Chemical Kinetics

Chemical Kinetics (b) Moderate reaction : This type of reactions proceed with a measurable rates at normal Chemical Kinetics : Chemical kinetics is the study temperature. In this a large number of bonds have of rates of reactions, the mechanism by which the to be broken in reactants molecules and a large reactions proceed and the effect of temperature, number of new bonds have to be formed in product pressure and concentration on the rates of reactions. molecules. On the basis of rate, chemical reaction are broadly * Mostly these reactions are molecular in nature. divided into three categories :Examples : (a) Very fast or instantaneous reactions : * Decomposition of H2O2 These reactions occur at a very fast rate. Generally 2H2O2  2H2O + O2 these reactions involve ionic species and known as ionic reactions. These reactions take about 1014 or 1016 * Decomposition of N2O5 2N2O5  2N2O4 + O2 seconds for completion. Its rate can be measured by employing special methods. So, It is almost * Hydrolysis of ester INTRODUCTION : impossible to determine the rates of these reactions. CH 3 COOC 2H 5 + NaOH  CH 3 COONa + C2H5OH Examples : AgNO3 + NaCl BaCl2 + H2SO4 HCl + NaOH acid base     AgCl + NaNO3 * Inversion of cane sugar in aqueous solution ppt. * Reaction of NO with chlorine NO + Cl2 BaSO4 + 2HCl  NOCl2 ppt. * NaCl + H2O (c) Very slow reaction : These reaction proceed very slowly. They may take months together to show any measurable change at room temperature. salt Moderate reactions are studied in chemical kinetics. 33 Chemical Kinetics Figure : The change in the concentrations of the reactants and products. Examples : * Rusting of iron * Reaction between H2 and O2 to form H2O at ordinary temperature in absence of catalyst. at room temperature CO + 2H2    CH3OH * * * The chemical reactions can be slowed down or speeded up by changing conditions under which they occur. For a reaction For example :  [A ]  [A ]1 R   2  t 2  t1 CO + 2H2  CH3OH The reaction can be speeded up by maintaining temperature around 4000C, pressure about 300 atm and using a catalyst containing ZnO and Cr2O3. RATE OF REACTION : Where time t1, AB  [A ]    t  [A]1 = Concentration of reactant A at [A]2 = Concentration of reactant A at time t2. * Rate of reaction defined as the change in concentration of reactant or product per unit time. It is always a positive quantity. Instantaneous Rate of Reaction : The rate of reaction at any particular instant during the course of reaction is called as instantaneous rate of reaction. Mathematically ; Instantaneous rate = (Averge rate)T  Rate of reaction = Average Rate of Reaction : The rate of reaction over a certain measurable period of time during the course of reaction, is called as average rate of reaction. It is denoted by R. Total change in concentration of reac tan t or product Change in time (in sec ond) v  [A]   [B]  R t       t  t 0  t  t 0 dc dt Where dc = change in concentration in a small interval dt or [+] sign is used when we refer to product concentration. Hence, Slope of the tangent at any point of curve indicates instantaneous rate of reaction. [–] sign is used when we refer to reactant concentration. * Rt   d[A ] d[B ]  dt dt The slope of the tangent at time t in plot of concentration with times gives rate at time t. Rate of reaction = tan  = dx dt (a) For an irreversible reaction * Initial Rate : The slope of the tangent at t = 0 in the plot of concentration vs time gives the initial rate. (b) For an reversible reaction Rate of reaction = tan  = dx dt 34 Chemical Kinetics Note : (m1 +m2+m3) = n(order of reaction) mole  mole  K  litre second  litre  * Rate of reaction in the form of stoichiometry of a chemical reaction : Let us consider a reaction : m1A + m2B + -----... n1P + n2Q + n3R + --------1 d[A ] 1 d[B ] 1 d[P ] 1 d[Q ] Rate of reaction      m 1 dt m 2 dt n 1 dt n 2 dt * Unit of Reaction Rate : Unit of reaction rate = mol L–1 time–1 i.e. (mol L–1 s–1 or mol L–1 min–1 or mol L–1 h–1) Note : For a chemical reaction A + B ––– P + Q Rate of disappearance of A = – Rate of disappearance of B = – Rate of formation of P = Rate of formation of Q = d[P ] dt d[Q ] dt second 1 The rate constant is constant for a particular reaction at a particular temperature and does not depend upon the concentration of the reactants. ORDER OF REACTION : The order of reaction may be defined as the sum of powers of concentration (or pressure) terms involved in rate low expression. i.e., number of molecules whose concentration determined the rate of the reaction is called order of the reaction. Rate Law : Reaction is : 2NO2 + F2  2NO2F (nitrogen fluoride) Rate Law : the expected rate expression is According to the law of mass action, the rate of a chemical reaction is directly proportional to active masses of reactants. rate = K[NO2]2 [F2] where K is proportionaly constant and is called velocity constant or rate constant or specific reaction rate. For general reaction : m1A + m2B + m3C  products m1 1 n Example : Rate of reaction  [A] [B] = K [A] [B] m2 The rate of reaction = K A  B  C  m3 but it is found experimentally that the rate of this reaction is proportional to the product of single conce ntration term of NO2 and F2. Thus the experimental rate of the reaction is given by : rate = K[NO2] [F2] Rate controlling step : The slowest step of the complex reaction is called the rate controlling step. Example : We have a reaction 2NO2 + F2  2NO2F The mechanism of this reaction is as follows – NO2 + F2  NO2F + F (slow) Unit of rate constant : NO2 + F  NO2F (fast) concentration m m  m  K concentration  1 2 3 time ..................................................................... m m r  K A  1 B  2 C  *  mole  K   litre  The mathematical expression that gives the true rate of reaction in terms of concentration of the reactants, which actually influence the rate, is called rate law. d[A ] dt d [B ] dt A + B  products * * m1  m 2  m 3 m3 2NO2 + F2  2NO2F (overall reaction) 35 * Order is a purely experimental quantity. It may be * a whole number or zero or even a fraction. * For the reaction : aA + dB  product reaction. * Order with respect to B = n The molecularity of a reaction cannot always be predicted from the stoichiometric reaction. It is total order = m + n obtained by compounding various elementry steps. * m and n may be equal to a and b. * Reactions are classified according to the order of reaction. Example : 2N2 O5  4NO2 + O2 r = K [N2O5] 2H2O2  2H2O + O2 First step H2O2   H2O + O(slow step) (a) A reaction is said to be of first order when the one concentration term determine the rate of reaction : Example : unimolecular, bimolecular, termolecular depending upon the number of molecules involved in the Experimental rate equation : r = K [A]m [B]n Order with respect to A = m, Chemical Kinetics Reactions are classified in terms of molecularity as Second step O + O  O2 (Fast step) * The rate determining step indicates that it involves only one molecule. So the molecularity is 1. (b) A reaction is said to be of second order when two (a) Unimolecular reaction : concentration terms determine the rate of reaction.   PCl3 + Cl2 PCl5  Example : 2N2O  2N2 + O2   2NO2 N2O4  2 r = K [N2O] 1 H2O2   H2O + O2 CH COO C H + NaOH  CH COONa + 3 2 5 2 3 C2H5OH (b) Bimolecular reaction : r = K [CH3COOC2H5] [NaOH] (c) A reaction is said to be of third order if three concentration terms determine its rate. Example : 2FeCl3 + SnCl2  2FeCl2 + SnCl4 CH3COO C2H5 + NaOH    CH3 COONa + C2H5OH Two molecules are involved in the reaction hence r = K [FeCl3]2 [SnCl4] The order of following reaction is fraction :- the molecularity of the reaction is 2. H2(g) + Br2(g)  2HBr(g) 2HI   H2 + I2 (molecularity is 2) r  K H 2  Br2  1 2  order = 1 + 1 2 = 3 2 (c) Trimolecular reaction : CHCl3 + Cl2  CCl4 + HCl 2FeCl3 + SnCl2  2FeCl2 + SnCl4 1 1 3 = r  K CHCl3  Cl2  2  order = 1 + 2 2 The molecularity of the reaction is 3 since three MOLECULARITY : Molecularity is defined as the number of molecules atoms or radicals that must collide simultaneously in order for the reaction to take place. * Saponification of ethyl acetate It is always a whole number and cannot be negative and fraction. molecules are involved in the reaction while the order of reaction is 2. 2NO + O2   2NO2 molecularity  3 H2O2 + 2H+ + 2I–  I2 + 2H2O molecularity 3 36 Chemical Kinetics Order of reaction Molecularity of reaction (1) It is the sum of the powers of concentration in terms of the reactants in the rate equation. (1) It is the number of chemical species taking part in the rate determining step. (2) It is empirical and must be determined experimentally. (2) It is concerned with the reaction mechanism and is a theoretical concept obtained from the balanced single step reaction (3) It is a whole number, zero or fraction (3) Molecularity is a whole number and never zero (4) It is same for the whole reaction no matter the reaction is simple or complex. (4) It represents a multistep reaction and is expressed for each step. (5) It refers to a reaction as a whole irrespective number of involved steps in the reaction (5) It depends upon the rate determining step in the reaction PSEUDO-UNIMOLECULAR REACTION : Such reactions whose order is one but the molecularity is two, are called pseudo-unimolecular reaction. Examples : H (a) CH 3COOC 2H 5 + H 2O   CH 3COOH + C2H5OH H (b) C12H22O11 + H 2O   C6H12O6 + C6H12O6 (c) C6H5N2Cl + H2O  C6H5OH + N2 + HCl (ii) Differential rate equation : A  Product – d[A ] dt dx  k 0 [A ]0 dt = k0[A]0 or Where x is molar concentration of product at time 't'. (iii) Calculation of rate constant : Let us take the reaction A  Product – d[A ] dt = k[A]0 = K t –  d[A]   k dt 0 – [A]t = kt + C (d) R–X + H2O  ROH + HX DIFFERENT ORDER OF REACTION (A) Zero Order Reaction : Reaction in which the concentration of reactant do not change with time are said to be zero order reaction. (i) Reaction : A  Product Example : hv (a) H2(g) + Cl2(g)   2HCl(g) hv (b) 2NH3(g)   N2(g) + 3H2(g) (c) Reaction between Acetone and Bromine. (d) Dissociation of HI on gold surface. (e) Decomposition of gas on surface of catalyst [A]t = [A]0 on t = 0 Put the value of C – [A]t = kt – [A]0 C = – [A]0 – [A]t = [A]0 – kt Note : For zero order reaction that rate of reaction is equal to rate constant Time for completion of reaction [A]t = [A]0 – kt For complete [A]t = 0 k= [A ]0 t 37 A t=0 After time t dx dt  a a–x Product 0 x = k(a – x)0 = k x = kt x = Amount of reactant that will change in product. (iv) Unit of rate constant : k = mol L–1 s–1 = unit of rate of reaction. (v) Half-life period : The half life period for a zero order reaction is directly proportional to the initial concentration of the reactants. x= a 2 x = kt t1/2 =  a  2k a 2 Chemical Kinetics (iii) Calculation of Rate Constant : Let us take the reaction. A  Product t=0 a 0 After time t (a – x) x Here, 'a' be the concentration of A at the start and (a – x) is the concentration of A after time taken t. i.e. x part of A has been changed in the product. So, the rate of reaction after time t is equal to. dx  (a – x) dt dx or (a  x) = k dx dt or = k (a – x) dt Where k is specific reaction rate for the first order reaction, 'a' is initial concentration of A and x moles/litre of A has been decomposed up to the time t. = kt t1/2  A Upon integration of above equation (vi) Graphical representation : dx  (a  x ) =  k dt or – n (a – x) = kt + C C = – n a x = 0 on t = 0 Put the value of 'C'  a (B) First Order Reaction : Reaction in which that rate of reaction depends only on one concentration term of reactant. Reaction : A  Product. (i) Example : (a) 2N2O5  4NO2 + O2 (b) NH4NO2  N2 + 2H2O  a  or k = 2 . 303 t log  a  x    kt = n a ax – kt = n Ct C0 = e–kt (a  x) a ax a = e–kt Ct = C0 e–kt (Willhalpi equation) Here C0  Concentration on time t = 0, Ct  Concentration on time t (iv) Unit of rate constant : k = second–1 (c) CH3COOC2H5  CH3COOH + C2H5OH (d) 2Cl2O7  2Cl2 + 7O2 (ii) Differential rate equation : – d[A ] = dt k[A] or dx dt = k(a – x)  or kt = n  a  x    – n (a – x) = kt – n a kt = n a ax kt = na – n(a – x) 38 (v) Half-life period : Putting a – x = a/2 and t = t1/2 in the integrated rate equation we get 1  a  t1/2 = k  n  a / 2    or t1 / 2  or t1/2 = n 2 k 0 . 693 k Half-life period for first order reaction is independent from the concentration of reactant. (vi) Time for 3/4 of the reaction (t3/4) Putting a – x = a 4 and t = t3/4 in the integrated rate equation we get t3/4 = 2 . 303 k log 4 = 1 . 386 k where x1 and x2 be the amounts decomposed up to the time t1 and t2 respectively. Sp e i c a l Note : Ti m e req ui re d f or t he completion of definite fraction of the first order reaction is independent of the initial concentration (viii) Graphical representation : (a) CH3COOC2H5 + H2O  CH3COOH + C6H12O6 k v  v0 2 . 303 log t v  vt where V0, V1 and V are the volumes of NaOH solution used for titrating a definite volume of the reaction mixture of zero time t and after time infinite respectively. H (b) C 12 H 22 O 11 + H 2 O   C 6 H 12 O 6 + C6H12O6 ; k t3 / 4 2 t1 / 2 2 . 303 a  x1 (vii) Interval formula : k  t  t log a  x 2 1 2 of the reactant. Chemical Kinetics A few examples of 1st order and their formula for rate constant are as under : r0  r  2 . 303 log t rt  r  where r0, rt and r are the polarimatric readings at zero time, after time t and after time infinitely respectively. (c) 2N 2 O 5  4NO 2 + O 2 and NH 4NO 2  2H2O + N2 k v 2 . 303 log t v  v t where Vt and V are the volume of O2 or N2 gas collected after time t and after time infinite respectively. (d) H2O2  H2O + k 1 2 O2 2 .303 v log 0 t vt where V0 and Vt are the volumes of KMnO4 solution used for titrating a definite volume of the reaction mixture in the begining (at t = 0) and at time t respectively. (C) Second Order Reaction : Reaction whose rate depends on two concentration terms of reactant. Reactions (i) 2A  Product (ii) A + B  Product 39 (i) Example : (a) CH3COOCH3 + NaOH  CH3COONa + CH3OH (b) 2CH3CHO  2CH4 + 2CO (c) 2NO2  2NO + O2 (d) S2O 26  + 2I–  2SO 23  + I2 (e) CH3COOH + C2H5OH  CH3COOC2H5 + H2O (ii) Differential rate equation : (a) When A and B have equal initial concentration [say 'a' mole/litre] OH ¯ dx dt = k (a – x)2 (b) When A & B have two different initial concentration (say 'a' & 'b' mole/litre) dx dt = k[A] (v) Half-life period (t1/2) : Putting a – x = a/2 and t = t1/2 in the integrated rate equation, we get a/ 2 1 1 t1/2 = k  a (a  a / 2 ) t1/2 = or t1 / 2  a ka Half-life of second order reaction is depend upon the concentraion of the reactant. (vi) Time for 3/4th of the reaction (t3/4) : Putting x = 3 4 a, a – x = a 4 and t = t 3/4 in integrated rate equation. = k(a – x) (b – x) (iii) Calculation of Rate Constant : (a) When concentration of A and B taking same. A + B  Product t=0 a a 0 t=t (a – x) (a – x) x dx dt dx dt Chemical Kinetics (iv) Unit of rate constant : k = mol–1 L s–1 [B] = k[a – x] [a – x] we get t3/4 = 3 ka or t3 / 4  1 a Note : Time required for the completion of definite fraction of the second order reaction is inversely proportional to the first power of initial concentration of reactant. (vii) Graphical representation : = k(a – x)2 On taking the integration of above equation. dx  (a  x)   k dt or 2 at t = 0 ;x=0 1 = kt + C, (a  x ) C=  1 a On putting the value of C on above equation 1 1 1 x  k t = (a  x) – or k  t  a(a  x)    a (b) When concentration of A and B are different Case I : a > b Then k 2.303  b(a  x)  log   t(a  b)  a(b  x)  Case II : b > a Then k 2 . 303  a (b  x )  log  t (b  a )  b (a  x )  (D) Third Order Reaction : Reaction in which rate of reaction is determined by three concentration term of reactant. Reaction : (i) A + B + C   Product (ii) A + 2B   Product (iii) 2A + B   Product (iv) 3A   Product 40 Chemical Kinetics (i) Example : (a) Reactio of nitiric oxide and chlorine 2NO + Cl2   2NOCl (b) Reaction of nitric oxide and oxygen 2NO + O2   2NO2 (c) 2FeCl3 + SnCl2   2FeCl2 + SnCl4 (ii) Differential rate equation : When the initial concentration of all the three reactants is same. dx dt = k (a – x)3 DETERMINATION OF ORDER OF REACTION : (a) Integration Method : In this method known quantities of the reactants are mixed and the reaction rate is determined by analysing the reaction mixture after different intervales of time. The values so obtained are subsituted, in the kinetic equations of first, second and third order . The equation which gives the most constant value for the K for a series of time intervales is the one corresponding to the order of reaction : K1  (iii) Integrated rate equation : 1 1 1 1 x(2 a  x ) k = t. 2 a 2 (a  x )2 (iv) Half-life period : Putting a – x = a 2 and t = t1/2 in the integrated rate equation, we get t1/2 = 2 ka 2 or 1 1   1 1    2 K2 = t  a  a  x  ; K3 = 2t  a  x 2 a      * This method can be used only for simpler reactions. (b) Half life Method : * General expression for the half life period for a reaction of nth order is : kt = 2 (a  x )2 – 2a 2 3 2.30 3 a log t ax t1 / 2 1  2 a Note : For a third order reaction, time required for the completion of definite fraction of the reaction is inversely proportional to the square of the initial concentration of the reactant. (v) Unit of rate constant : k = mol– 2 L2 s–1 (vi) Graphical representation : t1  2 2 n 1  1 K n a n 1 n  1 For first order For second order For third order In general   t1  2   t1  2 n 1 t1/2  a0 n=2 n=3 t1/2  1/an–1 t1/2  1/a t1/2  1/a2   n 1 a  1  2   a1   2 (c) Graphical Method :- If a straight line is obtained by plotting log (ax) against dx , it is a first order dt reaction. Similarly, if a straight line is obtained by plotting (ax)2 or (ax)3 against dx dt , the reactions are second and third order respectively. (d) The order of reaction can also be determined by Van't Hoff differential method : dx   dx  log10  1   log10  2   dt   dt  n log10 x1  log10 x 2 Here x1 and x2 different initial concentrations. 41 THEORY OF REACTION RATES : Important theories to explain reaction rates are : (a) Collision theory * Chemical Kinetics The energy changes during exothermic and endothermic reactions versus the process of reaction are shown below - (b) Transition state or absolute reaction theory. (a) Collision Theory : The important points of this theory are : (i) If two molecules are to react together they must collide with each other. (ii) The collision between all molecules do not lead to chemical reaction. Factors affecting Reaction rates : (iii) Only those collisions result in chemical reaction in (i) Nature of reactants : which the molecules acquire energy greater than the In reactions involving higher energy barrier, the activation energy. reactant molecules require higher energy of activation * The number of collisions taking place per second to cross the barrier and get converted into prodper unit volume of the reaction mixture is known ucts. A smaller number of molecules possesses this as collision frequency and its value is very high, of higher amount of energy hence the rate of reaction the order of 1025 to 1028 in case of binary collisions. is slow. In a reactions with lower energy barrier, the (iv) The colliding molecules must posses certain minimum reactant molecules require lesser amount of energy (threshold energy) to make the collision activation energy to pass the barrier. Hence large effective. number of molecules possess, this energy and the rate of reaction will be fast. (v) The additional energy required by the molecule to attain the threshold energy is called activation (ii) Concentration of reactants : energy. It is acquired by the molecules as a result The rate of reaction increases with the increase in of interchange of energies during collisions. concentration of reactant (except zero order reactions) In the reaction A + B AB, if the con* Thus, activation energy = Threshold energy. – centration of A and B are doubled the rate of Energy of colliding molecules. reaction will increase to 4 times. (b) Transition state theory : The important points are: (iii) Pressure : (i) The theory is based on the fact that bond formation For reaction involving gases an increase in pressure and bond dissociation involved in a chemical reaction increases the rate of reaction by increasing the probmust occur continuously. ability of collision among the molecules. (ii) For a reaction to occur, the reactants having (iv) Temperature : sufficient energy must approach each other to form Increase in temperature in most cases increases the activated complex. rate of the reaction and the rate constant. (iii) The energy of activated complex is higher than that The temperature coefficient of a chemical reaction of reactants and products. is defined as the ratio of the specific reaction rates (iv) The reactants must cross the energy barrier of a reaction at two different temperature differing (activated complex formation) before converting by 10oC. into products. K t  10 Temperature coefficient = K (v) The energy barrier has energy equal to the energy t of activation. Its value lies generally between 2 and 3. 42 When the temperature is increased, the kinetic 3 * * * *  energy  2 RT  of the reacting molecules increase.   This shall increase the number of collisions and the rate of reaction shall increase. * If temperature is raised from 10to n times then the rate of reaction will increase by 2n–1 times. If temperature is raised from 0° to n times then the * rate of reaction will increase by 2n times. Arrhenius suggested an equation which describes * K as a function of temperature K  Ae  Ea RT * Where K = rate constant A = a constant (Frequency factor) Ea = activation energy At two temperature T1 and T 2 taking log of Arrhenius equation log e K1  log A  log e K2  log A  for every reaction, a certain amount of energy is required. Higher the intensity of light greater will be the rate of reaction. For these reactions, the value of G may be positive or negative. Temperature have no significant effect. Mechanism : Cl2 + hCl2* (activated molecule) Primary step of reaction : Cl2*  Cl + Cl Secondary step of reaction : H2 + Cl  HCl + H Ea log e e RT1 Cl2 + H  HCl + Cl Ea loge e RT2 Quantum yield (or quantum efficiency) : Final step of reaction :  1 1  So, log e K1  log e K2  E a  RT  RT   2 1  log Chemical Kinetics In photochemical reaction, each molecule of the reacting substance absorbs one quantum (single photon) of radiation and is activated to form the product. K1 E a  T1  T2     K2 2 .3 03R  T1 T2  Cl + Cl  Cl2 The quantum efficiency yield (of a photochemical reaction may be expressed as  Number of molecules reacted or product formed Number of photon absorbed Application of photochemical reaction : (v) Catalyst : A catalyst is a substance which lowers the energy of activation of a reaction. The magnitude of energy barrier is reduced in presence of catalyst. Hence a greater number of reactant molecules can get over the energy barrier and as a result rate of reaction increases. (i) in photography PHOTO CHEMICAL REACTION : Reactions which occurs in presence of light are photochemical reaction i.e., reaction which do not occur in dark. Mechanism of Reactions : The most of the reactions occur in a series of steps, some of which may be quite fast while other may be slower. Example : h (i) H2 + Cl2   2HCl (ii) Photo synthesis (iii) Vision (ii) in photo printing (iii) in photosynthesis (iv) in vision (v) in electronic industries (vi) in many polymerization reactions. Vant Hoff proposed that whole kinetics of the reaction depends upon the rate deter mining step (the slowest step of a reaction is taken as the rate determining step) 43 Chemical Kinetics Examples : SPECIAL POINTS (i) Reaction between NO and H2 : 2NO + 2H2   N2 + 2H2O (1) Chemical kinetics predicts the rate and mechanism of chemical reaction. This reaction involves 4 molecules but it has found (2) Rate of reaction is the rate of change of to be third order because it occurs in the following concentration of reactants and products with time. steps: (3) Rate of reaction depends upon nature of reactants, (a) 2NO + H2  N2 + H2O2 (slow) concentration, temperature, catalyst, surface area (b) H2O2 + H2  2H2O (fast) and radiation. Thus (a) being slow is the rate determining step. The rate is given by: Rate = K[NO]2 [H2] (ii) Reaction between FeCl3 and KI : 2FeCl3 + 6KI   2FeI2 + 6KCl + I2 (4) Molecularity of a reaction is the total number of reacting molecules taking part in rate determining step. This reaction involves 8 molecules, but it has been (5) Order of reaction may be determined by integration, found to be of third order graphics or half period method. (a) FeCl3 + 2KI  FeI2 + 2KCl + Cl (slow) (6) First order reaction is one in which the rate is (b) 2KI + 2Cl   2KCl + I2 (fast) directly proportional to the concentration of reacting Rate = K [FeCl3] [KI]2 substance. This indicates that the reaction is of third order. (7) The additional energy required to attain threshold (iii) Reaction between H2 and I2: energy is called activation energy.  H2 + I2   2HI (8) Activated complex is the unstable complex formed (a) I2   2I (fast) in the transition state with effective collision. (b) H2 + I  H2I (fast) (9) The minimum energy which the molecules should (c) H2I + I  2HI (slow) possess so that their collisions result in chemical (iv) Combination of NO2 and F2 : reaction is called threshold energy. 2NO2 + F2  2NO2F (10) According to Arrhenius (a) NO + F   NO F + F (slow) 2 2 2 (b) NO2 + F  NO2 F (fast) Thus order = 2 rate of reaction  K  Ae  Ea RT 44