Chemical Kinetics and Nuclear Chemistry Mock Test

CHEMICAL KINETICS AND NUCLEAR CHEMISTRY MOCK TEST SECTION - I : STRAIGHT OBJECTIVE TYPE 1.  2NO2 + N2O5  1 O 2 2 When N2O5 decompose, its t1/2 does not change with its changing pressure during the reaction. So which one is the correct representation for ''pressure of NO2'' vs ''time'' during the reaction when initial PN2O5 is equals to P0. P0 PNO PNO 2 2 (A) (B) P0 time time 2P0 (C) 2P0 PNO (D) 2 PNO 2 time 2. time Choose the correct option : (a) Antineutrino can be detected during the emission of : (i) -rays (ii) -particles (iii) Protons (IV) X-rays (b) Which has magic number of neutrons : 27 (i) 13 Al 209 (ii) 83 Bi 210 84 Po (c)   206 4 82 Pb  2 (i) II A (iii) 238 92 U (iv) 56 26 Fe He . In this reaction predict the position of group of Po when lead is in the IVB group : (ii) IV B (iii) VIB (iv) VIA (d) Which of the following can show -radiations : (i) 81 37 Rb (iii) 11 7 N  0 1e  81 Kr  .... 36  24 He  17 O  ..... 8 11 11 B  ..... (ii) 6 C  5 (iv) 60m 27 Co  60 Co  .... 27 (A) a – i, b – ii, c – iii, d – iv (B) a – ii, b – i, c – iii, d – iii (C) a – ii, b – ii, c – iii, d – iv (D) a – ii, b – ii, c – iv, d – iv 3. In a hypothetical reaction A (aq) ƒ 2 B(aq) + C (aq) (1st order decomposition) 'A' is optically active (dextro-rotatory) 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 of plane polarised light or by measuring volume of H2O2 consumed in titration. In an experiment the optical rotation was found to be  = 30° at t 20 min and  = 15° at t = 50 min. from start of the reaction. If the progress would have been monitored by titration method, volume of H2O2 consumed at t = 30 min. (from start) is 3d ml then volume of H2O2 consumed at t = 90 min will be : (A) 60 ml (B) 45 ml (C) 52.5 ml (D) 90 ml 4. The elementary reaction A + B  products has k = 2 × 10–5 M–1 S–1 at a temperature of 27°C. Several experimental runs are carried out using stoichiometric proportion. The reaction has a temperature coefficient value of 2.0. At what temperature of should the reaction be carried out if inspite of halving the concentrations, the rate of reaction is l n6 = 2.585). l n2 (C) 57°C desired to be 50% higher than a previous run. (Given (A) 47°C (B) 53°C (D) 37°C 3B 5. A k1 2K1 1.5 4C K 1 5D All reactions are of 1st order At time t = t1 (t1 > 0) [B]   . Therefore at [C] t = t2 (where t2  t1) time [C]   which of the following is correct. [D] (A)  >  6. (B)  (C)  = 0.4 (D)  = 0.4 A graph between log t1/2 and log a (abscissa) a being the initial concentration of A in the reaction For reaction A  Product, is the rate law is log t1/2 Slope = –1 log a (A) 7. d[A] K dt (B) d[A]  K[A] dt (C) d[A]  K[A]2 dt (D) d[A]  K[A]3 dt For the reaction R – X + OH–  ROH + X– The rate is given of Rate = 5.0 × 10–5 [R – X] [OH–] + 0.20 × 10–5 [R – X] what percentage of R – X Reaction by SN2 mechanism when [OH–] = 1.0 × 10–2 M (A) 96.1 % (B) 3.9% (C) 80% (D) 20 % CHEMICAL KINETICS AND NUCLEAR CHEMISTRY 8. For a reaction the variation of the constant with temperature is given by the equation (In 3)t (t 0°C) 10 The value of the temperature coefficient of the reaction is (A) 0.1 (B) 1.0 (C) 10 In kt = In k0 + 9. (D) 3 The plot of log (V – V) versus t (where V is the volume of nitrogen collected under constant temperature and pressure conditions) for the decomposition of C6H5N2Cl is given at 50°C with an amount of C6H5N2Cl equivalent to 58.3 cc N2. The rate constant for the reaction in hr–1 expressing your answer in a single significant digit is (A) 1 (B) 2 (C) 4 (D) 8 10. The high temperature ( 1200 K) decomposition of CH3COOH(g) occurs as follows as per simultaneous 1st order reactions. CH3COOH k1 CH4 + CO2 k2 CH3COOH CH2CO + H2O What would be the % of CH4 by mole in the product mixture (excluding CH3COOH)? 50k1 (A) (k  k ) 1 2 100k1 (B) (k  k ) 1 2 200k1 (C) (k  k ) 1 2 (D) it depends on time 11. The inversion of cane sugar proceeds with half life of 600 minute at pH = 5 for any concentration of sugar, However at pH = 6, the half life changes to 60 minute, The rate 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]2 [H+]1 (D) r = k[sugar]0[H+]–1 12. A hypothetical reaction : A2 + B2   2AB Follows mechanism as given below : kc A2 A + B2 A+ B A + A .......... (fast) k1 k1 k2 (Kc – is equilibrium constant) AB + B .......... (slow) (k1 – rate constant) AB ............ (fast) (k2, k1 – are rate constant) The order of overall reaction is : (A) 2.5 (B) 1 13. For the system A k1 k–1 B, A (C) 3/2 k2 (D) Zero C which was started with only A the equilibrium concentration [A]eq is k–2 correctly related to [B]eq and [C]eq as : (A) k 1[B]eq  k 2 [C]eq (k1  k 2 ) (B) k 1[B]eq  k 2 [C]eq (k1  k 2 ) (C) k 1[B]eq  k 2 [C]eq (k1  k 2 ) (D) k 1[B]eq  k 2 [C]eq (k1  k 2 ) 14. A certain reaction A  B follows the given concentration (Molarity) – time graph. Which of the following statement is true? 0.5 0.4 0.3 [A] 0.2 0.1 0 20 40 60 80 100 Time (sec) (A) The reaction is second order with respect to A (B) The rate for this reaction at 40 second will be approximately 3.5 × 10–3 M s–1 (C) The rate for this reaction at 80 second will be 1.75 × 10–3 M s–1 (D) The [B] will be 0.25 M at t = 60 second SECTION - II : MULTIPLE CORRECT ANSWER TYPE 1 1 k1 (time ) k 2 (time ) For the consecutive reaction A   B   C following curves were obtained depending upon the relative values of k1 & k2 C B Conc. C Conc. 15. A B Time Figure 2 Time Figure 1 Now which of the following is the correct match (A) figure 1 – (k1 < k2) (C) figure 2 – (k1 >> k2) 16. A (B) figure 2 – (k1 < k2) (C) figure 1 – (k1 >> k2)  2NOBr follows the mechanism : The reaction 2NO + Br2  (A) NO + Br2 Fast NOBr2 Slow (B) NOBr2 + NO 2NOBr Which of the following is/are true regarding this : (A) The order of the reaction with respect to NO is two. (B) The molecularity of the steps (A) and (B) are two each. (C) The molecularity of the overall reaction is three. (D) The overall order of the reaction is three 17. The polarimeter readings in an experiment to measure the rate of inversion of cane sugar (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, loge 10 = 2.3 (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 dextroroatory (D) The angle would be 7.5° at half time CHEMICAL KINETICS AND NUCLEAR CHEMISTRY 18. Identify the true statement(s) (A) A catalyst is chemically unchanged at the end of a reaction (B) A catalyst may appear in the kinetic rate equation of the reaction (C) A catalyst will not affect the composition of an equilibrium mixture (D) A catalyst cannot cause a non-spontaneous (G > 0) reaction to proceed 19. Rate of a chemical reaction 2A (g)   B(g) is defined as 1 dn B V dt Where nB = number of moles of B formed and CB = concentration of B then which of the following relation is correct ‘rB’ = (A) rB = dCB CB dV  (If volume V is not constant) dt V dt (B) rB = dCB (If volume V is constant) dt (C) rB = dCB CB dV  ( If volume V is not constant) dt V dt (D) rB = dCB CB dV  (If volume V is not constant) dt V dt SECTION - III : ASSERTION AND REASON TYPE 20. Statement -1 : If the order reaction is zero than degree of dissociation will be independent upon initial concentration. Statement - 2 :The degree of dissociation of Zero order reaction is given by   kt c0 (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. 21.  C (rate = K[A]1[B]0, the half life time of reaction is only defined when conc of A Statement-1 : For A + 2B  and B are in stoichiometric ratio Statement - 2 : 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. (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 22. Statement-1 : Many reactions occurring on solid surface are zero order reactions. Au Statement-2 : N2O(g) N2(g) + 1/2O2 ; rate = k[N2O]0 = k = constant is a zero order reaction. (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 SECTION - IV : COMPREHENSION TYPE Read the following comprehensions carefully and answer the questions. Comprehension # 1 Consider the inter conversion of nitrosotriacetoamine into nitrogen phorone and water. O O C C (aq) ƒ N2(g) + H2O () + N NO (aq) – 20 kJ/mol H3C C C CH3 CH3 CH3 The reaction is 1st order in each direction, with an equilibrium constant of 104, the activation energy for the forward reaction is 57.45 kJ/mol. Assuming arrhenius pre exponential factor of 1012 s–1. 23. What is the expected forward constant at 300K, if we initiate this reaction starting with only reactant (A) 102 24. (B) 106 (D) 104 If the change in entropy of the reaction is 0.07 KJ. K–1 mol–1 at 1 atm pressure. Calculate up to which temperature the reaction would not be spontaneous. (For forward reaction) (A) T < 258.7 K 25. (C) 108 (B) T > 250 K (C) T < 340.2 K (D) T > 200K (C) 24.6 ×104 atm (D) 2.82×102 atm–1 Calculate Kp of the reaction at 300 K (A) 2.4 × 104 atm–1 (B) 104 atm Comprehension # 2 Reactions Tending To Equilibrium Not all chemical reactions proceed to a stage at which the concentrations of the reactants become vanishingly small. Here we consider the kinetics of such reactions. Let a reaction be represented in general terms by the scheme A k1 k–1 B where k1 and k–1 represent the rate constant for the forwards and reverse reactions, respectively. The equilibrium constant for this reaction may be written as K = [B] / [A] = k1/k–1 ...(1) where the subscript  refers to a time t, sufficiently long to establish equilibrium at the given temperature. The initial concentration of species A is [A]0, and that of B is [B]0. After a time t, let the concentration of species A be [A]t and that of B be [B]t. The total rate of change of [A]t is given by CHEMICAL KINETICS AND NUCLEAR CHEMISTRY d[A]t/dt = – k1[A]t + k–1[B]t If, as is usual, [B]0 is initially zero, it follows from a mass balance that at any time t, [B]t = [A]0 – [A]t , where d[A]t/dt = – k1[A]t + k–1([A]0 – [A]t)   k 1 [A]0  d[A]t / dt = – (k1 + k–1)  [A]t  k  k 1 1   or ...(2) Now, from (1) we have [B] / [A] = ([A]0 – [A]) / [A]= k1/ k–1, or [A] = [A]0 k–1 / (k1 + k–1) Introducing this result into (2), we obtain d[A]t/dt = – (k1 + k–1) ([A]t – [A]) Integrating In(A)t – [A]) = – (k1 + k–1) t + f For t = 0, f = In ([A]0 – [A]) ; hence,  [A]t  [A]  In   = – (k1 + k–1) t  [A]0  [A]  26. For the reaction A kf kb C (having both 1st order reactions), the concentration as a function of time are given for a certain experimental run along with a tangent to the graph at the origin. The ratio of the magnitude of the slopes of the graph of [A] and [C] at the origin would be 1.0 0.8 [A] Conc. 0.6 (M) 0.4 [C] 0.2 0 (A) 1 27. 28. (B) 5 3 4 10 15 Time (min.) (C) 20 4 3 (D) None of these At what time is the rate of change of concentration of A equal to rate of change of concentration of C in magnitude? (A) t = 0 only (B) t  equilibrium only (C) t = 10 only (D) at all times What is the KC of the reaction 2C ƒ (A) 2 3 (B) 3 2 2A ? (C) 1 (D) None of these Comprehension # 5 For first order parallel path reaction A k1 k2 B C t [A]0 k1 [1  e (k1  k 2 ) ] k1  k 2 [B]t = [A]t = [A]0 e–(k1 + k2)t. ; [A]0 k 2 [C]t = k  k [1 – e–(k1+k2)t] 1 2 two calculate the rate constant of first order reaction. 1 [A]0 k = t ln [A] . t B –1 –2 29. A k1 0 =1 k2 =2 se c ×1 0 –2 s ec – 1 2B If initial concentration of ‘A’ is 1M, the conc. of ‘A’ after 33.33 second is equal to – (A) 30. 1 e (B)  B+C A  time 2 e (C) 1 e2 (D) none of these is a first order reaction t  mole of reagent a b If A, B and C react with reagent and have ‘n’ factor in the ratio of 1 : 2 : 3 with reactant then the value of ‘k’ is – (A) k = (B) k = 1 4b ln t ba (C) k = 4b 1 ln 5(b  a) t (D) k =  nC A  (C) (A) Conc. 31. 1 b ln t ba 40 60 120 Time If half life of (A) is 20 sec. then 1 is n (A) 7 1 7 (B) (C) 7 8 (D) 1 8 4a 1 ln 5(b  a) t CHEMICAL KINETICS AND NUCLEAR CHEMISTRY SECTION - V : MATRIX - MATCH TYPE 32. Match Matrix  Alcohol + Salt (A) Ester + NaOH  (p) k = Ae–Ea/RT conc.  2H2O + O2 (B) 2H2O2  (q) t rate (C) 2NH3 Au N2 + 3H2  Glucose + Fructose (D) Sucrose + H2O  33. (r) time (s) t1/2 = x min at a constant temperature any time of the reaction.  C in column = II the graphs given can be from any of these four types. For A + B  (A)  dA Vs time (x axis) dt (B) t1/2 Vs initial conc. (x axis)  C0  C t   Vs time (x axis) (B)  C (D) Conc. Vs time (x axis) t   Match the graphs in Column - II for the given order of reactions in Column - I. Column - I Column - II (A) Ist order (p) (B) Zero order (q) (C) Second order (r) (D) Pseudo first order (s) 34. Match the following : Column - I (A) Constant boiling mixture (B) Kf Column - II (p) 2nd order reaction (q) Azeotrope (C) t3 t  2 1 (A  B, single run) 4 2 (r) 1st order reaction (D) t1 st t1 nd t1 rd (1 ): (2 ): (3 ) : : 1 : 2 : 4 (For the same run 2 2 2 (s) Solvent SECTION - VI : SUBJECTIVE TYPE 35. The rate constant of 1st order decomposition reaction of an organic compound (X) into another organic compound Y and an inorganic compound (gas) (Z) may be given by : log10k (sec–1) = 12.37 – 1.25  10 4 T(in kelvin) Find the activation energy and rate constant at 750 K and indicate your answer as the product of Ea (in Joules mol–1) and rate constant (in unit sec–1) and fill the answer as a nearest whole number. Take R = 8.3 JK–1mol–1, {antilog (–4.3) = 5 × 10–5} 36.  production, the following data is given for a particular run. For the reaction A  time (min.) : 0 5 15 35 1 (M 1 ) : [A] 1 2 4 8 Determine the order of the reaction. 37. Surface catalysed reactions that are inhibited by the products obey the rate equation (in same cases) dx K(a  x)  where a is the initial concentration of the reaction and K and b are constants. Intergrate this dt 1  bx  B. equation. Derive an expression for t1/2. x is the concentration of products an any time t and the reaction is A  38. (A) (B) 39. The decomposition of HI to yield H2 and I2 at 508°C has a half-life of 135 minutes when PHI = 0.1 atm (initial) which comes down to one-tenth of that value when the initial pressure is 1 ati. Calculate the rate constant. The viscosity of water changes by about 2% per degree at 25°C. What is the activation energy for the process? Decomposition of H2O2 (Ist order) 1 O (g) 2 2 Can be monitored by titration method by pressure measurement. It Time (min) t=0 30  Pressure (mm of hg) 200 375 400 While when progress of same reaction was monitored by titration method, the volume of titrant consumed after 10 min was found to be 20 ml. Then complete the following table. Time (min) t=0 10 30 Pressure (mm of Hg) 200 (B) (C) Volume of KMnO4 (A) 20 (D) consumed (ml)  H2O() + H2O2(aq)  MOCK TEST 1. C 2. C 14. B 15. B, C 24. A 25. C 3. C 4. B 5. C 6. C 7. D 8. 16. A, B, D 17. A, B, D 18. A, B, C, D 26. A 27. D 28. D 29. A 32. A  p, r ; B  p, r, s ; C  q ; D  p, r, s 34. A  q, B  s, C  r, D  p 30. C D 9. C 10. A 19. A,B 20. D 11. B 21. C 12. C 22. B 31. A 33. A  p, q ; B  p, r, s ; C  s ; D  p, q 13. A 23. B