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

Review of basic thermodynamics Matt Jacobson

Please never forget Thermodynamics Kinetics • Will something happen, • How fast will it happen? and to what extent? 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 transition state Kinetics is related to this reactants Thermo only tells you about this (diamond) G RT ln K Δ = − products (graphite)

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 ( ) H U PV Δ = Δ + Δ H U in sol' n Δ ≈ Δ

This is pretty much all you need to remember … at least for now G H T S Δ = Δ − Δ ( G 0 s pontaneous ) Δ < Δ G = Δ G ! + RT ln Q R = N A k b But keep in mind that enthalpy and entropy are themselves functions of temperature: " % ∂ H = C p Δ H ≈ C p Δ T $ ' dT # & P " % ∂ S = C P T Δ S ≈ C p ln T 2 $ ' # ∂ T & T P 1

This is pretty much all you need to remember … at least for now Enthalpy G H T S Entropy Δ = Δ − Δ Free energy ( G 0 s pontaneous ) Δ < Δ G = Δ G ! + RT ln Q At equilibrium, Q=K (equilibrium constant) and Δ G o = 0. R = N A k b So-called ‘gas constant’ Boltzmann Avogadro’s constant number But keep in mind that enthalpy and entropy are themselves functions of temperature: " % ∂ H = C p Δ H ≈ C p Δ T $ ' dT # & P Heat capacities " % ∂ S = C P T Δ S ≈ C p ln T 2 $ ' # ∂ T & T P 1

Oh, and one more equation … In modern notation, this is typically written S = k B 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 Δ S are favorable, not negative like Δ H. 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 Δ H, Δ G, 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? How about the reverse process? • What is Δ H for this process? [Relevance to drug • Zero (at least if molecules don’t interact, “ ideal solution ” ) distribution] So why does it happen?

How about the reverse process? Thought experiment: Molecules spontaneously moving to only one half of beaker? • Δ H 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 ~ 6 x 10 23 N = 1 N = 2 N p = (1/2) N = p = 1/2 p = 1/4 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 Δ S 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 S 1 = k B ln W 1 S 2 = k B ln W 2 $ ' ) = k B ln W 2 ( Δ S = S 2 − S 1 = k B ln W 2 − ln W 1 & ) W 1 % (

$ ' ) = k B ln W 2 ( Δ S = S 2 − S 1 = k B ln W 2 − ln W 1 & ) W 1 % ( 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. N = 1 Δ S = k B ln 2 = 1.38x10 -23 J/K * 0.693 ~ 10 -23 J/K To get units of energy, multiply by temperature (300 K), which gives 3x10 -21 J. TINY amount of energy.

$ ' ) = k B ln W 2 ( Δ S = S 2 − S 1 = k B ln W 2 − ln W 1 & ) W 1 % ( We assume that the 2 molecules are uncorrelated, i.e., the configuration of one is unrelated to the other. (This is again the definition of an “ ideal solution ” .) This then implies that the total number of configurations is the product (not N = 1 N = 2 sum) of the configurations for each Δ S = k B ln 2 Δ S = k B ln 4 molecule.

$ ' ) = k B ln W 2 ( Δ S = S 2 − S 1 = k B ln W 2 − ln W 1 & ) W 1 % ( N = 1 N = 2 N molecules Δ S = k B ln 2 Δ S = k B ln 4 Δ S = k B ln 2 N =Nk B ln 2

$ ' ) = k B ln W 2 ( Δ S = S 2 − S 1 = k B ln W 2 − ln W 1 & ) W 1 % ( N = 1 N = 2 N = N A = 1 mole Δ S = k B ln 2 Δ S = k B ln 4 Δ S = N A k B ln 2 = R ln 2 N = # of molecules N A = Avogadro ’ s number k B = Boltzmann ’ s constant R = N A k B = gas constant

$ ' ) = k B ln W 2 ( Δ S = S 2 − S 1 = k B ln W 2 − ln W 1 & ) W 1 % ( N = 1 N = 2 N = N A = 1 mole Δ S = k B ln 2 Δ S = k B ln 4 Δ S = N A k B ln 2 = R ln 2 Generalizing to any volume change, Molar entropy: N = # of molecules N A = Avogadro ’ s number " % Δ S = R ln V 2 k B = Boltzmann ’ s constant $ ' V # & R = N A k B = gas constant 1

We can also think of this in terms of concentrations State 1 State 2 Convert to concentrations, in molarity units: M = n/V, so V = n/M " % n M 2 " % " % Δ S = R ln V 2 = R ln M 1 $ ' V = volume ' = R ln $ $ ' $ ' n M 1 V M 2 N = # of molecules $ ' # & # & 1 # & n = # of moles R = N A k B = gas constant

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 2 nd law • At constant pressure, the correct criterion is Δ G < 0 where G = H – TS • If we additionally assume constant T, you get the famous equation Δ G = Δ H – T Δ S • Key point: note the sign on the entropy term • When Δ G = 0, this is called “ equilibrium ” • Molar free energy for a substance also referred to as “ chemical potential ” : µ = Δ G

Free energy of dilution/mixing State 1 State 2 Recall that … Δ H = 0 " % " % ' = R ln M 1 Δ S = R ln V 2 $ $ ' V M 2 # & # & 1 " % " % " % Δ G = − RT ln V 2 ' = − RT ln M 1 ' = RT ln M 2 $ $ $ ' V M 2 M 1 # & # & # & 1

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? ! $ ! $ K eq = M B = 10 & = 1 # & # M A " 1 % " % eq K ⋅ mol × 298 K × ln10 ≈ − 5.7 kJ J Δ G o = -RT ln K eq = − 8.31 mol

Notation/terminology for Drug- Binding and pH Equilibrium

Protein-ligand binding Association Dissociation P = macromolecule P + L ↔ PL PL ↔ P + L L = ligand (e.g., drug) [ P ][ L ] [ PL ] K a K 1 = K d = K a = d [ P ][ L ] [ PL ] Not to be confused with acid equilibrium constant

Acid-Base Equilibrium is just proton binding, but has a lot of specialized notation H 2 O ↔ H + + OH - HA ↔ H + + A - H 2 O + B ↔ HB + + OH - [ ] OH − [ ] = 10 − 14 K w = H + [ ] A − [ ] [ ] OH − [ ] H + HB + [ ] pH = − log H + K a = K b = [ ] [ ] HA B [ ] pOH = − log OH − pK a = − log K a pK b = − log K b pH + pOH = 14 pH = 7 (neutral) K K K = a b w pH < 7 (acidic) pK pK 14 + = pH > 7 (basic) a b " $ A − “Henderson Hasselbalch” # % pH = pK a + log Just the acid equilibrium [ ] HA condition written in log form

Binding curves are (usually) sigmoidal ( ) G 2 . 3 RT pH pKa Δ = − [ B ] ⎛ ⎞ log pH pKa ⎜ ⎟ = − ⎜ ⎟ [ BH ] ⎝ ⎠ 100% % prot 50% 0% 0.00 2.00 4.00 6.00 8.00 10.00 12.00 14.00 pH

Appendix 1 (Optional Material) Derivation of Δ G=-RTln K for a simple A è B reaction

At equilibrium (constant T, P): µ = µ A B Mº = 1 molar (to Assume ideal solution, make argument of the logarithm unitless) 0 ( aq ) + RT ln M A µ A ( aq ) = µ A M ° 0 ( aq ) + RT ln M B µ B ( aq ) = µ B M ° Equating these two: 0 ( aq ) + RT ln M A 0 ( aq ) + RT ln M B µ A M ° = µ B M °

0 ( aq ) + RT ln M A 0 ( aq ) + RT ln M B µ A M ° = µ B M ° 0 ( aq ) = RT ln M A M ° − RT ln M B 0 ( aq ) − µ A µ B M ° Constant that depends 0 ( aq ) − µ A 0 ( aq ) only on nature of Let Δ G ° = µ B reactants and products $ ' Δ G ° = − RT ln M B M ° − RT ln M A ) = − RT ln M B & % M ° ( M A DEFINE K M = (M B /M A ) eq − Δ G o Δ G o = -RT ln K M K M = e RT