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Zumdahl’s Chapter 16

Zumdahl’s Chapter 16. Spontaneity, Entropy, Free Energy, and Why All Things Happen … “The Universe Becomes Less Predictable”. Spontaneous Processes and Entropy.

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Zumdahl’s Chapter 16

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  1. Zumdahl’s Chapter 16 Spontaneity, Entropy, Free Energy, and Why All Things Happen … “The Universe Becomes Less Predictable”

  2. Spontaneous Processes and Entropy One of the main objectives in studying thermodynamics, as far as chemists are concerned, is to be able to predict whether or not a reaction will occur when reactants are brought together under a special set of conditions (for example, at a certain temperature, pressure, and concentration).

  3. Entropy is Probability • S = k ln W in modern symbolism. • W is an actual count of how many different ways the Universe could be arranged without being detectably different macroscopically. • And it is usually enormous! • For example, how many different poker hands might be in some player’s possession? • W (52)(51)(50)(49)(48) / 5! or 2,598,960. • For 4 players, that’s ~1.481024 different games. • Over twice Avogadro’s Number!

  4. Poker Microstates • One microstate in poker might be a flush; all cards of the same suit. • Wflush = 4(13)(12)(11)(10)(9) / 5! = 5148 as the number of ways to get a flush on the deal. • But Wflush/Wtotal gives ~505:1 odds against. • So flushes-on-the-deal are fairly ignorable.

  5. Chemical Microstates • Positional • In a solid, molecules are frozen in position. • But a liquid can swap molecular positions without macroscopic consequence: Sliq > Ssolid • A gas is far more chaotic: Sgas >> Sliquid!

  6. 2nd Law of Thermodynamics • “In any spontaneous process, the entropy of the Universe increases.” • We must include consideration of a system’s environment to apply this law. • For example, condensing a gas implies a large decrease in the system’s entropy! Ssys << 0 • Fortunately, the (latent) heat of vaporization gets released to force the surroundings to occupy higher energy levels, so Ssurr >> 0 and Suniv > 0! Suniv = Ssys + Ssurr  0

  7. haos Norse Mythology • Valhalla is the abode of the Norse gods. • But, contrary to many other mythologies, Norse gods are not immortal. • Valhalla is held up by a giant tree, the roots of which are being gnawed by a serpent. • The serpent will succeed, and when it does, Valhallaand the Universe will fall. • The serpent’s name is

  8. Universal Chaos, Suniv • The Norsemen were right! • There is Chaos growing in the Universe all the time at the expense of Order. It is now a fundamental principle of Science. • It’s called “entropy,” S, and is a state function that must always increase for the Universe as a whole, but some System’s Smay decrease.

  9. S order disorder S Entropy (S) is a measure of the randomness or disorder of a system. DS = Sf - Si If the change from initial to final results in an increase in randomness Sf > Si DS > 0 +DS; favors a spontaneous reaction for a chemical reaction -DS; favors a non-spontaneous reaction for a chemical reaction or does not happen 18.2

  10. nonspontaneous spontaneous Spontaneous Physical and Chemical Processes • A waterfall runs downhill • A lump of sugar dissolves in a cup of coffee • At 1 atm, water freezes below 0 0C and ice melts above 0 0C • Heat flows from a hotter object to a colder object • A gas expands in an evacuated bulb • Iron exposed to oxygen and water forms rust 18.2

  11. spontaneous nonspontaneous 18.2

  12. CH4(g) + 2O2(g) CO2(g) + 2H2O (l)DH0 = -890.4 kJ H+(aq) + OH-(aq) H2O (l)DH0 = -56.2 kJ H2O (s) H2O (l)DH0 = 6.01 kJ H2O NH4NO3(s) NH4+(aq) + NO3-(aq)DH0 = 25 kJ Enthalpy is a factor in whether a reaction is spontaneous; but is not the only factor. Spontaneous reactions 18.2

  13. An exothermic reaction (-DH) favors a spontaneous reaction since products are at a lower potential energy But does not Guarantee it. Reactants -DH Potential Energy Products

  14. H2O (s) H2O (l) Guidelines for Determining if DS is + or -. 1. For any substance, the solid state is more ordered than the liquid state and the liquid state is more ordered than gas state Ssolid < Sliquid << Sgas DS > 0

  15. Guidelines for Determining if DS is + or -. • Entropy often increases when one material dissolves in another. • Entropy increases as Temperature increases.

  16. What is the sign of the entropy change for the following reaction? 2Zn (s) + O2(g) 2ZnO (s) Guidelines for Determining if DS is + or -. 4. When gases are produced (or consumed) • If a reaction produces more gas molecules than it consumes, DS0 > 0. • If the total number of gas molecules diminishes, DS0 < 0. • If there is no net change in the total number of gas molecules, then DS0 may be positive or negative BUT DS0 will be a small number. The total number of gas molecules goes down, DS is negative. 18.3

  17. 4 moles gas 2 moles gas Higher entropy Guidelines for Determining if DS is + or -. 5. Entropy increases as number of moles of gases increases during chemical reaction. N2 (g) + 3 H2 (g) -> 2 NH3 (g) - DS 6. Entropy increases as the complexity of the molecule increases. C4H10 or CH4

  18. How does the entropy of a system change for each of the following processes? (a) Condensing water vapor Randomness decreases Entropy decreases (DS < 0) (b) Forming sucrose crystals from a supersaturated solution Randomness decreases Entropy decreases (DS < 0) (c) Heating hydrogen gas from 600C to 800C Randomness increases Entropy increases (DS > 0) (d) Subliming dry ice Randomness increases Entropy increases (DS > 0) 18.2

  19. Standard Entropy Values; S° 240

  20. Potential energy of hiker 1 and hiker 2 is the same even though they took different paths. Thermodynamics State functions are properties that are determined by the state of the system, regardless of how that condition was achieved. energy, enthalpy, pressure, volume, temperature , entropy 6.7

  21. The standard entropy of reaction (DS0 ) is the entropy change for a reaction carried out at 1 atm and 250C. rxn aS0(A) bS0(B) - [ + ] cS0(C) dS0(D) [ + ] = aA + bB cC + dD - S nS0(reactants) S nS0(products) = DS0 DS0 DS0 DS0 rxn rxn rxn rxn What is the standard entropy change for the following reaction at 250C? 2CO (g) + O2(g) 2CO2(g) S0(CO) = 197.9 J/K•mol S0(CO2) = 213.6 J/K•mol S0(O2) = 205.0 J/K•mol = 427.2 – [395.8 + 205.0] = -173.6 J/K•mol = 2 x S0(CO2) – [2 x S0(CO) + S0 (O2)] Entropy Changes in the System (DSsys)

  22. 3 Factors Involved in Determining Whether a Reaction is Spontaneous or Not • DH; a negative value (exothermic reactions) favors a spontaneous reaction • DS; a positive value (disorder increases) favors a spontaneous reaction • Temperature; If the above factors conflict; temperature determines if reaction is spontaneous or not

  23. Why do Chemical Reactions Occur? Free Energy • Free Energy

  24. Gibbs Free Energy For a constant-temperature process: Gibbs free energy (G) DG = DHsys -TDSsys DG < 0 The reaction is spontaneous in the forward direction. DG > 0 The reaction is nonspontaneous as written. The reaction is spontaneous in the reverse direction. DG = 0 The reaction is at equilibrium.

  25. DS DG=DH-TDS +DH +DS -DH +DS Spontaneous at all Temperatures Spontaneous at High Temperatures DH +DH -DS -DH -DS Spontaneous at Low Temperatures NonSpontaneous at all Temperatures

  26. Why do Chemical Reactions Occur? Free Energy

  27. DG; CHANGE IN GIBB’S FREE ENERGY DG = DHsys -TDSsys • Determine if reaction is spontaneous or not. • wmax; free energy available to do useful work. DG = wmaximum DHsys; represents energy available for useful work -TDSsys; represents energy that can’t Be harnessed to do useful work

  28. Ways to Obtain DGreaction • From DGfo Values • From Gibb’s Free Energy Equation DG = DHsys -TDSsys

  29. The standard free-energy of reaction (DG0rxn ) is the free-energy change for a reaction when it occurs under standard-state conditions. aA + bB cC + dD - [ + ] [ + ] = - nDG0 (reactants) S S = f Standard free energy of formation (DG0) is the free-energy change that occurs when 1 mole of the compound is formed from its elements in their standard states. DG0 DG0 rxn rxn f DG0 of any element in its stable form is zero. f dDG0 (D) nDG0 (products) aDG0 (A) bDG0 (B) cDG0 (C) f f f f f 18.4

  30. - mDG0 (reactants) S S = f 2C6H6(l) + 15O2(g) 12CO2(g) + 6H2O (l) DG0 DG0 DG0 - [ ] [ + ] = rxn rxn rxn [ 12 x –394.4 + 6 x –237.2 ] – [ 2 x 124.5 ] = -6405 kJ = Is the reaction spontaneous at 25 0C? 2DG0 (C6H6) 12DG0 (CO2) f f 6DG0 (H2O) f nDG0 (products) f What is the standard free-energy change for the following reaction at 25 0C? DG0 = -6405 kJ < 0 spontaneous 18.4

  31. CaCO3(s) CaO (s) + CO2(g) Determine the temperature range where the above reaction will be spontaneous. Temperature and Spontaneity of Chemical Reactions DH0 = 177.8 kJ; Does not favor spontaneous reaction DS0 = 160.5 J/K; Favors spontaneous reaction DG0 = DH0 – TDS0 0 = DH0 – TDS0 -DG; spontaneous DG=0;equilibrium +DG;nonspontaneous DH0 = TDS0 T=DH0/DS0 T = 177.8 kJ/ 0.1605 kJ/K =1108K

  32. H2O (l) H2O (g) Determine the boiling point of water given that DSvap = 109.3 J/mole K and DHvap = 40.62 kJ/mole. Gibbs Free Energy and Phase Transitions Ice melting; Liquid vaporizing Solid subliming Phase changes happen at equilibrium DG0 = 0 DG0 = DH0 – TDS0 T=DH0/DS0 T = 40.6 kJ/mole = 373 K = 100 °C 0.1093 kJ/mole K 18.4

  33. Gibbs Free Energy and Chemical Equilibrium -DG; spontaneous DG=0;equilibrium +DG;nonspontaneous Q<K; forward rxn spontaneous Q=K;equilibrium Q>K;forward rxn nonspontaneous DG = DG0 + RT lnQ(K) R is the gas constant (8.314 J/K•mol) T is the absolute temperature (K) (K)Q is the reaction quotient = (products/reactants) DGo is change in Gibbs Free Energy under standard state conditions DG is change in Gibbs Free Energy under experimental conditions

  34. Chemical equilibrium • Chemical Equilibrium predicts how reaction will proceed and how far can it go. • Chemical equilibrium does not indicate rates of reactions, just extent of reaction at equilibrium and how far the reaction is from equilibrium • Dynamic nature of chemical reaction aA + bB <------> cC + dD • A, B decrease with time, C, D increase with time until no change. • Equilibrium: rxn proceeds to a point where reactant to product balanced the reverse rxn, product to reactant.

  35. Equilibrium Equations and Equilibrium Constants • Consider the following general equilibrium reaction: aA + bB + …  mM + nN + … • Where A, B, … are the reactants; • M, N, …. are the products; • a, b, ….m, n, …. are coefficients in the balanced equation. • At equilibrium, the composition of the reaction mixture obeys an equilibrium equation.

  36. Equilibrium Equations and Equilibrium Constants • The value of K varies with temperature.

  37. The Law of Mass Action K = equilibrium constant = kforward/kreversefor single-step reactions • K depends on temperature ONLY • Don’t use solids or pure liquids (i.e. water) for the Law of Mass Action only gases or aqueous solutions • Gases have partial pressure equilibrium constants

  38. What about non-equilibrium reactions? Use the reaction quotient equation! • Can use Q to figure out which direction the reaction will go: Q = K reaction is at equilibrium Q > K reverse reaction rate is greater than forward reaction rate (leftward shift) Q < K forward reaction rate is greater than reverse reaction rate (rightward shift)

  39. 4 rules to determine what value to use for the concentrations in Q • Rule 1: Solvents (e.g. water): [l] is equal to the mole fraction of the solvent. Thus, [H2O] always equals 1. or (l)=1 so ignore • Rule 2: Pure solids in equilibrium with a solution (e.g. CaCO3(s), Fe(OH)3(s)): [s] always equals 1. • Rule 3:Gases in equilibrium with a solution (e.g. CO2(g), O2(g)): [i] equals the partial pressure of the gas (atm) • Rule 4: Compounds dissolved in water: [i] is always reported in units of moles/liter (molar).

  40. Homework # 58 • Using dat from Appendix 4, calculate ∆G for the reaction: 2H2S (g) + SO2 (g) 3S (s) + 2 H2O (g) For the following conditions at 25°C: PH2S = 1.0 x 10-4 atm PSO2 = 1.0 x 10-2 atm PH2O = 3.0 x 10-2 atm

  41. Answer • 58. ΔG° = 3(0) + 2(-229) - [2(-34) + 1(-300.)] = -90. kJ ΔG = ΔG° + RT ln = -90. kJ + ΔG = -90. kJ + 39.7 kJ = -50. kJ

  42. 66. Consider the following reaction at 298K 2 SO2(g) + O2(g) → 2 SO3(g); An equilibrium mixture contains O2(g) and SO3 (g) at partial pressures of .50 atm and 2.0 atm respectively. Using data from Appendix 4 determine the equilibrium partial pressure of SO2 in the mixture. Will this reaction be most favored at a high or low temp assuming standard conditions? • ΔG°= final -initial • ΔG° = 2(−371 kJ) − [2(−300. kJ)] = −142 kJ ΔG° = −RT ln K, ln K = = 57.311 K = e57.311 = 7.76 × 1024 K = 7.76 × 1024 = = 1.0 × 10-12 atm • From the negative value of ΔG°, this reaction is spontaneous at standard conditions. • There are more molecules of reactant gases than product gases, so ΔS° will be negative (unfavorable). • Therefore, this reaction must be exothermic (ΔH° < 0). When ΔH° and ΔS° are both negative, the reaction will be spontaneous at relatively low temperatures where the favorable ΔH° term dominates.

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