CH339K
A little thermodynamics
(which is probably more than anybody
wants)
Thermodynamics (Briefly)
• Systems est divisa in
partes tres
– Open
• Exchange energy and matter
– Closed
• Exchange energy only
– Isolated
• Exchange nothing
More Thermodynamics
• Energy can be exchanged as heat (q) or work
(w)
• By convention:
– q > 0: heat has been gained by the system from
the surroundings
– q < 0: heat has been lost by the system to the
surroundings
– w > 0: work has been done by the system on the
surroundings
– w < 0: work has been done on the system by the
surroundings
First Law of Thermo
• E
SYSTEM
= q – w or, alternatively, q = E + w
First law of Thermo (cont.)
Example: Oxidation of a Fatty Acid (Palmitic):
C16H32O2 + 23O2 (g)  16CO2 (g) + 16H2O (l)
• Under Constant Volume:
q = -9941.4 kJ/mol.
• Under Constant Pressure:
q = -9958.7 kJ/mol
First Law of Thermo (cont.)
• Why the difference?
• Under Constant Volume,
q = E + w = -9941.4 kJ/mol + 0 = -9941.4 kJ/mol
• Under Constant Pressure, W is not 0!
Used 23 moles O2, only produced 16 moles CO2
W = PΔV
ΔV = ΔnRT/P
W = ΔnRT = (-7 mol)(8.314 J/Kmol)(298 K) = -17.3 kJ
q = -9941.4 kJ/mol + (-17.3 kJ/mol) = -9958.7 kJ/mol
Enthalpy
• Technically speaking, most cells operate under
constant pressure conditions
• Practically, there’s not much difference most of the time
• Enthalpy (H) is defined as:
H = E + PV or
H = E + PV
• If H > 0, heat is flowing from the surroundings to the
system and the process is endothermic
• if H < 0, heat is being given off, and the process is
exothermic.
• Many spontaneous processes are exothermic, but not
all
Endothermic but spontaneous
• Ammonium Nitrate spontaneously dissolves
in water to the tune of about 2 kg/liter
• Ammonium nitrate has a Hsolution of +25.7
kJ/mol
• Remember positive enthalpy = endothermic
• This is the basis of instant cold packs
Second law of Thermo
• Any spontaneous process must be
accompanied by a net increase in entropy
(S).
• What the heck is entropy?
• Entropy is a measure of the “disorderliness”
of a system (and/or the surroundings).
• What the heck does that mean?
• Better, it is a measure of the number of states
that a system can occupy.
• Huh?...let me explain
Entropy
S = k x ln(W) where
• W is the number of possible states
• k is Boltzmann’s constant, = R/N
Two states of 5 “atoms” in 50 possible “slots.”
State 1…
State 2… etc…
X
X
X
X
X
X
X
X
X
X
What happens if the volume increases?
K
K
K
K
K
Adding volume increases the number of “slots,” therefore
increasing W, the number of states, thereby increasing
entropy.
• We can quantify that:
– Number of atoms dissolved = Na
– Number of original slots = no
– Number of original states = Wo
– Number of final slots = nf
– Number of final states = Wf
Wo = no (no 1)(no  2)...(n o  Na)
Wf = nf (nf 1)(nf  2)...(nf  Na)
• Since Na << Wo and Na << Wf (dilute solution), then:
n o  Na  n oand
n f  Na  n f
• So we can simplify the top equations to:
Wo = n
Na
o
and
Wf = n
Na
f
• Okay, so what (quantitatively) is the change in entropy from
increasing the volume?
S = Sf - So
• Substituting and solving:
S = k ln(Wf )  k ln(Wo )
 Wf 
S = k ln 

 Wo 
 n fNa 
S = k ln  Na 
 no 
 nf 
S = k ln  
 no 
Na
 nf 
S = Na  k ln  
 no 
So S is logarithmically related to the
change in the number of “slots.”
• Let’s make the assumption that we are dealing with 1 mole (i.e.
N atoms) of solute dissolved in a large volume of water.
• Since Boltzmann’s constant (k) = R/N, our equation resolves to:
 nf 
S = R  ln  
 no 
• Since the number of “slots” is directly related to the volume:
 Vf 
S = R  ln  
 Vo 
• And since the concentration is inversely related to the volume:
 Co 
S = R  ln  
 Cf 
Entropy (cont.)
• Entropy change tells us whether a reaction is
spontaneous, but…
• Entropy can increase in the System, the
Surroundings, or both, as long as the total is
positive.
• Can’t directly measure the entropy of the
surroundings.
• HOWEVER, the change in enthalpy of the
system is an indirect measure of the
change in entropy of the surroundings –
an exothermic reaction contributes heat
(disorder) to the universe.
Gibbs Free Energy
• We can coin a term called the Free Energy (G) of the
system which tells us the directionality of a reaction.
G = H – TS
ΔG = ΔH - T ΔS
If ΔG < 0, free energy is lost  exergonic – forward rxn
favored.
If ΔG > 0, free energy is gained  endergonic –
reverse rxn favored.
Different ΔG’s
• ΔG is the change in free energy for a
reaction under some set of real conditions.
• ΔGo is the change in free energy for a
reaction under standard conditions (all
reactants 1M)
• ΔGo’ is the change of free energy for a
reaction with all reactants at 1M and pH 7.
Partial Molar free Energies
• The free energy of a mixture of stuff is equal to the
total free energies of all its components
• The free energy contribution of each component is
the partial molar free energy:
Gx  Gx0  RT ln[G]
• Where:
Gx0  thestandardfree energyof thecomponent
[Gx ]  theactivity of thecomponentin themixtureor solution
• In dilute (i.e. biochemical) solutions,
• the activity of a solute is its concentration
• The activity of the solvent is 1
Free Energy and Chemical Equilibrium
Take a simple reaction:
A+B⇌C+D
Then we can figure the Free Energy Change:
G  Gproducts  Greactants
G  GoC  RTlnC  GoD  RTlnD - GoA - RTlnA - GoB - RTlnB
Rearranging


G  G oC  G oD  G oA  G oB  RTlnC  RTlnD - RTlnA - RTlnB
Combining
G  G o  RTlnC  lnD - lnA - lnB
Factoring
 CD 
G  G  RT ln

 AB 
o
Freee Energy and Equilibrium
(cont.)
 CD 
G  G o  RT ln

 AB 
Hang on a second!
[A][B] is the product of the reactant concentrations
[C][D] is the product of the product concentrations
  Products  
G  G  RTln 
  Reactants  


o
Remembering Freshman Chem, we have a word for
that ratio.
 CD 
K eq  

 AB 
Free Energy and Equilibrium (cont.)
SO: ΔGo for a reaction is related to the
equilibrium constant for that reaction.
ΔGo = -RTlnKeq
Or
Keq = e-ΔGo/RT
Note: things profs
highlight with
colored arrows are
probably worth
remembering
If you know one, you can determine the other.
Real Free Energy of a Reaction
As derived 2 slides previously:
G is related to Go’, adjusted for the
concentration of the reactants:
[Products]
ΔG  ΔG ' RT ln
[Reactants]
o
Example:
Glucose-6-Phosphate ⇄ Glucose + Pi ∆Go’ = -13.8 kJ/mol
At 100 μM Glucose-6-Phosphate
5 mM Phosphate
10 mM Glucose
[Glucose][ Pi]
ΔG  ΔG ' RT ln
[Glucose  6  Phosphate]
o
(.01M)(.005M)
ΔG  13800J/mol 8.315J/Kmol  310K ln
(.0001M)
ΔG  13800J/mol (1787J/mol) 15587J/mol
Measuring H, S, and G
We know
ΔG = ΔH - T ΔS
And
ΔGo = -RTlnKeq
So
ΔH - T ΔS = -RTlnKeq
Or
o
H  1  S
lnK eq     
R T R
o
Measuring H, S, and G
H  1  S
lnK eq     
R T R
o
•
•
•
•
o
This is the van’t Hoff Equation
You can control T
You can measure Keq
If you plot ln(Keq) versus 1/T, you get a line
– Slope = -ΔHo/R
– Y-intercept = ΔSo/R
Van’t Hoff Plot
1.00
0.90
y = -902.09x + 3.6084
0.80
0.70
ln(Keq)
0.60
0.50
0.40
0.30
0.20
0.10
0.00
0.0031
0.0032
0.0033
0.0034
0.0035
0.0036
0.0037
1/T (K-1)
ΔHo = -902.1* 8.315 = -7500 J/mol
ΔSo = +3.61 * 8.315 = 30 J/Kmol
Why the big Go’ for Hydrolyzing
Phosphoanhydrides?
• Electrostatic repulsion
betwixt negative
charges
• Resonance stabilization
of products
• pH effects
pH Effects –
o
G
vs.
o
G ’
Products 
G  G o  RT ln Reactants


ADPPiH  
G  G  RT ln
ATPH 2O 


H 
ADPPi
o
 RT ln
G  G  RT ln
H 2O 
ATP
o
 
At pH 7, H   10-7 M
ADPPi  RT ln107 
G o '  G o  RT ln
ATP
ADPPi  41.5 kJ
G o '  G o  RT ln
ATP
mol
G in kcal/mol)
WOW!
Cellular Gs are not Go’ s
Go’ for hydrolysis of ATP is about -31 kJ/mol
Cellular conditions are not standard, however:
In a human erythrocyte,
[ATP]≈2.25 mM, [ADP] ≈0.25 mM, [PO4] ≈1.65 mM
[ ADP][Pi]
[ ATP]
kJ
J
(.00025M )(.00165M )
 31
 8.315
 298K  ln
m ol
K  m ol
(.00225M )
kJ
kJ
kJ
 31
 (21
)  52
m ol
m ol
m ol
GHyd  G o ' RT ln
GHyd
GHyd
Unfavorable Reactions can be
Subsidized with Favorable Ones
Hydrolysis of Thioesters can also provide a lot of free energy
Acetyl Coenzyme A
Sample Go’Hydrolysis