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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

CH339K

(which is probably more than anybody wants)

A little thermodynamics


Thermodynamics briefly
Thermodynamics (Briefly)

  • Systems est divisa in partes tres

    • Open

      • Exchange energy and matter

    • Closed

      • Exchange energy only

    • Isolated

      • Exchange nothing


More thermodynamics
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
First Law of Thermo

  • ESYSTEM = q – w or, alternatively, q = E + w


First law of thermo cont
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 cont1
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
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
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
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
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.”


What happens if the volume increases
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:

and

  • So we can simplify the top equations to:

and


  • 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.”


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

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


Entropy cont
Entropy (cont.) (i.e. N atoms) of solute dissolved in a large volume of water.

  • 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
Gibbs Free Energy (i.e. N atoms) of solute dissolved in a large volume of water.

  • 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 g s
Different (i.e. N atoms) of solute dissolved in a large volume of water. 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
Partial Molar free Energies (i.e. N atoms) of solute dissolved in a large volume of water.

  • 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
Free Energy and Chemical Equilibrium (i.e. N atoms) of solute dissolved in a large volume of water.

Take a simple reaction:

A + B ⇌ C + D

Then we can figure the Free Energy Change:

Rearranging

Combining

Factoring


Freee energy and equilibrium cont
Freee Energy and Equilibrium (cont.) (i.e. N atoms) of solute dissolved in a large volume of water.

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
Free Energy and Equilibrium (cont.) (i.e. N atoms) of solute dissolved in a large volume of water.

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


Real free energy of a reaction
Real Free Energy of a Reaction (i.e. N atoms) of solute dissolved in a large volume of water.

As derived 2 slides previously:

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


Example
Example: (i.e. N atoms) of solute dissolved in a large volume of water.

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
Measuring H, S, and G (i.e. N atoms) of solute dissolved in a large volume of water.

We know

ΔG = ΔH - T ΔS

And

ΔGo = -RTlnKeq

So

ΔH - T ΔS = -RTlnKeq

Or


Measuring h s and g1
Measuring H, S, and G (i.e. N atoms) of solute dissolved in a large volume of water.

  • 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
Van’t Hoff Plot (i.e. N atoms) of solute dissolved in a large volume of water.

ΔHo = -902.1* 8.315 = -7500 J/mol

ΔSo = +3.61 * 8.315 = 30 J/Kmol


Why the big d g o for hydrolyzing phosphoanhydrides
Why the big (i.e. N atoms) of solute dissolved in a large volume of water. DGo’ for Hydrolyzing Phosphoanhydrides?

  • Electrostatic repulsion betwixt negative charges

  • Resonance stabilization of products

  • pH effects


Ph effects d g o vs d g o
pH Effects – (i.e. N atoms) of solute dissolved in a large volume of water. DGo vs. DGo’

(DG in kcal/mol)

WOW!


Cellular d gs are not d g o s
Cellular (i.e. N atoms) of solute dissolved in a large volume of water. 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
Unfavorable Reactions can be Subsidized with Favorable Ones (i.e. N atoms) of solute dissolved in a large volume of water.


Hydrolysis of (i.e. N atoms) of solute dissolved in a large volume of water. Thioesters can also provide a lot of free energy


Acetyl coenzyme a
Acetyl Coenzyme A (i.e. N atoms) of solute dissolved in a large volume of water.


Sample d g o hydrolysis
Sample (i.e. N atoms) of solute dissolved in a large volume of water. DGo’Hydrolysis


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