11/9/2012 CHEMICAL REACTIONS. 1. Will the reaction occur? 2. How far will the reaction proceed? 3. How fast will the reaction occur?

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1 CHEMICAL REACTIONS LECTURE 11: CHEMICAL KINETICS 1. Will the reaction occur? 2. How far will the reaction proceed? 3. How fast will the reaction occur? CHEMICAL REACTIONS C(s, diamond) C(s, graphite) G rxn = kj/mole KINETICS Studies the rate at which a chemical process occurs. Besides information about the speed at which reactions occur, kinetics also sheds light on the reaction mechanism (exactly how the reaction occurs). Is the reaction favorable? 1

2 OUTLINE : KINETICS Factors That Influence Reaction Rate Reaction Rates Rate Laws Integrated Rate Laws Half-life Arrhenius Equation Mechanisms How we measure rates. How the rate depends on amounts of reactants. How to calculate amount left or time to reach a given amount. How long it takes to react 50% of reactants. How rate constant changes with Temperature. Link between rate and molecular scale processes. Under a specific set of conditions, every reaction has its own characteristic rate, which depends upon the chemical nature of the reactants. Four factors can be controlled during the reaction: 1. Concentration - molecules must collide to react 2. Physical state - molecules must mix to collide 3. Temperature - molecules must collide with enough energy to react 4. The use of a catalyst FACTORS THAT AFFECT REACTION RATES REACTION RATES Concentration of Reactants As the concentration of reactants increases, so does the likelihood that reactant molecules will collide. Temperature At higher temperatures, reactant molecules have more kinetic energy, move faster, and collide more often and with greater energy. Catalysts Speed up the reaction by changing mechanism. Rates of reactions can be determined by monitoring the change in concentration of either reactants or products as a function of time. [A] vs t 2

3 REACTION RATES C 4 H 9 Cl(aq) + H 2 O(l) C 4 H 9 OH(aq) + HCl(aq) REACTION RATES C 4 H 9 Cl(aq) + H 2 O(l) C 4 H 9 OH(aq) + HCl(aq) [C 4 H 9 Cl] M In this reaction, the concentration of butyl chloride, C 4 H 9 Cl, was measured at various times, t. Average Rate, M/s The average rate of the reaction over each interval is the change in concentration divided by the change in time: REACTION RATES C 4 H 9 Cl(aq) + H 2 O(l) C 4 H 9 OH(aq) + HCl(aq) 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. REACTION RATES C 4 H 9 Cl(aq) + H 2 O(l) C 4 H 9 OH(aq) + HCl(aq) A plot of concentration vs. time for this reaction yields a curve like this. The slope of a line tangent to the curve at any point is the instantaneous rate at that time. 3

4 REACTION RATES C 4 H 9 Cl(aq) + H 2 O(l) C 4 H 9 OH(aq) + HCl(aq) The reaction slows down with time because the concentration of the reactants decreases. REACTION RATES AND STOICHIOMETRY C 4 H 9 Cl(aq) + H 2 O(l) C 4 H 9 OH(aq) + HCl(aq) In this reaction, the ratio of C 4 H 9 Cl to C 4 H 9 OH is 1:1. Thus, the rate of disappearance of C 4 H 9 Cl is the same as the rate of appearance of C 4 H 9 OH. Rate = - [C 4 H 9 Cl] t = [C 4 H 9 OH] t REACTION RATES AND STOICHIOMETRY REACTION RATES AND STOICHIOMETRY What if the ratio is not 1:1? To generalize, for the reaction H 2 (g) + I 2 (g) 2HI(g) Only 1/2 HI is made for each H 2 used. aa + bb cc + dd Reactants (decrease) Products (increase) 4

5 Sample Problem PROBLEM: Expressing Rate in Terms of Changes in Concentration with Time Because it has a nonpolluting product (water vapor), hydrogen gas is used for fuel aboard the space shuttle and in prototype cars with Earth-bound engines: 2H 2 (g) + O 2 (g) 2H 2 O(g) (a) Express the rate in terms of changes in [H 2 ], [O 2 ], and [H 2 O] with time. (b) When [O 2 ] is decreasing at 0.23 mol/l s, at what rate is [H 2 O] increasing? PROBLEM: SOLUTION: Expressing Rate in Terms of Changes in Concentration with Time Because it has a nonpolluting product (water vapor), hydrogen gas is used for fuel aboard the space shuttle and in prototype cars with Earth-bound engines: 2H 2 (g) + O 2 (g) 2H 2 O(g) (a) rate = [H 2 ] t = - [O 2 ] t = [H 2 O] t PLAN: Choose [O 2 ] as a point of reference since its coefficient is 1. For every molecule of O 2 which disappears, 2 molecules of H 2 disappear and 2 molecules of H 2 O appear, so [O 2 ] is disappearing at half the rate of change of H 2 and H 2 O. (b) - [O 2 ] = mol/l s = + t 1 2 [H 2 O] t [H ; 2 O] = 0.46 mol/l s t CONCENTRATION AND RATES CONCENTRATION AND RATE Each reaction has its own equation that gives its rate as a function of reactant concentrations. this is called its Rate Law To determine the rate law we measure the rate at different starting concentrations. Compare Experiments 1 and 2: when [NH 4+ ] doubles, the initial rate doubles. 5

6 CONCENTRATION AND RATE CONCENTRATION AND RATES Likewise, compare Experiments 5 and 6: when [NO 2 - ] doubles, the initial rate doubles. This equation is called the rate law, and k is the rate constant. RATE LAWS A rate law shows the relationship between the reaction rate and the concentrations of reactants. For gas-phase reactants use P A instead of [A]. k is a constant that has a specific value for each reaction. The value of k is determined experimentally. Constant is relative herek is unique for each rxn k changes with T RATE LAWS Exponents tell the order of the reaction with respect to each reactant. This reaction is First-order in [NH 4+ ] First-order in [NO 2 ] The overall reaction order can be found by adding the exponents on the reactants in the rate law. This reaction is second-order overall. 6

7 Sample Problem 16.2 Determining Reaction Order from Rate Laws PROBLEM: For each of the following reactions, use the given rate law to determine the reaction order with respect to each reactant and the overall order: (a) 2NO(g) + O 2 (g) 2NO 2 (g); rate = k[no] 2 [O 2 ] (b) CH 3 CHO(g) CH 4 (g) + CO(g); rate = k[ch 3 CHO] 3/2 (c) H 2 O 2 (aq) + 3I - (aq) + 2H + (aq) I 3- (aq) + 2H 2 O(l); rate = k[h 2 O 2 ][I - ] Determining Reaction Orders Using initial rates - Run a series of experiments, each of which starts with a different set of reactant concentrations, and from each obtain an initial rate. See Table 16.2 for data on the reaction O 2 (g) + 2NO(g) 2NO 2 (g) rate = k [O 2 ] m [NO] n Compare 2 experiments in which the concentration of one reactant varies and the concentration of the other reactant(s) remains constant. rate k [O 2 ] 2m [NO] n 2 2 = = [O m 2] 2 [O rate k [O 2 ] 1m [NO] n 1 1 [O 2 ] m = 2 ] m 2 1 [O 2 ] x 10-3 mol/l s 2.20 x 10-2 mol/l m = ; 2 = 2 m m = x 10-3 mol/l s 1.10 x 10-2 mol/l Do similar calculations for the other reactant(s). Sample Problem 16.3 Determining Reaction Orders from Initial Rate Data PROBLEM: Many gaseous reactions occur in a car engine and exhaust systems. One of these is NO 2 (g ) + CO (g ) N O (g ) + C O 2 (g ) rate = k[no 2 ] m [CO] n Use the following data to determine the individual and overall reaction orders: Experiment Initial Rate (mol/l s) Initial [NO 2 ] (mol/l) Initial [CO] (mol/l) PLAN: Solve for the order with respect to each reactant using the general rate law using the method described previously. SOLUTION: rate = k [NO 2 ] m [CO] n First, choose two experiments in which [CO] remains constant and the [NO 2 ] varies. 7

8 Sample Problem 16.3 continued rate 2 rate rate 3 rate 1 = = = = Determining Reaction Order from Initial Rate Data k [NO 2 ] m 2[CO] n 2 = k [NO 2 ] m 1 [CO] n m 0.10 k [NO 2 ] 2 3[CO] n 3 k [NO 2 ] 2 1 [CO] n = 1 n [NO 2 ] 2 [NO 2 ] 1 m 16 = 4.0 m and m = 2.0 [CO] 3 [CO] 1 n 1 = 2.0 n and n = 0 rate = k [NO 2 ] 2 [CO] 0 = k [NO 2 ] 2 The reaction is second order overall. The reaction is 2nd order in NO 2. The reaction is zero order in CO. SAMPLE PROBLEM: The initial rate of a reaction A + B C was measured for several different starting concentrations of A and B, and the results are as follows: Expt. No [A], M [B], M Initial Rate, M/s x x x 10-5 Using the data, determine a) rate law for the reaction, b) the rate constant c) rate of the reaction when [A] = 0.50M and [B] = 0.100M RATE LAWS Consider a simple zero order rxn: A B da rate k A in differential form - k dt How much A is left after time t? Integrate : d A kdt A 0 k 0 A A kt 8

9 RATE LAWS INTEGRATED RATE LAWS Consider a simple 1st order rxn: A B Differential form: How much A is left after time t? Integrate: The integrated form of first order rate law: Can be rearranged to give: [A] 0 is the initial concentration of A (t = 0). [A] t is the concentration of A at some time, t, during the course of the reaction. INTEGRATED RATE LAWS FIRST ORDER PROCESSES Manipulating this equation produces If a reaction is first-order, a plot of ln [A] t vs. t will yield a straight line with a slope of -k. So, use graphs to determine reaction order. which is in the form y = mx + b 9

10 SAMPLE PROBLEM Methyl methanoate is hydrolyzed when dissolved in excess hydrochloric acid at 298 K. The ester s concentration was 0.01 mol/li at the start of the reaction, but 8.09 x 10-3 after 21 min. What is the value of the first order rate constant k 1? What is the concentration of methyl methanoate after 30 minutes? FIRST ORDER PROCESSES CH 3 NC Consider the process in which methyl isonitrile is converted to acetonitrile. CH 3 CN How do we know this is a first order reaction? FIRST ORDER PROCESSES FIRST ORDER PROCESSES CH 3 NC CH 3 CN This data was collected for this reaction at C. Does rate=k[ch 3 NC] for all time intervals? When ln P is plotted as a function of time, a straight line results. The process is first-order. k is the negative slope: s

11 SECOND ORDER PROCESSES SECOND ORDER PROCESSES Similarly, integrating the rate law for a process that is second-order in reactant A: So if a process is second-order in A, a plot of 1/[A] vs. t will yield a straight line with a slope of k. Rearrange, integrate: also in the form y = mx + b First order: If a reaction is first-order, a plot of ln [A] t vs. t will yield a straight line with a slope of -k. SAMPLE PROBLEM The dimerization of methyl viologen, an active ingredient in weed killers is second order with respect to methyl viologen, MV. Calculate the value of the rate constant, k if the initial concentration of MV was mol/l and the concentration dropped to 4 x 10-4 mol/l after 0.02 s. SAMPLE PROBLEM Consider a second order reaction which consumes 15% of the initial material after 12 min and 23 s. If the initial concentration was mol/l calculate the rate constant k. 11

12 DETERMINING ORDER OF REACTION The decomposition of NO 2 at 300 C is described by the equation DETERMINING ORDER OF REACTION Graphing ln [NO 2 ] vs. t yields: and yields these data: NO 2 (g) NO (g) + 1/2 O 2 (g) The plot is not a straight line, so the process is not first-order in [A]. Time (s) [NO 2 ], M Time (s) [NO 2 ], M ln [NO 2 ] Does not fit: SECOND ORDER PROCESSES SECOND ORDER PROCESSES A graph of 1/[NO 2 ] vs. t gives this plot. The second type of second order reactions is represented by: A + B products Time (s) [NO 2 ], M 1/[NO 2 ] This is a straight line. Therefore, the process is second-order in [NO 2 ]. rate [ B ] 0 d[ A] dt 1 [ A] 0 d[ B ] dt [ B ][ A] ln [ A][ B ] 0 o kt k[ A][ B ] 12

13 SECOND ORDER PROCESSES For some reactions like: CH 3 COCl(aq) + H 2 O(l) CH 3 COOH(aq) + HCl(aq) rate k' [CH COCl ][ H O ] but if one reactant is always in excess, amount consumed is negligible, thus reaction rate is : rate k[ch COCl ] where k k' [ H O ] SAMPLE PROBLEM: Determining order of reaction from Rate Law For each of the following reactions, use the given rate law to determine the reaction order with respect to each reactant and the overall order: (a) 2NO(g) + O 2 (g) 2NO 2 (g); rate = k[no] 2 [O 2 ] (b) CH 3 CHO(g) CH 4 (g) + CO(g); rate = k[ch 3 CHO] 3/2 (c) H 2 O 2 (aq) + 3I - (aq) + 2H + (aq) I 3- (aq) + 2H 2 O(l); rate = k[h 2 O 2 ][I - ] also called pseudo first order reaction. SAMPLE PROBLEM HALF-LIFE A 1:1 reaction occurs between A and B, and is second order. The initial concentrations of A and B are 0.1 mol/l and 0.2 mol/l, respectively. What is the rate constant, k if the concentration of A after 0.5 hr is 0.05 mol/l? Half-life is defined as the time required for one-half of a reactant to react. Because [A] at t 1/2 is onehalf of the original [A], [A] t = 0.5 [A] 0. 13

14 HALF LIFE For a first-order process, set [A] t =0.5 [A] 0 in integrated rate equation: HALF-LIFE SECOND ORDER For a second-order process, set [A] t =0.5 [A] 0 in 2nd order equation. NOTE: For a first-order process, the half-life does not depend on [A] 0. SAMPLE PROBLEM The half-life of 60 Co is 10.5 min. If we start with 100g of 60 Co, how much remains after 42 minutes? SUMMARY OF RATE EQUATIONS: Order Differential Form Integrated Form Half-Life Units of the Rate Constants

15 DETERMINATION OF REACTION ORDER Reaction Orders For the reaction: A B, the rate law is: rate = k[a] m Order (m) [A] by a factor of: Effect on rate Zero (0) 2, 4, 15, ½, etc. None 2 2X 1 st (1) 3 3X 2 nd (2) 3 9X 2 4X ½ ¼X PROBLEM: Determining Reaction Orders from Initial Rate Data Many gaseous reactions occur in a car engine and exhaust systems. One of these is NO 2 (g ) + C O (g ) N O (g ) + C O 2 (g ) rate = k[no 2 ] m [CO] n Use the following data to determine the individual and overall reaction orders: Experiment Initial Rate (mol/l s) Initial [NO 2 ] (mol/l) Initial [CO] (mol/l) NH 4+ (aq) + NO 2 (aq) N 2 (g) + 2 H 2 O(l) Exp. # [NH 4+ ] o [NO 2 ] o Initial Rate (M s -1 ) x x x 10-7 Determine the rate law and calculate k. 15

16 2 NO(g) + Cl 2 (g) 2 NOCl(g) Exp. # [NO] o [Cl 2 ] o Initial Rate (M min -1 ) Determine the rate law and calculate k. BrO 3 (aq) + 5 Br (aq) + 6 H + (aq) Br 2 (l) + 2 H 2 O(l) Exp. # [BrO 3 ] o [Br ] o [H + ] o Initial Rate (M s -1 ) x x x x 10-3 Determine the rate law and calculate k. SAMPLE PROBLEM: Consider the following reaction: hydrogen peroxide decomposes in the presence of excess cerous ion (Ce +3 ). The following data were obtained: Time, s [H 2 O 2 ], mol/l Determine the order of reaction with respect to hydrogen peroxide. SAMPLE PROBLEM : Consider the data below which relate to the racemization of glucose in aqueous hydrochloric acid. Time, s [Glucose], mol/l What is the order of reaction with respect to glucose? 16

17 SAMPLE PROBLEM: Consider the following kinetic data for the reaction below at 25 o C. OCl - + I - OI - + Cl - [OI - ], Mol/L [Cl - ], Mol/L Initial rate, mol/l-s Using the differential method, determine the order of reaction with respect to OCI -. REACTION MECHANISM: A reaction mechanism is a sequence of molecular events, or reaction steps, that defines the pathway from reactants to products. OH - (aq) + CH 3 Cl(aq) CH 3 OH(aq) + Cl - (aq) REACTION MECHANISM: Reaction is broken into steps with intermediates being formed. some reactions occur in one step, but most occur in in multiple steps. Each step is called an elementary step, and the number of molecules involved in each step defines the molecularity of the step. Molecularity: is the number of reactants. REACTION MECHANISM: UNIMOLECULAR REACTIONS: A single reactant is involved. O 3 * O 2 + O BIMOLECULAR REACTIONS: elementary steps that involves two reactants (most common) HI + HI activated complex H 2 + I 2 TERMOLECULAR REACTIONS: Three reactants are involved (rare, due to probability of orientation and energy both being correct.) O(g) + O 2 (g) + N 2 (g) O 3 (g) + energetic N 2 (g) 17

18 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. 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. The sequence of elementary steps that leads to product formation is the reaction mechanism. 2NO (g) + O 2 (g) 2NO 2 (g) NET EQUATION REACTION MECHANISM N 2 O 2 is detected during the reaction! 1. Elementary step: NO + NO N 2 O Elementary step: N 2 O 2 + O 2 2NO 2 + Elementary step: NO + NO N 2 O 2 Elementary step: N 2 O 2 + O 2 2NO 2 Overall reaction: 2NO + O 2 2NO 2 Overall reaction: 2NO + O 2 2NO The rate-determining step is the slowest step in the sequence of steps leading to product formation. The reaction rate and rate law can be determined from the stoichiometry (coefficients) of the slowest elementary step in the mechanism *Know the difference between an elementary reaction and the net reaction! The Raschig process for the preparation of hydrazine (N 2 H 4 ) Overall RXN: 2NH 3 (g) + NaOCl(aq) N 2 H 4 (aq) + NaCl(aq) + H 2 O(l) Proposed Mechanism: Step 1: Step 2: Step 3: (Only from experiment) NH 3 (aq) + OCl - (aq) NH 2 Cl(aq) + OH - (aq) NH 2 Cl(aq) + NH 3 (aq) N 2 H Cl - (aq) N 2 H 5+ (aq) + OH - (aq) N 2 H 4 (aq) + H 2 O(l) Cancel intermediates and add steps to give overall RXN. 2NH 3 (g) + OCl - (aq) N 2 H 4 (aq) + Cl - (aq) + H 2 O(l) The overall rate law, mechanism, and the total order can t be predicted from the stoichiometry, only by experiment

19 The following is only true for individual steps: The rate law of an elementary step is given by the product of a rate constant and the conc. of the reactants in the step. Step Molecularity rate law A Product(s) uni rate = k[a] A + B Product(s) bi rate = k[a][b] A + A Product(s) bi rate = k[a] 2 2A + B Product(s) ter rate = k[a] 2 [B] The overall mechanism must match the observed rate law. Example: Overall RXN: 2NO 2 (g) + F 2 (g) 2NO 2 F(g) Observed Experimental rate law: rate = k[no 2 ][F 2 ] Question: Why does this rule out a single step reaction? Answer: rate law for single step process would be: rate = k[no 2 ] 2 [F 2 ] Let s try to work out a mechanism that matches the observed rate law. Usually one STEP is assumed to be the rate determining step Example: Overall RXN: 2NO 2 (g) + F 2 (g) 2NO 2 F(g) The rate law is dependent upon the slow step. Let s try making the 2 nd reaction slow and the first fast Overall RXN: 2NO 2 (g) + F 2 (g) 2NO 2 F(g) Observed Experimental rate law: rate = k[no 2 ][F 2 ] Observed Experimental rate law: rate = k[no 2 ][F 2 ] k slow: NO 2 (g) + F 2 (g) NO 1 2 F(g) + F(g) rate = k[no 2 ][F 2 ] k fast: NO 2 (g) + F 2 (g) NO 1 2 F(g) + F(g) fast: k NO 2 (g) + F(g) 2 NO 2 F(g) slow: k NO 2 (g) + F(g) 2 NO 2 F(g) rate = k[no 2 ][F ] 2NO 2 (g) + F 2 (g) 2NO 2 F(g) 2NO 2 (g) + F 2 (g) 2NO 2 F(g) The rate law is dependent upon the slow step

20 SAMPLE PROBLEM: The experimental rate law for the reaction between NO 2 and CO to produce NO and CO 2 is rate = k[no 2 ] 2. The reaction is believed to occur via two steps: STEP 1 NO 2 + NO 2 NO 3 + NO STEP 2 NO 3 + CO NO 2 + CO 2 What is the overall reaction? What is/are the intermediate/s? What can you say about the relative rates of reaction 1 and 2? SAMPLE PROBLEM: The following substitution reaction has a first order rate law: (Co(CN) 5 (H 2 O) 2- + I- Co(CN) 5 I 3- + H 2 O Rate = k[co(cn) 5 (H 2 O) 2- ] Suggest a mechanism in accord with the rate law. REACTION MECHANISM and RATE LAWS SAMPLE PROBLEM Determine rate law by experiment. Devise a mechanism. Predict the rate law for the mechanism If the predicted rate laws do not agree, retry devising a mechanism. If predicted and experimental rate laws agree, look for additional supporting evidence. Nitrogen oxide is reduced by hydrogen to give water and nitrogen, 2 H 2(g) + 2 NO (g) N 2(g) + 2 H 2 O (g) 20

21 One possible mechanism to account for this reaction is: 2 NO (g) N 2 O 2(g) N 2 O 2(g) + H 2(g) N 2 O (g) + H 2 O (g) N 2 O (g) + H 2(g) N 2(g) + H 2 O (g) What is the molecularity of each of the three steps? Show that the sum of these elementary steps is the net reaction. A MODEL FOR KINETICS TRANSITION STATE THEORY Generally reactions occur more rapidly at higher temperatures than at lower temperatures. The rate generally doubles for every 10 K rise in temperature. It s an exponential increase! energy barrier must be overcome reaction energy diagram [humpy diagrams] transition state energy--max of rxn. E diagram activated complex--deformed molecules in their transition state, formed at the E ts --unstable, can go either way! Activation energy, E*, Ea--energy a reacting molecule must absorb from its environment in order to react. 21

22 uncatalyzed catalyzed COLLISION THEORY Assumes molecules must collide in order to react! THE EFFECT OF TEMPERATURE OF REACTION RATE: ARRHENIUS EQUATION EFFECTIVE COLLISIONS ARE hindered by concentration, temperature and geometry of reactants and transition states 22

23 A is the frequency factor units of L/(mol s) & depends on the frequency of collisions and the fraction of these that have the correct geometry--# of effective collisions e -E*/RT is always less than 1 and is the fraction of molecules having the minimum energy required for reaction *Notice in the equation: As "Ea" increases, "k" gets smaller and thus, the rate would decrease. Also, notice that as "T" goes up, "k" increases and so the rate would also increase. 23

24 Taking this equation, plot 1/T vs. ln [k], and get a straight line. From the straight line, find the slope and then find the activation energy. slope = - Ea R so... Ea = - (R) (slope) Used to calculate -value of activation energy from temperature dependence of the rate constant -rate constant for a given temp - if the E* [also known as E a ] and A factor are known. Points to remember!! 1. Ea is smaller; k is greater; the reaction is faster. 2. Ea is greater; k is smaller; the reaction is slower. 24

25 SAMPLE PROBLEM The colorless gas dinitrogen tetroxide decomposes to the brown gas NO 2 in a first order reaction with a value of: k = 4.5 x 10 3 /s at 274K. If k is 1.00 x 10 4 /s at 283K, what is the energy of activation? SAMPLE PROBLEM Calculate the activation energy for the following set of data: T ( C) k (l/mol- s) x x x x 10-2 Exercise 12.8 The gas-phase reaction between methane and diatomic sulfur is given by the equation: CH 4 (g) + 2S 2 (g) CS 2 (g) + 2H 2 S (g) At 550 C the rate constant for this reaction is 1.1 M -1 s -1 and at 625 C the rate constant is 6.4 M -1 s -1. Using these values, calculate E a for this reaction. 25

26 COMPLEX REACTIONS REVERSIBLE REACTIONS 2H 2 (g) + O 2 (g) 2H 2 O (g) SAMPLE PROBLEM: For the reaction A + B C + D, various initial rate measurements were run using A and B only and C and D only. From the data below calculate the equilibrium constant for the reaction: Rate, (M/s) [A] [B] x x x Rate, (M/s) [C] [D] x x x THE COLLISION THEORY OF REACTION RATES Particles must collide. Only two particles may collide at one time. Proper orientation of colliding molecules so that atoms can come in contact with each other to become products. The collision must occur with enough energy to overcome the electron/electron repulsion of the valence shell electrons of the reacting species and must have enough energy to transform translational energy into vibrational energy in order to penetrate into each other so that the electrons can rearrange and form new bonds. 26

27 This new collision product is at the peak of the activation energy hump and is called the activated complex or the transition state. At this point, the activated complex can still either fall to reactants or to products. With all of these criteria met, the reaction may proceed in the forward direction. Amazing that we have reactions occurring at all! 27

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