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Chemical Kinetics Khadijah Hanim Abdul Rahman UniMAP

Chemical Kinetics Khadijah Hanim Abdul Rahman UniMAP. Subtopics. Experimental Chemical and Kinetics Reactions First Order Reactions Second Order Reactions Reaction Rates and Reaction Mechanisms Light Spectroscopy and Adsorption Chemistry (Experimental methods for fast reactions).

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Chemical Kinetics Khadijah Hanim Abdul Rahman UniMAP

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  1. Chemical KineticsKhadijahHanim Abdul RahmanUniMAP ERT 108 – Physical Chemistry Semester II- 2010/2011

  2. Subtopics • Experimental Chemical and Kinetics Reactions • First Order Reactions • Second Order Reactions • Reaction Rates and Reaction Mechanisms • Light Spectroscopy and Adsorption Chemistry (Experimental methods for fast reactions). ERT 108 – Physical Chemistry Semester II- 2010/2011

  3. Experimental Chemical and Kinetics Reactions • Rates of chemical Reactions: • the rate of speed with which a reactantdisappears or a productappears. • the rate at which the concentration of one of the reactants decreases or of one of the products increases with time. • typically, mol L-1 s-1. ERT 108 – Physical Chemistry Semester II- 2010/2011

  4. The rates of reactions • Depends on composition and temperature of reaction mixture. • Definition of rate • as the slope of the tangent drawn to the curve showing the variation of conc with time ERT 108 – Physical Chemistry Semester II- 2010/2011

  5. Rate of Reaction: A variable quantity • Rate of reaction is expressed as either: or [ Negative value ] [ Positive value ] ERT 108 – Physical Chemistry Semester II- 2010/2011

  6. Consider a reaction: A + 2B 3C + D • Rate of consumption – (one of the reactants, A or B) at a given time is d[R]/dt, R is A/B. • Rate of formation – (products, C/D denotes as P) at a given time is d[P]/dt. • These rates are positive values. The +/- sign- indicate that conc is increasing/decreasing. ERT 108 – Physical Chemistry Semester II- 2010/2011

  7. From the stoichiometry: • Rate of reaction is related to rates of change of concentration of products and reactants • Undesirability of having different rates to describe the same reaction- using the extent of reaction, ξ. • vJ is the stoichiometric coefficient for species J. • Rate of reaction, r ERT 108 – Physical Chemistry Semester II- 2010/2011

  8. For homogenous reaction the V can be taken inside the differential, [J]=nJ/V • For heterogeneous reaction, use the surface area (constant), A as substitution to V, σJ=nJ/A • Common units for r = mol dm-3 s-1 or related units for homogenous reaction • For heterogeneous reaction= mol m-2 s-1 ERT 108 – Physical Chemistry Semester II- 2010/2011

  9. Example 1 • The rate of change of molar conc of CH3 radicals in the reaction 2CH3(g) CH3CH3(g) was reported as d[CH3]/dt= -1.2 mol dm-3 s-1 under particular conditions. What is (a) the rate of reaction (b) the rate of formation of CH3CH3? ERT 108 – Physical Chemistry Semester II- 2010/2011

  10. Rate laws and rate constants • Rate of reaction is often proportional to conc of reactants raised to a power. r = k[A] [B] Each conc raised to first power. • k is the rate constant for the reaction • k- independent of conc but depends on temp. • Experimentally determined equation of this kind- rate law of the reaction ERT 108 – Physical Chemistry Semester II- 2010/2011

  11. Rate law- an equation that expresses the rate of reaction as a function of conc of all species present in overall chemical equation r = f([A], [B],….) • Rate law is determined experimentally- cannot be inferred from the stoichiometry of balanced chemical equation. • Application of rate law: • To predict the rate of reaction from the composition of mixture • Guide to the mechanism of the reaction- for any proposed mechanism- must be consistent with the observed rate law. ERT 108 – Physical Chemistry Semester II- 2010/2011

  12. Reaction order • Many reactions are found to have rate laws of the form r = k [A]a[B]b……. • The power to which the conc of a species is raised in a rate law- the order of the reaction with respect to that species. • A reaction with rate law r = k [A] [B] is first-order in A and first-order in B. • The overall order of a reaction, second-order overall-the sum of the individual orders. • Some reactions obey zero-order rate law- rate that is independent of conc of the reactant. Thus, r = k ERT 108 – Physical Chemistry Semester II- 2010/2011

  13. Integrated rate laws • First-order rate law is or [A] = [A]0e-kt Where [A]0 is the initial conc of A at t = 0 • If ln ([A]/[A]0) is plotted against t- first-order reaction will give a straight line of slope= -k. ERT 108 – Physical Chemistry Semester II- 2010/2011

  14. Half lives and time constants • Useful indication of the rate of a first-order chemical reaction- half-life,t1/2of a substance. • Half-life: time taken for the conc of reactant to fall to half its initial value. • Time for [A] to decrease from [A]0 to ½[A]0 in first-order reaction: kt1/2 = -ln = -ln ½ = ln 2 ∴ t1/2 = • Another indication of the rate of a first-order reaction- time constant, τ • time required for the conc of reactant to fall to 1/e of its initial value. kτ = -ln = -ln 1/e = 1 ∴time constant, τ = 1/k ERT 108 – Physical Chemistry Semester II- 2010/2011

  15. Second-order reactions • Second-order rate law: • is or • Where [A]0 is the initial conc of A (at t = 0) • To plot a straight line for second order reaction- plot 1/[A] against t. the slope= k. • The half life for second order reaction is ERT 108 – Physical Chemistry Semester II- 2010/2011

  16. Half-life of nth-order reaction • In general for nth-order reaction (with n > 1) of the form A  products, the half-life is related to the rate constant and the initial conc of A by ERT 108 – Physical Chemistry Semester II- 2010/2011

  17. Zero-order, First-order, Second-order Reactions ERT 108 – Physical Chemistry Semester II- 2010/2011

  18. Zero-order, First-order, Second-order Reactions ERT 108 – Physical Chemistry Semester II- 2010/2011

  19. Zero-order, First-order, Second-order Reactions Zero order First order Second order ERT 108 – Physical Chemistry Semester II- 2010/2011

  20. Example 2 (a) When [N2O5] =0.44M, the rate of decomposition of N2O5 is 2.6 x 10-4 mol L-1 s-1. • what is the value of k for this first-order reaction? (b) N2O5 initially at a concentration of 1.0 mol/L in CCl4, is allowed to decompose at 450C. At what time will [N2O5] be reduced to 0.50M? ERT 108 – Physical Chemistry Semester II- 2010/2011

  21. Example 3 • The data of the above table were obtained for the decomposition reaction: A → 2B + C. • Establish the order of the reaction. • What is the rate constant, k? ERT 108 – Physical Chemistry Semester II- 2010/2011

  22. Answer (Example 3) (a) Plot graph based on the data given in the Table. (b) The slope of the 3rd graph: ERT 108 – Physical Chemistry Semester II- 2010/2011

  23. Determination of the rate law • Experimental data gives species conc at various times during the reaction. • a few methods to determine the rate law from experimental conc vs. time data. • Consider the following: r = k [A]a[B]b……. it is usually the best to find the order of a, b, … first and then the rate constant, k. ERT 108 – Physical Chemistry Semester II- 2010/2011

  24. Determination of rate law- half-life method • Applies when the rate law has the form r = k[A]n. then this equation and apply. • If n = 1, then t1/2 is independent of [A]0. If n ≠1, then gives: • A plot of log10 t1/2 vs. log10 [A] – gives straight line of slope = n-1. ERT 108 – Physical Chemistry Semester II- 2010/2011

  25. To use the half-life method for determination of rate law: • Plot [A] vs. t • Pick any [A] value, eg. [A]’ and finds the point where [A] has fallen to ½ [A]’. The time interval between this 2 points is t1/2 for the initial conc [A]’. • Pick another point [A]” and determines the t1/2 for this A conc. • Repeat this process several times • Plot log10 t1/2 vs. the log of the corresponding initial A conc and measures the slope. ERT 108 – Physical Chemistry Semester II- 2010/2011

  26. Example 4 • Data for the dimerization 2A A2 of certain nitrile oxide (compound A) is ethanol solution at 40oC follow: Find the reaction order using the half-life method ERT 108 – Physical Chemistry Semester II- 2010/2011

  27. Method of Initial Rates • This simple method of establishing the exponents in a rate equation involves measuring the initial rate of reaction, ro for different sets of initial concentration. • Suppose we measure r0 for the 2 different initial A conc [A]0,1 and [A]0,2 while keeping [B]0, [C]o,… fixed. • With only [A]0 changed and with the rate law assumed to have form r = k[A]a[B]b…[L]l, the ratio of initial rates for run 1 and 2 is ro,2/ro,1 = ([A]o,2/[A]0,1)n

  28. Example 5 • The data of three reactions involving S2O82- and I- were given in the below table. (i) Use the data to establish the order of reaction: with respect to S2O82-, the order with respect to I- & the overall order. ERT 108 – Physical Chemistry Semester II- 2010/2011

  29. Example 5 (ii) Determine the value of k for the above reaction. (iii) What is the initial rate of disappearance of in a S2O82- reaction in which the initial concentrations are [S2O82- ] =0.050M & [I-]=0.025M? (iv) What is the rate of formation of SO42- in Experiment 1? ERT 108 – Physical Chemistry Semester II- 2010/2011

  30. The temperature dependence of reaction rates • Chemical reaction begins with a collision between molecules of A and molecules of B. • Chemical reactions tend to go faster at higher temperature. • slow down some reactions by lowering the temperature. • Increasing the temperatureincreases the fraction of the molecules that have energies in excess of the activation energy. • this factor is so important that for many chemical reactions it can lead to a doubling or tripling of the reaction rate for a temperature increase of only 100C. ERT 108 – Physical Chemistry Semester II- 2010/2011

  31. Reaction rates: Effect of temperature • In 1889, Arrhenius noted that the k data for many reactions fit the equation: where A & Ea are constants characteristics of the reaction & R = the gas constant. • Ea – the Arrhenius activation energy (kJ/mol or kcal/mol) • A – the pre-exponential factor (Arrhenius factor). • the unit of A is the same as those of k. • Taking log of the above equation: ERT 108 – Physical Chemistry Semester II- 2010/2011

  32. Reaction rates: Effect of temperature • If the Arrhenius equation is obeyed: • a plot of ln k versus 1/Tis a straight line with slope = (-Ea/R) and A is the intercept of the line at 1/T = 0. • This enables Ea and A to be found. • Another useful equation: (eliminate the constant A). • T2 and T1 - two kelvin temperatures. • k2 and k1 - the rate constants at these temperatures. • Ea – the activation energy (J/mol) • R – the gas constant (8.314 Jmol-1 K-1). ERT 108 – Physical Chemistry Semester II- 2010/2011

  33. The activation energy, Ea: the minimum kinetic energy that reactants must have in order to form products. • The pre-exponential factor, A: a measure of the rate at which collisions occur irrespective of their energy. ERT 108 – Physical Chemistry Semester II- 2010/2011

  34. Reaction Mechanisms • The mechanism of reaction is the sequence of elementary steps involved in a reaction • Most reactions occur in a sequence of steps called elementary reactions. • A mechanism is a hypothesis about the elementary steps through which chemical change occurs. • A typical elementary reaction is H + Br2 HBr + Br • Chemical equation for elementary reaction: equation only represents the specific process occurring to individual molecules • Molecularity of an elementary reaction is the no of molecules coming together to react in an elementary reaction. ERT 108 – Physical Chemistry Semester II- 2010/2011

  35. Reaction Mechanisms • Elementary processes in which a single molecule dissociates (unimolecular) or two molecules collide (bimolecular) much more probable than a process requiring the simultaneous collision of three bodies (termolecular). • All elementary processes are reversible and may reach a steady-state condition. In the steady state the rates of the forward & reverse processes become equal. The concentration of some intermediate becomes constant with time. • One elementary process may occur much more slower than all the others. In this case, it determines the rate at which the overall reaction proceeds & is called the rate-determining step. ERT 108 – Physical Chemistry Semester II- 2010/2011

  36. Rate laws and equilibrium constants for elementary reactions • An overall reaction occurs as a series of elementary steps • These steps constituting the mechanism of reaction • This section consider the rate law for elementary reaction. • Consider a bimolecular elementary reaction A + B  products, the rate of reaction, will be proportional to ZAB, the rate of A-B collisions per unit of time. • ∴r for an elementary bimolecular ideal-gas reaction will be r = k[A][B] ERT 108 – Physical Chemistry Semester II- 2010/2011

  37. For unimolecular ideal-gas reaction B products, fixed probability that any particular B molecule will decompose/isomerize to products per unit time. • The rate of reaction, r = k[B] • Similar considerations apply to reactions in ideally/ideally dilute solution. • In summary, in an ideal system, the rate law for the elementary reaction aA + bB  products is r = k[A]a[B]b, where a + b is 1,2 or 3. • For an elementary reaction, the orders in the rate law equal the coefficients of the reactants. ERT 108 – Physical Chemistry Semester II- 2010/2011

  38. Relation between the equilibrium constant for a reversible elementary reaction and rate constants for forward and reverse reactions. • Consider the reversible elementary reaction aA + bB⇌ cC + dD • rate laws for forward (f) and back (b) elementary reactions are rf = kf [A]a[B]b and rb = kb [C]c[D]d. • At equilibrium, these opposing rates are equal: rf,eq = rb,eq or kf([A]eq)a([B]eq)b = kb([C]eq)c([D]eq)d and • = Kc= kf/kb kf kb ERT 108 – Physical Chemistry Semester II- 2010/2011

  39. (b) The rate-determining-step approximation • In rate-determining step approximation- reaction mechanism assumed to consist 1 or more reversible reactions that stay close to equilibrium during most of the reaction. • followed by relatively slow rate-determining step then in turn followed by 1 or more rapid reactions. • As an example: A⇌B⇌C⇌D where step 2 (B⇌C)- assumed to be rate-determining step. • For this assumption to be valid- k-1 >> k2 k1 k3 k2 k-1 k-2 k-3 ERT 108 – Physical Chemistry Semester II- 2010/2011

  40. The slow rate of B  C compared with B  A – ensures that most B molecules go back to A rather than to C- ensuring that step 1 (A ⇌ B) remain close to equilibrium. • k3 >> k2 and k3 >> k-2, to ensure that step 2 acts as ‘bottleneck’ and product D is rapidly formed from C. • The overall rate is controlled by the rate-limiting step B  C. • Since we are examining rate of the forward reaction A  D, we further assume that k2[B] >> k-2[C]. • During early stage- the conc of C will be lower than B- this condition will hold. Thus, we neglect reverse reaction for step 2. ERT 108 – Physical Chemistry Semester II- 2010/2011

  41. The relative magnitude of k1 compared with k2 is irrelevant to the validity of the rate-determining-step approximation. • ∴ the rate constant k2 of the rate-determining step might be larger than k1. • However, the rate r2 = k2[B] of the rate-determing step must be smaller than r1=k1[A] of the first step. • This follows from k2<<k-1 and k1/k-1 [B]/[A] (the conditions for step 1 –near equilibrium). • For reverse overall reaction, the rate-determining step is the reverse of that for forward reaction. • For example: the rate-determining step is C  B. So, k-2 << k3 (ensures that step D⇌C is in equilibrium) and k-1>> k2 (ensures that B  A is rapid). ERT 108 – Physical Chemistry Semester II- 2010/2011

  42. The rate determining-step approximation • The rate law for the Br- - catalyzed aqueous reaction H+ +HNO2 + C6H5NH2 C6H5N2+ + 2H2O is observed to be r = k[H+][HNO2][Br-] …..(1) A proposed mechanism is H+ + HNO2⇌ H2NO2+ rapid equilib H2NO2+ + Br-  ONBr + H2O slow ……(2) ONBr + C6H5NH2  C6H5N2+ +H2O + Br- fast Deduced the rate law for this mechanism and relate the observed rate constant, k in (1)to the rate constants in assumed mechanism (2). k1 k-1 k2 k3 ERT 108 – Physical Chemistry Semester II- 2010/2011

  43. The second step in (2) is rate limiting. Since step 3 is much faster than 2, we can take d[C6H5N2+]/dt as = rate of formation of ONBr in step 2. therefore, the reaction rate is r = k2[H2NO2+][Br-] …..(3) (since step 2 is an elementary reaction, its rate law is determined by its stoichiometry. The species H2NO2+ in (3) is a reaction intermediate and we want to express r in terms of reactants and products. Since step 1 is in near equilib, gives: Kc,1 = k1/k-1 =[H2NO2+]/[H+][HNO2] And [H2NO2+] = (k1/k-1) [H+][HNO2] Substitute in (3) gives r = (k1k2/k-1)[H+][HNO2][Br-] k = k1k2/k-1 = Kc,1k2 ERT 108 – Physical Chemistry Semester II- 2010/2011

  44. The steady-state approximation ka kb • Multistep reaction mechanisms usually involve 1 or more species that do not appear in overall equation. A  I  P • After an initial induction period,[I] will start at 0, rise to max, [I]max, and then fall back to 0. • During the major part of the reaction, the rates of change of conc of all reaction intermediates are negligibly small, therefore d[I]/dt = 0 for each reaction intermediate. • The steady-state approximation assumes that rate of formation of reaction intermediate = rate of destruction. ERT 108 – Physical Chemistry Semester II- 2010/2011

  45. Steady-state approximation • Apply steady-state approximation to the mechanism for H+ + HNO2 + C6H5NH2 C6H5N2+ + 2H2O Given in the preceding exercise to find the predicted law. Br- ERT 108 – Physical Chemistry Semester II- 2010/2011

  46. To apply for rate-determining step approximation: (a) Take the reaction rate, r = rate of determining step (divided by the stoichiometric no srds of rate-determining step, if srds≠ 1. (b) Eliminates the conc of any reaction intermediates that occur in the rate expression obtained in (a) by using equilibrium-constant expressions. • To apply steady-state approximation: (a) Take the reaction rate, r = rate of formation of product (b) Eliminate the conc of any reaction intermediates that occur in (a) by using d[I]/dt = 0 to find the conc of each I (c) If step (b) introduces conc of other I, apply d[I]/dt = 0 to eliminate their conc. ERT 108 – Physical Chemistry Semester II- 2010/2011

  47. The Hydrogen-Iodine Reaction H2 (g) + I2 (g) → 2HI (g) • Rate of formation of HI = k [H2][I2] • The hydrogen-iodine reaction is proposed to be a two-step mechanism [Sullivan J. (1967). J.Chem.Phys.46:73]. • 1st step: iodine molecules are believed to dissociate into iodine atoms. • 2nd step: simultaneous collision of two iodine atoms and a hydrogen molecule. (this termolecular step is expected to occur much more slowly – the rate-determining step). ERT 108 – Physical Chemistry Semester II- 2010/2011

  48. The Hydrogen-Iodine Reaction 1st step: [Fast] 2nd step: [Slow] Net: • If the reversible step reaches a steady state condition: • rate of disappearance of I2 = rate of formation of I2 ERT 108 – Physical Chemistry Semester II- 2010/2011

  49. The Hydrogen-Iodine Reaction • For the rate-determining step: Rate of formation of HI = k3 [I]2[H2] = K[H2][I2] (K=k1k3/k2) ERT 108 – Physical Chemistry Semester II- 2010/2011

  50. Example 6 • The thermal decomposition of ozone to oxygen: 2O3 (g) → 3O2 (g) • The observed rate law: Rate of disappearance of O3 = • Show that the following mechanism is consistent with this experiment rate law. 1st: 2nd: ERT 108 – Physical Chemistry Semester II- 2010/2011

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