9ElementarySolution1dm3=1000cm3.i.e.1cm2=10-3dm3(i)1 mol = 6.02 × 1023 moleculemolecule-' = 6.02 × 1023 mol-lk = 2.50 ×10-°cm2molecule-ls-1= 2.50 × 10-9 (10-3 dm3)(6.02 × 1023 mol-l)s-l= 2.50 × 6.02 × 10-9-3+23 dm mo1-l s-1= 15.05 × 10ll dm3 mol-l s-lP-"RT=CRT(ii)Weknow(1 atm)0P1 atmor C =RT=0.0821atm dmmol-K-l×273K= 0.0446 mol dm-31 atm = 0.0446 mol dm-3Therefore,16 mol-" dm31 atm-lor0.0446k = 2.0 × 10-6 g-' atm-l= 2.0 ×10-6 s-lmol-'dm30.0446= 44.8 × 10- dm2 mol-l s-1Problem1.4For a certain reaction, the value of rate constant is 5.0×10-3dm’ mol-'sec-. Find the value of rate constant in (i) dm molecule-' sec-!(ii) cm"mol-' sec-" and (ii) cm"molecule-" sec-Solutionin dm’ mol-'sec-!(i)Rate constant = 5.0 ×10-3 dm2 mo1-'sec-11mol =6.02×1023moleculesRate constant= 5.0× 10-3dm3 (6.02×1023mol)- sec-l= 0.83 × 10-26 dm3 molecule-l sec-l)in cmmol'sec-!(i)1dm3=1000cm3
Elementary 9 Solution (i) 1 dm3 = 1000 cm3 , i.e. 1 cm3 = 10–3 dm3 1 mol = 6.02 × 1023 molecule molecule–1 = 6.02 × 1023 mol–1 k = 2.50 × 10–9 cm3 molecule–1s–1 = 2.50 × 10–9 (10–3 dm3 )(6.02 × 1023 mol–1)s–1 = 2.50 × 6.02 × 10–9–3+23 dm3 mol–1 s–1 = 15.05 × 1011 dm3 mol–1 s–1 (ii) We know P n RT CRT (1 atm) = = v or C P RT = = 1 atm 0.0821 atm dm mol K 273 K 3 –1 –1 × = 0.0446 mol dm–3 Therefore, 1 atm = 0.0446 mol dm–3 or 1 atm–1 = 1 0.0446 mol dm –1 3 k = 2.0 × 10–6 s–1 atm–1 = 2.0 × 10–6 s–1. 1 0.0446 ⎛ ⎝ ⎞ ⎠ mol–1dm3 = 44.8 × 10–6 dm3 mol–1 s –1 Problem 1.4 For a certain reaction, the value of rate constant is 5.0 × 10–3 dm3 mol–1sec–1. Find the value of rate constant in (i) dm3 molecule–1 sec–1 (ii) cm3 mol–1 sec–1 and (iii) cm3 molecule–1 sec–1. Solution (i) in dm3 mol–1sec–1 Rate constant = 5.0 × 10–3 dm3 mol–1sec–1 1 mol = 6.02 × 1023 molecules Rate constant = 5.0 × 10–3 dm3 (6.02 × 1023 mol)–1 sec–1 = 0.83 × 10–26 dm3 molecule–1 sec–1 (ii) in cm3 mol–1sec–1 1 dm3 = 1000 cm3
10ChemicalKineticsandReactionDynamicsRate constant = 5.0×10-' dmmol-'sec-!= 5.0 × 10-3(1000) cm2 mol-sec-l= 5.0 cm2mol-l sec-1(ii) in cm molecules-'sec-lRate constant = 5.0 cm (6.02×1023)-molecules-'sec-l= 0.83×10-23cmmolecules-'sec-l1.6ZeroOrderReactionsWhen noconcentrationterm affects the rateofreaction,ortherateofreactionremains same throughout the reaction, the reaction is known as zero-orderreaction.Let us consider a reactionA→ProductSince the rateof reaction remains same=kdtOnintegratingtheexpressionaswe getx=kt+zThe valueof integration constant z maybe obtained by taking the conditionsx =O,whent=O,the value ofz is zero and, therefore,rate equation becomesX=kt or k=(1.19)1which gives the unit of rate constant as mol dm-3 sec-' or conc. (time)-' ingeneral..The half-lifeperiodty/2 of a zero orderreaction can be calculated with thehelp of equation (1.19),takingt=tu2andx=a/2asTun=%(1.20)Thus,thehalf-lifeperiod ofzeroorder reaction isdirectlyproportional tothe initial concentration of the reactant.For example,on increasing theinitial concentrationbytwofold,thehalf-lifeperiod of the reaction wouldalso bedouble
10 Chemical Kinetics and Reaction Dynamics Rate constant = 5.0 × 10–1 dm3 mol–1sec–1 = 5.0 × 10–3(1000) cm3 mol–1sec–1 = 5.0 cm3 mol–1 sec–1 (iii) in cm3 molecules–1sec–1 Rate constant = 5.0 cm3 (6.02 × 1023) –1 molecules–1sec–1 = 0.83 × 10–23 cm3 molecules–1sec–1 1.6 Zero Order Reactions When no concentration term affects the rate of reaction, or the rate of reaction remains same throughout the reaction, the reaction is known as zero-order reaction. Let us consider a reaction A→ Product Since the rate of reaction remains same dx dt = k On integrating the expression as ∫ ∫ dx k dt = we get x = kt + z The value of integration constant z may be obtained by taking the conditions x = 0, when t = 0, the value of z is zero and, therefore, rate equation becomes x kt k x t = or = (1.19) which gives the unit of rate constant as mol dm–3 sec–1 or conc. (time)–1 in general. • The half-life period t1/2 of a zero order reaction can be calculated with the help of equation (1.19), taking t = t1/2 and x = a/2 as t a k 1/2 = 2 (1.20) Thus, the half-life period of zero order reaction is directly proportional to the initial concentration of the reactant. For example, on increasing the initial concentration by two fold, the half-life period of the reaction would also be double
Elementary11: According to equation (1.19) the slope of a plot of x or (a - x) (i.e. theconcentrationof productorconcentrationofreactant)versustimewillgivethe value of rate constant k (Fig.1.2).Slope = k(x D)Slope = kIntercept = a1Fig.1.2Concentration versus time plotfor zero order reaction.The combination of H2and Cl2 toform HCl inpresence of sunlight is azeroorderreactionH2+ Cl2 →2HCIThe rate of formation of HCl is not affected by a change in concentrationof either the reactant orproduct.However, it is influenced bythe intensity ofsun light.Problem1.5Azeroorderreactionis50%completein20min.Howmuchtimewillittaketo complete90%?Solution Let a = 100 mol dm-3. For a zero-order reaction50 (mol dm-3)k=岁=320×60(sec)1When reactionis90%completed,x=90.Therefore,50元=90Thus,1200=T= 90 ×1200 = 2160 sc = 36 minor50Problem1.6Areaction is 50% complete in20min.Howmuchtime will betakentocomplete75%reaction?SolutionForazeroorderreactionk=x/tx=a/2for50%ak=2t2×203afor 75% reactionx=4
Elementary 11 • According to equation (1.19) the slope of a plot of x or (a – x) (i.e. the concentration of product or concentration of reactant) versus time will give the value of rate constant k (Fig. 1.2). t t x (a – x) Slope = k Slope = k Intercept = a Fig. 1.2 Concentration versus time plot for zero order reaction. The combination of H2 and Cl2 to form HCl in presence of sunlight is a zero order reaction H2 + Cl2 → 2HCl The rate of formation of HCl is not affected by a change in concentration of either the reactant or product. However, it is influenced by the intensity of sun light. Problem 1.5 A zero order reaction is 50% complete in 20 min. How much time will it take to complete 90%? Solution Let a = 100 mol dm–3. For a zero-order reaction k x t = = 50 (mol dm ) 20 60 (sec) –3 × When reaction is 90% completed, x = 90. Therefore, Thus, 50 1200 = 90 t or t = 90 1200 50 = 2160 sec = 36 min × Problem 1.6 A reaction is 50% complete in 20 min. How much time will be taken to complete 75% reaction? Solution For a zero order reaction k = x/t x = a/2 for 50% k a t a = 2 = 2 20 × x a = 3 4 for 75% reaction
12ChemicalKineticsandReactionDynamics3ak=Therefore,4ta3aor2 × 204t40 ×3 = 30 min1475%reactionwillcompletein30min.1.7FirstOrderReactionsLet us consider a first-order reactionA>Products0Initially,aAttimet,xa-xWeknowthat in case of afirst-orderreaction, therate ofreaction,dx/dtis directly proportional to the concentration of the reactant.Therefore,dx=k(a-x)=kdtordt(a-x)Integrating, we get In (a - x) = kt + z.The integration constant z is determined by puttingt =O and x =0.Thusz=lnaand.therefore,therateconstantforafirstorderreaction is obtained ask=llnaa-x2.303 1ogak=3or(1.21)1a-x:The units of rate constant for afirst order reaction from equation (1.21)ismeasured as (time)-' and can be represented as sec-l, min-' or hour-!..The half-lifeperiodfor afirst-order reaction maybe obtainedfromequation(b) by substituting t=tin when x=al2, i.e.2.30310g0k=a-al211/22.303log2or(1.22)t1/2 =kThus, the half-life period of a first-order reaction is independent of initialconcentration of reactant. Irrespective of how many times the initialconcentration of reactant changes, the half-life period will remain same
12 Chemical Kinetics and Reaction Dynamics Therefore, k a t = 3 4 or a a 2 20 t = 3 × 4 t = 40 3 4 = 30 min × 75% reaction will complete in 30 min. 1.7 First Order Reactions Let us consider a first-order reaction A ⎯ → ⎯⎯ Products Initially, a 0 At time t, a – x x We know that in case of a first-order reaction, the rate of reaction, dx/dt is directly proportional to the concentration of the reactant. Therefore, dx dt = ( – ) ka x or dx a x kdt (–) = Integrating, we get ln (a – x) = kt + z. The integration constant z is determined by putting t = 0 and x = 0. Thus z = ln a and, therefore, the rate constant for a first order reaction is obtained as k t a a x = 1 ln – or k t a a x = 2.303 log – (1.21) • The units of rate constant for a first order reaction from equation (1.21) is measured as (time)–1 and can be represented as sec–1, min–1 or hour–1. • The half-life period for a first-order reaction may be obtained from equation (b) by substituting t = t1/2 when x = a/2, i.e. k t a a a = 2.303 log 1/2 – /2 or t k 1/2 = 2.303 log2 (1.22) Thus, the half-life period of a first-order reaction is independent of initial concentration of reactant. Irrespective of how many times the initial concentration of reactant changes, the half-life period will remain same
Elementary13·Further,equation(1.21)canberearranged ask(1.23)log(a- x) = -t+loga2.303which suggests that a plot of log (a-x)versus time will give a straight linewitha negative slope (k/2.303)andan intercept log a (Fig. 1.3).(xD) B0Thus, in case of a first-orderSlope = k/2.303reaction a plot between log [conc.]logaandtimewill alwaysbelinearandwith the help of slope, the value ofrateconstantcanbeobtained.Fig. 1.3 The log [conc.] versus time plotExamplesforfirst-orderreaction.1. Inversion of cane sugar (sucrose)C12H2O11 +H,0→C,H12O+C,H1206D-glucosD-fructoThe reaction is pseudo-first order and rate is proportional to [Sucrose]. Theprogress of the reaction can be studied by measuring the change in specificrotation of a plane of polarised light by sucrose.Let ro,r, and rare therotation at initially (when t= O),at any timet and final rotation,respectively.The initial concentration a is proportional to (ro-r.)and concentration atanytime t,(a-x)is proportional to (ro-r).Thus,therateconstant maybeobtained as(ro - ro)k = 2.303 1og (1.24)t(ri-T)2.The hydrolysis of ester in presence of acidCH,COOC,H,+H,OHCH,COOH +C,H,OHRate=k [ester]Since oneof theproduct isacetic acid, theprogressof reaction maybestudied by titrating a known volume of reaction mixture against a standardalkali solution using phenolphthalein as indicator.Let Vo.V,and V.bethevolumes of alkali required for titrating 10 ml of reaction mixture at zerotime, at anytime t and at the completion of thereaction, respectively.V。=AmountofH+(catalyst)presentin10mlof reactionmix.V,=AmountofH*(catalyst)in10ml ofreactionsmix+AmountofCH,COOHformed at anytimet.V=Amount of H* (catalyst) present in 10 ml of reaction mix + Amount ofCH,COOHformed at theend of reaction (oramountof esterpresentinitiallybecause1mol of estergives1mol of CH,COOOH)
Elementary 13 • Further, equation (1.21) can be rearranged as log ( – ) = – 2.303 a x + log k t a (1.23) which suggests that a plot of log (a – x) versus time will give a straight line with a negative slope (k/2.303) and an intercept log a (Fig. 1.3). Thus, in case of a first-order reaction a plot between log [conc.] and time will always be linear and with the help of slope, the value of rate constant can be obtained. Examples 1. Inversion of cane sugar (sucrose) log (a – x) log a t Slope = k/2.303 Fig. 1.3 The log [conc.] versus time plot for first-order reaction. C H O + H O C H O + C H O 12 22 11 2 6 12 6 D-glucose 6 12 6 D-fructose → The reaction is pseudo-first order and rate is proportional to [Sucrose]. The progress of the reaction can be studied by measuring the change in specific rotation of a plane of polarised light by sucrose. Let r0, rt and r∞ are the rotation at initially (when t = 0), at any time t and final rotation, respectively. The initial concentration a is proportional to (r0 – r∞) and concentration at any time t, (a – x) is proportional to (r0 – rt). Thus, the rate constant may be obtained as k t r r r r t = 2.303 log ( – ) ( – ) 0 ∞ ∞ (1.24) 2. The hydrolysis of ester in presence of acid CH COOC H + H O CH COOH + C H OH 3 2 (–) 5 2 H 3 ( ) 2 5 + a x x ⎯ →⎯ Rate = k [ester] Since one of the product is acetic acid, the progress of reaction may be studied by titrating a known volume of reaction mixture against a standard alkali solution using phenolphthalein as indicator. Let V0, Vt and V∞ be the volumes of alkali required for titrating 10 ml of reaction mixture at zero time, at any time t and at the completion of the reaction, respectively. V0 = Amount of H+ (catalyst) present in 10 ml of reaction mix. Vt = Amount of H+ (catalyst) in 10 ml of reactions mix + Amount of CH3COOH formed at any time t. V∞ =Amount of H+ (catalyst) present in 10 ml of reaction mix + Amount of CH3COOH formed at the end of reaction (or amount of ester present initially because 1 mol of ester gives 1 mol of CH3COOOH)