Reaction Thermodynamics Thermodynamics reflects the degree to which a reaction is favored or disfavored Recall: G = Gibbs free energy = the energy available to do work ΔG = change in G of the system as state A becomes state B ΔG = negative A B Energy downhill favorable spontaneous ΔG = positive Energy A B uphill disfavored ΔG = zero A B at equilibrium ΔG can change over time and is dependent on many factors including the reaction, concentrations, and temperature Thermodynamics of a reaction and free energy Two of the most important components of any chemical reaction are the reaction s thermodynamics and its kinetics. A reaction s thermodynamics and its kinetics are distinct and there is no necessary relationship between the two. A reaction s thermodynamics is the degree to which the reaction is favorable in other words, how much energy is released into the surroundings (or taken from the surroundings) as the substrates are transformed into products. You recently learned the basic thermodynamics underlying protein folding, and many of the same concepts can be used to study chemical reactions. Recall that free energy (G) is the energy available to do work and is expressed in units of energy per numbers of molecules (for example, calories per mole). The quantity of energy released or absorbed by a system as the starting materials of a reaction become products is called the change in Gibbs free energy of the system (ΔG).. A negative ΔG indicates that under a specified set of conditions, the system is releasing energy as the starting state begins to transform into the ending state. In this case, the free energy (G) of the starting state of the system is higher than that of the system some defined amount of time later. A process with a negative ΔG is favorable, or downhill under those specific set of conditions and takes places spontaneously. Conversely, a positive ΔG describes a process that increases the system s free energy. Such an uphill process is unfavorable under those specific set of conditions and does not take place spontaneously. ΔG for starting materials becoming products in a chemical reaction is dependent on many factors including the identity of the reactants, the concentrations of the substrates and products, and the temperature of the reaction. G and ΔG are constantly changing for a system that is not at equilibrium. When starting materials are allowed to react, the relative concentrations of substrates and products begin to change. These changes, in turn, lower G such that the total free energy for the system decreases; therefore ΔG is negative as a system heads toward equilibrium. The free energy of the system will continue to decrease as the reaction progresses until ΔG = 0. When ΔG = 0, the system is at equilibrium and there is no additional net conversion of substrates to products (or vice-versa). At this point, the free energy of the system is now constant. Professor David Liu and Brian Tse, Life Sciences 1a page 1
Thermodynamics: ΔG, ΔG rxn, and K eq To study and compare reactions, need a standard state, defined as 1 M concentrations, 1 atm pressure, constant temperature ΔG is defined as the change in G under standard state conditions for a hypothetical reaction that proceeds to 100% completion ΔG, unlike ΔG, depends only on the identity of the reactants A + B C + D ΔG = ΔG rxn + RT ln [C][D] [A][B] At equilibrium, 0 = ΔG rxn + RT ln K eq Therefore, ΔG rxn = RT ln K eq Because ΔG can change constantly and depends on so many factors including reactant concentrations, it cannot be easily used to study or compare two different reactions. To compare and interpret the thermodynamics of various reactions, scientists have defined a standard state set of reference conditions. This standard state is usually: substrate concentrations = 1 M, pressure = 1 atm, and a constant temperature (which must be specified). Under these standard state conditions, ΔG= ΔG by definition. ΔG is the change in free energy that would result from the hypothetical complete conversion of all the starting materials into products under standard state conditions. Because in the standard state all reaction conditions affecting free energy are pre-defined, the value of ΔG depends solely on the identities of the starting materials and products. Therefore, the concentrations of reactants or the extent to which a reaction is complete two factors that influence ΔG do not affect ΔG. Reactions of substrates high in free energy that generate products lower in free energy have negative values of ΔG ; conversely, reactions involving lower G substrates that give rise to higher G products have positive values of ΔG. When ΔG for a reaction is < 0 (negative), that reaction is favorable or downhill under the standard state conditions and can take place spontaneously. Conversely, a reaction with a ΔG > 0 (positive) is unfavorable or uphill and does not occur spontaneously to completion under standard state conditions (although at equilibrium this reaction would still consist of some product which would represent less than 50% conversion; see below). Keep in mind that when considering biological processes, however, ΔG is not a relevant value on its own because conditions in cells are very different from standard state conditions and because reactions rarely proceed to completion. Nevertheless, ΔG is very useful for comparing the reaction thermodynamics of different processes. For the representative reaction of A + B ---> C + D, ΔG and ΔG are related by a very useful equation: ΔG = ΔG + RT ln ([C][D] [A][B]). where T is temperature in degrees Kelvin and R is the ideal gas constant. When the system is at equilibrium, ΔG = 0 and [C][D] [A][B] = K eq ; therefore, at equilibrium this equation becomes ΔG = 1.4 log 10 (K eq ). A useful rule of thumb emerging from this equation is that at room temperature, ~1.4 kcal/mol of free energy corresponds to a 10-fold change in the equilibrium constant under standard state conditions. Professor David Liu and Brian Tse, Life Sciences 1a page 2
Thermodynamics of Coupled Reactions Small Energy A B ΔG > 0 ATP ADP (P) Large Energy ΔG < 0 ATP A B ADP (P) Small Energy ΔG < 0 Cells can enable uphill reactions to proceed efficiently by coupling them with favorable reactions (e.g., ATP hydrolysis) ne must be careful interpreting ΔG values. A positive ΔG indicates that a reaction is unfavorable under the stated reaction conditions (which may or may not correspond to a positive ΔG ) and must be coupled to a more favorable process in order to take place to a significant extent. In the reactions driving life processes, ATP hydrolysis, a highly favorable process (ΔG = -7.4 kcal/mol) is often coupled to reactions having positive ΔG values such that the overall process has a ΔG < 0 and therefore can take place. A negative ΔG indicates that a reaction is favorable, but does not necessarily mean that the substrates will become the products on a reasonable time scale. For example, as we demonstrated in class it is thermodynamically highly favorable (ΔG << 0) if hydrogen gas and oxygen gas reacted to form water. However, in the absence of a flame or spark, this reaction does not take place at any appreciable rate at room temperature. Professor David Liu and Brian Tse, Life Sciences 1a page 3
Reaction Thermodynamics Example concentration A K eq = 10 B start with 100% A time ΔG = RT ln K eq = 1.4 kcal/mol at 298 K concentration start with 100% B time [B]/[A] at equilibrium = 10 free energy (G) G decreases as a system approaches equilibrium ΔG 0 At equilibrium, ΔG = 0 & G is constant time time Let s integrate our understanding of reaction thermodynamics into an example. Imagine a chemical reaction in which molecule A is transformed into molecule B. In this example, K eq for the equilibrium between A and B as drawn above is 10. As you know, this K eq value means that at equilibrium, [B]/[A] = 10. Furthermore, based on the relationship between ΔG and K eq described earlier, you also should be able to deduce the difference in free energy under standard state conditions between A and B that is, the value of ΔG. At room temperature (25 C = 298 K), and using the value of R, the ideal gas constant, as approximately 2 cal/mol K, we can calculate that ΔG = 1.4 kcal/mol. This means that B is lower in free energy than A under standard state conditions by 1.4 kcal/mol. It s important to realize that this value of ΔG is constant it doesn t change regardless of whether the reaction has reached equilibrium, or even whether the reaction has taken place yet. ΔG depends only on the identity of A and B in this example. If we start this reaction with a test tube containing 100% A, molecules of A will begin to convert into molecules of B. The free energy of the system (G) decreases as the system moves toward equilibrium, and therefore ΔG starts out negative. nce the ratio of [B]:[A] reaches 10:1, the system is at equilibrium and no additional net changes in concentration take place. At this point, G is constant and ΔG is therefore zero. We can also start the reaction with a test tube containing 100% B. In this case, molecules of B will convert into molecules of A until the same equilibrium point is reached. Importantly, just as in the case of starting with 100% A, the free energy of the system decreases as the system moves toward equilibrium. Professor David Liu and Brian Tse, Life Sciences 1a page 4
The Thermodynamics Versus Kinetics of Blowing Stuff Up 2 H 2 + 2 (dilute) 2 H 2 + 2 (dilute) heat 2 H 2 + 2 (concentrated) heat very slow fast very fast All three of the above reactions are thermodynamically identical Heat increases the rate of a reaction Concentrating the substrates increases the rate of a reaction Why, and how do enzymes use these principles? Professor David Liu and Brian Tse, Life Sciences 1a page 5
Reaction Kinetics: Rates and Rate Constants Kinetics determines the rate at which reactions occur transition state S [TS] P Low energy TS = fast reaction High energy TS = slow reaction S 1 S 2 [TS] P For the above simple reaction, rate = k[s 1 ][S 2 ] k = the rate constant, a function of ΔG, the transition state free energy (larger ΔG = smaller k = slower reaction) Kinetics of a reaction and ΔG The reason why hydrogen and oxygen gas won t react (burn) at any significant rate at room temperature is explained by the second crucial component of a chemical reaction: the reaction s kinetics. While thermodynamics are most informative about the outcome of a reaction once it reaches equilibrium, the kinetics of a reaction describe the rate at which the reaction can take place under a given set of reaction conditions. Because living systems are never at equilibrium (indeed, a defining characteristic of life is that it manages to sustain itself far from equilibrium), an understanding of reaction kinetics is crucial to understanding the chemistry of life. In order for the substrates of a chemical reaction to become products, the free energy of the starting materials must be increased until it exceeds a certain threshold value. When the energy of a system reaches this threshold, the system is said to exist in the transition state, the state of the highest free energy during the course of a chemical reaction. The symbol is used to indicate that a descriptor refers to a transition state. Because the energy of the transition state is at a maximum, the system can spontaneously become products (or revert to starting materials) without requiring the infusion of any additional energy. The change in free energy between the substrates and the transition state is called ΔG, also known as the activation energy of a reaction. Professor David Liu and Brian Tse, Life Sciences 1a page 6
Reaction Kinetics: Rates and Rate Constants Kinetics determines the rate at which reactions occur transition state S [TS] P Low energy TS = fast reaction High energy TS = slow reaction S 1 S 2 [TS] P For the above simple reaction, rate = k[s 1 ][S 2 ] k = the rate constant, a function of ΔG, the transition state free energy (larger ΔG = smaller k = slower reaction) (notes continued from previous page) The difference between a fast reaction and one that occurs slowly is simply the value of Δ G. The larger the activation energy barrier, the more energy must be infused into the substrates (most commonly, by heating the reaction) before they can become products, and therefore the slower the reaction. Conversely, reactions with almost no ΔG proceed extremely quickly. The rate of a chemical reaction is dependent on k, the rate constant, as well as on the concentrations of each of the reactants. The rate constant (k) is related to e (-ΔG /RT) and as you probably have deduced, larger values of ΔG result in smaller rate constants. The rate of a simple one-step reaction is equal to k multiplied by the concentrations of each of the substrates. Doubling the concentration of each of two substrates in the simple reaction shown above will therefore increase the rate of the reaction by four-fold. Rates of reactions always have units of molecules per time or concentrations per time (for example, moles of product formed divided by seconds, or moles of substrate consumed per liter divided by seconds). It is crucial that you understand the distinction between K eq and k. The former is an equilibrium constant, a ratio of concentrations at equilibrium. The latter is a rate constant, a measure of how quickly a given substrate becomes a given product. Confusingly, researchers often abbreviate K eq using K, emphasizing the extreme importance of clearly notating uppercase K and lower-case k when describing reactions! Professor David Liu and Brian Tse, Life Sciences 1a page 7
Relationship Between K eq, k for, and k rev A + B k forward k reverse C + D Forward rate = k forward [A][B] Reverse rate = k reverse [C][D] At equilibrium, forward rate = reverse rate k forward [A][B] = k reverse [C][D] [C][D] k forward /k reverse = [A][B] Therefore, k forward /k reverse = K eq Although there is no necessary relationship between thermodynamics and kinetics i.e., you cannot infer K eq from k alone or vice-versa, there is a useful relationship between the equilibrium constant, the rate constant of the forward reaction (k forward ), and the rate constant of the reverse reaction (k reverse ). Recall that at equilibrium, the rate (not rate constant!) of the forward reaction equals the rate of the reverse reaction and no net transformation of substrates into products takes place. Since each rate equals the rate constant multiplied by the concentration of each substrate, at equilibrium you can equate in the example above k forward [A][B] = k reverse [C][D]. At equilibrium, [C][D]/([A][B]) is also equal to K eq. Therefore you can deduce that k forward /k reverse = K eq. Professor David Liu and Brian Tse, Life Sciences 1a page 8
Reaction Energy Diagrams reactants (substrates) transition state S [TS] P products Reaction energy diagrams depict changes in ΔG as one substrate molecule becomes one product molecule X-axis: the reaction coordinate Y-axis: free energy under defined conditions (does not change with concentration of S or P) G S [TS] ΔG P ΔG rxn Reaction energy diagrams Concepts in reaction thermodynamics and kinetics are best illustrated by drawing a reaction energy diagram. Such a diagram plots the free energy of a particular molecule undergoing a reaction on the Y- axis. The X-axis is the reaction coordinate, a measure of the degree of progress in the reaction (i.e., the extent to which a starting material molecule has changed into a product molecule. Different molecules (or collections of molecules) are represented in reaction energy diagrams as points in the diagram. These points are positioned vertically at their estimated free-energy levels, with more stable (lower energy) species closer to the bottom of the diagram. The points are placed horizontally to indicate where on the reaction coordinate they are thought to form, with earlier-forming species positioned to the left side of the diagram, and later-forming species to the right side. The transition state of a reaction is represented by a point reflecting its free energy and location along the reaction coordinate. By definition the highest point on the reaction energy diagram must be the transition state of the reaction. Because the products of a favorable reaction are lower in free energy than that of the substrates (ΔG < 0), the overall decrease in free energy for a favorable reaction is represented by an overall downward slope in the reaction energy diagram going from the starting materials to the products. f course the energy of the system during a chemical reaction does not simply change as a straight line connecting the starting materials and products instead, the reaction energy diagram must also reflect the higher energy of the transition state. Typically a smooth curve (occasionally called the energy landscape of a reaction) is used to connect the starting materials to the transition state to the products of a reaction in a reaction energy diagram. f course the exact free-energy values of all points on the curve other than the starting material and product are virtually impossible to ascertain. Nevertheless, the reaction energy diagram is a powerful tool for analyzing chemical reactions. It is important to realize that the Y-axis of a reaction energy diagram reflects the free energy under a specific set of conditions, for example, under standard state conditions. Even if those specific conditions are not standard state conditions (1 M, 1 atm, etc.), the Y-axis free energy values in a reaction energy diagram do not change as the concentrations of S or P change. Professor David Liu and Brian Tse, Life Sciences 1a page 9
Transition States and Intermediates intermediate A B C D E G A [B] Transition states are [D] energy maxima; C intermediates are local energy minima E Note: only molecules with the same combined molecular formula can be placed on the same energy diagram! Although the reactions we ve depicted above represent the simplest possible cases, more complex variants are quite common in the reactions of life. A reaction can have multiple transition states, although the transition state with the highest ΔG is the one that plays the largest role in determining the overall kinetics of the reaction. Likewise, many reactions proceed through intermediates that are neither starting materials nor products nor transition states. Instead, intermediates are resting points along a chemical reaction in which partially converted starting materials might exist for a period of time. In contrast, transition states by definition are fleeting. In a reaction energy diagram, the difference between an intermediate and a transition state is very clear: intermediates lie in minima (wells), while transition states are represented as maxima (hills). How do scientists know what transition states look like? Because a transition state is, by definition, a fleeting structure with no significant lifetime, it is extremely difficult for scientists to directly observe a transition state. Instead, scientists mostly use the outcomes of chemical reactions to infer the structure of the transition states. In many cases, key features such as the approximate shape, charge, and atom connectivity of a transition state can be deduced by examining reaction outcomes. A final, but very important note about reaction energy diagrams: it is only meaningful to put molecules on the same free energy diagram that have the same combined chemical formula. Since the starting materials and products of a simple chemical reaction must (under the principle of the conservation of matter) contain the same atoms, they can always be placed on the same energy diagram, assuming that the scientist has correctly depicted them! If you forget one of the starting materials or products or parts of a molecule when you depict a chemical reaction, however, creating a reaction energy diagram will be a confusing (and ultimately futile) exercise. Therefore it is crucial that all species depicted on the same energy diagram share the same combined chemical formula. Professor David Liu and Brian Tse, Life Sciences 1a page 10
Drawing (Simple) Transition States 1) Inventory the bonds being made and broken in one chemical step bond broken N H H bond made H N H 2) Draw a single structure in which each of these changing bonds is partially formed (dashed) H N H 3) Assign formal charges (e.g., assume partially formed bonds represent half a bond) making 2 bonds is neutral; making 1 bond (with 3 lone pairs) is 1; that makes ~1.5 bonds is δ δ δ H N H that makes ~1.5 bonds is δ Note overall charge conservation! Drawing Transition States The difficulty of their direct observation can make transition states challenging to draw in detail. Nevertheless, the essential features of transition states, including those features that are affected by enzymes, can often be depicted using the strategy shown here. To draw a simple transition state, start by taking inventory of the bonds that are made and broken over the course of the single chemical step of interest. When we draw starting materials, intermediates, and products of chemical reactions, these bonds are either 100% formed or 100% broken. In contrast, a transition state contains partially formed covalent bonds. Therefore, after identifying the bonds that are made and broken, draft a single structure beginning with the starting materials in which these changing bonds are partially formed. Partially formed bonds are drawn using dashed lines (not to be confused with hydrogen bonds, which are also often drawn using dashed lines). In the final step of drawing a transition state, examine each atom of your proposed structure and assign appropriate formal charges. Use the formal charge principles we previously discussed to assign partial charges (indicated by the symbol δ) to those atoms that participate in partial covalent bonds. For example, recall that an atom that makes two bonds (and has two lone pairs) is neutral, while an atom that makes one bond (and has three lone pairs) has a 1 charge. An atom that makes one full bond and one partial bond (roughly 1.5 bonds) will therefore have a δ charge. The process of drawing a transition state using the above strategy is applied to the first step of amide bond hydrolysis in this slide. Carefully go through this example, and then pick a different step from any reaction shown earlier in this course and try your own hand at drawing its transition state. Remember to bracket and place a around your finished structure to specify its fleeting, transition state status. Professor David Liu and Brian Tse, Life Sciences 1a page 11
Lectures 15-16: The Molecular Basis of Enzyme Catalysis: HIV Protease 1. The function and structure of HIV protease a. Introduction to proteases b. Discovery of HIV protease c. verview of the three-dimensional structure of HIV protease 2. Chemical reactions and the energies driving them a. Amide bond cleavage: the reaction catalyzed by HIV protease b. Thermodynamics of a reaction and free energy c. Kinetics of a reaction and ΔG d. Reaction energy diagrams e. Transition states, intermediates, and how to draw them 3. How enzymes accelerate chemical reactions: the case of HIV protease a. Catalysts alter a reaction s kinetics, but not its thermodynamics b. Chemical strategies behind enzyme catalysis i. Proximity and orientation effects ii. Nucleophilicity and electrophilicity iii. Acid and base catalysis 4. The molecular basis of substrate specificity a. Trypsin substrate specificity b. HIV protease substrate specificity Let s now apply what we ve learned about reaction thermodynamics and kinetics to the amide bond hydrolysis reaction catalyzed by HIV protease. While it is sometimes possible to predict if a reaction under standard state conditions is favorable (ΔG < 0) or unfavorable (ΔG > 0) with the assistance of several tables of empirical and theoretical data, it is usually not possible to predict ΔG of a reaction simply by inspection, especially when the reaction does not take place under standard state conditions. The cleavage of an amide bond in water under typical physiological conditions, however, has been empirically determined to be a favorable reaction. Despite the fact that amide hydrolysis is a favorable reaction, proteins are notoriously stable biological macromolecules, with very slow rates of spontaneous hydrolysis arising from high activation energy barriers along the amide hydrolysis reaction coordinate. In fact the half-life of a typical peptide bond in the absence of proteases is estimated to be ~1 to 100 years. This stability is crucial to life, as many of the structural and functional proteins in your body play roles that require long lifetimes. Yet on the other hand, many biological processes (including digestion of proteins and the processing of HIV s polyproteins into mature viral proteins) require the rapid hydrolysis of proteins. This apparent dilemma between the stability of proteins and the need for rapid protein hydrolysis is resolved by the ability of enzymes such as HIV protease to accelerate amide bond hydrolysis when needed. Professor David Liu and Brian Tse, Life Sciences 1a page 12
An Enzyme in Action Substrate Free enzyme Substrate binding Enzyme-substrate complex Product release Catalyzed reaction [TS] Free product(s) Enzyme-product complex 3. How enzymes accelerate chemical reactions: the case of HIV protease We are ready to analyze the specific chemical ways in which enzymes can accelerate reactions. At the molecular level, an enzyme-catalyzed reaction unfolds in the following way: the enzyme randomly encounters the substrate in solution. ccasionally such an encounter will take place in a manner that allows the enzyme to bind to the substrate. When bound to the enzyme, the substrate experiences a precisely tailored environment that facilitates the substrate s transformation into the transition state of the reaction. The enzyme-stabilized transition state can then undergo additional changes to become enzyme-bound product. The enzyme-product complex then dissociates into free product and free enzyme. The released enzyme is ready to catalyze the conversion of another molecule of substrate into product. Professor David Liu and Brian Tse, Life Sciences 1a page 13
Catalysts Lower ΔG Lower ΔG = Faster reaction (Fisher-Price version) Catalysts emerge unchanged and therefore never alter the ΔG for the reaction Enzymes catalyze reactions primarily by decreasing ΔG Catalysis affects a reaction s kinetics, not its thermodynamics Catalysts alter a reaction s kinetics, but not its thermodynamics Catalysts including enzymes accelerate chemical reactions by lowering the activation barrier of the reaction (i.e., by decreasing ΔG ) and/or by increasing the effective concentration of the substrates. Even though a catalyst may participate in a chemical reaction in a variety of ways, by the time the reaction is complete, a true catalyst by definition must be regenerated in exactly the same form that existed at the start of the reaction. Because a catalyst is not changed by a chemical reaction, it cannot alter the thermodynamics of the reaction. Therefore, a catalyst cannot make an unfavorable reaction favorable, or even make a favorable reaction more favorable the ΔG of a reaction is never changed by a catalyst. You can understand this very important point by appreciating that the free energy of the catalyst is identical in the substrate side of a reaction and in the product side because the catalyst emerges in the same form before and after a reaction; therefore the catalyst does not contribute to any change in free energy (ΔG) that occurs when the reaction takes place. Professor David Liu and Brian Tse, Life Sciences 1a page 14
ΔG (uncatalyzed) Free energy E + S E-S E + TS E-TS ΔG (enzyme catalyzed) Enzyme-catalyzed reaction (More complete version) Reaction coordinate Uncatalyzed reaction E-P Enzyme Catalysis (More Accurate) E + P S = substrate P = product TS = transition state E = enzyme The enzyme must stabilize the transition state more than the enzyme stabilizes the substrates in order to accelerate the reaction (so ΔG cat < ΔG uncat ) An enzyme most frequently lowers ΔG by stabilizing the transition state of a chemical reaction (in other words, by lowering G ) more than it stabilizes the substrates. Returning to our reaction energy diagram of an uncatalyzed and catalyzed chemical reaction, you can appreciate why an effective catalyst cannot stabilize both the substrate and the transition state equally if this were the case, the ΔG of the catalyzed reaction would be the same as the Δ G of the uncatalyzed reaction, and no rate acceleration of the reaction would result. An effective catalyst must preferentially stabilize the transition state. Professor David Liu and Brian Tse, Life Sciences 1a page 15
Breakout Question Chorismate mutase is an enzyme that catalyzes the conversion of chorismate to prephenate. Which of the following predicts the behavior of two identical solutions that start with 0.1 M chorismate, to which a tiny amount of chorismate mutase is added to only one solution? C 2 chorismate ΔG < 0 prephenate H H A) The [chorismate] in the solution with enzyme is always at least as high as the [chorismate] in the solution without enzyme. B) The solution with the enzyme will generate more free energy after any amount of time (short or long) than the solution without the enzyme. C) At equilibrium, [prephenate] is equal in the two solutions. D) The value of ΔG approaches zero as both reactions approach equilibrium Professor David Liu and Brian Tse, Life Sciences 1a page 16