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CHEMICAL KINETICS 92 NISHITH Multimedia India (Pvt.) Ltd., JEE ADVANCED - VOL - VI NISHITH Multimedia India (Pvt.) Ltd., SYNOPSIS * INTEGRATED RATE EQUATIONS To make the easier determination of instantaneous rate, differential rate equation is integrated to get a relation between directly measured experimental data and rate constant. * Zero Order Reactions : For zero order reaction the rate of reaction is independent of concentration of reactants. Let a zero order reaction is given as : A  Product d[A] 0 Rate = - = k[A] dt d[A] Rate = - = k dt d[A] = -k dt   dA = -k dt [A] = -kt + I where, I is the constant of integration.At t = 0, the concentration of the reactant [A] = [A]0 , where [A]0 is initial concentration of the reactant. Substituting in equation [A]0 = -k × 0 + I ; [A]0 = I Substituting the value of I in the equation [A] = -kt + [A]0 On solving 0 [A] - [A] k = t Comparing with equation of a straight line, y = mx + c, if we plot [A] against t, we get a straight line with slope = - k and intercept equal to [A]0 . * First Order Reactions Let us consider a first order reaction A  Products Suppose we start with ‘a’ moles per litre of the reactant A. After time t, suppose x moles per litre of it, have decomposed. Therefore, the concentration of A after time t becomes (a – x) moles per litre. Then according to Law of Mass Action: Rate of reaction  (a – x) i.e., dx dt  (a – x) or dx dt = k(a – x) where k is called the rate constant or the specific reaction rate for the reaction of the first order. dx a - x = kdt    dx k dt a - x or a k = a - x 1 ln t or a k = a - x 2.303 log t If the initial concentration of reactant A is [A]0 and the concentration after time t is [A] then putting a = [A]0 and (a – x) = [A], above equation becomes 0 [A] k = [A] 2.303 log t or 0 [A] kt = ln [A] which can be written in the exponential form as 0 kt -kt 0 [A] [A] = e [A] or = e [A] or [A] = [A]0 e-kt * Half-Life Period of a Reaction : The half-life period of a reaction is the time in which the concentration of a reactant is reduced to one half of its initial concentration. It is represented as t1/2. CHEMICAL KINETICS
NISHITH Multimedia India (Pvt.) Ltd., 93 JEE MAINS - CW - VOL - I JEE ADVANCED - VOL - VI CHEMICAL KINETICS NISHITH Multimedia India (Pvt.) Ltd., * Zero order reaction: For a zero order reaction, rate constant is given by equation 0 k = [A] - [A] t At t = t1/2, [A] = 0 1 [A] 2 The rate constant at t1/2 becomes 0 0 1/2 [A] -1/2[A] k = t 0 1/2 [A] t = 2k Thus for zero order reaction t1/2 [A]0 . * First order reaction: For a first order reaction rate constant is given by equation a x a log k 2.303 t   When half of the reaction is completed, x = a/2. The rate constant at t1/2 becomes t 1/2 = k 0.693 log 2 k 2.303 2 a a a log k 2.303    i .e, k 0.693 t1/ 2  Thus ‘a’ does not appear in this equation so that t 1/2 is independent of initial concentration a. Similarly, it can be seen that t1/2, t2/3 etc. will also be independent of the initial concentration. * n th Order reaction A  Product t = 0 a 0 t = t a – x x n (a x) dt dx   dx n = k(a - x) dt   1/2 a/2 t n 0 0 dx = kdt (a - x) Let a – x = z – dx = dz, when x7 = 0, z = a When x = a/2, z = a/2   1/2 a/2 t n a 0 -dz = kdt z or 1/ 2 a / 2 a n 1 kt n 1 Z                             1-n 1-n 1 - n 1/2 a 1 -1 =kt n -1 2 1/ 2 n 1 n 1 [2 1] kt a (n 1) 1      ,       n-1 n-1 1/2 1 (a ).t (2 -1) = k k(n -1) * Therefore for a nth order reaction, half life period in inversely related to initial amount t a constant n 1 1/ 2   ; t1/2 = 1 n n 1 1/ 2 constant or t constant a or a      1-n 1/2 t a METHODS FOR THE DETERMINATION OF ORDER: 1. Integration method: It is a hit and trial method to determine the order of a reaction. If this method the amount of reactants consumed after different time intervals are determined. Then these values, i.e, x and t are substituted in the first, second and third order rate equation. The equation which gives the most constant value of K corresponds the correct order of the reaction. (i) It is applicable for simple reactions but fails in case of complex reactions i.e. side or reversible reactions etc. (ii) In this method, it is necessary to study the reaction over a wide time intervals.
CHEMICAL KINETICS 94 NISHITH Multimedia India (Pvt.) Ltd., JEE ADVANCED - VOL - VI NISHITH Multimedia India (Pvt.) Ltd., 2. Method of fractional change : This method was emplyoed by Ostwald (1888) in determining the order of a reaction. It has already been pointed out that the time taken to complete a definite fraction of the reaction, is independent of initial concentration of the reactant in case of first order reaction. In case of second order reaction, it is inversely proportinal to the initial concentration and in case of third order reaction, it is inversely proportional ot the square of initial concentration. In general it may be said that the time (t) required to decompose the definite fraction of the reactants is inversely proportional to the initial concentration of reactants (a) raised to one power less than the order of the reaction. Thus 1 1 n t a   Where n is the orden of a reaction. At two time intervals:- 1 1 1 2 1 1 n n t and t a a     Then   1 1 2 2 1 ... 1 n t a t a         Taking logarithm of this equation, we get   1 2 2 1 1 t a ln n ln t a     2 1 1 2 1 a t n ln ln a t   n ln a ln a ln t ln t     1 2 1 1 2  1 2 1 2 2 1 2 1 1 1 ln t ln t ln t ln t n n ln a ln a ln a ln a          3. Graphical method: Here the order is determined by plotting graphs. We have different values of x at various time intervals t. Then by plotting x and t the dx/dt may be determined. Thus dx dt / tan  as evident from the figure (1) Now it is clear that in the reaction of first order, the rate is determind by the variation of only one concentration term. In second order the rate is determined by two concentration terms and so on.Thus dx dt d a x for first order reaction /          2 dx dt d a x for order reaction / second       3 dx dt d a x for third order reaction /   It is evident from the above three equations that
NISHITH Multimedia India (Pvt.) Ltd., 95 JEE MAINS - CW - VOL - I JEE ADVANCED - VOL - VI CHEMICAL KINETICS NISHITH Multimedia India (Pvt.) Ltd., the plot of graph between dx/dt aginst (a-x) will be a straight line in the case of first order reaction. Thus if a straight line is obtained by poltting dx/ dt against (a-x) then the order of the reaction is one. If a stright line is obtained by poltting dx/dt against   2 a x  then the reaction is of second order. If dx/dt agomts   3 a x  is a staight line then the order of the reaction will be three. 4. Van’t Hoff’s differential method: Thus method, as suggested by van’t Hoff (1884), is based on the fact that the rate of the reaction of nth order is proportional to nth power of the concentration, i.e , dc dc n n c or kc dt dt     c beings the initial concentration and n is the order, For two different concentrations 1 2 1 2 dc dc n n n kc and kc c dt dt      Where 1 2 1 2 / n dc dc c dt dt c          By taking logarithms we have   1 2 1 2 dc dc ln ln n ln c ln c dt dt                    1 2 1 2 or n ..... 3 dc dc ln ln dt dt ln c ln c                  Hence the order may be determined by eqution (3), For more accurate estimate of dc/dt, the various values of c and t are poltted. The slope of the curve obtained gives the value of dc/dt i.e. dc/dt = tan L.T. Reicher (1895)used an approximate method to determine the value of dc/dt. In his work the rate of change in c over an appreciable time interval t. i.e   c t / is determined and it is taken equal to dc/dt. * Some Examples of the First Order Reactions Decomposition of N2O5 : It follows first order kinetics. The compound, N2O5 , is a volatile solid which decomposes in the gaseous state as well as in the form of its solution in an inert solvent like carbon tetrachloride, chloroform etc. according to the equation: N2O5  N2O4 + 2 1 O 2  2NO2 When the reaction is carried out in the solution, N2O4 and NO2 remain in the solution and the volume of oxygen gas collected is noted at different intervals of time. It is obvious that Volume of oxygen gas collected at any time (Vt )  Amount of N2O5 decomposed (x) i.e., x  Vt Volume of oxygen gas collected at infinite time (V )  Amount of N2O5 initially taken (a) i.e., a  V  Substituting these values in the first order equation. a x a log t 2.303 k   V Vt V log t 2.303 k     The constancy in the value of k for given set of experimental data proves the reaction tobe first order. Example : Decomposition of benzene diazonium chloride was studied at a constant temperature by measuring the volume of N2 gas evolved at different inter- vals of time. Using the following data show that the reaction is of the first order Time in minutes: 0 20 50 70  Vol. of N2 measured in mL: 0 10 25 33 162 Solution: When=20,   162 10 152 V V      t (i) 2.303 162 3 1 log 3.22 10 min 20 162 10 K       (ii) 2.303 162 3 1 log 3.36 10 min 50 162 25 K       (iii) 2.303 162 3 1 log 3.26 10 min 70 162 33 K      