PHA111 Thermodynamics-2.pptx

MPharm Programme Thermodynamics MPharm PHA111 Overview • Bridging Kinetics and Thermodynamics • Law of Conservation of Energy • 1st, 2nd & 3rd Laws of TD • Internal Energy, Enthalpy & Entropy • Hess’ Law • Equilibria, Gibbs Free Energy • Non-covalent forces & Binding MPharm PHA111 Recommended Texts • why chemical reactions happen James Keeler & Peter Wothers Oxford University Press 2003 • Physical Chemistry for the Life Sciences Peter Atkins & Julio de Paula, OUP 2006 MPharm PHA111 Free Energy (G) has two components, Enthalpy (H) and Entropy (S) • G = H – TS ∆G = ∆H – T∆S ∆Gⱡ = ∆Hⱡ – T∆Sⱡ • Changes in enthalpy deal with changes in chemical bonding or non-covalent interactions. – -dH associated with forming new, stronger or more stable covalent bonds in a chemical reaction – Or/and, new, favorable solvent interactions (non-covalent) – Entropy has a higher contribution to free energy at higher temperatures (TdS) – Enthalpy has a larger contribution at low temperatures • Changes in entropy deal with changes in order or disorder associated with the process of interest. MPharm PHA111 Energy comes in two main types • Kinetic energy: energy associated with movement • Classically: EK = ½ mass .velocity2 • Potential energy: energy possessed due to position • EP = mass.gravity.height • Total Energy, E = EK +EP MPharm PHA111 Conservation of energy • Energy can be neither created or destroyed • Energy is transferred from one place to another, in varying forms until it is dissipated in the form of heat energy (non-useable, high entropy) • Q. What does the sun give us? Slide 7 of 29 MPharm PHA111 Thermodynamics: The First Law Two very important molecules • NADH is a highly reduced species, perfect for complementing oxidation reactions. • ATP releases energy during phosphate hydrolysis reactions. MPharm PHA111 Systems and Surroundings • System: the vessel of interest. • Reactions flasks, biological cells or whole animals. • Surroundings: place where we record our observations from. • Surroundings are thought have constant volume or pressure. • This means that surroundings remain MPharm PHA111 System Types – Which system are YOU? • Open: can exchange energy and matter with surroundings. • Closed: can exchange energy but not matter with surroundings. • Isolated: exchanges neither energy nor matter with surroundings. MPharm PHA111 Heating • Heating is energy transfer between system and surroundings. • Diathermic barriers eg metals, skin, and biological membranes allow efficient energy transfer. • Adiabatic barriers prevent transfer of energy even if significant temperature differences exist. MPharm PHA111 Zeroth Law A B A C B C If A is in thermal equilibrium with B, and B is in thermal equilibrium with C, then A will also be in thermal equilibrium with C MPharm PHA111 A molecular description of work and heat Useful , ordered energy Non-useful , random energy MPharm PHA111 Ok, back to basics for a minute...... • How much work does it take to move 10g of water to the leaves of a 20m tall tree from its roots? • Work = mgh = 0.01kg x 9.81ms-2 x 20m = 2.0J • How much work does it take to raise a 1kg textbook by 20cm? • Work = mgh = 1kg x 9.81ms-2 x 0.20m = 2.0J • In thermodynamics work has the symbol w, and heat q. MPharm PHA111 Sign Conventions for w and q • By convention if a system does work w has a negative value: eg raising a weight. • If a system has work done upon it w is positive: eg stretching a muscle or elastic band. • Similarly, q is negative if the system heats its surroundings. MPharm PHA111 Work w< 0 Heat q< 0 Hea t Work w> 0 q> 0 Work associated with gas expansion • Consider combustion of urea, driving up a piston..... • ΔV(volume) = height*area • Work = distance x opposing force • Work = h x pexA = hA x pex • System is doing work so sign is negative: • w = -pexΔV MPharm PHA111 Exhaling air • Work is required to push air from the lungs against atmospheric pressure, typically 12-20 times per minute. • Consider exhaling 0.5L (5.0 x 10-4 m3) against atmospheric pressure (typical Tidal Volume vs 6L total volume). • w = -pexΔV = -(1.01 x 105 Pa) x (5.0 x 10-4 m3) = -51 J • This is roughly equivalent to lifting 7 kg off the ground by 75 cm. • w = -mgh = -(7.0 kg) x (9.81 ms-2) x (0.75 m) = -52 J MPharm PHA111 Expansions and mechanical equilibrium • A system would do no work (w = 0) upon expansion if there were no opposing force. • A system does the most work possible when the opposing pressure is infinitesimally less than the internal pressure. Maximum work coincides with reversible change. • dw = -pexdV ≈ pdV (infinitesimal difference between internal and external). • w = -∫pdV, but p = nRT/V, so w = -∫(nRT/V)dV = -nRT∫(1/V)dV • w = -nRT ln(Vf/Vi), where Vf = final volume, and Vi = initial volume. MPharm PHA111 Measuring heat • A thermodynamic assessment of energy output during metabolic processes requires knowledge of ways to measure the energy transferred as heat. • Heat Capacity of a Substance, C = q/ΔT • therefore, if we already know the heat capacity, we can obtain a quantity of transferred heat via: q = CΔT Substance Cp,m (J K-1 mol-1) e.g. if the heat capacity of a beaker of water is known to be 0.50 kJ K-1 and a temperature rise is observed to be 4 K, then the heat transferred is: q = CΔT = (0.50 kJ K-1) x (4.0 K) = 2.0 kJ For common materials tables exist for the specific heat capacity, Cs (J K-1 g-1), or its molar heat capacity, Cm (J K-1 mol-1). Also be aware of Cp, heat capacity at constant pressure (open beaker), PHA111 MPharm Reversible Isothermal expansion of a Perfect Gas • Temperature of gas is same at beginning and end of expansion. • Since the speed of molecules in a gas is a function of temperature, this must also remain constant. This in turn means that the kinetic energy of the gas remains constant. • Since the potential energy of a perfect gas is negligible, the total energy = its kinetic energy. Therefore, the total energy of the gas must have remained constant. • This can only happen if heat from the surroundings has exactly replaced the energy lost in the form of work: • q = -w, and q = nRT ln(Vf/Vi) • This brings us on to internal energy… MPharm PHA111 This Photo by Unknown Author is licensed under CC BY-NC The Internal Energy, U • The total energy (Kinetic + Potential) of all the atoms, molecules and ions that constitute the system of interest. • For common materials we can describe a molar internal energy Um, as 1 kg of Iron will have half the internal energy of 2 kg of Iron. • Since the total energy would include all the kinetic and potential energies of all the subatomic particles in the system it is impossible to measure. • Instead, we look at changes in internal energy, ΔU, upon reaction or physical change as these can be measured by looking at changes in work and heat. • ΔU = w + q Slide 26 of 29 MPharm PHA111 Thermodynamics: The First Law back to perfect gases, again........ • Remember, for isothermal expansion of a perfect gas: q = -w. – They are equal but opposite • Therefore, for isothermal expansion of a perfect gas: ΔU = 0. • This means that the internal energy of a perfect gas is independent of the volume it occupies, at a given temperature. • On a molecular level, only the distance between molecules is changing, their speeds and therefore energies remain the same. • Internal energy is a State Function. It is a physical property which is dependent upon the current state of the system and is independent of the method by which that state was reached. MPharm PHA111 The First Law of Thermodynamics • Isolated systems can neither do work upon nor heat their surroundings. • This means that their internal energies can not change. • The internal energy of an isolated system is constant. • Consider “perpetual motion” devices. These fail as the internal energy always drops when they do work. • For any system that does work new sources of energy must be supplied at regular intervals e.g. food, petrol etc. MPharm PHA111 Measuring ΔU • Carry out reaction in vessel of fixed volume, so no expansion work is possible, w = 0. • Then, ΔU = qV (heat generated for system with fixed volume) • since: C = q/ΔT, then CV = ΔU/ΔT – CV = heat capacity @ constant volume MPharm PHA111 Enthalpy • Most biological systems operate at constant pressure rather than constant volume (open to the atmosphere), so we must make adjustments to account for this....... • Oxidation of tristearin: 2C57H110O6(s) + 163O2 ➞ 114CO2(g) + 110H2O(l) At 25 °C, burning 1g of fat causes a decrease of 600mL in gas volume. • As the system contracts the atmosphere does work upon the system. This can be accounted for by using the enthalpy, H, of the system, where: H = U + pV (pressure * volume) • Just as we usually deal in changes in internal energy, it is more usual to deal with changes in enthalpy: ΔH = ΔU + Δ(pV) • but since we are working at constant pressure: ΔH = ΔU + pΔV MPharm PHA111 Molar Enthalpies, Hm = H/n • Hm = Um + pVm • but, for 1 mole of a perfect gas: pV = RT • therefore, for a perfect gas: Hm = Um + RT • at 25 °C, RT = 2.5 kJ mol-1, so the enthalpy of a perfect gas is higher than its internal energy by that amount (at this temperature). • For an open system p = pex, the external pressure, so: ΔH = ΔU + pexΔV – but since we already know that i) ΔU = w + q, and that ii) w = -pexΔV, – then: ΔH = (-pexΔV + q) + pexΔV • ΔH = qp i.e. the change in enthalpy equals the heat transferred, at constant pressure. MPharm PHA111 Exothermic & Endothermic • Exothermic process: energy leaves the system as heat, q < 0, ΔH < 0. • Endothermic process: energy enters the system as heat, q > 0, ΔH > 0. • remembering that i) ΔH = qp, and that ii) C = q/ΔT, then the heat capacity at constant pressure must be: Cp = ΔH/ΔT • How does the enthalpy of 100g of water change if its temperature is raised from 20 °C to 80 °C? • ΔH = Cp ΔT = nCp,m ΔT = (5.55 mol) x (75.29 J K-1 mol-1) x (60 K) = +25 kJ • Cp,m - CV,m = R for ideal gases. For solids and liquids Cp,m ≈ CV,m Slide 6 of 24 MPharm PHA111 Thermodynamics: The Laws 1 Phase Transitions • Pure substances at exactly 1 bar are in their standard state. o is used as a superscript to denote standard conditions, with a temperature of 25 ℃unless otherwise noted. • Substances do not have to be in their most stable form under standard conditions, e.g. we can discuss ice or water vapour at 1 bar (po) and 25 ℃. Hmo is the standard molar enthalpy of a substance. • Also substances can have more than one phase in a particular state of matter e.g. calcite or aragonite for CaCO 3(s). Most phase transitions are accompanied by a change in enthalpy. • We sweat to keep cool. ΔvapHo (H2O) = +44 kJ mol-1. Vaporisation of sweat withdraws heat from within our bodies. MPharm PHA111 Standard enthalpies of transition at the transition temperature MPharm PHA111 Thermochemical equations • H2O(l) ➞ H2O(g) Ho = +44 kJ 2H2O(l) ➞ 2H2O(g) Ho = +88 kJ • Water has a high vaporisation enthalpy, due to hydrogen bonding. This is responsible for the low humidity of our climate and the presence of the oceans. • Melting ice is much easier, ΔfusHo = +6.01 kJ mol-1, as the molecules remain close in both states. • As H is a state function, freezing is equal but opposite in enthalpy to melting H2O(s) ➞ H2O(l) Ho = +6.01 kJ H2O(l) ➞ H2O(s) Ho = -6.01 kJ • Steam scalds because the enthalpy change accompanying condensation is passed as heat to the skin. • Compare to putting hand in oven briefly! MPharm PHA111 in general, ΔforwardHo = -ΔreverseHo MPharm PHA111 Sublimation and vapour deposition • Sublimation is direct conversion of a solid to a vapour. e.g. frost “vanishes”. • ΔsubHo = ΔfusHo + ΔvapHo • The overall enthalpy change of a process is the sum of the steps, observed or hypothetical, into which it may be divided. • To be valid, all enthalpy quantities must be determined at the temperature of interest. MPharm PHA111 Measuring Phase Changes: Differential Scanning Calorimetry • Measures difference in heat transferred to / from sample in comparison to reference material. • Protein, micelle, lipid, nucleic acids, pharmaceutical formulations. http://www.microcal.com/technology/ dsc.asp MPharm PHA111 Chemical Change: bond enthalpies • Fermentation of glucose by yeast, under standard conditions: C 6H12O6(s) ➞ 2C2H5OH(l) + 2CO2(g) ΔHo = -72 kJ • Change in energy associated with bond dissociation of 1 mole of a substance is the bond enthalpy: H2(g) ➞ 2H(g) ΔHo = +436 kJ mol-1 • In general, breaking bonds (dissociation) are endothermic processes, while making new bonds (association) is exothermic. • In most reactions some bonds are broken, while others are formed. We must develop ways to understand and measure overall enthalpy changes. MPharm PHA111 Selected bond enthalpies, ΔHA-B (kJ mol-1) • Diatomic Molecules H-H 436 Cl-Cl 242 H-Br 366 O=O 497 H-Cl 431 I-I 151 F-F 155 H-F 565 N≡N 945 O-H 428 Br-Br 193 H-I 299 C≡O 1074 • Polyatomic Molecules H-CH3 435 H-C6H5 469 H3C-CH3 368 H2C=CH2 699 HC≡CH 962 H-NH2 431 O2N-NO2 57 O=CO 531 H-OH 492 HO-OH 213 HO-CH3 377 Cl-CH3 452 Br-CH3 293 I-CH3 234 • Exact values depend on bond order and the electronic environment of the two atoms comprising the bond. • Many chemistry textbooks list mean bond enthalpy tables, which give typical values for compounds in the gas phase. MPharm PHA111 Using mean bond enthalpies • Hydrogen peroxide is formed in the body due to reactions involving oxygen. The enzyme catalase helps to decompose this toxic compound to oxygen and water. We will estimate the enthalpy change for this reaction. 2H2O2(l) ➞ 2H2O(l) + O2(g) • • Strategy: we need to do our bond breaking and forming steps in the gas phase to use table values. • This means we will have to take into account melting / vaporisation / sublimation steps to get things into this state, and condensation, fusion, vapour deposition steps to return substances to their final phase. MPharm PHA111 Solution (all values are ΔH / kJ) o • Vaporisation of 2 mol H2O2(l) ➞ H2O2(g) 2 x (+51.6) Dissociation of 4 mol O-H bonds: 4 x (+463) (exact) Dissociation of 2 mol O-O bonds: (+213) • Overall, so far: 2H2O2(l) ➞ 4H(g) + 4O(g) +2381 • Formation of 4 mol O-H bonds: 4 x (-492) (mean) Formation of 1 mol O2: Condensation of 2 mol H 2O(g) ➞ H2O(l) (-44) • Overall, stage 2: 4H(g) + 4O(g) ➞ O2(g) + H2O(l) 2x -497 2x -2553 • Overall enthalpy change, ΔHo = (+2381) + (-2553) = -172 kJ (exp. = 196 kJ) MPharm PHA111 Standard enthalpies of combustion, ΔcHo • Standard change in enthalpy per mole of combustible substance. • CH4(g) + 2O2(g) ➞ CO2(g) + 2H2O(l) ΔcHo(CH4,g) = -890 kJ mol-1 • A note of caution: heat released at constant volume, qV = ΔU • ΔH = ΔU + pVm, so we should really take into account the molar volumes. • However, for condensed phases molar volumes only require a correction of a few J mol-1 and can be ignored for most purposes. • For gases Vm is significant, and thus the correction must be included. ΔcH = ΔcU + ΔνgasRT (Δνgas is the change in the number of moles of gas) MPharm PHA111 Biological fuels • 18-20 year old males require ~12 MJ of energy input per day. • 18-20 year old females require ~9 MJ of energy input per day. • Carbohydrates provide 17 kJ g-1, while fats provide 38 kJ g-1. • Hydrocarbon fuels provide 48 kJ g-1. • The more oxidised a fuel is to begin with, the less energy we can get from its combustion. Body temperature is regulated by radiation and perspiration. • Fats are used for energy storage in mammals and to provide insulation. We burn carbohydrates whenever possible. Proteins are usually used to create new proteins instead of being employed as a fuel source. MPharm PHA111 Hess’s Law • The standard enthalpy of a reaction is the sum of the standard enthalpies of the reactions into which the overall reaction may be divided. • Really, this is just another way of saying that enthalpy is a state function. • We just need to expand upon the technique we used to look at the catalase reaction earlier. MPharm PHA111 Using Hess’s Law • During vigourous exercise muscle cells become deprived of oxygen. This means that glucose cannot be completely oxidised. Instead it is typically broken down to lactic acid (LA). • Using the following data: Glucose: C6H12O6(s) + 6O2(g) ➞ 6CO2(g) + 6H2O(l) ΔHo = 2808 kJ LA: CH3CH(OH)COOH(s) + 3O2(g) ➞ 3CO2(g) + 3H2O(l) ΔHo = -1344 kJ • We will calculate the enthalpy for glycolysis: C6H12O6(s) ➞ 2CH3CH(OH)COOH(s) • C6H12O6(s) + 6O2(g) ➞ 6CO2(g) + 6H2O(l) -2808 kJ 6CO2(g) + 6H2O(l) ➞ 2CH3CH(OH)COOH(s) + 6O2(g) 2 x (+1344 kJ) Overall: C6H12O6(s) ➞ 2CH3CH(OH)COOH(s) ΔHo = -120 kJ MPharm PHA111 Standard reaction enthalpies, ΔrHo • Difference between the standard molar enthalpies of the reactants and products, each weighted by the reaction stoichiometry (ν): • ΔrHo = ΣνHmo(products) ΣνHmo(reactants) • Each reactant is in its standard state (pure, 1 bar) and the products are in their standard states (pure, 1 bar). • Normally, we work indirectly by referencing each reactant and product to an imaginary reaction where one mole of it is formed from the constituent elements in their most stable states. MPharm PHA111 e.g. standard enthalpy of formation of liquid water • H2(g) + ½O2(g) → H2O(l)ΔfH∅ (H2O,l) = -286 kJ mol-1 • Using this trick we can now say: ΔrH∅= ΣνfH∅ (products) - ΣνfH∅ (reactants) • Standard enthalpies of formation of elements are zero. However, standard enthalpies of formation of an element in any other phase is not zero. • C(s, graphite) → C(s, diamond) ΔH∅ = +1.895 kJ • As the reference state of carbon is always graphite, then: ΔfH∅ (C,diamond) = +1.895 kJ mol-1 MPharm PHA111 Applying standard enthalpies of formation • We will use standard enthalpies of formation to estimate the standard enthalpy of combustion • C H O (s) + of 12O (g) → 12CO (g) + 11H O(l) sucrose. 12 22 11 2 2 2 • ΔrH∅ = {12ΔfH∅(CO2,g) + 11ΔfH∅(H2O,l)} – {ΔfH∅(C12H22O11,s) + 12ΔfH∅ (O2,g)} = {12 x (-393.51 kJ mol-1) + 11 x (-285.83 kJ mol-1)} - {(-2222 kJ mol-1) + 0} = -5644 kJ mol-1 (experimental value = -5645 kJ mol-1) Slide 23 of 24 MPharm PHA111 Thermodynamics: The Laws 1 Spontaneous & Non-spontaneous change • Gases expand to fill empty spaces. • Hydrogen and Oxygen combine violently to produce water. • Hot objects gradually cool to the same temperature as their surroundings. • All the above are spontaneous processes. • In order to reverse them we have to do work upon the system of interest. • We will look at why things can happen, this does not consider how quickly they will. MPharm PHA111 Spontaneity and energy • Spontaneous processes do not necessarily move in the direction of lower energy. • Isothermal expansion of a perfect gas into a vacuum has no overall change in energy. • The molecules move at the same speed as before, just the distance between them changes. • Even in a cooling process the energy lost by the system is transferred to the surroundings, it is not destroyed. • Consider two metal blocks, one hot and one cold, placed in contact within a vacuum. Eventually they will achieve thermal equilibrium. MPharm PHA111 The direction of spontaneous change • Energy and matter have a tendency to disperse. • Gas molecules move randomly, the probability that they will move into one corner of a container is negligible. • Hot atoms oscillate vigourously, collidling with neighbouring atoms. Energy is passed on with these collisions. MPharm PHA111 The Second Law of Thermodynamics • The entropy of an isolated system tends to increase. • Entropy, S, is the measure of how dispersed energy and matter is. • Isolated system e.g. reaction vessel and its surroundings, a mini universe. • As is usual in thermodynamics we will deal with changes rather than absolute values: ΔS = qrev/T • i.e. the change in entropy of a substance is equal to the heat transferred reversibly to it, divided by the temperature at which the transfer takes place. • Entropy is a state function. i.e. its value only depends upon the present state of the system. MPharm PHA111 Entropy – the tendency of energy to spread out MPharm PHA111 Entropy Hot Metal Bar 8 atoms, 7 energy packets MPharm PHA111 Cold Metal Bar 8 atoms, 3 energy packets Entropy Hot Metal Bar Cold Metal Bar What will happen when we touch the bars together? MPharm PHA111 Entropy Hot Metal Bar Cold Metal Bar What is the probability that we end up with 9 packets in the HOT bar, and 1 packet in the COLD bar? MPharm PHA111 0 566280 627264 1 2 3 4 5 6 7 8 9 19448 91520 231660 411840 If add up the scenarios where we end up with more than 7 packets in the hot bar (i.e., it gets hotter) we see there is a 10.5% chance of this occurring 91520 19448 231660 Unique combinations 411840 566280 Entropy 10 No. of energy packets in "hot" bar But why don’t we observe this in real life??? MPharm PHA111 Entropy 4.00E+72 Unique combinations 3.50E+72 3.00E+72 2.50E+72 2.00E+72 1.50E+72 1.00E+72 5.00E+71 0 0 3 6 9 12 15 18 21 24 27 30 33 36 39 42 45 48 51 54 57 60 63 66 69 72 75 78 81 84 87 90 93 96 99 No of energy packets in "hot" bar As we increase to 100 energy packets over 80 atoms the probability of the hot bar getting hotter drops drastically! Consider the number of atoms in a cup of water… (approx. 20 000 000 000 000 000 000 000 000) MPharm PHA111 The entropy of heating water. • If 100 kJ of heat is transferred to a large body of water at 0 °C, the resulting change in its entropy is: • ΔS = qrev/T = 100 x 103 J / 273 K = +366 J K-1 • For boiling water the same amount of heat would result in: • ΔS = qrev/T = 100 x 103 J / 373 K = +268 J K-1 • Assuming a large mass means that we can treat the temperature as constant. • Often molar entropies are quoted, the units for this are J K-1 mol-1. MPharm PHA111 Entropy changes accompanying heating. • ΔS = C ln (Tf/Ti), where C is the heat capacity of the system, Ti the initial temperature and Tf the final temperature. • when Tf > Ti the logarithm is positive, meaning that ΔS > 0, in line with what we would expect. • This equation also tells us that entropy changes are higher for materials with high heat capacities. • More energy is required to cause a change in temperature for these materials. • Think ‘loud sneeze’. • For most materials heat capacities are not constant w.r.t. temperature, more complex equations are needed. Especially solids at very low temperatures. Slide 8 of 39 MPharm PHA111 Thermodynamics: Laws 2 & 3 Entropy changes accompanying phase transitions • Molecules are able to move more freely upon melting or boiling therefore these process result in an increase in entropy. • Similarly phase transitions in biological macromolecules can be accompanied by entropy changes. • Think about frying an egg.... MPharm PHA111 Entropies of vaporisation of selected substances at 1 atm and their normal boiling point. • • • • • • • • • ΔvapS (J K-1 mol-1) Ammonia 97.4 Benzene 87.2 Bromine 88.6 Carbon Tetrachloride 85.9 Cyclohexane 85.1 Ethanol 104.1 Hydrogen Sulfide 87.9 Water 109.1 MPharm PHA111 Entropy of vaporization of water at 25 °C • Three calculations needed, first we need the entropy change associated with heating water to its boiling point..... • ΔS1 = Cp,m(H2O,l) ln(Tf/Ti) = (75.29 J K-1 mol-1) x ln(373 K/298 K) = +16.9 J K-1 mol-1 • Next, we calculate the transition entropy: • ΔS2 = ΔvapH(Tb)/Tb = (4.07 x 104 J mol-1) / 373 K = +1.09 x 102 J K-1 mol-1 • Finally, we need to cool the vapour from 100 °C back to 25 °C: • ΔS3 = Cp,m(H2O,g) ln(Tf/Ti) = (33.58 J K-1 mol-1) x ln(298 K/373 K) = -7.54 J K-1 mol-1 • Δvap MPharm S (298PHA111 K) = ΔS1 + ΔS2 + ΔS3 = +118 J K-1 mol-1 Metabolism and entropy • Humans, even at rest, constantly expend energy. • Typically, 100 J s-1 is lost by a resting human. • Assuming the person is in a room at 20 °C, then the entropy change of the surroundings after 1 day (24 x 60 x 60s) can be found by: • qsur = (86,400 s) x (100 J s-1) = 8,640,000 J • ΔSsur = qsur/T = 8640000 J / 293 K = +2.95 x 10 4 J K-1 • So, every human increases the entropy of the planet by ~30 kJ K-1 just by staying alive. Any activity dramatically increases this value. MPharm PHA111 Ok, but why should we be bothered about that? • The amount of heat passing to / from the surroundings is equal and opposite to the heat leaving / entering the system, which is what we are usually interested in: qsur = -q • This means we can say that: ΔSsur = -q/T • Therefore we can directly relate the energy change of the system to the entropy change of its surroundings and this is true regardless of reversibility. • At constant pressure we already know that q = ΔH, so we can say: ΔSsur = -ΔH/T • i.e. the change in entropy of the surroundings is tied to the enthalpy change of the system. MPharm PHA111 The Third Law of Thermodynamics • The absolute entropies of all perfectly crystalline substances is zero at absolute zero. S(0) = 0 • At absolute zero all atoms would be stationary. • Perfectly crystalline substances are the most ordered possible. • We can follow the changes in heat capacity of a given substance with temperature and obtain its standard molar entropy, Sm∅ MPharm PHA111 Standard molar entropies, Sm∅ (J K1 mol-1) of selected substances at 298.15 K MPharm PHA111 The standard reaction entropy, ΔrS∅ • This is the difference in molar entropy between the products and reactants of a reaction in their standard states: • ΔrS∅ = ΣνSm∅ (products) - ΣνSm∅ (reactants) • ν = stoichiometric coefficients in the MPharm PHA111 chemical equation. Standard reaction entropy of an enzyme catalysed reaction at 25 °C • Carbonic anhydrase catalyses hydration of carbon dioxide in red blood cells. • CO2(g) + H2O(l) ➞ H2CO3(aq) • ΔrS∅ = Sm∅ (H2CO3,aq) - [Sm∅ (CO2,g) + Sm∅ (H2O,l)] • ΔrS∅ = (187.4 J K-1 mol-1) {(213.74 J K-1 mol-1) + (69.91 J K-1 mol-1)} • ΔrS∅ = -96.3 J K-1 mol-1 MPharm PHA111 Glaucoma, benign intercranial hypertension, epilepsy, altitude sickness Acetazolamide Spontaneity and the entropy of the Universe • Processes occur spontaneously if the overall entropy of the universe is increased. • Thus we need to know the entropy changes for both the system and the surroundings. • Binding of NAD+ to lactate dehydrogenase has ΔrS∅ = -16.8 J K-1 mol1 at 25 °C and pH7, yet binding takes place spontaneously. • ΔSsur = -ΔrH∅/T = -(-24.2 kJ mol-1) / 298 K = +81.2 J K-1 mol-1 • ΔrSuniv = ΔrS∅ + ΔSsur = -16.8 J K-1 mol-1 + 81.2 J K-1 mol-1 = 64.4 J K-1 mol-1 • Binding takes place because the entropy of the universe is increased. MPharm PHA111 Summary • 2nd Law : The entropy of an isolated system tends to increase. • 3rd Law : The absolute entropies of all perfectly crystalline substances is zero at absolute zero. • Reactions take place spontaneously if they allow the overall entropy of the universe to be increased. MPharm PHA111 Where we got to in our discussion of Entropy • Processes occur spontaneously if the overall entropy of the universe is increased. • Thus we need to know the entropy changes for both the system and the surroundings. • Practically, this can be tricky. • Is there a way around this that relates spontaneity entirely to changes in the system? MPharm PHA111 Focusing on the System ΔSuniv = ΔSsys + ΔSsur For a spontaneous process, ΔSuniv > 0 But, we know that ΔSsur = -ΔH/T so, ΔSuniv = ΔSsys - ΔH/T, at constant pressure and temperature. • We have managed to relate the change in entropy of the universe entirely to the properties of the system! • from now on we will refer to ΔSsys as just ΔS for convenience. • • • • MPharm PHA111 The Gibbs Energy, G • Also commonly know as The Gibbs Free Energy. • G = H - TS • As H, T and S are all state functions we can say: • ΔG = ΔH - TΔS, where T is constant. • The Gibbs energy and Entropy have opposite signs. • So, since spontaneous processes lead to higher Entropy...... • ....... they also lead to lower Gibbs energy. MPharm PHA111 Josiah Willard Gibbs 1839-1903 1st PhD in Engineering (Yale) Work and the Gibbs energy change • The ΔG of any process is the maximum nonexpansion work that can be extracted from that process at constant temperature and pressure. • Non-expansion work: e.g. electrical, mechanical. Anything except expansion! • Redox chemistry within a cell, muscle contractions etc. • ΔG = wmax’ MPharm PHA111 Changes in Gibbs energy and metabolism • A bird weighs 30g. How much glucose will the bird metabolise in order to fly from the ground to a branch 10m above it? The change in Gibbs energy accompanying combustion of glucose at 25 °C is -2828 kJ mol-1. • Strategy: calculate work associated with raising the object to the required height. • As this is non-expansion work its value is the same as ΔG. • We can now calculate how many moles must be metabolised to produce the required amount of energy. • Finally, convert moles to mass using molar mass. MPharm PHA111 Solution • Energy required = Work = mgh = (30 x 10-3 kg) x (9.81 ms-2) x (10m) = 2.943 J • n = Energy req/Energy per mole = 2.943 J / 2.828 x 106 J mol1 = 1.041 x 106 mol • m = nM = (1.041 x 10-6 mol) x (180 g mol-1) = 1.9 x 10-4 g MPharm PHA111 ATP hydrolysis drives unfavourable biochemical reactions • At pH = 7.0 and T = 37 °C, ΔrH = -20 kJ mol-1 and ΔrG = -31 kJ mol-1 • Therefore we can extract up to 31 kJ of energy in the form of non-expansion work from hydrolysis of each mole of ATP. • If the energy is not utilised as work then 20 kJ mol-1 will be released as heat. MPharm PHA111 Coupling ATP hydrolysis to drive reactions • Formation of glutamine from glutamate is catalysed by the enzyme glutamine synthetase. • However, ΔrG = +14.2 kJ mol-1. • If we couple this to ATP hydrolysis (Δ rG = 31 kJ mol-1) we have more than enough energy available. MPharm PHA111 The standard reaction Gibbs energy, ΔrGo • In the same way we introduced standard enthalpies and entropies of reaction we can say: ΔrGo = ΣνGmo(products) - ΣνGmo(reactants) • However, absolute quantities are not known, so instead we can use: ΔrGo = ΔrHo – TΔrSo where: ΔrHo = ΣνfHo(products) - ΣνfHo(reactants) and: ΔrSo = ΣνSmo(products) - ΣνSmo(reactants) MPharm PHA111 Calculating the standard reaction Gibbs energy (at 25 °C) for a reaction catalysed by carbonic anhydrase • CO2(g) + H2O(l) ➞ H2CO3(aq) • ΔrHo = ΔfHo(H2CO3,aq) - {ΔfHo(CO2,g) + ΔfHo(H2O,l) • ΔrHo = -699.65 kJ mol-1 - {(-110.53 kJ mol-1) + (285.83 kJ mol-1)} • ΔrHo = -303.29 kJ mol-1 • Similarly, ΔrSo = -96.3 J K-1 mol-1 • ΔrGo = (-303.29 kJ mol-1) - {(298.15 K) x (-9.63 x 10-2 kJ K-1 mol-1)} • ΔrGo = -274.6 kJ mol-1 MPharm PHA111 Reaction Gibbs energies via standard Gibbs energies of formation, Δ fGo • Tabulated values obtained by combining standard enthalpies and entropies of formation at 298 K. • Each reactant and product is formed from its constituent elements from their reference states. • ΔrGo = ΣνΔfGo(products) - ΣνΔfGo(reactants) • The Gibbs energy of formation of a substance is a measure of its intrinsic stability. MPharm PHA111 Standard Gibbs energies of formation for selected substances at 298.15 K (ΔrGo / kJ mol-1) • Gases Carbon dioxide 394.36 Methane -50.72 Nitrogen oxide (NO) +86.55 Water -228.57 • Liquids Ethanol 174.78 Hydrogen peroxide -120.35 Water -237.13 MPharm PHA111 • Solids α-D-glucose 917.2 Glycine -532.9 Sucrose -1543 Urea -197.33 - • Solutes in Aq Solution Carbon dioxide -385.98 Carbonic acid -623.08 Phosphoric acid -1018.7 Example: standard reaction Gibbs energy for combustion of sucrose • C12H22O11(s) +12O2 ➞ 12CO2(g) +11H2O(l) • ΔrGo = {12ΔfGo(CO2,g) + 11ΔfGo(H2O,l)}{ΔfGo(C12H22O11,s) + 12ΔfGo(O2,g)} • ΔrGo = {12 x (-394 kJ mol-1) +11 x (-237 kJ mol-1)} {1543 kJ mol-1 + 0) • ΔrGo = -5790 kJ mol-1 • REMEMBER: O2 disappears from calculation because it is an element in its reference state – so by definition the value is zero! MPharm PHA111 The Gibbs energy and equilibria • ΔrGo = -RT ln K, where K is the equilibrium constant of the reaction. • For instance, the first step in the metabolic breakdown of glucose is a phosphorylation reaction to give glucose-6-phosphate (G6P): glucose(aq) + H2PO42-(aq) ➞ G6P(aq) • Inorganic phosphate is often abbreviated as Pi in biochemical reactions. • If the standard reaction Gibbs energy for this process is +14.0 kJ mol-1 at 37 °C what is the equilibrium constant? MPharm PHA111 Solution • ln K = -ΔrGo/RT • ln K = -(1.40 x 104 J mol-1) ÷ {(8.314 J K-1 mol-1) x (310 K)} • ln K = -5.43195698 • K = e-5.43195698 • K = 4.4 x 10-3 • always carry as many digits as possible when working with log functions MPharm PHA111 Temperature and reaction feasibility • Reactions are feasible, thermodynamically, when Δ rGo < 0. • Therefore we can define a temperature of feasibility for the critical point at which ΔrGo = 0. • Since ΔrGo = ΔrHo - TΔrSo, then this temperature will occur when: • 0 = ΔrHo - TΔrSo • TΔrSo = ΔrHo • therefore, T = ΔrHo/ΔrSo MPharm PHA111 Calculation example • Calculate the temperature at which the decomposition of calcium carbonate becomes spontaneous: • CaCO3(s) ➞ CaO(s) + CO2(g) kJ mol-1, ΔrSo = +161 J K-1 mol-1 ΔrHo = +178 • T = ΔrHo/ΔrSo = (178,000 J mol-1) ÷ (161 J K-1 mol-1) • T = 1105 K MPharm PHA111 Summary of thermodynamic reaction spontaneity • If a reaction is EXOTHERMIC and ENTROPICALLY favoured, then it can proceed at all temperatures. • If a reaction is EXOTHERMIC and ENTROPICALLY disfavoured, then it can proceed if: T < ΔrHo/ΔrSo • If a reaction is ENDOTHERMIC and ENTROPICALLY favoured, then it can proceed if: T > ΔrHo/ΔrSo • If a reaction is ENDOTHERMIC and ENTROPICALLY disfavoured, then it is unfeasible at any temperature. MPharm PHA111

PHA111 Thermodynamics-2.pptx

Document Details

Related

Full Transcript

Upgrade to continue