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Gibbs Free Energy & Biological Systems

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Gibbs Free Energy & Biological Systems GIBBS FREE ENERGY & BIOLOGICAL SYSTEMS Now that you have a better understand of the implications of Gibbs free energy, how NOTES: is it that human beings exist? 14 we’re made up of ~100 trillion (10 ) cells cells contain trillions of molecules, containing tens of thousands of atoms molecules and cells are arranged in structures, i.e., organs, bones, and skin molecules can synthesized on very short notice, i.e., adrenalin or insulin Insulin is a highly ordered molecule. It is a protein made up of 51 amino acids. Those amino acids are all connected in exactly the correct order and folded into exactly the molecular shape needed for its function in the metabolism of glucose. (Note: Hydrogen atoms are not shown for simplicity. Carbon atoms are black, oxygen atoms are red and nitrogen atoms are blue.) Thermodynamically speaking, we are very, very IMPROBABLE! So how can we exist?? The answer lies in the couplingof reactions. Your body extracts Gibbs free energy from the foods we eat. Consider the single nutrient glucose (also known as dextrose or blood sugar). A large quantity of Gibbs free energy can be released when glucose is oxidized, i.e., C6H12O6(aq) + 6O2(g)→ 6CO2(g)+ 6H2O(l); ΔG = –2870 kJ A reaction that releases Gibbs free energy is known as “exergonic”. When glucose is burned in the presence of air, all the Gibbs free energy is release as thermal energy. The same quantity of Gibbs free energy is available to the body when glucose is oxidised, but of course, if this amount of thermal energy were released all at once it would raise the temperature rapidly and kill many cells. Instead, your body makes use of a large number of reactions to release Gibbs free energy in small steps and the energy is stored in small quantities that can be used later. The most important method by which Gibbs free energy is stored in your body is through the formation of ATP from ADP. Within a typical human cell, this reaction takes place 32 times for each molecule of glucose that is oxidized. (In bacterial cells, it takes place 38 times.) The Gibbs free energy released by the exergonic glucose oxidation is then used to force the endergonic, reactant favoured process, of forming ATP from ADP. The overall process is 3– – C6H12O6(aq)+ 6O2(g)+ 32ADP (aq)+ H2PO4 (aq) → 4– 6CO2(g)+ 32ATP (aq)+ 38H2O(l); ΔG = -1894 kJ 58 NOTES: Because the overall reaction releases Gibbs free energy, appreciable quantities of products can be obtained. A spontaneous (exergonic) reaction is coupled with a non- spontaneous (endergonic) reaction to “drive” the non-spontaneous reaction to products. The words “exergonic” and “endergonic” have nearly the same prefixes as “exothermic” and “endothermic”. In both cases “ex” means out and “endo” means into. “Thermic” indicates that it is thermal energy that is released or taken up. “Ergonic” indicates that it is Gibbs free energy that is released or used up. ASIDE: The above equation simplifies the metabolic process by which Gibbs free energy stored in nutrients is transferred to ATP. The actual process takes place in three-stages. Stage 1 involves digestion where large molecules are broken down into smaller ones. Smaller molecules are then converted to acetyl groups attached to coenzyme A in stage 2. Finally, within the citric acid cycle, the acetyl groups are oxidized to carbon dioxide and water. ATP generated from glucose or other nutrients is a convenient and readily available Gibbs free energy resource. ATP can release Gibbs free energy in packets of 30.5 kJ for each ATP converted to ADP. This size is convenient for driving many biochemical processes in your body. When ATP reacts with water (hydrolysis) energy is released: – + H2O(l) → + H2PO4 (aq); (ATP) (ADP) ΔG = –30.5kJ/mol) Because ATP is high in Gibbs free energy, it is said to be a high-energy molecule (or ion). (Aside: Sometimes the bonds in ATP are called high-energy bonds, but this a misnomer because the bonds can be broken to form ADP and release Gibbs free energy.) 59 ATP and the metabolism of glucose: NOTES: When glucose is metabolized by the body the first step involves the attachment of a phosphate group to the glucose molecule via nucleophilic attack. + H PO –(aq) + 2 4 + H3O (aq); ΔG = +13.8 kJ glucose hexokinase glucose-6- phosphate (Hexokinase catalyzes the reaction between glucose and ATP where ATP binds to the enzyme as a complex with Mg2+.) How does this work? The reaction is not spontaneous and requires Gibbs free energy but the reaction can be forced to proceed if coupled to the conversion of ATP to ADP. 4– 3– – ATP (aq) + H2O(l) → ADP (aq) + H2PO4 (aq) ΔG = –30.5 kJ – 2– + glucose(aq) + H2PO4 (aq) → glucose-6-phosphate (aq) + H2O(l) + H (aq); ΔG = +13.8 kJ 4– 3– + 2– ATP (aq) + glucose(aq) → ADP (aq) + H (aq) + glucose-6- phosphate (aq) ΔG = (-30.5 kJ + 13.8) = – 16.7 kJ The overall negative value of ΔG means the reaction is overall spontaneous where the products are favoured. Thus the ATP to ADP transformation can force glucose to react with dihydrogen phosphate. The 16.7 kJ of Gibbs free energy released appears as heat and is transferred to surroundings (i.e., you!) In biochemistry/biological chemistry it is conventional to write these equations in a shorthand form that indicates that they are coupled. The first stage of glycolysis is thus represented by The curved arrow indicates that the ATP to ADP transformation occurs simultaneously with the glucose to glucose-6-phosphate reaction and that the two are coupled. 60 Another example: NOTES: Proteins are polymers made of amino acids. Consider the formation of the dipeptide (a two amino-acid unit) alanylglycine from alanine and glycine, which is the first step in the synthesis of a protein molecule: Alanine + Glycine → Alanylglycine ΔG = +29 kJ With the aid of an enzyme, the reaction is coupled to the hydrolysis of ATP as: ATP + H2O + Alanine + Glycine → ADP + H3PO4 + Alanylglycine The overall free-energy change, ΔG is (–31 kJ/mol + 29 kJ/mol) = –2 kJ/mol, which means that the coupled reaction now favours the formation of the product, and an appreciable amount of alanylglycine will be formed under this condition. A schematic representation of ATP synthesis and coupled reactions in living systems is shown below. PROBLEM 1: ATP undergoes hydrolysis with a release of Gibbs free energy. Other organophosphates undergo similar hydrolysis reactions: – creatine phosphate + H2O(l) → creatine + H2PO4 ; ΔG = –43.1 kJ glycerol 3-phosphate + H O(l) → glycerol + H PO –; ΔG = –9.7 kJ 2 2 4 Which of the reactions below are product-favoured? 1. ATP + creatine → creatine phosphate + ADP 2. ATP + glycerol → glycerol 3-phosphate + ADP 61 Answers: NOTES: – 1. ATP + H2O(l) → ADP + H2PO4 ; ΔG = -30.5 kJ – + creatine + H2PO4 → creatine phosphate + H2O(l) ; ΔG = kJ ATP + creatine → creatine phosphate + ADP; ΔG = kJ Product are (favoured or unfavoured)? – 2. ATP + H2O(l) → ADP + H2PO4 ; ΔG = -30.5 kJ – + glycerol + H2PO4 → glycerol 3-phosphate + H2O(l) ; ΔG = kJ ATP + creatine → creatine phosphate + ADP; ΔG = kJ Product are (favoured or unfavoured)? ΔG°′ Recall ΔG° corresponds to 1M solutions (i.e., pH = 1.00). Cells are typically at a pH of ~7 and body temperature is 37°C (310K). Biochemists/biological chemists use G° where the prime symbol ( ) indicates Δ ′ ′ –7 that the value of Gibbs free energy change is for pH = 7 (i.e., 1×10 ), the same + concentration of H3O (aq) as in a typical cell, i.e., e.g., C H O (aq) + 6O (g) → 6CO (g) + 6H O(l) 6 12 6 2 2 2 ΔG° = –2167 kJ ΔG°′ = –2870 kJ Note: Actually the values used in this section were really ΔG°′ where T = 37°C, [H+] = 1×10–7 and all other concentrations are at standard. PROBLEM 2: What is the ΔG°′ for the reaction: 4– 2– 3– + Glucose(aq) + ATP (aq)→ glucose 6-phosphate (aq) + ADP (aq) + H3O (aq); ΔG° = +24.8 kJ + The ΔG°′ value differs from ΔG° because one of the concentrations (i.e., H3O ) has the non- –7 standard value of 10 M. That is, ΔG°′ is ΔG for conditions such that every concentration is + 1M except for the [H3O ]. Therefore, set ΔG = ΔG°′, calculate Q and then ΔG°′. (Note the temperature would be body temperature, 37°C.) 62 ΔG°′ = ΔG = ΔG° + RT lnQ NOTES: ΔG°′ = +24.8 kJ + RT lnQ ΔG°′ = +24.8 kJ + (8.314×10-3 kJ/mol K)(273+37K) ln (1×10–7) ΔG°′ = +24.8 kJ/ – 41.5 kJ = –16.7 kJ Using LeChatelier’s principle, we predict more driving force towards products when the concentration of a product is small (e.g., 1×10–7 M versus 1M). Therefore ΔG°′ would be more negative than ΔG°. PROBLEMS 1. Is the rxn H2O(l) → H2O(g) spontaneous at 25°C under std. conditions (i.e., P{H2O, g} = 1 atm )? 2. Deduce the equilibrium vapour pressure of water at 298 K. 3. At what temperature will the equilibrium vapour pressure of water be 1 atm? [The same as asking "at what temperature does water boil at 1 atm?"] 4. Using the data available in tables, calculate a value for the solubility of AgCl in moles/liter. + – –14 5. For the reaction, 2H2O(l) ↔ H3O (aq) +OH (aq), Kw = 1.00 × 10 at 25°C, and ΔH° for the process at 25°C is +55.8 kJ. Use these data to deduce the pH of boiling water. 6. At constant T and P, a process at equilibrium is characterised by which set of equations: A.
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