Chemical Kinetics, Exams of Chemical Kinetics

Download Chemical Kinetics and more Exams Chemical Kinetics in PDF only on Docsity! Chemical Kinetics 4 Thermodynamics teaches us about the energetics and favorability of reactions but not whether they are fast or slow. For that we need to understand kinetics, the study of rates of reactions and the physical factors that influence those rates. Reactions that are thermodynamically favorable can take millions of years to reach equilibrium. In fact, the velocities of favorable reactions vary over about 15 orders of magnitude. The speed of reactions is critically important for living systems, which must ensure that they occur rapidly enough to be consistent with the rates at which cells grow and divide. Consequently, cells have evolved strategies to accelerate reactions so that they occur on a biological timescale. In this chapter, we will examine the factors that influence reaction rates, and in later chapters, we will explore the strategies that living systems use to increase and control those rates. Reaction rates are determined by the frequency, orientation, and energy with which molecules collide Every molecule has a characteristic size, shape, and distribution of electrons. As such, a reaction between two molecules can only occur when they physically contact each other; they must collide to react. The rate at which molecules react with each other to form a product (or deplete a reactant) is determined by the frequency with which they collide, the probability that they collide in an orientation that allows the reaction to occur, and the probability that the molecules that collide do so with enough energy for the reaction to take place. • describe the factors that determine reaction rates. • identify nucleophiles, electrophiles, and leaving groups. • use arrow pushing to represent reaction mechanisms. • draw and interpret reaction energy diagrams. • draw transition states. • write a rate expression for a simple chemical reaction and explain how the rate is related to ΔG‡. After this chapter, you should be able to To understand the physical factors that determine reaction rates. Objectives Goal Chemical KineticsChapter 4 2 Not every collision leads to a reaction, but the rate of a reaction depends on the frequency with which reactant molecules collide, which is influenced by the concentration of the reactants (Figure 1). The larger the number of reactant molecules that are confined within a given volume, the greater the probability of their colliding. The rate of reaction is also influenced by the velocity of the reactant molecules, which again influences the frequency of collisions. The rate of reaction also depends on the cross-sectional areas of the reactant molecules, a property determined by the shapes and sizes of the molecules involved in the reaction. Reactions can only occur between parts of molecules that are capable of forming new bonds; therefore, the rate of a reaction also depends on the probability that individual molecules collide in the appropriate orientation to react (Figure 1). However, colliding in a productive orientation is not sufficient. Individual molecules also need to collide with enough energy to react. As we have seen, breaking bonds requires energy, and because all reactions involve the breaking of bonds, an input of energy is required to overcome this energetic barrier (or “threshold of free energy”, as we will see below). Even for a favorable reaction in which the products are lower in energy than the reactants, some amount of energy is necessary to overcome this energetic barrier and initiate the bond-breaking reaction. The rate of a reaction is influenced by temperature, an insight that was formalized by the scientist Svante Arrhenius in the late 1800’s. Temperature increases the frequency with which molecules collide by influencing the velocity of the reactant molecules and the proportion of molecules with sufficient energy to overcome the energetic barrier to react (see Box 1). Therefore, temperature influences two of the five factors in Figure 1, namely the velocity of reactant molecules and the probability that molecules collide with enough energy to react. To better understand how the five factors influence the rate of reactions, we need to understand how reactions proceed, that is, the sequence of events or pathway by which reactions proceed. The reaction pathway can be represented in two complementary ways: by a molecular mechanism and by a reaction energy diagram. These representations allow us to derive rules that predict reaction outcomes. An arrow-pushing formalism describes reaction mechanisms Put simply, a chemical reaction is a rearrangement of valence electrons that leads to the cleavage and/or formation of bonds. The specific electron movements associated with each reaction are described by the reaction mechanism. Reaction mechanisms show where electrons originate in the reactant and where they ultimately reside in the product. This is represented Figure 1 Reaction rates depend on multiple factors reaction rate = concentration of reactant molecules velocity of reactant molecules cross-sectional area of reactant molecules probability that molecules collide in the right orientation probability that molecules collide with enough energy to react xxx x Chemical KineticsChapter 4 5 Let us apply these concepts to the reaction of carbonic acid (H2CO3) with water to yield bicarbonate and hydronium ion, as discussed in Chapter 3. The reaction mechanism is described using two arrows (Figure 3). One arrow points from the lone pair of electrons on the water molecule to a hydrogen atom on carbonic acid. This arrow conveys the idea that the oxygen from water is using its lone-pair electrons to form a bond with the hydrogen atom. Since hydrogen can only form one bond at a time, it must break its existing bond to oxygen. Breaking the existing O-H bond is represented by the second arrow, which points from the O-H bond in carbonic acid to the oxygen atom in the same O-H bond. This arrow conveys the idea that the O-H bond is breaking and that the electrons from that bond are being retained by carbonic acid’s oxygen as a lone pair. In this reaction, the oxygen in water is a nucleophile and the hydrogen in carbonic acid is an electrophile, as the oxygen in water is using its electrons to form a new bond and hydrogen is the recipient of those electrons. Finally, the remainder of the carbonic acid molecule (HCO3 −) serves as a leaving group, as it results from the cleavage of the carbonic acid O-H bond; it takes a pair of electrons with it. O C O O H H O H H O C O O H O H H H Leaving group Nucleophile ElectrophileFigure 3 Using arrow pushing to describe the mechanism of carbonic acid dissociation Box 3 As we will see in Chapter 5 and as shown below, proteins are made from individual amino acid monomers that are connected by peptide bonds (colored in red). The side chains of each amino acid are represented as “R1”, “R2”, etc. Proteins are broken down to amino acids in cells by hydrolysis. Hydrolysis is any reaction in which water reacts with another species to break an existing bond in that species while forming a new bond to the oxygen Applying arrow pushing to peptide bond hydrolysis O N HR4 H N R5 O O N HR2 H N R3 O R1 Breakout Shown below is the first step of peptide bond hydrolysis, a reaction used by all living systems to break down proteins. Which atom is the electrophile? A. The red oxygen. B. The black oxygen. C. Nitrogen. D. Carbon. E. The red hydrogen. N H O O H H Chemical KineticsChapter 4 6 of the water molecule. Peptide bond hydrolysis is a reaction that occurs when water reacts with the carbon atom of a peptide bond, resulting in the cleavage of the carbon-nitrogen bond and the formation of a new carbon-oxygen bond: Peptide bond hydrolysis is a multi-step reaction, and as such, we can think of peptide bond hydrolysis as a series of sequential chemical reactions, each with its own arrow-pushing mechanism. In the first step (below), the carbon of the peptide bond (shown in black) and the oxygen of the water molecule (shown in red) react because the carbon is electron-deficient (electrophilic) and the oxygen is electron-rich (nucleophilic). The curly arrows show the movement of electrons as the reaction proceeds. Arrow 1 shows that a lone pair on the oxygen of water is forming a new bond between the oxygen and the carbon. Carbon can only form a maximum of four bonds, so in order to form a new bond with oxygen, one of carbon’s other bonds must break. Arrow 2 shows that one of the carbon-oxygen bonds is breaking and forming a long pair on the oxygen (black). Note that the differences between the starting material and product are only those indicated by the arrows. Note also that the arrow-pushing mechanism involves a chain of curly arrows in which the atom at the beginning of the chain, the oxygen in the water molecule, acquires a positive charge by sharing a lone pair of electrons, and the atom at the end of the chain, the oxygen in the peptide bond, gains a negative charge by acquiring the two electrons it previously shared with the carbon. In the next step a proton is transferred from one atom in the molecule to another. You might imagine a direct mechanism for this transfer in which the lone pair on nitrogen reacts with the proton attached to positively charged oxygen shown in red. It is important to remember, however, that these molecules are bathed in water; consequently, this phase of the reaction occurs as a combination of the two reactions shown in the dashed box. In the first reaction, a lone pair from an oxygen atom in a water molecule forms a bond to the hydrogen atom (Arrow 3). Since hydrogen can only form one bond at a time, its other bond must break in order for the new bond to form (Arrow 4). In the second reaction, a lone pair from the nitrogen atom O +H+/-H+ N H O O HH N O OH H H -H+ N H O O HH O H H N H O OH O H H H 3 4 +H+ N O OH H H N H O O H O H H H H H 5 6 N H O O H H 1 2 O H HN H O + O O NH H H + Chemical KineticsChapter 4 7 forms a bond to a hydronium ion (Arrow 5). The bond that previously existed between this hydrogen and the oxygen in hydronium breaks, and the electrons from that bond form a new lone pair on the blue oxygen atom (Arrow 6). The next step involves the cleavage of the carbon-nitrogen bond. Note that the atoms in this mechanism have been recolored for clarity. Arrow 7 shows that a lone pair on the negatively charged oxygen forms a bond to the carbon atom, creating a double bond. Arrow 8 shows that the carbon-nitrogen bond breaks, with the electrons from that bond forming a new lone pair on nitrogen. At this point hydrolysis of the peptide bond has occurred, but in the cell the reaction is not yet complete. Completion involves another proton transfer. As before, this transfer occurs via two reactions involving water, as shown below. In one reaction, a lone pair from an oxygen atom in a water molecule forms a new bond to the red hydrogen atom (Arrow 9), and the bond between the red hydrogen and red oxygen breaks, with the electrons from the bond forming a new lone pair on oxygen (Arrow 10). In the other reaction, the lone pair on nitrogen forms a new bond with a hydrogen atom in hydronium (Arrow 11), and the bond between the hydrogen and oxygen in hydronium breaks, with the electrons from the bond forming a new lone pair on oxygen (Arrow 12). These final proton transfers are energetically highly favorable and drive the entire sequence of events that culminate in the hydrolysis of the peptide bond. N O OH H H N O O H H H 7 8 O O H O H H O O O H H H H N O H H H H H N O H H H H 9 10 11 12 Reaction energy diagrams describe changes in Gibbs free energy as reactants are converted to products Another way to represent a reaction mechanism is to depict the change in Gibbs free energy as individual reactant molecules react to produce a single set of products. The horizontal axis in the reaction energy diagram is the reaction coordinate, which is a measure of the extent to which the reactant molecule(s) has proceeded through energetically distinct steps in its conversion into a product molecule(s). The reaction coordinate should not be confused with the time it takes for the reaction to proceed or with progressive changes in the composition of a mixture of many molecules. Molecules are represented in reaction energy diagrams as points whose vertical positions indicate their Gibbs free energy, with more-stable (lower- energy) species closer to the bottom of the diagram. The difference between Chemical KineticsChapter 4 10 the concentration of the reactants. The greater the ΔG‡, the lower the concentration of the transition state. Since chemical reactions proceed to product by going through the transition state, the rate of a chemical reaction is directly proportional to the concentration of the transition state at any given moment. Thus, the reaction energy diagram is a powerful tool for analyzing chemical reactions because it allows us to assess a reaction’s rate as well as its thermodynamic favorability. Reactions with very large ΔG‡ values are said to have high energy barriers or activation energies, exemplifying the idea that the reaction occurs slowly because a lot of energy is needed for the reactants to transform into the transition state. In other words, a reaction with a highly favorable (highly negative) ΔG°rxn can be extremely slow if ΔG‡ for the transition state is very high. Box 4 Although transition states cannot be observed directly, their structures can be inferred from the reaction mechanism. To draw a transition state from an arrow-pushing mechanism: 1. Inventory the bonds that are being made and broken in one reaction step. 2. Draw a single structure in which each of these changing bonds is partially formed. Represent partially formed bonds using dashed lines. You should find that the dashed lines in your drawing form a continuous path from one atom to another in the transition state; they are never discontinuous, and they never branch. 3. Assign formal charges in which partially formed bonds represent half a bond, and use δ− and δ+ notation to represent “half ” charges. You will notice that atoms whose formal charges change during the reaction will have these partial charges. For example, an atom that is negative in the reactant and neutral in the product would be partially negative (δ−) in the transition state. This is intuitive given that transition states represent species that exist in between the reactant and product. Overall charge is conserved between the reactant, product, and transition state, so the “half ” charges in the transition state will always total the overall charge found in both the reactant and product. 4. Lastly, structures of transition states must be denoted as such by drawing them within brackets and marking them with the double dagger symbol (‡). The transition state for the first step of peptide bond hydrolysis is shown below. The bond that breaks during the reaction is shown in red and the bond that forms is shown in blue. Transition states can be represented as chemical structures N H O O H H N H O O HH transition state products δ+ δ− reactants N H O O HH bond breaking bond forming Chemical KineticsChapter 4 11 Reaction rate is determined by reactant concentrations and the rate constant We are now ready to describe the rate of a reaction in terms of reactant concentrations and a rate constant. We can express the rate of a single-step reaction that proceeds via a single transition state as follows: rate = k [R1][R2] Where [R1] and [R2] are the concentrations of reactants R1 and R2 and k is the rate constant. The rate constant is a term that describes all of the factors other than reactant concentrations that affect the rate of a reaction (Figure 6). The rate constant is related to ΔG‡ according to the Arrhenius equation (shown below as a proportionality) in which e is the base for the natural logarithm (≈ 2.718), R is the gas constant, and T is the temperature in Kelvin: We draw two important conclusions from this relationship. First, the rate constant is inversely related to ΔG‡; as ΔG‡ becomes larger, the rate constant becomes smaller. We can understand this relationship intuitively, as large values of ΔG‡ are associated with transition states that are difficult to attain (Figure 7). The second conclusion is that the rate constant varies with temperature. Figure 7 The rate constant k and ΔG‡ correlate with reaction rate Fast reactions Slow reactions k is large k is small ∆G‡ is small ∆G‡ is large rate constant (k) ∆G‡ 0 0 increasing rate ek −∆G‡ R T reaction rate = concentration of reactant molecules velocity of reactant molecules rate constant (k) cross-sectional area of reactant molecules probability that molecules collide in the right orientation probability that molecules collide with enough energy to react xxx reaction rate = concentration of reactant molecules x x Figure 6 The rate constant accounts for all factors affecting rate other than reactant concentration Chemical KineticsChapter 4 12 Finally, the rate constants for the forward and reverse reactions of a one- step process can be related to the equilibrium constant by a simple equation, which is derived in Figure 8: Keq = kforward kreverse Simply put, the equilibrium constant is directly related to the ratio of the forward and reverse rate constants. For example, if the forward rate constant is larger than the reverse rate constant, the equilibrium constant will be greater than one. [Note that the notations used to denote the equilibrium constant (Keq) and the rate constant (k) can be confused with each other. As a rule, the equilibrium constant is written with a capital letter K, and the rate constant is written with a lower-case letter k.] Favorable reactions with slow kinetics are common in living systems Thermodynamically favorable reactions that occur very slowly are common in living systems. For example, as we considered in Chapter 3, the cleavage of ATP is highly favorable but extremely slow. Living systems solve this problem by using protein catalysts known as enzymes to accelerate specific reactions. As we shall see in later chapters, these enzymes accelerate reactions by lowering the energy of the transition state, thus reducing ΔG‡. The use of enzymes to accelerate slow reaction rates has an important advantage for living systems; it enables cells to effectively turn reactions on and off by controlling when and where specific enzymes are present and active. The use of thermodynamically favorable but kinetically slow Keq = kforward kreverse kforward kreverse = [reactant] [product] kforward [reactant] = kreverse [product] kforward kreverse Keq = reactant product [reactant] [product] rateforward ratereverse = kforward = kreverse [reactant] [product] (1) (2) (3) (4) (5) Figure 8 Keq equals the ratio of the forward and reverse rate constants Shown is the derivation for the relationship between Keq and the forward and reverse rate constants. If we begin with a single- step reaction whose equilibrium constant equals the ratio of product to reactant (line 1), we can write expressions for the forward and reverse reaction rates (line 2). When the reaction is at equilibrium, we know that the forward and reverse reaction rates are equal, allowing us to set the rate expressions from line 2 equal to one another (line 3). Then we can divide both sides of the equation by reactant concentration and the reverse rate constant (line 4). Since the reaction is at equilibrium, we can substitute Keq for the ratio of product concentration over reactant concentration in line 4 to produce line 5.