THERMOCHEMISTRY

1. THERMOCHEMISTRY

2. Energy • The ability to do work or transfer heat. • Work: Energy used to cause an object that has mass to move. • Heat: Energy used to cause the temperature of an object to rise.

3. Definitions #1 Energy: The capacity to do work or produce heat Potential Energy: Energy due to position or composition Kinetic Energy: Energy due to the motion of the object

4. Definitions #2 Law of Conservation of Energy: Energy can neither be created nor destroyed, but can be converted between forms The First Law of Thermodynamics: The total energy content of the universe is constant

5. Definitions #3 Internal energy: The internal energy of a system is identified with the random, disordered motion of molecules; the total (internal) energy in a system includes potential and kinetic energy. The sample is at a height from the ground etc).  Symbol for Internal Energy Change: ΔU

6. 1) The internal energy of a system __________. • A) is the sum of the kinetic energy of all of its components • B) is the sum of the rotational, vibrational, and translational energies of all of its components • C) refers only to the energies of the nuclei of the atoms of the component molecules • D) is the sum of the potential and kinetic energies of the components • E) none of the above

7. Definitions #4 Enthalpy: Enthalpy is the amount of heat content used or released in a system at constant pressure

8. E = q + w E = change in internal energy of a system q = heat flowing into or out of the system -q if energy is leaving to the surroundings +q if energy is entering from the surroundings w = work done by, or on, the system -w if work is done by the system on the surroundings +w if work is done on the system by the surroundings

9. 2) Which one of the following conditions would always result in an decrease in the internal energy of a system? • A) The system loses heat and does work on the surroundings. • B) The system gains heat and does work on the surroundings. • C) The system loses heat and has work done on it by the surroundings. • D) The system gains heat and has work done on it by the surroundings. • E) None of the above is correct.

10. 3) The change in the internal energy of a system that releases 2,500 J of heat and that does 7,655 J of work on the surroundings is __________ J. • A) -10,155 B) -5,155 C) 1.9x107 D) 10,155 E) 5,155

11. Work problemsChapter 5 • 5.25 • 5.27 A and B • 5.31 All

12. Calorimetry The amount of heat absorbed or released during a physical or chemical change can be measured… …usually by the change in temperature of a known quantity of water 1 calorie is the heat required to raise the temperature of 1 gram of water by 1 C 1 BTU is the heat required to raise the temperature of 1 pound of water by 1 F

13. The Joule The unit of heat used in modern thermochemistry is the Joule 1 joule = 4.184 calories

14. A Bomb Calorimeter

15. A Cheaper Calorimeter

16. The amount of heat required to raise the temperature of one gram of substance by one degree Celsius. Specific Heat

17. Calculations Involving Specific Heat OR s = Specific Heat Capacity q = Heat lost or gained T = Temperature change

18. 4) The temperature of a 15-g sample of lead metal increases from 22 °C to 37 °C upon the addition of 29.0 J of heat. The specific heat capacity of the lead is __________ J/g-K. 7.8 1.9 29 0.13

19. Problems • 5.53 a and B

20. State Functions depend ONLY on the present state of the system ENERGYIS A STATE FUNCTION A person standing at the top of Mt. Everest has the same potential energy whether they got there by hiking up, or by falling down from a plane! WORKIS NOT A STATE FUNCTION WHY NOT???

21. State Functions Usually we have no way of knowing the internal energy of a system; finding that value is simply too complex a problem.

22. State Functions • However, we do know that the internal energy of a system is independent of the path by which the system achieved that state. • In the system below, the water could have reached room temperature from either direction.

23. State Functions • Therefore, internal energy is a state function. • It depends only on the present state of the system, not on the path by which the system arrived at that state. • And so, E depends only on Einitial and Efinal.

24. State Functions • However, q and w are not state functions. • Whether the battery is shorted out or is discharged by running the fan, its E is the same. • But q and w are different in the two cases.

25. Work When a process occurs in an open container, commonly the only work done is a change in volume of a gas pushing on the surroundings (or being pushed on by the surroundings).

26. Work We can measure the work done by the gas if the reaction is done in a vessel that has been fitted with a piston. w = −PV

27. Work, Pressure, and Volume Compression Expansion +V (increase) -V (decrease) -w results +w results Esystemdecreases Esystemincreases Work has been done on the system by the surroundings Work has been done by the system on the surroundings

28. Energy Change in Chemical Processes Endothermic: Reactions in which energy flows into the system as the reaction proceeds. + qsystem - qsurroundings Exothermic: Reactions in which energy flows out of the system as the reaction proceeds. - qsystem + qsurroundings

29. Endothermic Reactions

30. Exothermic Reactions

31. 5) Which one of the following is an exothermic process? • A) ice melting B) water evaporating C) boiling soup D) condensation of water vapor • E) Ammonium thiocyanate and barium hydroxide are mixed at 25 °C: the temperature drops

32. Enthalpy H = E + PV At constant pressure and volume the change in enthalpy is the heat gained or lost H = q

33. Enthalpies of Reaction The change in enthalpy, H, is the enthalpy of the products minus the enthalpy of the reactants: H = Hproducts−Hreactants

34. Enthalpies of Reaction This quantity, H, is called the enthalpy of reaction, or the heat of reaction.

35. Hess’s Law “In going from a particular set of reactants to a particular set of products, the change in enthalpy is the same whether the reaction takes place in one step or a series of steps.”

36. Hess’s Law

37. Hess’s Law Example Problem CH4 C + 2H2 +74.80 kJ Step #1: CH4 must appear on the reactant side, so we reverse reaction #1 and change the sign on H.

38. Hess’s Law Example Problem CH4 C + 2H2 +74.80 kJ C + O2 CO2 -393.50 kJ Step #2: Keep reaction #2 unchanged, because CO2 belongs on the product side

39. Hess’s Law Example Problem CH4 C + 2H2 +74.80 kJ C + O2 CO2 -393.50 kJ 2H2 + O2 2 H2O -571.66 kJ Step #3: Multiply reaction #3 by 2

40. Hess’s Law Example Problem CH4 C + 2H2 +74.80 kJ C + O2 CO2 -393.50 kJ 2H2 + O2 2 H2O -571.66 kJ CH4 + 2O2 CO2 + 2H2O -890.36 kJ Step #4: Sum up reaction and H

41. Calculation of Heat of Reaction Hrxn = Hf(products) - Hf(reactants) Hrxn = [-393.50kJ + 2(-285.83kJ)] – [-74.80kJ] Hrxn = -890.36 kJ

42. 13) Given the following reactions the enthalpy of the reaction in which sulfur dioxide is oxidized to sulfur trioxide 2SO2 (g) +O2 (g)  2SO3(g) is __________ kJ. 2SO2  2S + 2O2 ΔH = 594 kJ 2SO3  2S + 3 O2ΔH = 790 kJ

43. 6. 2SO2  2S + 2O2 ΔH = 594 kJkeep 2SO3  2S + 3 O2ΔH = 790 kJreverse 2S + 3 O2 2SO3 ΔH = -790 kJ Ans: -196kJ

44. Problems 5.63