1. Ch. 6: Thermochemistry 6.1 The Nature of Energy

2. Energy  Energy-  Law of conservation of energy- energy can be converted but not created or destroyed

3. Types of Energy  potential energy- (PE) ex. attractive or repulsive forces ex. attractive or repulsive forces  kinetic energy- (KE) KE = ½mv 2 :depends on mass and volume KE = ½mv 2 :depends on mass and volume

4. Types of Energy (a): PE A > PE B (b):  ball A has rolled down the hill  has lost PE to friction and PE in ball B

5. Transfer of Energy  Temperature-  Two Ways to Transfer Energy: Heat- (q) transfer of energy between two objects because of a temperature difference Heat- (q) transfer of energy between two objects because of a temperature difference Work- (w) force acting over a distance Work- (w) force acting over a distance

6. Pathway  the specific conditions of energy transfer  state function- depends only on current conditions, not past or future  energy change is independent of pathway because it is a state function  work and heat depend on pathway so are not state functions

7. Transfer of Energy Two parts of the Universe  system-  surroundings-  usually system: what is inside the container system: what is inside the container surroundings: room, container, etc. surroundings: room, container, etc.

8. Transfer of Energy  exothermic- exothermic- energy is produced in reaction energy is produced in reaction flows out of system flows out of system container feels hot to the touch container feels hot to the touch  endothermic- energy is consumed by the reaction energy is consumed by the reaction flows into the system flows into the system container feels cold to the touch container feels cold to the touch

9. Transfer of Energy Combustion of Methane Gas is exothermic

10. Transfer of Energy Reaction between nitrogen and oxygen is endothermic

11. Transfer of Energy  the energy comes from the ________________________ between the reactants and products  energy produced (or absorbed) by reaction must equal the energy absorbed (or produced) by surroundings  usually the molecules with higher potential energy have weaker bonds than molecules with lower potential energy

12. Thermodynamics  Thermodynamics- study of energy and its transfers  First Law of Thermodynamics-

13. Internal Energy  (E) sum of potential and kinetic energy in system  can be changed by work (w), heat (q), or both

14. Signs  signs are very important  signs will always reflect the system’s point of view unless otherwise stated ∆Eqw change in internal energy heatwork exothermic endothermic

15. Signs

16. Work  common types of work expansion- expansion- compression- compression- expansion compression P is external pressure – not internal like we normally refer to

17. Work

18. Example 1  Find the ∆E for endothermic process where 15.6 kJ of heat flows and 1.4 kJ of work is done on system Since it is endothermic, q is ___ and w is ___ Since it is endothermic, q is ___ and w is ___

19. Example 2  Calculate the work of expansion of a gas from 46 L to 64 L at a constant pressure of 15 atm. Since it is an expansion, ∆V is ___ and w is ___ Since it is an expansion, ∆V is ___ and w is ___

20. Example 3  A balloon was inflated from 4.00 x 10 6 L to 4.50 x 10 6 L by the addition of 1.3 x 10 8 J of heat. Assuming the pressure is 1.0 atm, find the ∆E in Joules. (1 L∙atm=101.3 J) Since it is an expansion, ∆V is ___ and w is ___ Since it is an expansion, ∆V is ___ and w is ___

21. Ch. 6: Thermochemistry 6.2 Enthalpy and Calorimetry

22. Enthalpy  definition:  since E, P and V are all state functions, then H is too  for the following, the process is at constant P and the only type of work allowed is PV work ∆E = q P + w = q P - P∆V  q P = ∆ E +P∆V H = E + PV  ∆H = ∆E+P∆V so at constant P

23. Enthalpy  heat of reaction and change in enthalpy are used interchangeably for a reaction at constant P ∆H = H products - H reactants endo: exo:

24. Calorimetry  science of measuring heat  calorimeter- device used to experimentally find the heat associated with a chemical reaction  substances respond differently when heated

25. Heat Capacity  (C) how much heat it takes to raise a substance’s T by ________________  the amount of energy depends on the amount of substance

26. Heat Capacity  specific heat capacity (s) heat capacity_________ (s) heat capacity_________ in J/°C*g or J/K*g in J/°C*g or J/K*g  molar heat capacity heat capacity __________ heat capacity __________ in J/°C*mol or J/K*mol in J/°C*mol or J/K*mol

27. Constant-Pressure Calorimetry  uses simplest calorimeter (like coffee- cup calorimeter) since it is open to air  used to find changes in enthalpy (heats of reaction) for reactions occurring in a solution since q P = ∆H  heat of reaction is an extensive property, so we usually write them per mole so they are easier to use

28. Constant-Pressure Calorimetry  when 2 reactants are mixed and T increases, the chemical reaction must be releasing heat so is _______________  the released energy from the reaction increases the _________ of molecules, which in turn increases the T

29. Constant-Pressure Calorimetry  If we assume that the calorimeter did not leak energy or absorb any itself (that all the energy was used to increase the T), we can find the energy released by the reaction: E released by rxn = E absorbed by soln

30. Constant-Volume Calorimetry  uses a bomb calorimeter  weighed reactants are placed inside the rigid, steel container and ignited  water surrounds the reactant container so the T of it and other parts are measured before and after reaction

31. Constant-Volume Calorimetry  Here, the ∆V = 0 so -P∆V = w = 0 ∆E = q + w = q V for constant volume E rxn = ∆T x C calorimeter

32. Example 1  When 1 mol of CH 4 is burned at constant P, 890 kJ of heat is released. Find ∆H for burning of 5.8 g of CH 4 at constant P.  890 kJ is released per mole of CH 4

33. Example 2  When 1.00 L of 1.00 M Ba(NO 3 ) 2 solution at 25.0°C is mixed with 1.00 L of 1.00 M Na 2 SO 4 solution at 25.0°C in a coffee-cup calorimeter, solid BaSO 4 forms and the T increases to 28.1°C. The specific heat capacity of the solution is 4.18 J/g*°C and the density is 1.0 g/mL. Find the enthalpy change per mole of BaSO 4 formed.

34. Example 2  Write the net ionic equation for the reaction:  Is the energy released or absorbed? What does that mean about ∆H and q?  How can we calculate ∆H or heat?  How can we find the m?

35. Example 2  Find the mass:  Find the change in T:  Calculate the heat created:

36. Example 2  since it is a one-to-one ratio and the moles of reactants are the same, there is no limiting reactant

37. Example 3  Compare the energy released in the combustion of H 2 and CH 4 carried out in a bomb calorimeter with a heat capacity of 11.3 kJ/°C. The combustion of 1.50 g of methane produced a T change of 7.3°C while the combustion of 1.15 g of hydrogen produced a T change of 14.3°C. Find the energy of combustion per gram for each.

38. Example 3  methane: CH 4  hydrogen: H 2

39. Ch. 6: Thermochemistry 6.3 Hess’ Law

40. Hess’ Law  since H is a state function, the change in H is independent of pathway  Hess’ Law- when going from a set of reactants to a set of products, the ∆H is the same whether it happens in _____________________________________

41. Example 1

42.  N 2 (g) + 2O 2 (g)  2NO 2 (g)∆H = 68 kJ OR  N 2 (g) + O 2 (g)  2NO(g)∆H = 180 kJ 2NO(g) + O 2 (g)  2NO 2 (g)∆H = -112 kJ N 2 (g) + 2O 2 (g)  2NO 2 (g)∆H = 68 kJ

43. Rules 1. If a reaction is reversed, the sign of ∆H must be reversed as well. because the sign tells us the direction of heat flow as constant P because the sign tells us the direction of heat flow as constant P 2. The magnitude of ∆H is directly proportional to quantities of reactants and products in reaction. If coefficients are multiplied by an integer, the ∆H must be multiplied in the same way. because ∆H is an extensive property because ∆H is an extensive property

44. Example 2  Using the enthalpies of combustion for graphite (-394 kJ/mol) and diamond (-396 kJ/mol), find the ∆H for the conversion of graphite to diamond. C graphite (s)  C diamond (s)∆H=?

45. Example 2 C graphite (s)  C diamond (s)∆H=? (1)(2)  to get the desired equation, we must reverse 2 nd equation: (1) C graphite (s) + O 2 (g)  CO 2 (g) ∆H=-394kJ/mol (2) ∆H=___________ C graphite (s)  C diamond (s)∆H= ∆H=

46. Example 3  Find ∆H for the synthesis of B 2 H 6, diborane: 2B(s) + 3H 2 (g)  B 2 H 6 (g) Given: (1) 2B(s) + 3/2O 2 (g)  B 2 O 3 (s) ∆H 1 =-1273kJ (2) B 2 H 6 (g) + 3O 2 (g)  B 2 O 3 (s) + 3H 2 O(g) ∆H 2 =-2035kJ (3) H 2 (g) + ½O 2 (g)  H 2 O(l) ∆H 3 =-286kJ (4) 3H 2 O(l)  3H 2 O(g) ∆H 4 =44 kJ

47. Example 3  Start by paying attention to what needs to be on reactants and products side (1)(2)(3)(4)

48. Example 3  Underline what you want to keep- that will help you figure out how to cancel everything else: (1)(2)(3)(4)

49. Example 3  Need 3 H 2 (g) so 3 x (3)  Need 3 H 2 O to cancel so 3 x (4) (1)(2)(3)(4) 2B(s) + 3H 2 (g)  B 2 H 6 (g) ∆H = -1273 + -(-2035) + 3(-286) + 3(44) = 36kJ

50. Ch. 6: Thermochemistry 6.4 Standard Enthalpies of Formation

51. Standard Enthalpy of Formation  ∆H f °  change in enthalpy that accompanies the formation of one mole of a compound from its elements in standard states  ° means that the process happened under standard conditions so we can compare more easily

52. Standard States  For a COMPOUND: for gas: P = 1 atm for gas: P = 1 atm pure liquid or solid state pure liquid or solid state in solution: concentration is 1 M in solution: concentration is 1 M  For an ELEMENT: form that it exists in at 1 atm and 25°C form that it exists in at 1 atm and 25°C O: O 2 (g)K: K(s)Br: Br 2 (l)

53. Writing Formation Equations  always write equation where 1 mole of compound is formed (even if you must use non-integer coefficients) NO 2 (g): ½N 2 (g) + O 2 (g)  NO 2 (g) ∆H f °= 34 kJ/mol CH 3 OH(l): C(s) + H 2 (g) + O 2 (g)  CH 3 OH(l) ∆H f °= -239 kJ/mol

54. Using Standard Enthalpies of Formation where n = ∑ means “sum of” ∆H f ° is the standard enthalpy of formation for reactants or products  ∆H f ° for any ______________ in standard state is ________ so elements are not included in the summation

55. Using Standard Enthalpies of Formation  since ∆H is a state function, we can use any pathway to calculate it  one convenient pathway is to break reactants into elements and then recombine them into products

56. Using Standard Enthalpies of Formation

58. Example 1  Calculate the standard enthalpy change for the reaction that occurs when ammonia is burned in air to make nitrogen dioxide and water 4NH 3 (g) + 7O 2 (g)  4NO 2 (g) + 6H 2 O(l)  break them apart into elements and then recombine them into products

59. Example 1

60.  can be solved using Hess’ Law: (1) 4NH 3 (g)  2N 2 (g) + 6H 2 (g)-4∆H f ° NH3 (2) 7O 2 (g)  7O 2 (g)0 (3) 2N 2 (g) + 4O 2 (g)  4NO 2 (g)4 ∆H f ° NO2 (4) 6H 2 (g) + 3O 2 (g)  6H 2 O(l)6 ∆H f ° H2O

61. Example 1  can also be solved using enthalpy of formation equation:  values are in Appendix 4: p. A21-A23

62. Example 2  Calculate the standard enthalpy change for the following reaction: 2Al(s) + Fe 2 O 3 (s)  Al 2 O 3 (s) + 2Fe(s)

63. Example 3  Compare the standard enthalpy of combustion per gram of methanol (CH 3 OH) with per gram of gasoline (C 8 H 18 ).  Write equations:

64. Example 3  Calculate the enthalpy of combustion per mole:

65. Example 3  Convert to per gram using molar mass: