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314 U5: MODULE 3 Understanding Mechanism We have discussed how to evaluate the thermodynamic stability of chemical systems and how to predict reaction directionality and extent. However, the values of DGorxn or the equilibrium constant for a chemical process tell us nothing about how long it will take for the reaction to happen. Thus, in this module we will shift our attention to issues related to the kinetic stability of chemical systems. The rate of a chemical reaction is influenced by a variety of internal and external factors. The chemical composition and structure of the species involved determine the value of the activation energy needed to initiate the process, as well as the configuration effectiveness of collisions between reacting substances. Concentration, temperature, and pressure affect the aver- Enzymes increase the rate of specific reactions. age rate at which reacting particles collide. To understand and predict the kinetic behavior of a chemical system, it is important to explore how the reaction actually happens. This is, what sequence of events, or mechanism, leads from reactants to products. Knowledge about the mechanism of a chemical process allows us to better understand and control the effect of variables such as concentration and temperature on the rate of reaction. As we will see in Unit 7, it also helps us predict the likely structure of the products of a chemical reaction. THE CHALLENGE Mirror Images Molecules of a given amino acid may exist in one of two forms, commonly called D and L isomeric forms. The D form is the mirror image of the L form. These types of molecules can transform into each other. The conversion process involves a two-step mechanism. • How would you expect the rate of conversion to depend on the concentration of the amino acid and on the external temperature? Share and discuss your ideas with one of your classmates. This module will help you develop the type of chemical thinking that is used to answer questions similar to those posed in the challenge. In particular, the central goal of Module 3 is to help you understand how to use information about reaction mechanism to predict rates of reaction. Chemical Thinking U5 How do we predict chemical change? 315 Reaction Mechanism There are chemical processes that are thermodynamically favored but take too long to happen. The reactant mixture in these types of cases tends to be kinetically stable although thermodynamically unstable. Consider, for example, the following two potential routes for the synthesis of the amino acid glycine on the primitive Earth: Figure 5.10 Schematic rep- CH2O(g) + HCN(g) + H2O(l) 2 CH4(g) + NH3(g) + 5/2 O2(g) C2H5NO2(s) C2H5NO2(s) + 3 H2O(l) DG DG o rxn o rxn = –154 kJ = –965 kJ resentation of the change in Gibbs free energy, DG, along the two reaction paths for the synthesis of glycine. Figure 5.10 shows a schematic representation of the changes in the Gibbs free energy of the reacting mixtures, DG, as each of these DG two reactions proceeds from reactants to products. As shown in this figure, although the second process represented above is more thermodynamically favored than the first one, it has such a large free energy of activation that it is likely that the first process will 2 CH4(g) + NH3(g) occur much faster and be a more common path for the synthesis + 5/2 O2(g) of glycine under standard conditions. Analyzing chemical systems from both the thermodynamic and kinetic point of view is crucial in evaluating the actual C2H5NO2(s) CH2O(g) + HCN(g) chemical stability of substances. For example, the transformation + H2O(l) of diamond into graphite has a DGo = –2.9 kJ/mol, which makes it thermodynamically favored, but the activation energy for the C2H5NO2(s) + 3H2O(l) process is so high (Ea = 728 kJ/mol) that the process takes millions of years to occur without external intervention. From this perspecReaction Path tive, diamond can be considered a chemically stable substance. To better understand the kinetic behavior of chemical systems it is useful to build models that describe how the reaction takes place at the submicroscopic level. These models recognize that most chemical reactions involve a sequence of steps in the path from reactants to products. This sequence of processes define the mechanism of the reaction. In this mechanistic approach, the overall reaction is Bimolecular Step viewed as the result of multiple elementary processes, or steps, occurring simultaneously across the system. The different elementary steps describe the likely order in which chemical bonds are broken and formed during the reaction. The most common elementary steps involve one of the following processes: a) Unimolecular step, corresponding to the direct transformation of a single molecule into another; b) Bimolecular step, involving the collision of two particles of the same or Transition State different type (Figure 5.11). c) Termolecular step, involving the simultaneous collision of three particles of the same or different types. As a result of these types of processes, atomic rearrangements will lead to the formation of new particles with different composition and structure. Some of these particles may suffer additional transformations in other reaction steps and will not be present in the final products. Such chemical species are known as reaction intermediates. Figure 5.11 Representation of a bimolecular step involving the collision of two particles. During any type of elementary step, there is only one transition state. 316 MODULE 3 Understanding Mechanism LET’S THINK Molecularity In chemistry, molecularity refers to the number of colliding particles that are involved in a single reaction step. Determining the molecularity of a reaction step thus mean to identify whether the process is unimolecular, bimolecular, etc. • Determine the molecularity of each of the two reaction steps associated with this reaction mechanism. A + A B B + C C + E • Identify the intermediate species in the mechanism and build a symbolic representation of the overall reaction that this mechanism is modeling. • Sketch the potential energy diagram for this reaction, showing the relative potential energy of reactants, products, and intermediates. (Hint: Analyze the number and nature of the chemical bonds broken and formed in each step of the mechanism). • Discuss how you would expect the rate of each step to change when the concentration of the reacting species in each step (i.e., A in the first step and B in the second step) is doubled. Share and discuss your ideas with a classmate, and clearly justify your reasoning. Concentration Effects If we know how many particles are involved in a reaction step (molecularity), we can predict the rate of the process will vary with changes in the concentration of the reacting species. The rate of the reaction is a measure of how the concentration of the reactants, or of the products that they form, changes per unit time. To begin our analysis, let us consider a unimolecular process in which a single particle A is involved. Imagine that we have nA moles of particles of the reactant A in a given volume V; the molar concentration would then be [A] = nA/V. For any reaction of this type, there is a probability k that a particle will undergo a transformation in a given unit of time. The larger this probability, the faster the rate of the reaction. The larger the concentration of A particles, the larger the number of such particles that will be transformed per unit time. In fact, we can expect the reaction rate to be directly proportional to [A]. Thus, the rate of a unimolecular step can be expressed as: (5.14) RATE = k [A] Unimolecular Process The constant k in this expression is commonly called the rate constant for the process. In this particular case, the rate constant has units of (1/time). Rates of reaction are commonly expressed in units of concentration per unit time (e.g., molarity/second). Given that the rate of reaction is directly proportional to the concentration of the reactant to the first power, we say that this is a “first order” reaction. Chemical Thinking U5 317 How do we predict chemical change? In bimolecular processes, the rate of the reaction depends on the probability that a collision between two particles A and B leads to a chemical transformation. This probability, k, depends on factors such as the activation energy of the reaction, the configuration effectiveness of the collision, and the temperature of the system. The concentration of each of the colliding species will also affect the rate. The higher the concentration of each of substance, the higher the likelihood that their particles will collide and the faster the reaction rate. In this case, the relationship between rate of reaction and concentration can be expressed as: (5.15) RATE = k [A] [B] Bimolecular Process The reaction rate depends on the product of the concentration of each species because the likelihood of a collision depends on the product of the probability of finding one particle of each type in the same location. Thus, if we double the concentration of each reactant, the reaction rate increases four times. The rate constant k for bimolecular processes has units of 1/(concentration x time). A bimolecular process is said to be first order with respect to each reacting species. However, overall, the process is a second order reaction. In general, the relationship between the rate of a reaction and the concentration of each reactant involved in the process is known as the Rate Law for the reaction. This rate law has the following general form: (5.16) RATE = k [A]a [B]b [C]c where the exponents a, b, and c are the order of the reaction with respect to each species (i.e, a = 1, first order; a = 2, second order, etc), and the overall reaction order is given by the sum a+b+c. Ozone Decomposition LET’S THINK With the development of photosynthetic organisms on the primitive Earth, the amount of O2(g) and O3(g) in the atmosphere increased considerably. The formation and decomposition of ozone molecules was induced by UV radiation from the Sun. The two-step reaction mechanism for the decomposition of O3(g) in the atmosphere can be represented as follows: O3(g) O2(g) + O(g) O3(g) + O(g) 2 O2(g) • • • Determine the molecularity of each step and identify all the reaction intermediates; Express the rate law for each step in the mechanism. Determine the rate order with respect to each species as well as the overall rate order for each step. Write the chemical equation that represents the overall process. The potential energy diagram for the overall reaction is shown to the right: • Ep Reactants Products Reaction Path Locate in the diagram the different reaction steps and the species involved in each of them. Identify the fastest step in the decomposition process. 318 MODULE 3 Understanding Mechanism As we will discuss in Module 4, there are well established methods to determine the rate law for an overall chemical reaction. This information can be combined with results from other experimental techniques used to identify reaction intermediates to propose a reaction mechanism. In the mechanistic model of chemical reactions, the overall rate of the reaction is determined by the rate of one or more of the individual steps. In particular, for reactions in which one of the steps is considerably slower than the others, the overall rate of reaction in determined by the slowest step (Figure 5.12). The speed at which chemical species are produced in this step sets the maximum speed at which the reaction can proceed. The relationship between the rate law of the slowest step in a reaction mechanism (rate determining step) and the rate law of the overall reaction is direct when the slowest step is the first step in the mechanism. Consider, for example, the following hypothetical mechanism for an overall reaction of the type A C: A + A B C Figure 5.12 In this analogy, the liquid passing through each filter represents a different reaction step. The rate at which the flask is filled is determined by the speed at which the liquid passes through the slowest filter. B + C SLOW (Rate Determining Step) FAST The rate law for the first (bimolecular) step in the mechanism can be expressed as: RATE1 = k1[A]2, where k1 is the associated rate constant. For the second step we have: RATE2 = k2[B]. Given that the first step is the slowest in the mechanism, the overall rate of the reaction should be equal to the rate of the first step: RATE = k1[A]2. If the above mechanism is correct, the experimental determination of the reaction order for the overall reaction should indicate that the process is second order with respect to the concentration of the reactant A. The above example can also be used to illustrate the problems that we face in deriving the rate law for an overall reaction for which the slowest process in the mechanism is not the first step. Imagine now that the first step in our example was faster than the second step. In this case, the overall rate law should be: RATE = k2[B]. The problem with this relationship is that it is expressed in terms of the concentration of the intermediate species B; a concentration that we cannot easily determine and control. To derive a more useful rate law we need to express [B] in terms of the concentrations of reactants, or of the reactants and products. Finding this relationship is not always straightforward but it can be more easily accomplished when we have a mechanism in which the reaction associated with the first step reaches chemical equilibrium at a rate much faster than the rate of the second step. In this case we can write: FAST SLOW A + A B B + C C K = [B][C]/[A]2 (Rate Determining Step) where K is the equilibrium constant for the first process. If we are able to determine the value of this constant, the concentration of the intermediate B can be expressed as [B] = K[A]2/[C]. Thus, the overall rate of the reaction takes the form: RATE = k2[B] = k2K[A]2/[C] = k[A]2/[C], where the overall rate constant k = k2K. The experimental analysis of this reaction should show a reaction rate that is second order with respect to A but also depends on the inverse of the concentration of product C. Chemical Thinking U5 Dimerization How do we predict chemical change? LET’S THINK Amino acids formed on the primitive Earth somehow combined to form dimers, trimers, and eventually polymers (proteins). Consider the overall reaction for the dimerization of alanine (Ala): Experimentally, it is found that the rate law for this process can be expressed as: RATE = k[Ala]2. Several possible mechanisms have been suggested for this reaction. Schematic representations of two of them are shown below: Ep Ep Ala-Ala* 2 Ala Reactants Ala-Ala + H2O Products Reaction Path • • • • 2 Ala Reactants Ala-Ala + H2O Products Reaction Path Write the individual steps that correspond to each mechanism; Identify the molecularity of each step and the associated rate law; Evaluate which of the two mechanisms would fit the experimental result for the overall rate law for the reaction; Propose a different reaction mechanism that involves a quick first step in which reacting species reach chemical equilibrium, followed by a slow step. Evaluate whether this type of mechanism would fit the experimental results for the overall rate law. Share and discuss your ideas with a classmate, and clearly justify your reasoning. Temperature Effects According to the kinetic model of chemical reactions, for a chemical process to occur colliding particles must be oriented in a manner that allows reacting groups to interact effectively. Additionally, colliding particles must have enough energy to reach a transition state that leads to the formation of the new products. The energy needed for colliding particles to adopt the structure of the transition state is the activation energy, Ea, of the reaction. The larger the value of Ea, the less likely that colliding particles will have enough energy to overcome the activation barrier and the lower the rate of the reaction will be. Increasing the temperature of the system leads to an increase in the average kinetic energy of the particles, increasing the probability of favorable collisions and raising the reaction rate. 319 320 MODULE 3 Understanding Mechanism The effect of configuration effectiveness, activation energy, and temperature on the rate of a reaction is represented in the value of the rate constant k in the associated rate law. This constant is a measure of the probability that collisions will lead to the formation of the transition state at a given temperature. This probability can be calculated using Arrhenius equation expressed as: k = A exp ( –Ea / RT) (5.17) where the pre-exponential factor A is a measure of the frequency of collisions with the proper configuration, and the term exp(-Ea/RT) is proportional to the fraction of colliding particles with enough energy to react at a given temperature. In general, we can assume that the frequency factor A and the activation energy Ea are constants for a given chemical process, independent of the value of T. Consequently, we can compare reaction rates at two different temperatures T1 and T2 by calculating the ratio of the corresponding rate constants k1 and k2: (5.18) k2 / k1 = exp ( –Ea/RT2) /exp ( –Ea/RT1) = exp[–Ea/R(1/T2– 1/T1)] LET’S THINK First Amino Acids The overall activation energy for the dimerization of alanine (Ala) that we studied in the previous problem is close to 90 kJ/mol. • Calculate how many times faster the dimerization of alanine is at a) 100 oC, and b) 400 oC, than at 25 oC. These temperatures represent the range of values suspected to exist in high pressure hydrothermal vents were life may have started in our planet. Share and discuss your ideas with a classmate, and clearly justify your reasoning. Figure 5.13 The catalyst reduces the overall activation energy by changing the mechanism from a single step to a two-step process. Ep Changing Mechanism It has been suggested that one crucial step in the origin of life in our planet was the synthesis of substances that could speed up the rate of certain chemical reactions. In chemistry, such substances are called catalysts and they act by directly reducing the activation energy Ea of a reaction or by changing the reaction mechaUncatalyzed nism, which often results in lower activation barriers (Figure 5.13). Catalyst may work by the action of intermolecular forces on the reacting particles or by actually forming chemical bonds with some species. These interactions facilitate the formation of a transition state that leads to the new products. Some catalysts may be solid, but they could also DH Catalyzed be species in solution. Catalysts participate in the reaction but they are not consumed during the process. They can be separated at the end of the reaction. Reaction Path Chemical Thinking U5 321 How do we predict chemical change? Catalysts are of great importance for living organisms because many metabolic processes would take too long in their absence. The substances that catalyze reactions in biological systems are called enzymes, which tend to be proteins with a composition and structure that varies depending on the process in which they participate. Most enzyme-catalyzed reaction rates are million times faster than those of the same uncatalyzed process. As is the case with all catalysts, enzymes participate in the chemical reactions but are not consumed in the process, nor they change the directionality and extent of a chemical reaction. However, enzymes differ from most chemical catalyst in their high specificity for the processes that they help accelerate. The action and high specificity of enzymes has been explained using different models that we will explore in the following activities. Lock and Key Model LET’S THINK Enzyme (E) + Substrate (S) Enzyme-Substrate Complex (ES) • • Enzyme-Substrate Complex (ES) Enzyme (E) + Product (P) FAST SLOW Build a potential energy diagram representing the reaction path from reactants to products; Write the rate law for each individual step in the reaction mechanism, as well as the equilibrium constant for the first step. Use these relationships to derive the overall rate law for the reaction in terms of concentrations that can be measured experimentally. The overall rate law that you derived in the previous step is only valid at low concentrations of substrate. At high concentrations, the reaction rate does not Maximum Rate (Vmax) change much with increasing [S} as shown in the graph. • • Use the lock and key model to explain the flattening of the reaction rate at high substrate concentrations. What is the order of the reaction under such conditions? Catalyzed Uncatalyzed Share and discuss your ideas with a classmate. By Jerry Crimson Mann (Own work) [Public domain] via Wikimedia Commons The process of enzyme catalysis begins with the binding of a substance, commonly called the substrate, to the active site of the enzyme. The active site is a region of the enzyme that interacts relatively strongly with the substrate. The binding process alters the electron distribution in the substrate molecules and induces structural changes that facilitate their transformation. In the “Lock and Key” model of enzyme action, the active site is assumed to have a unique geometric shape that is complementary to the shape of the substrate molecule. The enzyme-substrate system is thought to be analogous to a key-lock pair in which the lock is the enzyme and the key is the substrate. This model explains the high specificity of enzymes as only the correct key (substrate) can fit into the key hole (active site) of the lock (enzyme). The reaction mechanism for enzymatic action in this model can be represented as: MODULE 3 Understanding Mechanism LET’S THINK Induced Fit Model E + S ES ES EP EP E + P • • • FAST SLOW FAST Build a potential energy diagram representing the reaction path from reactants to products; Write the rate law for each individual step in the reaction mechanism, as well as the equilibrium constant for the first step. Use these relationships to derive the overall rate law for the reaction in terms of concentrations that can be measured experimentally. Compare the overall rate law predicted by the “Induced Fit Model” with that of the “Lock and Key Model.” Discuss whether their predictions could be used to identify the best model for enzymatic action. Share your ideas with a classmate, and clearly justify your reasoning. LET’S THINK Temperature Effects For a range of temperatures, the rate of an enzyme-catalyzed reaction increases as the temperature is raised. A ten degree rise in temperature will increase the activity of most enzymes by 50 to 100%. • Consider enzymes working at body temperature (37 oC). Estimate the range of values for the activation energy of reactions catalyzed by these types of enzymes. The graph to the right depicts the effect of temperature on the reaction rate of enzyme-catalyzed processes. • Build an explanation for the effect of temperature on the rate of reaction. Temperature By Tim Vickers (Own work) [Public domain] via Wikimedia Commons The “Lock and Key” model provides an inaccurate description of the behavior of most enzymes. The model has been modified to take into account changes in the structure of the enzyme as a result of its interaction with the substrate. In the “Induced Fit Model” the initial interactions between enzyme and substrate are assumed to be relatively weak, but these interactions induce conformational changes in the enzyme that strengthen the binding. The reaction mechanism for enzymatic action involves three steps represented by the following set of generic chemical equations: Reaction Rate 322 Chemical Thinking FACING THE CHALLENGE Mirror Images How do we know if two molecules are identical? In principle, they should have the same chemical composition and structure. However, sometimes making this judgment is not so easy. Consider the two molecules shown in the image. Are they the same? One way to answer this question is to explore whether the molecules are superimposable; this is, if we can make one lay on top of the other having the same distribution of atoms. If we do that with the amino acids in the image, it turns that the molecules are not identical. One is the mirror image of the other, like our two hands, and they are not superimposable. They are said to be enantiomers or optical isomers. When samples made up of these two types of molecules interact with polarized light, each of the samples rotates the polarization plane in different directions. Pairs of enantiomers are often designated with the letters D and L to differentiate them. Molecules that have at least one non-superimposable mirror image are said to be chiral. Molecular chirality is caused by the presence of carbon atoms that are bonded to four different groups. If all carbons in a molecule are bonded two only three or fewer different groups, the molecule will be achiral and it will not have enantiomers. Molecular chirality is relevant to our discusChiral sions in this module because many biologically active molecules are chiral and exist as only one type of enantiomer in living organisms. For example, biologically active amino acids are of type L but not D, while naturally occurring sugars are D but not L. Chirality is of central importance for many biological functions. Chiral molecules have most of the same properties, but they interact differently with L and D enantiomers of other chiral substances. Thus, for example, if the active site of an U5 How do we predict chemical change? enzyme has chiral components, it will not interact in the same way with the D or L isomers of a chiral substance. This property enhances the selectivity of the enzyme as a catalytic species. The origin in the asymmetry of the chirality of biologically molecules (homochirality) is not well understood. Many chemical reactions in which chiral molecules are produced generate the different enantiomers in the same proportion. Additionally, enantiomers can transform into each other through a chemical process known as racemization. The rate of interconversion varies from substance to substance, but for many naturally occurring compounds the rate is not negligible. Thus, it is not clear how living organisms developed to selectively use and produce only one type of enantiomet. There are several hypotheses about the origin of homochirality of naturally occurring molecules in our planet. For example, analysis of amino acids found in meteorites hitting our planet reveals the presence of preferentially one enantiomer. There is evidence that certain type of radiation to which these meteorites may have been exposed triggers the formation of optical isomers. These results suggest that homochirality on Earth may have had an extraterrestrial Achiral origin. However, there are other types of processes that result in the preferential formation of one optical isomer over the other. Stirring a solution as it crystallizes leads to one chiral form almost exclusively over the other. Similarly, boiling a supersaturated solution perturbs the crystallization process, leading to the formation of a single chiral phase. Once an enantiomeric imbalance is created, it may be amplified and propagated by chemical processes that favor the formation of one type of isomer over the other. Identifying the actual cause of homochirality thus demands a careful analysis of the mechanisms behind these types of chemical processes. 323 324 MODULE 3 Understanding Mechanism Let’s Apply ASSESS WHAT YOU KNOW Racemization Pure samples of L or D enantiomers eventually convert into a mixture of both forms. This process is called racemization and the mixture that is formed is a racemic mixture. Racemization in vitro or in vivo of some substances, such as chiral pharmaceutical drugs, may be a serious problem as one enantiomer may be the one that triggers the response that we want and the other may induce unwanted metabolic processes. Amino Acids The following potential energy diagram represents the racemization of the amino acid alanine. The process is characterized by having DHorxn = 0 and DSorxn = 0. Ep L L- + H+ + H+ H+ + D- D H+ + Reaction Path • • • • Represent the reaction mechanism for the racemization of alanine as a sequence of steps. Determine the molecularity of each step and identify all reaction intermediates. Establish the rate law for each reaction step and derive the rate law for the overall racemization process. Determine the order of the reaction with respect to the concentration of the amino acid. Discuss how a change in concentration would affect the rate of racemization. The activation energy for the first step of the racemization of alanine is close to 130 kJ/mol. Determine the temperature at which the rate of racemization would be double than that corresponding to standard conditions. Given the nature of this process, describe what you would expect to happen to a pure sample of L alanine as the time passes. What would be the value of ratio [D]/[L] when the system reaches chemical equilibrium? Share and discuss your ideas with a classmate, and clearly justify your reasoning. Chemical Thinking U5 How do we predict chemical change? 325 Thalidomide Thalidomide is a drug introduced in the 1950s to treat morning sickness in pregnant women and aid sleep. It was sold until 1961, when it was withdrawn from the market after being found that the drug caused birth defects. The thalidomide molecule is chiral and each enantiomer has a different effect in the human body. While the D isomer is a sedative, the L isomer can induce malformations (teratogenic). Unfortunately, the rate of racemization is relatively high, taking only a few hours for a pure sample of one of the enantiomer to generate a racemic mixture inside the body. The interconversion process can be represented in the following way: Thalidomide-D • • • • Thalidomide-L Write each of the reaction steps associated with the above process. Determine the molecularity of each step and identify all reaction intermediates. Build a potential energy diagram to represent the racemization process; Establish the overall rate law for the process and identify the order of the reaction with respect to the concentration of thalidomide. Measurements of the rate constant for the racemization process at two different temperatures generate the following values: k(15 oC) = 0.37 days-1; k(37 oC) = 0.30 hours-1. Use this information to determine the activation energy for the racemization. The rate of the racemization reaction in aqueous solution varies depending on the concentration of hydronium ion, H3O+ (which determines the acidity of the solution). If we represent the D-thalidomide molecule using the symbol T(D)=O, the reaction mechanism for the first half of the racemization process can be expressed as: T(D)=O + H3O+ T(D)=OH+ + H2O • • • T(D)=OH+ + H2O T(D)-OH + H3O+ FAST SLOW Determine the molecularity of each step and identify all reaction intermediates. Establish the rate law for each step, as well as the equilibrium constant for the first process. Use these relationships to find overall rate law for the process and identify the order of the reaction with respect to the relevant species. Identify the overall order of the reaction. The hydronium ion is said to “catalyze” the racemization process. Discuss any evidence you have based on the analysis of the reaction mechanism to support such a claim. Build an energy potential diagram that represents how the H3O+ ions may affect the reaction path and speed up the process. Share and discuss your ideas with a classmate, and clearly justify your reasoning. ASSESS WHAT YOU KNOW Chiral Carbon 326 MODULE 3 Understanding Mechanism Let’s Apply ASSESS WHAT YOU KNOW Photochemical Smog Photochemical smog is a type of air pollution produced when sunlight acts upon motor vehicle exhaust gases, such as nitrogen oxides (e.g., NO, NO2) to form harmful substances such as ozone (O3). The substances responsible for this phenomenon participate in a reaction cycle that determines the concentration of pollutants at different times during the day. Ozone Formation Nitrogen dioxide gas, NO2(g), is formed by chemical reactions involving N2(g) and O2(g) at the high temperatures inside internal combustion engines in our cars. In the presence of sunlight, NO2(g) reacts with O2(g) to generate O3(g) as described by the following overall reaction: NO2(g) + O2(g) NO(g) + O3(g) This chemical process occurs in two steps, as illustrated in the potential energy diagram. In the second step of the process, the radical species O reacts with O2. In order for the process to occur, the collision should involve a third particle M that can absorb excess energy from the collision (e.g., N2). • • • • • Ep Ea = 306.5 kJ/mol Ea = 107 kJ/mol O + NO O2 + M NO + O3 + M* NO2 Reaction Path Write the chemical equations for each reaction step and determine the molecularity of each of them. Identify all reaction intermediates. Express the rate law for each reaction step and derive the rate law for the overall process. The activation energy required to initiate the process is provided by sunlight. Calculate the minimum frequency n of the radiation needed to start the reaction. Estimate how many times faster this reaction is at 30 oC versus 10 oC. Based on the information provided, evaluate whether this process can be expected to occur to a large extent. Share and discuss your ideas with a classmate, and clearly justify your reasoning. Chemical Thinking U5 How do we predict chemical change? 327 Ozone formation is faster during the day due to the action of sunlight on NO2 molecules. Thus, O3(g) accumulates in the troposphere and its concentration peaks sometime in the afternoon. During the night, ozone reacts with NO(g) according to the following reaction: NO(g) + O3(g) PPB Ozone Consumption NO2(g) + O2(g) • • • Hour Build potential energy diagram that represents the transformation of reactants to products. Indicate on the diagram the activation energy and the heat of the reaction for this process. Derive the rate law for this process. Estimate how many times slower this process is at 0 oC than at 25 oC. Share your ideas with a classmate, and clearly justify your reasoning. Additional Reactions The oxygen atoms generated by the decomposition of NO2 molecules in the presence of light are highly reactive and interact with other substances to generate different types of pollutants. Consider the following sequence of reactions involving methane, CH4: • • Chemical Reaction NO2 NO + O• O• + H2O CH4 + OH• CH3• + O2 CH3O2• + NO CH3O• + O2 CH O + OH• Rate Constant k (25 oC) (molecules-1 cm3 s-1) Varies: Light Intensity 2 OH• H2O + CH3• CH3O2• NO2 + CH3O• 2.0 x 10-10 6.3 x 10-15 1.8 x 10-12 7.7 x 10-12 CH2O + HO2• 1.9 x 10-15 HCO• + H O 9.0 x 10-12 2 2 Analyze this reaction HCO • + O CO + HO • 5.2 x 10-12 mechanism. Identify 2 2 reactants, products, 2 HO2• H2O2 + O2 1.7 x 10-12 and reaction intermediates. Derive the overall chemical equation for this process. Use the data to evaluate which step may determine the overall rate of reaction. Share your ideas with a classmate, and clearly justify your reasoning. ASSESS WHAT YOU KNOW This is a single-step process with an activation energy Ea = 12.5 kJ/mol and a DHrxn = –199.5 kJ/mol.