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Chapter 19

Chapter 19. Chemical Thermodynamics. No Review Quiz No Lab. 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:

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Chapter 19

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  1. Chapter 19 Chemical Thermodynamics

  2. No Review Quiz • No Lab

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

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

  5. Exothermic reactions –∆H Endothermic reactions +∆H Energy Change

  6. Entropy Some call it entropy. I call it heaven • Entropy is the amount of disorder in a system.

  7. Energy Entropy

  8. Table 19.1 page 613

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

  10. 2NH3 + 3O2 + 2CH4 → 2HCN + 6H2O Use the information below to determine the ∆H. ½ N2 + 3/2 H2 → NH3∆H = - 46 kJ C + 2H2 → CH4∆H = -75 kJ ½ H2 + C + ½ N2 → HCN ∆H = +135.1 kJ H2 + ½ O2 → H2O ∆H = -242 kJ ∆H = -940 kJ

  11. Another method to determine ∆H for a reaction

  12. ∆H = ∑∆Hf(products) ─ ∑∆Hf(reactants) Appendix I in the back of the book has ∆Hfvalues

  13. ∆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 ∆H = -737.3 kJ

  14. ∆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

  15. 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

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

  17. Increase in Entropy (Driving Force)

  18. Increase in Entropy (+∆S) • Production of liquid or gas from a solid.

  19. Increase in Entropy (+∆S) • Production of gas from a liquid.

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

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

  22. Decrease in Entropy (-∆S) • Is simply a reverse of the previous processes.

  23. 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.

  24. A more complex molecule has a higher entropy.

  25. ∆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/K

  26. ∆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

  27. Spontaneous (-∆G) vs. Nonspontaneous (+∆G) • A spontaneous process proceeds without outside influence. • Spontaneous changes may be very slow or require activation energy. • A nonspontaneous process cannot proceed without constant outside influence.

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

  29. ∆G = 0 Reactants ↔ Products ∆G is negative = spontaneous = products favored ∆G is positive = nonspontaneous = reactants favored ∆G is 0 = equilibrium

  30. ∆G = ∆H – T∆S

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

  32. ∆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 ∆G = ∆H – T∆S C2H5OH(g) + 3O2(g) → 2CO2(g) + 3H2O(g)Determine ∆G given: ∆H = -1277.3 kJ ∆S = +95.64 J/K ∆G = -1306 kJ

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

  34. Solve for ∆S: ∆G = ∆H – T∆S or

  35. Solve for T: ∆G = ∆H – T∆S or

  36. CaO(s) + SO3(g) → CaSO4(s) Calculate ∆S at 298K using the data given below. ∆H = -401.5 kJ ∆G = -345.0 kJ ∆S = -0.1896 kJ/K or -189.6 J/K

  37. ∆G = ∆H – T∆S • Can be used to explain the driving forces and spontaneity

  38. C2H5OH(g) + 3O2(g) → 2CO2(g) + 3H2O(g) ∆H = -1277.3 kJ (exothermic) ∆S = +95.64 J/K (increased disorder) ∆G = ∆H – T∆S ∆G = (-) – [+(+)] negative – positive = always negative ∆G = -1305.2 kJ (spontaneous)

  39. ∆G = ∆H – T∆S What are the other possibilities?

  40. ∆G = ∆H – T∆S ∆H = + (endothermic) ∆S = - (decreased disorder) ∆G = ∆H – T∆S ∆G = (+) – [+(-)] positive – negative = always positive ∆G = + (nonspontaneous)

  41. ∆G = ∆H – T∆S ∆H = + (endothermic) ∆S = + (increased disorder) ∆G = ∆H – T∆S ∆G = (+) – [+(+)] positive – positive = ? ∆G = + or - (spontaneous or nonspontaneous)

  42. ∆G = ∆H – T∆S ∆H = - (exothermic) ∆S = - (decreased disorder) ∆G = ∆H – T∆S ∆G = (-) – [+(-)] negative – negative = ? ∆G = + or - (spontaneous or nonspontaneous)

  43. Spontaneity • Table 19.3 Page 621 gives some specific examples that support the previous slides.

  44. ∆G = ∆H – T∆S ∆G = (+) – [+(+)] Temperature ∆G = ∆H – T∆S ∆G = (-) – [+(-)] What is the determining factor for spontaneity when only one factor favors spontaneity?

  45. CaO(s) + SO3(g) → CaSO4(s) The data given below are for 298K. Calculate ∆G using the ∆H and ∆S values below at 2463K. ∆H = -401.5 kJ ∆G = -345.0 kJ ∆S = -189.6 J/K ∆G = +65.5 kJ at 2463K Conclusion: ∆S becomes a larger factor as temperature increases

  46. CaO(s) + SO3(g) → CaSO4(s) Estimate the temperature at which the reaction becomes spontaneous. ∆H = -401.5 kJ ∆S = -189.6 J/K T = 2118K = 1845°C

  47. CaO(s) + SO3(g) → CaSO4(s) Estimate the temperature at which the reaction becomes spontaneous. ∆H = -401.5 kJ ∆S = -189.6 J/K T = 2118K = 1845°C Why is this temperature only an estimate of the temperature?

  48. CaO(s) + SO3(g) ↔ CaSO4(s) Calculate ∆S at 298K using the data given below. ∆H = -401.5 kJ ∆G = -345.0 kJ ∆S = -0.1896 kJ/K As the temperature increases above ≈ 2118K the reaction becomes nonspontaneous as ∆G becomes positive. Why?

  49. Boiling & Equilibrium • Boiling is a reversible reaction: H2O(l)↔ H2O(g) • What is the value of ∆G for the boiling water?

  50. Estimate the boiling point of ethanol (C2H5OH) . BP = 350K = 77°C

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