Lecture 2: Biological Thermodynamics [PDF] Reading: Berg, Tymoczko & Stryer: pp. 11-14; pp. 208-210 problems in textbook: chapter 1, pp. 23-24, #4; and thermodynamics practice problems [PDF] Updated on: 1/11/07 at 3:30 pm 1/22/07: error in stated equilibrium ATP concentration for first sample calculation (on p. 6 of PDF) has been corrected -- it's 1 x 10-7 M, NOT 1.65 x 10-7 M. Key Concepts Transformation of energy and matter from surroundings --> complex, orderly structures Bioenergetics: the quantitative study of energy transformations in biological systems -- these concepts are the underpinnings for all of biochemistry. Change in Gibbs free energy (ΔG) for a reaction quantitates the energy available to do useful work is related to the change in enthalpy and the change in entropy: ΔG = ΔH - TΔS Actual free energy change (ΔG) depends on 2 parameters: standard free energy change for that reaction (ΔG ) (and thus to K eq, defining where equilibrium for this reaction lies), and actual mass action ratio, reflecting actual starting conditions, the actual concentrations of reactants and products ΔG (actual) = ΔG + RTln{actual mass action ratio} Standard free energy change for a reaction (ΔG ): the change in free energy for going from STANDARD CONDITIONS TO EQUILIBRIUM. related to equilibrium constant K eq by the equation ΔG = - RTlnK eq SIGN of ΔG tells us in which DIRECTION the reaction would have to go to reach equilibrium (the "spontaneous" direction), but ΔG gives NO information about RATE at which reaction will go. Reactions ALWAYS proceed in direction to go toward equilibrium. Negative value for ΔG means equilibrium lies to right (reaction will go as written, left to right). Positive value for ΔG means equilibrium lies to left (reaction will go right to left). Page 1 of 9
Free energy changes are ADDITIVE. For coupled or sequential reactions, overall free energy change for process is the SUM of ΔGs for component reactions or processes. FREE ENERGY COUPLING: an exergonic reaction can "drive" an endergonic reaction (sum of component ΔGs is negative) if the reactions can be coupled. Enzymes (which are proteins) are catalysts -- they increase rates of biological reactions to permit life on a biological timescale. Catalyst doesn't change K eq (position of equilibrium isn't altered) or ΔG. Catalyst increases rate at which reaction goes toward equilibrium. Objectives Briefly explain in conceptual terms what is represented in a change in free energy (ΔG), a change in enthalpy (ΔH), and a change in entropy (ΔS). Explain how ΔH and ΔS together determine ΔG. Explain the difference between the standard free energy change (ΔG ) for a reaction or process, and the actual free energy change (ΔG) for that process. What are the 3 simplifying assumptions made in biochemistry that are consistent with physiological conditions, and make "biochemical standard conditions" different from standard conditions normally referred to in chemistry? (These assumptions make the difference between ΔG and ΔG '.) Calculate any one of the following parameters from the other two: ΔG', ΔG ' and the actual mass action ratio. Calculate ΔG ' from K eq ' or vice versa, given the absolute temperature and the value of R, the gas constant. Explain the relationship of ΔG' to the direction in which a reaction will go spontaneously. General chemical reaction All reactions/processes proceed in direction TOWARD EQUILIBRIUM. For this reaction, the mass action ratio is given by mass action (m.a.) ratio at equilibrium = the equilibrium constant for the reaction = K eq FREE ENERGY Page 2 of 9
Bioenergetics: the quantitative study of energy transformations in biological systems (part of thermodynamics) essential for understanding how metabolic processes provide energy for the cell the structures of macromolecules how membrane transport processes occur all the fundamental processes that define biochemistry! bioenergetics useful for describing conditions under which processes occur spontaneously All reactions/processes proceed spontaneously in whatever direction is required to achieve, or at least go toward, equilibrium; "spontaneous" direction is always toward equilibrium. Bioenergetics determines whether a process will occur spontaneously (i.e., in what direction process will go), but bioenergetics does NOT determine how fast the process will occur. Gibbs Free Energy, G (the thermodynamic function that is most useful for biochemistry) G a function of Enthalpy, H, a measure of the energy (heat content) of the system at constant pressure, and Entropy, S, a measure of the randomness (disorder) of the system G = H TS G, H and S are state functions hard to measure as absolute values for a given state of a system. easier to measure CHANGES in the state functions for a change in state of the system: ΔG, ΔH, and ΔS for a reaction or a process Laws of Thermodynamics First Law of Thermodynamics: The total energy of a system and its surroundings is constant; energy can neither be created nor destroyed. Second Law of Thermodynamics: The total entropy of a system and its surroundings always INCREASES for a spontaneous process. For any process, If ΔH is negative, then heat is released (a favorable enthalpy change) making bonds: ΔH < 0, favorable (exothermic reaction) breaking bonds: ΔH > 0, unfavorable (endothermic reaction) If ΔS is positive, then the randomness of the system increases (a favorable entropy change). increased disorder: ΔS > 0, favorable increased order: ΔS < 0, unfavorable change in free energy for any process: ΔG = ΔH - TΔS. T = absolute temperature in units of K (T = o C + 273) If ΔG is negative (ΔG < 0) (exergonic reaction): process goes in direction written (left to right) If ΔG = 0: process is at equilibrium (no net reaction in either direction) If ΔG is positive (ΔG > 0) (reaction in direction written would be endergonic; process goes in reverse (right to left)) Value and sign of ΔG depend on interplay of enthalpy and entropy (Neg. ΔH doesn't necessarily neg. ΔG, and positive ΔS doesn't necessarily neg. ΔG.) Page 3 of 9
Example: Consider melting of ice and values of ΔG, ΔH, ΔS at various temperatures. ΔH positive (unfavorable) because hydrogen bonds are breaking ΔS positive (favorable) because H 2 O molecules more disordered in water than in ice Temperature ΔH TΔS +10 o C 0 o C -10 o C +6.4 kj/mol +6.0 kj/mol +5.6 kj/mol ΔG = ΔH - TΔS What Happens? +6.6 kj/mol - 0.2 kj/mol Ice melts +6.0 kj/mol 0 kj/mol Ice and water coexist +5.4 kj/mol + 0.2 kj/mol Water freezes Free energy diagrams Page 4 of 9
FREE ENERGY CHANGES AND CHEMICAL REACTIONS This important general equation gives the free energy change for any reaction to go to equilibrium from ANY starting conditions. Ratio [C] c [D] d / [A] a [B] b is the actual mass action ratio (m.a. ratio). NOTE: Thermodynamic calculations use natural logarithms, logarithms to the base e, symbolized ln; ph problems use logarithms to the base 10 (log), because the ph scale reflects factors of 10 in H + concentration. You can convert natural logs to logs to the base 10 if you wish: lnx = 2.303 log 10 x when doing thermodynamic calculations but it's very important that you be using the right kind of logarithm. Consider 2 different SPECIAL CASES of starting conditions, 1) when the starting mass action ratio = 1 ("standard conditions"), and 2) when the starting mass action ratio = K eq ΔG o : standard free energy change, the free energy change for the reaction when going from standard state of all components ("standard conditions") to equilibrium. Standard conditions ("standard state"): 1 M in each reactant and product (or 1 atm for gaseous reactants or products), with temperature = 25 C = 298 K. If concentrations of all products and all reactants are 1 M to start with, what is the numerical value of the last term in that equation (RT ln[m.a. ratio])? Thus, when starting from standard conditions (an actual m.a. ratio of 1), and reacting to get to the equilibrium m.a. ratio (K eq ), the free energy change (ΔG) = ΔG o. ΔG o is a constant for a particular reaction, a "reference" free energy change that defines where equilibrium lies for that reaction. Second special case: What if the starting conditions are at equilibrium? Actual mass action ratio in general ΔG equation = K eq What is ΔG if the reaction is at equilibrium? thus, ΔG o = - RTlnK eq K eq and ΔG o are different ways to express the same information, and are interconvertible. lnk eq = ΔG o /RT, so Page 5 of 9
K eq = e ΔGo/RT EQUILIBRIUM CONSTANTS AND FREE ENERGY CHANGES: One of the important reactions in biochemistry is the hydrolysis of ATP: (Note that the chemical equation, with compounds abbreviated, has been written with charge balance.) How do we calculate free energy changes for this reaction? The free energy change for this reaction is given by In biochemistry, we make two simplifying assumptions (justifiable under physiological conditions): The concentration of water [H 2 O] does not change during the reaction, i.e. [H 2 O] = 55.5 M. The ph = 7.0 and does not change during the reaction, i.e. [H + ] = 10 7 M. Under these conditions the (constants) [H 2 O] and [H + ] are incorporated into ΔG o to give a new "biochemical" standard free energy change, ΔG o '. The "biochemical" equilibrium constant, related to ΔG o ', is designated K eq '. The concentrations of water and of protons are left completely out of all calculations. NOTE: "Biochemical" free energy changes are always supposed to be designated with "prime" symbols. However, in this course we'll ALWAYS be dealing with biochemical free energy changes, so if the primes are left out sometimes, assume that they're supposed to be there! Using these assumptions, and for simplicity, using ADP instead of ADP 3- ATP, instead of ATP 4- P i, instead of HPO 4 2-, the free energy equation becomes: Sample Calculations Now we can do some calculations (complete answers [PDF]). 1. If the equilibrium concentrations of ATP = 1.0 x10-7 M, ADP = 0.165 M and P i = 0.1 M, what are the Page 6 of 9
equilibrium constant and ΔG o ' for the hydrolysis of ATP at 37 o C? a) Use the data to calculate K eq. b) Use the value of K eq to calculate ΔG o '. Use the data to calculate K eq. K eq = 1.65 x 10 5 M (note units!) Use the value of K eq to calculate ΔG o '. ΔG o' = - RTln K eq ΔG o' = - 8.314x10-3 kj/k mol x 310K x ln (1.65 x 10 5 M) ΔG o' = - 31 kj/mol Note: UNITS of the mass action ratio or K eq in this case "disappear" when you take the ln, but they have to be either unitless (if conc. units cancel for the reaction under study), or if units don't cancel, they have to be in M, M 1, etc. -- NOT in mm or µm, etc. 2. In a typical cell at 37 o C the concentration of ATP = 8 x 10-3 M, ADP = 1 x 10-3 M, and P i = 8 x 10-3 M. What is the actual free energy change (ΔG') for ATP hydrolysis under these conditions? ΔG' = - 31 kj/mol + RT ln {[1x10-3 M][8x10-3 M]/[8x10-3 M]} = - 49 kj/mol COUPLED REACTIONS: Many biological processes are endergonic -- they require the input of free energy to make them go in an otherwise unfavorable direction. endergonic reaction: unfavorable (positive) free energy change for going in direction of equilibrium How can a reaction that "wants" to go BACKWARD be "driven" forward? by coupling it to an exergonic reaction (one with a negative, favorable, free energy change) Nelson & Cox, Lehninger Principles of Biochemistry, 4th ed. (2004), Fig. 1-26a: Mechanical example Downward motion of an object releases potential energy (pink side, exergonic) that can be used to do mechanical work, moving another object upward (blue side, endergonic). coupling mechanism (rope in this example) required to enable exergonic process to drive endergonic one Page 7 of 9
Lehninger Principles, 4th ed., Fig. 1-26b: Chemical example phosphorylation of glucose to produce glucose-6-phosphate very important reaction in the cell first reaction in metabolism of glucose that enters a cell from the blood Reaction 1: condensation of glucose (alcohol) with inorganic phosphate ion (acid) to make glucose-6-phosphate (an ester) Glucose + P i <=> glucose-6-phosphate + H 2 O (ΔG o ' = + 13.8 kj/mol, endergonic) Reaction 2: hydrolysis of ATP, a phosphoanhydride, to generate ADP and inorganic phosphate ATP + H 2 O <=> ADP + P i (ΔG o ' = - 30.5 kj/mol, exergonic) To couple the 2 reactions (which requires some chemical mechanism, of course), add reactants on left, add products on right, and add ΔG o' values to get ΔG o' for coupled reaction: Glucose + ATP <=> glucose-6-phosphate + ADP (ΔG o ' = - 16.7 kj/mol) Page 8 of 9
The coupled reaction is exergonic; it will go spontaneously (forward, left to right) in the cell (i.e., equilibrium lies to the right), but will it proceed at a rate consistent with cellular needs? There's NO information about rates in the value of a ΔG -- we can't answer this question from bioenergetics. Most biological reactions would proceed at a very slow rate indeed if they were not catalyzed. The biological catalyst enabling the coupled reaction above to proceed on a biological timescale (as opposed to a geological timescale!) is an enzyme, hexokinase. Free energy coupling, with enzymes as catalysts, is the strategy used in metabolic pathways. zieglerm@u.arizona.edu Department of Biochemistry & Molecular Biophysics The University of Arizona Copyright ( ) 2007 All rights reserved. Page 9 of 9