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Chemical Thermodynamics: Energy Transformations in Reactions

Explore the energy transformations in chemical and physical changes, driven by enthalpy and entropy. Learn to calculate ∆H, ∆S, and ∆G for various reactions and understand their spontaneity. Discover the relationship between energy, disorder, and spontaneity in chemical reactions.

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Chemical Thermodynamics: Energy Transformations in Reactions

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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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