Chemical Kinetics and Nuclear Chemistry Solved Examples

SOLVED EXAMPLES Ex.1 For a reaction A k1 (A) – k2 [B] Sol. k2 B C. If the reactions are of 1st order then (B) +k [A] (C) k1[A] – k2[B] d[B] is equal to dt (D) k1[A] + k2[B] Rate of increase in [B] = k1[A] Similarly rate of decrease in [B] = k2[B] d[B] = k1[A] – k2[B] dt Thus, Hence the answer is [C] Ex.2 The half life period t1/2 is independent of initial concentration of reactant when the order of reaction is (A) Negative Sol. (B) 0 (C) 1 (D) Fractional t½ of a reaction of an order n is related to initial concentration by the expression t½  1 n –1 C0 (Here, n = order of reaction) for n = 1, t½ is independent of concentration term. Hence the answer is [C]. Ex.3 For a first order reaction, A B, the rate of reaction at [A] = 0.1 M is 1.0 × 10–1 mol L–1 min–1. The half-life period for the reaction is (A) 42 sec Sol. (B) 21 sec (C) 20 sec (D) 28 sec r = k[A] k= r 10 –1 = =1 [A] 0.1 0.693 0.693 = = 0.693 min = 0.693 × 60  42 sec k 1 Hence, (A) is the correct answer. t 12 = Ex.4 Find the activation energy [kJ/mol] for the reaction, A(g) + B(g)  C(g) + D(g). From the plot given below : 100 C+D 80 60 40 A +B 20 0 (A) 20 Sol. Ea = 100 – 40 = 60 kJ mol–1 Hence the answer is [B]. (B) 60 Reaction course (C) 40 (D) 80 CHEMICAL KINETICS AND NUCLEAR CHEMISTRY Ex.5 In an endothermic reaction, H represents the enthalpy of reaction in kJ/mol, the minimum value for the energy of activation will be. (A) less than H (B) zero (C) more than H (D) equal to H H Sol. Hence, (C) is the correct answer. Ex.6 Progress of reaction Consider the following first order competing reactions A B, C D, the ratio of (A) 0.415 Sol. Ea> H k1 k , if only 25% of A have been reacted whereas 50% of C has been reacted, calculate the ratio of 1 k2 k2 (B) 0.246 (C) 2.06 (D) 0.06 k1 = 100 2.303 log for 25% (A) reacted 75 t1 k2 = 100 2.303 log for 50% (C) reacted 50 t2 k1 t 2 0.1249  k = t × 0.3010 2 1 Since t2 = t1 k1 0.1249  k = 0.3010 = 0.415 2 Hence, (A) is the correct answer. Ex.7 Sol. A catalyst lowers the activation energy of a reaction from 30 kJ mol–1 to 15 kJ mol–1. The temperature at which the uncatalysed reaction will have the same rate as that of the catalysed at 27°C is (A) –123°C (B) 327°C (C) –327°C (D) +23°C E'a Ea 15 20 = = T1 T2  300 T2  T2 = 600 K = 327°C Hence, (B) is the correct answer. Ex.8 –5 –1 SO2Cl2 ‡ˆ ˆˆ ˆ† ˆˆ SO2 + Cl2, is the first order gas reaction with k = 2.2 × 10 sec at 270°C. The percentage of SO2Cl2 decomposed on heating for 50 minutes is (A) 1.118 (B) 0.1118 (C) 18.11 (D) 6.39 Sol. k= 2.303 a a kt log =  log t (a – x) (a – x) 2.303 2.2 ×10 –5 × 50 × 60 a = = 0.0286 2.303 a–x a–x a = 0.936 Hence, = 1.068  a (a – x) x x  1 – = 0.936  = 0.068 = 6.39 % a a Hence, (D) is the correct answer.  log Ex.9 In the decomposition of N2O5, the plot between the reciprocal of concentration of the reactant and the time was found to be linear as shown in figure. Determine the order of reaction. 1 Conc. Time Sol. The reaction is of second order, because for II order, x 1 . t a(a – x) k= or Ex.10 time (t) vs. or x 1 . a(a – x) k 1 graph is linear conc. The activation energies of two reactions are Ea and Ea' with Ea > Ea'. If the temperature of the reacting systems is increased from T1 and T2 predict which alternative is correct. k1' and k2' are rate constants at higher temperature. Assume A being the same for both the reactions. k'1 k'2 (A) k = k 1 2 Sol. t= (B) k1 < k2 and k1' < k2' (C) k1 > k2 and k1' < k2' (D) k1' 2k 2' < k1 k2 (D) 141 54 More is energy of activation lesser is rate constant. k = Ae–Ea/RT k1 < k2 and k1' < k2' Hence, (B) is the correct answer. Ex.11 Fill in the blank : 23 5 92 1 U  10 n  ?  92 36 Kr  3 0 n (A) Sol. 141 56 (B) Ba 139 56 (C) Ba 139 54 Ba Ba 92 + 0 = Z + 36 + 0  Z = 56 235 + 1 = A + 92 + 3  A = 141 Missing nucleide is 141 56 Ba Hence, (A) is the correct answer. Ex.12 In the nuclear reaction, 234 92 (A) 1, 3 Sol. 234 92 U   U   (B) 2, 3 222 86 Rn + x 24 α + y0–1β 234 = 222 + 4x  x = 3 92 = 86 + 2x – y or y = 0 Hence, (C) is the correct answer. 222 86 Rn , the number of  and -particles lost would be : (C) 3, 0 (D) 0, 3 CHEMICAL KINETICS AND NUCLEAR CHEMISTRY Ex.13 Sol. Ex.14 The inversion of cane sugar proceeds with half-life of 250 minutes at pH = 4 for any concentration of sugar. However, if pH = 5, the half-life changes to 25 minutes. The rate law expression for the sugar inversion can be written as (A) r = k [sugar]2 [H]6 (B) r = k [sugar]1 [H+]0 (C) r = k [sugar]1 [H+]1 (D) r = k [sugar]0 [H+]1 At pH = 4, the half-life is 250 minutes for all concentrations of sugar that is t1/2 [sugar]0. The reaction is first order with respect to sugar. Let, rate = k[sugar]1 [H+]x For [H+] t1/2  [H+]1–x  250  (10–4)(1–x) ......(1) At pH = 5, the half life is 25 minutes so 50  (10–5)1–x ......(2) (1 – x) 10 = 10  (1 – x) = 1  x = 0 Therefore, rate = k [sugar]1 [H+]0 Hence, (B) is the correct answer. In a hypothetical reaction x  y, the activation energies for the forward and backward reactions are 13 and 8 kJ/mol respectively. The potential energy of x is 10 kJ/mol, then (A) the threshold energy of the reaction is 23 kJ/mol (B) potential energy of y is 15 kJ (C) heat of reaction is 5 kJ (D) the reaction is endothermic Ef Energy Sol. 8kJ 13kJ (z) 10kJ (y) (x) Hence, (A), (B), (C) and (D) are correct answer. Sol. The reactions, CH3COOC2H5 + NaOH  CH3COONa + C2H5OH, is (A) biomolecular reaction (B) second order reaction (C) third order reaction (A) and (B) Ex.16 The rate constant of a reaction is given by k = 3.2 × 1010 e Ex.15 – (D) none of the above 2700 2.303R It means that Sol. (A) log k vs 1 (B) log k vs 1 –2700 2.303R T will be straight line with slope = T will be a straight line with intercept on log k axis = log 3.2 × 1010 (C) the number of effective collisions are 3.2 × 1010 cm–3 sec–1 (D) half-life of the reaction increases with increase in temperature (A) and (B) are correct, (C) is wrong because frequency factor gives total number of collisions and not the effective collision cm–3 sec–1, (D) is wrong because half-life of the reaction decreases with increase in temperature (as reaction becomes faster). Hence, (A) and (B) are correct answer. Ex.17 Two substances x and y are present such that [x0] = 2[y0] and half-life of x is 6 minutes and that of y is 18 minutes. If they start decaying at the same time following first order kinetics how much time later will the concentration of both of them would be same? (A) 15 minutes (B) 9 minutes (C) 5 minutes (D) 12 minutes n Sol. 1 1 Amount of x left in n1 half-lives =   [x 0 ] 2 n 1 2 Amount of y left in n2 half-lives =   [y0 ] 2 At the end, [x 0 ] [x 0 ] = n2 2n1 2  2 1 = n 2 , {[x0] = 2[y0]} n1 2 2  2n = 2  2 n1 n 2 = (2)1 2n  n1 – n2 = 1 n2 = (n1 – 1) ...(1) Also, t = n1 × t1/2(x) ; t = n2 × t1/2(y) (Let, concentration of both become equal after time t)  n1 × t1/2(x) n 2 × t1/ 2(y) =1 n1 n1 × 6 =3 =1 n2 n 2 ×18 ...(2) From Eqs. (1) and (2), we get n2 = 0.5, n' = x1.5 t = 0.5 × 18 = 9 minutes Hence (B) is the correct answer. Ex.18 Consider a gaseous reaction, the rate of which is given by k[x] [y], the volume of the reaction vessel containing these gases is suddenly increased to 3rd of the initial volume. The rate of reaction relative to the original rate would be (A) 9/1 Sol. (B) 1/9 (C) 6/1 By increasing volume to 3rd the concentration will become (D) 1/6 1 1 times, hence rate times. 3 9 Hence, (B) is the correct answer. Ex.19 The rate of reaction is doubled for every 10° rise in temperature. The increase in reaction rate as a result of temperature rise from 10° to 100° is. (A) 112 Sol. (B) 512 (C) 400 Hence, (B) is the correct answer. Ex20. Sol. (D) 614 Increase in steps of 10° has been made 9 times. Hence, rate of reaction should increase 29 times i.e., 512 times. Van't Hoff equation is (A) (d/dT) lnK = (–E/RT2) (B) (d/dT) lnK = + (E/RT2) (C) (d/dT) lnK = –(E/RT) (D) K = Ae–Ea/RT (B) and (D). CHEMICAL KINETICS AND NUCLEAR CHEMISTRY Ex.21 Sol. Ex.22 Sol. The rate of chemical reaction (except zero order) (A) decreases from moment to moment (C) depends upon the order of reaction (A) and (C) (B) remains constant throughout (D) none of the above The rate constant for the reaction, 2N2O5  4NO2 + O2 , is 4.0 × 10 –5 sec–1. If the rate of reaction is 4.80 × 10–5 molL–1 sec–1 , the concentration of N2O5 (molL–1) is : (A) 1.4 (B) 1.2 (C) 0.04 (D) 0.8 r = k [N2O5] r 4.80 ×10 –5 = = 1.2 molL–1 k 4.0 ×10 –5 Hence, (B) is the correct answer.  [N2O5] = Ex.23 Sol. Ex.24 The rate constant, the activation energy and the Arrhenius parameter of a chemical reaction at 25°C are 2 × 10– 4 , s–1, 114.4 kJ mol–1 and 6.0 × 1014 s–1 respectively, the value of the rate constant at T   is. (A) 2.0 × 1018 s–1 (B) 3.6 × 1030 s–1 (C)  (D) 6.0 × 1014 s–1 k = Ae–Ea/RT When T  kA A = 6 × 1014 s–1 Hence, (D) is the correct answer. If a reaction A + B  C, is exothermic to the extent of 40 kJ/mol and the forward reaction has an activation energy 60 kJ/mol, the activation energy for the reverse reaction is (A) 30 kJ/mol (B) 40 kJ/mol (C) 70 kJ/mol (D) 100 kJ/mol 60kJ Energy Energy 100kJ 40kJ Sol. Progress of reaction Activation energy for backward reaction = 100 kJ Hence, (D) is the correct answer. Ex.25 The accompanying figure depicts the change in concentration of species A and B for the reaction A  B, as a function of time the point of inter section of the two curves represents. conc. [B] [A] Time Sol. (A) t1/2 (B) t3/4 (C) t2/3 (D) data insufficient to predict The intersection point indicates that half life of the reactant A is converted into B. Hence the answer is [A]. Ex.26 Sol. For the reaction, A + B  C, the following data were obtained. In the first experiment, when the initial concentrations –4 –1 –1 of both A and B are 0.1 M, the observed initial rate of formation of C is 1 × 10 mol litre minute . In second experiment –4 –1 –1 when the initial concentrations of (A) and (B) are 0.1 M and 0.3 M, the initial rate is 3.0 × 10 mollitre minute . In the third experiment, when the initial concentrations of both A and B are 0.3 M, the initial rate is –3 –1 –1 2.7 × 10 mol litre minute . (A) Write rate law for this reaction. (B) Calculate the value of specific rate constant for this reaction. x y Let, Rate = k[A] [B] –4 x y r1 = 1 × 10 = k(0.1) (0.1) ...(1) –4 x y r2 = 3 × 10 = k(0.1) (0.3) ...(2) –3 x y r3 = 2.7 × 10 = k(0.3) (0.3) ...(3) y By Eqs. (1) and (2), –4 r1 1×10 1 = –4 =    y = 1 3 r2 3×10 By Eqs. (2) and (3), x 3×10–4  1  r2 = =   x = 2 r3 27 ×10–4  3  2 1  Rate = k[A] [B] –4 2 1 Also, 1 × 10 = k (0.1) (0.1) –1 2 –1 –1  k = 10 = 0.1 L mol min Ex.27 Sol. 2O3  3O2 proceeds as follows : O3 O2 + O .... (Fast) O + O3  2O2 .... (Slow) The rate law expression should be (A) r = k[O3]2 (B) r = k[O3]2 [O2]–1 (C) r = k[O3][O2] Rate of reaction (r) = k [O] [O3] The rate of formation if [O] depends on first step The chemical reaction, (D) Unpredictable [O 2 ][O] [O3 ]  Since Keq =  [O3 ] [O] = Keq [O ] 2 or Ex.28 Sol. r = kkeq [O3 ][O3 ] = k . [O3]2 [O2]–1 [O2 ] 13 Disintegration of radium takes place at an average rate of 1.42 × 10 -particles per minute. Each -particle takes up 2 electrons from the air and becomes a neutral helium atom. After 420 days, the He gas collected –3 was 0.5 × 10 L measured at 300 K and 750 mm of mercury pressure. From the above data, calculate Avogadro's number. 13 –1 No. of -particles (or) He formed = 1.42 × 10 min 13 18  No. of He particles formed in 420 days = 1.42 × 10 × 420 × 1440 = 8.588 × 10 Also at 27°C and 750 mm ; He = 0.5 mL Using PV = nRT 750 0.5 –5 × = n × 0.0821 × 300  n = 2.0 × 10 moles 760 1000 –5 18 2.0 × 10 moles of He = 8.588 × 10 particles of He 8.588×1018 22  4.294 × 10 particles 2.0 ×10–5 22  Avogadro's number = 4.294 × 10 particles/mol  1 mole of He = CHEMICAL KINETICS AND NUCLEAR CHEMISTRY Ex.29 For the non-equilibrium process, A + B  Products, the rate is first order with respect to A and second order with respect to B. If 1.0 mol each of A and B are introduced into a 1 litre vessel, and the initial rate were 1.0 × 10–2 mol/litre-sec, calculate the rate when half of the reactants have been used. Sol. Rate1 = k [A] [B]2  10–2 = k [1] [1]2 or k = 10–2 litre2 mol–2 sec–1 Now or RateII = 10–2 × 0.5 × (0.5)2 New rate = 1.2 × 10–3 mol/litre-sec Ex.30 The energy of activation for a certain reaction is 100 kJ/mol. Presence of catalyst lowers the energy of activation by 75%. What will be effect on rate of reaction at 25°C, other things being equal ? Sol. k = A e RT , k1 = A e RT , k2 = A e RT –E a –100 –25 –100 -75 k e RT  1 = –25 = e RT k2 e RT loge 75 k2 = log ee RT k1  log 75 ×103 k 2 75 = = k1 RT 8.314 × 298 k2 13 n = 1.4020 × 10 , As r = k(conc) k1 r2 k 2 13 = = 1.4020 × 10 r1 k1 Ex.31 Show that for a first order reaction, time required for 99% completion is twice for the time required for the completion of 90% of the reaction. Sol. t99% = 2.303 100 log10 k 100 – 99 .....(1) ( a = 100; x = 99) t90% = 2.303 100 log10 k 100 – 90 .... (2) ( a = 100; x = 90)  By Eqs. (1) and (2), log10 100 t 99% = =2 t 90% log10 10  t99% = 2 × t90% Ex.32 The optical rotations of sucrose in 0.5 N HCl at 35°C at various time intervals are given below. Show that the reaction is of first order : Time (minutes) Rotation (degrees) Sol. 0 10 20 30 40  +32.4 +28.8 +25.5 +22.4 +19.6 –11.1 The inversion of sucrose will be first order reaction if the above data confirm to the equation, k1 = r0 – r 2.303 log r – r where r0, rt and r represent optical rotations initially, at the commencement of the reaction t t  after time t and at the completion of the reaction respectively. In this case, a0 = r0 – r = +32.4 – (–11.1) = +43.5 The value of k at different times is calculated as follows : Time rt rt – r  k 10 min +28.8 39.9 2.303 43.5 –1 log = 0.008625 min 10 39.9 20 min +25.5 36.6 2.303 43.5 –1 log = 0.008625 min 20 36.6 30 min +22.4 33.5 2.303 43.5 –1 log = 0.008694 min 30 33.5 40 min +19.6 30.7 2.303 43.5 –1 log = 0.008717 min 40 30.7 The constancy of k1 indicates that the inversion of sucrose is a first order reaction. Ex.33 For A + B  C + D; H = 20 kJ mol–1 ; the activation energy of the forward reaction is 85 kJ/mol. Calculate activation energy of the reverse reaction. Sol. H of forward reaction = 20 kJ mol–1. Energy of activation for forward reaction (Ea) = 85 kJ mol–1  Energy of activation for backward reaction = Ea – H = 85 – 20 = 65 kJ mol–1 CHEMICAL KINETICS AND NUCLEAR CHEMISTRY Ex.34 Sol. –3 –1 The reaction given below is observed to be first order with rate constant 5.26 × 10 sec . Calculate the time required for the total pressure in a system containing A at an initial pressure of 0.1 atm to rise 0.145 atm and also find the total pressure after 100 sec. 2A (g) 4B (g) + C (g) 2A (g)  4B (g) + C(g) Initial P0 0 0 At time t P0 – P' Ptotal = P0 – P' + 2P' + P'/2 = P0 + P' = 2P' P'/2 3 P' 2 2 (0.145 – 0.1) = 0.03 atm 3 k= P0 2.303 log P – P' t 0 t= 2.303  0.1   = 67.82 sec –3 log  5.26 ×10  0.07  Also, k = 2.303  0.1  log   t  P0 – P'  –3 5.26 × 10 = 2.303  0.1  log   100  0.1 – P'  0.1 – P' = .059 P' = 0.041 Ptotal = 0.1 + Ex.35 Sol. 3 (0.041)  0.162 atm. 2 The time required for 10% completion of first order reaction at 298 K is equal to that required for its 76% completion 9 –1 at 308 K. If the pre-exponential factor for the reaction is 3.56 × 10 s , calculate its energy of activation. For first order reactions, t= N0 2.303 log N1 k At 298 K ; t = 2.303 100 log k 298 90 At 309 k ; t = 100 2.303 log 76 k 308 Since time is same 2.303 100 100 2.303 log = log k 298 76 k 308 90 or 0.0458 0.1191 = k 308 k 298 or k 308 0.1191 = 2.60 = k 298 0.0458 According to Arrhenius equation, 2.303 log 1  k 308 E  1 – = a  k 298 8.314  298 308  or 2.303 log 2.60 = E a  10   298 × 308   8.314 Ea = 72.93 kJ Ex.36 In a reaction, the decrease in reactant’s concentration is 20% in 20 minute and 40% in 40 minute. Calculate order of reaction and rate constant. Sol. For zero order reaction : t = Ex.37 Sol. If t = t20% = 20 minute, Then k= If t = t40% = 40 minute, Then k= k= x t x = 20 20 = 1 mol litre–1 minute–1 20 x = 40 40 = 1 mol litre–1 minute–1 ; Thus, reaction is of zero order.. 40 –3  Ea = Ea = –1 –1 k 2 E a  T2 – T1  = k1 R  T1 × T2  2.303R T1 T2 k log10 2 (T2 – T1 ) k1  1.38×10 –2  2.303× 8.314 ×10-3 × 303× 323  = 64.91 kJ mol–1 log10   2.80 ×10 –3  (323 – 303) Also, k = A e or Sol. or The rate constant for the decomposition of a certain substance is 2.80 × 10 m s at 30°C and –1 –1 –1 –3 –1 –1 1.38 × 10 m s at 50°C. Evaluate the Arrhenius parameters of the reaction. (R = 8.314 × 10 kJmol K ) Energy of activation (Ea) and pre-exponential factor A are Arrhenius parameters. –3 –1 –1 k1 = 2.80 × 10 m s at 303 K –2 –1 –1 k2 = 1.38 × 10 m s at 323 K As log10 Ex.38 x k –Ea/RT A=ke Ea/RT –3 = 2.80 × 10 e 64 .9 1 ×1 0 3 / 8 .31 4 × 3 03 8 –1 = 4.34 × 10 s For a reaction 3A  Products, it is found that the rate of reaction doubles if concentration of A is increased four times, calculate order of reaction. Rate = k [Reactant]n if [Reactant] = a; rate = r1 r1 = k[a]n 2r1 = k[4a]n  1 1  = 2  4  if [Reactant] = 4a; rate = 2r1 n  n= 1 2 CHEMICAL KINETICS AND NUCLEAR CHEMISTRY 14 Ex.39 The amount of 6C isotope in a piece of wood is found to be one fifth of that present in a fresh piece of wood. 14 Calculate the age of the piece of wood (half-life of 6C = 5577 year). Sol. t= 2.303 N log 0 λ N t = 2.303 × t1/2 N log 0 0.693 N N0 2.303× 5577 2.303× 5577 × 0.6989 log  N0 / 5 0.693 0.693 = 12.953 years t= Ex.40 In a reaction, 2A  Products, the concentration of A decreases from 0.5 mol litre–1 to 0.4 mol litre–1 in 10 minute. Calculate rate during this interval. Sol. Rate of reaction = = Ex.41 Sol. 1 × rate of disappearance of A 2 1 2  d[A]  1 – =  dt  2  0.5 – 0.4  –1 –1  10  = 0.005 mol litre minute Calculate the order of reaction for which rate becomes half if volume of container having same amount of reactant is doubled. (Assume gaseous phase reaction) Rate = k[a]n For Case I : Let a mole of reactant in vessel of V litre n  a r1 = k   V For Case II : The volume is doubled, rate becomes half ......(1) n Ex.42 Sol. r1  a   =k  .....(2) 2  2V   By Eqs. (1) and (2), or 2 = (2)n  n=1 –1 Rate constant of a first order reaction, A B, is 0.0693 min . Calculate rate (i) at start and (ii) after 20 minutes. Initial concentration of A is 1.0 M. –1 k1 = 0.0693 min t1 = 2 0.693 0.693 = = 10 min 0.0693 k1 1 Since C = C0   2 20 =2 n= 10 n  t  n =  t 1/2   C0 = 1 M 2 1 1 C=1×   = M 4 2 Rate of the reaction at the start of the reaction = k1 × C0 –1 = 10 × 0.0693 × 1 = 0.693 M min Rate after 30 min. = k1C = 0.0693 × 1 –3 –1 = 17.33 × 10 M min 4 Ex.43 Sol. Two reactants A and B separately show two chemical reactions. Both reactions are made with same initial concentration of each reactant. Reactant A follows first order kinetics whereas reactant B follows second order kinetics. If both have same half-lives, compare their rates at the start of reactions. For A : rate = kA [A] ...(i) t 12(A) = 0.693 kA ...(ii) For B : rate = kB [B] t 12(B) = 2 ...(iii) 1 kB × a ...(iv) Initial rate of A Initial rate of B rA = kA × a 2 rB = kB × a rA kA × a kA  r = k × a2 = k × a B B B If t 1 2 (A)  t 1 2 (B ) , then ...(v) 1 0.693 = kB × a kA  kA = 0.693 × a kB  rA 0.693 × a = = 0.693 rB a Ex.44 A first order reaction takes 69.3 minute for 50% completion. How much time will be needed for 80% completion ? Sol.  Now k= 0.693 0.693 = minute–1 69.3 t1/2 k= 100 2.303 log 100 – x t k= 2.303 100 log10 t 20 ( t½ = 69.3 min) ; [if a = 100, x = 80 and a – x = 20] 0.693 2.303 = log10 5 ; 69.3 t Ex.45 The half-life of the nucleide Sol. t 12 = 220 Rn is 693 s. What mass of radon is equivalent to a 1 millicurie (mci) ? 0.693 0.693 0.693 –3 –1  λ= = × 10 s t 12 693 λ 7 –1 1 mci = 3.7 × 10 disintegration s = – N=– t = 160.97 minute dN dN  – = N dt dt dN / dt 3.7 ×107 s –1 10 = = 3.7 × 10 λ 10 –3 s Mass of 220 Rn = 3.7 ×1010 × 220 –11 –14 = 1.35 × 10 g = 1.06 × 10 kg 6.022 ×1023 CHEMICAL KINETICS AND NUCLEAR CHEMISTRY Ex.46 For the reaction A  B+C Time t  Total pressure of (B + C) P2 P3 Calculate k. Sol. A  B + C At t = 0 P1 0 0 At t = t P1 – x x x At t =  0 P1 P1  2P1 = P3  P1 = P3 2 2x = P2  x =  P1 – x = P2 2 P3 P2 P3 – P2 – = 2 2 2 1 [A]0 1 P3 = ln  k = ln t [A]t t (P3 – P2 ) Ex.47 The rate of a certain reaction depends on concentration according to equation – k1[A] d[A] = 1 + k 2 [A] dt What will be the order of reaction when (i) concentration is very high (ii) very low? Sol. Given, – (i) k1[A] –d[A] k1 d[A] =  = 1 1 + k [A] dt dt 2 + k2 [A] When [A] is very high  – 1 is very small, and thus negligible [A] d[A] k1 = constant = k2 dt Thus, order of reaction is zero. (ii) When [A] is very low [1 + k2 [A] = k'  – d[A] k1[A] = k"[A] = k' dt Thus, order of reaction is one. Ex.48 Pseudo first order rate for the reaction, A + B Product, when studied in 0.1 M of B is given by – Sol. where, k = 1.25 × 10 sec , calculate the value of second order rate constant. A + B Product 4 –1 d[A] d[A] 4 = k[A]  – = 1.25 × 10 × [A] dt dt Assuming the reaction to be of second order – ....(i) d[A] d[A] = k' [A] [B]  – = k' [A] [0.1] dt dt Dividing Eq. (i) by (ii), we get – 1= d[A] = k[A] , dt ....(ii) 1.25 ×104 k' × (0.1) 5 –1 –1  k' = 1.25 × 10 L mol s Ex.49 N2O5 decomposes according to equation, N2O5  2NO2 + 1 O 2 2 Sol. –d[N 2O5 ] d[O2 ] denote ? (B) What does denote ? (C) What is the units of rate of this reaction? dt dt (A) Rate of decomposition of N2O5. (B) Rate of formation of O2. (C) Unit of rate = mol litre–1 time–1. Ex.50 Derive a relation between t 1 and temperature for an n order reaction where n > 2 ? 2 Sol. lnk = lnA – (A) What does th t12 = 2 n –1 Ea RT (Arrhenius equation)  –1 ...(ii) k(n – 1)a 0n –1 2n –1 – 1 – lnk (n – 1)a 0n –1 From the Eqs. (i) and (iii)  ln  t 1 2  = ln ln  t 1 2  = ln ...(iii) 2n –1 – 1 E – lnA + a n –1 RT (n – 1)a 0  ln  t 1 2  = lnA + where A = ....(i) E RT 2 n –1 – 1 (n – 1)a n0 –1 × A That is t 1 2 decreases with increase in temperature. A plot of t 1 2 vs 1 gives a straight line with slope Ea. T