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Introduction to Entropy

Introduction to Entropy. by Mike Roller. THIS MEANS THAT THINGS FALL. THEY FALL FROM HEIGHTS OF ENERGY AND STRUCTURED INFORMATION INTO MEANINGLESS, POWERLESS DISORDER. MATTER IS ENERGY. ENERGY IS INFORMATION. EVERYTHING IS INFORMATION.

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Introduction to Entropy

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  1. Introduction to Entropy by Mike Roller

  2. THIS MEANS THAT THINGS FALL. THEY FALL FROM HEIGHTS OF ENERGY AND STRUCTURED INFORMATION INTO MEANINGLESS, POWERLESS DISORDER. MATTER IS ENERGY. ENERGY IS INFORMATION. EVERYTHING IS INFORMATION. PHYSICS SAYS THAT STRUCTURES... BUILDINGS, SOCIETIES, IDEOLOGIES... WILL SEEK THEIR POINT OF LEAST ENERGY. THIS IS CALLED ENTROPY. Entropy (S) = a measure of randomness or disorder

  3. Entropy: Time’s Arrow

  4. Second Law of Thermodynamics occurs without outside intervention  In any spontaneous process, the entropy of the universe increases. ΔSuniverse > 0 Another version of the 2nd Law: Energy spontaneously spreads out if it has no outside resistance Entropy measures the spontaneous dispersal of energy as a function of temperature How much energy is spread out How widely spread out it becomes Entropy change = “energy dispersed”/T

  5. Entropy of the Universe ΔSuniverse = ΔSsystem + ΔSsurroundings Positional disorder Energetic disorder ΔSuniverse > 0  spontaneous process Both ΔSsys and ΔSsurr positive Both ΔSsys and ΔSsurr negative ΔSsys negative, ΔSsurr positive ΔSsys positive, ΔSsurr negative spontaneous process. nonspontaneous process. depends depends

  6. Entropy of the Surroundings(Energetic Disorder) System Entropy Heat ΔSsurr > 0 ΔHsys < 0 Surroundings System Surroundings Heat Entropy ΔSsurr < 0 ΔHsys > 0 Low T  large entropy change (surroundings) High T  small entropy change (surroundings)

  7. Positional Disorder and Probability Probability of 1 particle in left bulb = ½ " 2 particles both in left bulb = (½)(½) = ¼ " 3 particles all in left bulb = (½)(½)(½) = 1/8 " 4 " all " = (½)(½)(½)(½) = 1/16 " 10 " all " = (½)10 = 1/1024 " 20 " all " = (½)20 = 1/1048576 " a mole of " all " = (½)6.021023 The arrangement with the greatest entropy is the one with the highest probability (most “spread out”).

  8. Entropy of the System: Positional Disorder Ssolid < Sliquid << Sgas Ludwig Boltzmann Ordered states Low probability (few ways) Ludwig Boltzmann Low S Disordered states High probability (many ways) High S Ssystem Positional disorder S increases with increasing # of possible positions

  9. The Third Law of Thermodynamics The Third Law: The entropy of a perfect crystal at 0 K is zero. Everything in its place No molecular motion

  10. Entropy Curve Solid Liquid Gas  vaporization S (qrev/T) (J/K)  fusion 0 Temperature (K) 0 S° (absolute entropy) can be calculated for any substance

  11. Entropy Increases with... • Melting (fusion) Sliquid > Ssolid ΔHfusion/Tfusion = ΔSfusion • Vaporization Sgas > Sliquid ΔHvaporization/Tvaporization = ΔSvaporization • Increasing ngas in a reaction • Heating ST2 > ST1 if T2 > T1 • Dissolving (usually) Ssolution > (Ssolvent + Ssolute) • Molecular complexity more bonds, more entropy • Atomic complexity more e-, protons, neutrons

  12. Recap: Characteristics of Entropy • S is a state function • S is extensive (more stuff, more entropy) • At 0 K, S = 0 (we can know absolute entropy) • S > 0 for elements and compounds in their standard states • ΔS°rxn = nS°products - nS°reactants • Raise T  increase S • Increase ngas  increase S • More complex systems  larger S

  13. Entropy and Gibbs Free Energy by Mike Roller

  14. Entropy (S) Review • ΔSuniverse > 0 for spontaneous processes • ΔSuniverse = ΔSsystem + ΔSsurroundings  positional  energetic • We can know the absolute entropy value for a substance • S° values for elements & compounds in their standard states are tabulated (Appendix C, p. 1019) • For any chemical reaction, we can calculate ΔS°rxn: • ΔS°rxn = S°(products) - S°(reactants)

  15. ΔSuniverse and Chemical Reactions • ΔSuniverse = ΔSsystem + ΔSsurroundings • For a system of reactants and products, • ΔSuniverse = ΔSrxn – ΔHrxn/T • If ΔSuniverse > 0, the reaction is spontaneous • If ΔSuniverse < 0, the reaction is not spontaneous • The reverse reaction is spontaneous • If ΔSuniverse = 0, the reaction is at equilibrium • Neither the forward nor the reverse reaction is favored

  16. C6H12O6(s) + 6 O2(g) 6 CO2(g) + 6 H2O(g) Compound C6H12O6(s) O2(g) CO2(g) H2O(g) ΔH°f (kJ/mol) -1275 0 -393.5 -242 S° (J/mol K) 212 205 214 189 ΔSuniverse = ΔSrxn – ΔHrxn/T ΔS°rxn = S°(products) - S°(reactants) = [6 S°(CO2(g)) + 6 S°(H2O(g))] – [S°(C6H12O6(s)) + 6 S°(O2(g))] = [6(214) + 6(189)] – [(212) + 6(205)] J/K ΔS°rxn= 976 J/K ΔH°rxn = ΔH°f (products) - ΔH°f(reactants) = [6 ΔH°f(CO2(g)) + 6 ΔH°f(H2O(g))] – [ΔH°f(C6H12O6(s)) + 6 ΔH°f(O2(g))] = [6(-393.5) + 6(-242)] – [(-1275) + 6(0)] kJ ΔH°rxn = -2538 kJ

  17. C6H12O6(s) + 6 O2(g) 6 CO2(g) + 6 H2O(g) Compound C6H12O6(s) O2(g) CO2(g) H2O(g) ΔH°f (kJ/mol) -1275 0 -393.5 -242 S° (J/mol K) 212 205 214 189 ΔSuniverse = ΔSrxn – ΔHrxn/T ΔS°rxn = 976 J/K (per mole of glucose) ΔH°rxn = -2538 kJ (per mole of glucose) At 298 K, ΔS°universe = 0.976 kJ/K – (-2538 kJ/298 K) ΔS°universe = 9.5 kJ/K

  18. Gibbs Free Energy (G) ΔSuniverse = ΔS – ΔH/T ΔG = ΔH – TΔS Divide both sides by –T -ΔG/T = -ΔH/T + ΔS G = H – TS At constant temperature, ΔG = ΔH – TΔS (system’s point of view) –ΔGmeans+ΔSuniv A process (at constant T, P) is spontaneous if free energy decreases Josiah Gibbs

  19. ΔG and Chemical Reactions • ΔG = ΔH – TΔS • If ΔG < 0, the reaction is spontaneous • If ΔG > 0, the reaction is not spontaneous • The reverse reaction is spontaneous • If ΔG = 0, the reaction is at equilibrium • Neither the forward nor the reverse reaction is favored • ΔG is an extensive state function

  20. Ba(OH)2(s) + 2NH4Cl(s) BaCl2(s) + 2NH3(g) + 2 H2O(l) ΔH°rxn = 50.0 kJ (per mole Ba(OH)2) ΔS°rxn = 328 J/K (per mole Ba(OH)2) ΔG = ΔH - TΔS ΔG° = 50.0 kJ – 298 K(0.328 kJ/K) ΔG° = – 47.7 kJ Spontaneous At what T does the reaction stop being spontaneous? The T where ΔG = 0. ΔG = 0 = 50.0 kJ – T(0.328 J/K) 50.0 kJ = T(0.328 J/K) T = 152 K not spontaneous below 152 K

  21. Effect of ΔH and ΔS on Spontaneity ΔG = ΔH – TΔS ΔG negative  spontaneous reaction ΔH – + – + ΔS + + – – • Spontaneous? • Spontaneous at all temps • Spontaneous at high temps • Reverse reaction spontaneous at low temps • Spontaneous at low temps • Reverse reaction spontaneous at high temps • Not spontaneous at any temp

  22. Ways to Calculate ΔG°rxn 1.ΔG° = ΔG°f(products) - ΔG°f(reactants) • ΔG°f = free energy change when forming 1 mole of compound from elements in their standard states 2.ΔG° = ΔH° - TΔS° 3.ΔG° can be calculated by combining ΔG° values for several reactions • Just like with ΔH° and Hess’s Law

  23. 2H2(g) + O2(g) 2 H2O(g) 1.ΔG° = ΔG°f(products) - ΔG°f(reactants) ΔG°f(O2(g)) = 0 ΔG°f(H2(g)) = 0 ΔG°f(H2O(g)) = -229 kJ/mol ΔG° = (2(-229 kJ) – 2(0) – 0) kJ = -458 kJ 2.ΔG° = ΔH° - TΔS° ΔH° = -484 kJ ΔS° = -89 J/K ΔG° = -484 kJ – 298 K(-0.089 kJ/K) = -457 kJ

  24. 2H2(g) + O2(g) 2 H2O(g) 3.ΔG° = combination of ΔG° from other reactions (like Hess’s Law) 2H2O(l) 2H2(g) + O2(g)ΔG°1 = 475 kJ H2O(l) H2O(g)ΔG°2 = 8 kJ ΔG° = - ΔG°1 + 2(ΔG°2) ΔG° = -475 kJ + 16 kJ = -459 kJ Method 1: -458 kJ Method 2: -457 kJ Method 3: -459 kJ

  25. What is Free Energy, Really? • NOT just “another form of energy” • Free Energy is the energy available to do useful work • If ΔG is negative, the system can do work (wmax = ΔG) • If ΔG is positive, then ΔG is the work required to make the process happen • Example: Photosynthesis • 6 CO2 + 6 H2O  C6H12O6 + 6 O2 • ΔG = 2870 kJ/mol of glucose at 25°C • 2870 kJ of work is required to photosynthesize 1 mole of glucose

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