Entropy. ΔS = q rev T. CLASSICAL THERMODYNAMICS study of macroscopic/thermodynamic properties of systems: U, T, V, P,

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1 Entropy Rudolf Clausius 1 st law: ΔU = q - w A new State Func,on: ΔS = q rev CLASSICAL HERMODYNAMICS study of macroscopic/thermodynamic properties of systems: U,, V, P, Ludvig Boltzmann S = k lnn SAISICAL HERMODYNAMICS establishing relationships between microstates and macrostates

2 Entropy as a measure of number of possibili,es

3 Entropy : coun,ng microstates the observed macrostate looks like a typical microstate with the same U macro- state (eg phase) consists of N microstates Microstates are combined from molecular microstates (n micro ) S = k B ln N microstates N forna molecules = n N A Phase A n micro =1 Phase B n micro =10 1 molecule 1 molecule lna x = x ln a ΔS mole = Rln10 AB S mole = k B ln N = k B N A ln n = Rlnn

4 Classical Entropy and Units ΔS = q Entropy Change of Revers. Process = Small Heat over Absolute emperature Units: Joule/K (or cal/k ) Warnings: he units of Entropy are the same as Heat Capacity, however ΔS and C are totally different: C = q Δ Heat Capacity of material = Heat over Small emperature Change

5 Entropy Change in an Irreversible Process the Second Law of hermodynamics ΔS 0 : in an isolated system S is increasing as it is reaching its equilibrium maximum value B dq < For an irreversible transi,on from state A to state B: A Irreversible spontaneous no work required ΔU = q Clausius: No process is possible whose sole result is the transfer of heat from a body of lower temperature to a body of higher temperature Kelvin: No process is possible in which the sole result is the absorp,on of heat from a reservoir and its complete conversion into work B A dq rev = ΔS

6 Entropy Changes in Specific Processes Changes in volume or pressure (e.g. in isothermal expansion of ideal gas). Larger volume, more space microstates, N~V, S ~ lnv Changes in temperature (illustrated by isobaric hea,ng of ideal gas) PV=nR. More microstates for energy values ~. N ~ S ~ ln Phase changes (at the transi,on temperature) Entropy of mixing (bigger volume for each molecule)

7 Entropy. Gas. Classical method: S ( Volume ) at =const V A =const ΔU=0, w=q Increase in entropy V B S = nr lnv # ΔS = nr ln V B % $ V A & ( ' Classical Deriva,on (op,onal) Consider ideal gas isothermally expanding from V A to V B. ΔU = 0 because the internal energy for ideal gas only depends on the temperature As ΔS does not depend on path, choose a reversible path VB VB VB VB Entropy change: ΔS = rev = dqrev PdV dv = = = V V V V V A V A dq 1 A 1 A A V A nr nr ln( V ) " q rev = w = nr ln V % " B $ ' ΔS = nr ln V % " B $ ' = -nr ln P % B $ ' # & # & # & P A V V B A

8 A Entropy. Gas. S(emperature), P=const Increase in entropy Classical Deriva,on (op,onal) Consider ideal gas heasng from A to B at constant pressure or any other system where C p does not change between A and B As ΔS does not depend on path, choose a reversible path: Entropy change: B dq rev = C P d ΔS = B A dq rev = B A C P S = C P ln # ΔS = C P ln & B % ( $ ' d A dx note : = ln x x If C P ~ constant Note for n moles C P = nc m P ΔS P = C P B A d = C P ln B A

9 Kirchhoff rules extended to ΔS Now we can calculate both H and S from C p at 2 (Kirchhoff) ) ( ) ( ) ( C H H P ln ) ( ) ( C S S P ) ( ln C H C d C dq S P P P rev Δ = = Δ he same formula is used to measure entropy in a calorimeter ) ( ) ( C S S P For Δ << 1

10 Entropy in Sta,s,cal hermodynamics Entropy is a measure of disorder, or randomness S = k Σ p i ln p i, where: p i is the probability of the microstate i k (k B ) is the Boltzmann constant If the microstates are equiprobable: S = k ln N where N is the number of microstates for the ensre system (dissnguishable ways the system can be put together with given U and V), N mole = n Navogadro S mole =R ln (n ) if n is the number of microstates for ONE MOLECULE. ΔS m 1->2 = R ln (n 2 / n 1 )

11 Entropy and Growing Crystals - I HCl hydrogen chloride Low entropy perfect crystals Once the 1 st molecule is in place, there is only one way to put the crystal together: At = 0, N = 1 S = k ln N = k ln 1 = 0

12 Entropy and Growing Crystals - II CO carbon oxide Higher entropy imperfect (rota,onally disordered) crystals Once the 1 st molecule is in place, there are two ways to posi,on the 2 nd molecule, for each of these, two ways to posi,on the 3 rd molecule etc: N = 2 n At = 0, S = k ln N = k ln 2 n = nk ln 2 For a mole of CO, S = N A k ln 2 = R ln 2 (R gas constant)

13 Entropy Changes: Examples Configura,onal Entropy of one molecule (per mole): Example: the number of states N in one molecule, with n rotatable bonds with 3 equiprobable states S = R ln (N) N = 3 n S = Rn ln(3)

14 Entropy is Addi,ve Entropy (disorder) of two parts, A and B S A = k ln n A, S B = k ln n B, Entropy of both parts, A + B: he number of states: n AB = n A n B S AB = k ln( n A n B ) = k ln n A + k ln n B = S A +S B Entropy is ADDIIVE, since lnxy=lnx+lny For the Avogadro number of molecules with n states: S mole = N A k ln n = R ln n

15 Entropy Summary S = k ln N for any number of molecules, N is the total number of combina,ons of states for all molecules. N = n Na for 1 mole (Na) of molecules, n is the number of states per MOLECULE S for one mole: S m = k ln N = k Na ln(n)=rln(n) S for c moles is cs m ΔS m 1- >2 = R ln (n 2 /n 1 )

16 Es,ma,ng ln(1+x) and more In many problems you need to es,mate expressions looking like ln(300k/280.) ln(1.07) or 1/1.2, etc. o evaluate them use a simple technique based on the aylor expansion: 2 ' x '' f ( a + x) = fa + xf a + fa n x n! f ( n) a ln(1 + e x (1 + x) x) 1+ x n x 1+ nx Examples : n = 1, 1 1 x 1+ x 1 (1 + x) 1+ x 2 2 (1 + x) 1+ 2x 1 2,2 E. g. ln(1.07) 0.7

17 Reaction/transition thermochemistry In a reaction: Reactants (R) Products (P) ΔH heat of reaction v ΔH = H(products) H(reactants) v ΔH < 0 means is exothermic (produces heat) v ΔH > 0 means is endothermic (absorbs heat) Heat can be measured in a (micro)calorimeter, and used to: v Characterize the reaction (Endothermic/ exothermic? Molar ΔH? Which forces are likely involved?) v Find the extent of reaction (how much product formed?) IC = isothermal titration calorimetry

18 Exothermic vs endothermic Problem: Calorimetric data on binding of several anti-hiv drugs to their common target, HIV protease, at =300K is presented in the table. For how many drugs in the list, the binding reaction is endothermic? v 1 v 2 v 3 v 4 v 5 v 6 Answer: 3 drugs. Endothermic means ΔH > 0, which is the case for Indinavir, Saquinavir, and Nelfinavir Drug ΔH binding (kcal/mol) Nelfinavir 3.1 Indinavir 1.8 Saquinavir 1.2 ipranavir -0.7 Lopinavir -3.8 Atazanavir -4.2 Ritonavir -4.3 Amprenavir -6.9 Darunavir -12.7

19 (Quantitative) ΔH of reaction/transition Reaction Equation ΔH m (react) = molar ΔH(react) = heat produced/absorbed when Dissolution D solid + H 2 O D aq 1 mole of drug/substance dissolves in infinite amount of water Neutralization OH - + H + H 2 O 1 mole of OH - interacts with 1 mole of H + to form 1 mole of H 2 O 1:1 protein/ligand binding P + L PL 1 mole of L binds to 1 mole of P to form 1 mole of PL complex ΔH(react) = ΔH m (react) n n is the extent of reaction (# of moles): Dissolution: n = amount of substance dissolved Neutralization: n = amount of limiting reactant Protein/ligand binding: n = amount of complex (not necessarily equal to limiting reactant!) n = ΔH/ ΔH m

20 Acid-base neutralization heat Problem: Molar enthalpy of neutralization reaction between HCl and NaOH at 25 C is kj/mol. When 25 mmol HCl was added to 10 ml of NaOH solution in a calorimeter, the evolved heat was measured to be 1 kj. What was the concentration of NaOH in the solution? v 25 mm v 25 M v 0.55 M v 1.79 M v 17.9 mm Solution: If 25 mmol of HCl were completely neutralized by NaOH, ΔH would be equal to mol x kj/mol ~ -1.4 kj. However, we only observed heat of 1 kj. v NaOH was a limiting reactant v Only 1/55.9 = mol of reaction product (water) was formed v herefore 17.9 mmol NaOH was present in the solution in the beginning v his corresponds to NaOH concentration of 1.79 M. Answer: 1.79M Note: Acid/base neutralization involves formation of H2O from OH- and H+; covalent bonds are formed, exothermic.

21 Protein-ligand binding equilibrium binding dissociation 1 mole of protein 1 mole of ligand unbound 1 mole of protein/ligand complex bound

22 Protein-ligand binding equilibrium binding dissociation 1 mole of protein 1 mole of ligand unbound 0.4 mole of unbound 1 mole protein of protein/ligand complex 0.4 mole of unbound ligand bound 0.6 mole of bound complex EQUILIBRIUM

23 Enthalpy of drug binding - I Problem: 0.3 ml of 1mM solution of a drug is added to a tube with a protein solution. Some of the drug reacts with the protein to form 1:1 protein-drug complex. he reaction produces a heat of 2.4 x 10 3 cal. Estimate the molar enthalpy of binding. v 8 kcal/mol v 12 kcal/mol v 24 kcal/mol v 3.6 x 10 3 cal/mol v impossible to tell because the extent of reaction is not known Solution: he total molar amount of drug is 0.3 µmol but how much of it reacted? And how much product (complex) was formed, in moles? v If n = 0.3 µmol, ΔH m = ΔH/n = 2.4 x 10 3 cal / (0.3 x 10 6 mol) = 8 kcal/mol v If n = 0.2 µmol, ΔH m = ΔH/n = 2.4 x 10 3 cal / (0.2 x 10 6 mol) = 12 kcal/mol v If n = 0.1 µmol, ΔH m = ΔH/n = 2.4 x 10 3 cal / (0.1 x 10 6 mol) = 24 kcal/mol Answer: impossible to tell because the extent of reaction is not known v However, we know that the reaction is exothermic v And that ΔH m 8 kcal/mol

24 Enthalpy of drug binding - II Problem: 0.3 ml of 1mM solution of a drug is added to a tube with a protein solution, and the drug reacts entirely with the protein to form a 1:1 protein-drug complex (no unbound drug is left). he reaction produces a heat of 2.4 x 10 3 cal. Estimate the molar enthalpy of binding. v 8 kcal/mol v 12 kcal/mol v 12 kcal/mol v 3.6 x 10 3 cal/mol v impossible to tell because the extent of reaction is not known Solution: v he total molar amount of drug is 0.3 µmol v All of it reacted, so 0.3 µmol of complex was formed (n = 0.3 µmol) v ΔH m = ΔH/n = 2.4 x 10 3 cal / (0.3 x 10 6 mol) = 8 kcal/mol Answer: 8 kcal/mol Note: It is important that all 0.3 µmol of drug was bound

25 ΔH, C p, and heating matter C p is sample heat capacity at constant P J/K or cal/k C p,m is molar heat capacity J/(K mol) or cal/(k mol) C p = n C p,m where n is the number of moles Molar ΔH m = C p,m Δ otal ΔH = C p Δ = n C p,m Δ Specific heat capacity (C p per unit mass) may be useful. WHA IS I FOR H 2 O? Calculation assumes that C p is not changing within the temp. range of the experiment. Δ = ΔH/(n C p,m ) C p,m = ΔH/(n Δ)

26 Dissolution heat and changes Problem: After dissolving some amount of Lidocaine HCl in 1 ml pure water at room temperature, the ampule cooled down by 1.04 K. How much drug was dissolved? Molar enthalpy of Lidocaine HCl dissolution is 43.5 kj/mol. Consider the density and the heat capacity of the solution approximately equal to that of pure water. v 1 mole v 100 mmoles v 10 mmoles v 1 mmole v 100 µmoles Solution: wo processes: (1) dissolution of Lidocaine and (2) heating water v Specific heat capacity of water is J/(g K) v o decrease temperature of 1 ml of water by 1.04K, heat of J/(g K) x 1.04K x 1g ~ 4.35 J must be absorbed v his corresponds to dissolution of 4.35 J / 43.5 kj/mol ~ 0.1 mmole of the drug Answer: 100 µmoles

27 Heat capacity of gases C p is (molar/specific/sample) heat capacity at constant P PV = nr: v isochoric/isovolumetric heating: V = const, P increases (e.g. heating in a closed container) v isobaric heating: P = const, V increases Useful to have C v (molar/specific/sample) at constant volume For liquid and solid samples, C p ~ C v For gaseous samples, C p ~ C v + nr v Molar: C p,m ~ C v,m + R

28 Heat capacities Problem: Estimate the specific heat capacity at constant pressure and room temperature of nitrogen gas, N 2, in J/(g K). Assume that Cv =20.8 J /(mol K). MW(N 2 ) = 28 g/mol. v 0.44 J /(g K) v ~1 J /(g K) v 20.8 J /(g K) v 29.1 J /(g K) v impossible to tell Solution: v N 2 is a diatomic gas, so each molecule has 5 degrees of freedom (bond vibrations are not excited). v Molar heat capacity at constant pressure is C p,m = C v,m + R = 20.8 J /(mol K) J /(mol K) = 29.1 J /(mol K) v Specific heat capacity at constant pressure is C p,m / MW = 29.1 / 28 = 1.04 J/(g K). Answer: ~1 J/(g K)

29 hermodynamic cycle & Hess law he enthalpy change for a reaction carried out in a series of steps is equal to the sum of the enthalpy changes for the individual steps. In a reaction: Reactants (R) Products (P) Enthalpy change for the reaction can be found from enthalpies of formation of R and P: ΔH f (R) + ΔH r = ΔH f (P) Reactants ΔH f (R) ΔH r Products ΔH f (P) Elements

30 hermodynamic cycle Problem: Standard enthalpy of formation of ammonia at 25 C is kj/mol for its aqueous form and kj/mol for its gaseous form. What is the enthalpy of dissolution of 2 moles of gaseous ammonia in water at 25 C? v kj v 34.9 kj v kj v 69.8 kj v kj v kj Solution: Molar enthalpy of dissolution is the difference between standard enthalpies of formation: v (-45.9) = kj/mol. v he enthalpy of dissolution of 2 moles is 2 x = kj. v Ammonium dissolution is exothermic! Answer: kj.

31 Kirchhoff s Law Relates enthalpies at different temperatures H 2 H 1 = C p ( 2 1 ), v assuming that C p is approx. constant within Δ H 2 is H 1 plus energy needed to heat the sample. Find enthalpy at a different temp: v H 2 = H 1 + C p ( 2 1 ) Find heat capacity within the given temp. range: v C p = (H 2 H 1 ) / ( 2 1 ) Find a new temp given enthalpy: v 2 = 1 + (H 2 H 1 ) / C p

32 Kirchhoff s Law Problem: Molar enthalpies of formation of gaseous CO 2 are -393 kj/mol at 293K and kj/mol at 323K. Approximately how much heat is needed to raise the temperature of a 3 mole sample of CO 2 from 30 C to 40 C at constant pressure? v 1.11 kj v ~1 J v 37 J v -10 kj v 1179 kj Solution 1: Δ f H 1, Δ f H 2, Kirchhoff's Law molar hear capacity of CO 2 at constant pressure and C: v C p,m = ( kj/mol+393 kj/mol)/30k ~ 37 J/(mol K) v Heating a 3 mole sample by 10 C will require: Δ x n x C p,m = 10 K x 3 mol x 37 J/(mol K) ~ 1.11 kj of heat Solution 2: Heating 3 moles by 10 C is ~ the same as heating 1 mole by 30 simply take the difference Δ f H 2 - Δ f H 1 = 1.11 kj Answer: 1.11 kj