Lec18221

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Lec18221

  1. 1. Thermochemistry: Hess’s Law and Standard Heats of Reaction Lecture 18
  2. 2. It is very difficult to carry out a chemical reaction separately from others.
  3. 3. However, even if we can’t run a reaction in the lab, it is still possible to find its enthalpy change.
  4. 4. Just imagine: we can find ΔH of any reaction for which we can write an equation, even if it is impossible to carry out. How can we?
  5. 5. H, E, P, and V are state functions. It means that their values depend only on the difference between X final and X initial – and do not on the path of change. And it is awesome!
  6. 6. <ul><li>Germain Henri Hess (1802-1850), Russian scientist </li></ul>
  7. 7. Hess’s law of heat summation: The enthalpy change of an overall process is the sum of enthalpy changes of its individual steps.
  8. 8. Hess’s law of heat summation
  9. 9. Hess’s law of heat summation
  10. 10. Hess’s law of heat summation ΔH 1 = ΔH 2 + ΔH 3 -890 kJ = -802 kJ + (-88 kJ)
  11. 11. We imagine an overall reaction as the sum of a series of reaction steps whether or not it really occurs that way. Each step is chosen because its ΔH is known.
  12. 12. A sample problem on using Hess’s law to calculate an unknown ΔH.
  13. 13. Application of Hess’s law
  14. 14. How to calculate an unknown ΔH: <ul><li>Identify the target equation, the step whose ΔH is unknown, and note the number of moles of reactants and products; </li></ul><ul><li>Manipulate the equations with known ΔH values so that the target numbers of moles of reactants and products are on the correct sides; </li></ul>
  15. 15. How to calculate an unknown ΔH: <ul><li>Remember to change the sign of ΔH when you reverse an equation; </li></ul><ul><li>Remember to multiply numbers of moles and ΔH by the same factor; </li></ul><ul><li>Add the manipulated equations to obtain the target equation; </li></ul><ul><li>All substances except those in the target equation must cancel. Add their ΔH values to obtain the unknown ΔH. </li></ul>
  16. 16. Another sample problem on using Hess’s law to calculate an unknown ΔH.
  17. 17. <ul><li>Problem: Find the enthalpy change in the process of oxidation of sulfur to sulfur trioxide. </li></ul><ul><li>Plan : Imagine this process in two steps: 1)formation of sulfur dioxide from sulfur and oxygen; 2) formation of sulfur trioxide from sulfur dioxide and oxygen  write and balance equations of the consecutive reactions  write the overall equation  f ind the overall enthalpy change by summation of the heats. </li></ul><ul><li>Solution: S (s) + O 2(g)  SO (g) ΔH 1 = -296.8 kJ </li></ul><ul><li>2SO 2(g) + O 2(g)  2SO 3 (g) ΔH 2 = -198.4 kJ </li></ul><ul><li>SO 2(g) + 1/2O 2(g)  SO 3 (g) ½ΔH 2 = -99.2 kJ </li></ul><ul><li>S (s)   3/2O 2(g)  2SO 3 (g) </li></ul><ul><li>ΔH 3 =ΔH 1 + 1/2ΔH 2 = -296.8kJ + ( -90.29 kJ ) = -396.0 kJ </li></ul>
  18. 18. However, in the preceding problem, where did we take ΔH for S (s) + O 2(g)  SO (g) ?
  19. 19. There are standard heats of reactions. They have been tabulated in chemistry reference books.
  20. 20. Standard heat of a reaction is the change of enthalpy that accompanies this reaction and has been measured with all the reactants and products in their standard states.
  21. 21. Why are the standard heats necessary? Because thermodynamic variables, such as ΔH, vary somewhat with conditions.
  22. 22. The standard state for a gas: <ul><li>Pressure 1 atm , 101325 Pa with the gas behaving ideally. </li></ul>
  23. 23. The standard state for a substance in aqueous solution: <ul><li>Concentration 1 mol/L , 1M. </li></ul>
  24. 24. The standard state for a pure substance (element or compound): <ul><li>Most stable form of the substance at 1 atm and the temperature of interest, usually 298K (25°C). </li></ul>
  25. 25. Standard heat of reaction
  26. 26. Heat of formation
  27. 27. Standard heat of formation is the standard heat of a reaction, in which 1 mol of a compound forms from its elements.
  28. 28. Standard heats of formation C (graphite) + 2H 2(g)  CH 4(g) ΔH 0 f = -74.9 kJ Na (s) + 1/2Cl 2(g)  NaCl (s) ΔH 0 f = -411.1 kJ 2C (graphite) + 3H 2(g) + 1/2O 2(g) = C 2 H 5 OH (l) ΔH 0 f = -277.6 kJ 
  29. 30. Some points <ul><li>An element in its standard state is assigned a ΔH 0 f of zero. </li></ul><ul><li>Most compounds have a negative ΔH 0 f . </li></ul>
  30. 31. A sample problem on writing formation equations.
  31. 32. A sample problem on calculating heats of reaction from heats of formation.
  32. 33. Application of Hess’s law
  33. 34. The heat of reaction: ΔH 0 rxn = Σ(ΔH 0 products ) - Σ(ΔH 0 reactants )
  34. 35. THE END

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