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First Law of Thermodynamics

First Law of Thermodynamics. energy cannot be created or destroyed. Therefore, the total energy of the universe is a constant. Energy can, however, be converted from one form to another or transferred from a system to the surroundings or vice versa. Enthalpy/Entropy.

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First Law of Thermodynamics

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  1. First Law of Thermodynamics • energy cannot be created or destroyed. • Therefore, the total energy of the universe is a constant. • Energy can, however, be converted from one form to another or transferred from a system to the surroundings or vice versa. 14-אנרגיה חופשית וספונטניות

  2. Enthalpy/Entropy • Enthalpy is the heat absorbed by a system during a constant-pressure process. • Entropy is a measure of the randomness in a system. • Both play a role in determining whether a process is spontaneous. 14-אנרגיה חופשית וספונטניות

  3. Spontaneous Processes • Spontaneous processes proceed without any outside assistance. • The gas in vessel A will spontaneously effuse into vessel B, but it will not spontaneously return to vessel A. • Processes that are spontaneous in one direction are nonspontaneous in the reverse direction. 14-אנרגיה חופשית וספונטניות

  4. Experimental Factors Affect Spontaneous Processes • Temperature and pressure can affect spontaneity. • An example of how temperature affects spontaneity is ice melting or freezing. 14-אנרגיה חופשית וספונטניות

  5. Reversible and Irreversible Processes Irreversible processes cannot be undone by exactly reversing the change to the system or cannot have the process exactly followed in reverse. Also, any spontaneous process is irreversible! Reversible process: The system changes so that the system and surroundings can be returned to the original state by exactly reversing the process. This maximizes work done by a system on the surroundings. 14-אנרגיה חופשית וספונטניות

  6. Entropy • Entropy can be thought of as a measure of the randomness of a system. • It is a state function: • It can be found by heat transfer from surroundings at a given temperature: 14-אנרגיה חופשית וספונטניות

  7. Second Law of Thermodynamics • The entropy of the universe increases in any spontaneous processes. • This results in the following relationships: 14-אנרגיה חופשית וספונטניות

  8. Entropy on the Molecular Scale • Boltzmann described entropy on the molecular level. • Gas molecule expansion: Two molecules are in the apparatus above; both start in one side. What is the likelihood they both will end up there? (1/2)2 • If one mole is used? (1/2)6.02×1023! (No chance!) • Gases spontaneously expand to fill the volume given. • Most probable arrangement of molecules: approximately equal molecules in each side 14-אנרגיה חופשית וספונטניות

  9. Statistical Thermodynamics • Thermodynamics looks at bulk properties of substances (the big picture). • We have seen what happens on the molecular scale. • How do they relate? • We use statistics (probability) to relate them. The field is called statistical thermodynamics. • Microstate: A single possible arrangement of position and kinetic energy of molecules 14-אנרגיה חופשית וספונטניות

  10. Boltzmann’s Use of Microstates • Because there are so many possible microstates, we can’t look at every picture. • W represents the number of microstates. • Entropy is a measure of how many microstates are associated with a particular macroscopic state. • The connection between the number of microstates and the entropy of the system is: • k = Boltzmann constant = 1.38 × 10−23 J/K 14-אנרגיה חופשית וספונטניות

  11. Entropy Change • Since entropy is a state function, the final value minus the initial value will give the overall change. • In this case, an increase in the number of microstates results in a positive entropy change (more disorder). 14-אנרגיה חופשית וספונטניות

  12. W These are energetically equivalent states for the expansion of a gas. It doesn’t matter, in terms of potential energy, whether the molecules are all in one flask or evenly distributed. But one of these states is more probable than the other two. 14-אנרגיה חופשית וספונטניות

  13. Macrostates → Microstates This macrostate can be achieved through several different arrangements of the particles. 14-אנרגיה חופשית וספונטניות

  14. Macrostates → Microstates These microstates all have the same macrostate. So there are six different particle arrangements that result in the same macrostate. 14-אנרגיה חופשית וספונטניות

  15. Macrostates and Probability • There is only one possible arrangement that gives state A and one that gives state B. • There are six possible arrangements that give state C. • The macrostate with the highest entropy also has the greatest dispersal of energy. • Therefore, state C has higher entropy than either state A or state B. • There is six times the probability of having the state C macrostate than either state A or state B. 14-אנרגיה חופשית וספונטניות

  16. ההגדרה הסטטיסטית של אנטרופיה FIGURE 13-1 Enumeration of microstates 14-אנרגיה חופשית וספונטניות

  17. Effect of Volume and Temperature Change on the System • If we increase volume, there are more positions possible for the molecules. This results in more microstates, so increased entropy. • If we increase temperature, the average kinetic energy increases. This results in a greater distribution of molecular speeds. Therefore, there are more possible kinetic energy values, resulting in more microstates, increasing entropy. 14-אנרגיה חופשית וספונטניות

  18. Molecular Motions • Molecules exhibit several types of motion. • Translational: Movement of the entire molecule from one place to another • Vibrational: Periodic motion of atoms within a molecule • Rotational: Rotation of the molecule about an axis • Note: More atoms means more microstates (more possible molecular motions). 14-אנרגיה חופשית וספונטניות

  19. Entropy on the Molecular Scale • The number of microstates and, therefore, the entropy tend to increase with increases in • temperature. • volume. • the number of independently moving molecules. 14-אנרגיה חופשית וספונטניות

  20. Entropy and Physical States • Entropy increases with the freedom of motion of molecules. • S(g) > S(l) > S(s) • Entropy of a system increases for processes where • gases form from either solids or liquids. • liquids or solutions form from solids. • the number of gas molecules increases during a chemical reaction. 14-אנרגיה חופשית וספונטניות

  21. Entropy DS= SfinalSinitial 14-אנרגיה חופשית וספונטניות

  22. Entropy 14-אנרגיה חופשית וספונטניות

  23. Entropy 14-אנרגיה חופשית וספונטניות

  24. Entropy Changes in Surroundings • Heat that flows into or out of the system changes the entropy of the surroundings. • For an isothermal process • At constant pressure, qsys is simply H°for the system. 14-אנרגיה חופשית וספונטניות

  25. 14-אנרגיה חופשית וספונטניות ΔfusH° T ΔfusS= = qrev T nAl = = 0.9266 mol 25.0 g 26.98 g mol-1 0.9266 mol(10.7 kJ mol-1) 933.5 K = 0.01062 kJ K-1 ΔfusS= Measuring Dispersal of E: Entropy • Calculate ΔfusS when 25.0 g of Al(s) melts at its normal melting point (660.3 °C; 1 atm). ΔfusH°(Al)=10.7 kJ mol-1 qp = ΔH Reversible if normal melting T = 660.3 °C = 933.5 K (system) ΔfusS= 10.6 J K-1

  26. qrev T ΔS = Sfinal – Sinitial= The Third Law of Thermodynamics: Absolute Entropy A perfect crystal, at absolute zero (0 K) has: • Minimum molecular motion (min. E dispersal). S = 0 (perfect crystal at 0 K) (The 3rd law of thermodynamics) Measure the heat to change from 0 K  room-T (reversibly). Gives ΔS and the absolute S at room T. Sroom T = ΔS – Sinitial =ΔS 14-אנרגיה חופשית וספונטניות

  27. 14-אנרגיה חופשית וספונטניות Absolute Entropy Values • Standard molar entropies (S°), are measured by heating 1 mol of substance from 0 K to: • A specified T (often 25 °C = 298.15 K) • At constant P = 1 bar • S° has J K-1mol-1 units. • S° is always positive, for all materials.

  28. 14-אנרגיה חופשית וספונטניות Standard Molar Entropies (298.15 K)

  29. 14-אנרגיה חופשית וספונטניות Qualitative Guidelines for Entropy • Sgas >> Sliquid > Ssolid • Gas molecules have few restrictions on motion. • Liquid molecules can slide past each other. • Solid molecules vibrate about fixed lattice points. • S[Cl2(g)] = 223 S[Br2(ℓ)] = 152 S[I2(s)] = 116 J K-1 mol-1

  30. 14-אנרגיה חופשית וספונטניות Qualitative Guidelines for Entropy • Larger, more complex, molecules have larger S: • S°complex-molecule > S°simple-molecule Larger molecules have more ways to distribute E (more bonds to hold potential and kinetic E).

  31. 14-אנרגיה חופשית וספונטניות Cl- Na+ Q1Q2 d2 d Ionic force, F = k F- Na+ Qualitative Guidelines for Entropy • Ionic solids Weaker ionic forces = larger S • (Less tightly bound = more extensive motion)

  32. 14-אנרגיה חופשית וספונטניות Qualitative Guidelines for Entropy • Solid or liquid dissolving • Larger matter dispersal. S usually* increases: J K-1 mol-1 units *Strong solvation may increase order. ΔS < 0 is possible.

  33. 14-אנרגיה חופשית וספונטניות Qualitative Guidelines for Entropy • Gas dissolving • Gas-molecule motion becomes restricted. ΔS < 0.

  34. 14-אנרגיה חופשית וספונטניות Predicting Entropy Changes • CaCO3(s) → CaO(s) + CO2(g) ΔS = 161 J K-1mol-1 • Large entropy increase: • Solid → gas, ΔS > 0. • 1 molecule → 2 molecules, ΔS > 0. H2(g) + F2(g) → 2 HF(g) ΔS = 14 J K-1mol-1 • All gases, S ≈ constant. • 2 reactant gases, 2 product gases, S ≈ constant. Actual ΔS shows a small increase.

  35. 14-אנרגיה חופשית וספונטניות 3 gas molecules →2 gas molecules. Decrease NaCl(s) → Na+(aq) + Cl-(aq) Increase Ions locked in a crystal → ions dispersed in water. Predicting Entropy Changes H2O(ℓ) → H2O(s) Decrease Liquid molecules less restricted than solid molecules. • Predict whether S will increase or decrease for: • 2 CO(g) + O2(g) → 2 CO2(g)

  36. 14-אנרגיה חופשית וספונטניות Calculating Entropy Changes • Entropy is a state function (like H). For a reaction: • ΔrS° = (nproductS°product) – S(nreactantS°reactant) Example Methanol is a common fuel additive.Calculate ΔrS° at 298 K for its combustion: 2 CH3OH(ℓ) + 3 O2(g) → 2 CO2(g) + 4 H2O(ℓ) Look up S° values (J K-1 mol-1): CH3OH(ℓ) = 126.8 O2(g) = 205.138 CO2(g) = 213.74 H2O(ℓ) = 69.91

  37. 14-אנרגיה חופשית וספונטניות Calculating Entropy Changes Example: ΔrS° at 298 K for: 2 CH3OH(ℓ) + 3 O2(g) → 2 CO2(g) + 4 H2O(ℓ) ΔrS° = (nproductS°product) – (nreactantS°reactant) = (2S°CO2+ 4S°H2O) – (2S°CH3OH+ 3S°O2) = [2(213.74) + 4(69.91)] – [2(126.8) + 3(205.14)] = 707.12 – 869.02 = –161.90 J K-1 mol-1 Sdecreases as expected: 3 gas + 2 ℓ→ 2 gas + 4 ℓ

  38. חישוב האנטרופיה בהתפשטות איזותרמית הפיכה עבור גז אידאלי 14-אנרגיה חופשית וספונטניות

  39. Vapor-Pressure Lowering of Solutions: Raoult’s Law 14-אנרגיה חופשית וספונטניות

  40. Boiling-Point Elevation and Freezing-Point Depression of Solutions 14-אנרגיה חופשית וספונטניות

  41. Boiling-Point Elevation and Freezing-Point Depression of Solutions 14-אנרגיה חופשית וספונטניות

  42. 14-אנרגיה חופשית וספונטניות Entropy & the 2nd Law of Thermodynamics E dispersal is accompanied by an increase in disorder of a system: Increased disorder = increased S • Second Law of Thermodynamics • “The total entropy of the universe is continuously increasing.” • The universe is slowly becoming more disordered.

  43. 14-אנרגיה חופשית וספונטניות ΔrHsurr T -ΔrHsys T qrev T Entropy & the 2nd Law of Thermodynamics The entropy of a system can increase or decrease. • Suniversealways goes up. • Suniverse = Ssystem + Ssurr (No subscript? S = Ssystem ). • If Ssystem falls, Ssurr must increase by a larger amount. • ΔrS°surr = = =

  44. 14-אנרגיה חופשית וספונטניות ° Calculate ΔrSunivfor the combustion of octane in air: 2 C8H18(g) + 25 O2(g)  16 CO2(g) + 18 H2O(ℓ) For the reaction ΔrH° = -11,024 kJ mol-1 and ΔrS° = -1383.9 J mol-1K-1at 298.15 K. ΔrSuniv= ΔrSsys+ ΔrSsurr ΔrSuniv= -1383.9 J/K + (+11,024 x 103 J)/298.15 K = -1383.9 J/K + 36.975 kJ/K = +35,591 J/K ° ° ° -ΔHsys / T given Entropy & the 2nd Law of Thermodynamics °

  45. 14-אנרגיה חופשית וספונטניות Gibbs Free Energy • Neither entropy (S), nor enthalpy (H ), alone can predict whether a reaction is product favored. • Spontaneous (product favored) reactions can: • Be exothermic or endothermic. • Increase or decrease Ssystem. The Gibbs free energy (G), combines H and S. ΔrGdoes predict if a reaction is product favored or not.

  46. 14-אנרגיה חופשית וספונטניות Gibbs Free Energy Josiah Willard Gibbs • For a constant T change, ΔG is equal to: • ΔrG = ΔrH - TΔrS If G: Decreases (ΔrG< 0) a reaction is product favored Increases (ΔrG> 0) a reaction is reactant favored (at constantP and T).

  47. 14-אנרגיה חופשית וספונטניות Effect of T on Reaction Direction ΔrG= ΔrH– TΔrS

  48. 14-אנרגיה חופשית וספונטניות Gibbs Free Energy • ΔrG° can be calculated from ΔrH° and ΔrS° values… • … or from Gibbs free energies of formation (ΔfG°). ΔrG° = (nproductΔfG°product) – (nreactantΔfG°reactant) ΔfG° equals: • ΔrGto make 1 mol of compound from its elements. • 0 for an element in its most stable state (like ΔfH°)

  49. 14-אנרגיה חופשית וספונטניות Gibbs Free Energy • Calculate ΔrG° for the following reaction at 25°C: • CO(g) + 2 H2(g) → CH3OH(ℓ) ΔfG°values: CO(g) = -137.2 kJ/mol CH3OH(ℓ) = -166.3 kJ/mol ΔG° = ΔfG°CH3OH– (2ΔfG°H2+ 1ΔfG°CO) = -166.3 kJ/mol – [2(0) + 1(-137.2kJ/mol)] = -29.1 kJ/mol Product favored element in its standard state

  50. 14-אנרגיה חופשית וספונטניות 0 = ΔrH° − TΔrS° ΔrH°= TΔrS° T= ΔrH° ΔrS° Effect of T on Reaction Direction A reactions will change from reactant to product-favored (or visa versa) when ΔrG° = 0. ΔrG° = 0 = ΔrH°− TΔrS° ΔrH° and ΔrS° typically remain ≈constant over large T ranges and thus the “switchover” T can be calculated:

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