Review of basic thermodynamics Matt Jacobson Please never forget - - PowerPoint PPT Presentation

Review of basic thermodynamics

Matt Jacobson

Please never forget

Thermodynamics

• Will something happen,

and to what extent? Kinetics

• How fast will it happen?

These are fundamentally different issues A simple example: It is thermodynamically favorable for diamonds to change into graphite. However, this of course takes an incredibly long time to happen.

Free energy

reactants (diamond) products (graphite) transition state

K RT G ln − = Δ

Thermo only tells you about this Kinetics is related to this

Thermodynamics in water

• Because water is so incompressible, there is really

very little difference between “constant pressure” and “constant volume” ensembles

• Thus no need to distinguish between Helmholtz vs.

Gibbs free energies

• “Enthalpy” and “energy” can be used interchangably

( )

n sol' in U H PV U H Δ ≈ Δ Δ + Δ = Δ

This is pretty much all you need to remember … at least for now

) pontaneous ( s G S T H G < Δ Δ − Δ = Δ

ΔG = ΔG! + RT lnQ R = N Akb

But keep in mind that enthalpy and entropy are themselves functions of temperature:

∂H dT " # $ % & '

P

= Cp ΔH ≈ CpΔT ∂S ∂T " # $ % & '

P

= CP T ΔS ≈ Cp ln T2 T

1

This is pretty much all you need to remember … at least for now

) pontaneous ( s G S T H G < Δ Δ − Δ = Δ

But keep in mind that enthalpy and entropy are themselves functions of temperature:

∂H dT " # $ % & '

P

= Cp ΔH ≈ CpΔT ∂S ∂T " # $ % & '

P

= CP T ΔS ≈ Cp ln T2 T

1

Free energy Enthalpy Entropy At equilibrium, Q=K (equilibrium constant) and ΔGo = 0. So-called ‘gas constant’ Avogadro’s number Boltzmann constant Heat capacities

ΔG = ΔG! + RT lnQ R = N Akb

Oh, and one more equation …

In modern notation, this is typically written S = kB ln W W is the number of configurations available to a system, and is related to the probabilities we just calculated.

Note that S is a state function, and positive values of ΔS are favorable, not negative like ΔH. Units on S are J/K.

Racemization is purely entropic

Thalidomide This is the bad one

From view of statistical thermodynamics, no different than flipping coins. What are ΔH, ΔG, K for R->S? How fast will this happen?

How about entropy of continuous variables like position/concentration?

Thought experiment: Consider a beaker with a partition right in the middle that starts out holding molecules in only one half. What happens when you remove the partition?

• What is ΔH for this process?
• Zero (at least if molecules don’t interact, “ideal solution”)

So why does it happen? How about the reverse process? [Relevance to drug distribution]

How about the reverse process?

Thought experiment: Molecules spontaneously moving to only one half of beaker?

• ΔH is still zero
• This seems unlikely, except maybe if there are very few particles
• Let’s calculate the probabilities quantitatively

Probability of being on one side

1 mol N = 1 p = 1/2 N = 2 p = 1/4 N p = (1/2)N N ~ 6 x 1023 p = (1/2)N = very very small This fits with our intuition (and the second law) which says that things tend toward disorder, i.e., higher entropy [unless there is an input of energy to counteract …]

Let’s compute ΔS for this process

This is not just a thought experiment. This is equivalent to entropy change upon dilution, or mixing two solutions. State 1 State 2 S1 = kB ln W1 S2 = kB ln W2

ΔS = S2 − S1 = kB lnW2 − lnW1

( ) = kB ln W2

W1 $ % & ' ( )

N = 1 ΔS = kB ln 2 = 1.38x10-23 J/K * 0.693 ~ 10-23 J/K

ΔS = S2 − S1 = kB lnW2 − lnW1

( ) = kB ln W2

W1 $ % & ' ( )

We can define the number of “configurations” of the system in different ways, but regardless, W clearly doubles for a single molecule when we double the volume. To get units of energy, multiply by temperature (300 K), which gives 3x10-21 J. TINY amount of energy.

N = 1 ΔS = kB ln 2 N = 2 ΔS = kB ln 4 We assume that the 2 molecules are uncorrelated, i.e., the configuration of

• ne is unrelated to the other.

(This is again the definition of an “ideal solution”.) This then implies that the total number

• f configurations is the product (not

sum) of the configurations for each molecule.

ΔS = S2 − S1 = kB lnW2 − lnW1

( ) = kB ln W2

W1 $ % & ' ( )

N = 1 ΔS = kB ln 2 N = 2 ΔS = kB ln 4 N molecules ΔS = kB ln 2N =NkBln 2

ΔS = S2 − S1 = kB lnW2 − lnW1

( ) = kB ln W2

W1 $ % & ' ( )

N = 1 ΔS = kB ln 2 N = 2 ΔS = kB ln 4 N = NA = 1 mole ΔS = NAkBln 2 = R ln 2 N = # of molecules NA = Avogadro’s number kB = Boltzmann’s constant R = NAkB = gas constant

ΔS = S2 − S1 = kB lnW2 − lnW1

( ) = kB ln W2

W1 $ % & ' ( )

N = 1 ΔS = kB ln 2 N = 2 ΔS = kB ln 4 N = NA = 1 mole ΔS = NAkBln 2 = R ln 2 Generalizing to any volume change, Molar entropy:

ΔS = Rln V2 V

1

" # $ % & '

N = # of molecules NA = Avogadro’s number kB = Boltzmann’s constant R = NAkB = gas constant

ΔS = S2 − S1 = kB lnW2 − lnW1

( ) = kB ln W2

W1 $ % & ' ( )

V = volume N = # of molecules n = # of moles R = NAkB = gas constant

We can also think of this in terms of concentrations

State 1 State 2

ΔS = Rln V2 V

1

" # $ % & ' = Rln n M2 n M1 " # $ $ $ % & ' ' ' = Rln M1 M2 " # $ % & '

Convert to concentrations, in molarity units: M = n/V, so V = n/M

There are other flavors of entropy

• We focused here on “translational” entropy,

which is equivalent to entropy of dilution, or entropy of mixing

• There is also “rotational” entropy and

“vibrational entropy”, very important in drug binding as well

Free Energy

• Best understood as a statement of the 2nd law
• At constant pressure, the correct criterion is

ΔG < 0 where G = H – TS

• If we additionally assume constant T, you get the famous

equation ΔG = ΔH – TΔS

• Key point: note the sign on the entropy term
• When ΔG = 0, this is called “equilibrium”
• Molar free energy for a substance also referred to as

“chemical potential”: µ = ΔG

Free energy of dilution/mixing

State 1 State 2

ΔH = 0 ΔS = Rln V2 V

1

" # $ % & ' = Rln M1 M2 " # $ % & ' ΔG = −RT ln V2 V

1

" # $ % & ' = −RT ln M1 M2 " # $ % & ' = RT ln M2 M1 " # $ % & '

Recall that …

Slightly more complicated

A(aq) ↔ B(aq)

Example: tautomers of allopurinol (drug used to treat gout)

↔

• It turns out that the one on the right is a bit more stable. So

what do we expect to happen?

• 100% B? No, because that would be unfavorable for entropy
• 50% A and 50% B? No, that made sense for enantiomers, but if

B is a little more stable, there should be more of it at

• equilibrium. (like an unfair coin being tossed)

From equilibrium to free energy

A(aq) ↔ B(aq) ↔

At equilibrium, there is 10x more B than A. What is the free energy of tautomerization from A to B?

Keq = M B M A ! " # $ % &

eq

= 10 1 ! " # $ % & =1

ΔGo = -RT lnKeq = −8.31 J K ⋅mol ×298K × ln10 ≈ −5.7 kJ mol

Notation/terminology for Drug- Binding and pH Equilibrium

Protein-ligand binding

] ][ [ ] [ L P PL Ka =

P + L ↔ PL PL ↔ P + L Association Dissociation

] [ ] ][ [ PL L P Kd =

P = macromolecule L = ligand (e.g., drug)

1 =

d aK

K

Not to be confused with acid equilibrium constant

Acid-Base Equilibrium

is just proton binding, but has a lot of specialized notation

HA ↔ H+ + A- H2O + B ↔ HB+ + OH-

Ka = H +

[ ] A− [ ]

HA

[ ]

pKa = −logKa Kb = HB+

[ ] OH− [ ]

B

[ ]

pKb = −logKb

H2O ↔ H+ + OH- Kw = H +

[ ] OH− [ ] =10−14

pH = −log H +

[ ]

pOH = −log OH−

[ ]

pH + pOH =14 pH = 7 (neutral) pH < 7 (acidic) pH > 7 (basic)

14 = + =

b a w b a

pK pK K K K

pH = pKa + log A− " # $ % HA

[ ]

“Henderson Hasselbalch” Just the acid equilibrium condition written in log form

Binding curves are (usually) sigmoidal

0% 50% 100% 0.00 2.00 4.00 6.00 8.00 10.00 12.00 14.00

pH % prot

( )

pKa pH BH B pKa pH RT G − = ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ − = Δ ] [ ] [ log 3 . 2

Appendix 1 (Optional Material) Derivation of ΔG=-RTln K for a simple AèB reaction

B A

µ µ =

At equilibrium (constant T, P): Assume ideal solution,

µA(aq) = µA

0 (aq) + RTln MA

M°

Mº = 1 molar (to make argument of the logarithm unitless)

µB(aq) = µB

0 (aq) + RTln MB

M°

Equating these two:

µA

0 (aq) + RT ln MA

M° = µB

0 (aq) + RTln MB

M°

µA

0 (aq) + RT ln MA

M° = µB

0 (aq) + RTln MB

M°

µB

0 (aq) − µA 0 (aq) = RTln MA

M° − RTln MB M°

ΔG° = − RTln MB M° − RTln MA M° $ % & ' ( ) = −RTln MB MA

Let ΔG° = µB

0 (aq) − µA 0 (aq)

Constant that depends

• nly on nature of

reactants and products

DEFINE KM= (MB/MA)eq

ΔGo = -RTlnKM

KM = e

− ΔG o RT

Extension to more general equilibria:

aA + bB ↔ cC + dD

B A D C

b a d c µ µ µ µ + = +

ΔG0 = −RTln MC

c MD d

MA

aMB b (aRTln MA = RTln MA a)

KM = MC

c MD d

MA

aMB b

" # $ % & '

eq

ΔG0 = cµC

0 (aq) + dµD 0 (aq) − aµA 0 (aq) + bµB 0 (aq)

( )

At equilibrium (constant T, P; closed system):

eq b a d c C

B A D C K ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ = ] [ ] [ ] [ ] [

Alternate Notation

Other concentration units

Derivations are extremely similar. For molality (mol/kg solvent), the chemical potential can be written as

µi(aq) = µi

• (aq) + RTln mi

m°

where the standard state is now defined by mº = 1 molal. The equilibrium constant winds up looking like

A B m

m m K =

m

K

• RT

G ln = = Δ

Expression for gasses is also very similar, with partial pressures replacing concentrations.

Appendix 2

Review of solving simple equilibrium problems analytically

Protein Folding Example

Consider the denaturation, or unfolding, reaction of a protein in aqueous solution to be described by the simple reaction mechanism F↔U, where F represents the folded protein and U the unfolded, or denatured, protein. At 298 K, the free energy of unfolding is 5 kJ/mol.

• a. What is the value of the equilibrium constant, Keq, for this reaction at 298 K?

ΔG0 = 5 kJ/mol = 5000 J/mol

Keq = e

−ΔG0 RT = e − 5000J mol 8.314J K⋅mol*298K = 0.133

• b. What fraction of protein is unfolded at equilibrium at 298 K?

Keq = U

[ ]

F

[ ]

= x 1− x = 0.133 x = 0.117

Drug binding example

Bosutinib is a second-generation Bcr-Abl inhibitor (imatinib, aka “Gleevec”, being the first generation), with 1.4 nM binding affinity to Bcr-Abl. Suppose that you mix together 10 mL of 10 nM Bcr-Abl and 10 mL of 10 nM bosutinib. At equilibrium, what fraction of the drug is bound to Bcr-Abl? What fraction of Bcr-Abl has drug bound?

Drug binding example

Bosutinib is a second-generation Bcr-Abl inhibitor (imatinib, aka “Gleevec”, being the first generation), with 1.4 nM binding affinity to Bcr-Abl. The 1.4 nM is a binding constant, but is it Kd or Ka? Kd = [P][L] [PL] =1.4nM =1.4×10−9M PL ↔ P + L

Drug binding example

Suppose that you mix together 10 mL of 10 nM Bcr-Abl and 10 mL of 10 nM bosutinib.

P L PL I 5 nM 5 nM C

• x
• x

+x E 5e-9-x 5e-9-x x

Initial Change End reactants product Kd = [P][L] [PL] = 5⋅10−9 − x

( ) 5⋅10−9 − x ( )

x =1.4⋅10−9M

A little algebra + quadratic formula:

x = 2.96x10-9 ~ 3 nM

P L PL I 5 nM 5 nM C

• 3 nM
• 3 nM

+3 nM E 2 nM 2 nM 3 nM

Initial Change End reactants product At equilibrium, what fraction of the drug is bound to Bcr-Abl? What fraction of Bcr-Abl has drug bound?

There is a total of 5 nM drug, and 3 nM is bound to the protein, so that’s 60% of the drug bound. Also, in this case, there is 5 nM protein total, so it’s also 60% of the Bcr-Abl that has drug bound.

pH example: Estimate the fraction of morphine that is protonated in the blood?

Morphine is a weak base with pKb=6.13. Blood has pH=7.4, and it is heavily buffered, so the pH of the blood will not change upon adding morphine. We can solve this using the pKb, or we can switch things around and use the pKa of the conjugate acid. pKa + pKb = 14 pKa = 14 - 6.13 = 7.87

pH example: Estimate the fraction of morphine that is protonated in the blood?

pH = pKa + log A−

[ ]

HA

[ ]

7.4 = 7.87 + log [M] [HM] log [M] [HM] = 7.4 − 7.87 = −0.47 log[HM] [M] = 0.47 [HM] [M] =100.47 = 2.95 %prot = [HM] [M]+[HM] = 2.95[M] [M]+ 2.95[M] = 2.95 1+ 2.95 = 75%