* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Download Sign of enthalpy changes Exothermic vs endothermic Acid
Host–guest chemistry wikipedia , lookup
Chemical thermodynamics wikipedia , lookup
Multi-state modeling of biomolecules wikipedia , lookup
Cooperative binding wikipedia , lookup
Electrolysis of water wikipedia , lookup
Lewis acid catalysis wikipedia , lookup
Chemical equilibrium wikipedia , lookup
Click chemistry wikipedia , lookup
Copper in heat exchangers wikipedia , lookup
Clinical neurochemistry wikipedia , lookup
Heat transfer wikipedia , lookup
Transition state theory wikipedia , lookup
Ligand binding assay wikipedia , lookup
Drug design wikipedia , lookup
Heat capacity wikipedia , lookup
Stability constants of complexes wikipedia , lookup
Organosulfur compounds wikipedia , lookup
Sign of enthalpy changes Reaction/transition thermochemistry • In I a reaction: ti Reactants (R) Products (P) • H heat of reaction H = H(products) – H(reactants) H < 0 means “” is exothermic (produces heat) H > 0 means ““”” iis endothermic d th i (absorbs ( b b h heat) t) H – + –TS TS heat is evolved exothermic heat is absorbed endothermic – • Heat can be measured in a ((micro)calorimeter, ) , and used to: Characterize the reaction (Endothermic/exothermic? Molar H? Which forces y involved?)) are likely Find the extent of reaction (how much product formed?) + • ITC = isothermal titration calorimetry Surroundings warm up • • • • • Freezing Condensation Acid/base neutralization Burning Drug/protein binding (typically) Surroundings cool down • • • Melting Evaporation Crystal dissolution (typically) http://microcal.com/technology/itc.asp Exothermic vs endothermic • Problem: Calorimetric data on binding of several anti-HIV drugs to their common target, target HIV protease, protease at T=300K is presented in the table. For how many drugs in the list, the binding reaction is endothermic? Drug Nelfinavir Indinavir 1 Saquinavir 2 Tipranavir 3 L i Lopinavir i 4 Atazanavir 5 Ritonavir 6 Amprenavir • Answer: 3 drugs. Endothermic means H > 0, which is the case for Darunavir I di Indinavir, i S Saquinavir, i i and dN Nelfinavir lfi i H binding (kcal/mol) 3.1 18 1.8 1.2 -0.7 -3.8 38 -4.2 -4.3 -6.9 -12.7 Acid-base neutralization heat • • • Reaction Equation Hm(react) = molar H(react) = heat produced/absorbed when… Dissolution Dsolid + H2O Daq 1 mole of drug/substance g dissolves in infinite amount of water Neutralization OH- + H+ H2O 1 mole of OH- interacts with 1 mole of H+ to form 1 mole of H2O 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 • H(react) = Hm(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 = H/ Hm Protein-ligand binding equilibrium Problem: Molar enthalpy of neutralization reaction between HCl and NaOH at 25C is -55.9 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 kJ. What was the concentration of NaOH in the solution? • (Quantitative) H of reaction/transition 25 mM 25 M 0.55 M 1.79 M 17.9 mM binding Solution: If 25 mmol of HCl were completely neutralized by NaOH, H would be equal to 0.025 mol x -55.9 kJ/mol ~ -1.4 kJ. However, we only observed heat of 1 kJ. NaOH was a limiting reactant Only 1/55.9 1/55 9 = 0.0179 0 0179 mol of reaction product (water) was formed Therefore 17.9 mmol NaOH was present in the solution in the beginning This corresponds to NaOH concentration of 1.79 M. Answer: 1.79M 1 79M Note: Acid/base neutralization involves formation of H2O from OH- and H+; covalent bonds are formed, exothermic. dissociation 1 mole of protein 1 mole of ligand unbound 1 mole of protein/ligand complex bound Enthalpy of drug binding - I Protein-ligand binding equilibrium • binding dissociation 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. The reaction produces a heat of –2.4 x 10–3 cal. Estimate the molar enthalpy of binding binding. • –8 kcal/mol –12 kcal/mol 24 kcal/mol 3.6 x 10–3 cal/mol impossible to tell because the extent of reaction is not known Solution: The total molar amount of drug is 0.3 mol – but how much of it reacted? And how much product (complex) was formed, in moles? If n = 0.3 mol, Hm = H/n = –2.4 x 10–3 cal / (0.3 x 10–6 mol) = –8 kcal/mol If n = 0.2 mol, Hm = H/n = –2.4 x 10–3 cal / (0.2 x 10–6 mol) = –12 kcal/mol If n = 0.1 mol, Hm = H/n = –2.4 x 10–3 cal / (0.1 x 10–6 mol) = –24 kcal/mol • 1 mole of protein 1 mole of ligand unbound mole of protein protein/ligand complex 0.4 0 4 mole of 1unbound bound 0.4 mole of unbound ligand 0.6 mole of bound complex EQUILIBRIUM 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). The reaction produces a heat of –2 2.4 4 x 10–3 cal. cal Estimate the molar enthalpy of binding. • –8 kcal/mol –12 kcal/mol 12 kcal/mol 3.6 x 10–3 cal/mol impossible to tell because the extent of reaction is not known Solution: The total molar amount of drug is 0.3 mol All of it reacted, so 0.3 mol of complex was formed (n = 0.3 mol) Hm = H/ H/n = –2.4 2 4 x 10–33 call / (0.3 (0 3 x 10–66 mol) l) = –8 8 kkcal/mol l/ l • • Answer: –8 kcal/mol Note: It is important that all 0.3 mol of drug was bound Dissolution heat and T 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 43.5 5 kJ/mol kJ/mol. Consider the density and the heat capacity of the solution approximately equal to that of pure water. • 1 mole 100 mmoles 10 mmoles 1 mmole 100 moles Solution: Two processes: (1) dissolution of Lidocaine and (2) heating water Specific heat capacity of water is 4.184 J/(g K) To decrease temperature of 1 ml of water by 1 1.04K, 04K heat of 4 4.184 184 J/(g K) x 1.04K x 1g ~ 4.35 J must be absorbed This corresponds to dissolution of 4.35 J / 43.5 kJ/mol ~ 0.1 mmole of the drug • Answer: 100 moles Answer: impossible to tell because the extent of reaction is not known However, we know that the reaction is exothermic And that Hm –8 kcal/mol H, Cp, and heating matter • Cp is sample heat capacity at constant P J/K or cal/K • Cp,m is molar heat capacity J/(K mol) or cal/(K mol) Cp = nCp,m p m where n is the number of moles • Molar Hm = Cp,m T • Total H = Cp T = nCp,m p m T • Specific heat capacity (Cp per unit mass) may be useful. WHAT IS IT FOR H2O? • Calculation assumes that Cp is not changing within the temp. range of the experiment. • T = H/(nC H/(n Cp,m) • Cp,m= H/(nT) Heat capacity of gases • Cp is (molar/specific/sample) heat capacity at constant P • PV = nRT: isochoric/isovolumetric heating: V = const, P increases (e.g. (e g heating in a closed container) isobaric heating: P = const, V increases • Useful to have Cv (molar/specific/sample) at constant volume • For liquid and solid samples, Cp ~ Cv • For gaseous samples, Cp ~ Cv + nR Molar: Cp,m ~ Cv,m + R DF vs Cv,m Experimentally measured Cv,m at 298K Gas CV,m, J/(mol·K) CV,m/R # DF He ● 12.5 1.5 3 Ne ● 12.5 1.5 3 Ar ● 12.5 1.5 3 Kr ● 12.5 1.5 3 Xe ● 12.5 1.5 3 H2 ●–● 20.18 2.43 5 CO ●–● 20.2 2.43 5 N2 ●–● 20.8 2.50 5 O2 ●–● 21.03 2.53 5 Cl2 ●–● 24.1 3.06 ~6 ~7 Br2 (v) ●–● 28.2 3.39 H2O (v)* ●●● 28.49 3.43 ~7 CO2 ●-●-● 28.5 3.43 ~7 CH4 ●●● ● 27.1 3.26 ~ 6-7 DF and heat capacities • # of DF determines heat capacity of a substance at constant volume: – Each rotational/translational DF contribute R/2 to Cv,m – Each (excited) vibrational mode contributes up to R to Cv,m • Monoatomic gases have 3 DF: Cv,m = 3/2R • Diatomic gases have 5 DF b l below Tvib • P Problem: bl E ti t the Estimate th specific ifi heat h t capacity it att constant t t pressure and room temperature of nitrogen gas, N2, in J/(g K). Assume that vibrational modes of N2 molecules are not excited it d att room ttemperature. t MW(N2) = 28 g/mol. / l • Solution: N2 is a diatomic gas, so each molecule has 5 degrees of freedom (bond vibrations are not excited). Molar heat capacity at constant volume is Cv,m v m = 5/2R ~ 20.8 J /(mol K). Molar heat capacity at constant pressure is Cp,m = Cv,m + R = 20.8 J /(mol K) + 8.314 J /(mol K) = 29.1 J /(mol K) Specific heat capacity at constant pressure is Cp,m / MW = 29.1 / 28 = 1.04 J/(g K). – for H2, CO, N2, Tvib >> 298K • When T Tvib, vibrational DF’s appear * Value for H2O at 373K – for Cl2, Br2, I2, Tvib 298K 0.44 J /(g K) ~1 J /(g K) 20 8 J /(g K) 20.8 29.1 J /(g K) impossible to tell • Answer: ~1 J/(g K) Thermodynamic cycle Thermodynamic cycle & Hess law • The 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 py change g for the reaction can be found from enthalpies of formation of R and P: Hf(R) + Hr = Hf(P) Products • Problem: Standard enthalpy of formation of ammonia at 25C is -80.8 kJ/mol for its aqueous form and -45.9 kJ/mol for its gaseous form. What is the enthalpy of dissolution of 2 moles of gaseous ammonia in water at 25C? +34.9 kJ –34.9 kJ +69.8 kJ –69.8 kJ +126.7 kJ –126.7 kJ Hr Reactants Hf (P) Hf (R) • Solution: Molar enthalpy of dissolution is the difference between standard enthalpies of formation: -80.8-(-45.9) = -34.9 kJ/mol. The enthalpy of dissolution of 2 moles is 2 x -34.9 = -69.8 kJ. Ammonium dissolution is exothermic! Elements • Answer: -69.8 kJ. Kirchhoff’s Law Kirchhoff’s Law • Relates enthalpies at different temperatures • HT2 – HT1 = Cp(T2 –T1), • assuming that Cp is approx. constant within T • HT2 is HT1 plus energy needed to heat the sample. • Find enthalpy at a different temp: HT2 = HT1 + Cp(T2 –T1) • • Find heat capacity within the given temp. range: 1.11 kJ ~1 J 37 J -10 kJ 1179 kJ Solution 1: fHT1, fHT2, Kirchhoff's Law molar hear capacity of CO2 at constant pressure and 20 20-50C: 50C: Cp,m = (-391.89 kJ/mol+393 kJ/mol)/30K ~ 37 J/(mol K) Heating a 3 mole sample by 10C will require: T x n x Cp,m = 10 K x 3 mol x 37 J/(mol K) ~ 1.11 1 11 kJ of heat Cp = (HT2 – HT1) / (T2 –T1) = H / T • Find a new temp given enthalpy: Problem: Molar enthalpies of formation of gaseous CO2 are -393 393 kJ/mol at 293K and -391.89 kJ/mol at 323K. Approximately how much heat is needed to raise the temperature of a 3 mole sample of CO2 from 30C to p 40C at constant pressure? • Solution 2: Heating 3 moles by 10C is ~ the same as heating 1 mole by 30 • Answer: 1.11 kJ T2 = T1 + (HT2 – HT1) / Cp simply p y take the difference fHT2 - fHT1 = 1.11 kJ