9. Solid Liquid Gas MeltingVaporization Condensation Freezing

10. A plot of temperature vs. time that represents the process in which energy is added at a constant rate Courtesy Christy Johannesson www.nisd.net/communicationsarts/pages/chem Heating Curves Animation

11. Temperature ( o C) 40 20 0 -20 -40 -60 -80 -100 120 100 80 60 140 Time Melting - PE  Solid - KE  Liquid - KE  Boiling - PE  Gas - KE 

12. A plot of temperature vs. time that represents the process in which energy is added at a constant rate

13. The temperature doesn’t change during a phase change. If you have a mixture of ice and water, the temperature is 0ºC At 1 atm, boiling water is 100ºC You can’t get the temperature higher until it boils

14. Chemical Energy 2 parts of the universe as it relates to a chemical reaction: System – the reactants and the products Surroundings Surroundings – everything else in the universe (such as container, the room, etc.) Law of Conservation of Energy: the total energy of the universe is constant and can neither be created nor destroyed; it can only be transformed.

15. The First Law of Thermodynamics: The total energy content of the universe is constant Chemical energy lost by combustion = Energy gained by the surroundings system surroundings Signs (+/-) will tell you if energy is entering or leaving a system + indicates energy enters a system - indicates energy leaves a system

16. Two types of processes based on energy flow: Exothermic Exothermic – produces energy (heat flows out of the system) Endothermic Endothermic – absorbs energy (heat flows into the system) Chemical Energy

17. Conservation of Energy in a Chemical Reaction Surroundings System Surroundings System Energy In this example, the energy of the reactants and products decreases, while the energy of the surroundings increases. In every case, however, the total energy does not change. Myers, Oldham, Tocci, Chemistry, 2004, page 41 Exothermic Reaction Reactant Product + Energy Before reactionAfter reaction

18. Conservation of Energy in a Chemical Reaction Surroundings System Surroundings System Energy Before reactionAfter reaction In this example, the energy of the reactants and products increases, while the energy of the surroundings decreases. In every case, however, the total energy does not change. Myers, Oldham, Tocci, Chemistry, 2004, page 41 Endothermic Reaction Reactant + Energy Product

20. 20 Thermochemistry Every reaction has an energy change associated with it Energy is stored in bonds between atoms Making bonds gives energy Breaking bonds takes energy

21. To more easily measure and study the energy changes that accompany chemical reactions, chemists have defined a property called enthalpy. EnthalpyEnthalpy (H) is the heat content of a system at constant pressure.

22. Although you cannot measure the actual energy or enthalpy of a substance, you can measure the change in enthalpy, which is the heat absorbed or released in a chemical reaction. The change in enthalpy for a reaction is called the enthalpy (heat) of reaction (∆H rxn ). You have already learned that a symbol preceded by the Greek letter ∆ means a change in the property.

23. Thus, ∆H rxn is the difference between the enthalpy of the substances that exist at the end of the reaction and the enthalpy of the substances present at the start. Because the reactants are present at the beginning of the reaction and the products are present at the end, ∆H rxn is defined by this equation.

24. Enthalpy (H) is used to quantify the heat flow into or out of a system in a process that occurs at constant pressure.  H = H (products) – H (reactants)  H = heat given off or absorbed during a reaction at constant pressure H products < H reactants  H < 0 H products > H reactants  H > 0 6.4

25. reaction Exothermic, heat given off & temperature of water rises Endothermic, heat taken in & temperature of water drops Energy Change in Chemical Processes Exothermic process:  H < 0 (at constant pressure) Endothermic process:  H > 0 (at constant pressure)

26. Exothermic process Exothermic process is any process that gives off heat – transfers thermal energy from the system to the surroundings. Endothermic process Endothermic process is any process in which heat has to be supplied to the system from the surroundings. 2H 2 (g) + O 2 (g) 2H 2 O (l) + energy H 2 O (g) H 2 O (l) + energy energy + 2HgO (s) 2Hg (l) + O 2 (g) 6.2 energy + H 2 O (s) H 2 O (l)

28. Endothermic Reactions

29. Exothermic Reactions

30. exothermic endothermic exothermic endothermic

31. Effect of Catalyst on Reaction Rate reactants products Energy activation energy for catalyzed reaction Reaction Progress No catalyst Catalyst lowers the activation energy for the reaction. What is a catalyst? What does it do during a chemical reaction?

32. Temperature ( o C) 40 20 0 -20 -40 -60 -80 -100 120 100 80 60 140 Time  H = mol x  H fus  H = mol x  H vap Heat = mass x  t x C p, liquid Heat = mass x  t x C p, gas Heat = mass x  t x C p, solid

33. 33 The heat that is absorbed by one mole of a substance in melting at a constant temperature is the molar heat of fusion  H fus The heat lost when one mole of a liquid solidifies at a constant temperature is the molar heat of solidification  H sol H 2 O (s) H 2 O (l)  H fus = 6.01 kJ/mol H 2 O (l)H 2 O (g)  H sol = - 6.01 kJ/mol

34. 34 The molar heat of vaporization: –Heat needed to change one mol of a liquid to gas  H vap The molar heat of condensation: –Heat needed to change one mol of a gas to liquid  H con H 2 O (l) H 2 O (g)  H vap = 40.7 kJ/mol H 2 O (g) H 2 O (l)  H con = - 40.7 kJ/mol

35. 35 The heat that is released or absorbed in a chemical reaction is equivalent to  H C + O 2 (g) CO 2 (g) +394 kJ C + O 2 (g) CO 2 (g)  H = -394 kJ In thermochemical equation it is important to say what state H 2 O(g) H 2 (g) + ½ O 2 (g)  H = 241.8 kJ H 2 O(l) H 2 (g) + ½ O 2 (g)  H = 285.8 kJ Difference = 44.0 kJ

36. “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.”

38. 38 The change in heat that accompanies the formation of a mole of a compound from its elements at standard conditions Standard conditions 25°C and 1 atm. Symbol is  H ◦ f u The standard heat of formation of an element at its most stable form is 0 u This includes the diatomics

39. 39 There are tables of heats of formations (pg. 316) For most compounds it is negative –Because you are making bonds –Making bonds is exothermic The heat of a reaction can be calculated by subtracting the heats of formation of the reactants from the products  H =  H◦ f (products) -  H◦ f (reactants)

40. 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 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

41. 41 If H 2 (g) + 1/2 O 2 (g) H 2 O(l)  H=-285.5 kJ/mol then H 2 O(l) H 2 (g) + 1/2 O 2 (g)  H =+285.5 kJ/mol If you turn an equation around, you change the sign 2 H 2 O(l) 2 H 2 (g) + O 2 (g)  H =+571.0 kJ/mol If you multiply the equation by a number, you multiply the heat by that number. –Twice the moles, twice the heat

42. 42 You make the products, so you need their heats of formation You “unmake” the reactants so you have to subtract their heats. https://www.youtube.com/watch?v=_NLAgSnqNOE&noredirect=1

43. Calculate the heat of combustion of methane, CH 4 CH 4 (g) + 2 O 2 (g) CO 2 (g) + 2 H 2 O(g )  H◦ f CH 4 (g) = -74.86 kJ/mol  H◦ f O 2 (g) = 0 kJ/mol  H◦ f CO 2 (g) = -393.5 kJ/mol  H◦ f H 2 O(g) = - 241.8 kJ/mo l 43 pg. 316 Step #1: since 2 moles of water are produced by each mole of methane, we multiply the  H ◦ f. of water by 2. H 2 (g) + ½ O 2 (g)  H 2 O(g ) 2x(- 241.8)= - 483.6kJ/mol

44. Calculate the heat of combustion of methane, CH 4 CH 4 (g) + 2 O 2 (g) CO 2 (g) + 2 H 2 O(g )  H◦ f CH 4 (g) = +74.86 kJ/mol  H◦ f O 2 (g) = 0 kJ/mol  H◦ f CO 2 (g) = -393.5 kJ/mol  H◦ f H 2 O(g) = -483.6 kJ/mo l 44 pg. 316 Step #2: sum up all the  H ◦ f. :  H◦ f = [-393.5 kJ/mol + (-483.6 kJ/mol)]- [-74.86 kJ/mol + (0 kJ/mol )]  H◦ f = [-393.5 -483.6] + 74.86 = -877.1 + 74.86 = -802.2 kJ/mol  H rxn =   H f (products) -   H f (reactants)

45. temperature ( o C) Specific Heat Capacity Specific Heat Capacity = heat required to raise the temperature of 1 gram of a substance object by 1 o C Affected by −What the substance is −Mass of the object

46. 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

47. – A device used to experimentally determine the amount of heat released or absorbed during a physical or chemical change heat gained = heat lost

48. Most common units of energy 1. S  unit of energy is the joule (J), defined as 1 (kilogrammeter 2 )/second 2, energy is also expressed in kilojoules (1 kJ = 10 3 J). 2. Non-S  unit of energy is the calorie. One cal = 4.184 J or 1J = 0.2390 cal. Units of energy are the same, regardless of the form of energy

49. The amount of heat required to raise the temperature of one gram of substance by one degree Celsius. Page 296

50. 50 The higher the specific heat the more energy it takes to change its temperature. Pizza burning the roof of your mouth The same amount of heat is released when an object cools down

51. C = Specific Heat (J/ ºC.g) Q = Heat lost or gained ( J)  T = Temperature change = T f – T i (ºC) Q = m. C.  T

52. Change in energy = mass * specific heat * change in temp. How much heat is need to raise 5 g of water 10 ̊ C? (Water’s specific heat is 4.18 J/( ̊ C. g ) 1. Q = m * C *  T 2. Q = m * C *  T 5g * C *  T 4.18 J/(g- ̊ C) *  T 10 ̊ C Known m= 5 Kg  T = 10 ̊ C C = 4.18 J/(g- ̊ C) Unknown heat needed? 3. Q = 209 J

53. 28,875 J of energy are added to a 5-kg piece of copper that has an initial temperature of 293 K. What will be the final temperature of this piece of copper? (Copper’s specific heat:385 J/(kg-K)) 1. Q = m * C *  T Known m= 5 Kg T i = 293 K C = 385 J/(g- C) Unknown Final temperature? 1. Q = m * C *  T 2. 28,875 J = m * C *  T 5 kg * C *  T385 J/(kg-K) *  T 3.  T = 15 K 293 K + 15 K = 308 K

54. Temperature ( o C) 40 20 0 -20 -40 -60 -80 -100 120 100 80 60 140 Time  H = mol x  H fus  H = mol x  H vap Heat = mass x  t x C p, liquid Heat = mass x  t x C p, gas Heat = mass x  t x C p, solid

66. Choose all that apply... C(s) + 2 S(g) CS 2 (l)  H = 89.3 kJ Which of the following are true? A)This reaction is exothermic B) It could also be written C(s) + 2 S(g) + 89.3 kJ CS 2 (l) C) The products have higher energy than the reactants D) It would make the water in the calorimeter colder