To apply chemistry you must know the reactants and products to write the rxn, then stoich, spontaneity (thermodynamics) and speed (kinetics).

LO 4.1 The student is able to design and/or interpret the results of an experiment regarding the factors (i.e., temperature, concentration, surface area) that may influence the rate of a reaction. (Sec 12.2-12.4) LO 4.2 The student is able to analyze concentration vs. time data to determine the rate law for a zeroth-, first-, or second-order reaction. (Sec 12.4) LO 4.3 The student is able to connect the half-life of a reaction to the rate constant of a first-order reaction and justify the use of this relation in terms of the reaction being a first-order reaction. (Sec 12.4) LO 4.4 The student is able to connect the rate law for an elementary reaction to the frequency and success of molecular collisions, including connecting the frequency and success to the order and rate constant, respectively. (Sec 12.6) LO 4.5 The student is able to explain the difference between collisions that convert reactants to products and those that do not in terms of energy distributions and molecular orientation. (Sec 12.6)

LO 4.6 The student is able to use representations of the energy profile for an elementary reaction (from the reactants, through the transition state, to the products) to make qualitative predictions regarding the relative temperature dependence of the reaction rate. (Sec 12.6) LO 4.7 The student is able to evaluate alternative explanations, as expressed by reaction mechanisms, to determine which are consistent with data regarding the overall rate of a reaction, and data that can be used to infer the presence of a reaction intermediate. (Sec 12.5) LO 4.8 The student can translate among reaction energy profile representations, particulate representations, and symbolic representations (chemical equations) of a chemical reaction occurring in the presence and absence of a catalyst. (Sec 12.7) LO 4.9 The student is able to explain changes in reaction rates arising from the use of acid-base catalysts, surface catalysts, or enzyme catalysts, including selecting appropriate mechanisms with or without the catalyst present. (Sec 12.7)

In chemical kinetics we study the rate (or speed) at which a chemical process occurs. Besides information about the speed at which reactions occur, kinetics also sheds light on the reaction mechanism, a molecular-level view of the path from reactants to products.

Factors that Affect Reaction Rates Physical state of the reactants Reactant concentrations Reaction temperature Presence of a catalyst

Physical State of the Reactants The more readily the reactants collide, the more rapidly they react. Homogeneous reactions are often faster. Heterogeneous reactions that involve solids are faster if the surface area is increased; i.e., a fine powder reacts faster than a pellet or tablet.

Reactant Concentrations Increasing reactant concentration generally increases reaction rate. Since there are more molecules, more collisions occur.

Reaction rate generally increases with increased temperature. Kinetic energy of molecules is related to temperature. At higher temperatures, molecules move more quickly, increasing numbers of collisions and the energy the molecules possess during the collisions.

Catalysts affect rate without being in the overall balanced equation. Presence of a Catalyst Catalysts affect rate without being in the overall balanced equation. Catalysts affect the kinds of collisions, changing the mechanism (individual reactions that are part of the pathway from reactants to products). Catalysts are critical in many biological reactions.

Reaction Rate Change in concentration of a reactant or product per unit time. [A] means concentration of A in mol/L; A is the reactant or product being considered. [A]/T can be positive if referring to products or negative if referring to reactants. Always adjust to give a positive answer. Copyright © Cengage Learning. All rights reserved

The Decomposition of Nitrogen Dioxide Rates can be given in terms of products or reactants and depend on stoich. Rate laws given in terms of reactants only and NO stoich determines exponents. Must be determine experimentally. Copyright © Cengage Learning. All rights reserved

The Decomposition of Nitrogen Dioxide 22NO2(g) → 2NO(g) +O2 (g) Rate = -[NO2] = [NO] = [O2] 2[T] 2[T] [T] NO and NO2 curves are the same b/c same stoich (tangents and any point are the same magnitude but opposite signs) ** rates decrease with time because [reactant] decreases unless it’s a zero order reaction. Copyright © Cengage Learning. All rights reserved

Reaction Rate Instantaneous Rate Types of rate measured: average rate instantaneous rate initial rate Instantaneous Rate Value of the rate at a particular time. Can be obtained by computing the slope of a line tangent to the curve at that point. Copyright © Cengage Learning. All rights reserved

Following Reaction Rates Rate of a reaction is measured using the concentration of a reactant or a product over time. In this example, [C4H9Cl] is followed.

Following Reaction Rates The average rate is calculated by the –(change in [C4H9Cl]) ÷ (change in time). The table shows the average rate for a variety of time intervals.

A plot of the data gives more information about rate. The slope of the curve at one point in time gives the instantaneous rate. The instantaneous rate at time zero is called the initial rate; this is often the rate of interest to chemists. This figure shows instantaneous and initial rate of the earlier example. Although not always – zero order has a constant rate regardless of the [reactant] Note: Reactions typically slow down over time!

Relative Rates As was said, rates are followed using a reactant or a product. Does this give the same rate for each reactant and product? Rate is dependent on stoichiometry. If we followed use of C4H9Cl and compared it to production of C4H9OH, the values would be the same. Note that the change would have opposite signs—one goes down in value, the other goes up.

Relative Rates and Stoichiometry What if the equation is not 1:1? What will the relative rates be for: 2 O3 (g) → 3 O2 (g)

AP Learning Objectives, Margin Notes and References LO 4.1 The student is able to design and/or interpret the results of an experiment regarding the factors (i.e., temperature, concentration, surface area) that may influence the rate of a reaction. AP Margin Notes Appendix 7.6 “Distinguishing Between Chemical and Physical Changes at the Molecular Level” Additional AP References LO 4.1 (see APEC 10, “Investigating Reaction Rates”)

Differential Rate Law Shows how the rate depends on the concentrations of reactants. For the decomposition of nitrogen dioxide: 2NO2(g) → 2NO(g) + O2(g) Rate = k[NO2]n: k = rate constant n = order of the reactant K depends on temperature and catalyst, not on concentration. Also depends on how the rate is defined in terms of products or reactants. Larger k value (109), fast rxn. Smaller k (101), slow rxn. Copyright © Cengage Learning. All rights reserved

Rate Law Rate = k[NO2]n The concentrations of the products do not appear in the rate law because the reaction rate is being studied under conditions (initial rates) where the reverse reaction does not contribute to the overall rate. Copyright © Cengage Learning. All rights reserved

Rate Law Rate = k[NO2]n The value of k and the exponent n must be determined by experiment; it cannot be written from the balanced equation (due to potential multiple elementary steps). n is usually equal to 0, 1, or 2, but can be a fraction. Copyright © Cengage Learning. All rights reserved

Types of Rate Laws Differential Rate Law (rate law) – shows how the rate of a reaction depends on concentrations. Integrated Rate Law – shows how the concentrations of species in the reaction depend on time. Once you have one, you can write the other. Just choose based on the data provided or easiest to collect. Copyright © Cengage Learning. All rights reserved

Rate Laws: A Summary Because we typically consider reactions only under conditions where the reverse reaction is unimportant, our rate laws will involve only concentrations of reactants. Because the differential and integrated rate laws for a given reaction are related in a well–defined way, the experimental determination of either of the rate laws is sufficient. Copyright © Cengage Learning. All rights reserved

Rate Laws: A Summary Experimental convenience usually dictates which type of rate law is determined experimentally. Knowing the rate law for a reaction is important mainly because we can usually infer the individual steps involved in the reaction from the specific form of the rate law. Copyright © Cengage Learning. All rights reserved

Consider a reaction A + B → C for which rate = k[A][B]2 Consider a reaction A + B → C for which rate = k[A][B]2. Each of the following boxes represents a reaction mixture in which A is shown as red spheres and B as purple ones. Rank these mixtures in order of increasing rate of reaction. 2 < 1 < 3

AP Learning Objectives, Margin Notes and References LO 4.1 The student is able to design and/or interpret the results of an experiment regarding the factors (i.e., temperature, concentration, surface area) that may influence the rate of a reaction. Additional AP References LO 4.1 (see APEC 10, “Investigating Reaction Rates”)

Determine experimentally the power to which each reactant concentration must be raised in the rate law. Copyright © Cengage Learning. All rights reserved

Method of Initial Rates The value of the initial rate is determined for each experiment at the same value of t as close to t = 0 as possible. Several experiments are carried out using different initial concentrations of each of the reactants, and the initial rate is determined for each run. We keep every concentration constant except for one reactant and see what happens to the rate. Then, we change a different reactant. We do this until we have seen how each reactant has affected the rate. Copyright © Cengage Learning. All rights reserved

Overall Reaction Order The sum of the exponents in the reaction rate equation. Rate = k[A]n[B]m Overall reaction order = n + m k = rate constant [A] = concentration of reactant A [B] = concentration of reactant B Units for k vary such that the rate lands on molarity/ time (usually sec) Copyright © Cengage Learning. All rights reserved

An Example of How Concentration Affects Rate Experiments 1–3 show how [NH4+] affects rate. Experiments 4–6 show how [NO2−] affects rate. Result: The rate law, which shows the relationship between rate and concentration for all reactants: Rate = k [NH4+] [NO2−]

More about Rate Law The exponents tell the order of the reaction with respect to each reactant. In our example from the last slide: Rate = k [NH4+] [NO2−] The order with respect to each reactant is 1. (It is first order in NH4+ and NO2−.) The reaction is second order over all (1 + 1 = 2; we just add up all of the reactants’ orders to get the reaction’s order). What is k? It is the rate constant. It is a temperature-dependent quantity. Once you have the order of the reaction, then just plug in data to solve for k value.

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: Using these data, determine (a) the rate law for the reaction, (b) the rate constant, (c) the rate of the reaction when [A] = 0.050M and [B] = 0.100 M. Rate=k[A]2[B]0 K=4x10-3M-1s-1 rate=1x10-5M/s See Example 12.1 on page 562

CONCEPT CHECK! How do exponents (orders) in rate laws compare to coefficients in balanced equations? Why? The exponents do not have any relation to the coefficients (necessarily). The coefficients tell us the mole ratio of the overall reaction. They give us no clue to how the reaction works (its mechanism). Copyright © Cengage Learning. All rights reserved

AP Learning Objectives, Margin Notes and References LO 4.1 The student is able to design and/or interpret the results of an experiment regarding the factors (i.e., temperature, concentration, surface area) that may influence the rate of a reaction. LO 4.2 The student is able to analyze concentration vs. time data to determine the rate law for a zeroth-, first-, or second-order reaction. LO 4.3 The student is able to connect the half-life of a reaction to the rate constant of a first-order reaction and justify the use of this relation in terms of the reaction being a first-order reaction. Additional AP References LO 4.1 (see APEC 10, “Investigating Reaction Rates”) LO 4.2 (see APEC 11, “Rate Law Determination”) Single reactant laws

First-Order aA → products Rate = k[A] = -[A] t Integrated: ln[A] = –kt + ln[A]o [A] = concentration of A at time t k = rate constant t = time [A]o = initial concentration of A Copyright © Cengage Learning. All rights reserved

Plot of ln[N2O5] vs Time Copyright © Cengage Learning. All rights reserved

Finding the Rate Constant, k Besides using the rate law, we can find the rate constant from the plot of ln [A] vs. t. Remember the integrated rate law: ln [A] = − k t + ln [A]o The plot will give a line. Its slope will equal −k. Also can express as ln ([A]o/[A]) = k t

The decomposition of a certain insecticide in water at 12 °C follows first-order kinetics with a rate constant of 1.45 yr–1. A quantity of this insecticide is washed into a lake on June 1, leading to a concentration of 5.0 × 10–7 g/cm3. Assume that the temperature of the lake is constant (so that there are no effects of temperature variation on the rate). (a) What is the concentration of the insecticide on June 1 of the following year? (b) How long will it take for the insecticide concentration to decrease to 3.0 × 10–7 g/cm3? 1.2x10-7 g/cm3 0.35yr

k = rate constant First-Order Time required for a reactant to reach half its original concentration (at constant temperature) Half–Life: k = rate constant Half–life does not depend on the concentration of reactants. Plug in A = A/2 and solve for t1/2 to derive equation Copyright © Cengage Learning. All rights reserved

Copyright © Cengage Learning. All rights reserved

A first order reaction is 35% complete at the end of 55 minutes A first order reaction is 35% complete at the end of 55 minutes. What is the value of k? k = 7.8 × 10–3 min–1 ln(0.65) = –k(55) + ln(1) k = 7.8 x 10-3 min-1. If students use [A] = 35 in the integrated rate law (instead of 65), they will get k = 1.9 x 10-2 min-1. Note: Use the red box animation to assist in explaining how to solve the problem. Copyright © Cengage Learning. All rights reserved

Second-Order Rate = k[A]2 = = -[A] Integrated: [A] = concentration of A at time t k = rate constant t = time [A]o = initial concentration of A Double A will 4xrate, triple A, will 9xrate Common particularly with gases. Copyright © Cengage Learning. All rights reserved

Plot of ln[C4H6] vs Time and Plot of 1/[C4H6] vs Time Second order First order

Second-Order Half–Life: [A]o = initial concentration of A k = rate constant [A]o = initial concentration of A As the reaction progresses and the concentration of reactants decrease, half–life gets longer Each successive half–life is double the preceding one. Inverse relationship between [Ao] vs. t1/2 Copyright © Cengage Learning. All rights reserved

EXERCISE! For a reaction aA  Products, [A]0 = 5.0 M, and the first two half-lives are 25 and 50 minutes, respectively. Write the rate law for this reaction. rate = k[A]2 b) Calculate k. k = 8.0 × 10-3 M–1min–1 Calculate [A] at t = 525 minutes. [A] = 0.23 M a) rate = k[A]2 We know this is second order because the second half–life is double the preceding one. b) k = 8.0 x 10-3 M–1min–1 25 min = 1 / k(5.0 M) c) [A] = 0.23 M (1 / [A]) = (8.0 x 10-3 M–1min–1)(525 min) + (1 / 5.0 M) Copyright © Cengage Learning. All rights reserved

Zero-Order Rate = k[A]0 = k = -[A] Integrated: [A] = –kt + [A]o [A] = concentration of A at time t k = rate constant t = time [A]o = initial concentration of A Rate is independent of [A] Not very common Copyright © Cengage Learning. All rights reserved

Plot of [A] vs Time Copyright © Cengage Learning. All rights reserved

Zero-Order Half–Life: k = rate constant [A]o = initial concentration of A As the reaction progresses and the concentration of reactants decrease, half–life gets shorter direct relationship between [Ao] vs. t1/2 Copyright © Cengage Learning. All rights reserved

Figure 12.7 | The decomposition reaction 2N2O(g)  2N2(g) + O2(g) takes place on a platinum surface. Although [N2O] is three times as great in (b) as in (a), the rate of decomposition of N2O is the same in both cases because the platinum surface can accommodate only a certain number of molecules. As a result, this reaction is zero order. Figure 12-7 p572

CONCEPT CHECK! How can you tell the difference among 0th, 1st, and 2nd order rate laws from their graphs? For the zero-order reaction, the graph of concentration versus time is a straight line with a negative slope. For a first-order graph, the graph is a natural log function. The second-order graph looks similar to the first-order, but with a greater initial slope. Students should be able to write a conceptual explanation of how the half-life is dependent on concentration (or in the case of first-order reactions, not dependent). Copyright © Cengage Learning. All rights reserved

CONCEPT CHECK! How can you tell the difference among 0th, 1st, and 2nd order rate laws from their graphs? For the zero-order reaction, the graph of concentration versus time is a straight line with a negative slope. For a first-order graph, the graph is a natural log function. The second-order graph looks similar to the first-order, but with a greater initial slope. Students should be able to write a conceptual explanation of how the half-life is dependent on concentration (or in the case of first-order reactions, not dependent). Copyright © Cengage Learning. All rights reserved

Zero order – reaction rate is constant (independent of [A]), therefore as [A] decreases, less time is needed to react half of a small amount of A. 1st order – reaction rate is directly related to [A], so as [A] decreases, the reaction rate decreases, so half life remains constant. 2nd order – reaction rate is directly proportional to the square of [A], so a decrease in [A] causes a large decrease in reaction rate which results in an increase in half-life. Students should be able to write a conceptual explanation of how the half-life is dependent on concentration (or in the case of first-order reactions, not dependent). Copyright © Cengage Learning. All rights reserved

Rate Laws To play movie you must be in Slide Show Mode PC Users: Please wait for content to load, then click to play Mac Users: CLICK HERE Copyright © Cengage Learning. All rights reserved

Summary of the Rate Laws Copyright © Cengage Learning. All rights reserved

Zero order First order Second order 4.7 M 3.7 M 2.0 M EXERCISE! Consider the reaction aA  Products. [A]0 = 5.0 M and k = 1.0 × 10–2 (assume the units are appropriate for each case). Calculate [A] after 30.0 seconds have passed, assuming the reaction is: Zero order First order Second order 4.7 M 3.7 M 2.0 M a) 4.7 M [A] = –(1.0×10–2)(30.0) + 5.0 b) 3.7 M ln[A] = –(1.0×10–2)(30.0) + ln(5.0) c) 2.0 M (1 / [A]) = (1.0×10–2)(30.0) + (1 / 5.0) Copyright © Cengage Learning. All rights reserved

Pseudo-first order rate law Pseudo-first order rate law - rate law obtained by simplifying a more complicated one.

AP Learning Objectives, Margin Notes and References LO 4.7 The student is able to evaluate alternative explanations, as expressed by reaction mechanisms, to determine which are consistent with data regarding the overall rate of a reaction, and data that can be used to infer the presence of a reaction intermediate. Additional AP References LO 4.7 (see Appendix 7.8, “Mechanisms with Fast Forward and Reverse First Steps”)

Reaction Mechanism Most chemical reactions occur by a series of steps known as elementary reactions/steps. An intermediate is formed in one step and used up in a subsequent step and thus is never seen as a product in the overall balanced reaction. A balanced chem equation does not tell how reactants become products, ie the order of the bonds broken, formed, change in relative position of atoms. Copyright © Cengage Learning. All rights reserved

NO2(g) + CO(g) → NO(g) + CO2(g) A Molecular Representation of the Elementary Steps in the Reaction of NO2 and CO NO2(g) + CO(g) → NO(g) + CO2(g) k1 K1 and k2 are rate constants for the individual elementary rxns k2 Copyright © Cengage Learning. All rights reserved

Rate laws and relative speeds of elem rxns dictate the overall rate law of a rxn. This is why the overall rate law must be determined experimentally. Overall rate law can be determined via a mechanism and compared to the experimentally determined rate law to validate a possible mechanism. The molecularity of an elementary reaction tells how many molecules are involved in that step of the mechanism.

These are rare, if they indeed do occur. Termolecular? Termolecular steps require three molecules to simultaneously collide with the proper orientation and the proper energy. These are rare, if they indeed do occur. These must be slower than unimolecular or bimolecular steps. Nearly all mechanisms use only unimolecular or bimolecular reactions.

What Limits the Rate? The overall reaction cannot occur faster than the slowest reaction in the mechanism. We call that the rate-determining step. The rate-determining step (slowest step) determines the rate law and the molecularity of the overall reaction.

Reaction Mechanism Requirements The sum of the balanced elementary steps must give the overall balanced equation for the reaction. All intermediates are made and used up. The mechanism must agree with the experimentally determined rate law. Any catalysts are used and regenerated. Copyright © Cengage Learning. All rights reserved

An intermediate is not equal to a transition state An intermediate is not equal to a transition state. Intermediates can be stable and sometimes isolated an identified. Transition states are inherently unstable and hard to characterize.

A Mechanism With a Slow Initial Step Overall equation: NO2 + CO → NO + CO2 Rate law: Rate = k [NO2]2 If the first step is the rate-determining step, the coefficients on the reactants side are the same as the order in the rate law! So, the first step of the mechanism begins: NO2 + NO2 → Second order in NO2 and zero order in CO

A Mechanism With a Slow Initial Step (continued) The easiest way to complete the first step is to make a product: NO2 + NO2 → NO + NO3 We do not see NO3 in the stoichiometry, so it is an intermediate, which needs to be used in a faster next step. NO3 + CO → NO2 + CO2 CO is not in the rate law because it reats in the second step that follows the rate limiting step

A Mechanism With a Slow Initial Step (completed) Since the first step is the slowest step, it gives the rate law. If you add up all of the individual steps (2 of them), you get the stoichiometry. Each step balances. This is a plausible mechanism. Can never prove mechanism only say if it is consistent with the experimental data.

a) 2N2O → 2N2 +O2 b) rate=k[N2O] 1st step is slow

A Mechanism With a Fast Initial Step Equation for the reaction: 2 NO + Br2 → 2 NOBr The rate law for this reaction is found to be Rate = k [NO]2 [Br2] Because termolecular processes are rare, this rate law suggests a multistep mechanism.

A Mechanism With a Fast Initial Step (continued) The rate law indicates that a quickly established equilibrium is followed by a slow step. Step 1: NO + Br2 ⇌ NOBr2 Step 2: NOBr2 + NO → 2 NOBr NOBr2 is an intermediate and these are usually unstable. Will enter equilibrium to return into reactants over building in concentration

The rate law for that step would be Rate = k2[NOBr2] [NO] What is the Rate Law? The rate of the overall reaction depends upon the rate of the slow step. The rate law for that step would be Rate = k2[NOBr2] [NO] But how can we find [NOBr2]?

[NOBr2] (An Intermediate)? NOBr2 can react two ways: With NO to form NOBr. By decomposition to reform NO and Br2. The reactants and products of the first step are in equilibrium with each other. For an equilibrium (as we will see in the next chapter): Ratef = Rater

Substituting for the forward and reverse rates: The Rate Law (Finally!) Substituting for the forward and reverse rates: k1 [NO] [Br2] = k−1 [NOBr2] Solve for [NOBr2], then substitute into the rate law: Rate = k2 (k1/k−1) [NO] [Br2] [NO] This gives the observed rate law! Rate = k [NO]2 [Br2] If there is a fast reversible then a slow, substitute the reactants in the fast for the intermediate in the slow rate law

d

CONCEPT CHECK! The reaction A + 2B  C has the following proposed mechanism: A + B D (fast equilibrium) D + B  C (slow) Write the rate law for this mechanism. rate = k[A][B]2 rate = k[A][B]2 The sum of the elementary steps give the overall balanced equation for the reaction. Copyright © Cengage Learning. All rights reserved

AP Learning Objectives, Margin Notes and References LO 4.4 The student is able to connect the rate law for an elementary reaction to the frequency and success of molecular collisions, including connecting the frequency and success to the order and rate constant, respectively. LO 4.5 The student is able to explain the difference between collisions that convert reactants to products and those that do not in terms of energy distributions and molecular orientation. LO 4.6 The student is able to use representations of the energy profile for an elementary reaction (from the reactants, through the transition state, to the products) to make qualitative predictions regarding the relative temperature dependence of the reaction rate.

Factors That Affect Reaction Rate Concentration (see rate laws) Temperature Frequency of collisions Orientation of molecules Energy needed for the reaction to take place (activation energy)

Temperature and Rate Generally, as temperature increases, rate increases. The rate constant is temperature dependent: it increases as temperature increases. Rate constant doubles (approximately) with every 10 ◦C rise.

Frequency of Collisions The collision model is based on the kinetic molecular theory. Molecules must collide to react. If there are more collisions, more reactions can occur. So, if there are more molecules, the reaction rate is faster. Also, if the temperature is higher, molecules move faster, causing more collisions and a higher rate of reaction. Only a small fraction of the collisions actually produce a reaction, so more to the story.

In a chemical reaction, bonds are broken and new bonds are formed. 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.

Orientation of Molecules Molecules can often collide without forming products. Aligning molecules properly can lead to chemical reactions. Bonds must be broken and made and atoms need to be in proper positions.

Activation Energy, Ea The minimum threshold energy needed for a reaction to take place is called activation energy. An energy barrier must be overcome for a reaction to take place. The collision model postulates that the energy needed comes from the KE possessed by the reacting molecules before the collision. The KE is changed to PE as the molecules are distorted during a collision to break bonds and rearrange atoms. If molecules don’t have enough KE, then they just bounce off of each other. Ea varies from one reaction to another. Copyright © Cengage Learning. All rights reserved

Reactions can be endothermic or exothermic after this. Plots are made to show the energy possessed by the particles as the reaction proceeds. At the highest energy state, the transition state (or activated complex) is formed. Reactions can be endothermic or exothermic after this. E has no effect on the rate of the reaction. Rate depends on the size of Ea Smaller Ea, faster the rxn at a given temp.

Change in Potential Energy Forward vs. backward reaction Copyright © Cengage Learning. All rights reserved

Transition States and Activation Energy To play movie you must be in Slide Show Mode PC Users: Please wait for content to load, then click to play Mac Users: CLICK HERE Copyright © Cengage Learning. All rights reserved

For Reactants to Form Products Collision must involve enough energy to produce the reaction (must equal or exceed the activation energy). Relative orientation of the reactants must allow formation of any new bonds necessary to produce products. Copyright © Cengage Learning. All rights reserved

Distribution of the Energy of Molecules Gases have an average temperature, but each individual molecule has its own energy. At higher energies, more molecules possess the energy needed for the reaction to occur. The fraction of effective collisions increases exponentially with temperature

The fraction of molecules possessing energy Ea or greater Arrhenius Found that most reaction rate data obeyed an equation based on the following: The fraction of molecules possessing energy Ea or greater The number of collisions/ sec The fraction of collisions that have the appropriate orientation. Copyright © Cengage Learning. All rights reserved

Arrhenius Equation k = rate constant A = frequency factor Ea = activation energy R = gas constant (8.3145 J/K·mol) T = temperature (in K) “A” takes into account the frequency of collisions and the fraction of collisions with the proper orientations The rest represents the fraction of collisions with sufficient energy to produce a reaction. Generally as Ea increases, rxn rate decreases for constant A and T. Copyright © Cengage Learning. All rights reserved

Linear Form of Arrhenius Equation Most rate constants obey this equation, so if know k, then can determine Ea ot A for a rxn. Copyright © Cengage Learning. All rights reserved

Linear Form of Arrhenius Equation Can also find Ea by using two data points in the following equation: Copyright © Cengage Learning. All rights reserved

Rank the reverse reactions from slowest to fastest. 2<3<1 2<1<3 Rank the reverse reactions from slowest to fastest.

AP Learning Objectives, Margin Notes and References LO 4.8 The student can translate among reaction energy profile representations, particulate representations, and symbolic representations (chemical equations) of a chemical reaction occurring in the presence and absence of a catalyst. LO 4.9 The student is able to explain changes in reaction rates arising from the use of acid-base catalysts, surface catalysts, or enzyme catalysts, including selecting appropriate mechanisms with or without the catalyst present. Additional AP References LO 4.9 (see Appendix 7.9, ” Acid Catalysis”)

Catalyst A substance that speeds up a reaction without being consumed itself. Provides a new pathway for the reaction with a lower activation energy. Can speed up reactions by raising the temperature, but often not effective energetically or biologically. Copyright © Cengage Learning. All rights reserved

Energy Plots for a Catalyzed and an Uncatalyzed Pathway for a Given Reaction Lowers the Ea but does not affect E. Copyright © Cengage Learning. All rights reserved

Effect of a Catalyst on the Number of Reaction-Producing Collisions Figure 12.14 | Effect of a catalyst on the number of reaction-producing collisions. Because a catalyst provides a reaction pathway with a lower activation energy, a much greater fraction of the collisions is effective for the catalyzed pathway (b) than for the uncatalyzed pathway (a) (at a given temperature). This allows reactants to become products at a much higher rate, even though there is no temperature increase. Copyright © Cengage Learning. All rights reserved

Heterogeneous Catalyst Catalyst that is in a different phase than reacting molecules. Most often involves gaseous reactants being adsorbed on the surface of a solid catalyst. Adsorption – collection of one substance on the surface of another substance. Adsorbed- collection of one substance on the surface of another Absorbed- penetration of one substance into another Copyright © Cengage Learning. All rights reserved

Heterogeneous Catalysis To play movie you must be in Slide Show Mode PC Users: Please wait for content to load, then click to play Mac Users: CLICK HERE Copyright © Cengage Learning. All rights reserved

Heterogeneous Catalyst Adsorption and activation of the reactants. Migration of the adsorbed reactants on the surface. Reaction of the adsorbed substances. Escape, or desorption, of the products. Figure 12.15 | Heterogeneous catalysis of the hydrogenation of ethylene. (a) The reactants above the metal surface. (b) Hydrogen is adsorbed onto the metal surface, forming metal–hydrogen bonds and breaking the H−H bonds. The p bond in ethylene is broken and metal–hydrogen bonds are formed during adsorption. (c) The adsorbed molecules and atoms migrate toward each other on the metal surface, forming new C−H bonds. (d) The C atoms in ethane (C2H6) have completely saturated bonding capacities and so cannot bind strongly to the metal surfaces. The C2H6 molecule thus escapes. Copyright © Cengage Learning. All rights reserved

Homogeneous Catalyst Exists in the same phase as the reacting molecules. Enzymes are nature’s catalysts. Catalysts are put in and come out the same vs an intermediate that is formed in one reaction and used up in a subsequent reaction. Copyright © Cengage Learning. All rights reserved

Homogeneous Catalysis To play movie you must be in Slide Show Mode PC Users: Please wait for content to load, then click to play Mac Users: CLICK HERE Copyright © Cengage Learning. All rights reserved