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