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- 1. 12/01/2011 Course Title: General and Inorganic Chemistry TOPIC 3: CHEMICAL KINETICS Lecturer and contacts Mr. Vincent Madadi Department of Chemistry, University of Nairobi P. O. Box 30197-00100, Nairobi, Kenya Chemistry Dept. Rm 114 Tel: 4446138 ext 2185 Email: vmadadi@uonbi.ac.ke, madadivin2002@yahoo.com Website: http://www.uonbi.ac.ke/staff/vmadadi1/12/2011 mov 1 Introduction• Kinetics is the study of rates of chemical reactions and the mechanisms by which they occur.• The reaction rate is the increase in concentration of a product per unit time or decrease in concentration of a reactant per unit time.• A reaction mechanism is the series of molecular steps by which a reaction occurs.1/12/2011 mov 2 1
- 2. 12/01/2011 Thermodynamic vs kinetics of reaction• Thermodynamics determines if a reaction can occur. Kinetics determines how quickly a reaction occurs• Some reactions that are thermodynamically feasible are kinetically so slow as to be imperceptible• The Rate of a Reaction Cdiamond + O2(g)→ CO2(g) ΔG°= -396kJ Very Slow H+(aq) + OH-(aq) → H2O(l) ΔG°= -79kJ1/12/2011 Very Fastmov 3 Rate of chemical reaction• 2 N2O5 → 4 NO2 + O2• 2 moles of N2O5 disappear for every 4 moles of NO2 and 1 mole of O2 formed.• Reaction rates are the rates at which reactants disappear or products appear.• This movie is an illustration of a reaction rate.1/12/2011 mov 4 2
- 3. 12/01/2011 Reaction rate• Reaction rate is the change of concentration of a reactant or product per unit time aA + bB → cC + Dd Reaction rate = ΔConcentration Δtime• Rate is expressed either as rate of appearance of product or rate of disappearance of reactant• E.g. aA + bB → cC + dD1/12/2011 mov 5 Reaction rate 2 NO2(g) → 2 NO(g) + O2(g)1/12/2011 mov 6 3
- 4. 12/01/2011 Reaction rate• Mathematically, the rate of a reaction can be written as:• Square brackets [ ] are often used to express molarity (i.e.[HCl] means Molarity of HCl)• The relative rates of consumption of reactants and formation of products depend on the reaction stoichiometry 1/12/2011 mov 7 Reaction rate • The relative rates of consumption of reactants and formation of products depend on the reaction stoichiometry • For the reaction 2HBr (g) → H2 (g) + Br2 (g) • two moles of HBr are consumed for every one mole of H2 which is formed 1/12/2011 mov 8 4
- 5. 12/01/2011 Reaction rate• Experimental Rate Law: the rate of a reaction is proportional to the product of the concentrations of the reactants raised to some power.• For a reaction aA + bB → products, the rate law is the equation rate = k[A]x[B]y• Relationships Between Rate and Concentration1/12/2011 mov 9 Reaction rate• x and y are the orders of the reaction in [A] and [B] respectively• The overall order of the reaction is x + y• x and y are usually small integers, but may be zero, negative, or fractions• k is the specific rate constant1/12/2011 mov 10 5
- 6. 12/01/2011 Rate constant k • Units depend on overall reaction order • Value does not change with concentration • Value does not change with time • Valid for a specific temperature • Dependent on presence or absence of a catalyst • Value must be determined experimentally 1/12/2011 mov 11 Factors that affect rate of reaction• They are six key factors that affect rate of reaction: 1) Nature of reactants and products 2) Concentration 3) Temperature 4) Catalyst 5) Surface area 6) Light radiation 1/12/2011 mov 12 6
- 7. 12/01/2011 Factors that affect rate of reaction 1. Nature of reactants and products • Chemical reactions involve rearrangement of bonds: Bonds in reactants are broken and new bonds are formed in products • In organic or molecular reactions, large number of bonds are broken in reactants and formed in products-hence reactions are slow • E.g. Hydrolysis of cane sugar: C12H22O11 + H2O C6H12O6 + C6H12O6 Cane sugar glucose fructose 1/12/2011 mov 13 Factors that affect rate of reaction...• Inorganic reactions involve ions, hence no bonds to be broken in reactants . The reactions are faster.• E.g. Ag+(aq) + NO3-(aq) + Na+Cl-(aq) AgCl(aq) + NaNO3(aq)• Concentration• Based on the rate law of mass action, the rate of reaction is directly proportional to the product of the concentration of the reactants at a particular temperature• For reaction aA + bB cC + dC 1/12/2011 Rate = K[A]a[B]b mov 14 7
- 8. 12/01/2011 Factors that affect rate of reaction ...3) Temperature• In most reactions, the rate of reaction doubles with 10 K increase in temperature• But with 10 k increase in temperature, collision frequency (Z) increases by a factor of 1.016 I.e. Z α T½ Z2/Z1 = (T2/T1)1/2 = (310/300)1/2 = 1.016• However, the rate increases by almost 100%• This is because increasing temperature by 10 K increases the number of active molecules (molecules with E > Ea) which increases the rate drastically mov 1/12/2011 15 Factors that affect rate of reaction ... Effect of increase in temperature by 10 K 1/12/2011 mov 16 8
- 9. 12/01/2011 Factors that affect rate of reaction...• 4) Catalyst• A catalyst is a substance that alters the rate of reaction without itself getting consumed• There are two types of catalysts:• Positive catalyst: It increases the rate of reaction e.g. MnO2 2KClO3 2KCl + 3 O2 300 ⁰C• MnO2 acts as a positive catalyst since uncatalysed reaction takes place at 700 ⁰C and is slower1/12/2011 mov 17 Factors that affect rate of reaction...• Negative catalyst: This is the catalyst which retards the rate of reaction e.g. Oxidation of chloroform is retarded by ethanol 1% ethanol 4CHCl3 + 3 O2 4COCl2 + 2Cl2 + H2OMechanism of catalysis• A catalyst alters the rate of reaction by providing a path with lower or higher activation energy1/12/2011 mov 18 9
- 10. 12/01/2011 Factors that affect rate of reaction...5) Surface area• The rate of homogeneous reaction is influenced by the surface area of the reactants• Particle size decreases, surface area increases for the same mass because of creation of new surfaces. Hence the rate of reaction increases• E.g. Powdered zinc reacts faster with dilute HCl than a block of zinc1/12/2011 mov 19 Factors that affect rate of reaction...6) Light radiation• The rate of photochemical reaction is affected by light radiation• Photons (E = hv) supply the necessary energy of activation to the reactant molecules to form products λ = 400 nm• E.g. H2 + Cl2 2HCl1/12/2011 mov 20 10
- 11. 12/01/2011 Order of reaction• This is the sum of the powers of the concentration term in an experimentally established rate law• Example aA + bB → cC + dD• The theoretical rate law is: Rate = K[A]a[B]b• But experimentally determined rate expression is: Rate = K[A]α[B]β1/12/2011 mov 21 Order of reaction cont.• Where α and β are the actual moles of A and B and may not be necessarily be equal to “a” and “b”• Overall order of reaction is given by the sum of the individual orders i.e.• Overall order = α + β• Order of reaction can be zero, +ve, -ve or fraction. But higher orders of reaction are rare1/12/2011 mov 22 11
- 12. 12/01/2011 Order of reaction cont.• In complex reactions the order of reaction is determined by the slowest step of the reaction which is also called the rate determining step1/12/2011 mov 23 Molecularity of reaction• This is the total number of molecules in the step leading to chemical reaction• For any reaction, the least number of molecules is one• Thus molecularity cannot be zero or fraction• Types of molecularity:1) Unimolecular reactions:•1/12/2011 Reactions involving one molecule of the reactants mov 24 12
- 13. 12/01/2011 Molecularity of reaction • Example: PCl5 PCl3 + Cl2 2) Bimolecular reactions CH3COOC2H5 + H2O CH3COOH + C2H5OH 3) Thermolecular reactions 2NO + O2 2NO2 1/12/2011 mov 25 Pseudo uni-molecular reactions• The reaction in which the order is one but molecularity is 2• E.g. CH3COOC2H5 + H2O CH3COOH + C2H5OH• Order is one because water is in excess, hence concentration does not change during the reaction• Thus, the rate is independent of the conc. Of water but only dependent on conc. of ester 1/12/2011 mov 26 13
- 14. 12/01/2011 Rate law• Exercise1) The rate of reaction 2NO + O2 2NO2 follows the rate law Rate = K[NO]2[O2] If K = 2x 10-6 mol-1L2, What is the rate of the reaction when [NO] = 0.04 molL-1 and [O2] = 0.2 molL-1 [ Ans = 6.4 x 10-10 molL-1s-1]1/12/2011 mov 27 Rate law2. The rate of the reaction 2NO + O2 2NO2is doubled when the concentration of O2 is doubled, but increases by factor of 8 when the concentration of both reactants is doubled.Determine the:a) Order of reaction wrt NO and O2b) Overall order of reaction1/12/2011 mov 28 14
- 15. 12/01/2011 Activation energy• This is the additional amount of energy that reactant molecules must acquire in order to react and form products• It is defined as the amount of energy that the reactants must absorb to pass over the activation energy barrier to form products• Activation energy diagram Ea Reactant ER E Th Product 1/12/2011 mov 29 Activation Energy cont.• E = Eth –ER = Threshold Energy –Energy possessed by molecules• The activation energy is related to the rate constant K and temperature T according to Arrhenius equation K = Ae-Ea/RTWhere A = Frequency factor K = rate constant T = temperature in Kelvin R = Gas constant E = Activation energy e = base of natural logarithm 1/12/2011 mov 30 15
- 16. 12/01/2011 Derivation of Arrhenius equation • Arrhenius equation is derived from the Vant’ Hoffs reacton isochore dlnKc/dT = ΔE/RT2 and the reaction dynamic equilibrium A+B=C+D • Kc for the reaction, Kc = Kf/Kb = [C][D]/[A][B] Kf[A][B] = Kb[C][D] • Hence, Kc = Kf/Kb 1/12/2011 mov 31 Derivation of Arrhenius equation cont.• If ΔE is written as Ef –Eb, then from equation 1 and 2 then,• dlnKf/dT – dlnKb/dT = Ef/RT2 – Eb/RT2• or dlnk/dT = E/RT2 dlnk = EdT/RT2• Integrating the equation gives, lnk = -E/RT + C 1/12/2011 mov 32 16
- 17. 12/01/2011 Derivation of Arrhenius equation cont.• Provided E is a constant, the equation can be written as: k = Ae-E/RT• This is the Arrhenius equation• Application of activation energy1) To determine activation energy1/12/2011 mov 33 Determination of activation Energy• There are two methods for determining of activation energy Ea 1) Graphical method 2) Rate constant method1) Graphical method• From Arrhenius equation, K = Ae-E/RT• Applying lo to both sides of the equation,1/12/2011 mov 34 17
- 18. 12/01/2011 Determination of activation Energy cont.• lnK = lnA – Ea/RT• >2.303logK = 2.303logA – Ea/RT• >logK = logA – Ea/2.303RT• >logK = -Ea/2.303RT + logA Ξ y = mx + C• Plotting logK against 1/T give a straight line and Ea can be calculated1/12/2011 mov 35 Graphical Determination of activation energy• Graph Log A Slope = -Ea/2.303RT Log K Ea = slope x -2.303 R1/12/2011 mov 36 18
- 19. 12/01/2011 Graphical Determination of activation energy...• Slope = -Ea/2.303xR• Ea = -2.303 x R x Slope• Log A = Intercept A = Antilog (Intercept)1/12/2011 mov 37 2. Determination of activation energy from rate constant method• From K = Ae-Ea/RT• lnK = lnA – Ea/RT• Let at Temperature T1 and rate constant K1; and at T2 and rate constant K2• For small change in temperature, the change Ea and A do not significantly change1/12/2011 mov 38 19
- 20. 12/01/2011 Determination of activation energy from rate constant method ...• Thus, lnK1 = lnA – Ea/RT1 3 lnK2 = lnA – Ea/RT2 4• Subtracting 3 from 4,• LnK2 – lnk1 = Ea/RT1 – Ea/RT2 5• logK2/K1 = Ea [1/T1 – 1/T2] 2.303• logK2/K1 = Ea [(T2-T1)/T1T2] 2.303R 1/12/2011 mov 39 Determination of activation energy from rate constant method... • Log K2/K1 = Ea [(T2-T1)/T1T2] Eq. 6 2.303R Since K1, K2, T1 and T2 are known, Ea can be calculated from the equation 6 1/12/2011 mov 40 20
- 21. 12/01/2011 Rate Laws• A rate law shows the relationship between the reaction rate and the concentrations of reactants. Exponents tell the order of the reaction with respect to each reactant.• This reaction is First-order in [NH4+] First-order in [NO2−]• The overall reaction order can be found by adding the exponents on the reactants in the rate law.• This reaction is second-order overall.• For gas-phase reactants use PA instead of [A] 1/12/2011 mov 41 Integrated Rate Laws• These are expressions which relate the concentration of the reactants with time• Application:1) Used to predict amount of reactant or product at a particular time2) Predict how long the reaction will take3) To predict when a toxic chemical can be disposed• Can be classified into Zero, first, second and third order reactions 1/12/2011 mov 42 21
- 22. 12/01/2011 Integrated Rate Laws cont.1) Zero order reactions• This is a reaction where the rate of reaction is independent of the concentration of the reactants• Derivation:• Let A Product• Initial conc. (mol/l) t = 0, a 0• At time t = t, a-x x• Where x is the concentration of the reactant (A) undergoing decomposition1/12/2011 mov 43 Zero order reactions cont.• Rate expression for zero order reaction:• dx/dt α (a-x)0 ........................................................Eq.1• Or dx/dt = Ko(a-x)o = ko .......................................Eq. 2• K0 is the rate constant for zero order reaction• Integrating the equation,• ∫dx = ∫Kodt = Ko ∫dt x = Kot + C, .............................................................Eq. 3• Where C is the integration constant1/12/2011 mov 44 22
- 23. 12/01/2011 Zero order reactions cont. • But when t = 0 and x = 0 • Thus, 0 = Ko x 0 + C .............................................Eq. 4 • Hence C = 0 • Substituting the value of C into equation 3 gives, x = Kot • Therefore, Ko = x/t ...........................................Eq. 5 • Eq.5 is the rate constant equation for zero order reaction 1/12/2011 mov 45 Zero order reactions cont.• Units for rate constant:• Ko = x/t = conc/time = molL-1/s = molL-1s-1• Half-life –This is the time duration in which half the concentration of reactants is transformed into products• Thus at t = t½, x = a/2,• Substituting these values into equation 5 K0 = a/2t½ t ½ = a/2k0 t½ α a• This shows that half-life of a zero order reaction is directly proportional to the initial concentration 1/12/2011 mov 46 23
- 24. 12/01/2011 Zero order reactions cont.• Graph for zero order reaction The rate of reaction is independent of the concentration of reactants Rate Concentration of reactants1/12/2011 mov 47 First Order Reactions• First Order Reactions• These are reactions where the sum of the powers of concentration of the exponential term in an experimentally established rate law is one• Thus the rate of reaction is dependent on the single power of the concentration term of the reactants• Example:• For the reaction, A Product1/12/2011 mov 48 24
- 25. 12/01/2011 First Order Reactions cont. • Initial conc (molL-1) at t=0, a o ......Eq. 1 • Conc (molL-1) at t=t a-x x .......Eq. 2 • Where x moles of A have decomposed into products in time “t” • Deferential, -d(a-x)/dt or dx/dt α(a-x) ...............EQ. 3 • Or dx/dt = k1(a-x) ...............................................Eq. 4 • Where, K1 is the rate constant for first order reaction. 1/12/2011 mov 49 First Order Reactions cont.• Rearrangement,• dx/(a-x) = k1dt ....................................................Eq. 5• Integrating, [identity ∫dx/x = ln x] ∫dx/(a-x) = k1 ∫dt -ln(a-x) = k1t + C ..............................................Eq. 6• Where C = constant of integration• But at t=0, x = o• Substituting the values into equation 6, 1/12/2011 mov 50 25
- 26. 12/01/2011 First Order Reactions cont.• -lna = C .............................................................Eq. 7• Substitute the value of C into equation c,• -ln(a-x)= k1t –lna ................................................Eq. 8• Rearranging• k1t = lna – ln(a-x)• K1 = ln[a/(a-x)]x 1/t• Change ln to log10 [lnx = 2.303logx]• K1 = (2.303/t)log(a/a-x) .......................................Eq.8• This is the expression for rate constant for first order reaction 1/12/2011 mov 51 First Order Reactions cont.• Unit of the rate constant• From Eq. 8• K1 = (2.303/t)log(a/a-x) = conc/(time xconc) k1 = 1/time = s-1 ...................................................................Eq. 9Graph for first order reaction• From equation 8• -ln(a-x)= k1t –lna• Multiply though by -1•1/12/2011 ln(a-x)= -k1t + lna mov 52 26
- 27. 12/01/2011 First Order Reactions cont. • Introduce log10 • 2.303log(a-x) = -k1t + 2.303 x log(a) • Or log(a-x) = -k1 x t + log(a) ...........................Eq. 10 2.303 Ξ y =mx +c Log(a) Slope = -k1/2.303Log(a-x) 1/12/2011 t mov 53 Second order reactions cont. • From the graph, • K1 = -2.303 x slope • Half-life of first order reaction • From Eq. 8, K1 = 2.303 Log (a/a-x) t • At t = t ½ , x = a/2 • Substituting the values into the equation • K1 = 2.303 Log (a/a - a/2) = 2.303 log 2 t½ t½ 1/12/2011 mov 54 27
- 28. 12/01/2011 First Order Reactions cont.• Making t ½ the subject,• t½ = 2.303 log 2 = 2.303 x 0.301 = 0.693 k1 k1 k1• Thus, t½ = 0.693/k1• This means that for 1st order reactions, half-life is independent of the initial concentration of reactants1/12/2011 mov 55 3. Second order reaction• These are reactions in which the sum of the powers of the of the concentration term in an experimentally established rate law is 2• There are two cases of second order reaction• Case 1: 2A Product• Thus, Rate law = dx/dt = K[A]2 Order = 2• Case 2: A+B Product• Rate law = dx/dt = k[A][B] order = 21/12/2011 mov 56 28
- 29. 12/01/2011 Second order reaction cont.• Case 1: Occurs when the concentration of both reactants is the same• Case 2: Occurs when the concentration of both reactants is NOT the same• 1) Case 1: Concentration of both reactants is the same• Let the reaction, A + A Product• Initial conc. t=0 a a 0 ..Eq. 1• Conc. At t = t a-x a-x x ...Eq. 2 1/12/2011 mov 57 Second order reaction cont.• Conc. is in mol L-1• Thus, x molL-1 of the reactant A decomposes in time t• Deferential,• dx/dt α (a-x)(a-x) or dx/dt α (a-x)2 ..................Eq. 3• Introduce rate constant k2• dx/dt = k2 (a-x)2 ......................................Eq. 4• Where k2 is the rateconstant for the second order reaction 1/12/2011 mov 58 29
- 30. 12/01/2011 Second order reaction cont.• Rearranging,• dx/ (a-x)2 = k2 dt .....................................................Eq. 5• Integrating,• ∫dx/(a-x)2 = k2∫dt note [∫xndx = xn+1/n+1]• Hence,• 1/(a-x) = k2t + C ...................................................Eq. 6• Where C = constant of integration• But when t= 0, x= 0 1/12/2011 mov 59 Second order reaction cont.• Substituting the values into the equation,• 1/a = C• Substitute the value of C into equation 6• 1/(a-x) = k2t + 1/a .............................Eq. 7• Thus, K2t = 1/(a-x) - 1/a• Hence, K2 = (1/at) x [x/(a-x)] ...................Eq. 8• Eq.8 is the expression for rate constant for second order reaction where initial conc. of the reactants is the same 1/12/2011 mov 60 30
- 31. 12/01/2011 Second order reaction cont.• Case 2: When concentration of both reactants is different• Let the reaction, A + B Product• Initial conc. t=0 a b 0 ..Eq. 1• Conc. At t = t a-x b-x x ...Eq. 2• Where x mol/L of A and B decomposed in time t to form product1/12/2011 mov 61 Second order reaction cont.• Differential• dx/dt α (a-x)(b-x)• dx/dt = K2(a-x)(b-x)• Where K2 is the rate constant for second order reaction• Separating the variables,• dx/[(a-x)(b-x)] = K2 dt ..........................................Eq. 31/12/2011 mov 62 31
- 32. 12/01/2011 Second order reaction cont.• Integration by parts• ∫ dx = ∫ A dx + ∫B dx = ∫k2dt ...Eq.4 (a-x)(b-x) (a-x) (b-x)• 1 = A + B ......................Eq.5 (a-x)(b-x) (a-x) (b-x)• Multiply through by (a-x)(b-x)• 1 = A(b-x) + B(a-x) ..............................................Eq. 61/12/2011 mov 63 Second order reaction cont.• When x = a, 1 = A(b-a) Thus, A = 1/(b-a)• When x = b, 1 = B(a-b) Thus, B = 1/(a-b)Or• ∫ dx = 1 ∫dx + 1 ∫dx = k2 ∫ dt ..Eq. 7 (a-x)(b-x) (b-a) (a-x) (a-b) (b-x)• Factorise the middle term by 1/(a-b)• = ∫ dx = 1 [ ∫dx - ∫dx ] = k2 ∫ dt ...Eq. 8 (a-x)(b-x) (a-b) (b-x) (a-x)1/12/2011 mov 64 32
- 33. 12/01/2011 Second order reaction cont.• ∫ dx = 1 [ -ln(b-x) + ln(a-x)] = k2 ∫ dt ..Eq. 9 (a-x)(b-x) (a-b)Therefore,• = 1 [ ln (a-x) ] = k2t + C ......................Eq. 10 (a-b) (b-x)When t = 0, x = 0,Substituting into the equation 14,1/12/2011 mov 65 Second order reaction cont.• = 1 [ ln (a ] = C .....................................Eq. 11 (a-b) (b)• Substitute the value of C in eq. 15 into eq. 14• = 1 [ ln (a-x) ] = k2t + 1 [ ln (a ]..........Eq. 12 (a-b) (b-x) (a-b) (b)Thus• K2 = 1 [ ln b(a-x) ] ............................Eq. 13 (a-b)t a(b-x)1/12/2011 mov 66 33
- 34. 12/01/2011 Second order reaction cont.• Applying log10• K2 = 2.303 log b(a-x) .............................................Eq. 14 (a-b)t a(b-x)Characteristics:Unit : K2 = 1 x x at (a-x) = 1/(conc x time) K2 = [conc. X time]-1 .............................................................Eq. 15 1/12/2011 mov 67 Second order reaction cont.• K2 = [mol-1s]-1 = mol-1Ls-1• When one of the reactants is in excess,• i.e. Let a>> b• a-b = a, a-x =a• Thus b and x can be neglected in comparison to a• Hence, K2 = 2.303 logb(a-x) ................................Eq. 16 (a-b)t a(b-x)• k2 = = 2.303 log(b/b-x)) .....................................Eq. 17 at 1/12/2011 mov 68 34
- 35. 12/01/2011 Second order reaction cont.• k2a = = 2.303 log(b/b-x)) ................................Eq. 18 t• k’ = = 2.303 log(b/b-x)) .................................Eq. 20 tWhere K2a = k’ = rate constant for the first order reaction1/12/2011 mov 69 Second order reaction cont.• Graph for second order reactions 1 lnb(a-x)(a-b) a(b-x) t1/12/2011 mov 70 35
- 36. 12/01/2011 Second order reaction cont.• Half life• From equation 8, case 1• K2 = 1 x x at (a-x)• t = t 1/2 , x = a/2• K2 = 1 x a/2 at ½ (a – a/2)• t ½ = 1/k2.a .....................................................Eq. 211/12/2011 mov 71 Second order reaction cont.• Thus t ½ is directly proportional to 1/a1/12/2011 mov 72 36

Full NameComment goes here.Lumala Philimonthanks, u make my studies very easyRick Middleton Jr, Chemistry Instructor 8 months agoNikom Klomkliang, Lecturer at Naresuan University 1 year ago