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CH4003 Physical Chemistry 2 Lecture Notes 7-10 Dr. Erzeng Xue

CH4003 Lecture Notes7-10 (Erzeng Xue) . CH4003 Physical Chemistry 2 Lecture Notes 7-10 Dr. Erzeng Xue. CH4003 Lecture Notes7 (Erzeng Xue) . C A. L. C A,1. C A,2. C A. t. t 1. t 2. Reaction kinetics. Forms of Rate Equations.

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CH4003 Physical Chemistry 2 Lecture Notes 7-10 Dr. Erzeng Xue

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  1. CH4003 Lecture Notes7-10 (Erzeng Xue) CH4003 Physical Chemistry 2 Lecture Notes 7-10Dr. Erzeng Xue

  2. CH4003 Lecture Notes7 (Erzeng Xue) CA L CA,1 CA,2 CA t t1 t2 Reaction kinetics Forms of Rate Equations • Differential rate equation for a reaction vAA + vBB = vCC + vDD General rate equation, based on A • It tells the change of the concentration of any component at any time • Differential rate for the same reaction are different at different stages • The conc. of reactants varies at different point (time, location) For steady state reaction, the conc. of each component vary with location For a batch reaction, the conc. of each component varies with time • The effect of products is different at the beginning from that at later At the beginning (low conversion) there is no or little products present At later stage (or high conversion), the conc. of products are high so do their effect. • To get relation between concentration of each component with time • the equation need to be integrated Þ Integrated form of rate equation

  3. CH4003 Lecture Notes7 (Erzeng Xue) Reaction kinetics Forms of Rate Equations • Integrated rate equation for a reaction vAA + vBB = vCC + vDD Differential rate equation Integration [ A] = f(t) • The integrated form of rate equation gives the conc. of a component as a function of t • Some differential rate equations can be integrated mathematically, giving a definite f(t), whereas the others can only be solved numerically, • Analytical solution - the result, usually a function, of integration of a feasible rate equation • Numerical solution - a set of number / values calculated based on certain initial conditions • Analysis of those rate equations having mathematic solutions can provide some useful information which can characterise the reactions having these type rate eqns.

  4. CH4003 Lecture Notes7 (Erzeng Xue) Reaction kinetics Zero Order Rate Equation • Differential rate equation zero order, i.e. a=b=l=d=0, so that integration with boundary conditions at t=0 [A]=[A]0, and t=t [A]=[A] Graph produced by setting [A]0=100 mol, k=10,20 mol/s & t in s. • Concentration of reactant decreases linearly with time from initial value, with the slope=k • The larger k value is, the faster the decrease of the reactant concentration

  5. CH4003 Lecture Notes7 (Erzeng Xue) Reaction kinetics First Order Rate Equation • Differential rate equation 1st order, i.e. a+b+l+d=1, assume a=1 so that integration with boundary conditions at t=0 [A]=[A]0, and t=t [A]=[A] Graph produced by setting [A]0=100 mol, k=1, 2 mol/s & t in s. • Concentration of reactant decreases exponentially with time from initial value • When using semi-ln graph, a straight line is obtained with the slope=k • The larger k value is the faster the decrease of the reactant concentration

  6. CH4003 Lecture Notes7 (Erzeng Xue) Reaction kinetics Second Order Rate Equation (1) • Differential rate equation 2ndt order, i.e. a+b+l+d=2, assume a=2 so that integration with boundary conditions at t=0 [A]=[A]0, and t=t [A]=[A] Graph produced by setting [A]0=100 mol, k=1, 2 mol/s & t in s. • Concentration of reactant decreases with time in a complicated way from initial value • When plotting 1/[A] with time, a straight line is obtained with the slope=k • The larger k value is the faster the decrease of the reactant concentration

  7. CH4003 Lecture Notes7 (Erzeng Xue) Reaction kinetics Second Order Rate Equation (2) • Differential rate equation 2nd order, i.e. a+b+l+d=2, assume a=b=1 so that if at t=0 [A]=[A]0 and [B]=[B]0, and t=t [A]=[A] and[B]=[B] reaction vAA + vBB = vCC + vDD initial conc. [A]0 [B]0 at t converted x (vB/vA)x conc. at t [A]0-x [B]0-(vB/vA)x so that If initially if initially [A] - t relation becomes more complicated. This is the same form as the one on the previous page

  8. CH4003 Lecture Notes7 (Erzeng Xue) Reaction kinetics nth Order Rate Equation • For a nth order reaction vAA + vBB + … … = vLL + vNN (1) Rate: (2) If initially (3) following the same treatment as that in the previous page, (4) bring (4) into rate equation (2) (5) Usually reaction order is £ 3.

  9. CH4003 Lecture Notes7 (Erzeng Xue) Reaction kinetics Summary of Rate Equations order differential form integrated form linear relation 0 1 2 3 n

  10. CH4003 Lecture Notes8 (Erzeng Xue) Reaction vAA + vBB = vCC + vDD Rate CA [A]1 [A]2 r1 r2 [A]t [A]0 Þ Differential rate equations useful for continuous flow reactor, especially when looking at localised environ. & modelling reactor [A]0 Continuous flow reactor [A]0 t=0® t [A] Þ Integrated rate equations good for batch reactor Batch reactor t t 0 Reaction kinetics More about Rate Equations • Differential rate equations tell the change of the concentration of a component at any given time; the rate varies with t. • Integrated rate equations tell the concentration of each component at time or the accumulated change of concentration.

  11. CH4003 Lecture Notes8 (Erzeng Xue) Order 0 1 2 n Reaction kinetics More about Determining Rate Equations • Determination of the rate of a reaction involves • determining the values of all kinetic parameters, the rate constant, k, including A0 and Ea as in Arrhenius equation, k=A0exp(-Ea/RT)), and the reaction orders. • finally producing a set of reaction rate equation either for the purpose of reactor design or analysis of reaction mechanism. Using the integrated form of rate equations • plotting appropriate forms of [A], as derived from integrated rate equations against t. However, we may have difficulty when • rxn order >1 and the ratio of [A]0 : [B]0 does not follow that of their stoichiometry coeff.’s. • reaction order is a fraction. What can you do then?

  12. CH4003 Lecture Notes8 (Erzeng Xue) Reaction vAA + vBB = vCC + vDD Rate Y intersect slope=a b X Reaction kinetics More about Determining Rate Equations • Mathematical transformation of the above differential rate equation We apply the same strategy as that used for integrated rate eqn’s Þ get linear relation y=ax+b Examine the parameters in the eqn.(2). Constants: A0, Ea, R, a, b, l, d(these are all kinetic parameters - need to be determined) Variables: rA, T, [A], [B], [C], [D] • To define a linear relation, y=ax+b, only two variables can be varied, which must also be the values experimentally measurable, and the rest have to be, or to be kept, constant.

  13. CH4003 Lecture Notes8 (Erzeng Xue) lnrA slope=a ln [A] Reaction kinetics More about Determining Rate Equations • Determine the reaction orders • Rearrange the eqn.(1) Let • rA is the rxn rate which can be measured, ln(rA) can be calculated. • a will be the slope if a straight line is obtained, • [A] is the concentration, which must be a variable. Giving [A], ln([A]) can be calculated. • b contains many variables. To make it constant, we can do followings • carrying out reaction at a constant temperature • keep [B], [C] and [D] in large excess, or keep the consumption of very small (e.g. <10%) Thus by measuring the reaction rate, rA, at different [A], at constant T and at a low conversion level, the reaction order w.r.t. A, a, can be determined from the slope of line.

  14. CH4003 Lecture Notes8 (Erzeng Xue) lnrA slope=b ln [B] Reaction kinetics More about Determining Rate Equations • Determine the reaction orders (cont’d) • In a similar way, rearrange the eqn.(1) Let Thus by measuring the reaction rate, rA, at different [B], at constant T and at a low conversion level, the reaction order w.r.t. B, b, can be determined from the slope of line. l and d can be determined in turn in the same way

  15. CH4003 Lecture Notes8 (Erzeng Xue) lnrA 1/T Reaction kinetics More about Determining Rate Equations • Determine the reaction activation energy, Ea, • Rearrange the eqn.(1) Let Similar to the way the reaction orders being determined, by measuring the reaction rate, rA, at different temperatures (not the concentration [A] !), and at a low conversion level, the reaction activation energy, Ea, can be determined from the slope of line. Note: 1) The slope of the plot of lnrA vs. 1/T should be negative as Ea value is always positive. 2) Temperature is in Kelvin scale.

  16. CH4003 Lecture Notes8 (Erzeng Xue) Reaction kinetics More about Determining Rate Equations • Determine the pre-exponential factor, A0, • Once all other kinetic parameters (Ea, R, a, b, l, d) have been determined as shown in the previous slides, A0 value can be calculated from any set of exp’l data (assume that all values of rA, T, [A], [B], [C], [D] have been taken in those exp’s). You have now your experimentally determined reaction rate equation. • You can determine the rate expressions based on other components of the reaction by simply repeating the same procedure as above.

  17. CH4003 Lecture Notes8 (Erzeng Xue) Reaction kinetics Important Notes on Rate Equations • Before starting the determination of the reaction rate, one must make sure that the reaction rates measured are not diffusion controlled (or mass transfer controlled). There are well-established techniques to check if the reaction is diffusion control. • The procedure for determining the kinetic parameters described in this lecture is based on the mathematical approximation, I.e. b»constant, which holds only if the conversion levels are low (called differential conversions). This also means that, the rate equations may need to be modified when being applied to situations where high conversions are demanded (e.g. including a term containing equil. constant). • The rate equations determined in this way apply only within the range in which exp’s were carried out. Extrapolation to beyond may lead to error. As a good practice, one should try, whenever possible, to carry out the kinetics study in a broader range of reaction conditions (T, [i] etc.) than that practically encountered. • The type of rate equations we discussed here are called Engineering Kinetics (the other type rate equations is called Mechanistic Kinetics, often more complicated). • Kinetics study requires measuring low conversions, which often runs into the detection limit of analytical instruments. Any inaccuracy in measurement should be indicated in the rate equation determined, e.g. k=3.5±0.7 x 102 (unit).

  18. CH4003 Lecture Notes9 (Erzeng Xue) Reaction Kinetics Other Method of Rate Law determination • Isolation method - to simply a complicated rate law to a simple one The concentration of all the reactants except one under concern are in large excess. e.g. For a rate equation of form: ri=k[A]a[B]b[C]l [D]d 1). If we keep [B]b[C]l [D] in large excess, we may re-write the rate equation as ri=k [B]b[C]l [D]d [A]a=k’ [A]awhere k’= k [B]b [C]l [D]d »constant 2). by taking logarithm ln(ri)=ln k’ + aln[A] 3). measuring ri with different [A] and plotting ln(ri) against ln[A] Þ a straight line slope=a and intersect=ln k’ 4). The values of other orders, b, l , d, can be determined in turn. • Advantage: convert certain complicated rate equation to simple form e.g. [B] can be initial conc. or average conc.

  19. CH4003 Lecture Notes9 (Erzeng Xue) Order 0 1 2 n Reaction Kinetics Other Method of Rate Law determination • Half-time method The time it takes for a reactant concentration to fall to half of its initial value. • Integrate the differential form of rate equation with special boundary conditions at t=0, [A]=[A]0, at t=t1/2, [A]=1/2[A]0 (or replace the integrated rate equations with these values) Thus measuring the time required (t1/2) for concentration change from [A]0 to 0.5[A]0 enables one to determine the rate constant of reactions of different orders

  20. CH4003 Lecture Notes9 (Erzeng Xue) Reaction Kinetics Other Method of Rate Law determination • Relaxation method Measuring the time constant for a reaction changing from one equili. state to another. 1). at equilibrium state 1, 2). A sudden change (e.g T) is introduced at t=0, after time t, a new equil.(2) is achieved 3). at equilibrium state 2, 4). The overall change of [A] between equil.1 and equili.2 (t=0®t) is x0, x0=[A]equil.2-[A]equil.1 The rate of [A] change during t=0®t 5). Integrate and let t=1/(kforw,2+krev,2), we have: x=x0e-t/t in which the time to reach the new equilibrium is t, which is measurable and equal to t. 6). Combine equilibrium constant Kequil.2=ka/kb and t=1/(kforw,2+krev,2), both kforw,2 and krev,2can then be determined Note: the temperature change can be introduced easily such by using laser to heat system within very short time (m or nano seconds); the disturbance can also be a pressure.

  21. CH4003 Lecture Notes9 (Erzeng Xue) Chemical Reaction Experimental Techniques for Kinetic Study • Determine the concentrations of reactants • Most of kinetic measurements involve monitoring the concentrations of reactants/products at different time. • In some kinetic studies the conversions of reactants (again by measuring the change of concentration) at constant time are measured. • Many analytical techniques can be employed to detect the concentrations • Volumetric measurement (titration etc) • Instrumental measurement (widely used) • Spectrophotometry (visible light absorption) • Spectroscopes (MS, IR, XPS, XRD, NMR etc) • Chromatography (GC, HPLC etc) • pH, electroconductivity • Temperature, pressure • specialised techniques This detection techniques are usually used in combination with different types of reactor (PFR, CSTR, Recirc-R, Thermo-scan Reactor etc) and modelling softwares.

  22. CH4003 Lecture Notes9 (Erzeng Xue) t2 t1 A B A B no light no reaction flash light reaction no light no reaction reaction time Chemical Reaction Experimental Techniques for Kinetic Study • Various techniques are in use • Flow method constant flow the measurement at different location reflects different time • Stopped-flow method • Flash photolysis • Many other methods

  23. CH4003 Lecture Notes9 (Erzeng Xue) Reaction Kinetics Study of Chemical Reactions • Tasks • In a chemical process, one of final products failed to meet specifications… • One of the raw materials (feed stock) exceed specifications… • The process parameters exceed the required level… • The Environmental standard is strengthened, one of the waste streams needs to be treated before emitted… • Dead fish is found in river… • Residences report an unusual smell in the air, you need to find out what… You are asked to investigate and provide ‘hard’ information on how to handle these situations. What do you do as a chemist?

  24. CH4003 Lecture Notes9 (Erzeng Xue) Chemical Reaction Study of Chemical Reactions • Identify the reactions • What are suspected reactants? • Proof by detecting the presence of certain chemicals. • What kind of reactions may be involved? • Are the reactions you thought possible? (Chemical thermodynamics) Note that the same reactants may undergo different reactions & the same products can be formed from different reaction. • What is the characteristics of the reactions you proposed? • Info. of diffusion rate in the situations concerned - mass transfer behaviour • To what extent the reactions may occur - equilibrium controlled or not • How fast the reaction may proceed under the conditions you are investigating • Kinetics rate eqn’s indicating the conc. and their changes with time & further trend. • Factors that affect the reaction rate and the extent of the influence. • Making your proposal on how to handle the situation • To make confident and effective proposal on the measures taken • you will need a good understanding of the reactions - reaction mechanism.

  25. CH4003 Lecture Notes9 (Erzeng Xue) Chemical Reaction Reaction Mechanism • What is reaction mechanism? Mechanism explains the whole process of a reaction in terms of elementary steps, routes and intermediates involved, as well as the relative speed of each step. • Many reactions do not complete in one step and may have many steps • Each of these steps involves a single transformation of molecules (elementary reaction) • An intermediate state or Intermediates are formed but not all of these are detectable • The route of reaction process may be complicated and often condition-dependent. Example; Overall reaction: Br2(g) + 2 NO(g) ® 2 BrNO elementary reaction: Br2(g) + NO ® Br2NO(g) B2NO(g) + NO(g) ® 2 BrNO intermediate sequential reaction

  26. CH4003 Lecture Notes10 (Erzeng Xue) Chemical Reaction Reaction Mechanism • Questions to be answered in reaction mechanistic study • Why some reactions are faster than others? • Why rate laws have different appearances? • What are the meanings of kinetic parameters (A0, Ea, orders) in a rxn process? • Can we make an ‘educated guess’ of certain aspects of a reaction process from a rate law observed? • Can we generalise some types of reaction mechanism and gain a better understanding of various types of reactions? Note: 1). Various theories have been explored and developed trying to explicate what happens in a reaction process. We will limit ourselves to the main stream ones. 2). A mechanism describe a chemical process in a molecular level, which in most cases cannot be visualised directly, it has to be experimentally approved and has to be able to be extended to similar type reactions.

  27. CH4003 Lecture Notes10 (Erzeng Xue) Chemical Reaction Reaction Mechanism • The collision theory In the collision theory of reaction rate, we consider that the reacting molecules • must collide with one another • must collide with sufficient energy, and • must collide in an orientation that can lead to rearrangement of atoms • Accounting for the dependence reaction rate on the kinetic parameters Observations: • An increase of concentrations Þ more collisions, rµ [i]a • Molecules of different components may have different effectiveness leading to reaction, hence different values of reaction order w.r.t. different component. • An increase temperature Þ molecules move faster and carry more energy, T­ Þk­ Þr ­ • Pre-exponential factor, A0, reflects the frequency as well as the orientation of collisions, thus it has direct effect on the reaction rate constant, k.

  28. CH4003 Lecture Notes10 (Erzeng Xue) uncatalytic catalytic transition state reactants products reactant energy product Ea reaction process reactant energy H3C CH3 H CH3 C=C C=C product Uncatalysed Ea=262kJ/mol Catalysed by I2Ea=116kJ/mol H H H3C H reaction process cis-2-butene trans-2-butene Chemical Reaction Reaction Mechanism • The activation energy, Ea • Ea represents the energy barrier for a reaction to occur. Molecules of reactants have to collide, form intermediate (a high energy transition state) then become products (intermediate molecules may also return to reactants). • Because the molecules at transition state are at a max in potential energy they are very unstable, so intermediate cannot be isolated nor can its structure be determined experimentally. • When a reaction is catalysed, the rate of reaction is raised due to the presence of a catalyst. Different intermediates may formed with the help of catalysts and the energy of intermediate may be greatly lowered. e.g.

  29. CH4003 Lecture Notes10 (Erzeng Xue) Chemical Reaction Reaction Mechanism • Overall reaction & elementary reactions (steps) • Elementary reactions • A reaction process may consist of a sequence of more than one elementary reactions ( steps) • Reaction equation • for overall reaction - It represents the stoichiometry of overall rxn (no intermediate) • for elementary rxn - It represents a single event in the sequence of a rxn process • Rate law • for overall reaction - It shows the rate constant and reaction order w.r.t. reactants & products of overall reaction, not related to the species appeared in each step. • for elementary rxn - There is a rate law associated with each step. The power index of each conc. term is the stoichiometry number of that component and has to be an integral. k k1 k2 Ov NO2(g)+F2(g) ® 2FNO2(g) E.1 NO2(g)+F2(g) ® FNO2(g)+F(g) E.1 NO2(g)+F(g) ® FNO2(g) For rxn above general Ov r =k [NO2][F2] r =k [A]a[B]b E.1 r1=k1[NO2][F2] r1=k1[A]vA[B]vB E.1 r2=k2[NO2][F] r2=k2[A]vA[P]vP st. coeff. vA, vB,VP,… must be integral a, b are not necessarily equal to vA,vB

  30. CH4003 Lecture Notes10 (Erzeng Xue) Chemical Reaction Reaction Mechanism • The rate determining step (r.d.s.) The rate of the slowest elementary rxn will determine the rate of overall reaction rate. • Steady-state approximation The conc. & the rate of change of all rxn intermediates are small (in the major part of a rxn). ka kb Rxn A ® B ® C (assume r.d.s. is A ® B, i.e. kb >> ka) Proof of steady-state approximation if kb>>ka kbt>>1

  31. CH4003 Lecture Notes10 (Erzeng Xue) ka kb Rxn A ® B ® C Chemical Reaction Reaction Mechanism • Use of steady-state approximation (s.s.a.) Simplify the analysis of the rate of reaction based on elementary steps. e.g. Apply s.s.a. to eqn (2) The same results as shown in the previous slide but much easier to obtain.

  32. CH4003 Lecture Notes10 (Erzeng Xue) ka k’a kb Chemical Reaction Reaction Mechanism • Michaelis-Menten mechanism (of enzyme action) Observation: The rate of an enzyme (E) catalysed reaction in which a substrate (S) is converted into product (P) depends on the enzyme concentration [E]. e.g. E + S (ES) P + E (r.d.s. ES ® P + E) Write the rate law for each elementary step, Apply s.s.a. as [E] + [ES] = [E]totalÞ [E] = [E]total-[ES], and [S] = [S]total we have Bring (2) into (1) in which Michaelis constant

  33. CH4003 Lecture Notes10 (Erzeng Xue) Chemical Reaction Reaction Mechanism • Lindemann-Hinshelwood mechanism - To explain ‘unimolecular reaction’ • If collision between two molecules of the same component as elementary step A + A ®P the rate law for this elementary step: r = k[A][A] Þ order should be 2 Observation: This type reaction can be first order. How can this be? • The mechanism scheme proposed by Lindemann-Hinshelwood (1) A + A ® A* + A (A* - activated A molecule) d[A*]/dt=ka[A]2 (2) A* + A ® 2A (A* lost energy by collision with another) d[A*]/dt=k’a[A*][A] (3) or A* ® P (A* transform to product) d[A*]/dt=-kb[A*] & d[P]/dt=kb[A*] The net rate of formation of A*: d[A*]=ka[A]2 - k’a[A*][A] - kb[A*] = 0 (apply s.s.a.) solve it for [A*], we have thus When k’a[A*][A]>>kb[A*], or k’a[A]>>kb We then have the overall reaction is first order

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