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Thermodynamics

Thermodynamics. Chapter 18. 1 st Law of Thermodynamics. Energy is conserved. E = q + w. SPONTANEOUS : occur without any outside intervention. Example: drop an egg. The REVERSE is not spontaneous!!.

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Thermodynamics

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  1. Thermodynamics Chapter 18

  2. 1st Law of Thermodynamics Energy is conserved. E = q + w

  3. SPONTANEOUS: occur without any outside intervention Example: drop an egg The REVERSE is not spontaneous!!

  4. REVERSIBLE PROCESS: change can be restored to its’ original state by exactly reversing the change. Example: ice water at 0o C

  5. IRREVERSIBLE PROCESS: cannot simply be reversed to original state. Example: gas expanding

  6. Processes in which the disorder of the system increases tend to occur spontaneously. Ex: gas expanding, ice melting, salt dissolving

  7. ENTROPY: (S) the change in disorder. (Change in randomness) The more disorder, the larger the entropy. S = Sfinal - Sinitial

  8. S = > 0 when the final state is in more disorder S = < 0 when the final state is more ordered than original

  9. H2O (l) H2O (s) Ag+(aq) + Cl-(aq) AgCl(s)

  10. - a solid melts - a liquid vaporizes - a solid dissolves in water - a gas liquefies

  11. For a process at constant temperature, the entropy change is the value of qrev divided by the absolute temperature. S = qrev/T

  12. Example: Calculate the entropy change when 1 mol of water is converted into 1 mol of steam at 1 atm pressure. (Hvap = 40.67 kJ/mol)

  13. (1 mole)(40.67 kJ/ mol)(1000 J/1 kJ) 373 K S = 109 J/K

  14. The normal freezing point of mercury is -38.9oC, an its molar enthalpy of fusion is Hfus = 2.331 kJ/mol. What is the entropy change when 50.0 g of Hg(l) freezes at the normal freezing point?

  15. -2.48 J/K The answer is negative because the process brings more order

  16. The normal boiling point of ethanol, is 78.3oC and its molar enthalpy of vaporization is 38.56 kJ/mol. What is the change in entropy when 25.8 g of C2H5OH(g) condenses to liquid at the normal boiling point?

  17. -61.4 J/K

  18. The element gallium, Ga, freezes at 29.8oC, and its enthalpy of fusion is 5.59 kJ/mol. Calculate the value of S for the freezing of 90.0 g of Ga(l).

  19. S = -23.8 J/K

  20. 2nd Law of Thermodynamics: In any reversible process, Suniverse = 0. In any irreversible (spontaneous) process, Suniverse > 0. Suniverse = Ssystem + Ssurroundings

  21. On a Molecular Level TRANSLATIONAL MOTION: movement of molecules VIBRATIONAL MOTION: the movement of atoms within the molecule. ROTATIONAL MOTION: the molecules spinning

  22. Increasing Temperature Increases Entropy

  23. 3rd Law of Thermodynamics: the entropy of a pure crystalline substance at absolute zero is zero. S(0K) = 0

  24. In general, the entropy increases when: Liquids or solutions are formed from solids Gases are formed from either solids or liquids The number of molecules of gas increases during a chemical reaction.

  25. CaCO3(s) CaO(s) + CO2(g) N2(g) + 3H2(g)  2NH3(g)

  26. Standard molar entropies: (So) absolute entropies for substances in their standard state. (J/mol-K) 1. Unlike enthalpies of formation, the standard molar entropies of elements are not zero. 2. The So of gases are greater than those of liquids and solids. 3. The So generally increases with increasing molar mass. 4. The So generally increase with the number of atoms in the formula.

  27. So = nSo(products) - mSo(reactants) Calculate So for the synthesis of ammonia: N2(g) + 3H2(g)  2NH3(g)

  28. So = (2 mol)(192.5 J/mol-K) - [(1 mol)(191.5 J/mol-K) + (3 mol)(130.6 J/mol-K)] = -198.3 J/K

  29. Using Appendix C, calculate the standard entropy change, for the following reaction: Al2O3(s) + 3H2(g) 2Al(s) + 3H2O(g)

  30. 180.4 J/K

  31. C2H4(g) + H2(g) C2H6(g) NH3(g) + HCl(g) NH4Cl(s)

  32. -120.5 J/K -284.6 J/K

  33. GIBBS FREE ENERGY The spontaneity of a reaction involves both enthalpy and entropy. The relationship is known as free energy. G = H - TS

  34. 1. If G is negative, the reaction is spontaneous in the forward direction. 2. If G is zero, the reaction is at equilibrium. 3. If G is positive, the reaction in the forward direction is nonspontaneous; work must be supplied from the surroundings to make it occur. However, the reverse reaction will be spontaneous.

  35. G

  36. Go = nGfo(products) - mGfo(reactants) N2(g) + 3H2(g) 2NH3(g)

  37. -33.32 kJ

  38. CH4(g) + 2O2(g) CO2(g) + 2H2O(g)

  39. CH4(g) + 2O2(g) CO2(g) + 2H2O(g) -800.7 kJ

  40. Assuming no change for Ho and So, what happens to Go with an increase in temperature? N2(g) + 3H2(g) 2NH3(g)

  41. Calculate G at 298K for a reaction mixture that consists of 1.0 atm N2, 3.0 atm H2 and 0.50 atm NH3. N2(g) + 3H2(g) 2NH3(g)

  42. Use standard free energies of formation to calculate the equilibrium constant K at 25oC for the reaction involved in the Haber process.

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