Lecture 3: Thermodynamics

Transcription

1 3 LAWS OF THERMODYNAMICS Lecture 3: Thermodynamics Matter and energy are conserved Margaret A. Daugherty Fall 2004 Entropy always increases Absolute zero is unattainable System and Surroundings 1st Law of Thermodynamics Relationship of heat,work, internal energy & Enthalpy Bookkeeping function to keep track of: heat transfers work expenditures Total E of a system & its surroundings is a constant E = E final - E start = Q + W Q: heat absorbed by system; W: work done on system Energy cannot be created or destroyed.

2 Biological Systems and Enthalpy, H: Calorimetry & van t Hoff plots: Determination of enthalpy Definition: H = E + PV N <--> U Calorimetry For a biological system: Temperature Pressure Volume constant! K eq =[U]/[N] Measure K as a function of T H = -R dlnkeq d(1/t) Work is a function of V & P W = V P + P V; pressure and volume are constant W = O H = +533 kj/mol H = E = Q Enthalpy is equal to the heat absorbed in a biological process H < 0 exothermic H > 0 endothermic 2nd Law of Thermodynamics Systems of molecules have a natural tendency to randomization or disorder; The degree of randomness or disorder is measured by a function of state called the entropy (S); 1M NaCl 0.5M NaCl The entropy of an isolated system will tend to increase to a maximum value. For an isolated system, the favored direction: S = S final - S initial > 0 Diffusion of a solute Second Law: Entropy and Disorder S = entropy 1). Systems proceed from an ordered to a disordered state 2). Reversible processes - entropy of Sys + Sur is unchanged Irreversible process = entropy of Sys + Sur increases 3). All process tend to equilibrium - minimum potential energy Three cases S = positive S = zero S = negative Disorder! Reaction will tend to occur Reaction is reversible; at equilibrium Order! Reaction is unfavorable

3 Entropy: the equations S = k log W k = Boltzmann s constant 1.38 x J/K W = # of possible ways to arrange a system at a given temperature Third Law: Absolute zero is unattainable What is absolute zero? A temperature where entropy is zero! Absolute zero = 0 K Conversion to Celsius = degrees K C The entropy of any perfectly crystalline substance approaches zero as absolute zero is approached ds rev = dq/t Relates entropy to heat absorbed The heat capacity, Cp, allows us to have an absolute entropy scale T S = CpdlnT or Cp = dh/dt 0 If Cp < 0, molecules become more restricted; if Cp > 0, molecules aquire new ways to move Gibbs Free Energy, G: Is a reaction feasible? Three Ways To Have Thermodynamically Favored Reactions ( G = H - T S; G < 0) For a reaction A <--> B H negative S positive H very negative S negative H positive S very positive Keq = [B]/[A] G = -RTlnKeq Gibbs Free Energy Relates 1 rst and 2 nd Laws of Thermodynamics J. Willard Gibbs G = H - T S

4 State Functions Functions of State depend only on the initial and final states - not on the path taken. initial G = G init - G final final State Functions G H S Volume Temperature Pressure Not State Functions Work Heat What do thermodynamic quantities tell us? G H S Cp 1). Favorable v. favorable unfavorable 2). Unfavorable favorable unfavorable H-bonds less labile 3). Unfavorable unfavorable unfavorable H-bonds more labile Standard State Free Energy Standard state: 25C,1 atm, concentration of all solutes = 1 M A + B < --> C + D G = G o + RTln [C][D] constant for a specific reaction [C][D] Keq = At equilibrium G = 0 and [C][D] = Keq our criteria for spontaneity is G Standard State Free Energy: ph* G o = standard state free energy at ph 7.0 For reactions that evolve protons: A --> B + H + G o = G o + RTln[H + ] For reactions that absorb protons: A - + H + --> AH G o = G o - RTln[H + ] G o = - RTln Keq *Note that 1 M H + = ph 0, which is unphysiological

5 Physiologically, we don t operate at 1 M solution conditions A + B < --> C + D G = G o + RTln [C][D] Phosphocreatine + H > creatine + Pi G o = kj/mol at 37C Coupled processes: How we actually survive! Problem: Formation of ATP is energetically unfavorable ADP + P i ---> ATP G = + 55 kj/mol Solution: Couple this reaction to a favorable reaction PEP + H > pyruvate + P i G = -78 kj/mol G = RT ln (0.001 x 0.001) (0.001) What we end up with: G = kj/mol Muscle uses p-creatine to regenerate ATP from ADP G = -23 kj/mol Note: coupling occurs via enzymes; in this case, pyruvate kinase (see CH 18) Chemically, why is ATP so good? Reactants: electrostatic repulsion causes bond strain (4 neg charges) --- destabilizes molecule Products: stabilized by ionization and resonance + Pi Entropically favorable: 1 reactant --> 2 products 4 resonance hybrids REVIEW 1). Living organisms are thermodynamically open systems. 2). Living organisms operate under the laws of thermodynamics. 3). First law: H = Q: H < 0 for spontaneous reaction 4). 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 phosphate molecules, which positions it as an energy donor and energy acceptor. Thus various chemical reactions can be coupled in a controlled manner. 12). Various chemical factors contribute to the large G o for ATP hydrolysis.