Exothermic vs endothermic

Transcription

1 Exothermic vs endothermic Problem: Calorimetric data on binding of several anti-hiv drugs to their common target, HIV protease, at T=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 Tipranavir -0.7 Lopinavir -3.8 Atazanavir -4.2 Ritonavir -4.3 Amprenavir -6.9 Darunavir -12.7

2 (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

3 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 Therefore 17.9 mmol NaOH was present in the solution in the beginning v This 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.

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

5 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

6 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. The 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: The 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

7 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 2.4 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 The 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

8 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 ΔT Total ΔH = C p ΔT = n C p,m ΔT Specific heat capacity (C p per unit mass) may be useful. WHAT IS IT FOR H 2 O? Calculation assumes that C p is not changing within the temp. range of the experiment. ΔT = ΔH/(n C p,m ) C p,m = ΔH/(n ΔT)

9 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 43.5 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: Two processes: (1) dissolution of Lidocaine and (2) heating water v Specific heat capacity of water is J/(g K) v To 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 This corresponds to dissolution of 4.35 J / 43.5 kj/mol ~ 0.1 mmole of the drug Answer: 100 µmoles

10 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 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

11 Thermodynamic 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 The enthalpy of dissolution of 2 moles is 2 x = kj. v Ammonium dissolution is exothermic! Answer: kj.

12 Kirchhoff s Law Relates enthalpies at different temperatures H T2 H T1 = C p (T 2 T 1 ), v assuming that C p is approx. constant within ΔT H T2 is H T1 plus energy needed to heat the sample. Find enthalpy at a different temp: v H T2 = H T1 + C p (T 2 T 1 ) Find heat capacity within the given temp. range: v C p = (H T2 H T1 ) / (T 2 T 1 ) Find a new temp given enthalpy: v T 2 = T 1 + (H T2 H T1 ) / C p