Elementary29=2kt+11(1.53)(a- x)2a2Therefore, a plot of i/(a-x)?versus timewill give a straight linewithapositive intercept (l/a)and a slope (2k). Thus, the value of rate constantmaybyobtainedfrom the slope of thestraight line.冷Slope = 2k_D)B 1/a2TimeFig.1.6Third orderplot.(b)Where three reactants are involved with different initialconcentrationsLet us consider the reactioncAB++→ProductsInitially at timetbc0a(a-x)(b-x)(c-x)xThe rate expression is given bydx=k(a-x)(b-x)(c-x)dtThis equation can be integrated with help of partial fractions.The integratedrateequationisobtainedas2.303k=i(a-b)(b-c)(c-a)ba(b - c) 10g(a--) + (c - a) 10g (b--) + (a -b) log(c-x)(1.54)(c)Wheretworeactants are involvedand stoichiometry is 2:1Initially at timet2Axb+→ Productsboa(a-2x)x(b-x)The differential rate equation is given by=k(a -2x)2(b-x)dt
Elementary 29 1 ( – ) = 2 + 1 2 2 a x kt a (1.53) Therefore, a plot of 1/(a – x) 2 versus time will give a straight line with a positive intercept (1/a2 ) and a slope (2k). Thus, the value of rate constant may by obtained from the slope of the straight line. 1 ( – )2 a x Time Slope = 2k 1/a2 Fig. 1.6 Third order plot. (b) Where three reactants are involved with different initial concentrations Let us consider the reaction Initially at time t A (–) + B (–) + C (–) Products a 0 a x b b x c c x x → The rate expression is given by dx dt = ( – )( – )( – ) ka x b x c x This equation can be integrated with help of partial fractions. The integrated rate equation is obtained as k ta b b c c a = 2.303 ( – )( – )( – ) ( – ) log ( – ) + ( – ) log ( – ) + ( – ) log ( – ) b c a a x c a b b x a b c c x ⎛ ⎝ ⎞ ⎠ (1.54) (c) Where two reactants are involved and stoichiometry is 2:1 Initially at time t 2A ( –2 ) + B (–) + Products a 0 a x b b x x → The differential rate equation is given by dx dt = ( – 2 ) ( – ) ka x b x 2
30ChemicalKineticsandReactionDynamicsAfterrearrangingandintegrating,theintegratedformofequation is obtained asb(a-2x)2.303(1.55)kt:1oga(b-x)(2b-a)(a-2r)(2b-a)2(d) Where two reactants have same initial concentrations and third hasdifferentconcentrationInitiallyattimetBCA→ Products++aanx(a-x)(a-x)(c-x)The differential rate equation is given by=k(α -x)2(c -x)dtand integralformof equationfortherateconstantmaybeobtained asb(a-x)2.303111kt =(1.56)r(b-a)((a-x)a(b-x)a(a-b)2The differential rate equations, corresponding integral rate equations andrateconstantsforvarious reactions (havingorderzerotothree)underdifferentsets of conditionsaresummarized inTable1.1.1.11DeterminationofOrderofReactionThekineticinvestigationofareactioniscarriedouttoestablishtheratelawandtomeasuretheratecoefficients.Thefirst stepisto identify orexaminetherole of each component,i.e.to determinetheorder of reaction withrespectto each reactantorproduct.Therearevarious methods,whichcanbeusedtodeterminetheorderofreactionwithrespecttoareactant.However,everymethodrequiresessentiallythemeasurementofconcentrationofthereactant orproduct at various time intervals1.11.1IntegrationMethodInthis method,thedifferentrateequations intheirintegrated forms (giveninTableI)areused.Theamountofreactanta-xorproductxatdifferenttimeintervals t is first experimentally determined.Then the values of x, a -x andtime are introduced into the different rate equations and the value of rateconstant k is calculated at different time intervals.The equation which givestheconstantvalueof rateconstant indicates theorderofreaction.Forexample,thevalues of rateconstants at differenttimeintervalsare same inequationk=-2ta-xthe order of reaction will be 2.Thus, this is the method of trial and can beusedforsimplehomogenousreactions
30 Chemical Kinetics and Reaction Dynamics After rearranging and integrating, the integrated form of equation is obtained as kt b a a x a b a ba x ab x = 1 (2 – ) – 1 ( – 2 ) – 1 + 2.303 (2 – ) log ( – 2 ) ( – ) 2 ⎛ ⎝ ⎞ ⎠ ⎛ ⎝ ⎞ ⎠ (1.55) (d) Where two reactants have same initial concentrations and third has different concentration Initially at time t A (–) + B (–) + C (–) Products a a x a a x c c x x → The differential rate equation is given by dx dt = ( – ) ( – ) ka x c x 2 and integral form of equation for the rate constant may be obtained as kt tb a a x a a b ba x ab x = 1 ( – ) 1 ( – ) – 1 + 2.303 ( – ) log ( – ) ( – ) 2 ⎛ ⎝ ⎞ ⎠ ⎛ ⎝ ⎞ ⎠ (1.56) The differential rate equations, corresponding integral rate equations and rate constants for various reactions (having order zero to three) under different sets of conditions are summarized in Table 1.1. 1.11 Determination of Order of Reaction The kinetic investigation of a reaction is carried out to establish the rate law and to measure the rate coefficients. The first step is to identify or examine the role of each component, i.e. to determine the order of reaction with respect to each reactant or product. There are various methods, which can be used to determine the order of reaction with respect to a reactant. However, every method requires essentially the measurement of concentration of the reactant or product at various time intervals. 1.11.1 Integration Method In this method, the different rate equations in their integrated forms (given in Table 1) are used. The amount of reactant a – x or product x at different time intervals t is first experimentally determined. Then the values of x, a – x and time are introduced into the different rate equations and the value of rate constant k is calculated at different time intervals. The equation which gives the constant value of rate constant indicates the order of reaction. For example, the values of rate constants at different time intervals are same in equation k t ax a = 1 2 1 – – 1 { } the order of reaction will be 2. Thus, this is the method of trial and can be used for simple homogenous reactions
31Elementary(lo)o=aaa-10a=S00a=adorsD)eeeM1od o.ecOEY=osenorrsed(x p)o=ooso1=dos=ados1V+n(r=p)(x-D) B01(xq)p/(xD)q Bolz(r-D)/(X= D)/Ie"Pznlouupa-1ou-s.uPar-iour-sg-uplou-s,up,-1owo1-8tn"98s33S-层爱aCOIFroeaaa[or-D)-(r5a(x-q)EOE19-D)1-D)O0(x1C-XI2-121-13132/-----=Y1S191I1I=YBo(+qx-kEKkaaSuonent2r-1(x -)(X-(X-)Y=I1""IS0OS8专KK+A2opioe0-1YY7TY234181A3AA0A30A0
Elementary 31 Table 1.1 Differential rate equations, corresponding integral rate equations and rate constants for various reactions Reaction Order Differential Integral equation t1/2 Units of k Nature of plot equation A Product (a) k ⎯ →⎯⎯ 0 dxdt k = k x t = a k 2 mol dm–3s–1 Slope = k or Intercept = a A Product (a) k ⎯ →⎯⎯ 1/2 dxdt = k t = 2 0.586 1/2 k a mol1/2dm–3/2s–1 Slope = k/2 Intercept = a/2 k (a – x)1/2 [a1/2 – (a – x1/2] A Product (a) k ⎯ →⎯⎯ 1 dxdt = k t = 2.303 0.693 k s–1 Slope = k/2.303, Intercept = log a k (a – x) log ( – ) a a x A Product (a) k ⎯ →⎯⎯ 3/2 dxdt = k t = 2 0.828 1/2 ka mol–1/2dm3/2s–1 Slope = k/2, Intercept = 11/2 a k (a – x)3/2 1 ( – ) – 1 1/2 1/2 ax a ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ A + B Product (a) (a) k ⎯ →⎯ 2 dxdt = k t = 1 1ka mol–1/2dm3 s–1 Slope = k Intercept = 1 a k (a – x) 2 1 ( – ) – 1 a x a ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ A + B Product (a) (b) k ⎯ →⎯ 2 dxdt = k ta b = 2.303 ( – ) – mol–1dm3 s–1 Passing through Slope = k (a – x)(b + x) log ba x ab x ( – ) ( – ) the origin, ka b ( – ) 2.303 Time ) x – a(Time x Time ) x – a( 1/2 Time ) x–a log ( Time ) x – a 1/( Time b(a – x)/a(b – x) log Time ) x – a 1/( 1/2 (Contd.)
32ChemicalKinetics and ReactionDynamics-1E0t22%K1oid jo anenenor2ao=adoBused311z(x = D)/l(r-D))(x +) B01-s,upe-lowup-ou-,upe1oujosunS1-5051U(-)(0-9)T-80(0-)+10oenaeauE0EDBo1(qD)1al!EOE0-2-+03D)!2-121-u1e+=Y=yIB01 (r + 2)(rK(xq)(x-enuauojenbarR1II1130apo332个A4UoU3++on茗aBao+++0TAASM2
32 Chemical Kinetics and Reaction Dynamics Reaction Order Differential Integral equation t1/2 Units of k Nature of plot equation A + B Autocatalytic k → 2 dxdt = k ta c = 2.303 ( + ) – mol–1dm3 s–1 Passing through Slope = Product k (a – x)(c + x) log ac x ca x ( + ) ( – ) the origin, ka c ( + ) 2.303 A + B + C (a) (a) (a) k → 3 dxdt = k t = 1 2 3 2 2 ka mol–2dm6 s–1 Slope = 2k, Intercept = 1 2 a Product k(a – x) 3 1 ( – ) – 1 2 2 ax a ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ A + B + C (a) (b) (c) k ⎯ →⎯ 3 dxdt = k ta b b c c a = 2.303 ( – ) ( – ) ( – ) – mol–2dm6 s–1 Product b c a a x c a b b x a b c c x – ) log (–) + ( – ) log (–) + ( – ) log (–) ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ Time k (a – x)(b – x) (c – x) Time )/ x + c( a log ) x – a( c ) x – a 1/( 2
33Elementary-卡do1--(x- )/lsg,uPentou-s(-xwp(m)[ou(-w)py(I-u)1 (1-)21下(XZ-D)-18(X-4)(x-D))DD)301-(1-u)N10xx一14x220-(x-AI1SSSAn1zu一"+0aBEnB++.ASAG2
Elementary 33 2 A + B Product (a) (b) k ⎯ →⎯ 3 dxdt = 1 1 2–) 1 ( –2 ) – 1 + 2 – 303 (2 – ) log ( –2 ) (–) 2 t b a a x a b a ba x ab x × ⎧⎨⎩ ⎫⎬⎭ × ⎛⎜⎜⎜⎜⎜⎜⎜⎜⎝ ⎞⎟⎟⎟⎟⎟⎟⎟⎟⎠ – mol1/2dm–6s–1 k (a – 2x)2 (b – x) A + B + C + (a) (a) (a) . k → n ≥ 2 dxdt = 1 ( – 1) t n 2 – 1 ( – 1) ( –1) ( –1) n n n Ka mol–(n–1) Slope = (n – 1)k, Intercept = Product k (a – 2x)2 × 1 ( – ) – 1 –1 –1 ax a n n ⎛⎜⎝ ⎞⎟⎠ dm3(n–1)s–1 1 –1 a n ) x – a 1/( –1 n