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A little thermodynamicsPowerPoint Presentation

A little thermodynamics

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Thermodynamics (Briefly)

- Systems est divisa in partes tres
- Open
- Exchange energy and matter

- Closed
- Exchange energy only

- Isolated
- Exchange nothing

- Open

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

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 + PV

- 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

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?

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.”

- 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.) (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

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