Lecture 3:ThermodynamicsMargaret A. DaughertyFall 2003BIOC 205The three laws of thermodynamics Matter and energy are conservedBIOC 205Entropy always increases Absolute zero is unattainableSystem and SurroundingsBIOC 2051st Law of ThermodynamicsRelationship of heat,work, internal energy & EnthalpyTotal E of a system & its surroundings is a constant∆E = Efinal - Estart = Q + WQ: heat absorbed by system; W: work done on systemBIOC 205Biological Systems and Enthalpy, ∆H:For a biological system:TemperaturePressureVolumeconstant!Work is a function of V & P W = V∆P + P∆V; pressure and volume are constant W = ODefinition: H = E + PV∆H = ∆E = QEnthalpy is equal to the heatabsorbed in a biological processBIOC 205Calorimetry & van’t Hoff plots: Determination of enthalpyA <--> B Keq = [B]/[A]∆H = -RdlnKeqd(1/T)CalorimetryBIOC 205∆H = +533 kJ/mol2nd Law of Thermodynamics• The degree of randomness or disorder of a system ismeasured by a function of state called the entropy (S);1M NaCl0.5M NaClDiffusion ofa soluteFor an isolated system, the favored direction: ∆S = Sfinal - Sinitial > 0• The entropy of an isolated system will tend toincrease to a maximum value.BIOC 205Second Law:Entropy and Disorder1). Systems proceed from an ordered to adisordered state2). Reversible processes - entropy of Sys + Suris unchanged Irreversible process = entropy of Sys +Sur increases3). All process tend to equilibrium - minimumpotential energy∆S = entropyBIOC 205Entropy: the equationsS = k log Wk = Boltzmann’s constant 1.38 x 10-23 J/KW = # of possible ways to arrange asystem at a given temperaturedSrev = dq/T1844-1906Relates entropy to heat absorbedBIOC 205Third Law:Absolute zero is unattainableThe entropy of any perfectly crystalline substanceapproaches zero as absolute zero is approachedWhat is absolute zero? A temperature where entropy is zero!Absolute zero = 0 KConversion to Celsius = degrees K - 273.15CIf ∆Cp < 0, molecules become more restricted;if ∆Cp > 0, molecules aquire new ways to moveBIOC 205The heat capacity, Cp, allows us to have an absolute entropyscaleS = CpdlnTor Cp = dH/dT 0TGibbs Free Energy, ∆G: Is a reaction feasible?J. Willard Gibbs1839-1903Gibbs Free EnergyRelates 1rst and 2nd Laws of Thermodynamics∆G = ∆H - T∆SFor a reaction A <--> BKeq = [B]/[A]∆G = -RTlnKeqBIOC 205Three Ways To Have Thermodynamically Favored Reactions(∆G = ∆H - T∆S; ∆G < 0)∆H negative∆S positive∆H very negative∆S negative∆H positive∆S very positiveBIOC 205State FunctionsFunctions of State depend only on the initial and final states - not onthe path taken.initialfinalState Functions∆G∆H∆SVolume TemperaturePressureNot State FunctionsWorkHeat∆G = Ginit - GfinalStandard State Free Energy, ∆GoA + B < --> C + D∆G = ∆Go + RTln [C][D][A][B]At equilibrium ∆G = 0 and [C][D][A][B]= Keq∆Go = - RTln Keq Standard state: 25C,1 atm, concentration of all solutes = 1 M[C][D][A][B]Keq =our criteria for spontaneity is ∆G∆Go’ = standard state free energy at pH 7.0Note that 1 M H+ = pH 0, which is unphysiologicalBIOC 205Physiologically, we don’t operate at 1 M solution conditionsA + B < --> C + D∆G = ∆Go + RTln [C][D][A][B]Phosphocreatine + H20 --> creatine + Pi∆Go = -42.8 kJ/mol at 37CMuscle uses p-creatine to regenerate ATP from ADP∆G = -42.8 + RT ln (0.001 x 0.001)(0.001)∆G = -60.5 kJ/molBIOC 205Coupled processes: How we actually survive!Problem: Formation of ATP is energetically unfavorableADP + Pi ---> ATP ∆G = + 55 kJ/molSolution: Couple this reaction to a favorable reactionPEP + H20 ----> pyruvate + Pi ∆G = -78 kJ/molWhat we end up with:∆G = -23 kJ/molNote: coupling occurs via enzymes; in thiscase, pyruvate kinase (see CH 15)BIOC 205What do thermodynamic quantities tell us?1231). Favorable2). Unfavorable3). Unfavorable∆G ∆H ∆S ∆Cp1). Favorable v. favorable2). Unfavorable favorable3). Unfavorable unfavorable1). Favorable v. favorable unfavorable2). Unfavorable favorable unfavorable3). Unfavorable unfavorable unfavorable1). Favorable v. favorable unfavorable2). Unfavorable favorable unfavorable H-bonds less labile3). Unfavorable unfavorable unfavorable H-bonds more labileBIOC 205Chemically, why is ATP so good?+ PiReactants: electrostatic repulsion causes bond strain(4 neg charges) --- destabilizes moleculeProducts: stabilized by ionization and resonanceEntropically favorable: 1 reactant --> 2 productsBIOC 205REVIEW1). Living organisms are thermodynamically open systems.2). Living organisms operate under the laws of thermodynamics.3). First law: ∆H = Q: ∆H < 0 for spontaneous reaction4). Second law: Systems tend to maximize entropy, ∆S > 0 for spont. rxn.5). Third law: ∆Cp provides information on molecular order in a reaction.6). ∆G provide information on spontaneity; relates ∆H & ∆S.7). Thermodynamic quantities provide chemical information on reactions.8). Standard state free energy allows us to compare biochemical reactions.9). Metabolically, we can couple unfavorable reactions to favorable reactions.10). High energy phosphate molecules drive metabolic reactions.11). ATP has an intermediate energy among the high energy phosphatemolecules, which positions it as an energy donor and energy acceptor. Thusvarious chemical reactions can be coupled in a controlled manner.12). Various chemical factors contriubte to the large ∆Go’ for ATPhydrolysis.BIOC