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Chapter 19

Chapter 19. Chemical Thermodynamics. Chemical Thermodynamics. The study of the energy transformations that accompany chemical and physical changes. The Driving Forces. All reactions (changes) in nature occur because of the interplay of two driving forces: (1) The drive toward lower energy.

wilma-bentley
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Chapter 19

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  1. Chapter 19 Chemical Thermodynamics

  2. Chemical Thermodynamics • The study of the energy transformations that accompany chemical and physical changes.

  3. The Driving Forces • All reactions (changes) in nature occur because of the interplay of two driving forces: (1) The drive toward lower energy. Decrease in enthalpy (2) The drive toward increased disorder. Increase in entropy

  4. Enthalpy • Enthalpy is the internal energy of a system at constant pressure. • We cannot measure enthalpy directly but we can measure changes in the enthalpy of a system. • Changes in enthalpy normally are in the form of heat.

  5. Exothermic reactions –∆H Endothermic reactions +∆H Enthalpy (Energy) Change

  6. Energy Entropy

  7. Table 19.1 page 613

  8. P4O10 + 6H2O(l) → 4H3PO4 Use the information below to determine the ∆H. 4P + 5O2 → P4O10 ∆H = -2984 kJ/mol H2 + ½ O2 → H2O(l) ∆H = -285.83 kJ/mol 3/2 H2 + P + 2O2 → H3PO4 ∆H = -1267kJ/mol ∆H = -369 kJ/mol

  9. 2NH3 + 3O2 + 2CH4 → 2HCN + 6H2O Use the information below to determine the ∆H. ½ N2 + 3/2 H2 → NH3∆H = - 46 kJ/mol C + 2H2 → CH4∆H = -75 kJ/mol ½ H2 + C + ½ N2 → HCN ∆H = +135.1 kJ/mol H2 + ½ O2 → H2O ∆H = -242 kJ/mol ∆H = -940 kJ/mol

  10. Another method to determine ∆H for a reaction

  11. ∆H = ∑∆Hf(products) ─ ∑∆Hf(reactants) Appendix I in your notebook has ∆Hfvalues

  12. ∆H = ∑∆Hf(products) ─ ∑∆Hf(reactants) Determine ∆H for the reaction using the above formula. Na(s) + O2(g) + CO2(g) → Na2CO3(s) ∆Hf for Na2CO3(s) = -1130.8kJ/mol ∆H = -737.3 kJ/mol

  13. ∆H = ∑∆Hf(products) ─ ∑∆Hf(reactants) Determine ∆H for the reaction using the above formula. C2H5OH(g) + 3O2(g)→ 2CO2(g) + 3H2O(g) ∆H = -1277.4 kJ/mol

  14. Calculate the amount of energy released when 100.0g of C2H5OH is burned. C2H5OH(g) + 3O2(g)→ 2CO2(g) + 3H2O(g) ∆H = -1277.4 kJ -2772 kJ = 2772 kJ released

  15. Change in Entropy (∆S) (+∆S) = increase in disorder (entropy) (-∆S) = decrease in disorder (entropy)

  16. An Increase in Entropy (+∆S) Is a Force that Drives Physical and Chemical Changes.

  17. Second Law of Thermodynamics Every time a change occurs it increases the entropy of the universe.

  18. Second Law of Thermodynamics The second law implies that whenever a change occurs, some of the energy will be wasted which will lead to an increase in the disorder of the universe.

  19. Entropy • Entropy is a complicated concept to truly understand. • It may be best to think of entropy as the degree of dispersion. • As matter or energy disperses (becomes more free to move or becomes more spread out) entropy will increase.

  20. Entropy 1 Video ≈ 2:25

  21. Entropy increases when matter is dispersed. • Production of liquid or gas from a solid “or” production of gas from a liquid results in the dispersal of matter. • The individual particles become more free to move and generally occupy a larger volume which causes entropy to increase.

  22. Increase in Entropy (+∆S) Expansion of a gas.

  23. Increase in Entropy (+∆S) • Formation of a mixture.

  24. Increase in Entropy (+∆S) • Dissolution of a crystalline solid in water.

  25. Increase in Entropy (+∆S) • More particles are created.

  26. Increase in Entropy (+∆S) More particles are created.

  27. Decrease in Entropy (-∆S) • Is simply a reverse of the previous processes. A gas dissolves in a liquid A precipitate forms

  28. Third Law of Thermodynamics The entropy of any pure substance at 0 K is zero. We can interpret this to mean that as temperature increases entropy increases.

  29. A more complex molecule has a higher entropy.

  30. Calculating Entropy Change

  31. ∆S = ∑S(products) ─ ∑S(reactants) Determine ∆S for the reaction using the above formula. C2H5OH(g) + 3O2(g) → 2CO2(g) + 3H2O(g) ∆S=+95.64 J/mol K

  32. “Gibbs” Free Energy • Allows use to determine whether a reaction is spontaneous or not spontaneous. • We will say a process is or is not “thermodynamically favored” instead of using the terms spontaneous or not spontaneous. • This avoids common confusion with associating the term spontaneous with the idea that something must occur immediately or without cause.

  33. “Gibbs” Free Energy • If a process is “thermodynamically favored” means the products are favored at equilibrium. • The term “notthermodynamically favored” means the reactants are favored at equilibrium.

  34. Change in Free Energy (ΔG) • It is the change in free energy (ΔG) that determines whether a reaction is thermodynamically favorable or not.

  35. ∆G = ∑∆Gf(products) ─ ∑∆Gf(reactants) Determine ∆G for the reaction using the above formula. C2H5OH(g) + 3O2(g) → 2CO2(g) + 3H2O(g) ∆G = -1306 kJ/mol

  36. (-∆G) vs (+∆G) • If -∆G the process is “thermodynamically favored” and the products are favored at equilibrium. • If +∆G the process is “notthermodynamically favored” and the reactants are favored at equilibrium.

  37. Understanding ΔG • Just because a process is spontaneous or “thermodynamically favored” (-∆G) does not mean that it will proceed at any measurable rate. • A reaction that is thermodynamically favored, (spontaneous), may occur so slowly that in practice it does not occur at all.

  38. Understanding ΔG • Thermodynamically favored reactions that do not occur at any measurable rate are said to be under “kinetic control”. • These reactions often have a very high activation energy. • The fact that a process does not proceed at a noticeable rate does not mean that the reaction is at equilibrium. • We would conclude that such a reaction is simply under kinetic control.

  39. Example The reaction of diamond (pure carbon) with oxygen in the air to form carbon dioxide is thermodynamically favored (spontaneous) but has an extremely slow rate.

  40. Understanding ΔG • A reaction with a with a +ΔG may be forced to occur with the application of energy or by coupling it to thermodynamically favorable reactions.

  41. +ΔG • Energy can be used to cause a process to occur that is not thermodynamically favored. • Using electricity to charge a battery.

  42. +ΔG • Energy can be used to cause a process to occur that is not thermodynamically favored. • Photoionization of an atom by light.

  43. +ΔG • A thermodynamically unfavorable reaction may be made favorable by coupling it to a favorable reaction or series of favorable reactions. • This process involves a series of reactions with common intermediates, such that the reactions add up to produce an overall reaction that is thermodynamically favorable (–ΔG).

  44. C2H5OH(g) + 3O2(g) → 2CO2(g) + 3H2O(g) ∆H = -1277.3 kJ/mol ∆S = +95.64 J/mol K ∆G = -1306 kJ/mol

  45. ∆G = 0 Reactants ↔ Products -∆G = thermodynamically favored = products favored +∆G = not thermodynamically favored = reactants favored ∆G is 0 = equilibrium

  46. ∆G = ∆H – T∆S

  47. ∆G = ∆H – T∆S C2H5OH(g) + 3O2(g) → 2CO2(g) + 3H2O(g)Determine ∆G given: ∆H = -1277.3 kJ/mol ∆S = +95.64 J/mol K ∆G = -1306 kJ/mol or -1,306,000J/mol

  48. ∆G = ∑∆Gf(products) ─ ∑∆Gf(reactants) Determine ∆G for the reaction using the above formula. C2H5OH(g) + 3O2(g) → 2CO2(g) + 3H2O(g) ∆G = -1306 kJ/mol ∆G = ∆H – T∆S C2H5OH(g) + 3O2(g) → 2CO2(g) + 3H2O(g)Determine ∆G given: ∆H = -1277.3 kJ/mol ∆S = +95.64 J/mol K ∆G = -1306 kJ/mol

  49. Solve for ∆H: ∆G = ∆H – T∆S ∆H = ∆G + T∆S

  50. Solve for ∆S: ∆G = ∆H – T∆S or

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