Thermodynamics and Its Application in Biological Systems PDF

Cilvēka fizioloģijas un bioķīmijas katedra Videolecture Thermodynamics and Its Application in Biological Systems Assist. prof....

Cilvēka fizioloģijas un bioķīmijas katedra Videolecture Thermodynamics and Its Application in Biological Systems Assist. prof. Agnese Brangule 2020 Rīga CONTENT OF THE LECTURE 1. What is a thermodynamics process? 2. Terms and Laws of thermodynamics 3. Exothermic and endothermic reactions. Enthalpy 4. Entropy 5. Spontaneous and nonspontaneous process. Gibbs free energy Cilvēka fizioloģijas un bioķīmijas katedra Videolecture Thermodynamics and Its Application in Biological Systems What is a thermodynamics process? What is thermodynamics? All organisms require energy to stay alive. Organisms are energy transformers. Organisms take in energy and transduce it to new forms. All chemical reactions in cells involve energy transformations. Energy and biological needs Biological perspectives of thermodynamics principles Processes: Light Chemical PHOTOSYNTHESIS Chemical Electrical NERVOUS SYSTEM Chemical Mechanical MUSCLES Energy and biological needs Cells need energy to do all their work BIOLOGICAL ENERGY NEEDS To generate and maintain its structure To generate all kinds of movements To generate concentration and electrical gradients across cell membranes To maintain body temperature To generate light in some animals  Examples Green plants transform radiant energy into Humans are „energy parasites‟ chemical energy.  What is thermodynamics? Thermodynamics is the study of energy. The science deals with energy in its various forms and the conversion of one form of energy into another. Thermodynamics Heat Power What is thermodynamics? Objectives of thermodynamics 1. To understand the relationship between quantities of heat and work in biological systems. 2. To understand the influence of energy changes in biological phenomena. 3. To predict the effect of temperature on a variety of physico-chemical and biological phenomena in systems at equilibrium. e.g. bioreactors. 4. To understand the biochemical processes. Cilvēka fizioloģijas un bioķīmijas katedra Videolecture Thermodynamics and Its Application in Biological Systems Terms and Laws of thermodynamics Energy and biological needs What is Energy? Energy is defined as the ability to do work. Organisms take in energy and transduce it to new forms. The flow of energy maintains order and life. Energy and biological needs Basic forms of energy Kinetic Potential Light Energy in Stored motion energy Mechanical Heat Energy forms Electrical Chemical Thermodynamic systems What is a system? What is surroundings? An assemblage of matter, which can interact with energy is called a system. A system is separated from its surroundings by a boundary. Thermodynamic systems An open system exchanges matter and energy with its environment. A closed system exchanges only energy with its environment. An isolated system exchanges neither matter nor energy with its environment. System Energy Matter open √ √ closed √ X isolated X X Laws of thermodynamics The First Law of Thermodynamics Law of conservation of energy – this law was put forward by Robert Mayer in 1941. The law states: the total energy of a system plus its environment remains constant. This law declares: energy is neither created nor destroyed in the universe and it allows to be exchanged between a system and its surroundings. Robert Mayer  The Second Law of Thermodynamics Also called law of the degradation of energy or law of entropy (chaos). This law was developed in 1850s by German Physicist Rudolf Clausius. This law states that “a system and its surroundings always proceed to a state of maximum disorder or maximum entropy”.  The Second Law of Thermodynamics Living systems are ordered, while the natural tendency of the universe is to move toward systems of disorder with unavailable energy. The second law is an important indicator of the direction of the reaction. All reactions proceed in a direction with increase in entropy and decrease in free energy.  Key Concepts of Laws of Thermodynamics The sum of the energy before the conversion is equal to the sum of the energy after conversion. The total quantity of energy in the universe remains constant. The energy conversion is never 100% efficient. Ecological efficiencies vary from 1% to 56% depending on organisms Some energy is wasted in increasing the disorder or entropy. Cilvēka fizioloģijas un bioķīmijas katedra Videolecture Thermodynamics and Its Application in Biological Systems Exothermic and endothermic reactions. Enthalpy.  Main «elements» or Variables for the characterization of thermodynamic processes Name Symbol Unit Enthalpy ΔH° Entropy S° Gibbs free energy ΔG° Absolute temperature T K Heat Q or q J, kJ  Concept of enthalpy ΔH° (kJ/mol) Definition: Enthalpy is a change in heat content or heat of formation of a system. The enthalpy change of a reaction is roughly equivalent to the amount of energy lost or gained during the reaction. A reaction is favored if the enthalpy of the system decreases over the reaction. https://chem.libretexts.org Concept of enthalpy ΔH° (kJ/mol) ΔH° < 0 ΔH° > 0  Concept of enthalpy ΔH° (kJ/mol) Definition: Enthalpy is a change in heat content or heat of formation of a system. Reactants → Products  Concept of enthalpy ΔH° (kJ/mol) Enthalpy of Solution Enthalpy of Combustion Enthalpy of Vaporisation Enthalpy of Neutralisation Enthalpy of Formation Enthalpy of Neutralisation Concept of enthalpy ΔH° (kJ/mol) Enthalpy of Formation Standard conditions are: 298 K (25o C) A pressure 100 kPa Concentration 1 mol/L Standard states  The Thermochemical equation Exothermic reactions CH4(g) + 2O2(g) → CO2(g) + 2H2O(ļ) + 890 kJ CH4(g) + 2O2(g) → CO2(g)+ 2H2O(ļ) ∆H = - 890 kJ/mol Endothermic reactions N2(g) + O2(g) → 2NO(g) - 180 kJ N2(g) + O2(g) → 2 NO(g) ∆H = +180 kJ/mol -∆H N2(g) + O2(g) → 2 NO(g) +∆H  The Thermochemical equation Glucose metabolism equation: C6H12O6(s)+6O2(g)→6CO2(g)+6H2O(l) ∆H=-2803 kJ 2C6H12O6(s)+12O2(g)→12CO2(g)+12H2O(l) ∆H=-5606 kJ 6CO2(g) + 6H2O(l) → C6H12O6(s) + 6 O2(g) ∆H=2803 kJ  Calculation of enthalpy changes a A+b B+... → e E +f F +... ∆Horxn =[e ·∆Hof(E) + f ·∆Hof(F) +...]–[a ·∆Hof (A) + b ·∆Hof (B) +...] ∆Horxn = Σ∆Hof(products) – Σ∆Hof(reactants) Standard enthalpy change of formation The enthalpy change that occurs when 1 mol of Standard conditions: substance is formed from its elements in their T=298 K standard states under standard conditions. P=101.325 kPa  The Thermochemical equation The Direct Method C (graphite) + O2(g)→ CO2(g) ∆Hrxn = -393.5 kJ/mol The Thermochemical equation The InDirect Method C (graphite) + ½ O2(g)→ CO (g) C (graphite) + O2(g)→ CO2(g) ∆Hrxn = -393.5 kJ/mol CO(g) + ½ O2(g) → CO2(g) = -283.0 kJ/mol rxn – abbrevation - Hess’s law The InDirect Method  Hess’s law The energy change in an overall chemical reaction is equal to the sum of the energy changes in the individual reactions comprising it.  Application of thermodynamics Calorimetry Calorimetry is the process of measuring the amount of heat released or absorbed during a chemical reaction. 1 cal = 4.18 J Video: https://www.youtube.com/watch?v=jNKVylDT7rc Application of thermodynamics Calorimetry q = Cp m ΔT m - mass of the solution; g Cp - the specific heat of the water; J/g K ΔT - the change in temperature; K q = the heat released by the reaction. J ΔH = -q/n n – amount of the limiting reactant; mol  Strong acid with Strong alkali ∆H neutralization rxn Rxn slow – lose heat to surrounding Plot Temp/vol – extrapolation done, Temp correction HCI + NaOH → NaCI + H2O Coffee cup calorimeter Data collection Thermometric Titration Vol/m 0 5 10 15 20 25 30 35 40 45 50 2.M HCI – 25 ml added and temp recorded l Temp 2 24. 26 28 30 31 30 30 29 28 27 V, NaOH = 50 ml m = 25 ml+50 ml = 75 ml /C 2 c = 1M T initial = 22 oC T final = 31 oC by extrapolation. Temp correction – using heating/cooling ∆H rxn = Heat absorbed by water curve = (mc∆T) = 75 x 4.18 x (31 – 22) final Temp = 75 x 4.18 x 9 = - 2821J = 31 oC Limiting reactant n (OH-) = cV 0.05 mol = - 2821 J used = 1 x 0.05 1 mol = - 56 kJ mol-1 = 0.05 mol ∆H neutralization = 1 mol H+ react 1 mol OH- form 1 mol H2O HCI + NaOH → NaCI + H2O ∆H = - 56 kJ mol -1 Lit value = - 57.3 kJ mol -1 Strong acid vs Strong alkali Weak acid vs strong alkali initial Temp Vol, acid = 25 ml = 22oC Max Temp = 31oC 34 ∆T = ( 31 – 22) = 9oC  Application of thermodynamics Phase Changes: Exothermic or Endothermic? Hot packs Dissolution/crystallization Video: https://www.youtube.com/watch?v=6vgXUeY7tgg  Application of thermodynamics We obtain our energy primarily from carbohydrates, fats, and proteins in our diet. As we eat, our foodstuffs are digested and absorbed. The products of digestion circulate in the blood, enter various tissues, and are eventually taken up by cells and oxidized to produce energy. Cilvēka fizioloģijas un bioķīmijas katedra Videolecture Thermodynamics and Its Application in Biological Systems Entropy  Main «elements» or Variables for the characterization of thermodynamic processes Name Symbol Unit Enthalpy ΔH° Entropy S° Gibbs free energy ΔG° Absolute T K temperature Heat Q or q J, kJ Entropy S, Entropy is a measure of the randomness or disorder of a system. The value of entropy depends on the mass of a system. Entropy can have a positive or negative value. According to the second law of thermodynamics, the entropy of a system can only decrease if the entropy of another system increases.  Entropy, S ∆S > 0 Disorder increases Disorder decreases ∆S < 0 ∆Sorxn = ΣSo(products) – ΣSo(reactants)  Entropy, S Question. Where do molecules have the higher entropy? A B Correct answer : A  Entropy Ssolid < Sliquid 0 if Snp > Snr Example-1: C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(g), (Snp > Snr) Sorxn = {(3 x SoCO2) + (4 x SoH2O)} – {(SoC3H8) + (5 x SoO2)} = {(3 x 214) + (4 x 189)} J/K mol – {270 + (5 x 205)}J/K mol = (642 + 756) J/K – (270 + 1025) J/K mol = 103 J/K mol Summary ∆So ∆Sorxn = ΣnSo(products) – ΣmSo(reactants) ∆So >0, ∆So - T ∆S - T ∆S > ∆H The effect of enthaly and entropy on Gibbs free energy Chemistry 4th Edition by John Green, Sadru Damji IBID Press  Effect of Temperature on Go Go = Ho – TSo Example-1: For the reaction: N2(g) + 3H2(g) → 2NH3(g), Ho = -92 kJ/mol and So = -199 J/K mol = -0.199 kJ/K mol At 25oC, TSo = 298 K x (-0.199 J/K mol) = -59.3 kJ/mol Go = Ho - TSo = -92 kJ/mol – (-59.3 kJ/mol) = -33 kJ/mol; ➔ reaction is spontaneous at 25oC At 250oC, TSo = 523 K x (-0.199 J/K mol) = -104 kJ/mol; Go = Ho - TSo = -92 kJ/mol – (-104 kJ/mol) = 12 kJ/mol; ➔ reaction is nonspontaneous at 250oC Effect of Temperature on Go Go = Ho – TSo Example-2: For the reaction: CH4(g) + H2O(g) → CO(g) + 3H2(g), Ho = 206 kJ/mol and So = 216 J/K mol = 0.216 kJ/K mol At 25oC, TSo = 298 K x (0.216 J/K mol) = 64.4 kJ/mol Go = Ho - TSo = 206 kJ/mol – 64.4 kJ/mol = 142 kJ/mol; ➔ reaction is nonspontaneous at 25oC. At 1200 K, TSo = 1200 K x (0.216 J/K mol) = 259 kJ/mol; Go = Ho - TSo = 206 kJ/mol – 259 kJ/mol) = -53 kJ/mol; ➔ reaction is spontaneous at 1200 K Application of thermodynamics Hydrolysis of ATP to ADP and inorganic phosphate (Pi). Releases energy, approximately -7.3 kcal/mole Application of thermodynamics If the ∆Go for conversion of glucose G6-P to glucose G1-P is +1.65 kcal/mole, what is the ∆Go of the reverse reaction? The ∆Go for the reverse reaction Is -1.65 kcal. Application of thermodynamics  Application of thermodynamics Coupling Reactions A nonspontaneous reaction can be coupled to a spontaneous one to make it happen. Example: Fe2O3(s) → 2Fe(s) + 3/2 O2(g); Go = 740 kJ (eq-1) CO(g) + ½ O2(g) → CO2(g); Go = -283 kJ 3CO(g) + 3/2 O2(g) → 3CO2(g); Go = -849 kJ (eq-2) Combining eq-1 and eq-2, Fe2O3(s) + 3CO(g) → 2Fe(s) + 3CO2(g); Go = -109 kJ Application of thermodynamics Coupling Reactions in Biological System The formation of ATP from ADP and H2PO4- is nonspontaneous, but it can be coupled to the hydrolysis of creatine-phosphate that has a negative Go. ADP + H2PO4- → ATP + H2O; Go = +30 kJ Creatine-phosphate → creatine + phosphate; Go = -43 kJ Combining the two equations yields a spontaneous overall reaction: Creatine-phosphate + ADP → Creatine + ATP; Go = -13 kJ Summary ∆Ho , ∆So , ∆Go ∆Horxn = ΣnHfo(products) – ΣmHfo(reactants) ∆Sorxn = ΣnSo(products) – ΣmSo(reactants) ∆Gorxn = ΣnGo(products) – ΣmGo(reactants) ∆ G = ∆H - T ∆S ∆Ho >0, ∆Ho 0, ∆So 0, ∆Go