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 = Sfinal - Sinitial) S should be an extensive property (S=kSk)

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 Ssys and Ssur under various conditions?

Entropy change

system

surroundings: T=const.

heat (q<0)

endothermic process exothermic process

time

tstart tend

T

temperature

Tsystem

Tsurroundings

S for reversible processes1. 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!

Tk

Ssys;k=qk/Tk

system

surroundings

time

tstart tend

T

temperature

Tsystem

Tsurroundings

2.1. Spontaneous processes: endothermic (q>0)

S for spontaneous processes2. Spontaneous processes (irreversible):During spontaneous processes system and surroundings fall out of equilibrium temporarily. Temperature inhomogeneities arise.

Entropy is created!

Tk

Ssys;k=qk/Tk

system

surroundings

time

tstart tend

T

temperatureTsystem

Tsurroundings

S for spontaneous processes2. Spontaneous processes (irreversible):During spontaneous processes system and surroundings fall out of equilibrium temporarily. Temperature inhomogeneities arise.

2.2. Spontaneous processes: exothermic (q<0)

Entropy is created!

The second law of thermodynamicsFor a spontaneous process the total entropy of the system plus its surroundings always increases:

Stotal > 0 (spontaneous process)

Spontaneous processes are irreversible.

For reversible processes the total entropy does not change:

Stotal = 0 (reversible process)

Or: Ssystem > q/T

Or: Ssystem = 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 insmall reversible steps to change itstemperature from T to T+dT:

T T+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=25oC.)

Standard entropies

Substance S˚(J/(mol˚K))

C(graphite) 5.7H2(gas) 130.6CH4(gas) 186.1

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 H2(g).

Example: calculate "standard entropy of formation" for methane: C(s)+2 H2(g)CH4(g)

S˚(J/(mol˚K)): 5.7 2·130.6 186.1

"S˚f" (CH4(g)) = [186.1-(5.7+2·130.6)] J/(mol˚K) = = -80.8 J/(mol˚K)

Gibbs free energyFocusing on the system, the criterion for spontaneity (second law) at constant pressure can be written asS > qP/T = H/T (both S and H refer to the system).

Or: 0 > H - TS (spontaneous process)

Let us define a new quantity: Gibbs free energy (G):

G =H - TSAt constant p and T the change in free energy ofthe system determines the spontaneity of the process:

G =H - TS

G < 0 (spontaneous process)Therefore:

and 0 = H - TS (reversible process)

G = 0 (reversible process)and

G =H - TS = (U + pV) - TS

Gibbs free energyInterpretation 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 - TS = (q + w) - TS

If p, V, and T are constant:

Quasi-equilibrium: q=TS G = w wsystem = -w = |G|

Irreversible: q<TS G < w wsystem = -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=25oC, p=1atm) is formed from its elements in their reference forms.

Standard free energy changeStandard 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=25oC.)

Substance G˚f (kJ/mol) Reaction of formation .

water -237 H2(g)+1/2 O2(g)H2O(l)

methane -50 C(s)+2 H2(g)CH4(g)Note: CH4: G˚f = H˚f -TS˚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

C2H5OH(l)+3 O2(g) 2 CO2(g) + 3 H2O (l)

G˚f(kJ/mol): -175 0 -394 -237

G˚ = [2· (-394)+3·(-237) -1·(-175) - 3·0] kJ/mol = = -1324 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 25oC (298oK) use the approximation:

G˚T H˚298 - 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 >0 spontaneous at all T C6H12O6+6O26CO2+6H2O

>0 <0 nonspontaneous at all T 3O22O3

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 ofsubstance in solution("chemical potential")

free energy ofsubstance in a

1M solution

actualconcentrationof substance

in solution

dW= – (V)·dV

G for non-standard conditions

m·g=·Am·g=·Am·g=·Am·g=·Am·g=·Am·g=·Am·g=·Am·g=·Am·g=·Am·g=·Am·g=·A

Vi Vf

i

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 Vi=1liter to Vf 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 ofsubstance in solution("chemical potential")

free energy ofsubstance in a

1M solution

actualconcentrationof 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˚<–5.9 kJ/mol lgK>1 K>10 at equilibrium [B]/[A]>10

A simple rule of thumb

At body temperature:

2.576 kJ/mol 2.303·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 couplingHow does the body make endergonic processes happen?What happens to energy released in exergonic processes?

A

CGAC>0endergonic

doesn't happen

B

DA

C

GA+BC+D ==GAC+ GBD<0

The coupled reaction isspontaneous,less heat is released.

GBD<<0exergonicB

D

GBD releasedas heat

heat

B

DA

C

heat

G =- 48 kJ/molATPADP+P

Thermodynamic couplingATP ADP + Pi G˚=-30.5 kJ/mol

In the cell:[ATP]=5mM, [ADP]=1mM, [Pi]=5mM G-48 kJ/mol

glucose+6O2

6CO2+6H2O32 ADP+32 Pi

32 ATP

G =-2866 kJ/molgl+6O26CO2+6H2O

G =+1536 kJ/mol32(ADP+P)32 ATP

G =-1330 kJ/molcoupled

heat

ATP

ADP + Pi

pyruvate+CO2

oxalo-acetate

G =+34 kJ/molpyr+CO2OA

G =-14 kJ/molcoupled

heat

ATP hydrolysis is highly exergonic, ATP synthesis is endergonic!

Back to the body...

ATP

ADP

precursors

macromolecules

mechanicalwork

muscle

heat

food

CO2+H2O