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# A little thermodynamics - PowerPoint PPT Presentation

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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## PowerPoint Slideshow about 'A little thermodynamics' - meris

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(which is probably more than anybody wants)

A little thermodynamics

• Systems est divisa in partes tres

• Open

• Exchange energy and matter

• Closed

• Exchange energy only

• Isolated

• Exchange nothing

• 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

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

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

• 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

• 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

• 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.”

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.) (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 (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 (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 (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 (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.) (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.) (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 (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: (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 (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 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 (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 (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 – (i.e. N atoms) of solute dissolved in a large volume of water. DGo vs. DGo’

(DG in kcal/mol)

WOW!

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

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