1. Chapter 6 Energy and Chemical Reactions

2. Macroscale Kinetic Energy energy that something has because it is moving Potential Energy energy that something has because of its position Molecular Kinetic Energy energy of motion of molecules = temperature Chemical Potential Energy reactive molecules have the potential to form stable ones and release energy Nanoscale

3. Energy Units 1J = 1 kg m 2 /sec 2 1 cal = 4.184 J 1kcal = 1 Cal thus 1 Cal = 1 kcal = 1000 cal = 4.184 kJ = 4184 J

4. First Law of Thermodynamics energy can neither be created nor destroyed the total amount of energy in the universe is a constant energy can be transformed from one form to another

5. Measuring Temperature K.E. of motion  Temperature

6. Energy Transfer Energy is always transferred from the hotter to the cooler sample

7. Heat Transfer Energy is always transferred from the hotter to the cooler sample

8. Internal Energy E the sum of the individual energies of all nanoscale particles (atoms, ions, or molecules) in that sample including all chemical bonds and intermolecular interactions we cannot measure the internal energy accurately but chemistry depends only on changes in internal energy  E

9. Thermochemistry Divide universe into two parts system  that part of the universe under investigation surroundings  the rest of the universe universe = system + surroundings consider everything from the system’s viewpoint

10. Internal Energy

11. First Law of Thermodynamics System can exchange energy with surroundings via heat or work heat  q work  w internal energy change   E  E = q + w

12. Combustion of gasoline in your car engine results in the production of 124 kJ of energy. If the engine does 31 kJ of work to move the car how much heat must be released by the radiator? 1.31 kJ 2.124 kJ 3.93 kJ 4.155 kJ

13. Internal Energy, Heat, and Work

14. Which of these changes always results in a decrease in the internal energy of the system? 1.The system absorbs heat and does work on the surroundings. 2.The system releases heat and does work on the surroundings. 3.The system absorbs heat and work is done on it by the surroundings. 4.The system releases heat and work is done on it by the surroundings.

15. Specific Heat c the amount of heat necessary to raise the temperature of 1 gram of the substance 1 ° C independent of mass substance dependent Specific Heat of Water = 4.184 J/g ° C

16. Specific Heat Capacity

17. Heat q = m  c   T whereq  heat, J m  mass, g c  specific heat, J/g.  C  T = change in temperature,  C

18. Molar Heat Capacity the heat necessary to raise the temperature of one mole of substance by 1  C substance dependent c m metals: c m ≈ 25 J/mol.  C

19. Hot and Cold Iron

20. Freezing and Melting

21. Heat Transfer Two substances in an isolated system – no energy exchange with the surroundings  E = 0, no work q lost + q gained =  E = 0, q lost = -q gained (m  c   T) lost = - (m  c   T) gained  T = final temperature – initial temperature  T = T f – T i T f is same for both substances

22. EXAMPLE If 100. g of iron at 100.0 o C is placed in 200. g of water at 20.0 o C in an insulated container, what will the temperature, o C, of the iron and water when both are at the same temperature? The specific heat of iron is 0.106 cal/g o C. (100.g  0.106cal/g o C  (T f - 100.) o C) = q lost - q gained = - (200.g  1.00cal/g o C  (T f - 20.0) o C) 10.6(T f - 100. o C) = - 200.(T f - 20.0 o C) 10.6T f - 1060 o C = - 200.T f + 4000 o C (10.6 + 200.)T f = (1060 + 4000) o C T f = (5060/211.) o C = 24.0 o C

23. At 25 º C, the same quantity of heat is added to 1.00-g samples of gold and iron. MetalSpecific Heat Capacity (J g -1 º C -1 ) Gold0.128 Iron0.451 Identify the correct statement. 1.The temperatures of both will remain the same. 2.The temperatures of both will increase by the same amount. 3.The temperatures of both will decrease by the same amount. 4.The temperature of the gold sample will rise higher than that of iron. 5.The temperature of the iron sample will rise higher than that of gold.

24. Heating, Temperature Change, and Phase Change

25. Vaporization and Condensation

26. Which of the following is an exothermic process? 1.Water boiling 2.Ice melting 3.Heating soup 4.Steam condensing

27. EXAMPLE: How much heat is required to heat 10.0 g of ice at -15.0 o C to steam at 127.0 o C? q ice = m  c   T ice c ice = 2.09J/g o C q water = m  c   T water c water = 4.18J/g o C q steam = m  c   t steam c steam = 2.03J/g o C q fusion = m  heat of fusion  333J/g q boil. = m  heat of vaporization 2260J/g q overall = q ice + q fusion + q water + q boil. + q steam

28. EXAMPLE: How much heat is required to heat 10.0 g of ice at -15.0 o C to steam at 127.0 o C? q overall = q ice + q fusion + q water + q boil. + q steam q= (10.0g  2.09J/g o C  15.0 o C) + (10.0g  333J/g) + (10.0g  4.18J/g o C  100.0 o C) + (10.0g  2260J/g) + (10.0g  2.03J/g o C  27.0 o C) q = (314 + 3.33×10 3 + 4.18×10 3 + 2.26×10 4 + 548)J = 23.3 kJ

29. Heat Flow in Reactions exothermic –reaction that gives off energy q < 0 isolated system  E=0 heat released by reaction raises the temperature of the system constant T, heat is released to the surroundings endothermic – reaction that absorbs energy q > 0

30. Expansion Type Work w = -P  V system does work P P V initial VV  V = V final - V initial q p = +2kJ

31. Do 250 J of work to compress a gas, 180 J of heat are released by the gas What is  E for the gas? 1.430 J 2.70 J 3.-70 J 4.-180 J 5.-250 J

32. Enthalpy H  E = q + w at constant V, w expansion = 0  E = q v at constant P, w expansion = -P  V  E = q p - P  V Define  PV) =  P  V  H = q p

33. Enthalpy Enthalpy  heat at constant pressure q p =  H = H products - H reactants Exothermic Reaction  H = (H products - H reactants ) < 0 H 2 O (l)  H 2 O (s)  H < 0 Endothermic Reaction  H = (H products - H reactants ) > 0 H 2 O (l)  H 2 O (g)  H > 0

34. State Functions  H and  E along with  P,  T,  V are state functions. They are the same no matter what path we take for the change. q and w are not state functions, they depend on which path we take between two points. initial final EE q w q w  E=E final -E initial q and w can be anything