2)_10_1_&_2_lecture PDF

 Structure and stability of pharmaceutical systems Structure – Equilibrium constants – Dynamic equilibria Stability – System at or not at equilibrium? – System and surrounding respond to reach equilibrium The energy of the system is the key determinant of e...

 Structure and stability of pharmaceutical systems Structure – Equilibrium constants – Dynamic equilibria Stability – System at or not at equilibrium? – System and surrounding respond to reach equilibrium The energy of the system is the key determinant of equilibrium, structure and stability  Energy and equilibrium Need to find the relationship between the energy of a system and the equilibrium constant K for any process First need to clarify what is meant by the energy of a system Pharmaceutical systems very complex; to develop theory better to start from the simplest systems that can be imagined Theory can subsequently be modified to allow for more complex systems  Ideal gases Simplest possible systems: molecules moving at random through space not interacting No chemical interaction between them No defined spatial relationships Hence no need to allow for the energy contributions of interactions between molecules Such systems are known as ideal gases [also assumed that the molecules in ideal gases occupy no space] Some gases behave close to ideal under certain conditions  Studies on ideal gases Studies on the variation of pressure (P), volume (V) and temperature (T) of a fixed amount of an ideal gas Boyles’s law: P vs. V at constant T Charles’ law: V vs. T at constant P Allowed proposal of the ideal gas law  Boyle’s law and Charles’ law ‘∝’ means ‘proportional to’ Robert Boyle (1627-1691) Jacques Charles (1746-1823) 1 𝑃∝ 𝑉∝𝑇 𝑉 at constant P at constant T  Pressure (P), volume (V) and temperature (T) studies on fixed amounts of (assumed ideal) gases 1 Boyle’s law 𝑃∝ at constant T 𝑉 Charles’ law 𝑉∝𝑇 at constant P 𝑃𝑉 Combining gives: = constant 𝑇 For a fixed quantity of ideal gas  Avogadro’s Hypothesis (1811) Samples of different gases at the same pressure (P), volume (V) and temperature (T) contain the same amount of substance Could be generalised to allow a definition of unit of amount of substance: the mole (mol) Definition: 12.0 g of pure 12C isotope contains one mole of carbon atoms Contains 6.022 x 1023 mol−1 of atoms, known as Avogadro’s number Note: is a vast number 1 mol of any pure chemical substance contains Avogadro’s number of entities The quantity (n) of substance is usually given as the number of moles  Relationship between pressure (P), volume (V) and temperature (T) of n moles of an ideal gas 𝑃𝑉 The gas constant = constant = 𝑅 = 8.314 J mol−1 K−1 𝑛𝑇 (value of R depends on the units) PV = nRT the ideal gas law Note: T in Kelvin (K) [0 °C = 273.15 K]  Mixtures of ideal gases Mixture of two ideal gases, A and B, total pressure (P) nA moles of gas A; nB moles of gas B 𝑛𝐴 + 𝑛𝐵 𝑅𝑇 𝑃= 𝑉 PA and PB are known as the partial pressures of A and B  Mole fraction For component A, for example: xA is the mole fraction of component A In general, mole fraction of component i of any mixture is ‘∑’ means the sum of all the terms ni (i.e., the sums of the numbers of moles of all components of the mixture)  PV = nRT problem A 1L cylinder contains 5.00 g of a gaseous anaesthetic (MW 42.1 g mol−1). The cylinder will leak if the pressure exceeds 1.0 x 106 Pa. From what temperature (in ◦C) will the cylinder leak? R = 8.314 Pa m3 K−1 mol−1 1L = 10−3 m3  Tutorial If the mole fraction of oxygen (O2) in air is 0.22, what is the mass of oxygen in 1.0 L of air at 20 °C and 1.0 atm (atmosphere) pressure, assuming ideal behaviour? Use R = 0.082 L atm mol−1 K−1 Need PV = nRT, but solving for n Need to convert °C to K Need Molecular Mass of O2 (32 g mol−1 )  First Law of Thermodynamics “Energy can neither be created nor destroyed” Based on observation and experience Internal energy (U) – An assembly of atoms, ions and/or molecules constitute the system – For an ideal gas, energy due to: translational, rotational and vibrational motion; ‘storage’ of electrons – (No energy due to interactions between molecules of the gas) “The internal energy, U, of an isolated system is constant”  Changes to the internal energy of a system If system in contact with surroundings, when changes in U occur U = Uafter change − Ubefore change] Usystem + Usurroundings = 0 (first law) i.e. Usystem = −Usurroundings Changes to the internal energy of a system Several ways of changing the internal energy of a system But assuming no changes to the phases or components of the system… – (i.e., no reactions generating new components, no processes, such as crystallisation, generating new phases) …energy of the system can only change through: Heat being transferred to/from the system, or Work being done on/by the system Changes to the internal energy of a system Heat and work both forms of energy The amount of energy transferred through heat exchange quantified by q The amount of energy transferred through work being on quantifed by w Leads to another formulation of the First Law: U = q + w  U = q + w Work done on the system (+ve) Change in internal Heat absorbed by the system (+ve) energy of the system ▪Heat: energy transferred between system and surroundings as a consequence of temperature differences between system and surroundings ▪Quantified by value of q ▪Work: energy transferred between system and surroundings capable, in principle, of lifting a weight ▪Quantified by value of w  Relating work (w) to P and V Imagine an ideal gas in a cylinder trapped by a piston Imagine an infinitely small volume change dV Pressure P unchanged Volume change dV at constant pressure P Expansion; work done by the system (-ve) Work done = force × distance = pressure × change in volume, i.e. PdV Expansion of gas from volume Vi to volume Vf at constant pressure P Expansion; work (w) done by the system (-ve) System reaches equilibrium after each infinitismal change dV Work done = force × distance = pressure × change in volume The total work done the summation of all the PdV terms  U = q + w U = q − PV At constant volume, V = 0, so U = qv (subscript ‘v’ implies constant volume ) Most pharmaceutical processes occur at constant pressure U = qp − PV (subscript ‘p’ implies constant pressure)  Enthalpy H U = qp − PV i.e. qp = U + PV Define a new energy function Enthalpy (H) H = U + PV  Enthalpy H Enthalpy H = U + PV For a process at constant pressure (Hf − Hi) = (Uf −Ui) + P(Vf − Vi) H = U + PV [H = qp] The change in enthalpy equals the heat transferred at constant pressure  Thermochemistry Examines heat transfers (enthalpy changes) during several important process E.g. melting of drug substance, binding of drug to receptor, dilution of drug solution, reaction of precursors to form drug substance, etc. E.g. enthalpy of fusion (melting), enthalpy of binding, etc. Enthalpy of formation (Hf): heat absorbed at constant pressure when 1 mole of a compound is formed (at some specified temperature) from its element in their most stable forms Hf⁰ : value of Hf at 25 ⁰C and 1 atmosphere pressure Heat lost from system to surroundings: exothermic process (H -ve) Heat gained by system from surroundings: endothermic process (H +ve)  Thermodynamic state functions The value of a state function depends only on the present condition (state) of the system and not on the history of the system Independent of the pathway by which the present state was obtained U and H are thermodynamic state functions w and q are not (are pathway dependent)  Intensive vs. extensive properties Intensive properties: independent of the size of the system. E.g. pressure (P), temperature (T) Extensive properties: dependent on the size of the system E.g. internal energy (U), enthalpy (H). [units: kJ mol−1] Is the basis of thermal analysis  Hess’s Law “If a process is carried out in several steps, H for the process will equal the sum of the enthalpy changes for the individual steps” E.g. determining Hf⁰ for CS2 (l) given Hf⁰ CS2 (l) = H1⁰ + 2 H2⁰ − H3⁰ Hf⁰ CS2 (l) = (−393.3) + 2(−292.9) − (−1108.8) kJ mol−1 = 129.7 kJ mol−1  Enthalpies of reaction Hrxn = ∑Hf° (products) − ∑ Hf° (reactants) At a specified temperature (298 K in the following example) E.g. C2H4 (g) + Cl2 (g) → C2H4Cl2 (l) Given: Hf° (C2H4 (g) ) 52.5 kJ mol−1, Hf° (C2H4Cl2(l) ) −167.0 kJ mol−1, Hf° (Cl2 (g) ) 0 kJ mol−1 (element) [Hf° for an element in its standard state = 0] Calculate Hrxn Hrxn = (−167.0 ) − [52.5 + 0] = −219.5 kJ mol−1 Tutorial. Thermochemistry problems A. Given the enthalpies of formation (at 298 K) of fumaric acid and maleic acid are −807.6 and −784.4 kJ mol−1 respectively, determine Hrxn for the following process. maleic acid → fumaric acid (at 298 K) B. Glucose (C6H12O6) is metabolised in the presence of oxygen (O2) to give carbon dioxide (CO2) and water (H2O). (i) Write a balanced equation for the reaction of glucose and oxygen to give carbon dioxide and water (ii) Using the following data (305 K), calculate the enthalpy of the process. Hf⁰ C6H12O6 (s) = −1273.0 kJ mol−1 Hf⁰ O2 (g) = 0 kJ mol−1 (element) Hf⁰ H2O (l) = −285.8 kJ mol−1 Hf⁰ CO2 (g) = −393.5 kJ mol−1  The direction of spontaneous change The first law of thermodynamics describes the energy balance in a given process (Key parameter is the change in enthalpy ΔH) But it does not indicate the preferred direction of the process i.e., the direction of spontaneous change To predict the position of equilibrium for a process, need ΔH, but also need to know the direction of spontaneous change  Spontaneous processes By observation: many processes only occur in one direction – Heat flows from a hotter body to a colder body, not vice versa – Gas molecules initially separated mix randomly but not the reverse, i.e.: Partition removed Reverse process not observed spontaneously Process can occur in the reverse direction but requires intervention  Entropy Entropy (S): degree of disorder or randomness of a system Higher S implies greater disorder; lower S greater order (or organisation) Entropy is a thermodynamic state function The change in entropy (ΔS) for a process is the key parameter for determining the direction of spontaneous change Entropy (S) vs. temperature for a single component Note entropy decreases as temperature decreases i.e. going from less ordered to more ordered phases Large drops in entropy at phase transitions Within phases, some decrease in entropy (increasing order) with decreasing temperature Entropy is a thermodynamic state function S for a particular process is independent of path S at 0 K = 0 J mol−1 K−1 (3rd Law of Thermodynamics) Energy; relating temperature, heat and entropy Any form of energy can by described as an intensity factor × a capacity factor The intensity factor is the driving force The capacity factor is the extent over which the force is conveyed E.g., for mechanical energy: force (intensity) × distance (capacity) or pressure (intensity) × volume (capacity) For electrical energy: potential (intensity) × quantity of charge (capacity)  Heat (q) as a form of energy Likewise can be described as an intensity factor multiplied by a capacity factor The intensity factor is the temperature (T) The capacity factor is the entropy (S) Hence: q = T × S For a given process, the change in entropy is given by S 𝑞 ∆𝑆 = 𝑇  Reversible and irreversible processes Spontaneous processes: system and surroundings not at equilibrium Process continues until equilibrium reached: irreversible process Reversible process: system and surroundings at equilibrium Dynamic equilibrium [system and surroundings at equilibrium after each infinitesimal change]  Reversible transfer of heat E.g. ice and water in equilibrium at 0 ◦C Temperature of ice and water in contact both at 273.15 K during melting or freezing Hmelting = 6.01 kJ mol−1 (equals the amount of heat q absorbed from the surroundings when 1 mol of ice melts reversibly) (H = q for a process at constant pressure) 𝑞𝑟𝑒𝑣 ∆𝑆 = = (6.01 kJ mol−1)/(273.15 K) = 22.0 J mol−1 K−1 𝑇 Heat gained by the ice equals the heat lost by the surrounding water, so S for both the system and surroundings are equal No overall change in entropy of the universe  Irreversible transfer of heat Transfer of heat from a hot object (Thot) to a cold object (Tcold) Hot object loses heat (−q), cold object gains heat (q) 𝑞 −𝑞 ∆𝑆𝑐𝑜𝑙𝑑 = ∆𝑆ℎ𝑜𝑡 = 𝑇𝑐𝑜𝑙𝑑 𝑇ℎ𝑜𝑡 Thot > Tcold, so 𝑞 −𝑞 ∆𝑆𝑢𝑛𝑖𝑣𝑒𝑟𝑠𝑒 = ∆𝑆𝑐𝑜𝑙𝑑 + ∆𝑆ℎ𝑜𝑡 = + = +𝑣𝑒 𝑛𝑢𝑚𝑏𝑒𝑟 𝑇𝑐𝑜𝑙𝑑 𝑇ℎ𝑜𝑡  Second law of thermodynamics 𝑞𝑟𝑒𝑣 Reversible process: ∆𝑆 = 𝑇 𝑞𝑖𝑟𝑟𝑒𝑣 Irreversible process: ∆𝑆 > 𝑇 𝑞 ∆𝑆 ≥ 𝑇 This is the second law of thermodynamics In words: “In a spontaneous process, the entropy of the universe increases”