Chemical Kinetics and Nuclear Chemistry Exercise # 1

CHEMICAL KINETICS AND NUCLEAR CHEMISTRY Exercise # 1 1. [Single Correct Choice Type Questions] aA + bB  Product, dx/dt = k [A]a [B]b . If concentration of A is doubled, rate is four times. If concentration of B is made four times, rate is doubled. 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 2. For the reaction, 2NO(g) + 2H2(g)  N2(g) + 2H2O(g) the rate expression can be written in the following ways: {d [N2] / dt} = k1 [NO][H2] ; {d[H2O] / dt} = k[NO][H2] ; {– d[NO] / dt} = k1 [NO] [H2] ; {–d[H2] / dt} = k1 [NO][H2] The relationship between k, k1 , k1 and k1. is : (A) k = k1 = k1 = k1 (B) k = 2k1 = k1 = k1 (C) k = 2k1 = k1 = k1 (D) k = k1 = k1 = 2 k1 3. Rate of formation of product at t = 20 seconds is (A) 0.5 MS–1 4. (B) 1 M S-1 (D) 2MS–1 xA  yB In the following reaction :  d[ A ]  (C) 1.5 M S–1  d[B]  log  dt  = log  dt  + 0.3     where –ve sign indicates rate of disappearance of the reactant. Thus, x : y is : (A) 1 : 2 (B) 2 : 1 (C) 3 : 1 (D) 3 : 10 5. Rate of formation of SO3 in the following reaction 2SO2 + O2  2SO3 is 100 g min–1. 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 6. For a reaction pA + qB  products, the rate law expression is r = k[A]1 [B]m , then : (A) (p+1) < (1+ m) (B) (p + q) > (1 + m) (C) (p + q) may or may not be equal to (1+ m) (D) (p + q) = (1 + m) 7. If rate constant is numerically the same for the three reactions of first, second and third order respectively. Assume all the reactions of the kind A  products. 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 8. For the irreversible process, A + B  products, the rate is first–order w.r.t. A and second–order w.r.t. B. If 1.0 mol 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 (B) 1.0 × 10–2 mol L–1 s–1 (C) 2.50 × 10–3 mol L–1 s–1 (D) 2.0 × 10–2 mol L–1 s–1 9. What will be the order of reaction and rate constant for a chemical change having log t50% vs log concentration of (A) curves as : log t50% 45° log a (A) 0, 1/2 10. (B) 1, 1 For a reaction 2A + B  product, rate law is – (C) 2, 2 (D) 3, 1 d[A] 1 = k[A]. At a time when t = , concentration of the reactant is dt k (C0 = initial concentration) (A) C0 e (B) C0e (C) 1 (D) C 0 C0 e 2 11. 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 12. The rate constant of the reaction A  2B is 1.0 × 10–3 mol lit–1 min–1, if the initial concentration of A is 1.0 mole lit– 1 what would be the concentration of B after 100 minutes. (A) 0.1 mol lit–1 (B) 0.2 mol lit–1 (C) 0.9 mol lit–1 (D) 1.8 mol lit–1 13. A drop of solution (volume 0.05 mL) contains 3.0 × 10–6 moles of H+. If the rate constant of disappearance of H+ is 1.0 × 107 mole litre–1 sec–1. 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 14. Graph between concentration of the product and time of the reaction A  B is of the type X I Hence graph time between – d[A]/dt and time will be of the type : (–d[A]/dt) (–d[A]/dt) (A) (B) Time Time (–d[A]/dt) (–d[A]/dt) (C) (D) Time 15. A reaction follows the given concentration (M)–time graph. The rate for this reaction at 20 seconds will be : (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 Time 0.5 0.4 0.3 0.2 0.1 0 20 40 60 80 100 Time/second 16. In a first order reaction the reacting substance has half-life period of ten minutes. 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 CHEMICAL KINETICS AND NUCLEAR CHEMISTRY 17. A certain reaction A + B C, the first order with respect to each reactant k = 10–3. Determine the final concentration of A after 100 s, if the initial concentration of A was 0.1 M and that of B was 0.2 M. (A) 0.098 M (B) 0.088 M (C) 0.078 M (D) 0.068 M 18. For the reaction 2NO2  N2O2 + O2, rate expression is as follows d[NO2 ] =K [NO2]n , where K = 3 × 10–3 mol–1 L sec–1. If the rate of formation of oxygen is 1.5 × 10–4 mol L–1 sec–1, then dt the molar concentration of NO2 in mole L–1 is (A) 1.5 × 10–4 (B) 0.0151 (C) 0.214 (D) 0.316 – 19. At the point of intersection of the two curves shown, the conc. of B is given by.......for, A  nB : A0 Conc. nA 0 (A) 2 A0 (B) n–1 time (C) nA 0 n 1  n – 1  A0 (D)   n  1 20. A reaction, which is second order, has a rate constant of 0.002 L mol–1 s–1. If the initial conc. of the reactant is 0.2 M. how long will it take for the concentration to become 0.0400 M ? (A) 1000 s (B) 400 s (C) 200 s (D) 10, 000 s 21. 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. 22. A graph between log t 1/ 2 and log a (abscissa) a being the initial concentration of A in the reaction For reaction A  Product, is the rate law is : (A) – d[A] =K dt – d[A] (B) = K [A] dt (C) – d[A] = K [A]2 dt (D) – d[A] = K [A]3 dt log t 1/2 Slope = – 1 log a t3 / 4 for third order reaction is equal to 5 : 1 t1 / 2 23. Statement-1 : The ratio of 24. Statement-2 : t1/2 & t3/4 of a reaction depends only on order, not on the concentration of reactant. (A) Statement 1 and statement 2 are correct and statement 2 is the correct explanation of statement 1. (B) Statement 1 and statement 2 are correct but the statement 2 is not correct explanation of statement 1. (C) statement 1 is correct but statement 2 is false. (D) statement 1 is false but statement 2 is correct. Match the graphical study with the order of the reactions : I : Rate II : Half life I A C 25. 26. time [initial conc] time (A) (C) A : First B : Second C : Zero III : (a – x)–1 II B B III C A (B) (D) The data for the reaction A + B  C is Exp. [A]0 [B]0 1 0.012 0.035 2 0.024 0.035 3 0.012 0.070 4 0.024 0.070 (A) r = k [B]3 (B) r = k [A]3 I B C II C A III A B initial rate 0.10 0.80 0.10 0.80 (C) r = k [A] [B]4 (D) r = k [A]2 [B]2 . In three different reactions, involving a single reactant in each case, a plot of rate of the reaction on the yaxis, versus concentration of the reactant on the xaxis, yields three different curves shown below. dC dC (i) – dt dC (ii) – dt C (iii) – dt C C What are the possible orders of the reactions (i), (ii), (iii). (A) 1, 2, 3 27. (B) 2, 1, 1/2 (C) 0, 1, 2 (D) 0, 1, 1/2 Which integrated equation is correct for the following Ist order reaction started with only A(g) in a closed rigid vessel. A (g)  B (g) + C (g) + D (g) Pi = initial pressure ; Pt = total pressure at time t (A) K =  Pi  2.303 log10  P  t  t (B) K =  Pt  2.303 log10  P  t  i (C) K =  2Pi  2.303 log10  3P – P  t t   i (D) K =  3Pi  2.303 log10  2P – 3P  t t   i CHEMICAL KINETICS AND NUCLEAR CHEMISTRY 28. The decompostion NH3 gas on a heated tungsten surface gave the following results : Initial pressure (mm) 65 105 y 185 Half-life (sec) 290 x 670 820 Calculate approximately the values of x and y. (A) x = 410 sec (B) x = 467 sec (C) x = 490 sec (D) x = 430 sec y = 115 mm y = 150 mm y = 120 mm y = 105 mm 29. In the reaction NH4NO2 (aq.)  N2 (g) + 2 H2O (l) the volume of N2 after 20 min and after a long time is 40 ml and 70 ml respectively. The value of rate constant is : (A) (1/20) In (7/4) min–1 (B) (2.303 /1200) log (7/3) sec–1 –1 (C) (1/20) log (7/3) min (D) (2.303 / 20) log (11/7) min–1 30. At 373 K, a gaseous reaction A  2B + C is found to be of first order. Starting with pure A, the total pressure at the end of 10 min. was 176 mm and after a long time when A was completely dissociated, it was 270 mm. The pressure of A at the end of 10 minutes was : (A) 94 mm (B) 47 mm (C) 43 mm (D) 90 mm 31. The reaction A(s)  2 B(g) + C(g) is first order. The pressure after 20 min. and after very long time are 150 mm Hg and 225 mm Hg. The value of rate constant and pressure after 40 min. are : (A) 0.05 In 1.5 min–1 ,200 mm (B) 0.5 ln 2 min–1 ,300 mm –1 (C) 0.05 In 3 min , 300 mm (D) 0.05 In 3 min–1 , 200 mm 32. Half life of reaction : H2O2(aq)  H2O(l) + 1 O (g) is independent of initial concentration of H2O2 volume of 2 2 O2 gas after 20 minute is 5L at 1 atm and 27°C and after completion of reaction 50L. The rate constant is (A) 33. 1 log 10 min –1 20 (B) 2.303 log 10min –1 20 (C) 2.303 50 log min –1 20 45 (D) 2.303 45 log min –1 20 50 The decomposition of N2O5 in chloroform was followed by measuring the volume of O2 gas evolved : 2N2O5 (CCI4)  2N2O4 (CCI4) + O 2 (g). The maximum volume of O2 gas obtained was 100 cm3. In 500 minutes, 90 cm3 of O2 were evolved. The first order rate constant (in min–1) for the disappearance of N2O5 is : (A) 2.303 500 (B) 2.303 100 log 500 90 (C) 2.303 90 log 500 100 (D) 100 10×500 34. For a reaction A  B+ C, it was found that at the end of 10 minutes from the start the total optical rotation of the system was 50° and when the reaction is complete, it was 100°. Assuming that only B and C are optically active and dextrorotatory. calculate the rate constant of this first order reaction. (A) 0.693 min–1 (B) 0.0693 sec–1 (C) 0.0693 min–1 (D) 0.00693 sec–1 35. In the above question solution is optically inactive when : (A) rt = a (B) rt = 0 (C) rt = x (D) rt = (a + x) 36. The following data were obtained in an experiment on inversion of cane sugar (a first order kinetics) Time (min) 0 10 After a long time Total angle of rotation (degree) +40 +15 – 10 The rate constant (in second–1) is [ln 2 = 0.693] (A) 0.0693 (B) 1.15  10–3 (C) 0.693 (D) 1.15  10–2 37. The rate constant for two parallel reactions were found to be 1.0 × 10–2dm3 mol-1 s–1 and 3.0 × 10–2 dm3 mol–1s–1. If the corresponding energies of activation of the parallel reactions are 60.0 kJ mol–1 and 70.0 kJ mol–1 respectively, what is the apparent overall energy of activation ? (A) 130.0 kJ mol–1 (B) 67.5 kJ mol–1 (C) 100.0 kJ mol–1 (D) 65.0 kJ mol–1 38. The substance undergoes first order decomposition. The decomposition follows two parallel first order reactions as : K1 B K2 C A K1 = 1.26 × 10–4 sec–1 and K2 = 3.8 × 10–5 sec–1 The percentage distribution of B and C (A) 80% B and 20% C (C) 90% B and 10% C 39. (B) 76.83% B and 23.17%C (D) 60% B and 40% C For the follwing parallel chain reaction if the sum of the concentration of B and C at any time is 2M then what will be [B] t and [C] t respectively? (A) 40. (B) k1 For the system A 3 5 M, M 4 4 (C) B,A (C) (D) 8 18 M, M 13 13 k2 k –1 [B]eq + k –2 [C]eq (B) (k1 + k 2 ) k –1 [B]eq + k –2 [C]eq (D) (k1 – k 2 ) k –1 [B]eq – k –2 [C]eq (k1 – k 2 ) k –1 [B]eq – k –2 [C]eq (k1 + k 2 ) Consider the elementary reaction sequence shown in figure. Which of the following equations are correct? k1 A B k4 D 42. 4 6 M, M 5 5 C which was started with only A the equilibrium concentration [A]eq is k –1 k–2 correctly related to [B]eq and [C]eq as : (A) 41. 11 13 M M 12 12 k2 k3 C (A) d[A] = k1[A] + k4[D] dt (B) d[C] = k2[B]  k3[C] dt (C) d[D] = k4[D] + k3[D] dt (D) Nothing can be said about order of reactions in this problem At a given temperature, k1 = k2 for the reaction A+B C+D If (A) (C)  dx   dt  = k1 [A] [B] – k2[C] [D] in which set of the concentration reaction ceases?   [A] 0.1 M 0.2 M [B] 0.2 M 0.2 M [C] 0.3 M 0.3 M [D] 0.4 M 0.2 M (B) (D) [A] 0.4 M 0.2 M [B] [C] 0.25 M 0.2 M 0.2 M 0.4 M [D] 0.5 M 0.2 M CHEMICAL KINETICS AND NUCLEAR CHEMISTRY 43. k1 Following reaction can take place in both direction A k2 Graph is given for the forward reaction : and for the backward reaction following data are given [B] Rate 0.01 M 0.02 M 1 × 10–2 Ms–1 2 × 10–2 Ms–1 B, Rate (M s)–1 [A],M Hence, net reaction rate of B is : (A) = k1 [A] – k2 [B] (C) = k1 [A] – k2 44. For the complex (B) = k1 – k2 [B] (D) = k1 – k2 Ag+ + 2NH3 [Ag(NH3)2+]  dx   dt  = 2 × 107 L2 mol–2 s–1 [Ag+] [NH3]2 – 1 × 10–2 s–1 [Ag(NH3)2+]   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 (D) 0.5 × 10–9 L2 mol–2 45. The rate constant K1 of a reaction is found to be double that of rate constant K2 of another reaction. The relationship between corresponding activation energies of the two reactions at same temperature (E 1 and E2) can be represented as : (A) E1 > E2 (B) E1 < E2 (C) E1 = E2 (D) None of these 46. When a graph between log K and 1/T is drawn a straight line is obtained. The point at which line cuts y -axis and x -axis respectively correspond to the temp : (A) 0, Ea / 2.303 R log A (B) , Ea / (R ln A) (C) 0, log A (D) None of these. 47. How much faster would a reaction proceed at 25°C than at 0°C if the activation energy is 65 kJ? 48. (A) 2 times (B) 5 times (C) 11 times (D) 16 times The rate constant, the activation energy and the frequency factor of a chemical reaction at 25°C are 3.0 × 10–4 s–1, 104.4 KJ mol–1 and 6.0 × 1014 s–1 respectively. The value of the rate constant as T   is : (A) 2.0 × 1018 s–1 49. (B) 6.0 × 1014 s–1 (C) infinite (D) 3.6 × 1030 s–1 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. 50. 51. For a reaction A  B, Ea = 10 kJ mol–1 , H = 5 kJ mol–1 . Thus, potential energy profile for this reaction is : (A) (B) (C) (D) For a certain reaction the variation of the rate constant with temperature is given by the equation  ln 3  ln kt = ln k0 +   t  10  (t  0°C) The value of the temperature coefficient of the reaction rate is therefore (A) 4 (B) 3 (C) 10 52. The Ea of a reaction in presence of a catalyst is 4.15 kJ mol–1 and in absence of a catalyst is 8.3 kJ mol–1. What is slope of the plot of lnk vs 1 in the absence of catalyst T (A) + 1 53. (D) 2 (B) –1 (C) + 1000 (D) –1000 For the decomposition of H the following logarithmic plot is shown : [R = 1.98 cal/mol-K] The activation energy of the reaction is about 1 (A) 45600 cal logk (B) 13500 cal 0 –1 (C) 24600 cal –2 (D) 32300 cal –3 1.2 1.3 1.4 1.5 1.6 103/T 54. The reaction of hydrogen, and iodine monochloride is represented by the equation : H2(g) + 2Cl(g)  2HCl(g) + 2(g) This reaction is first–order in H2(g) and also first–order in Cl(g). Which of these proposed mechanism can be consistent with the given information about this reaction ? Mechanism  : H2(g) + 2Cl(g)  2HCl(g) + 2(g) Mechanism  : (A)  only H2(g) + Cl(g) Slow HCl(g) + H(g) HI(g) + Cl(g) fast HCl(g) + I2(g) (B)  only (C) both  and  (D) neither  nor  CHEMICAL KINETICS AND NUCLEAR CHEMISTRY 55. Reaction A + B  D + E take place as K1 A+B 2C ; C+B K2 2D K3 C +A 2E The rate of disapperance of C is given by 56. (A) – d[C] = K2 [B][C] + K3[A][C] – 2K1[A][B] dt (B) – d[C] = K2[B][C] + K3 [E] – K1[C] dt (C) – d[C] = K2[D]O + K3[E] – K1[C] dt (D) – d [C] = 2K1[A][B] – K2 [B][C] – K3[A][C] dt The steady state concentration of the activated molecule [A] in the following sequence of steps is given by: A+A (A) K1 A+A K 2 [A] K1 ; K 1[ A ] (B) K 2 A+A K2 2A (C) K1K2[A] (D) K 1K 2 [A] 57. 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, 3/ 2 (B) 3/2 , 3/2 (C) Not defined, 3/2 (D) 1,1/2 58. The reaction of NO2 (g) and O3 (g) is first-order in NO2 (g) and O3 (g) 2 NO2 (g) + O3 (g)  N2O5 (g) + O2 (g) The reaction can take place by mechanism : I : NO2 + O3 Slow NO3 + O2 NO3 + NO2 fast N2O5 II : O3 ka fast kb O2 + [O] NO2 + O Slow NO3 NO3 + NO2 fast N 2O 5 Select correct mechanism. (A) I only (B) II only 59. (D) None of I and II A hypothetical reaction X2 + Y2  2XY follows the mechanism given below.. X2 X +X X + Y2  XY + Y X + Y  XY [Fast] The order of overall reaction is (A) 2 (B) 1 60. (C) both I and II [Fast] [Slow] (C) 1.5 (D) Zero A radioactive element has a half life of one day. After three days the amount of the element left will be : (A) 1/2 of the original amount (B) 1/4 of the original amount (C) 1/8 of the original amount (D) 1/16 of the original amount 61. The activity per ml of a solution of radioactive substance is x. How much water be added to 200 ml of this solution so that the acitivity falls to x/20 per ml after 4 half-lives ? (A) 100 ml (B) 150 ml (C) 80 ml (D) 50 ml. 62. The  activity in 1g sample of 226Ra (t1/2 = 1600 years) is equal to : (A) 2.19 × 1012 dpm (B) 2.19 × 1014 dpm (C) 2.19 × 1016 dpm 63. Radioactivity is affected by : (A) temperature (C) electric and magnetic field (D) 4.96 × 1012 dpm (B) Pressure (D) none of these 64. Half life period of 10 gm radioactive element is 20 days. Then half life period of 100 gm of this element is : (A) 20 days (B) 200 days (C) 100 days (D) 10 days 65. Determine the average life of U238 having t0.5 = 138.5 day. (A) 138.5 day (B) 199.86 day (C) 238 day (D) None of these A sample of rock from moon contains equal number of atoms of uranium and lead (t1/2 for U = 4.5 × 109 years). The age of the rock would be : (A) 9.0 × 109 years (B) 4.5 × 109 years (C) 13.5 ×109 years (D) 2.25 × 109 years 66. 67. 68. The half-life of 99Tc is 6.0 h. The total residual activity in a patient after receiving an injection containing 99Tc must not be more than 0.01 Ci , after 36.0 hours. What is the maximum activity (in Ci) that the sample injected can have? (A) 0.06 (B) 0.16 (C) 0.32 (D) 0.64 A radioactive isotope X with half - life of 6.93 × 109 years decay to Y which is stable. A sample of rock from the moon was found to contain both the element X and Y in the mole ratio 1 : 7.What is the age of the rock? (A) 2.079 × 1010 years (B) 1.94 × 1010 years (C) 1.33 × 109 years (D) 1010 years 69. The half  life of a radioactive isotope is 3 hours . If the initial mass of the isotope were 256 gm , the mass of it remaining undecayed after 18 hours would be : (A) 4 gm (B) 8 gm (C) 12 gm (D) 16 gm 70. The radioactive decay is first order reaction. Its rate, on increasing 10ºC temperature (A) increases 2-3 times (B) decreases 2-3 times (C) does not change (D) increases 10 times 71. If for a reaction in which A(g) converts to B(g) the reaction carried out at const. V & T results into the following graph. 72. (A) then the reaction must be A(g)  3B(g) and is a first order reaction. (B) then the reaction must be A(g)  3B(g) and is a second order reaction. (C) then the reaction must be A(g)  3B(g) and is a zero order reaction. (D) then the reaction must be A(g)  3B(g) and is a first order reaction. A graph plotted between log t50% vs. log concentration is a straight line. What conclusion can you draw from this graph. (A) n = 1 ; t1/2  a (C) n = 1 ; t1/2 = (0.693 / k) 73. (B) n = 2, t1/2  1/a (D) None of these Concentration of the reactant in first-order is reduced to (A) one natural life-time (C) three natural life-time 1 e2 after : (Natural life = (B) two-natural life-time (D) four natural life-time 1 ) K CHEMICAL KINETICS AND NUCLEAR CHEMISTRY 74. For a chemical reaction A  products, the rate of disappearance of A is given by : K1CA –dCA = 1+ K C At low CA the reaction is of the ..... order with rate constant.....(Assume K1, K2 are lesser than 1) dt 2 A K1 (A) , K 2 75. (C) II, K1/K2 K1 (D) II, K + K 1 2 The reaction A(g) + 2B(g)  C(g) is an elementary reaction. In an experiment involving this reaction, the initial partial pressures of A and B are P A = 0.40 atm and P B = 1.0 atm respectively. When pressure of C becomes 0.3 atm in the reaction the rate of the reaction relative to the initial rate is : (A) 76. (B) I, K1 1 12 (B) 1 50 (C) 1 25 (D) none of these 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 77.  C   0 In a reaction involving one single reactant, the fraction of the reactant consumed may be defined as f =  1 – C   where C0 and C are the concentrations of the reactant at the start and after time, t. For a first order reaction (A) 78. df = k(1  f) dt df = kf dt (C)  df = k(1  f) dt (D) df = kf dt  dx   dx  A  Product and   = k[A]2 . If log   is plotted against log [A], then graph is of the type :  dt   dt  (A) (B) (C) (D) For a first order reaction, nA B whose concentration vs time curve is as shown in the figure. If half life for the reaction is 24 minutes . Find out the value of n. B concentration 79. (B)  A 20 48 72 Time (min.) (A) 1 (B) 2 (C) 3 (D) 4 80. Acid catalysed hydrolysis of ester is first-order reaction and rate constant is given by k= V  V0 2.303 log V  V t  t where V0, Vt and V are the volume of standard NaOH required to neutralise acid present at a given time ; if ester is 50% hydrolysed then : (A) V = Vt (B) V = (Vt – V0) (C) V = 2Vt – V0 (D) V = 2Vt + V0 81. 82. (1st order decomposition) In a hypothetical reaction, A(aq) 2B(aq) + C(aq) 'A' is optically active (dextro-rototory) while 'B' and 'C' are optically inactive but 'B' takes part in a titration reaction (fast reaction) with H2O2. Hence, the progress of reaction can be monitored by measuring rotation of plane polarised light or by measuring volume of H2O2 consumed in titration. In an experiment the optical rotation was found to be  = 40° at t = 20 min and  = 10° at t = 50 min. from start of the reaction. If the progress would have been monitored by titration method, volume of H 2O2 consumed at t = 15 min. (from start) is 40 ml then volume of H2O2 consumed at t = 60 min will be: (A) 60 ml (B) 75 ml (C) 52.5 ml (D) 90 ml Inversion of sucrose (C12H22O11) is first-order reaction and is studied by measuring angle of rotation at different instant of time + C12H22O11 + H2O H C6H12O6 + C6H12O6 Sucrose d Glucose d Fructose l If (r – r0)  a and (r – rt)  (a – x) (where r, rt and r are the angle of rotation at the start, at the time t and at the end of the reaction respectively, then there is 50% inversion when : (A) r0 = 2rt – r (B) r0 = rt – r (C) r0 = rt – 2r (D) r0 = rt + r 83. The inversion of cane sugar proceeds with half life of 50 minute at pH = 5 for any concentration of sugar. However if pH = 6, the half life changes to 500 minute of any concentration of sugar. The rate law expression for the inversion of cane sugar is written as : r = K [sugar]1 [H+]y. Determine the value of y. (A) r = K [sugar]1 [H+]-1 (B) r = K [sugar]1 [H+]2 (C) r = K [sugar]1 [H+]1 (D) r = K [sugar]1 [H+]0. 84. The following data is for the decomposition of ammonium nitrite in aqeous solution. Volume of N2 in cc. 6.25 9.00 11.40 13.65 35.05 The order of the reaction is (A) zero (B) one 85. Time (minutes) 10 15 20 25 infinity (C) minus one (D) minus two Formation of NO2F from NO2 and F2 as per the reaction 2NO2(g) + F2(g)  2NO2F(g) is a second order reaction, first order with respect to NO2 and first order with respect to F2. 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, what will be the total pressure in the vessel after the reaction is complete? (A) 1atm (B) 2 atm (C) 2.5 atm (D) 3 atm CHEMICAL KINETICS AND NUCLEAR CHEMISTRY 86. 87. For a first order reaction A B + 2C + 3D (A is optically inactive and B, C and D are dextrorotary), the optical rotation at time t and  are rt and r respectively, the expression for rate constant is (A) K = rt 1 ln t r  rt (B) k = (C) k = 1 r  rt ln rt t (D) none of these r 1 ln r  rt t For the decomposition of H2O2(aq) it was found that VO2 (t = 15 min.) was 100 mL (at 0ºC and 1 atm) while VO2 (maximum) was 200 mL (at 0ºC and 2 atm). If the same reaction had been followed by the titration method ( c M) ( c M) and if VKMnO 4 (t = 0) had been 40 mL, what would VKMnO 4 (t = 15 min) have been ? (A) 30 mL (B) 25 mL (C) 20 mL (D) 15 mL EXERCISE - 1 1. B 2. B 3. A 4. B 5. D 6. C 7. D 8. A 9. A 10. A 11. B 12. B 13. C 14. C 15. D 16. B 17. A 18. D 19. C 20. D 21. C 22. C 23. C 24. D 25. B 26. B 27. C 28. B 29. B 30. B 31. D 32. C 33. A 34. C 35. B 36. B 37. B 38. B 39. D 40. A 41. B 42. B 43. B 44. B 45. D 46. B 47. C 48. B 49. B 50. B 51. B 52. D 53. A 54. B 55. A 56. B 57. C 58. C 59. C 60. C 61. D 62. A 63. D 64. A 65. B 66. B 67. D 68. A 69. A 70. C 71. C 72. C 73. B 74. B 75. C 76. B 77. A 78. C 79. C 80. C 81. B 82. A 83. A 84. B 85. B 86. B 87. A