1. Hess’ Law

2. 12-2 Practice Problems 1)-79.6 kJ 2)-155 kJ 3)17.4 kJ 4)-3.70 kJ 5)-142 kJ 6) 88.1 kJ 7) -79.0 kJ 8) -2660 kJ or -2630 kJ 9) -256 kJ, exo 10) 3.96 kJ, endo

3. 12-2 apply 1.-306 kJ 2.-255 kJ 3.Propane is more per mole, but methane is more per gram 4.-1937 BTU and -844 BTU 5.Want methane b/c you get more heat per gram.

4. Hess’ Law Last set of notes for chemistry!!

5. Hess’ Law Sometimes a reaction does not occur in just one step; it’s actually the sum of two or more intermediate equations. – Sometimes you can’t take a direct flight to London; you have to have a layover.

6. There are actually 3 rxns occurring one after the other. The net result the combination of all 3 rxns. Notice that the activation energy of the first reaction is that greatest, that is why that reaction is labeled slow. Potential energy diagram with multiple steps (equations)

7. The heat of reaction (H) for the overall equation can be determined by summing (adding) the heats of reaction for the intermediate equations.

8. When given the intermediate equations, you may have to flip &/or multiply them by a factor in order to get them to equal the overall equation. Whatever is done to the intermediate equation must be done to H!

9. Steps Cancel substances that occur in more than one intermediate reaction Cuts down on some of the information so we can determine what we need to do. Determine if you will have to flip &/or multiply that equation by a factor by comparing to overall equation. You may only change the intermediate equations! Rewrite H, multiplying by factor and/or flipping the sign. Add H.

10. Overall Equation C diamond  C graphite Intermediate equation C graphite + O 2  CO 2 H = -393.5 kJ C diamond + O 2  CO 2 H = -395.4 kJ

11. Overall Equation C diamond  C graphite Intermediate equation C graphite + O 2  CO 2 H = -393.5 kJ C diamond + O 2  CO 2 H = -395.4 kJ Cancel substances that occur in both intermediate equations as they will not help us determine what to do.

12. Overall Equation C diamond  C graphite Intermediate equation C graphite + O 2  CO 2 H = +393.5 kJ C diamond + O 2  CO 2 H = -395.4 kJ Since C graphite in the intermediate equation is a reactant, but a product in the overall equation, this equation must be flipped. When an equation is flipped, the sign on  H is flipped!

13. Overall Equation C diamond  C graphite Intermediate equation C graphite + O 2  CO 2 H = +393.5 kJ C diamond + O 2  CO 2 H = -395.4 kJ Since C diamond in the intermediate equation is a reactant and a reactant in the overall equation, nothing must be done to this equation.

14. Overall Equation C diamond  C graphite Intermediate equation C graphite + O 2  CO 2 H = +393.5 kJ C diamond + O 2  CO 2 H = -395.4 kJ  H for the overall equation is the sum of  H’s of the intermediates, after we have made the changes, in this case  H of the overall equation is -1.9 kJ

15. Overall Equation PbCl 2 + Cl 2  PbCl 4 Intermediate equations Pb + 2Cl 2  PbCl 4 H = -392.2 kJ Pb + Cl 2  PbCl 2 H = -359.4 kJ

16. Overall Equation PbCl 2 + Cl 2  PbCl 4 Intermediate equations Pb + 2Cl 2  PbCl 4 H = -392.2 kJ Pb + Cl 2  PbCl 2 H = -359.4 kJ Cancel substances that occur in both intermediate equations. Even though the coefficients do not match, that is ok. Do not use either Pb or Cl 2 to determine what needs to be done.

17. Overall Equation PbCl 2 + Cl 2  PbCl 4 Intermediate equations Pb + 2Cl 2  PbCl 4 H = -392.2 kJ Pb + Cl 2  PbCl 2 H = -359.4 kJ PbCl 4 is a product in both the intermediate equation and the overall equation. The coefficients are the same also so nothing is done to the first equation.

18. Overall Equation PbCl 2 + Cl 2  PbCl 4 Intermediate equations Pb + 2Cl 2  PbCl 4 H = -392.2 kJ Pb + Cl 2  PbCl 2 H = +359.4 kJ PbCl 2 is a product in the intermediate equation and a reactant in the overall equation, so the second equation must be flipped. When the equation is flipped the sign of  H is also flipped.

19. Overall Equation PbCl 2 + Cl 2  PbCl 4 Intermediate equations Pb + 2Cl 2  PbCl 4 H = -392.2 kJ Pb + Cl 2  PbCl 2 H = +359.4 kJ  H for the overall equation is the sum of the  H’s of the intermediates after the changes have been made. The change in enthalpy or the heat of the overall equation is -32.8 kJ.

20. Overall Equation 2Cu + O 2  2CuO Intermediate Equations CuO + Cu  Cu 2 O H = -11.3 kJ Cu 2 O + ½O 2  2CuO H = -114.6 kJ

21. Overall Equation 2Cu + O 2  2CuO Intermediate Equations CuO + Cu  Cu 2 O H = -11.3 kJ Cu 2 O + ½O 2  2CuO H = -114.6 kJ Cancel substances that occur in both equations, even though the coefficients do not match. These substances will not help determine what to do.

22. Overall Equation 2Cu + O 2  2CuO Intermediate Equations 2CuO + 2Cu  2Cu 2 O H = 2(-11.3 kJ) Cu 2 O + ½O 2  2CuO H = -114.6 kJ Cu is a reactant in both the intermediate equation and the overall equation so the reaction does not need to be flipped. However the coefficients do not match. We must change the intermediate equation to match the overall equation. Therefore we must double the first equation.

23. Overall Equation 2Cu + O 2  2CuO Intermediate Equations 2CuO + 2Cu  2Cu 2 O H = 2(-11.3 kJ) 2Cu 2 O + O 2  4CuO H = 2(-114.6 kJ) O 2 is a reactant in both the intermediate equation and the overall equation so the reaction does not need to be flipped. However the coefficients do not match. We must change the intermediate equation to match the overall equation. Therefore we must double the second equation.

24. Overall Equation 2Cu + O 2  2CuO Intermediate Equations 2CuO + 2Cu  2Cu 2 O H = 2(-11.3 kJ) 2Cu 2 O + O 2  4CuO H = 2(-114.6 kJ)  H for the overall equation is the sum of the  H’s of the intermediates after the changes have been made. The change in enthalpy or the heat of the overall equation is -251.8 kJ.

25. Overall Equation C graphite + ½O 2  CO Intermediate equations: C graphite + O 2  CO 2 H = -393.5 kJ 2CO + O 2  2CO 2 H = -566 kJ

26. Overall Equation C graphite + ½O 2  CO Intermediate equations: C graphite + O 2  CO 2 H = -393.5 kJ 2CO + O 2  2CO 2 H = -566 kJ Cancel substances that occur in both intermediate equations

27. Overall Equation C graphite + ½O 2  CO Intermediate equations: C graphite + O 2  CO 2 H = -393.5 kJ 2CO + O 2  2CO 2 H = -566 kJ Cancel substances that occur in both intermediate equations. Even though the coefficients do not match, these substances will not help us determine what to do.

28. Overall Equation C graphite + ½O 2  CO Intermediate equations: C graphite + O 2  CO 2 H = -393.5 kJ 2CO + O 2  2CO 2 H = -566 kJ Look at C graphite. What needs to be done to the first equation?

29. Overall Equation C graphite + ½O 2  CO Intermediate equations: C graphite + O 2  CO 2 H = -393.5 kJ 2CO + O 2  2CO 2 H = -566 kJ Look at C graphite. What needs to be done to the first equation? NOTHING. It is on the same side of the equations and they have the same coefficients.

30. Overall Equation C graphite + ½O 2  CO Intermediate equations: C graphite + O 2  CO 2 H = -393.5 kJ 2CO + O 2  2CO 2 H = -566 kJ Look at CO. Determine what must be done to the second equation.

31. Overall Equation C graphite + ½O 2  CO Intermediate equations: C graphite + O 2  CO 2 H = -393.5 kJ 2CO + O 2  2CO 2 H = +566 kJ Look at CO. Determine what must be done to the second equation. Since CO is a reactant in the intermediate equation but a product in the overall equation, the second equation must be flipped. When the equation is flipped, the sign on  H is flipped.

32. Overall Equation C graphite + ½O 2  CO Intermediate equations: C graphite + O 2  CO 2 H = -393.5 kJ CO + ½ O 2  CO 2 H =(+566 kJ)/2 Look at CO. Determine what must be done to the second equation. Since CO is a reactant in the intermediate equation but a product in the overall equation, the second equation must be flipped. When the equation is flipped, the sign on  H is flipped. Since the coefficient is 2 in the intermediate equation but only 1 in the overall equation, the second equation must also be divided by 2. Whatever you do to the equation, do the same thing to  H.

33. Overall Equation C graphite + ½O 2  CO Intermediate equations: C graphite + O 2  CO 2 H = -393.5 kJ CO + ½ O 2  CO 2 H =(+566 kJ)/2  H for the overall equation is the sum of the  H’s of the intermediates after the changes have been made. The change in enthalpy or the heat of the overall equation is -110.5 kJ.

34. Overall Equation NO + ½ O 2  NO 2 Intermediate equations N 2 + O 2  2NOH = 180.8 kJ N 2 + 2O 2  2NO 2 H = 33.6 kJ

35. Overall Equation NO + ½ O 2  NO 2 Intermediate equations N 2 + O 2  2NOH = 180.8 kJ N 2 + 2O 2  2NO 2 H = 33.6 kJ Cancel substances that occur in both intermediate equations.

36. Overall Equation NO + ½ O 2  NO 2 Intermediate equations N 2 + O 2  2NOH = 180.8 kJ N 2 + 2O 2  2NO 2 H = 33.6 kJ Cancel substances that occur in both intermediate equations. N 2 and O 2 are not going to help us determine what to do.

37. Overall Equation NO + ½ O 2  NO 2 Intermediate equations N 2 + O 2  2NOH = 180.8 kJ N 2 + 2O 2  2NO 2 H = 33.6 kJ Determine what needs to be done to the first equation.

38. Overall Equation NO + ½ O 2  NO 2 Intermediate equations ½ N 2 + ½ O 2  NO H = (-180.8 kJ)/2 N 2 + 2O 2  2NO 2 H = 33.6 kJ Since NO is a product in the intermediate equation but a reactant in the overall equation, the first equation must be flipped. ALSO since the coefficient in the intermediate is 2, but only one in the overall equation it must also be divided by 2.

39. Overall Equation NO + ½ O 2  NO 2 Intermediate equations ½ N 2 + ½ O 2  NO H = (-180.8 kJ)/2 N 2 + 2O 2  2NO 2 H = 33.6 kJ Determine what must be done to the second equation.

40. Overall Equation NO + ½ O 2  NO 2 Intermediate equations ½ N 2 + ½ O 2  NO H = (-180.8 kJ)/2 ½ N 2 + O 2  NO 2 H = (33.6 kJ)/2 Since NO 2 is a product in both the intermediate and overall equations, the reaction does not need to be flipped. However, the coefficients do not match. The intermediate must be changed to match the overall equation so the second equation must be divided by 2.

41. Overall Equation NO + ½ O 2  NO 2 Intermediate equations ½ N 2 + ½ O 2  NO H = (-180.8 kJ)/2 ½ N 2 + O 2  NO 2 H = (33.6 kJ)/2  H for the overall equation is the sum of the  H’s of the intermediates after the changes have been made. The change in enthalpy or the heat of the overall equation is -73.6 kJ.