1. THERMODYNAMICS - an area of physical chemistry the interconversion of different forms of energy the determination of the direction of spontaneous processes consists of 3 fundamental laws using mathematical functions to describe the systems bioenergetics - subsection within thermodynamics; providing a quantitative analysis of how organisms process energy Review of basic concepts: system- part of universe we are interested in, has defined boundaries; e.g. cell, petri dish, earth surroundings- all else outside the system Types of systems: 1) isolated system: unable to exchange mass or energy with surroundings 2) closed system: energy flow, no mass flow with surroundings 3) open system: can exchange both mass and energy Internal Energy - represented by E - any system contains a certain amount of energy - kinetic and potential energy (all non-nuclear forms of E) - at atomic and/or molecular level: - translational energy = mv 2 /2 = (3/2)RT - rotational - quantized E states > translational E - vibrational - quantized E states > rotational E - electronic - largest spacings between quantized states (E in chemical bonds) - energy in noncovalent interactions; inter & intra molecular as discussed in Ch 1

2. Internal Energy - is a function of the state of a system - the value of a “State Function” does not depend on the path of a process or on how a system arrived at that state, but depends only on the state of the system e.g. E final – E initial =  E is independent of path - the thermodynamic state of a system can be defined by the amount of all substances in the system and 2 out of 3 of the following variables: T, P, V (these are also state variables) There are two main ways a system can exchange energy with its surroundings: transfer of heat (q): (+) heat absorbed by the system; increase in E (–) heat flows from the system; decrease in E work (w) done by or on the system (+) work done by system on surroundings; decrease in E (–) surroundings do work on the system; increase in E work can take many forms- PV work- e.g. expansion of lungs against external P work = P  V electrical work - pumping ions across a membrane q and w are not state variables - can be thought of as energy in transit - we speak of the energy of a system, but not of the heat or work of a system

3. 1st Law of Thermodynamics - describes the “bookkeeping” involved when a system exchanges energy  E = q – w - conservation of energy - E can be gained and spent by a system but not destroyed -  E depends only on the initial and final states of a system - in contrast, the amount of q and w exchanged depends on the path taken to get from initial to final state, depends on conditions e.g. of  E in presence and absence of catalyst

4. see e.g. from book p.59 - palmitic acid oxidation under 2 different sets of conditions: 1) reax carried out in bomb calorimeter immersed in water bath at constant volume - no PV work done -ignite mixture -measure heat transferred from system to water bath as increase in T - we know mass of water and heat capacity of water  E = q v = –9941.4 kJ/mol (unit:Joules;1 cal = 4.184 J) - we are measuring energy released (–) that was stored in the chemical bonds of the system 2) oxidation of palmitic acid under conditions of constant P - run at P = 1 atm - from balanced equation can see decrease in number of moles of gas: (  n = 16 final – 23 initial ) - measure decrease in V proportional to a decrease in the number of moles of gas present - decrease in V = work done on the system (-) w = P  V - assume initial and final temperatures are 25°C (298K) and the gases are ideal so we can use PV = nRT  V =  n (RT/P) R = 8.314 J/Kmol and w = – 17.3 kJ/Kmol

5. now consider the q evolved for constant P conditions: -q is not a state function and depends on conditions - from constant V conditions:  E = –9941.4 kJ/mol -  E is a state function; the sane for both sets of conditions - use 1st Law;  E = q – w = –9941.4 q =  E + w = –9941.4 + (–17.3) = –9958.7 kJ/mol - more heat is evolved under 2nd set of conditions (constant P); (in the 2nd case, the PV work done on the ends up as extra heat released by the system) Essentially all biochemical processes take place under constant P conditions, (P = 1 atm ) Thus, another state function has been defined to describe the heat evolved under constant P conditions: Enthalpy - represented by H; function of state H = E + PV = q p  H =  E + P  V - the value of  H is the same as the heat of a reaction measured at constant pressure - the differences between  H &  E in biological systems are very small (because most reax are carried out in solution and amount of PV work done is small relative to  H &  E) as a result,  H is often referred to as the energy change

6. 2nd Law of Thermodynamics - - In addition to E and H, there are other factors which appear to be fundamental to biological processes - consider the criterion for a favorable process. It was thought that reactions with  H < 0 used to all be spontaneous - there are many counterexamples. - many processes without energy changes always go in a specific direction e.g. 2 beakers with water, one at 75°, one at 25°, when brought into thermal contact, both end up at 50°. The reverse never happens. -- Nothing in the 1st Law prevents the reverse from happening. Two tendencies are at work in the universe: 1) tendency towards lowest energy state 2) tendency towards randomness or disorder 2nd tendency is described by the state function Entropy (S) - we also must consider whether or not a process is reversible - the reversibility of a process depends on the given conditions - reversible processes occur when conditions are close to the equilibrium state for the system - irreversible processes occur when conditions are far from equilibrium, then the process drives the system towards equilibrium

7. Different terms are used - spontaneous vs favored direction of a reaction - spontaneous implies “rapid” rate, but thermodynamics does not tell us anything about the rate of a reaction, only which direction is favored Entropy (S) - is a measure of the degree of randomness of a system - consider a system in a given thermodynamic state - this state may have many substates of equal energy e.g. molecules making up a system with a particular amount of energy can be arranged in different ways (substates) S = k ln W wherek = Boltzmann’s constant k = R/ Avogadro’s number W = number of substates so, the molar entropy for a system is: S (molar) = R ln W - as number of substates (W) increases, so does S in addition: S ordered state < S disordered state low Shigh S ice at 0°C water at 0°C protein in native, folded stateprotein unfolded (due to the number of substates that have equal energy)

8. There are several ways to state the 2nd Law: 1) Isolated systems tend to a state of maximum entropy 2) The efficiency of a real process (fraction of heat that can be converted to work) < 1. There are always frictional losses. 3) For an isolated system, an increase in entropy is the driving force towards equilibrium. 3rd Law of Thermodynamics creates a scale for entropies. Whereas the scale for energy is arbitrary, the scale for entropy is not. S = 0 only for a perfect crystal at 0 K (-273°C) Biochemists never deal with isolated systems. In open or closed systems, there are 2 opposing tendencies. Systems tend to - low energy and - high entropy, and the physical state that dominates depends on T. To encompass both factors, Gibb’s free energy (G) is defined G = H – T S G is a state function T is in K At constant T,  G =  H – T  S Now we have a criterion for favorable (spontaneous) processes that encompasses both governing factors: (  G) T,P < 0 exergonic

9. Overall, consider: favorable process; (  G) T,P < 0; exergonic at equilibrium;  G = 0; reversible process reverse of reax is favorable;  G T,P > 0; endergonic There is a balance between S and H. Favorable processes typically occur when there is a decrease in energy (–  H ) and /or an increase in entropy (+  S ). At low T: T  S term is small; solid is the stable state. At high T: T  S dominates, the gaseous state is stable. (your book has some good examples of the interplay of H and S) 2 points that can be confusing: 1) whether or not a process is favorable has nothing to do with the rate; a process may have a large -  G and be very slow 2)  S of an open system can decrease this often happens is biological systems (food -- becomes complex, ordered proteins which is accompanied by a large decrease in  S) however, energy is expended in the process to pay for this increase in organization; living organisms spend energy to overcome the decrease in entropy

10.  G called “free” energy because it represents the portion of  H that is available to do useful work (beyond PV work) - the amount of E in T  S is not available in other words, we measure the efficiency of a process as: work accomplished / maximum work that can be done by  G -the sign of  G tells us whether a process or its reverse is favorable, and - the magnitude tells us how far a process is from equilibrium -we want to express these ideas quantitatively for solution thermodynamics; to do so, we need to define a relationship between concentration and G: Chemical Potential: partial molar free energy -represents the contribution per mole of a particular component to the total free energy of a system -so for a mixture containing a moles of component A and b moles of component B, etc. G = a G A + b G B + c G C where G A or  a represents the chemical potential per mole of A -for dilute solution, chemical potentials depend only on the concentration of the substance but at higher concentrations we need to consider the activity of the substance

11. activity- represented by a; a dimensionless quantity that measures the effective concentration of a substance component A’s contribution to the free E of the system can be described by: G A = G° A + RT ln a A T (K) R = 8.314 J / K mol for dilute solutions: a A ~ [A] [ ] represents molar concentration moles / liter and: G A = G° A + RT ln [A] when the concentration of substance A is 1 molar then: G A = G° A so you can see that: G° A represents a reference or standard state defined as: - 1M concentration for each solute in solution and as - pure solvent for the solvent - chemical potential is always expressed with respect to a standard state -chemical potential is the driving force to equilibrium

12. - chemical reactions have certain enthalpy and free energy changes associated with them - in combining chemical reactions and their enthalpies or free energies, we treat the products and reactants as algebraic quantities, for example aA + bB  cC + dD  H =  H  products) -  H (reactants)  H = cH c + dH d – aH a – bH b H = enthalpy / mole - we are concerned with H differences, the scale is arbitrary and we need to define one point - by convention a value has been assigned to enthalpies for elements in their most stable forms at 1 atm pressure - this condition is referred to as the standard state and  H° f = 0 for elements in the standard state (also referred to as heat of formation) - then the enthalpies of compounds are related to this arbitrarily set zero value - the standard enthalpy / mole of a compound equals the enthalpy of formation of 1 mole of the compound at 1 atm from its elements in their standard states - many standard enthalpies / mole of a compound H° (compound) have been tabulated at 25 C - the same convention exists for standard free energies of elements and compounds in their standard states

13. - now apply this algebraic treatment to chemical potentials to quantitatively describe the free energy changes accompanying chemical reactions - this enables us to predict the favored direction of a reaction under certain conditions aA + bB  cC + dD  G =  G  products) -  G (reactants) write the free energies in terms of chemical potentials:  G = cG c + dG d – aG a – bG b the driving force for the reaction is the total free energy of the products minus that of the reactants - in order to calculate  G, we substitute using: G A = G° A + RT ln[A] (and similarly for each component) - we group terms and rearrange to yield:  G =  G° + RT (ln[C] c + ln [D] d - ln [A] a - ln [B] b )  G° represents the standard state free energy change in the reaction which would be observed if a moles of A and b moles of B, at 1 molar concentration, formed c moles of C and d moles of D, each at 1 M.

14. We can combine these terms one step further:  G =  G° + RT ln ( [C] c [D] d / [A] a [B] b ) in general, this can be written as:  G =  G° + RT ln ( [products] / [reactants] ) remember, this is with each concentration raised to the power equal to its stoichiometry in the reaction - so when all concentrations are equal to 1 M:  G =  G° - suppose the reaction has come to equilibrium, then 2 things must be true: 1) the concentration terms must be at equilibrium concentrations, where the equilibrium constant (K), K = ([products] / [reactants]) (also raised to stoichiometric powers) 2) at equilibrium,  G = 0 so, now  G = 0 =  G° + RT ln K or:-  G° = RT ln K

15. At equilibrium: –  G° = RT ln K can also be stated as: K = e –  G° / RT -this last equation allows us to use tabulated standard state free energy changes to predict equilibrium constants  G° can be used as a reference value to compare intrinsic  G° values for different reactions under equivalent conditions BUT- it is  G (in vivo conditions) that tells us if a reaction goes forward not  G° (under standard state conditions) Consider the following important biochemical reaction: the isomerization of glucose-6-phosphate into fructose-6- phosphate (second step in glycolytic pathway) G6P  F6P  G° = + 1.7 kJ/mol - endergonic under standard condition - reverse reaction is favored at equilibrium:[G6P] > [F6P] - let’s express this quantitatively

16. G6P  F6P  G° = + 1.7 kJ/mol K = e –  G° / RT at 25°C K = e – (1700 J/mol) / (8.314 J/Kmol)(298K) K = 0.504 = ([F6P] / [G6P]) equilibrium conc you can see K < 1, equilibrium lies to the left let’s represent concentrations as fractions of total material K = [F6P] / total concentration (f F6P ) eq ---------------------------------- = -------------- [G6P] / total concentration (f G6P ) eq Now, replace (f G6P ) = 1– (f F6P ) and express in terms of one component: K = (f F6P ) / 1 – (f F6P ) = 0.504 solve for (f F6P ): (f F6P ) eq = 0.335 at equilibrium: 33.5% sugar exists as F6P 66.5% as G6P we are now able to calculate  G at any concentration using:  G =  G° + RT ln { (f F6P ) / 1 – (f F6P )}

17. Standard States: in biochemistry the standard state usually includes one additional condition: pH = 7; [H + ] = 10 -7 (which is much less than [H + ] = 1M) this is denoted by  G°’ as opposed to just  G° chemists’ standard state =  G° biochemists standard state =  G°’ (the additional condition alters the value of K ) what is important is  G which is independent of the standard state, this is what tells us if a reaction is favored the difference in standard state becomes important in reactions where [H + ] is produced or consumed by the reaction consider the following where [H + ] is produced (this is the case for hydrolysis reactions which are common in biochemistry): A + B  C + xH + chemists’ standard state: K = [C] [H + ] x / [A] [B]  G° for all [ ] = 1M biochemist has  G° ‘ for [A], [B], [C] = 1 M and [H + ] = 10 -7 M to compensate for the difference in [H + ], K’ (biochemist) > K (chemist) can show that K / K’ = 10 -7x ;  G° –  G°’ = 39.9 kJ/mole of H + ----low [H + ] (pH~7) contributes enormously to the favorability of a reaction when a proton is being produced

18. Coupling of reactions and spontaneity: -when looked at alone, many biochemical reactions do not have favorable  G’ values (transport against a concentration gradient) - these intrinsically unfavorable reactions are made thermodynamically favorable by coupling them to strongly favorable reactions -as an example, one step on the glycolytic pathway; the conversion of glucose to glucose-6-phosphate (G6P) is unfavorable: glucose  G6P  G°’ = + 3.2 kcal/mole - this reaction won’t go under standard conditions - the reaction is driven by coupling it with ATP hydrolysis ATP + H 2 O  ADP + H + + P i  G°’= –7.3 Overall, glucose + ATP + H 2 O  G6P + ADP + H + + P i  G°’ = + 3.2 – 7.3 = – 4.1 kcal (This reaction without the enzyme, hexokinase, is slow, in spite of –  G°’. This enzyme aids in providing the means for the coupling to take place.) Coupling is very common in biochemical processes. Cells contain a number of compounds that undergo reactions with large –  G’s and often enzymes facilitate the coupling of these reactions with unfavorable reactions. Enzymes also accelerate reactions.

19. Factors that contribute to large free energy changes for high energy phosphate compounds (see fig 3.7 in text): 1) resonance stabilization or tautomerization of product molecules: eg. phosphate products P i HPO 4 2– orthophosphate, inorganic phosphate - resonance forms of equal energy (substates) contribute to the high entropy of the product, P i - so release of P i results in an increase in entropy 2) additional hydration of charged hydrolysis products -recall hydration of ions in aqueous solution is favored 3) electrostatic repulsion between charged products - reaction is favored when repulsion between 2 anionic products takes place 4) release of a proton in buffered solution -we just saw that when [H + ] is produced under conditions where [H + ] is kept low (pH= 7);  G°’ <  G° by 39.9 kJ/mole - your text shows you how to arrive at this number from  G°’ =  G° + RT ln (10 -7 )