A little thermodynamics
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CH339K. (which is probably more than anybody wants). A little thermodynamics. Thermodynamics (Briefly). Systems est divisa in partes tres Open Exchange energy and matter Closed Exchange energy only Isolated Exchange nothing. More Thermodynamics.

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A little thermodynamics

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(which is probably more than anybody wants)

A little thermodynamics

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

  • ESYSTEM = 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


  • 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 DHsolution 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


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.”

What happens if the volume increases?

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

  • Since Na << Wo and Na << Wf (dilute solution), then:


  • So we can simplify the top equations to:


  • Substituting and solving:

  • Okay, so what (quantitatively) is the change in entropy from increasing the volume?

So DS 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:

  • Since the number of “slots” is directly related to the volume:

  • And since the concentration is inversely related to the volume:

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.

  • ΔGois 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:

  • Where:

  • 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:




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.

Free Energy and Equilibrium (cont.)

SO: ΔGo for a reaction is related to the equilibrium constant for that reaction.

ΔGo = -RTlnKeq


Keq = e-ΔGo/RT

If you know one, you can determine the other.

Note: things profs highlight with colored arrows are probably worth remembering

Real Free Energy of a Reaction

As derived 2 slides previously:

DG is related to DGo’, adjusted for the concentration of the reactants:


Glucose-6-Phosphate ⇄ Glucose + Pi ∆Go’ = -13.8 kJ/mol

At 100 μM Glucose-6-Phosphate

5 mM Phosphate

10 mM Glucose

Measuring H, S, and G

We know

ΔG = ΔH - T ΔS


ΔGo = -RTlnKeq


ΔH - T ΔS = -RTlnKeq


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

Van’t Hoff Plot

ΔHo = -902.1* 8.315 = -7500 J/mol

ΔSo = +3.61 * 8.315 = 30 J/Kmol

Why the big DGo’ for Hydrolyzing Phosphoanhydrides?

  • Electrostatic repulsion betwixt negative charges

  • Resonance stabilization of products

  • pH effects

pH Effects – DGo vs. DGo’

(DG in kcal/mol)


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

Unfavorable Reactions can be Subsidized with Favorable Ones

Hydrolysis of Thioesters can also provide a lot of free energy

Acetyl Coenzyme A

Sample DGo’Hydrolysis

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