rate constants? 5. H.50 Rajeev Gandhi Nagar.2 mol L–1. i.COM India's No. Rate = k (PCH3OCH3)3/2 If the pressure is measured in bar and time in minutes. (i) Write the differential rate equation. [B] = 0. Calculate the rate of reaction after [A] is reduced to 0. Mention the factors that affect the rate of a chemical reaction.No. 6. What are the rates of production of N2 and H2 if k = 2. Calculate the initial rate of the reaction when [A] = 0.e.06 mol L–1. so the rate can also be expressed in terms of the partial pressure of dimethyl ether. From the rate expression for the following reactions. For the reaction : 2A + B A2B the rate = k[A][B] 2 with k = 2. 9. How is the rate of reaction affected if the concentration of the reactant is (i) doubled (ii) reduced to half ? 7. Rajasthan 324005 CHEMICAL KINETICS # 39 HelpDesk : Tel. The decomposition of dimethyl ether leads to the formation of CH4. Kota.0 × 10 –6 mol –2 L 2 s –1.1 mol L–1. 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. 092142 33303 ..5 × 10–4 mol–1 L s–1? 4. A reaction is first order in A and second order in B. (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. the following results were obtained : (i) Calculate the average rate of reaction between the time interval 30 to 60 seconds. The decomposition of NH3 on platinum surface is zero order reaction. (ii) Calculate the pseudo first order rate constant for the hydrolysis of ester. A reaction is second order with respect to a reactant. 15. An archaeological artifact containing wood had only 80% of the 14C found in a living tree. 092142 33303 .50 Rajeev Gandhi Nagar.COM India's No. 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. The experimental data for decomposition of N2O5 [2N2O5 4NO2 + O2] in gas phase at 318K are given below : (i) Plot [N2O5] against t.1 Online Coaching for JEE Main & Advanced CHEMICAL KINETICS # 40 3rd Floor. The rate constant for a first order reaction is 60 s–1. ETOOSINDIA. 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.11.1 years. (ii) Find the half-life period for the reaction. 16. The reaction between A and B is first order with respect to A and zero order with respect to B. The half-life for radioactive decay of 14C is 5730 years. (vi) Calculate the half-life period from k and compare it with (ii). one of the products is 90Sr with half-life of 28. During nuclear explosion.No. 12. (iii) Draw a graph between log[N2O5] and t. Kota. If 1µg of 90Sr was absorbed in the bones of a newly born baby instead of calcium. How much time will it take to reduce the initial concentration of the reactant to its 1/16th value? 17. H. how much of it will remain after 10 years and 60 years if it is not lost metabolically. Estimate the age of the sample. Fill in the blanks in the following table: 13. Rajasthan 324005 HelpDesk : Tel. (iv) What is the rate law ? (v) Calculate the rate constant. 30.34 – 1. The time required for 10% completion of a first order reaction at 298K is equal to that required for its 25% completion at 308K. If the value of A is 4 × 1010 s–1. Kota. The following data were obtained during the first order thermal decomposition of SO2Cl2 at a constant volume. ETOOSINDIA. 23.00 hours. A first order reaction takes 40 min for 30% decomposition.418 × 10–5 s–1 at 546 K. Consider a certain reaction A Products with k = 2.0 × 10–2 s–1.5 × 1011s–1) e–28000K/T. Calculate k at 318 K and Ea. Calculate t1/2. 22. The rate constant for the decomposition of hydrocarbons is 2. k = (4. For the decomposition of azoisopropane to hexane and nitrogen at 543 K. 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. Calculate Ea. the following data are obtained. Calculate the concentration of A remaining after 100 s if the initial concentration of A is 1.COM India's No. At what temperature would k be 1.18. H. For a first order reaction. The decomposition of hydrocarbon follows the equation k = (4.5 × 1011s–1) e–28000K/T 27. 25. 21.9 kJ/mol.No.1 Online Coaching for JEE Main & Advanced 3rd Floor. SO2 Cl2 (g) SO2 (g) + Cl2 (g) Calculate the rate of the reaction when total pressure is 0. 20.25 × 104K/T Calculate Ea for this reaction and at what temperature will its half-period be 256 minutes? 28. If the energy of activation is 179. Predict the rate constant at 30° and 50°C. with t1/2 = 3. Sucrose decomposes in acid solution into glucose and fructose according to the first order rate law. Calculate the energy of activation of the reaction assuming that it does not change with temperature.65 atm. What fraction of sample of sucrose remains after 8 hours ? 26. The decomposition of A into product has value of k as 4. show that time required for 99% completion is twice the time required for the completion of 90% of reaction. The rate of a reaction quadruples when the temperature changes from 293 K to 313 K.0 mol L–1.50 Rajeev Gandhi Nagar. 19. The rate constant for the first order decomposition of H2O2 is given by the following equation : log k = 14. what will be the value of pre-exponential factor.5 × 104 s–1? 29. 092142 33303 . 24. Rajasthan 324005 CHEMICAL KINETICS # 41 HelpDesk : Tel. Calculate the rate constant.5 × 103 s–1 at 10°C and energy of activation 60 kJ mol–1. r. (a) 4. (B) D-2. (A) 15. q. Rajasthan 324005 CHEMICAL KINETICS # 42 HelpDesk : Tel. (D) 5. or for zero Ea 79. (D) 37. (C) C-11.5 gm min–1 (b) 76. (B) E-4. (D) C-5. (AD) 41.019 M1 s1 . (B) B-10. T 53. infinte temp. T 47. T 39. (D) F-13. F 36. T 60. (B*) 12. (C) D-11. (B) 11. 81. (A) F-12. (B) C-21. (b) Rate of formation of N2 and rate of formation of H2 (c) 2. (B) E-3. (D) B-4. (B) 6. (B) 7.5 × 10–4 M sec–1 (d) (i) 1 . (D) B-6. (B) H-3. (C) 38. (CD) 45. (C) G-2. (C) D-12. (A) 30. (B) B-7. T 41. (D) G-5. (A) E-5. (C) 35. F 54. (A) C-1. (C) 39. T 59. (D) G-3. (B) B-9. (D) 22. (D) B-3. (a) Rate of decomposition of NH3. s . (A) 25. (A) 33. rare 67. (A) 27. (B) 29. (C) C-24. H. (A) p. (B) F-1. (a) 40. T 51. (B) s . r. (A) 27. (C) F-14. F 44. (B) A-2. (C) p. unchanged 2. 7. (B) C-14. (C) 20. (C) G-6. (D) C-3.2 × 105 M/s ETOOSINDIA. (C) 13. (B) 12.1 Online Coaching for JEE Main & Advanced 3rd Floor. (B) C-9. (C) F-7. (D) A-6. (B) C-4.50 Rajeev Gandhi Nagar. sum. (D) 36. (D) 18. (C) C-23. (A*) 7. (C) p . (B) D-3. lowering 68. (B) 20.5 × 10–3 M sec–1 (b) 3. EXERCISE # 1 PART # I A-1. (AC) 43. (B*) 4.038 M1 s1 5. F 45. (D) D-4. (D) D-5. (A) r . (A) C-20.5 gm min–1 3. –k 77. (D) p. (B) E-6. (B) 28. (C) 23. equal to H 64. (B) PART # II 1. (D) q 19. F 61. (B) 16. rate = k[M]2 75. T 48. s 17. (C) C-12. (A*) 2. (B) F-11. (a) 0. (B) A-8. Kota. (C) B-5. (C) C-26. 2 and 3 1 74. rate constant 73. (B) H-5. (A*) 5. (C*) 15. threshold energy 62. T 33. (C) 26.0 × 10–3 M sec–1 2. (D) q 16. (B*) 11. F 50. q. r. (C*) 10. (B) p . (C) 24. (B) 31. (A) G-7. first 80.32 × 10–4 M sec–1 4. (C*) 8. (C*) 14. (D) 40. (B) r. (B) 34. (B) C-16. (B) C-19. F 57. F 34. (C) s. (C*) 9. (B) A-4. increases 71. (A) E-2. (C) 32.* (ABCD) H-1. (B) 17.No. (B) G-4. T 35. (B) H-6. (A) F-9. (B) C-17. (D) D-10. (B) H-4. (C) 31. (D) C-10. (a) 5 × 105 M/s (b) 4. 1 76. (A) D-6. (C) 4. (C) p. (C) 21. (C) C-13. (A) C-27. Activation energy for forward reaction 2 k 78. (B) B-2. (B) F-15.303 82. (D) F-8. r. (D) 22. (D) 14. (A) D-8. (B) H-2. (C) F-3. fast 69. (ACD) 42. (C) 26. (C) 3. (D) 9. (B) C-6. slowest 70. T 49. (A) 28. (ii) 0 7.COM India's No. F 58. (C) B-1. exponents EXERCISE # 2 1. F 40. (A) s. F 46. directly proportional 65. (D) 25. T 52. T 38. (B) B-8. (B) E-1. s . (D) C-8. (D) G-1. (B) C-7. T 42. s . T 43. (B) q. (D) 30. half 63. F 32. (D) q 18. (A) F-2. (A*) 13. (D) D-9. (B) 21. (C) A-3. (A) F-4. (D*) 3. (A) r . 092142 33303 . (A) 23. (D) 19. (C) D-1. (B) C-18. (C) 10. T 56. (D) B-11. (C) 24. (D) B-12. mol L–1s–1 66. (C) D-7. (C) 2. (D) A-7. (A) A-5. (C) D-13. (A) F-10. (A) F-5. (C*) 6. 4.5 × 10–4 M sec–1 . (C) C-22. (A) 8. (C) C-15. (ABC) SUBJECTIVE 1. (D) A-10. (A) C-2. (C) F-6. 64 6. F 37. (B) A-9. (C) C-25. (b) 0. 3 72. T 55. (A) 29. (ABC) 44. Rate of reaction = rate of diappearance of A = 0. t = 444 s 6.4 t1/2 = 24 min.4 (4) 2. 13. (b) A.001607 22.16 (1) 2. (iii) Third order.3 (4) 2.11 (A) 1. Kota.897 kJ mol–1 9. B.19 (4) 2. 1. 1.693 21.10% 17. (c) C. Rajasthan 324005 CHEMICAL KINETICS # 43 HelpDesk : Tel.8 (4) 2. 1. k [N2O5] t 3 2 18.1 Online Coaching for JEE Main & Advanced 3rd Floor.925 × 10–4 s–1 8. 2P0 14. (ii) 1st order. 1.8 (a) R0 = k[A0].1 (D) 1.5 (A) 1.21 (2) EXERCISE # 4 1. ln ( V V ) 16.58 × 10–2.15 3 1.18 (3) 2. (e) rate = k'[A][B] (f) 2.9 (3) 2.2 (1) 2. (iv) r = K[A]2[B] (v) k = 3.50 Rajeev Gandhi Nagar. n2 Rate constant K = = 6. (a) n = 2.13 (A) 1.13 (4) 2.7 (1) 2. Now when Px = 700 = 800 – 2x so x = 50 mm of Hg so total pressure = 800 + 3x = 950 mm of Hg 1. Since half left is independent of initial concentration so reaction must Ist order with respect to X.27 × 10–3 min–1 P0 1 ( V3 V1) 15. (b) First Order (2 P0 P)2 12.5 4. (d) rate = k [B][C].20 (3) 2.93 × 10–3 min–1.15 (2) 2. 092142 33303 . 3. 0.6 (B) 1.12 (D) 1.005 mol litre–1min–1 3. t1/ 2 Time taken for 75% completion = 2 × t1/2 = 200 min.18 9 PART-II 2. Ea = 52.5 sec–1.5 (3) 2.33 × 105 mole–2 +2 sec–1 9.471 × 10–19 ETOOSINDIA. 0.1 (1) 2. XY Rate = k[X]2 The rate will increase 9 times 5. rav = 6.COM India's No.33 atm.17 (2) 2. order with respect to A = 1. (i) 2nd order.16 (A) 1. % decomposition = 67.17 (A. 3 11. Order of the reaction is 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.8.D.14 (2) 2. H.No. D) 1.7 (A) 1.6 (3) 2. 2.3 (B) 1.14 0 1.66 × 10–6 Ms–1 2. Rate = k [NO]2 [Br2] 19.10 (1) 2. 1. order with respect to B = 0 10. Also in next 100 min Px decreases from 400 to 200 to again t1/2 = 100 min.11 (1) 2.10 (D) 1.12 (3) 2. t= K f Kb EXERCISE # 3 PART-I 1.2 EA = 100 KJ/mol 1. (b) 0. (a) 2 B+F 2E.