Unit 15 Chemical Kinetics

1. CHM 1046: General Chemistry and Qualitative Analysis Unit 15Chemical Kinetics Dr. Jorge L. Alonso Miami-Dade College – Kendall Campus Miami, FL • Textbook Reference: • Chapter # 16 • Module # 4

2. Thermodynamics vs Kinetics {Kinetics: paper, Fe, C} Rusting of Iron: 2Fe (s) + O2 (g) + 2H2O (l) → 2Fe(OH)2 (s) (with limited O2, magnetite Fe3O4 is formed: FeO·Fe2O3) Horxn = -884.6 kJ {Thermite.Rxn} Thermite Reaction: Fe2O3 (s) + 2 Al (s) Mg ignition Al2O3 (s) + 2 Fe (l) Horxn = -847.6 kJ {FSH1} {FSH2}

3. Kinetics • Studies the rate (speed) at which a chemical process occurs. • Kinetics also sheds light on the reaction mechanism (exactly how the reaction occurs). Factors That Affect Reaction Rates: • Physical State of the Reactants • Concentration of Reactants • Temperature • Presence of a Catalyst

4. Factors That Affect Reaction Rates Which will react faster? • Physical State of the Reactants (surface area) • In order to react, molecules must come in contact with each other: • The more homogeneous the mixture of reactants, the faster the molecules can react. • Gases, liquids or solutions react faster than solids. (Higher pressure and concentration also affects rate.) (1) Gases, Liquids, Solutions (High P & Conc.) • Finely ground substances have more surface areas and react faster than chunk pieces. {RxRate.LicopodiumPowder} (2) Solids

5. Factors That Affect Reaction Rates • Concentration of Reactants • As the concentration of reactants increases, so does the likelihood that reactant molecules will collide. {*Rxn.withConcOxy} 0.3 M 6 M {*RxRate&Conc.Mg+HCl} {RxRate&Conc.Mg+HClGraph}

6. A B Reaction Rates [A]& [B] [A] [B] [B] t -[A] t Rate = = {RxRateIntro} determined by monitoring the change in concentration of either reactants or products as a function of time. Spectrometer

7. Reaction Rates C4H9Cl(aq)+ H2O(l) C4H9OH(aq)+ HCl(aq) butanol butyl chloride -[A] t -[A] t [B] t -[A] t Rate = =

8. Ave. rate = -[C4H9Cl] t Reaction Rates C4H9Cl(aq) + H2O(l)  C4H9OH(aq) + HCl(aq) The average rate of the reaction over each interval is the change in concentration divided by the change in time: • Note that the average rate decreases as the reaction proceeds. • This is because as the reaction goes forward, there are fewer collisions between reactant molecules.

9. Reaction Rates C4H9Cl(aq) + H2O(l) C4H9OH(aq) + HCl(aq) Concentration vs. Time Graph • The slope of a line tangent to the curve at any point is the instantaneous rate at that time. • All reactions slow down over time. • Therefore, the best indicator of the rate of a reaction is the instantaneous rate near the beginning.

10. -[C4H9Cl] t Rate = = [C4H9OH] t Reaction Rates and Stoichiometry C4H9Cl(aq) + H2O(l) C4H9OH(aq) + HCl(aq) • In this reaction, the ratio of C4H9Cl to C4H9OH is 1:1 • Thus, the rate of disappearance of C4H9Cl is the same as the rate of appearance of C4H9OH.

11. a A + b B c C + d D 1 2 [HI] t [HI] t [HI] t Rate = − Rate1 = − Rate = − = = Rate2 = 2 [I2] t [I2] t [I2] t Reaction Rates and Stoichiometry • What if the ratio is not 1:1? 2 HI(g) H2(g) + I2(g) • To generalize, then, for the reaction How do rates compare? ≠

12. Practice Problems

14. N2(g) + 2 H2O(l) NH4+(aq) + NO2−(aq) How does Concentration affect Rate? • The data demonstrates: Rate  [NH4+] Rate  [NO2−] Rate  [NH4+] [NO2−] Data shows the relationship between the reaction rate and the conc. of reactants. This equation is the rate law, and k is the rate constant @ particular temp. or Rate = k[NH4+] [NO2−]

15. c C a A + b B Generalized Rate Laws Rate = k [A]x [B]y • The exponents, x and y, express the order of reaction and bear no necessary relationship to the coefficients of the balanced equation* – they must be determined experimentally! • This reaction is: x - order in [A] y - order in [B] Overall rate = x + y • The overall reaction order can be found by adding the exponents on the reactants in the rate law. • The previous reaction is second-order overall. * Only if reaction occurs in one step mechanism will x and y equal coefficients of balanced equation.

16. * Determination of Rate Law from Reaction Rate Data Possibilities for x and y: Zero order = no effect 1st order = linear effect 2nd order = exponential Rate = k [A]x [B]y What are the possible values for x and y? If rate not affected by [A], then order with respect to [A] is x = 0 [2A]  rate= k, [3A]  rate= ketc…. thesame applies to [B] If the rate affected by [A] in linear fashion, then order [A] is x = 1, [2A]1 rate= 2x,[3A]1 rate= 3x,etc…. the same applies to [B] If rate affected by [A] in exponential fashion, order [A] is x = 2, [2A]2 rate =4x, [3A]2 rate =9x,etc…. the same applies to [B] Rate= k [A]2 [B]0 = k [A]2

17. Products For reaction: a A 1 [A]0 1 [A]t = kt + Integrated Rate Laws -[A] t Rate = = k [A]x Reaction rate can be defined in two mathematical ways: (1) empirically, as change in conc. over time, or (2) as a function of concentration (rate law). Using calculus we can integrate the rate law equation to gives us a mathematical relationship that shows us how the concentration varies over a period of time. Rate expressions are then rearranged into linear equations. For zero order rxn (x=0) For first order rxn (x=1) For second order rxn (x=2) -[A] t -[A] t -[A] t = k = k [A]0 = k [A]1 = k [A]2  Integrate Integrate Integrate   ln [A]t = −kt + ln [A]0 [A]t = −kt + [A]0 y = mx + b y = mx + b y = mx + b

18. 1 [A]0 1 [A]t = kt + Integrated Rate Laws For zero order rxn (x=0) For first order rxn (x=1) For second order rxn (x=2) -[A] t -[A] t -[A] t = k [A]0 = k = k [A]1 = k [A]2 Integrate Integrate Integrate ln [A]t= −kt + ln [A]0 [A]t= −kt + [A]0 y = mx + b [A] [A] ln[A] [A] 1 [A] ln[A]

19. How many moles of X were initially in the flask? How many molecules of Y were produced in the first 20 minutes of the reaction? What is the order of this reaction with respect to X? Justify your answer. Write the rate law for this reaction. X(g) 2 Y(g) + Z (g)

20. X(g) 2 Y(g) + Z (g) Calculate the specific rate constant for this reaction. Specify units. Calculate the concentration of X in the flask after a total of 150 minutes of reaction. ln [A]t= −kt + ln [A]0

21. Practice Problems

22. 1 k[A]0 [A]0 2k 0.693 k * Half-Life {1stOrder&½Life} For a zero-order process, • Half-life is defined as the time required for one-half of a reactant to react. [A]0 [A]½ • Because [A] at t1/2 is one-half of the original [A], [A]t = 0.5 [A]0.  • t1/2 = For a first-order process, • t1/2 =  For a second-order process,  • t1/2 =

23. 0.5 [A]0 [A]0 1 k[A]0 1 [A]0 1 [A]0 1 [A]0 2 [A]0 2 [A]0 1 0.5 [A]0 = kt1/2 + ln = −kt1/2 1 [A]0 1 [A]t = kt1/2 + = kt + 0.693 k = t1/2 - = kt1/2 = t1/2 Half-Life For a second-order process, ln [A]t = −kt + ln [A]0 ln 0.5[A]0 = −kt1/2 + ln [A]0 For a first-order process, ln 0.5 = −kt1/2 −0.693 = −kt1/2 NOTE: For a first-order process, the half-life does not depend on [A]0.

24. 0.693 k = t1/2 Practice Problems ln [A]t = −kt + ln [A]0 ln 0.5[A]0 = −kt½ + ln [A]0 For each of the following rate expression, determine the units of the rate constant, k. -[A] t -[A] t -[A] t = k [A]0 = k [A]1 = k [A]2 -[M] t -[M] t -[M] t = k [M]2 = k [M]1 = k [M]0

25. Kinetics Factors That Affect Reaction Rates: Rate (s)  (l ) (g) • Physical State of the Reactants -[A] t • Concentration of Reactants Rate = = k [A]x • Temperature • Activation Energy (Transition State Theory) • Reaction Mechanisms • Presence of a Catalyst

26. Factors That Affect Reaction Rates • Temperature At higher temperatures, reactant molecules have more kinetic energy, move faster, and collide more often and with greater energy. {RxRate&Temp} ln [A]t= −kt + ln [A]0 • Generally, as temperature increases, so does the reaction rate. • This is becausekis temperature dependent. • k is also dependent on activation energy.

27. Activation Energy: The Collision Model • In a chemical reaction, bonds are broken and new bonds are formed. • Molecules can only react if they collide with each other with sufficient (activation) energy (Ea). O3 + NO O2 + NO2 ( ) Reactants Activated Complex Products + + {Ea.CollisionEnergy} Furthermore, molecules must collide with the correct orientation and with enough energy to cause bond breakage and formation. {Ea.Orientation} {Ea.Ener+Orient}

28. Transition State Theory Potential Energy - DH Reactants Products Reaction Coordinate * EnergyReaction Coordinate Diagrams: Reactants ( Activated Complex ) Products + + - DH Transition state (Energy Level)) X3-YZ Activated Complex (the molecule) Ea = Activation Energy {Ea&TransS tate}

29. Maxwell–Boltzmann Distributions • At any temperature there is a wide distribution of kinetic energies. • Temperature is defined as a measure of the average kinetic energy of the molecules in a sample. • As the temperature increases, the curve flattens and broadens. • Thus at higher temperatures, a larger population of molecules has higher energy. • If the dotted line represents the activation energy, as the temperature increases, so does the fraction of molecules that can overcome the activation energy barrier. • As a result, the reaction rate increases.

30. Maxwell–Boltzmann Distributions This fraction of molecules can be found through the expression where R is the gas constant and T is the Kelvin temperature. f = e−Ea/RT

31. Arrhenius Equation Svante Arrhenius developed a mathematical relationship between the rate constantk , the temperature (T) at which the reaction occurs and the activation energyEa: k = A e−Ea/RT where A is the frequency factor, a number that represents the likelihood that collisions would occur with the proper orientation for reaction.

32. 1 T k = A e−Ea/RT Arrhenius Equation Taking the natural logarithm of both sides, the equation becomes • ln k = -Ea() + ln A  R y = m x + b Therefore, if k is determined experimentally at several temperatures, Eacan be calculated from the slope of a plot of ln k vs. 1/T. Problem: Calculate the activation energy (in J/mol) for the reaction in plot above. R= 8.31 J/mol·K

33. Reaction Mechanisms The detailed sequence of events that describes the actual pathway by which reactants become products. OH- (aq) + CH3Cl (g)  CH3OH (aq) + Cl-(aq) Methyl alcohol Methyl chloride Transition State Reactants Activated Complex Products {RxMecha.Bimolecular.Intro}

34. Reaction Mechanisms * NO2(g) + CO (g) NO (g) + CO2(g) Bimolecular mechanism: conc. of both reactants affects rate. Consider the following reaction: {RxMechanism.NO2+CO.Prop1} Experimental Evidence: reaction rate is second order in[NO2] & does not depend on[CO] at all, even though CO is required for reaction to occur. Rate = k [NO2]2 • A proposed mechanism for this reaction is • Step 1: NO2 + NO2 NO3* + NO (slow) • Step 2: NO3* + CO  NO2 + CO2 (fast) {Movie1} {Movie 2} NO3* = intermediate reactant • The overall reaction cannot occur faster than this slowest, rate-determining step.

35. Determining Rx Mechanisms Using radioactive isotope labeling can help us to experimentally determine the reaction mechanism. The simplest proposed mechanism is NO2(g) + CO (g) NO(g) + CO2(g) ½ labeled • Better proposed mechanism is: • Step 1: NO2 + NO2 NO3 + NO (slow) • Step 2: NO3 + CO  NO2 + CO2 (fast) {DeterRxMechanismIsotopLabel1.NO2+CO} {DeterRxMechanismIsotopLabel2.NO2+CO}

36. Factors That Affect Reaction Rates • Presence of a Catalyst • Catalysts speed up reactions. • Catalysts are not consumed during the course of the reaction. MnO2 2 H2O2 (l) 2 H2O (l) + O2 (g) {*Catalysis of H2O2 by MnO2} H2O {Catalyst of SO2 + H2S} SO2 + 2 H2S 2 H2O(l) + 3 S(aq)

37. Catalysts Increase the reaction rateby changing the mechanism, thus also changing (decreasing) the activation energy by which the process occurs.. Ea Add catalyst: Ea

38. Surface Catalysis Some Reactions an in Internal Combustion Engine: 2 C8H18 (l) + 25 O2 (g) 16 CO2 (g) + 18 H2O (g) (+heat) N2 (g) + O2 (g)  2 NO (g)(causes acid rain & ozone depletion)) Pt {Pt Catalytic Converter: 2 NO(g) O2(g) + N2(g)} NO N2 O2 NO Pt Pt Surface Atoms recombine to form product, which are then released from surface Reactant molecules attach to Catalytic Surface Bonds of attached molecules are Broken

39. Catalytic Converters The catalyst (in the form of platinum and palladium) is coated onto a ceramic honeycomb or ceramic beads that are housed in a muffler-like package attached to the exhaust pipe. The catalyst helps to convert carbon monoxide into carbon dioxide. It converts the hydrocarbons into carbon dioxide and water. It also converts the nitrogen oxides back into nitrogen and oxygen.

40. Catalysis Ni H2 + H2C=CH2 H3C-CH3 Ethylene Ethane One way a catalyst can speed up a reaction is by holding the reactants together and helping bonds to break. {Surface.Catalysis.Hydrogenation}

41. Enzymes: biological catalysts substrate • Lock and Key Theory: the substrate (reactant) fits into the active siteof the enzyme much like a key fits into a lock. enzyme

44. Additional Practice Problems Where are the answers?

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