Transcript Chemical thermodynamics I.

Chemical thermodynamics II.
Medical Chemistry
László Csanády
Department of Medical Biochemistry
Entropy change
Experience:
 heat does not flow from a cold towards a hot object
 energy tends to become dissipated into less ordered
forms ("disorder" increases)
Consequence:
Some processes happen spontaneously, others do not.
Let us define a quantity which reflects randomness:
Entropy (S):
 S should be a state function (DS = Sfinal - Sinitial)
 S should be an extensive property (S=kSk)
Entropy change
How should we define S?
Experience:
 heat absorption increases randomness in a system:
DS~q
 "more randomness" is generated by heat absorption
at lower temperature: DS~1/T
Let us define entropy through its change:
Entropy change:
DS=q/T (J/˚K)
Entropy change
Consider a process in which heat flows between a
system and its surroundings. What can we say about
DSsys and DSsur under various conditions?
endothermic process
exothermic process
surroundings: T=const.
surroundings: T=const.
system
system
heat (q>0)
heat (q<0)
DS 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).
temperature
surroundings
system
T
Tsurroundings
Tsystem
time
tstart
tend
Entropy is only redistributed!
DS for spontaneous processes
2. Spontaneous processes (irreversible):
During spontaneous processes system and surroundings fall out
of equilibrium temporarily. Temperature inhomogeneities arise.
2.1. Spontaneous processes: endothermic (q>0)
temperature
surroundings
system
Tk
T
Tsurroundings
Tsystem
time
tstart DS =q /T tend
sys;k
k k
Entropy is created!
DS for spontaneous processes
2. Spontaneous processes (irreversible):
During spontaneous processes system and surroundings fall out
of equilibrium temporarily. Temperature inhomogeneities arise.
2.2. Spontaneous processes: exothermic (q<0)
temperature
surroundings
system
Tk
Tsystem
T
Tsurroundings
time
tstart DS =q /T tend
sys;k
k k
Entropy is created!
The second law of thermodynamics
For a spontaneous process the total entropy of the
system plus its surroundings always increases:
DStotal > 0
Or:
(spontaneous process)
DSsystem > q/T
Spontaneous processes are irreversible.
For reversible processes the total entropy does not
change:
DStotal = 0 (reversible process)
Or:
DSsystem = q/T
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:
T T+dT
Add DS for phase transitions:
For fusion: For evaporation:
Standard entropies
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.)
Substance S˚(J/(mol˚K))
C(graphite)
5.7
H2(gas)
130.6
CH4(gas)
186.1
Standard entropy change for a reaction (DS˚): the
entropy change for a reaction in which reactants in
their standard states yield products in their standard
states. DS˚ = S˚final – S˚initial
(S is a state function)
DS˚ = S˚(products) - S˚(reactants)
Standard entropies
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
"DS˚f" (CH4(g)) = [186.1-(5.7+2·130.6)] J/(mol˚K) =
= -80.8 J/(mol˚K)
Note: "DS˚f" is zero for C(s) and H2(g).
General rules of thumb:
DS 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
Gibbs free energy
Focusing on the system, the criterion for spontaneity
(second law) at constant pressure can be written as
DS > qP/T = DH/T (both DS and DH refer to the system).
Or:
0 > DH - TDS
(spontaneous process)
and
0 = DH - TDS
(reversible 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:
DG = DH - TDS
Therefore:
DG < 0
(spontaneous process)
and
DG = 0
(reversible process)
Gibbs free energy
Interpretation of DG: the sign of DG determines
spontaneity, but what is the physical meaning of DG?
(Compare: DH is heat, DS is change in disorder...)
DG 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 = H - TS = (U + pV) - TS
If p, V, and T are constant:
DG = DU - TDS = (q + w) - TDS
Quasi-equilibrium: q=TDS  DG = w  wsystem = -w = |DG|
Irreversible:
q<TDS  DG < w  wsystem = -w < |DG|
Standard free energy change
Standard free energy change (DG˚): 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.)
Standard free energy of formation (DG˚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.
Substance DG˚f (kJ/mol)
water
-237
methane
-50
Reaction of formation
H2(g)+1/2 O2(g)H2O(l)
C(s)+2 H2(g)CH4(g)
.
Note: CH4: DG˚f = DH˚f -TDS˚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 (DG˚):
DG˚ = DG˚f (products) - DG˚f (reactants)
Example: calculate DG˚ for combustion of ethanol
C2H5OH(l)+3 O2(g)  2 CO2(g) + 3 H2O (l)
DG˚f(kJ/mol):
-175
0
-394
-237
DG˚ = [2· (-394)+3·(-237) -1·(-175) - 3·0] kJ/mol =
= -1324 kJ/mol
The reaction is spontaneous under standard conditions.
DG for non-standard conditions
Dependence of free energy on temperature
At a first approximation DH and DS do not change with
temperature. To get DG˚ at some temperature (T) other
than 25oC (298oK) use the approximation:
DG˚T  DH˚298 - T·DS˚298
The signature of DH˚ and DS˚ determines temperature
dependence of spontaneity:
DH˚
<0
<0
>0
>0
DS˚
<0
>0
<0
>0
T-dependence
Example
spontaneous at low T
freezing, condensation
spontaneous at all T
C6H12O6+6O26CO2+6H2O
nonspontaneous at all T 3O22O3
spontaneous at high T
melting, vaporization
.
DG 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
DG 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. 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?
p
m·g=p·A
dW= – p(V)·dV
pi
pf
V
Vi
Vf
DG 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
DG for non-standard conditions
DG 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
DG and the equilibrium constant
We have shown:
DG = DG˚+ RT·lnQ
At equilibrium:
Q = K (equilibrium constant)
DG = 0
Thus:
0 = DG˚+ RT·lnK
DG˚ = – RT·lnK
K = e–DG˚/(RT)
If Q<K  DG<0  reaction is spontaneous as written
("exergonic")
If Q>K  DG>0  reaction not spontaneous as written
("endergonic")
(spontaneous in the reverse direction)
DG and the equilibrium constant
A simple rule of thumb
At body temperature:
DG˚ = – RT·lnK = –5.9 kJ/mol · lgK
2.576 kJ/mol
2.303·lgK
Consider the simple process A
B
If DG˚<–5.9 kJ/mol  lgK>1  K>10
 at equilibrium [B]/[A]>10
If DG˚>+5.9 kJ/mol  lgK<–1  K<0.1
 at equilibrium [B]/[A]<0.1
If –5.9 kJ/mol<DG˚=+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?
B
endergonic
DGAC>0
C
heat
A
doesn't happen
B
A
C
DGA+BC+D =
=DGAC+ DGBD<0
heat
exergonic
DGBD<<0
D
DGBD released
as heat
D The coupled reaction is
spontaneous,
less heat is released.
Thermodynamic coupling
ATP  ADP + Pi
In the cell:
DG˚=-30.5 kJ/mol
[ATP]=5mM, [ADP]=1mM, [Pi]=5mM  DG-48 kJ/mol
ATP hydrolysis is highly exergonic, ATP synthesis is endergonic!
glucose+6O2
ATP
32 ATP
heat
32 ADP+32 Pi
6CO2+6H2O
oxaloacetate
heat
pyruvate+CO2
ADP + Pi
DG gl+6O26CO2+6H2O=-2866 kJ/mol
DGATPADP+P =-48 kJ/mol
DG 32(ADP+P)32 ATP=+1536 kJ/mol
DGpyr+CO2OA =+34 kJ/mol
DG coupled =-1330 kJ/mol
DG coupled =-14 kJ/mol
Back to the body...
macromolecules
ATP
food
precursors
mechanical
work
muscle
heat
CO2+H2O
ADP