- By
**meris** - Follow User

- 86 Views
- Uploaded on

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

**An Image/Link below is provided (as is) to download presentation**

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.

- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -

Presentation Transcript

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

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.)

- 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:

Rearranging

Combining

Factoring

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

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

As derived 2 slides previously:

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

Example:

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

- 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

Why the big DGo’ for Hydrolyzing Phosphoanhydrides?

- Electrostatic repulsion betwixt negative charges
- Resonance stabilization of products
- pH effects

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

Download Presentation

Connecting to Server..