14Chemical KineticsandReactionDynamicsThus,we can takeVo-Vo= aV,- Vo = xor(V- Vo) - (V,- Vo) = V- V,= a- xTherefore, the rate constant for the reaction may be obtained as(V - Vo)k= 2.303(1.25)1og (Ve- V)13.DecompositionofN,OsN,05→2 NO2+/202Nitrogen pentaoxide in carbon tetrachloride solution decomposesto giveO2The progress of reaction is monitored bymeasuring the volume of O,atdifferenttimeintervals and using therelationVo2.303k=(1.26)ogV.-V,twhere Vis the final value of O, when reaction is complete and correspondsto initial concentration of N,Os,V,is thevalue of O at anytimetand(V.-V) corresponds to (a -x).4.DecompositionofH,O2inaqueoussolutionH,02P>H,0+0Theconcentration of H,O2 at differenttime intervals is determinedby titratingtheequal volumeof reaction mixture against standardKMnO4.Problem 1.7The specific rotation of sucrose in presence of hydrochloricacidat35°Cwasmeasuredandisgivenasfollows:02040Time (min)801805008032.428.825.519.610.36.1Rotation (°C)-14.1Calculate the rate constant at various time intervals and showthatthe reactionis first orderSolution02.3032.303ro -rok=loglog-1axr, Too32.4 - (-14.1)2.303.00403 min-lk20=log2028.8 - (14.1)32.4 - (-14.1)2.303k40=.00406 min-1log25.5 - (-14.1)40
14 Chemical Kinetics and Reaction Dynamics Thus, we can take V∞ – V0 = a Vt – V0 = x or (V∞ – V0) – (Vt – V0) = V∞ – Vt = a – x Therefore, the rate constant for the reaction may be obtained as k t V V V Vt = 2.303 log ( – ) ( – ) ∞ 0 ∞ (1.25) 3. Decomposition of N2O5 N2O5 → 2 NO2 + 1 /2 O2 Nitrogen pentaoxide in carbon tetrachloride solution decomposes to give O2. The progress of reaction is monitored by measuring the volume of O2 at different time intervals and using the relation k t V V Vt = 2.303 log – ∞ ∞ (1.26) where V∞ is the final value of O2 when reaction is complete and corresponds to initial concentration of N2O5, Vt is the value of O2 at any time t and (V∞ – Vt) corresponds to (a – x). 4. Decomposition of H2O2 in aqueous solution H O H O + O 2 2 Pt → 2 The concentration of H2O2 at different time intervals is determined by titrating the equal volume of reaction mixture against standard KMnO4. Problem 1.7 The specific rotation of sucrose in presence of hydrochloric acid at 35°C was measured and is given as follows: Time (min) 0 20 40 80 180 500 ∞ Rotation (°C) 32.4 28.8 25.5 19.6 10.3 6.1 –14.1 Calculate the rate constant at various time intervals and show that the reaction is first order. Solution k t a ax t r r r r t = 2.303 log – = 2.303 log – – 0 ∞ ∞ k20 –1 = 2.303 20 log 32.4 – (–14.1) 28.8 – (–14.1) = – .00403 min k40 –1 = 2.303 40 log 32.4 – (–14.1) 25.5 – (–14.1) = – .00406 min
15Elementary32.4 - (-14.1)2.303.004025 min-lkso = 1og 19.6 - (-14.1)8032.4 - (-14.1)2.303=-.004002min-1k200 =log2006.1-(-14.1)Average=0.0040295min-lSince the first order rate constant remains same,the reaction is of first orderProblem1.8Afirstorder reactionis25%completein50min.What wouldbeconcentrationattheendof another5ominif the initial concentrationofthe reactant is5.0×103mol dm-3?SolutionReactionis25%consumedin50min.After50mintheconcentrationcwill be5.0×103×25a-x=5.0×103=3.75×103100(5.0 × 103)Co2.30312.303k=5.375×103min-1Nowlog109tC503.75×103Again,concentrationafteranother50min,i.e.(50+50=100min)c= Co e-kt = 5.0 ×103e-(5.35×10-3/100) = 2.81 ×103mol dm-3Problem1.9Following observations were made for decomposition of areactant at 35°C:[A]Rate of decompositionmol dm-3[d[A]/dt] (mol dm- sec-')0.050.150.300.100.200.60Find the order of reaction.Calculate rate constant and therate of decompositionof A, when [A] = 0.45 mol dm-3.Solution It can be observed from the data that the rate of decomposition indirectly proportional to [A], i.e.d[Al α[A]dtdIAl k[A]ordt
Elementary 15 k80 –1 = 2.303 80 log 32.4 – (–14.1) 19.6 – (–14.1) = – .004025 min k200 –1 = 2.303 200 log 32.4 – (–14.1) 6.1 – (– 14.1) = – .004002 min Average = 0.0040295 min–1 Since the first order rate constant remains same, the reaction is of first order. Problem 1.8 A first order reaction is 25% complete in 50 min. What would be concentration at the end of another 50 min if the initial concentration of the reactant is 5.0 × 103 mol dm–3? Solution Reaction is 25% consumed in 50 min. After 50 min the concentration c will be a x – = 5.0 10 – 5.0 10 25 100 = 3.75 10 3 3 3 × × × × Now k t C = 2.303 log C = 2.303 50 log (5.0 10 ) 3.75 10 = 5.375 10 min 0 3 3 × 3 –1 × × Again, concentration after another 50 min, i.e. (50 + 50 = 100 min) c = c0 e–kt = 5.0 103 –(5.35 10 /100) –3 × × e = 2.81 × 103 mol dm–3 Problem 1.9 Following observations were made for decomposition of a reactant at 35°C: [A] Rate of decomposition mol dm–3 [–d[A]/dt] (mol dm–3 sec–1) 0.15 0.05 0.30 0.10 0.60 0.20 Find the order of reaction. Calculate rate constant and the rate of decomposition of A, when [A] = 0.45 mol dm–3. Solution It can be observed from the data that the rate of decomposition in directly proportional to [A], i.e. – [A] [A] d dt ∝ or – [A] [A] d dt k
16Chemical Kinetics and Reaction Dynamicswhere k is the rate constant. Hence, the reaction is first order.- d[A/d = 00 = 0.33; 0.100.20k=0.33 and= 0.33 sec-1[A]0.300.150.60k (average)=0.33sec-lAgain,rate of decomposition when [A]=0.45mol dm-3 isdIA = k [A] = 0.33 (sec-l) × 0.45 mol dm3dt=0.1485mol dm-3sec-l=0.15mol dm-3sec-Problem1.10o Thekinetics of a reaction was followedby measuringtheabsorbance dueto a reactant at its2maxat 25°C.The log(absorbance)versustime (min)plot was a straight line with a negative slope (0.30 × 10-2) and apositiveintercept.Find thehalf-lifeperiod ofreaction.Solution Whenlog (a-x)versustimeplot is straight line,thereaction is firstorderand slopegives the value ofk/2.303while intercept gives the value oflog a.Rate constant k = Slope = 0.30 × 10-22.303k = 0.6909 × 10-2 min-lTherefore,2.303log20.6939Half-life period t1/2=30.6939 × 10-= 100 mink(min-l)Problem1.11Ina firstorderreactionthelog (concentrationofreactant)versustime plot was a straight line with a negative slope 0.50 ×10 sec-l.Findtherateconstantand half-lifeperiod of reaction.Solution Theequation of first order rate constant may be written asklog(a-x)=-2.303 I + log aThus,plot of log (a-x)versustime will be a straight linewith a negativekslope equal to 2.303:k2.303 = Slope = 0.50 × 10* sec-Therefore,k=2.303×0.5×104=1.1515×104sec-or
16 Chemical Kinetics and Reaction Dynamics where k is the rate constant. Hence, the reaction is first order. k d dt = – [A]/ [A] = 0.05 0.15 = 0.33; 0.10 0.30 = 0.33 and 0.20 0.60 = 0.33 sec –1 k (average) = 0.33 sec–1 Again, rate of decomposition when [A] = 0.45 mol dm–3 is – [A] = [A] d dt k = 0.33 (sec–1) × 0.45 mol dm–3 = 0.1485 mol dm–3 sec–1 = 0.15 mol dm–3 sec–1 Problem 1.10 The kinetics of a reaction was followed by measuring the absorbance due to a reactant at its λmax at 25°C. The log (absorbance) versus time (min) plot was a straight line with a negative slope (0.30 × 10–2) and a positive intercept. Find the half-life period of reaction. Solution When log (a – x) versus time plot is straight line, the reaction is first order and slope gives the value of k/2.303 while intercept gives the value of log a. Rate constant 2.303 = Slope = 0.30 10 k –2 × Therefore, k = 0.6909 × 10–2 min–1 Half-life period t1/2 = 2.303 log 2 (min ) = 0.6939 0.6939 10 = 100 min –1 –2 k × Problem 1.11 In a first order reaction the log (concentration of reactant) versus time plot was a straight line with a negative slope 0.50 × 104 sec–1. Find the rate constant and half-life period of reaction. Solution The equation of first order rate constant may be written as log ( – ) = – 2.303 a x + log k t a Thus, plot of log (a – x) versus time will be a straight line with a negative slope equal to k 2.303 . Therefore, k 2.303 = Slope = 0.50 × 104 sec–1 or k = 2.303 × 0.5 × 104 = 1.1515 × 104 sec–1
Elementary172.303log 20.693Again half-lifeperiod ty2k1.1515×104= 0.60 × 10- sec-l = 60 μsec1.8RadioactiveDecayas aFirstOrderPhenomenonWhen a radioactive substance is separated from itsparent and its activity ismeasured fromtimetotime,therate ofdecayfollows the law as=(1.27)dtwhereI is activity at any time t and a constant called the radioactivedisintegration constant which is characteristic of the radioactive elementirrespectiveof itsphysicalconditionor stateof chemical combination.Thisequation,onintegration,givesI = lo e-N(1.28)where lo in the initial activity, i.e. when t = O.Since the activityIis proportional to number of atoms that have not yetdisintegrated,wecanuse relationn= noe-n(1.29)入=2.303orfwherenoandnarethenumberof atoms of radioactive substanceat timet=Oand at any timet,respectively.Equation (1.29)suggests that thereis adefiniteprobability ofany particularatomdisintegrating ata givenmomentandthisis proportional tothenumberof atomspresentatthatmoment.Half-life period gives the timeTwhich must elapse for the radioactivityto decayto half its value at anyinstant and maybe obtained by pullingnoinequation (1.29).ThusIno=noe-^7log20.693orT=(1.30)入入In timeTthe activity is reduced to one-half of its initial value and,thereforeinnTtime,theactivityisdecreased to(1-Theoreticallythereforetheactivity never falls to zero
Elementary 17 Again half-life period t1/2 = 2.303 log 2 = 0.693 1.1515 104 k × = 0.60 × 10–4 sec–1 = 60 µsec 1.8 Radioactive Decay as a First Order Phenomenon When a radioactive substance is separated from its parent and its activity is measured from time to time, the rate of decay follows the law as – = dI dt λ I (1.27) where I is activity at any time t and λ a constant called the radioactive disintegration constant which is characteristic of the radioactive element irrespective of its physical condition or state of chemical combination. This equation, on integration, gives I = I0 e–λt (1.28) where I0 in the initial activity, i.e. when t = 0. Since the activity I is proportional to number of atoms that have not yet disintegrated, we can use relation n = n0 e–λt (1.29) or λ = 2.303 log 0 t n n ⎛ ⎝ ⎞ ⎠ where n0 and n are the number of atoms of radioactive substance at time t = 0 and at any time t, respectively. Equation (1.29) suggests that there is a definite probability of any particular atom disintegrating at a given moment and this is proportional to the number of atoms present at that moment. Half-life period gives the time T which must elapse for the radioactivity to decay to half its value at any instant and may be obtained by pulling n = 1 2 n0 in equation (1.29). Thus 1 2 n0 = n0 e–λT or T = log 2 = 0.693 λ λ (1.30) In time T the activity is reduced to one-half of its initial value and, therefore, in nT time, the activity is decreased to 1 – 1 2 ( ) n . Theoretically, therefore the activity never falls to zero
18Chemical KineticsandReactionDynamicsAverage-lifeperiod iofanyradioactiveatomisthelengthof time,whichcanexist beforetheatom disintegratesand can becalculated as followsSince the activity is proportional to number of atoms, which have notdisintegrated,equation(1.27)maybewrittenas-dnldt = n-dn=Anoe-ndt; n = noe-α (from eq.(1.29)(1.31)orThe number of atoms disintegrating in the interval between t and t+ dt isequal to dn. Sincedt is very small, dnmay betaken as number of atomdisintegrating attimet.Theperiod of average lifeiis obtained bymultiplyingeverypossiblelifeperiodtfromzeroto infinity,bythenumber of atoms dnand then dividing the product by the total number of atoms no present at thebeginning of thetime.Thus,tdn(1.32)noSubstituting the value of dn fromequation (1.31)and ignoring the sign of dn,we haveAd(1.33)1 = 1/^(1.34)orThe average-life1 of a radioactive atom is thus equal to the reciprocal of itsradioactive constant and is a measureofmean expectation of future life ofthe atoms present at any instant.RadioactiveEquilibriumThe rate of decay of a given radioactive substance must also represent therate at which its disintegration product is being formed. The product willalso disintegrate at a rate depending on its amount present.This will be smallinitially,but will increase with time.Thus, in any given series a state ofequilibrium will eventuallybereached,whentherateof formation of anyelement from its parent is equal to the rate at which it itself disintegrated.Mathematically,it can be represented asdnidn2_dn3(1.35)dtdtdtwhere n, n2, n3represent the number of atoms of different radioactiveelements in the series present at equilibrium.Again, since-dnldt=n,wemayhave2ini=2n2=23n3
18 Chemical Kinetics and Reaction Dynamics Average-life period ι of any radioactive atom is the length of time, which can exist before the atom disintegrates and can be calculated as follows. Since the activity is proportional to number of atoms, which have not disintegrated, equation (1.27) may be written as –dn/dt = λn or –dn = λn0 e–λt dt; n = n0 e–λt (from eq. (1.29) (1.31) The number of atoms disintegrating in the interval between t and t + dt is equal to dn. Since dt is very small, dn may be taken as number of atom disintegrating at time t. The period of average life ι is obtained by multiplying every possible life period t from zero to infinity, by the number of atoms dn and then dividing the product by the total number of atoms n0 present at the beginning of the time. Thus, ι = 0 0 ∞ ∫ t dn n (1.32) Substituting the value of dn from equation (1.31) and ignoring the sign of dn, we have ι λ λ = 0 – ∞ ∫ t e dt t (1.33) or ι = 1/λ (1.34) The average-life ι of a radioactive atom is thus equal to the reciprocal of its radioactive constant λ and is a measure of mean expectation of future life of the atoms present at any instant. Radioactive Equilibrium The rate of decay of a given radioactive substance must also represent the rate at which its disintegration product is being formed. The product will also disintegrate at a rate depending on its amount present. This will be small initially, but will increase with time. Thus, in any given series a state of equilibrium will eventually be reached, when the rate of formation of any element from its parent is equal to the rate at which it itself disintegrated. Mathematically, it can be represented as dn dt dn dt dn dt 1 2 3 = = . . . (1.35) where n1, n2, n3 . . . represent the number of atoms of different radioactive elements in the series present at equilibrium. Again, since –dn/dt = λn, we may have λ1n1 = λ2n2 = λ3n3 . .