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Thermodynamics Chemical reactions proceed according to the rules of thermodynamics The law of conservation of energy – energy can be converted from one.

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Presentation on theme: "Thermodynamics Chemical reactions proceed according to the rules of thermodynamics The law of conservation of energy – energy can be converted from one."— Presentation transcript:

1 Thermodynamics Chemical reactions proceed according to the rules of thermodynamics The law of conservation of energy – energy can be converted from one form to another but the total amount of energy is constant Entropy – the universe is becoming more chaotic ACK!

2 Gas constant: R = 8.315 Joules/K* mol or
Thermodynamics Some constants Gas constant: R = Joules/K* mol or cal/K.mol Faradays constant: F = Joules/Volt.mol or 23062 cal/Volt* mol

3 Energy: definitions Energy – ability to do work Energetics – energy transfer Types of energy Potential – trapped energy Kinetic – energy of movement

4 Energy Categories: more definitions
Radiant energy – energy released from one object to another Mechanical energy – energy to move objects from place to place Electrical energy – energy that results from the movement of charged particles down a charge gradient Thermal energy – reflected in the movement of particles and serves to increase temperature Chemical energy – energy that is held within chemical bonds

5 Energy Categories, Cont.
Animals rely on all five types of energy, which are interconvertible

6 Food Webs are Transfers of Energy
Figure 2.3

7 Thermodynamics in a biological setting
Free Energy (G) 1. Change in free Energy (ΔG) ΔG = Products – Reactants ΔG negative – reaction will proceed forward → ΔG positive – reaction will proceed backward ← ΔG zero – reaction at equilibrium ↔ 2. Standard free Energy – ΔGo: 298 K (25oC), 1 atm pressure, pH 7.0 and 1M [initial] for all reactants and products

8  Thermal energy   movement of molecules
Most chemical reactions involve changes in thermal energy Exothermic reactions – release heat Endothermic reactions – absorb heat

9 Chemical Reactions and Thermal Energy
Enthalpy Enthalpy – average thermal energy of a collection of molecules i.e. bond energy Change in enthalpy (DH) = Hproducts – Hsubstrates Exothermic: DH is negative i.e. C6H12O6 + 6O → 6CO2 + 6H2O + energy Endothermic: DH is positive i.e. ADP + Pi → ATP

10 Chemical Reactions and Thermal Energy
Enthalpy and Entropy together Entropy (S) – measure of randomness or disorder Exothermic: DH is negative, increase in DS → reaction will occur spontaneously – negative DG Endothermic: DH is positive, DS is positive → reaction will occur spontaneously. It has to overcome the positive DH

11 Free Energy: calculations
Free energy changes of reactions are additive (coupled reactions): Consider the phosphorylation of glucose to glucose 6-phosphate: DGo: glucose + Pi ↔ glucose-6-phosphate + H2O = 3.3 kcal/mol DGo: ATP + H2O ↔ ADP + Pi = -7.3 kcal/mol Summing these reactions together: ATP + glucose ↔ ADP + glucose 6-phosphate   DG° = (-7.3) = - 4kcal/mol (favourable)

12 DG = DGo + RTln ([products]/[reactants])
Biological reactions DG = DGo + RTln ([products]/[reactants]) Where R = gas constant, T = temperature in Kelvin Example: glucose + ATP ↔ glucose-6-phosphte + ADP DGo: glucose + Pi ↔ glucose-6-phosphate + H2O = 3.3 kcal/mol DGo: ATP + H2O ↔ ADP + Pi = -7.3 kcal/mol Glucose: [5mM]; ATP: [2mM]; ADP: [0.15mM]; glucose-6-phosphate: [0.05mM] So, DG = kcal/mol cal/K mol)(298K)ln((0.05*0.15)/(5*2)) = -8.26kcal/mol Recall DG = products - reactants

13 ΔG for reactions that don’t make or break bonds
DGo is zero - Examples: glucose transport, ion transport across membranes DG = RTln ([inside]/[outside]) Or for charged ions: DG = RTln ([inside]/[outside]) + zFEm where z = valence of the ion; F = Faraday constant and Em = membrane potential

14 Transport across membranes
DG = RTln ([inside]/[outside]) + zFEm where z = valence of the ion; F = Faraday constant and Em = membrane potential Example: Diffusion of Cl- from out to in Cl- outside cell: 120mM; Cl- inside cell: 10mM; Em = -80mV DG = (1.987cal/K mol)(298K)(ln(10/120) + (-1)(23062 cal/V mol)(-0.08V) = 376 cal/mol


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