22 Nov. 2010 1 Objective: SWBAT write rate expressions and calculate reaction rates for chemical reactions. Do now: Describe one very slow reaction that you’ve seen, and one very fast reaction. Agenda Do now II. Kinetics notes III. Reaction Rates Demonstrations IV. Rate constant and reaction rates problems. Homework: p. 602 #2, 3, 5, 7, 12, 13, 15, 17, 19 I. 3 How can we predict whether or not a reaction will take place? Thermodynamics Once started, how fast does the reaction proceed? Chemical kinetics: this unit! How far will the reaction go before it stops? Equilibrium: next unit Chemical Kinetics The area of chemistry concerned with the speeds, or rates, at which a chemical reaction occurs. reaction rate: the change in the concentration of a reactant or product with time (M/s) Why do reactions have such very different rates? Steps in vision: 10-12 to 10-6 seconds! Graphite to diamonds: millions of years! In chemical industry, often more important to maximize the speed of a reaction, not necessarily yield. Chemical Kinetics Reaction rate is the change in the concentration of a reactant or a product with time (M/s). A B D[A] rate = Dt D[A] = change in concentration of A over time period Dt D[B] rate = Dt D[B] = change in concentration of B over time period Dt Because [A] decreases with time, D[A] is negative. A D[A] rate = Dt D[B] rate = Dt 6 B red-brown 2Br- (aq) + 2H+ (aq) + CO2 (g) Br2 (aq) + HCOOH (aq) time t 1< t 2 < t 3 393 nm light 7 Detector D[Br2] a D Absorption Br2 (aq) + HCOOH (aq) 2Br- (aq) + 2H+ (aq) + CO2 (g) slope of tangent slope of tangent slope of tangent [Br2]final – [Br2]initial D[Br2] average rate = =Dt tfinal - tinitial 8 instantaneous rate = rate for specific instance in time rate a [Br2] rate = k [Br2] rate = rate constant k= [Br2] = 3.50 x 10-3 s-1 2H2O2 (aq) 2H2O (l) + O2 (g) PV = nRT n P= RT = [O2]RT V 1 [O2] = P RT D[O2] 1 DP rate = = RT Dt Dt measure DP over time 11 Reaction Rates and Stoichiometry 2A B Two moles of A disappear for each mole of B that is formed. 1 D[A] rate = 2 Dt aA + bB D[B] rate = Dt cC + dD 1 D[A] 1 D[B] 1 D[C] 1 D[D] rate = == = a Dt b Dt c Dt d Dt 12 Example Write the rate expression for the following reaction: CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (g) Write the rate expression for the following reaction: CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (g) D[CH4] D[CO2] 1 D[O2] 1 D[H2O] rate = = == Dt Dt Dt 2 Dt 2 14 Practice Problems Write the rate expressions for the following reactions in terms of the disappearance of the reactants and appearance of products. a. I-(aq) + OCl-(aq) Cl-(aq) + OI-(aq) b. 4NH3(g) + 5O2(g) 4NO(g) + 6H2O(g) Using Rate Expressions Consider the reaction: 4NO2(g) + O2(g) 2N2O5(g) Suppose that, at a particular moment during the reaction, molecular oxygen is reacting at the rate of 0.024 M/s. a. At what rate is N2O5 being formed? b. At what rate is NO2 reacting? Consider the reaction: 4PH3(g) P4(g) + 6H2(g) Suppose that, at a particular moment during the reaction, molecular hydrogen is being formed at the rate of 0.078 M/s. a. At what rate is P4 being formed? b. At what rate is PH3 reacting? The Rate Law The rate law expresses the relationship of the rate of a reaction to the rate constant and the concentrations of the reactants raised to some powers. aA + bB cC + dD Rate = k [A]x[B]y Reaction is xth order in A Reaction is yth order in B Reaction is (x +y)th order overall x and y are determined experimentally! F2 (g) + 2ClO2 (g) 2FClO2 (g) rate = k [F2]x[ClO2]y Double [F2] with [ClO2] constant Rate doubles rate = k [F2][ClO2] x=1 Quadruple [ClO2] with [F2] constant Rate quadruples y=1 Rate Laws • Rate laws are always determined experimentally. • Reaction order is always defined in terms of reactant (not product) concentrations. • The order of a reactant is not related to the stoichiometric coefficient of the reactant in the balanced chemical equation. F2 (g) + 2ClO2 (g) 2FClO2 (g) rate = k [F2][ClO2] 1 20 Determine the rate law and calculate the rate constant for the following reaction from the following data: S2O82- (aq) + 3I- (aq) 2SO42- (aq) + I3- (aq) Experiment [S2O82-] [I-] Initial Rate (M/s) 1 0.08 0.034 2.2 x 10-4 2 0.08 0.017 1.1 x 10-4 3 0.16 0.017 2.2 x 10-4 Determine the rate law and calculate the rate constant for the following reaction from the following data: S2O82- (aq) + 3I- (aq) 2SO42- (aq) + I3- (aq) Experiment [S2O82-] [I-] Initial Rate (M/s) 1 0.08 0.034 2.2 x 10-4 2 0.08 0.017 1.1 x 10-4 3 0.16 0.017 2.2 x 10-4 rate = k [S2O82-]x[I-]y y=1 x=1 rate = k [S2O82-][I-] Double [I-], rate doubles (experiment 1 & 2) Double [S2O82-], rate doubles (experiment 2 & 3) 22 2.2 x 10-4 M/s rate k= = = 0.08/M•s 2[S2O8 ][I ] (0.08 M)(0.034 M) Practice Problems The reaction of nitric oxide with hydrogen at 1280oC: 2NO(g) + 2H2(g) N2(g) + 2H2O(g) From the following data collected at this temperature, determine (a) the rate law, (b) the rate constant and (c) the rate of the reaction when [NO] = 12.0x10-3 M and [H2] = 6.0x10-3 M 23 Experiment [NO] M [H2] M Initial Rate (M/s) 1 5.0x10-3 2.0x10-3 1.3x10-5 2 10.0x10-3 2.0x10-3 5.0x10-5 3 10.0x10-3 4.0x10-3 10.0x10-5 23 Nov. 2010 24 Objective: SWBAT write rate expressions and calculate reaction rates for chemical reactions. Do now: Calculate the rate constant (k) from the data below: Agenda 25 Do now II. Practice Problems Homework: Revisit last night’s assignment #19 hint: Graph lnP vs. time. If linear, it is 1st order. slope = -k Excel works great. I. Write the reaction rate expressions for the following in terms of the disappearance of the reactants and the appearance of products: a) 2H2(g) + O2(g) 2H2O(g) b) 4NH3(g) + 5O2(g) 4NO(g) + 6H2O(g) Consider the reaction N2(g) + 3H2(g) 2NH3(g) Suppose that at a particular moment during the reaction molecular hydrogen is reacting at a rate of 0.074 M/s. a) At what rate is ammonia being formed? b) At what rate is molecular nitrogen reacting? Calculate the rate of the reaction at the time when [F2] = 0.010 M and [ClO2] = 0.020 M. F2(g) + 2ClO2(g) 2FClO2(g) [F2] (M) [ClO2] (M) Initial Rate (M/s) 0.10 0.010 1.2x10-3 0.10 0.040 4.8x10-3 0.20 0.010 2.4x10-3 Consider the reaction X + Y Z From the following data, obtained at 360 K, a) determine the order of the reaction b) determine the initial rate of disappearance of X when the concentration of X is 0.30 M and that of Y is 0.40 M 29 Initial Rate of Disappearance of X (M/s) [X] (M) [Y] (M) 0.053 0.10 0.50 0.127 0.20 0.30 1.02 0.40 0.60 0.254 0.20 0.60 0.509 0.40 0.30 30 Consider the reaction A B. The rate of the reaction is 1.6x10-2 M/s when the concentration of A is 0.35 M. Calculate the rate constant if the reaction is a. first order in A b. second order in A 24 Nov. 2010 Objective: SWBAT determine reaction order graphically and relate concentration of 1st order reactions to time. Do now: What is the overall order of reaction for the data shown below? What is the rate constant? Run # Initial [A] ([A]0) Initial [B] ([B]0) Initial Rate (v0) 1 1.00 M 1.00 M 1.25 x 10-2 M/s 2 1.00 M 2.00 M 2.5 x 10-2 M/s 3 2.00 M 2.00 M 2.5 x 10-2 M/s Agenda Do now II. Homework solutions III. First order reactions IV. Time calculations for first-order reactions. Homework: p. 603 #19, 21, 22, 23, 25-29 I. The rate laws can be used to determine the concentrations of any reactants at any time during the course of a reaction. 29 Nov. 2010 34 Take Out Homework p. 603 #19, 21, 22, 23, 2529 Objective: SWBAT compare 1st order, 2nd order, and zero order reactions, and describe how temperature and activation energy effect the rate constant. Do now: Calculate the half life of the reaction F2(g) + 2ClO2(g) 2FClO2(g), with rate data shown [ClO2] (M) Initial Rate (M/s) below: [F2] (M) 0.10 0.010 1.2x10-3 0.10 0.040 4.8x10-3 0.20 0.010 2.4x10-3 Agenda 35 Homework solutions st II. Review 1 order reactions III. Second and zero order reactions IV. Activation Energy V. Problem Set Quiz Weds. p. 31, 32, 35, 37, 39, 42 + Problem set I. First-Order Reactions A k= product D[A] rate = Dt rate M/s = = 1/s or s-1 M [A] [A] = [A]0e−kt rate = k [A] D[A] = k [A] Dt [A] is the concentration of A at any time t [A]0 is the concentration of A at time t=0 ln[A] = ln[A]0 - kt Graphical Determination of k 2N2O5 37 4NO2 (g) + O2 (g) The reaction 2A B is first order in A with a rate constant of 2.8 x 10-2 s-1 at 800C. How long will it take for A to decrease from 0.88 M to 0.14 M ? The reaction 2A B is first order in A with a rate constant of 2.8 x 10-2 s-1 at 800C. How long will it take for A to decrease from 0.88 M to 0.14 M ? [A]0 = 0.88 M ln[A] = ln[A]0 - kt [A] = 0.14 M kt = ln[A]0 – ln[A] ln[A]0 – ln[A] = t= k 39 ln [A]0 [A] k ln = 0.88 M 0.14 M 2.8 x 10-2 s-1 = 66 s The conversion of cyclopropane to propene in the gas phase is a first order reaction with a rate constant of 6.7x10-4 s-1 at 500oC. a) b) c) If the initial concentration of cyclopropane was 0.25 M, what is the concentration after 8.8 minutes? How long, in minutes, will it take for the concentration of cyclopropane to decrease from 0.25 M to 0.15 M? How long, in minutes, will it take to convert 74% of the starting material? Practice Problem The reaction 2A → B is first order in A with a rate constant of 2.8×10-2s-1 at 80oC. How long, in seconds, will it take for A to decrease from 0.88 M to 0.14 M? The rate of decomposition of azomethane (C2H6N2) is studied by monitoring partial pressure of the reactant as a function of time: CH3-N=N-CH3(g) → N2(g) + C2H6(g) The data obtained at 300oC are shown here: Time (s) Partial Pressure of Azomethane (mmHg) 0 284 100 220 150 193 200 170 250 150 300 132 Are these values consistent with first-order kinetics? If so, determine the rate constant. 43 The following gas-phase reaction was studied at 290oC by observing the change in pressure as a function of time in a constant-volume vessel: ClCO2CCl3(g) 2COCl2(g) Determine the order of the reaction and the rate constant based on the following data, where P is the total pressure Time (s) P (mmHg) 0 400 2,000 316 4,000 248 6,000 196 8,000 155 10,000 122 Ethyl iodide (C2H5I) decomposes at a certain temperature in the gas phase as follows: C2H5I(g) → C2H4(g) + HI(g) From the following data, determine the order of the reaction and the rate constant: Time (min) [C2H5I] (M) 0 0.36 15 0.30 30 0.25 48 0.19 75 0.13 First-Order Reactions The half-life, t½, is the time required for the concentration of a reactant to decrease to half of its initial concentration. t½ = t when [A] = [A]0/2 ln t½ = [A]0 [A]0/2 k ln 2 0.693 = = k k What is the half-life of N2O5 if it decomposes with a rate constant of 5.7 x 10-4 s-1? How do you know decomposition is first order? First-Order Reactions The half-life, t½, is the time required for the concentration of a reactant to decrease to half of its initial concentration. t½ = t when [A] = [A]0/2 ln t½ = [A]0 [A]0/2 k ln 2 0.693 = = k k What is the half-life of N2O5 if it decomposes with a rate constant of 5.7 x 10-4 s-1? 0.693 t½ = ln 2 = = 1200 s = 20 minutes -4 -1 k 5.7 x 10 s How do you know decomposition is first order? 46 units of k (s-1) First-order reaction A product # of half-lives [A] = [A]0/n 1 2 47 2 4 3 8 4 16 48 The decomposition of ethane (C2H6) to methyl radicals is a first-order reaction with a rate constant of 5.36x10-4 s-1 at 700oC: C2H6(g) 2CH3(g) Calculate the half-life of the reaction in minutes. 49 Calculate the half-life of the decomposition of N2O5: 2N2O5 4NO2(g) + O2(g) t (s) [N2O5] (M) ln [N2O5] 0 0.91 -0.094 300 0.75 -0.29 600 0.64 -0.45 1200 0.44 -0.82 3000 0.16 -1.83 Second-Order Reactions A product D[A] rate = Dt rate M/s = = 1/M•s k= 2 2 M [A] 1 1 = + kt [A] [A]0 50 D[A] = k [A]2 Dt [A] is the concentration of A at any time t [A]0 is the concentration of A at time t=0 t½ = t when [A] = [A]0/2 1 t½ = k[A]0 rate = k [A]2 Iodine atoms combine to form molecular iodine in the gas phase: I(g) + I(g) I2(g) This reaction follows second-order kinetics and has the high rate constant 7.0x109/M·s at 23oC. a. If the initial concentration of I was 0.086 M, calculate the concentration after 2.0 minutes. b. Calculate the half-life of the reaction if the initial concentration of I is 0.60 M and if it is 0.42 M. The reaction 2A → B is second order with a rate constant of 51/M·min at 24oC. a. Starting with [A]o = 0.0092 M, how long will it take for [A]t = 3.7x10-3 M? b. Calculate the half-life of the reaction. Zero-Order Reactions A product D[A] rate = Dt D[A] =k Dt rate = M/s k= 0 [A] [A] = [A]0 - kt [A] is the concentration of A at any time t [A]0 is the concentration of A at time t = 0 t½ = t when [A] = [A]0/2 [A]0 t½ = 2k 53 rate = k [A]0 = k Summary of the Kinetics of Zero-Order, First-Order and Second-Order Reactions Order 0 54 Rate Law rate = k 1 rate = k [A] 2 [A]2 rate = k Concentration-Time Equation [A] = [A]0 - kt ln[A] = ln[A]0 - kt 1 1 = + kt [A] [A]0 Half-Life t½ = [A]0 2k t½ = ln 2 k 1 t½ = k[A]0 55 Activation Energy and Temperature Dependence of Rate Constants Reaction rates increase with increasing temperature Ex: Hard boiling an egg Ex: Storing food How do reactions get started in the first place? Collision Theory 56 Chemical reactions occur as a result of collisions between reacting molecules. reaction rate depends on concentration But, the relationship is more complicated than you might expect! Not all collisions result in reaction (p. 582-583) 57 In order to react, colliding molecules must have a total KE ≥ activation energy (Ea) Ea: minimum amount of energy required to initiate a chemical reaction activated complex (transition state): a temporary species formed by the reactant molecules as a result of the collision before they form the product. A+B Exothermic Reaction + AB+ C+D Endothermic Reaction The activation energy (Ea ) is the minimum amount of energy required to initiate a chemical reaction. =a barrier that prevents less energetic molecules from reacting Temperature Dependence of the Rate Constant k A e( Ea / RT ) (Arrhenius equation) Ea is the activation energy (J/mol) R is the gas constant (8.314 J/K•mol) T is the absolute temperature A is the frequency factor (constant across wide range of temps.) Alternate format (y=mx+b): Ea 1 ln k = + lnA R T 59 Alternate Form of the Arrhenius Equation At two temperatures, T1 and T2 or 60 The rate constants for the decomposition of acetaldehyde: CH3CHO(g) → CH4(g) + CO(g) were measured at five different temperatures. The data are shown below. Plot lnk versus 1/T, and determine the activation energy (in kJ/mol) for the 3 reaction. (Note: the reaction is 2 order in CH3CHO, 1 so k has the units of 1 / M 2 s ) 1 2 k (1 / M s) 0.011 0.035 T (K) 700 730 0.105 0.343 0.789 760 790 810 0.00 0.0012 0.00125 0.0013 0.00135 0.0014 0.00145 -0.50 -1.00 -1.50 slope = -2.19x104 K slope = E a ln K -2.00 -2.50 -3.00 R -3.50 y = -21881x + 26.662 -4.00 -4.50 -5.00 1/T (K-1) The second order rate constant for the decomposition of nitrous oxide (N2O) into nitrogen molecule and oxygen atom has been measured at different temperatures. Determine graphically the activation energy for the reaction. 63 k (1 / M s) 1.87x10-3 0.0113 T (oC) 600 650 0.0569 0.244 700 750 30 Nov. 2010 Take Out Homework p. 605# 31, 32, 35, 37, 39 Objective: SWBAT use the Arrhenius equation to solve for rate constants and temperatures, and solve practice problems on kinetics. Do now: Match 64 Order Rate Law 2 rate = k[A] Conc-Time Eq. Half Life Eq. [A]=[A]0-kt t1/2=1/k[A]o 1 rate = k[A]2 1/[A]=1/[A]0 + kt 0 rate = k ln[A]=ln[A]0 –kt t1/2=ln2/k t1/2=[A]0/2k Agenda 65 Homework solutions II. Using the Arrhenius equation part 2 III. Molecular orientation IV. Problem Set work time Homework: Complete problem set and p. 605 #40, 42 Quiz tomorrow I. 66 The rate constant of a first order reaction is 3.46x10-2 /s at 298 K. What is the rate constant at 350 K if the activation energy for the reaction is 50.2 kJ/mol? 67 The first order rate constant for the reaction of methyl chloride (CH3Cl) with water to produce methanol (CH3OH) and hydrochloric acid (HCl) is 3.32x10-10/s at 25oC. Calculate the rate constant at 40oC if the activation energy is 116 kJ/mol. Homework 68 Quiz Weds. p. 604 #40, 42 + Problem set #2, 18, 20, 25, 29, 32, 37, 42, 55, 57: Weds. Problem Set 69 Handout #2, 18, 20, 25, 29, 32, 37, 42, 55, 57: Weds. Quiz Weds. 2 Dec. 2010 70 Objective: SWBAT describe a reaction in terms of elementary steps and determine the rate law. Do now: Which graph represents an exothermic reaction? Which is endothermic? Why? Agenda 71 Do now II. Elementary steps in reactions III. Catalysts Homework: p. 605 #45, 47, 48, 52, 54, 55, 56, 58, 62: Mon. I. 72 For simple reactions (between atoms, for example) the frequency factor (A) is proportional to the frequency of collision between the reacting species. “Orientation factor” is also important. Importance of Molecular Orientation effective collision ineffective collision 73 74 A balanced chemical equation doesn’t tell us much about how the reaction actually takes place. It may represent the sum of elementary steps Reaction mechanism: the sequence of elementary steps that leads to product formation. Reaction Mechanisms The overall progress of a chemical reaction can be represented at the molecular level by a series of simple elementary steps or elementary reactions. The sequence of elementary steps that leads to product formation is the reaction mechanism. 2NO (g) + O2 (g) 2NO2 (g) N2O2 is detected during the reaction! 75 Elementary step: NO + NO N 2O 2 + Elementary step: N2O2 + O2 2NO2 Overall reaction: 2NO + O2 2NO2 2NO (g) + O2 (g) Mechanism: 76 2NO2 (g) Intermediates are species that appear in a reaction mechanism but not in the overall balanced equation. An intermediate is always formed in an early elementary step and consumed in a later elementary step. Elementary step: NO + NO N 2O 2 + Elementary step: N2O2 + O2 2NO2 Overall reaction: 2NO + O2 2NO2 The molecularity of a reaction is the number of molecules reacting in an elementary step. • Unimolecular reaction – elementary step with 1 molecule • Bimolecular reaction – elementary step with 2 molecules • Termolecular reaction – elementary step with 3 molecules 77 Rate Laws and Elementary Steps Unimolecular reaction A products rate = k [A] Bimolecular reaction A+B products rate = k [A][B] Bimolecular reaction A+A products rate = k [A]2 Writing plausible reaction mechanisms: • The sum of the elementary steps must give the overall balanced equation for the reaction. • The rate-determining step should predict the same rate law that is determined experimentally. The rate-determining step is the slowest step in the sequence of steps leading to product formation. 78 Sequence of Steps in Studying a Reaction Mechanism 79 The experimental rate law for the reaction between NO2 and CO to produce NO and CO2 is rate = k[NO2]2. The reaction is believed to occur via two steps: Step 1: NO2 + NO2 NO + NO3 Step 2: NO3 + CO NO2 + CO2 What is the equation for the overall reaction? What is the intermediate? What can you say about the relative rates of steps 1 and 2? The experimental rate law for the reaction between NO2 and CO to produce NO and CO2 is rate = k[NO2]2. The reaction is believed to occur via two steps: Step 1: NO2 + NO2 NO + NO3 Step 2: NO3 + CO NO2 + CO2 What is the equation for the overall reaction? NO2+ CO NO + CO2 What is the intermediate? NO3 What can you say about the relative rates of steps 1 and 2? 81 rate = k[NO2]2 is the rate law for step 1 so step 1 must be slower than step 2 82 rate determining step: the slowest step in the sequence of steps leading to product formation. Decomposition of Hydrogen Peroxide 83 2H2O2(aq) 2H2O(l) + O2(g) Can be catalyzed using iodide ions (I-) rate = k[H2O2][I-] Why?! Determined experimentally. Step 1: H2O2 + I- H2O + IOStep 2: H2O2 + IO- H2O + O2 + I- 7 Dec. 2010 84 Objective: SWBAT describe catalysis and how ammonia is synthesized through the Haber process. Do now: The rate for the reaction 2H2O2(aq) 2H2O(l) + O2(g) is rate = k[H2O2][I-] Step 1: H2O2 + I- H2O + IOStep 2: H2O2 + IO- H2O + O2 + IWhich is the rate determining step? Why? Agenda 85 Do now II. Homework solutions III. Catalysts and catalyzing mechanisms IV. AP Kinetics problem set Homework: p. 605 #55, 56, 58 Bring your book on Thursday! I. Example The gas-phase decomposition of nitrous oxide (N2O) is believed to occur via two elementary steps: Step 1: N2O N2 + O Step 2 N2O + O N2 + O2 Experimentally the rate law is found to be rate = k[N2O]. a) Write the equation for the overall reaction. b) Identify the intermediates. c) What can you say about the relative rates of steps 1 and 2? 86 87 For the decomposition for H2O2, the reaction rate depends on the concentration of I- ions, even though Idoesn’t appear in the overall equation. I- is a catalyst for the reaction. A catalyst is a substance that increases the rate of a chemical reaction without itself being consumed. k A e( Ea / RT ) Ea Catalyzed k Uncatalyzed ratecatalyzed > rateuncatalyzed 88 Ea′ < Ea Catalysts 89 forms an alternative reaction pathway for example, it might form an intermediate with the reactant. Ex: 2KClO3(s) 2KCl(s) + 3O2(g) Very slow, until you add MnO2, a catalyst. The MnO2 can be recovered at the end of the reaction! In heterogeneous catalysis, the reactants and the catalysts are in different phases (usually, catalyst is a solid, reactants are gases or liquids). • Haber synthesis of ammonia • Ostwald process for the production of nitric acid • Catalytic converters In homogeneous catalysis, the reactants and the catalysts are dispersed in a single phase, usually liquid. 90 • Acid catalysis • Base catalysis Haber Process Synthesis of Ammonia Extremely slow at room temperature. Must be fast and high yield! Process occurs on the surface of the Fe/Al2O3/K2O catalyst, which weakens the covalent N-N and H-H bonds. Fe/Al2O3/K2O N2 (g) + 3H2 (g) catalyst 91 2NH3 (g) Ostwald Process 4NH3 (g) + 5O2 (g) Pt catalyst 2NO (g) + O2 (g) 2NO2 (g) + H2O (l) 92 4NO (g) + 6H2O (g) 2NO2 (g) HNO2 (aq) + HNO3 (aq) Pt-Rh catalysts used in Ostwald process Catalytic Converters catalytic CO + Unburned Hydrocarbons + O2 converter CO2 + H2O catalytic 2NO + 2NO2 converter 2N2 + 3O2 93 Enzyme Catalysis biological catalysts 94 Binding of Glucose to Hexokinase 95 96 p. 605 #45, 47, 48, 52, 54, 55, 56, 58, 62