Chemical Thermodynamics

Transcription

1 Chemical Thermodynamics 1

2 Thermodynamics Thermodynamics is a Greek term which means, heat power. Thermodynamics is the study of energy and its transformations. 2

3 Thermodynamics Thermochemistry how we observe, measure and predict energy changes for physical changes chemical reactions Thermodynamics how to predict if a chemical reaction will occur or not. 3

4 Collision Theory Used to Explain Reaction Rates Atoms, ions, and molecules can form a chemical bond when they collide, provided the particles have enough kinetic energy. Particles lacking the necessary kinetic energy to react still collide, but simply bounce apart. Activation energy - the minimum energy colliding particles must have in order to react. 4

5 Factors that Affect Reaction Rates Temperature Concentration Particle size Catalyst 5

6 Factors that Affect Reaction Rates Temperature Increasing temp increases the number of particles that have enough kinetic energy to react when they collide. Concentration changes Cramming more particles into a fixed volume increases the collision frequency. 6

7 Factors that Affect Reaction Rates Particle Size the smaller the particle size, the larger the surface area for a given mass of particles. Decreasing particle size will increase the rate of reaction. Catalyst A catalyst is a substance that increases the rate of a reaction without being used up itself in the reaction. 7

8 The Nature of Energy Energy is the capacity to do work or to transfer heat. Heat is the energy transferred from one object to another because of a difference in temperature. 8

9 The 1st Law of Thermodynamics Energy is neither created nor destroyed; it can only be transformed. energy can change from potential energy to kinetic energy, or exchanged between the system and the surroundings. The total amount of energy in the universe is constant. 9

10 Energy Diagrams exothermic endothermic 10

11 Energy Diagram (exothermic) CH 4 (g) + 2O 2 (g) CO 2 (g) + 2H 2 O (l) E ner gy CH 4 (g) + 2O 2 (g) Amount of energy needed to activate the reaction Reactants Heat given off!h CO 2 (g) + 2H 2 O (l) Products Reaction Progress 11

12 Energy Units Joule calorie 1 cal = 4.18 J 12

13 Change in Enthalpy 13

14 Heat and Enthalpy Changes Enthalpy is the amount of heat energy possessed by substances. ΔH rxn = H final - H initial 14

15 Exothermic Equations Reactions with enthalpy changes that are negative, have heat flowing out of the system (energy is a product) and are called exothermic processes. CH 4 (g) + 2O 2 (g) CO 2 (g) + 2H 2 O (g) kj ΔH = -890 kj 15

16 Endothermic Reactions Endothermic reactions are those in which heat is added (energy is a reactant), or flows, into the system, the enthalpy change of the process has a positive value. N 2 (g) + O 2(g) + 43kcal 2 NO (g) H = +43 kcal Why is it a good thing that the above rxn requires energy? 16

17 Endothermic Reactions Ba(OH) 2 8H 2 O (s) + 2NH 4 NO 3 (s) + Energy Ba(NO 3 ) 2 (s) + 2NH 3 (g) + 10H 2 O (l) Freezes to the board Endothermic reactions take heat, so they get cold. 17

18 Heat and Enthalpy Changes Enthalpy is the amount of heat energy possessed by substances. ΔH rxn = H final - H initial 18

19 Hess s Law Hess s law states that if a reaction is carried out in a series of steps, ΔH for the reaction will be equal to the sum of the enthalpy changes for each step. ΔH 19

20 Hess s Law The overall enthalpy change in a reaction is equal to the sum of the enthalpy changes of the individual steps. ΔH rxn = ΔH 1 + ΔH 2 + ΔH

21 Hess s Law C (graphite) + O 2 (g) CO 2(g) ΔH = kj/mol rxn H 2 + 1/2 O 2 (g) H 2 O (l) ΔH = kj/mol rxn CH 4 (g) + 2O 2 (g) CO 2 (g) + 2H 2 O (l) ΔH = kj/mol rxn Use the above information to find H for the reaction below C (graphite) + 2H 2 (g) CH 4 (g) H =? 21

22 C (graphite) + 2H 2 (g) CH 4 (g) C (graphite) + O 2 (g) CO 2(g) H 2 + 1/2 O 2 (g) H 2 O (l) CO 2 (g) + 2H 2 O (l) CH 4 (g) + 2O 2 (g) 22

23 C (graphite) + 2H 2 (g) CH 4 (g) C (graphite) + O 2 (g) CO 2(g) H 2 + 1/2O 2 (g) H 2 O (l) (2) ΔH = kj/mol rxn ΔH = kj/mol rxn CO 2 (g) + 2H 2 O (l) CH 4 (g) + 2O 2 (g) ΔH = kj/mol rxn ΔH = (2)(-285.8) ΔH = -74.8kJ 23

24 Hess s Law 24

25 Enthalpy Changes The enthalpy change for a reaction is equal in magnitude but opposite in sign to ΔH for the reverse reaction. The enthalpy change for a reaction depends on the states of the reactants and the products. 2HBr (g) H 2(g) + Br 2(l) ΔH = +72.8kJ/mol rxn H 2(g) + Br 2(l) 2HBr (g) ΔH = -72.8kJ/mol rxn 25

26 Calorimetry Calorimetry is the measurement of heat flow. These measurements are made using an apparatus called a calorimeter. 26

27 Heat Capacity The amount of heat it takes to change an object s temperature by exactly 1 C. calorie - the quantity of heat that raises the temperature of 1 g of pure water 1 C. Calorie - dietary calorie, always refers to food. 1 Cal = 1000 cal = 1 kcal Joule - the SI unit for heat and energy. 1 J will raise the temp of 1 g of pure water C. 1 J = cal 1 cal = 4.18 J 27

28 Specific Heat Capacity (C) The amount of heat it takes to change 1 g of a substance by 1 C. Specific Heat Capacity has three units cal/g C or J/g C 28

29 Specific Heat Capacity The value of Specific Heat is different for various types of matter. Water has the highest specific heat. Heat affects the temperature of those things with a high specific heat much less than those with a low specific heat. Metals have low specific heats. A small amount of energy is needed to raise the temperature of metals. 29

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31 Calculating Heat Energy Heat = Specific Heat x Mass x Change in Temperature Heat(q) = C(cal/g C) x Mass (g) x T ( C) q = CmΔT Mass - the mass of the matter being measured. (g) Change in Temperature - the increase or decrease in the temperature of the matter. ( C) Specific Heat (C) - the amount of heat energy required to raise 1g of a substance 1 C. (cal/g 0 C) or (J/g 0 C) 31

32 Summary q gained = -q lost Heat gained is equal and opposite to the heat lost. q = CmΔT Heat = Specific heat Capacity x mass x change in Temp 32

33 Do the algebra C = q/mδt q = m = ΔT = 33

34 Do the algebra C = q/mδt q = C m ΔT m = q/cδt ΔT = q/mc 34

35 Sample Problem The temperature of a piece of copper with a mass of 95.4 g changes from 25.0 C to 48.0 C when the metal absorbs 849 J of heat. What is the specific heat of copper? 35

36 Sample Problem The temperature of a piece of copper with a mass of 95.4 g changes from 25.0 C to 48.0 C when the metal absorbs 849 J of heat. What is the specific heat of copper? q =CmΔT Basic formula 36

37 Sample Problem The temperature of a piece of copper with a mass of 95.4 g changes from 25.0 C to 48.0 C when the metal absorbs 849 J of heat. What is the specific heat of copper? q =CmΔT C =q/m ΔT Do Algebra to solve for specific heat 37

38 Sample Problem The temperature of a piece of copper with a mass of 95.4 g changes from 25.0 C to 48.0 C when the metal absorbs 849 J of heat. What is the specific heat of copper? q =CmΔT C =q/m ΔT C = 849 J/[95.4 g x (48.0 C C)] Fill in values from the problem 38

39 Sample Problem The temperature of a piece of copper with a mass of 95.4 g changes from 25.0 C to 48.0 C when the metal absorbs 849 J of heat. What is the specific heat of copper? q =CmΔT C =q/m ΔT C = 849 J/(95.4 g x 23.0 C) C = J/g C Solve for the specific heat. Notice that specific heat has 3 units 39

40 Sample Problem #2 What is the specific heat of a substance that absorbs 2.5 x 10 3 joules of heat when a sample of 1.0 x 10 3 g of the substance increases in temperature from C to C? 40

41 Sample Problem #2 What is the specific heat of a substance that absorbs 2.5 x 10 3 joules of heat when a sample of 1.0 x 10 3 g of the substance increases in temperature from C to C? q =CmΔT C =q/m ΔT C = 2.5 x 10 3 J/(1.0 x 10 3 g x 60.0 C) C = J/g C 41

42 Heat and Changes of State Heat of fusion =(ΔH fus ) the heat absorbed by a substance in melting from a solid to a liquid at constant temperature. (s) (l) Heat of solidification =(ΔH solid ) the heat lost when a liquid changes to a solid at a constant temp. (l) (s) Heat of vaporization =(ΔH vap ) the heat absorbed by a substance in vaporizing from a liquid to a gas at constant temperature. (l) (g) Heat of condensation =(ΔH con ) the heat lost when a gas changes to a liquid at a constant temp. (g) (l) 42

43 Heat and Changes of State Energy is conserved in all physical and chemical changes. ΔH fus = - ΔH solid ΔH vap = - ΔH con 43

44 Changes of state for water Melting (fusion) ΔH fus = 80 cal/g Solidification ΔH solid = - 80 cal/g ΔH fus = - ΔH solid Vaporization ΔH vap = 540 cal/g Condensation ΔH con = cal/g ΔH vap = - ΔH con Melting and Vaporization are endothermic. They require energy. Solidification and Condensation are exothermic. They give off energy. 44

45 Heating Curve for Water 45

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47 C steam=.4 cal/g C C water = 1.0 cal/g C C ice =.5 cal/g C 47

48 ΔH vap = 540 cal/g.4 cal/g C ΔH fus = 80 cal/g 1.0 cal/g C.5 cal/g C 48

49 ΔH vap = 540 cal/g.4 cal/g C ΔH fus = 80 cal/g 1.0 cal/g C.5 cal/g C To calculate the heat required, we must use the Specific heat capacity of water in its different states and the heat required for the change of state. 49

50 How much heat is required to change 25.0 g of solid water from C to gaseous water at 140 C? 50

51 How much heat is required to change 25.0 g of solid water from C to gaseous water at 140 C? Solid Fusion Liquid Vaporization Gas.5 cal/g C x mass x ΔT 80 cal/g x mass 1 cal/g C x mass x ΔT 540 cal/g x mass.4 cal/g C x mass x ΔT 51

52 How much heat is required to change 25.0 g of solid water from C to gaseous water at 140 C? Solid.5 cal/g C x 25.0 g x 20 C Fusion 80 cal/g x 25.0 g Liquid 1 cal/g C x 25.0 g x 100 C Vaporization 540 cal/g x 25.0 g Gas.4 cal/g C x 25.0 g x 20 C 52

53 How much heat is required to change 25.0 g of solid water from C to gaseous water at 140 C? Solid Fusion Liquid Vaporization Gas.5 cal/g C x 25.0 g x 20 C = 250 cal 80 cal/g x 25.0 g = 2000 cal 1 cal/g C x 25.0 g x 100 C = 2500 cal 540 cal/g x 25.0 g = 13,500 cal.4 cal/g C x 25.0 g x 40 C = 400 cal Total heat = 250 cal cal cal+ 13,500 cal cal = 18,650 cal = kcal Notice that most of the energy is used for Vaporization. 53

54 Water vapor will condense back into liquid water on a cold surface: condensation. 54

55 Phase Changes for Water Vaporization ΔH = 540 cal/g Condensation ΔH=-540 cal/g Deposition ΔH = -620 cal/g Hess s Law: ΔH Sub = ΔH Fus + ΔH Vap 620 = Sublimation ΔH = 620 cal/g Fusion ΔH =80 cal/g Solidification ΔH = -80 cal/g 55

56 Thermochemistry Stoichiometry Hydrazine, N 2 H 4, is used as a rocket fuel. When 1.00g of hydrazine is burned how much heat is given off? N 2 H 4 + O 2 N 2 + 2H 2 O ΔH = -618 kj 56

57 Hydrazine, N 2 H 4, is used as a rocket fuel. When 1.00g of hydrazine is burned how much heat is given off? N 2 H 4 + O 2 N 2 + 2H 2 O ΔH = -618 kj/mol 1.00gN H 2 4 " 1moleN H 2 4 " 618kJ 32.0gN H 1mole =

58 Hydrazine, N 2 H 4, is used as a rocket fuel. When 1.00g of hydrazine is burned how much heat is given off? N 2 H 4 + O 2 N 2 + 2H 2 O ΔH = -618 kj/mol 1.00gN H 2 4 " 1moleN H 2 4 " 618kJ 32.0gN H 1mole = kj 58

59 Gaseous butane, C 4 H 10, is burned in cigarette lighters. 2 C 4 H O 2 8 CO H 2 O ΔH = x 10 3 kj/mol How many kj of heat would be released by the combustion of 10.0 g butane? 59

60 Gaseous butane, C 4 H 10, is burned in cigarette lighters. 2 C 4 H O 2 8 CO H 2 O ΔH = x 10 3 kj/mol How many kj of heat would be released by the combustion of 10.0 g butane? 10.0gC 4 H 10 " 1moleC 4H gC 4 H 10 " 5.756x103 kj 2moleC 4 H 10 = 60

61 Gaseous butane, C 4 H 10, is burned in cigarette lighters. 2 C 4 H O 2 8 CO H 2 O ΔH = x 10 3 kj/mol How many kj of heat would be released by the combustion of 10.0 g butane? 10.0gC 4 H 10 " 1moleC 4H gC 4 H 10 " 5.756x103 kj 2moleC 4 H 10 = 4.96 x 10 2 kj 61