1. A little thermodynamics
CH339K
A little thermodynamics
(which is probably more than anybody wants)

3. Thermodynamics (Briefly)
Systems est divisa in partes tres
Open
Exchange energy and matter
Closed
Exchange energy only
Isolated
Exchange nothing

4. 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

5. First Law of Thermo
ESYSTEM = q – w or, alternatively, q = E + w

6. First law of Thermo (cont.)
Example: Oxidation of a Fatty Acid (Palmitic):
C16H32O2 + 23O2 (g)  16CO2 (g) + 16H2O (l)
Under Constant Volume:
q = kJ/mol.
Under Constant Pressure:
q = kJ/mol

7. First Law of Thermo (cont.)
Why the difference?
Under Constant Volume,
q = E + w = kJ/mol + 0 = 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) = kJ
q = kJ/mol + (-17.3 kJ/mol) = kJ/mol

8. 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

9. Endothermic but spontaneous
Ammonium Nitrate spontaneously dissolves in water to the tune of about 2 kg/liter
Ammonium nitrate has a DHsolution of kJ/mol
Remember positive enthalpy = endothermic
This is the basis of instant cold packs

10. 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

11. 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

12. What happens if the volume increases?K
Adding volume increases the number of “slots,” therefore increasing W, the number of states, thereby increasing entropy.

13. Number of atoms dissolved = Na Number of original slots = no
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
Since Na << Wo and Na << Wf (dilute solution), then:
and
So we can simplify the top equations to:
and

14. Okay, so what (quantitatively) is the change in entropy from increasing the volume?
Substituting and solving:
So DS is logarithmically related to the change in the number of “slots.”

15. Let’s make the assumption that we are dealing with 1 mole (i. e
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:
Since the number of “slots” is directly related to the volume:
And since the concentration is inversely related to the volume:

16. 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.

17. Gibbs Free Energy ΔG = ΔH - T ΔS
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.

19. 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.

20. 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:
Where:
In dilute (i.e. biochemical) solutions,
the activity of a solute is its concentration
The activity of the solvent is 1

21. Free Energy and Chemical Equilibrium
Take a simple reaction: A + B ⇌ C + D Then we can figure the Free Energy Change:
Rearranging
Combining
Factoring

22. Freee Energy and Equilibrium (cont.)
Hang on a second! [A][B] is the product of the reactant concentrations [C][D] is the product of the product concentrations Remembering Freshman Chem, we have a word for that ratio.

23. 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
If you know one, you can determine the other.
Note: things profs highlight with colored arrows are probably worth remembering

24. Real Free Energy of a Reaction
As derived 2 slides previously: DG is related to DGo’, adjusted for the concentration of the reactants:

25. Example: Glucose-6-Phosphate ⇄ Glucose + Pi ∆Go’ = -13.8 kJ/mol
At 100 μM Glucose-6-Phosphate
5 mM Phosphate
10 mM Glucose

26. Measuring H, S, and G We know ΔG = ΔH - T ΔS And ΔGo = -RTlnKeq So
ΔH - T ΔS = -RTlnKeq Or

27. Measuring H, S, and G 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

28. Van’t Hoff Plot ΔHo = -902.1* 8.315 = -7500 J/mol
ΔSo = * = 30 J/Kmol

30. Why the big DGo’ for Hydrolyzing Phosphoanhydrides?
Electrostatic repulsion betwixt negative charges
Resonance stabilization of products
pH effects

31. pH Effects – DGo vs. DGo’
(DG in kcal/mol)
WOW!

32. Cellular DGs are not DGo’ s
DGo’ 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

33. Unfavorable Reactions can be Subsidized with Favorable Ones

35. Hydrolysis of Thioesters can also provide a lot of free energy

36. Acetyl Coenzyme A

37. Sample DGo’Hydrolysis