Chemical Kinetics and Nuclear Chemistry Exercise # 2

Exercise # 2 1. [Multiple Correct Choice Type Questions] Part # I A reaction takes place in three steps. The rate constant of the three steps are K 1, K2 and K3 respectively. The overall rate constant K = K 1K 3 K2 . The energy of activation for the three steps are 40, 30 and 20 KJ respectively. Therefore : (A) Overall energy of activation is 10 KJ (B) Overall energy of activation is 30 KJ K1 (C) The reaction mechanism is 2A A* + A ; A * K1 (D) The reaction mechanism is A 2. K2 B;B K2 product and overall order is one. C;C K3 product and overall order is one For the reaction CH4 + Br 2  CH 3Br + HBr the experimental data require the following rate equation : k1[CH4 ][Br2 ] d [CH3Br] = 1+ k [HBr] / [Br ] dt 2 2 Which of the following is/are true regarding this ? (A) The reaction is a single step reaction (B) The reaction is 2nd order in the initial stages {[HBr]  0} (C) The reaction is 2nd order in the final stages {[Br2]  0} (D) The molecularity of the reaction is two 3. Rate of radioactive disintegration (–dN/dt) is also known as : (A) Activity (B) Radioactivity (C) Half-life period (D) Average life period 4. Decomposition of 3 A(g)  2 B(g) + 2C(g) follows first order kinetics. Initially only A is present in the container.. Pressure developed after 20 min. and infinite time are 3.5 and 4 atm respectively. Which of the following is true. (A) t50% = 20 min (B) t75% = 40 min (C) t99% = 64/3 min (D) t87.5% = 60 min 5. 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 Identify the true statement (s) log 2 = 0.3, log 3 = 0.48, log 7 = 0.84 (A) the half life of the reaction is 75 min. (B) the solution is optically inactive at 120 min. (C) The equimolar mixture of the products is dextrorotatory. (D) The angle would be 7.5° at half time. 6. Consider the decay of P to A and B by two parallel first order reactions as shown in Fig. Given Reaction H Rate constant Energy of activation P A HA kA EA PB HB kB EB Which of the following is(are) true? (A) a = EB (B) b = EA (C) HA = b  d (D) HB = c  a CHEMICAL KINETICS AND NUCLEAR CHEMISTRY 7. 2000 Rate constant k varies with temperature by equation , log k(min–1) = 5 – T(K) . We can conclude: (A) pre–exponential factor A is 5 (C) pre–exponential factor A is 10 8. (B) Ea is 2000 kcal 5 (D) Ea is 9.212 kcal Activation energy of a chemical reaction can be determined by : (A) evaluating rate constant at standard temperature (B) evaluating velocities of reaction at two different temperatures (C) evaluating rate constant at two different temperatures (D) changing concentration of reactants 9. Which of the following statements are correct ? (A) Law of mass action and rate law expressions are same for single step reactions. (B) Order of the slowest elementary reaction of a complex reaction gives the order of the complex reaction (C) Both order and molecularity have normally a maximum value of 3 (D) Molecularity of a complex reaction, A + 2B  C is 3. 10. Table I (reactions) and table II (methods) have been matched Correct matching is : 11. P Q R (A) X Y Z (C) Z X Y P Q R (B) Y X Z (D) Z Y X Which of the following statements are correct about half–period. (A) It is proportional to initial concentration for zeroth order (B) Average life = 1.44 half–life for first order reaction. (C) Time of 75% completion of reaction is thrice of half–life(initial half life) period in second order reaction (D) 99.9% reaction takes place in 100 minutes for the case when rate constant is 0.0693 min–1 12. Zn + 2 H+  Zn2+ + H2 Half–life period is independent of concentration of zinc at constant pH. For the constant concentration of Zn, rate becomes 100 times when pH is decreased from 3 to 2. Hence , (A) dx = k [Zn]0[H+]2 dt  dx  (B)  dt  = k [Zn][H+]2   K  Ae Ea / RT (A) Plot P shows that the energy of activation is independent of temperature. (B) Plot Q describes the behaviour of temperature dependence of energy of activation (C) Arrhenius behaviour is described by P (D) The slope of curve P gives the value – 14. If the rate of reaction, 2SO2(g) + O2(g) log10K 13. (C) rate is not affected if concentration of zinc is made four times and that of H+ ion is halved (D) rate becomes four times if concentration of H+ ion is doubled at constant Zn concentration Which of the following statement are true regarding the log K vs. 1/T plot shown in the given diagram ? P Q 1/T Ea R Pt 2SO3(g) is given by : Rate = K [SO 2 ] [SO 3 ]1/ 2 which statements are correct : (A) The overall order of reaction is –1/2 (B) The overall order of reaction is +1/2 (C) The reaction slows down as the product SO3 is build up (D) The rate of reaction does not depend upon concentration of SO3 formed 15. 1 For a second order reaction plots are made for [ A ] vs time for the reaction, 2A  Product. Pick up the correct sentences. (A) the graph will show straight line with slope K (B) the graph will show straight line with intercept [A]0 (C) the graph will show straight line with slope [A]0 1 16. (D) the graph will show straight line with intercept [ A ] 0 A (aq)  B (aq) + C (aq) is a first order reaction. Time t  moles of reagent n1 n2 Reaction progress is measure with the help of titration of reagent ‘R’.If all A, B and C reacted with reagent and have   ‘n’ factors  n factors; eq.wt. = 1  n2  (A) k = t ln  n – n  1   2 mol.wt  in the ratio of 1 : 2 : 3 with the reagent.The k in terms of t, n1 and n2 is : n  1  2n 2  (B) k = t ln  n – n  1   2 1  4n 2  (C) k = t ln  n – n  1   2  1  4n 2 (D) k = t ln  5(n – n )  2 1   CHEMICAL KINETICS AND NUCLEAR CHEMISTRY 17. A 5 K= 2×10 s 1 B 6 K= 8×10 s 1 3 1 K= 3×10 s C D The rate determing step of the reaction is : (A) A  B 18. (B) C D (C) B  C (D) A D The Arrhenius relationship of two different reactions is shown below. Which reaction is faster at a lower temperature and which is more sensitive to changes of temperature ? A (A) B faster, A more sensitive (B) B in both cases (C) A in both cases (D) A faster, B more sensitive log k B 1/T 19. Consider the following reactions at 300 K. A  B (uncatalysed reaction) catalyst A   B (catalyst reaction) The activation energy is lowered by 8.314 KJ mol–1 for the catalysed reaction. How many times the rate of this catalysed reaction greater than that of uncatalysed reaction ? (Given e3.33 = 28) (A) 15 times 20. (B) 38 times (C) 22 times (D) 28 times For the following parallel chain reaction –3 K1 = 2× 10 A –1 s 4B 3C  [B]t 16  what will be that value of overall half-life of A in minutes?  Given that [C] = 9  t   (A) 3.3 21. (B) 6.3 For the follwing parallel chain reaction (C) 3.6 (D) None the overall half life of A is 12 hours If rate of formation of C is 60% of a rate of decomposition of A then what will be half life of A while it is converting into B ? (A) 40 hours 22. (B) 60 hours (C) 50 hours (D) 30 hours At what time will B be present in greatest concentration ? k1 (A) K  K 2 1 k1 1 (B) K  K ln k 1 2 2 k1 1 (C) K  K ln k 2 1 2 (D) None of these For the same mechanism as Problem refer to the given Fig. and choose the correct set of identifications. E + (enzyme) ES (intermediate-1) S (reactant) ES (intermediate-1) EP (intermediate-2) E (enzyme) EP (intermediate-2) (1) + P (product) (2) (1) (4) (2)(3) Reaction coodinate (3) (4) (A) E for E + S  ES Ea for ES  EP E overall for S  P Ea for EP  E + P (B) Ea for E + S  ES E for E + S  ES Ea for ES  EP  overall for S  P (C) Ea for ES  EP Ea for E + S  ES Ea for EP  E + P Ea for ES  EP E overall for S  P Ea for EP  E + P E for EP  E + P E overall for S  P E for E overall E for l Ea for E + S  ES for S  P EP  E + P EP  E + P (D) (E) 24. Potential energy 23. A simple mechanism for enzyme-catalyzed reaction is given by the following set of equations E + (enzyme) S (reactant) ES (intermediate-1) ES (intermediate-1) EP (intermediate-2) E (enzyme) EP (intermediate-2) + P. E. (3) (2) P (product) (4) (1) Reaction coodinate This is known as the Michaelis–Menten mechanism. The potential energy diagram is shown in the fig. Which of the following sets of identifications is correct ? (Assume that the temperature and pressure are constant). (1) (2) (3) (4) 25. (A) E + P. EP (B) ES Activated complex EP Activated complex (C) EP Activated complex ES Activated complex (D) E+S ES Rate law of the reaction A  Product is, rate = k[A]. Graphically it is represented as Hence, rate constant is : (A) 3 × 10–4 s–1 (B) 1 × 10–2 s–1 (C) 3 × 10–2 s–1 (D) 1 × 10–4 s–1 ES EP E+S E+P CHEMICAL KINETICS AND NUCLEAR CHEMISTRY 26. For a given reaction A  Product, rate is 1 × 10–4 M s–1 when [A] = 0.01 M and rate is 1.41 × 10–4 M s–1 when [A] = 0.02 M. Hence, rate law is : (A)  27. d [A] = k [A]2 dt (B)  d [A] = k[A] dt (C)  Following reaction can take place in both direction A It is given that for the forward reaction : k1 k2 and for the backward reaction [B] (D)  d [A] = k[A]1/2 dt B, –1 Rate (M s) Rate 0.01 M 1 × 10–2 Ms–1 0.02 M 2 × 10–2 Ms–1 Hence, net reaction rate of B is : (A) = k1 [A] – k2 [B] (B) = k1 – k2 [B] 28. d [A] k  [A] dt 4 For the complex Ag+ + 2NH3 [A],M (C) = k1 [A] – k2 (D) = k1 – k2 [Ag(NH3)2+]  dx   dt  = 2 × 107 L2 mol–2 s–1 [Ag+] [NH3]2 – 1 × 10–2 s–1 [Ag(NH3)2+]   29. Hence, ratio of rate constants of the forward and backward reactions is : (A) 2 × 107 L2 mol–2 (B) 2 × 109 L2 mol–2 (C) 1 × 10–2 L2 mol–2 In the reaction N2 + 3H2 2NH3 (D) 0.5 × 10–9 L2 mol–2 [N2 ] [H2 ]3  dx  2 3 –3 2 = 10–5 M2  dt  = 1 × 10 [N2] [H2] – 1 × 10 [NH3] and at some instant if [NH3 ]2    dx  then at this instant value of   is :  dt  (A) 0 30. (B) 1 × 105 (C) 1 × 10–5 In the following graphical representation for the reaction A  B (A) I and II both represent kinetic region at different interval (B) I and II both represent equilibrium regions at different time interval (C) I represents kinetic while II represents equilibrium region (D) I represents equilibrium while II represents kinetic region (D) 1 × 10–3 there are two types of regions : 31. For the following reaction  dx  (CH3)3CCl + H2O  (CH3)3 COH + HCl ,  dt  = k[CH3)3CCl], hence, rate determining step can be:   (A) (CH3)3CCl  (CH3)3C+ + Cl– (C) (CH3)3C+ + H2O  (CH3)3COH + H+ 32. (B) (CH3)3CCl + H2O  (CH3)3COH + HCl (D) H+ + Cl–  HCl For the non-equilibrium process, A + B products, the rate is first-order w.r.t. A and second-order w.r.t. B. If 1.0 mole each of A and B introduced into a 1.0 L vessel and the initial rate was 1.0 × 10–2 mol L–1 s–1, rate when half reactants have been turned into products is : (A) 1.25 × 10–3 mol L–1 s–1 (C) 2.50 × 10–3 mol L–1 s–1 33. Select the correct statements out of I, II and III for zero order reaction I : Quantity of the product formed is directly proportional to time II : Larger the initial concentration of the reactant, greater the half-life period III : If 50% reaction takes place in 100 minutes, 75% reaction take place in 150 minutes. (A) I only 34. (B) I and II only (C) II and III only (D) I, II and III For the first-order reaction Tav (average life), T50 and T75 in the increasing order are : (A) T50 < Tav < T75 35. (B) 1.0 × 10–2 mol L–1 s–1 (D) 2.0 × 10–2 mol L–1 s–1 (B) T50 < T75 < Tav (C) Tav < T50 < T75 (D) Tav = T50 < T75 Which is not true for a second order reaction ? (A) It can have rate constant 1 × 10–2 L mol–1 s–1 (B) Its half-life is inversely proportional to its initial concentration (C) Time to complete 75% reaction is twice of half-life 1 (D) T50 = K × Initial conc. 36. N KI Cl  I2 COCH3 Kinetics can be studied by titration using : (A) Na2S2O3 37. (B) KMnO4 (C) H2C2O4 (D) OH A certain zero order reaction has k = 0.025 M s–1 for the disappearance of A. What will be the concentration of A after 15 seconds if the initial concentration is 0.50 M ? (A) 0.50 M 38. (B) 0.375 M (C) 0.125 M (D) 0.060 M A reaction of first-order completed 90% in 90 minutes, hence, it is completed 50% in approximately : (A) 50 min (B) 54 min (C) 27 min (D) 62 min CHEMICAL KINETICS AND NUCLEAR CHEMISTRY 39. T50 (Half-life period) of first-order reaction is 10 minute. Starting with 10 mol L–1, rate after 20 minute is : (A) 0.0693 mol L–1 min–1 (B) 0.0693 × 2.5 mol L–1 min–1 (C) 0.0693 × 5 mol L–1 min–1 (D) 0.0693 × 10 mol L–1 min–1 40. M Z A(g)  M 8 Z4 B(g) + ( - particles) (-particles are helium nuclei, so will form helium gas by trapping electrons) The radioactive disintegration follows first-order kinetics Starting with 1 mol of A in a 1-litre closed flask at 27ºC pressure developed after two half-lives is approximately : (A) 25 atm (B) 12 atm (C) 61.5 atm (D) 40 atm 41. A.G.M. counter is used to study the radioactive process of first-order. In absence of radioactive substance A, it counts 3 disintegration per second (dps). When A is placed in the G.M. counter, it records 23 dps at the start and 13 dps after 10 minutes. It records x dps after next 10 minutes and A has half-life period y minutes. x and y are : (A) 8 dps, 10 min (B) 5 dps, 10 min (C) 5 dps, 20 min (D) 5 dps, 5 min 42. Saponification of ethyl acetate (CH3COOC2H5) by NaOH (Saponification of ethyl acetate by NaOH is second order Rxn) is studied by titration of the reaction mixture initially having 1:1 molar ratio of the reactants. If 10 mL of 1 N HCl is required by 5 mL of the solution at the start and 8 mL of 1 N HCl is required by another 5 mL after 10 minutes, then rate constant is : (A) k = 2.303 10 log 10 8 (B) k = 2.303 10 log 10 2 (C) k = 1 1 1   10  8 10  (D) k = 1 1 1   10  2 10  43. Following are the values of Ea and H for three reactions carried out at the same temperature : I : Ea = 20 kJ mol–1 , H = – 60 kJ mol–1 II : Ea = 10 kJ mol–1 , H = – 20 kJ mol–1 III: Ea = 40 kJ mol–1 , H = + 15 kJ mol–1 If all the three reaction have same frequency factor then fastest and slowest reactions are : Fastest Slowest (A) I II (B) II III (C) I III (D) can’t be predicted. 44. Milk turns sour at 400C three times as faster as 00C. Hence Ea in cal of process of turning of milk sour is : 45.  1 2.303  2  313  273 log  3    40 (A) 2.303  2  313  273 log3 40 (B) (C) 2.303  2  40 log3 273  313 (D) None of the above Graph between log k and 1 (k is rate constant in s–1 and T is the temperature in K) T is a straight line. As shown in figure if OX = 5 and slope of the line = – log k  1 then Ea is : 2.303 2 cal 2.303 (A) 2.303 × 2 cal (B) (C) 2 cal (D) None of these O 1 T 46. For the reaction A (g) + 2B(g)  C(g) + D(g) dx = k [A] [B]2 dt Initial pressure of A and B are respectively 0.60 atm and 0.80 atm. At a time when pressure of C is 0.20 atm, rate of the reaction, relative to the initial value is : (A) 47. 1 6 (B) (C) 1 4 (D) 1 24  dx  If   = k [H+]n and rate becomes 100 times when pH changes from 2 to 1. Hence, order is :  dt  (A) 1 48. 1 48 (B) 2 (C) 3 (D) 0 Consider following reaction, Zn + 2 H+  Zn2+ + H2 Half–life period is independent of concentration of Zn at constant pH. At constant Zn concentration, half–life is 10 minutes at pH = 2 and half–life is 100 minutes at pH = 3. Hence , rate law is : (A) k [Zn][H+] 49. (B) k [Zn][H+]2 (B) 2 (B) 0.002 (D) 0 (C) 0.001 (D) 2 For the first-order reaction (C = C0 e k1t ) and Tav = k1–1. After two average lives concentration of the reactant is reduced to : (A) 25% 52. (C) 3 A reaction is catalysed by H+ ion;and in the rate law the dependence of rate is of first order with respect to the concentration of H+ ions, in presence of HA rate constant is 2 × 10–3 min–1 and in presence of HB rate constant is 1 × 10–3 min–1. HA and HB have relative strength as : (A) 0.5 51. (D) k [Zn]0 [H+]2 If the fermentation of sugar in an enzymatic solution that is 0.12 M , the concentration of the sugar is reduced to 0.06 M in 10 h and to 0.03 M in 20 h. What is the order of the reaction ? (A) 1 50. (C) k [Zn]0 [H+] (B) 75% (C) 100 % e (D) 100 e2 % A(g)  B(g) + C(g) – d [A] = k [A] dt At the start pressure is 100 mm and after 10 min , pressure is 120 mm. hence , rate constant (min–1) is : (A) 53. 2.303 120 log 10 100 (B) 2.303 10 0 log 10 20 (C) 2.303 10 0 log 10 80 (D) The rate law for the dimerisation of NO2 is – d [NO 2 ] = k [NO2]2 dt which of the following changes will change the value of the specific rate constant , k : (A) Doubling the total pressure on the system (B) Doubling the temperature (C) Both of (A) and (B) (D) None of the above 2.303 10 0 log 10 12 0 CHEMICAL KINETICS AND NUCLEAR CHEMISTRY 54. In the following reaction, how is the rate of appearance of the underlined product related to rate of disappearance of the underlined reactant : BrO3– (aq) + 5 Br– (aq) + 6 H+ (aq)  3 Br2 () + 3 H2O () – (A) 5 d[Br ] d [Br2 ] =– 3 dt dt (B) d[Br ] d [Br2 ] =– dt dt (C) d[Br ] d [Br2 ] =– dt dt (D) d [Br2 ] 3 d[Br ] =– dt 5 dt – – – 55. Number of natural life times (Tav) required for a first-order reaction to achieve 99.9% level of completion is : (A) 2.3 (B) 6.9 (C) 9.2 (D) 0.105 56. Match the graphical study with the order of the reactions : I : Rate (A) (C) 57. II : Half life I A C II B B III C A III : (a – x)–1 (B) (D) I B C II C A III A B The bromination of acetone that occurs in acid solution is represented by CH3COCH3 (aq) + Br2 (aq)  CH3COCH2 Br (aq) + HBr (aq) Given : Initial Rate of Disapperance of Br2 , Ms–1 Initial concentrations Br2 0.050 0.10 0.10 0.050 [CH3COCH3] 0.30 0.30 0.30 0.40 M [H+] 0.050 0.050 0.10 0.20 5.7 × 5.7 × 1.2 × 3.1 × 10–5 10–5 10–4 10–4 Based on the above data , rate law is : 58.  dx  (A)   = k [CH3COCH3] [Br2] [H+]  dt   dx  (B)   = k [CH3COCH3] [H+]  dt   dx  (C)   = k [CH3COCH3] [Br2]  dt   dx  (D)   = k [CH3COCH3] [Br2] [H+]2  dt  3 A  B + C, It would be a zero order reaction when : (A) the rate of reaction is proportional to square of concentration of A (B) the rate of reaction remains same at any concentration of A (C) the rate remains unchanged at any concentration of B and C (D) the rate of reaction doubles if concentration of B is increased to double 59. 60. The potential energy diagram for a reaction R  P is given below : H0 of the reaction corresponds to the energy – (A) x (B) y (C) z (D) (x+y) Read the following industrial methods for the preparation of H2SO4 and answer the question at the end. Professor Molina of the Massachusetts Institute of Technology won the 1995 Nobel Prize in Chemistry for his work on atmospheric chemistry. One reaction that he has studied in detail is the acid rain reaction which produces H2SO4 in the atmosphere. He has proposed two possible stoichiometric reactions : Proposal A : H2O (g) + SO3 (g)  H2SO4 (g) Proposal B : 2 H2O (g) + SO3 (g)  H2SO4 (g) + H2O (g) Using simple collision theory , what reaction orders would be expected for proposal B ? Proposal B is thought to proceed by the following two-step process : k1 SO3 + 2 H2O k 1 SO3.2 H2O (fast) 61. SO3.2 H2O k H2SO4 + H2O (slow) (SO3.2 H2O is a complex which is stabilized by hydrogen bonds and k2 << k1 or k–1 ). (A) k [H2O] [SO3] (B) k [H2O]2 [SO3] (C) k [SO3] (D) k [H2O] In gaseous reactions important for the understanding of the upper atmosphere H2O and O react bimolecularly to form two OH radicals. H for this reaction is 72 kJ at 500 K and Ea is 77 kJ mol–1, then Ea for the bimolecular recombination of two OH radicals to form H2O and O is : (A) 3 kJ mol–1 (B) 4 kJ mol–1 (C) 5 kJ mol–1 (D) 7 kJ mol–1 62. A reaction takes place in three steps having rate constants K1, K2, K3 respectively. The overall rate constant 2 K1 K 3 K = K . If energies of activations for the three steps are 40, 30, 20 kJ respectively, the overal energy of activation 2 is : (A) 10 (B) 15 (C) 30 (D) 60 63. 64. Two reaction, A  products and B  products have rate constants Ka and Kb at temperature T and activation energies Ea and Eb respectively. If Ka > Kb and Ea < Eb and assuming that A for both the reactions is same then (A) At higher temperatures Ka will be greater than Kb (B) At lower temperature Ka and Kb will differ more and Ka > Kb (C) As temperature rises Ka and Kb will be close to each other in magnitude (D) All of the above At room temperature, the reaction between NO and O2 to give NO2 is fast, while that between CO and O2 is slow. It is due to : (A) CO is smaller in size that of NO (B) CO is poisonous (C) The activation energy for the reaction, 2NO + O2  2NO2 is less then 2CO + O2  2CO2. (D) None of the above CHEMICAL KINETICS AND NUCLEAR CHEMISTRY 65. The rate law for a reaction between the substances A and B is given by rate = K[A]n [B]m. On doubling the concentration of A and halving the concentration of B, the ratio of the new rate to the earlier rate of the reaction will be as : (A) 1/2m + n (B) (m + n) (C) (n – m) (D) 2(n – m) 66. Which of the following statement is/are correct ? (A) A second order reaction must be a bimolecular elementary reaction (B) A bimolecular elementary reaction must be a second order reaction (C) Zero order reaction must be a complex reaction (D) First order reaction may be complex or elementary reaction 67. For the reaction 2A + B  C with the rate law d [C ] = k [A]1 [B]–1 and started with A and B in dt stoichiometric proportion. Which is/are true? (A) unit of k is Ms–1 (B) [A], [B] and [C] all will be linear functions of time (C) [C] = 2kt (D) [C] = kt 68. 1 For a second order reaction plots are made for [ A ] vs time for the reaction, 2A  Product. Pick up the correct sentences. (A) the graph will show straight line with slope K (B) the graph will show straight line with intercept [A]0 (C) the graph will show straight line with slope [A]0 1 (D) the graph will show straight line with intercept [ A ] 0 69. Which of the following are example of Pseudo uni molecular reaction? (A) CH 3 – C – OC2 H5 + H2O || H CH3COOH + C2H5OH O (B) C12H22O11 + H2O H (C) CH 3 – C – OC2 H5 + H2O C6H12O6 (glucose) + C6H12O6 (fructose) OH  CH3 – C – OH || || O O (D) CH 3 – C – Cl + H2O  CH3COOH + HCl || O 70. In the following gaseous phase first order reaction A(g)  2B(g) + C(g) initial pressure was found to be 400 mm and it changed to 1000 mm after 20 min. Then : (A) Half life for A is 10 min (B) Rate constant is 0.0693 min–1 (C) Partial pressure of C at 30 min is 350 mm (D) Total pressure after 30 min is 1100 mm For the consecutive reaction A values of k1 & k2 1 k1 (time ) B k2(time1) C following curves were obtained depending upon the relative B A Conc. C C Conc. 71. A B Time Figure 2 Time Figure 1 Now which of the following is the correct match 72. 73. (A) figure 1 – (k1 < k2 ) (B) figure 2 – (k1 < k2) (C) figure 2 – (k1 >> k2) (D) figure 1 – (k1 >> k2) 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 Which of the following statements about the arrhenius equation are correct ? (A) The pre exponential factor become equal to the rate constant of the reaction at temperature T   (B) When the activation energy of the reaction  zero, the rate becomes independent of temperature. (C) The term e–Ea/RT represents the fractions of the molecules having energy is excess of the threshold value. (D) On raising temperature , rate constant of the reaction of greater activation energy increases less rapidly than that of the reaction of smaller activation energy. 74. Select the correct statements : (A) the molecularity of an elementary reaction indicates how many reactant molecules take part in the step. (B) the rate law of an elementary reaction can be predicted by simply seeing the stoichiometry of reaction. (C) the slowest elementary step in sequence of the reactions governs the overall rate of formation of product. (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. 75. Which of the following are true about rate of a reaction? (A) Rate of a reaction cannot be negative (B) Rate of a reaction is change in concentration of the reactant per unit time per mol (C) Average rate and instantaneous rate can never be equal. (D) Rate depends upon surface area of the reactants. CHEMICAL KINETICS AND NUCLEAR CHEMISTRY [Assertion & Reason Type Questions] Part # II Each question has 5 choices (A), (B), (C), (D) and (E) out of which only one is correct. (A) Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1. (B) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1. (C) Statement-1 is True, Statement-2 is False. (D) Statement-1 is False, Statement-2 is True. (E) Statement-1 and Statement-2 both are False. 1. Statement-1 : Statement-2 : If the activation energy of reaction is zero temperature will have no effect on the rate constant Lower the activation energy faster is the reaction 2. Statement-1 : In A k1 B k2 C If half life of A is very less as compared to B, so the net reaction is A  C with rate constant (k1 × k2) 3. Statement-2 : Slowest step is the rate determining step so B  C is rate determining step. Statement-1 : Statement-1 : For A + 2B  C (rate= K[A]1[B]0), the half life time of reaction is only defined when conc. of A and B are in stoichiometric ratio For above given order half life of reaction is directly proportional to conc. of A and not to conc. of B due to its zero order. Many reactions occurring on solid surface are zero order reactions. Statement-2 : Au N2 O(g)   N2 (g)  1/ 2O 2 ; rate = k[N2O]0 = k = constant is a zero order reaction. Statement-1 : Half life of a certain radioactive element is 100 days. After 200 days, fraction left undecaye will be 25%. Statement-2 : 4. 5. n Statement-2 : Ct  1     , where symbols have standard meaning. C0  2  6. Statement-I : Statement-II : The pre-exponential factor A has the same units for all reactions. –E /RT e a has no unit. 7. Statement-I : Disintegration 1H (tritium) is accompained by -emission. Statement-II : Tritium has high n/p ratio. Statement-I : Neutrons are the best bombarding particles. Statement-II : Neutrons are neutral particles. Statement-I : The order of reaction can have fractional value. Statement-II : For an elementary reaction, the partial orders are determined by the reaction stoichiometry. Statement-I : Catalyst can increase that rate constant to a large extent. Statement-II : By using suitable catalyst, we can significantly increase yield. Statement-I : Product can form only when the required orientation and energy conditions are met. Statement-II : All collisions between reactants yield the desired product. 8. 9. 10. 11. 3 CHEMICAL KINETICS AND NUCLEAR CHEMISTRY EXERCISE - 2 : PART # I 1. B, C 2. 10. B, C B, C 3. 11. A, B, C, D A, B 4. A, B, D 5. A, B, D 6. A, B, C 7. 12. B, C, D 13. A, B, C 14. B, C 15. A, D C, D 8. B, C 9. A, B, C 16. D 17. C 18. A 19. D 20. A 21. D 22. B 23. B 24. D 25. B 26. D 27. B 28. B 29. A 30. C 31. A 32. A 33. D 34. A 35. C 36. A 37. C 38. C 39. B 40. C 41. A 42. C 43. B 44. A 45. C 46. A 47. B 48. A 49. A 50. D 51. D 52. C 53. B 54. D 55. B 56. D 57. B 58. B 59. A 60. B 61. C 62. C 63. C 64. C 65. D 66. B, D 70. A, B, C 71. B, D 72. A, B 73. A, B, C 74. A, B, C, D 67. A, B, C 68. A, D 75. A, D PART # II 1. B 2. D 3. C 4. B 5. A 6. A 7. A 8. A 9. B 10. C 11. C 69. A, B