Chemical thermodynamics II. Medical Chemistry László Csanády Department of Medical Biochemistry

Entropy change Consequence: Some processes happen spontaneously, others do not. Let us define a quantity which reflects randomness: Entropy (S):  S should be a state function (  S = S final - S initial )  S should be an extensive property (S=  k S k ) Experience:  heat does not flow from a cold towards a hot object  energy tends to become dissipated into less ordered forms ("disorder" increases)

Entropy change  "more randomness" is generated by heat absorption at lower temperature:  S~1/T Let us define entropy through its change: Entropy change:  S=q/T (J/˚K) How should we define S? Experience:  heat absorption increases randomness in a system:  S~q

system surroundings: T=const. heat (q>0) Consider a process in which heat flows between a system and its surroundings. What can we say about  S sys and  S sur under various conditions? Entropy change system surroundings: T=const. heat (q<0) endothermic processexothermic process

time t start t end T temperature T system T surroundings  S for reversible processes 1. Reversible processes: Slow processes during which the system and its surroundings remain at quasi-equilibrium at all times. Thus, thermal equilibrium is maintained (no temperature inhomogeneities arise). system surroundings Entropy is only redistributed!

TkTk  S sys;k =q k /T k system surroundings time t start t end T temperature T system T surroundings 2.1. Spontaneous processes: endothermic (q>0)  S for spontaneous processes 2. Spontaneous processes (irreversible): During spontaneous processes system and surroundings fall out of equilibrium temporarily. Temperature inhomogeneities arise. Entropy is created!

TkTk  S sys;k =q k /T k system surroundings time t start t end T temperature T system T surroundings  S for spontaneous processes 2. Spontaneous processes (irreversible): During spontaneous processes system and surroundings fall out of equilibrium temporarily. Temperature inhomogeneities arise Spontaneous processes: exothermic (q<0) Entropy is created!

The second law of thermodynamics For a spontaneous process the total entropy of the system plus its surroundings always increases:  S total > 0 (spontaneous process) Spontaneous processes are irreversible. For reversible processes the total entropy does not change:  S total = 0 (reversible process) Or:  S system > q/T Or:  S system = q/T

For evaporation: The third law and absolute entropy For a perfectly crystalline pure substance at 0 ˚K S=0 (perfect order). Calculate absolute entropies (S): take 1 mole substance, add heat in small reversible steps to change its temperature from T to T+dT: TT+dT Add  S for phase transitions: For fusion:

The standard (absolute) entropy (S˚) of a substance is the entropy value of 1 mol of the substance in its standard state. (Liquids, solids: p=1 atm. Gases: partial p =1 atm. Solutions: c=1M. Usually T=25 o C.) Standard entropies Substance S˚(J/(mol˚K)) C(graphite) 5.7 H 2 (gas) CH 4 (gas) Standard entropy change for a reaction (  S˚): the entropy change for a reaction in which reactants in their standard states yield products in their standard states.  S˚ = S˚ final – S˚ initial (S is a state function)  S˚ =  S˚(products) -  S˚(reactants)

Standard entropies General rules of thumb:  S is positive for reactions in which  a molecule is broken into smaller molecules  there is an increase in moles of gas  solid changes to liquid or gas  liquid changes to gas Note: "  S˚ f " is zero for C(s) and H 2 (g). Example: calculate "standard entropy of formation" for methane: C(s)+2 H 2 (g)  CH 4 (g) S˚(J/(mol˚K)): 5.7 2· "  S˚ f " (CH 4 (g)) = [186.1-(5.7+2·130.6)] J/(mol˚K) = = J/(mol˚K)

Gibbs free energy Focusing on the system, the criterion for spontaneity (second law) at constant pressure can be written as  S > q P /T =  H/T (both  S and  H refer to the system). Or: 0 >  H - T  S (spontaneous process) Let us define a new quantity: Gibbs free energy (G): G =  H - TS At constant p and T the change in free energy of the system determines the spontaneity of the process:  G =  H - T  S  G < 0 (spontaneous process)Therefore: and 0 =  H - T  S (reversible process)  G = 0 (reversible process)and

G =  H - TS = (U + pV) - TS Gibbs free energy Interpretation of  G: the sign of  G determines spontaneity, but what is the physical meaning of  G? (Compare:  H is heat,  S is change in disorder...)  G is the maximum non-volumetric (osmotic, electric) work that can be gained from a process at constant p and T. This requires quasi-equilibrium conditions.  G =  U - T  S = (q + w) - T  S If p, V, and T are constant: Quasi-equilibrium: q=T  S   G = w  w system = -w = |  G| Irreversible: q<T  S   G < w  w system = -w < |  G|

Standard free energy of formation (  G˚ f ): the free energy change that occurs when 1 mol of a substance in its standard state (T=25 o C, p=1atm) is formed from its elements in their reference forms. Standard free energy change Standard free energy change (  G˚): the free energy change for a reaction in which reactants in their standard states yield products in their standard states. (Liquids, solids: p=1 atm. Gases: partial p =1 atm. Solutions: c=1M. Usually T=25 o C.) Substance  G˚ f (kJ/mol) Reaction of formation. water -237 H 2 (g)+1/2 O 2 (g)  H 2 O(l) methane -50 C(s)+2 H 2 (g)  CH 4 (g) Note: CH 4 :  G˚ f =  H˚ f -T  S˚ f = -74 kJ/mol - 298˚K·(-80.8 J/(mol˚K)) = -50 kJ/mol

Standard free energy change Calculation of standard free energy change for a reaction (  G˚):  G˚ =  G˚ f (products) -  G˚ f (reactants) Example: calculate  G˚ for combustion of ethanol C 2 H 5 OH(l)+3 O 2 (g)  2 CO 2 (g) + 3 H 2 O (l)  G˚ f (kJ/mol):  G˚ = [2· (-394)+3·(-237) -1·(-175) - 3·0] kJ/mol = = kJ/mol The reaction is spontaneous under standard conditions.

 G for non-standard conditions Dependence of free energy on temperature At a first approximation  H and  S do not change with temperature. To get  G˚ at some temperature (T) other than 25 o C (298 o K) use the approximation:  G˚ T   H˚ T·  S˚ 298 The signature of  H˚ and  S˚ determines temperature dependence of spontaneity: >0 >0 spontaneous at high T melting, vaporization  H˚  S˚ T-dependence Example. <0 <0 spontaneous at low T freezing, condensation 0 spontaneous at all T C 6 H 12 O 6 +6O 2  6CO 2 +6H 2 O >0 <0 nonspontaneous at all T 3O 2  2O 3

 G for non-standard conditions Dependence of free energy on concentration G of a substance in solution depends on its concentration: a high concentration means a potential to do osmotic work. free energy of substance in solution ("chemical potential") free energy of substance in a 1M solution actual concentration of substance in solution

dW= –  (V)·dV  G for non-standard conditions m·g=  ·A ViVi VfVf ii ff V  Dependence of free energy on concentration G of a substance in solution depends on its concentration: a high concentration means a potential to do osmotic work. What is the maximum, reversible, work done by 1 mol dissolved substance while the solution volume expands from V i =1liter to V f due to osmosis? Dependence of free energy on concentration G of a substance in solution depends on its concentration: a high concentration means a potential to do osmotic work.

 G for non-standard conditions Dependence of free energy on concentration G of a substance in solution depends on its concentration: a high concentration means a potential to do osmotic work. free energy of substance in solution ("chemical potential") free energy of substance in a 1M solution actual concentration of substance in solution

 G of a reaction under non-standard conditions Consider the reaction A+B C+D Can we predict whether the reaction will be spontaneous in the written direction even if the actual concentrations |A|, |B|, |C|, and |D| are arbitrary (not 1 M)? Q: reaction quotient  G for non-standard conditions

We have shown:  G =  G˚+ RT·lnQ  G and the equilibrium constant At equilibrium: Q = K (equilibrium constant)  G = 0  G˚ = – RT·lnK K = e –  G˚/(RT) If Q>K   G>0  reaction not spontaneous as written ("endergonic") (spontaneous in the reverse direction) Thus: 0 =  G˚+ RT·lnK If Q<K   G<0  reaction is spontaneous as written ("exergonic")

 G and the equilibrium constant  G˚ = – RT·lnK = –5.9 kJ/mol · lgK  If  G˚ 1  K>10  at equilibrium [B]/[A]>10  A simple rule of thumb At body temperature: kJ/mol2.303·lgK Consider the simple process A B  If  G˚>+5.9 kJ/mol  lgK<–1  K<0.1  at equilibrium [B]/[A]<0.1  If –5.9 kJ/mol<  G˚=+5.9 kJ/mol  –1<lgK<1  0.1<K<10  at equilibrium [A] and [B] are comparable

Thermodynamic coupling How does the body make endergonic processes happen? What happens to energy released in exergonic processes? A C  G A  C >0 endergonic doesn't happen B D A C  G A+B  C+D = =  G A  C +  G B  D <0 The coupled reaction is spontaneous, less heat is released.  G B  D <<0 exergonic B D  G B  D released as heat heat B D A C

 G =- 48 kJ/mol ATP  ADP+P Thermodynamic coupling ATP  ADP + P i  G˚=-30.5 kJ/mol In the cell: [ATP]=5mM, [ADP]=1mM, [P i ]=5mM   G  -48 kJ/mol glucose+6O 2 6CO 2 +6H 2 O 32 ADP+32 P i 32 ATP  G =-2866 kJ/mol gl+6O 2  6CO 2 +6H 2 O  G = kJ/mol 32(ADP+P)  32 ATP  G =-1330 kJ/mol coupled heat ATP ADP + P i pyruvate+CO 2 oxalo- acetate  G = + 34 kJ/mol pyr+CO 2  OA  G =-14 kJ/mol coupled heat ATP hydrolysis is highly exergonic, ATP synthesis is endergonic!

Back to the body... ATP ADP precursors macromolecules mechanical work muscle heat food CO 2 +H 2 O