A little thermodynamics

(which is probably more than anybody wants)

CH339K

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.)

C16H32O2 + 23O2 (g) 16CO2 (g) + 16H2O (l)

• Under Constant Volume:

q = -9941.4 kJ/mol.• Under Constant Pressure:

q = -9958.7 kJ/mol

Example: Oxidation of a Fatty Acid (Palmitic):

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 = PΔVΔV = ΔnRT/PW = 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 Hsolution 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.”

State 1… State 2… etc…

X

X

X

X X

X

X

X

X

X

What happens if the volume increases?

K

K

K K

K

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

o o o o oW = n (n 1)(n 2)...(n Na)

f f f f fW = n (n 1)(n 2)...(n Na) • Since Na << Wo and Na << Wf (dilute solution), then:

o on Na n f fn Na n and

• So we can simplify the top equations to:Na

o oW = n Naf fW = n

and

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

f oS = S - S• Substituting and solving:

f oS = k ln(W ) k ln(W )

f

o

WS = k ln

W

NafNao

nS = k ln

n

Na

f

o

nS = k ln

n

f

o

nS = Na k ln

n

So S 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:

f

o

nS = R ln

n

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

f

o

VS = R ln

V

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

o

f

CS = R ln

C

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 – TSΔG = 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.

• ΔGo is 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:

]ln[0 GRTGG xx • Where:

solutionor mixture in thecomponent theof the][

component theofenergy free standard the0

activityG

G

x

x

• 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 EquilibriumTake a simple reaction:

A + B C + D⇌Then we can figure the Free Energy Change:

reactantsproducts GG G BRTln - G - ARTln - G - DRTln G CRTln G G o

BoA

oD

oC

Rearranging BRTln - ARTln - DRTln CRTln G G G G G o

BoA

oD

oC

Combining

Factoring Bln - Aln - Dln ClnRT G G o

B A

D CRTln G G o

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.

B A

D CRTln G G o

o ProductsG G RTln

Reactants

B A

D C K eq

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:

G is related to Go’, adjusted for the concentration of the reactants:

][Reactants

[Products]RTln'ΔGΔG o

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

At 100 μM Glucose-6-Phosphate 5 mM Phosphate 10 mM Glucose

15587J/mol1787J/mol)(13800J/molΔG

(.0001M)

5M)(.01M)(.00ln310Kl8.315J/Kmo13800J/molΔG

Phosphate]6[Glucose

Pi] [Glucose][RTln'ΔGΔG o

Measuring H, S, and G

We know

ΔG = ΔH - T ΔS

And

ΔGo = -RTlnKeq

So

ΔH - T ΔS = -RTlnKeq

Or

R

S

T

1

R

H- Kln

oo

eq

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

R

S

T

1

R

H- Kln

oo

eq

Van’t Hoff Plot

ΔHo = -902.1* 8.315 = -7500 J/molΔSo = +3.61 * 8.315 = 30 J/Kmol

y = -902.09x + 3.6084

0.00

0.10

0.20

0.30

0.40

0.50

0.60

0.70

0.80

0.90

1.00

0.0031 0.0032 0.0033 0.0034 0.0035 0.0036 0.0037

1/T (K-1)

ln(K

eq)

Why the big Go’ for Hydrolyzing Phosphoanhydrides?

• Electrostatic repulsion betwixt negative charges

• Resonance stabilization of products

• pH effects

pH Effects – Go vs. Go’

mol

kJ 5.41ln'

10lnln'

M10 7,pH At

lnln

ln

ln

7

7-

2

2

ReactantsProducts

ATP

PiADPRTGG

RTATP

PiADPRTGG

H

OH

HRT

ATP

PiADPRTGG

OHATP

HPiADPRTGG

RTGG

oo

oo

o

o

o

G in kcal/mol)

WOW!

Cellular Gs are not Go’ sGo’ 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

mol

kJ

mol

kJ

mol

kJG

M

MMK

molK

J

mol

kJG

ATP

PiADPRTGG

Hyd

Hyd

oHyd

52)21(31

)00225(.

)00165)(.00025(.ln298315.831

][

]][[ln'

Unfavorable Reactions can be Subsidized with Favorable Ones

Hydrolysis of Thioesters can also provide a lot of free energy

Acetyl Coenzyme A

Sample Go’Hydrolysis