Reaction kinetics

1. Reaction kinetics

2. R1. The rate of reaction Consider the following reaction (R1) Aand B are the reactants, C and D are the products, the (greek)n-s are the stoichiometric coefficients. For the definition of the rate of consumption for reactant and rate for formation for product we take the derivative of the number of moles with respect to time,

3. dni/dt (amount of substance converted in unit time) decreases for reactants. (R2a) For reactants like (R2b) For products like A simple reaction without parallel reactions and further (consecutive) reaction of the products has the form (S maybe product or reactant): (R3) According to this equation the stochiometic coefficients of reactants are defined negative, those of products are defined positive.

4. The quantities defined in (R2a) and (R2b)depend on volume, as well. Therefore in the definition of rate of reaction (v) we divide these type expressions both by volume and by the stoichiometric coefficient: [mol dm-3 s-1] (R4) The values of the stoichiometric coefficients influence the measure of the conversion. Since the signs of dni/dt and are always the same, v is always positive.

5. The rate of reaction can be expressed with the help of any component: (R5) This expression is unambigous, as mentioned, if - there is strict connection among the stoichiometric coefficients, - there are no side or parallel reactions, - the products do not react further.

6. In case ofparallelandconsecutive reactions(see later) we can only define the concentration changes of the different components separately. For parallel reactions (supposing volume V=constant) (R6) For consecutive (further) reactions, supposing the volume as constant (R7) In these cases a different rate of reaction is defined for each component.

7. The rate of reaction can be expressed with • amount of substance • concentration • extent of reaction • conversion The following formulas are all for reactant A (nA is negative), a)Amount of substance (R8)

8. b) Concentration (R9) As it was already mentioned, v is positive because both nAand d[A]/dt are negative (concentration of reactant decreases in time).

9. (R10) c) Extent of reaction The extent of reaction (x) is also defined as a positive quantity. Its unit is mol (both the numinator and the denominator are negative). (R11)

10. d)Conversion The conversion of a reactant shows what fraction of that reactant has been converted(0wA1). If the reactants are not in stoichiometric ratio, the conversion is different for the different reactants. Remember the definition ofv (R4)! (R12) The change of reactant conversion (DwA) is positive!

11. R2. The concept of order of reaction It has been found that the rate of most of the reactions depend on the concentrations of the reactants.Usually the rate can be expressed in a power equation: (R13) Such kind equations express the rate laws, they arerate equations k is called the rate constant, n is the order of reaction. (R14) First order reaction: n=1; E.g. decomposition of sulfuryl chloride: SO2Cl2→SO2 +Cl2

12. The order of reaction is not necessarily equal to the stoichiometric coefficient.E.g. the decomposition of nitrogen pentoxide: 2N2O5→4NO2 +O2is found to be a first order reaction. Explanation: the reaction takes place in two steps: N2O5→ N2O3 +O2 - slow N2O3 + N2O5 → 4NO2 - fast The first slow step determines the overall reaction rate. (R15a) (R15b)

13. Consider the following reaction with two reactants: nAA + nBB → Products The following rate equation can be set up: (R16) Note that the rate equation is not always normalized with the stoichiometric coefficient. Here n is the order with respect to A, m is the order with respect to B and n+mis the overall reaction order. The following rate equation shows a reaction which is first order with respect to A, first order with respect to B and second order overall. (R17)

14. Second order reaction with respect to one component: (R18) E.g. the thermal decomposition of nitrogen dioxide: 2NO2 → 2NO + O2 The rate equation: Or if the rate isnormalized with the stoichiometric coefficient: (The value of the rate constant differs by a factor of 2.)

15. The order of reaction is not always an integer. E.g. the decomposition of acetaldehyde: CH3CHO → CH4 + CO The rate equation: In this case the order of reaction is 3/2. Thefraction value of the reaction orderrefers that thereaction proceeds in several steps or parallel reactions run.

16. The unit of rate constant depends on the order of reaction. First order [k] = s-1 Second order [k] = dm3mol-1s-1 n order [k] = (time)-1(concentration)1-n

17. R3.Reaction molecularity and reaction mechanism Many reactions proceed through a number of steps from initial reactants to final products. Each of the individual steps is called anelementary reaction. The molecularity indicates how many molecules of reactants are involved in the elementary reaction. Unimolecular e.g. SO2Cl2 → SO2 + Cl2Bimolecular e.g. NO + O3 → NO2 + O2Trimolecular reactions are very rare. The molecularity is integer, 1, 2, (higher are very rare).

18. Reaction mechanism On the one hand reaction mechanism means the sequence of elementary reactions that gives the overall chemical change. On the other hand reaction mechanism means the detailed analysis of how chemical bonds in the reactants rearrange to form the activated complex. An activated complex is an intermediate state that is formed during the conversion of reactants into products. 1. example:decomposition of ozone 2O3→ 3O2 The following rate equation was found experimentally

19. Mechanism: 1. Rapid decomposition of ozone to O2 and atomic oxygen → equilibrium. 2. The slow, rate determining step is the reaction of atomic oxygen with ozone. 1. O3 → O2 + O. 2. O. + O3 → 2O2 k

20. 2. example: decomposition of N2O5 (see example below equation R18) 3. example:formation of hydrogen bromide from hydrogen and bromine H2 + Br2 → 2HBr This is not a simple bimolecular reaction but achain reaction of radicals. Chain initiation Br2 → 2Br. Chain propagation Br. + H2 → HBr + H. H. + Br2→ HBr + Br. Chain inhibition H. + HBr → Br. +H2 Chain termination 2Br.→ Br2

21. The reaction is initiated by bromine radicals from the thermal dissociation of Br2. The chain propagation steps regenerate the bromine radicals, ready for another cycle. In this step HBr is generated and also hydrogen radicals for the inhibition step. The chain inhibition step removes the H radicals, slowing down the chain propagation.

22. Considering the mechanism desribed above, the following rate equation can be derived (no need to memorize): (R19) Notice that [HBr] occurs in the denominator, so that the rate is inhibited by the [HBr] while [H2] (in the numinator) initiate the reaction.

23. R4. First order reactions In a first order reaction the rate is proportional to the concentration of the reactant. A → B (+ C +…) Examples: SO2Cl2 → SO2 + Cl2 2N2O5→ 4NO2 +O2 (R20) The rate equation: where k is the rate constant, (its dimension is time-1.)

24. Separate the variables and integrate from t = 0 (conc. = [A0]) to time t (conc. = [A]). (R21a) (R21b)

25. Theconcentration of the reactantAdecreases exponentially with time.Theconcentration of the product (B)if it is formed in stoichiometric ratio (one molecule of B is formed from one molecule of A): (R22)

26. The concentrations as functions of time. concentration [A]0 [B] [A] time

27. Linear plot (R21b) ln[A] experimental data ln[A]0 tga = -k a time

28. If we plot ln[A] against time and the data points lie on a staight line (within experimental errors), the reaction is of first order. The slope of the line gives the rate constant. Half lifetis the time required to reduce the concentration of reactant to half its initial value. (R23) In case of first order reaction the half life is independent of initial concentration.

29. R5. Second order reactions We study two cases: (R24) 1. The rate equation is This rate equation applies when a) The reaction issecond orderwithrespect toone component. E.g. 2NO2→ 2NO + O2 See example below equation R18!

30. b) The reactants of a bimolecular reaction are in stoichiometric ratio. A + B → Products but [A]0 = [B]0 so [A] = [B] at any time. (R25) We have the same rate equation in cases a) and b). We integrate from t = 0 (conc. = [A]0) to time t (conc. = [A]).

31. (The integral of 1/[A]2 is -1/[A] +const.) The integrated form of the rate equation is (R26) If we plot the reciprocal of the concentration of the reactant against time, we obtain a straight line in case of a second order reaction. The slope is the rate constant.

32. Linear plot (R26) 1/[A] experimental data a tga = k 1/[A]0 time

33. The dimension of k is (concentration)-1·(time)-1, Its unit is usually dm3/mol.s-1 Half lifet: [A] = [A]0/2 (R27) The half life of second order reactions depends on the initial concentration.

34. Note: The stoichiometric equation for second orderreactions 2A  Products is written sometimes in the form Comparing this with the equation R25, the relation-ship between the two rate constants (it is the same reaction!): k = 2k´.

35. 2.The second order is the sum of two first orders(A + B  Products) and the reactants arenot in stoichiometric ratio. (R28a) (R28b) It is difficult to solve this differential equation because both [A] and [B] are variables (but their changes are not independent). The next 4 slides contain the derivation (Deriv1) of a complicate equation. It is not compusory to memorize it. However, the result is important.

36. Deriv1. Solution of the differential equation R28. We introduce a variable x, which is the difference of [A]0 and [A]. Hence it is also the difference of [B]0 and [B]. x = [A]0 - [A], [A] = [A]0 - xx = [B]0 - [B], [B] = [B]0 - x The differential equation: Since d[A] = -dx,

37. To alter the left hand side consider the following identity: If we substitute [A]0-x for a and [B]0-x for b, the differential equation takes the form

38. We integrate from 0 to t and from 0 to x. The two terms of the left hand side can be integrated separately. The integral of the right hand side

39. So the solution of the differential equation Rearrange this equation considering that [A]0-x =[A] and [B]0-x =[B] If we multiply both the first and the second factor of the left hand side by -1, we get to the final form.

40. (R29) We can rearrange this equation to make is suitable for a linear plot (R30)

41. Plot ln([A]/[B]) as a function of time experimental data a time

42. R6. Determination of reaction order and rate constant If a reaction of unknown kinetic parameters is studied, the primary experimental data are concentrations as function of time. 1. Integral methods.We assume an order of reaction (1 or 2) andcheck if the experimental data fulfil the corresponding rate equation.

43. First order (R21b) ln[A] If the experimental data sit on this like straight line, the reaction is of first order. time

44. First order (R21b) ln[A] If the experimental data do not sit on the R21b like straight line, the reaction is not of first order. time

45. Second order (R26) 1/[A] If the experimental data sit on a this kind straight line, the reaction is of second order. time

46. Second order (R26) 1/[A] In this case the experimental data do not sit on a straight line, the reaction is not of second order. time

47. The integral method can be used e.g. for deciding if the reaction is of first order or second order. The linear plot also produces the rate constant. In more complicated cases (when the rate depends on more than one concentration or the order of the reaction is not an integer) the integral method cannot be used. 2. Differential methods.From the concentration - time data we determine the reaction rates by graphical or numerical derivation.

48. The concentration of a reactant as a function of time. concentration [A]0 (R31) [A] a t time

49. Experimentally we obtain reaction rates at different concentrations. Consider the following general equation. nAA + nBB  Products The rate equation (R32) How to simplify this expression?

50. We want to determine n (the order with respect to A). We apply B in great excess so that its concentration can be regarded as constant : • [B] ~ [B]0 (R33) (R34) where In this case the rate equation has only two constants (k´ and n) to be determined. If we want to determinem, we apply A in great excess