1. Chapter 10 Energy

2. Energy: Types Objectives: 1)To understand the general properties of energy 2)To understand the concepts of temperature and heat 3)To consider the direction of energy flow as heat

3. Energy: Types Energy: the ability to do work or produce heat TWO TYPES Potential energy (energy due to position or composition) PE = mgh Kinetic energy: energy is due to the motion of an object KE=1/2 mv 2 Law of Conservation of Energy: energy can be converted from one form to another but can be neither created nor destroyed.

4. Energy: Types Work: force acting over a distance. As energy is transferred some of it is converted to heat (due to friction). STATE FUNCTION: property of the system that changes independently of its pathway. Examples: ENERGY is a state function work and heat are not.

5. Energy: Temperature and Heat Temperature: measure of the random motion of the components of a substance. Insulated Box and a thin metal sheet What will happen to the temperature of the water on both sides?

6. Energy: Temperature and Heat Heat: flow of energy due to a temperature difference. T final = T hot (initital) + T cold (initial) 2

7. Exothermic and Endothermic System: part of the universe on which we wish to focus Surroundings: include everything else in the universe. Exothermic: a process that results in the evolution of heat. ENERGY FLOWS OUT Endothermic: process that absorbs energy from the surroundings ENERGY FLOWS IN.

9. 10.4 Thermodynamics Objective: To understand how energy flow affects internal energy

10. 10.4 Thermodynamics Thermodynamics: study of energy First Law of Thermodynamics: The energy of the universe is constant (same as Law of Conservation of Energy) Internal energy (E)= sum of kinetic and potential energy of the system –This can be changed by a flow of work, heat or both

11. 10.4 Thermodynamics Change in Internal energy  q+w q= heat w= work Thermodynamic quantities consists of 2 Parts: a number and a sign. The sign reflects the system’s point of view.

12. 10.4 Thermodynamics Change in Internal energy  q+w System Surroundings Energy  E<0  E>0 q is +, endothermic, heat into system q is -, exothermic, heat out w is +, if surroundings do work on system w is -, system does work on surroundings

13. 10.5 Energy Changes Objective: 1) To understand how heat is measured

14. 10.5 Energy Changes Calorie: (metric system) the amount of energy (heat) required to raise the temperature of one gram of water by one Celsius degree. In food, kilocalorie. C The joule (SI unit) can be most conveniently defined in terms of the calorie. 1 calorie=4.184 joule 1 cal=4.184 J

15. 10.5 Energy Changes Express 60.1 cal of energy in units of joules. 60.1 cal x 4.184 J = 251 J 1 cal Calculating energy requirements Determine the amount of energy(heat) in joules required to raise the temperature of 7.40 g water from 29.0  to 46.0  4.184 J x 7.40 g x 17.0 C = 526 J g  C

16. 10.5 Energy Changes Another important factor is IDENTITY OF SUBSTANCE. Specific heat capacity: amount of energy required to change the temperature of one gram of a substance by one Celsius degree. Q=s x m x  Q= energy (heat) required s=specific heat capacity m= mass of the sample in grams  T=change in temperature in Celsius degrees. Table 10.1 p. 297

17. 10.5 Energy Changes A 1.6 g sample of a metal that has the appearance of gold requires 5.8 J of energy to change its temperature from 23 C to 41 C. Is the metal pure gold.

18. 10.5 Energy Changes 5.8= s x 1.6 x (41-23) 5.8 = s s=0.20 J/g C 1.6 x 18pure gold s=0.13 J/g C

19. Section 10.6 Thermochemistry (Enthalpy) Objective: To consider the heat (enthalpy) of chemical reactions.

20. Section 10.6 Thermochemistry (Enthalpy) Enthalpy: a special energy function (  H) Under constant pressure, the change in enthalpy is equal to the energy that flows as heat.  H p =heat

21. Section 10.6 Thermochemistry (Enthalpy) When 1 mole of methane(CH 4 ) is burned at constant pressure, 890kJ of energy is released as heat. Calculate  H for a process in which a 5.8 g sample of methane is burned at constant pressure. q p =  H= -890kJ/mol CH 4 5.8 g CH 4 x 1mol= 16.0 g

22. Section 10.6 Thermochemistry (Enthalpy) When 1 mole of sulfur dioxide reacts with excess oxygen to form sulfur trioxide at constant pressure, 198.2kJ of energy is released as heat. Calculate  H for a process in which a 12.8 g sample of sulfur dioxide reacts with oxygen at constant pressure. 39.6kJ

23. Section 10.6 Thermochemistry (Enthalpy) Calorimeter is a device used to determine the heat associated with a chemical reaction.

24. Figure 10.6: A coffee-cup calorimeter. NCSSM Distance Learning T.I.G.E.R. - Chemistry page 4#thermo

25. 10.7 Hess’s Law Objective: To understand Hess’s law

26. 10.7 Hess’s Law Enthalpy is a state function Independent of path soooo –The change in enthalpy is the same whether the reaction takes place in one step or in a series of steps. (Hess’s Law) –Important: because it allows us to calculate heats of reaction that might be difficult or inconvenient to measure directly in a calorimeter.

27. 10.7 Hess’s Law N 2 (g) + O 2 (g) 2NO 2 (g)  H=68kJ N 2 (g) + O 2 (g) 2NO (g)  H 2 =180kJ 2NO (g) + O 2 (g) 2NO 2  H 3 = -112kJ N 2 (g) + O 2 (g) 2NO 2 (g)  H=68kJ

28. 10.7 Hess’s Law Characteristics of Enthalpy Changes 1)If a reaction is reversed, the sign of  H is also reversed. 2)The magnitude of  H is directly proportional to the quantities of reactants and products in a reaction. If the coefficients in a balanced reaction are multiplied by an integer, the value of  H is multiplied by the same integer.

29. 10.7 Hess’s Law Given the following data 4CuO(s) 2Cu 2 O(s) + O 2 (g)  H=288kJ Cu 2 O(s) Cu(s) + CuO(s)  H=11 kJ Calculate  H for the following reaction 2Cu(s) + O 2 (g) 2CuO(s) Cu(s) + CuO(s) Cu2O DH= -(11kJ) 2 Cu2O + O2 4CuO(s) DH=-(288kJ) 2(Cu(s) + CuO(s) Cu 2 O  H= -2(11kJ) 2 Cu 2 O + O 2 4CuO(s)  H=-(288kJ) 2Cu(s) + O 2 (g) 2CuO(s)  H= -310kJ

30. 10.10 Energy as a Driving Force Objective: To understand energy as a driving force for natural processes.

31. 10.10 Energy as a Driving Force Energy spread: in a given process, concentrated energy is dispersed widely. This distributions happens every time an exothermic process occurs. Matter spread: molecules of a substance spread out and occupy a larger volume.

32. 10.10 Energy as a Driving Force Entropy: natural tendency of the universe to become disordered. S (as things become more disordered), the value of S increases Second law of thermodynamics: The entropy of the universe is always increasing.

33. Putting it all together! Gibbs Free Energy G=H-T  S Temperature in Kelvin Change in Free Energy  G=  H-T  S (constant temperature) Related to system If  G is negative (spontaneous reaction),  S (+)

34. Calculations Let’s predict the spontaneity of the melting of ice H 2 O(s) H 2 O(l)  H =6.03 x 103 J/mol and  S=22.1 J/K mol What is  G at -10 , 0  and 10  C? Which is spontaneous?  G=2.2x 10 2  G=0  G= -2.2 x 10 2

35. More calculations Br 2 (l) Br 2 (g)  H=31.0 kJ/mol and  S=93.0 J/K mol What is the normal boiling point?  G=  H- T  S 0=  H- T  S T=  H/  S=333K

36. Figure 10.7: Energy sources used in the United States.

37. Figure 10.8: The earth’s atmosphere.

38. Figure 10.9: The atmospheric CO 2 concentration over the past 1000 years.

39. Figure 10.10: Comparing the entropies of ice and steam.