S and G Entropy and Gibbs Free Energy

Transcription

1 Week 3 problem solving + equa'ons, + applica'ons S and G Entropy and Gibbs Free Energy

2 Classical defini,on of ΔS = q/t Problem: A calorimeter measured heat of 120J absorbed by a drug specimen while T increased from 300K to 304K. What is the approximate entropy change during that process? 8 kcal/k 36 kj K 0.4 J/K 2.5 SI units of entropy Solu*on 1: ΔS = q/t = 120J/300K = 0.4 J/K Or: ΔS = C p ln(t 2 /T 1 )=C p ln(1+δt/t) C p (ΔT/T)=(q/ΔT) (ΔT/T)=q/T Answer: 0.4 J/K

3 Micro-defini,on of S m =k ln (N) = R ln(n) Problem: One mole of molecules can exist in 10 Na diswnct equiprobable states under the given temperature and pressure condiwons (here N A is the Avogadro number, 6x10 23 ). EsWmate the entropy of the system. 19 kj/k 183 J/K -5 J/(K mol) 85 kj/k -2 J/K Solu*on: Boltzmann definiwon: entropy of one mole equals S = k ln N = R ln n, where n is the number of states of a single molecule 10 Na states for a mole 10 states for a single molecule S = ln 10 = 19 J/K Hint: ln 10 ~ 2.3; ln 20 ~ 3

4 Problem: The standard entropy of formawon of a compound at 0 C equals 150 J/(mol K), and at 100 C 200 J/(mol K). Which expression best approximates the molar heat capacity of this compound in the C temperature range, in J/(mol K) (assume C P is constant)? C P = 50 / 100 C P = 50 x 100 C P = 50 / ln 1.37 C P = 50 x ln 1.37 C P = 50 / ln 100 C P = 50 x ln 100 Solu,on: Kirchhoff s Law: entropy vs T Using Kirchhoff's Law, ΔS = C p ln(t 2 /T 1 ) C P =( )(mol K)/ ln(373/273) = 50/ln 1.37 [J/(mol K) ] H( T2 ) H( T1 ) + CP( T2 T1 ) T 2 S( T + 2) S( T1 ) CP ln T1

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7 Entropy upon protein-ligand binding depends on ligand movement and water shell changes E.g. binding of anw-hiv drugs (HIV protease inhibitors) to their target: Generic name ΔH (kcal/ mol) -TΔS (kcal/mol) ΔS (cal/ mol K) Nelfinavir Indinavir Saquinavir Tipranavir Lopinavir Atazanavir Ritonavir Amprenavir Darunavir

8 Sign of entropy and G changes ΔG= TΔS ΔS>0 More states + ΔS<0 Fewer states MelWng VaporizaWon HeaWng Gas expansion DissociaWon (= increase in the number of parwcles ) Ideal mixing DissoluWon (typically) Freezing CondensaWon Cooling AssociaWon (= decrease in the number of parwcles ) CrystallizaWon (typically)

9 Gibbs Free energy, ΔG, has enthalpic and entropic components ΔG = ΔH TΔS ΔH heat of reacwon v ΔH = H(products) H(reactants) v ΔH < 0 means is exothermic (produces heat) v ΔH > 0 means is endothermic (absorbs heat) ΔS reacwon entropy v ΔS = S(products) S(reactants) v ΔS > 0 means increases disorder (lower ΔG) v ΔS < 0 means increases order (higher ΔG)

10 Spontaneity: ΔG < 0 (not ΔH or ΔS) Problem: The standard entropy of dissoluwon of sodium naproxen in water at 293 K equals 240 J/(mol K). Will the drug spontaneously precipitate from a saturated soluwon at this temperature? the result is unknown because entropy alone does not determine the direcwon of processes the drug will precipitate because the crystallizawon enthalpy is negawve the drug will not precipitate because the dissoluwon entropy is posiwve the drug will not precipitate because the system is at equilibrium Answer: Entropy, ΔS, alone does not determine the direcwon of processes, need to know ΔG = ΔH - TΔS

11 Gibbs free energy and spontaneity Problem: An ice cube is taken from the freezer (-4 C) and placed in a cup with room temperature water (20 C), which iniwates acwve melwng. Which statement about molar Gibbs free energy, G m, of the two phases is correct? G m of water is equal to T G m of ice and entropy gain compensates the enthalpy loss G m of water is equal to G m of ice G m of water is lower than G m of ice G m of water is higher than G m of ice Answer: AcWve spontaneous melwng indicates negawve ΔG, hence G m of water is lower than G m of ice Liquid is indeed the most stable phase of H 2 O at 20 C and P=1atm

12 Binding of drugs to targets: spontaneous with negawve ΔG Signs of molar ΔH bind and ΔS bind may vary ΔG bind = ΔH bind TΔS bind is always negawve Otherwise they would not be drugs Typically -12 to -15 kcal/mol Generic name ΔG (kcal/ ΔH (kcal/ mol) mol) Nelfinavir Indinavir Saquinavir Tipranavir Lopinavir Atazanavir Ritonavir Amprenavir Darunavir TΔS (kcal/ mol)

13 Using equawon ΔG=ΔH TΔS Problem: Lopinavir binding to HIV protease at 25 C is characterized by 1:1 stoichiometry, molar ΔG of kcal/mol and molar ΔH of -3.8 kcal/mol. How does molar entropy change in the binding process? Decreases by 37.9 cal/(k mol) Increases by 37.9 cal/(k mol) Decreases by 11.3 cal/(k mol) Increases by 11.3 cal/(k mol) Remains unchanged Impossible to tell Solu,on: TΔS = ΔG ΔH = kcal/mol (entropy in favor of binding) TΔS = 11.3 kcal/mol > 0 ΔS > 0 ΔS = 11.3 kcal/mol / 298 K = 37.9 cal/(k mol)

14 Phase transiwons: pure substances ΔG trs = 0 ΔH trs = T trs ΔS trs s/l boundary for H 2 O s/l boundary for most substances P = 1 atm T f (1 atm) T b (1 atm) Triple point 3 equally stable phases Normal T boiling and T freezing are where P = 1atm intersects the s/l and l/g boundaries

15 Equilibrium phase transiwon Problem: The melwng temperature of 1:3 mixture of Estradiol- Norethindrone is 25 C, and the crystallizawon enthalpy at this temperature is -8 kj/mol. Which number is closer to the crystallizawon entropy at 25 C? 25 kcal/mol K 2.3 cal/mol K 0.1 cal/k -27 J/(mol K) -90 J/K Answer: Phase transiwon at equilibrium: ΔG trs = 0, so TΔS trs = ΔH trs ΔS trs = ΔH trs /T ~ -8 kj/mol / 298K ~ -27 J/(mol K) CrystallizaWon entropy is negawve, disorder