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Chapter 61 Principles of Reactivity: Energy and Chemical Reactions Chapter 6.

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Presentation on theme: "Chapter 61 Principles of Reactivity: Energy and Chemical Reactions Chapter 6."— Presentation transcript:

1 Chapter 61 Principles of Reactivity: Energy and Chemical Reactions Chapter 6

2 2 Energy: Some Basics From Physics: Force – a kind of push or pull on an object. Energy – the capacity to do work. Work – force applied over a distance w = F d Heat – energy transferred from a warmer object to a cooler object.

3 Chapter 63 Kinetic and Potential Energy Kinetic Energy (Thermal Energy) – energy due to motion. Energy: Some Basics

4 Chapter 64 Kinetic and Potential Energy Potential Energy (Stored Energy) – the energy an object possesses due to its position. -Potential energy can be converted into kinetic energy. Example: a ball of clay dropping off a building. Energy: Some Basics

5 Chapter 65 First Law of Thermodynamics The total amount of energy in the universe is fixed. Also referred to as the Law of Conservation of Energy Energy: Some Basics

6 Chapter 66 Temperature and Heat Temperature is a measure of heat energy Heat is not the same as temperature. The more thermal energy a substance has the greater its molecular motion (kinetic energy). The total thermal energy in an object is the sum of the energies of all the bodies in the object. Energy: Some Basics

7 Chapter 67 Systems and Surroundings System – portion of the universe we wish to study. Surroundings – everything else. Universe = System + Surroundings Energy: Some Basics

8 Chapter 68 Directionality of Heat Heat energy always flows from the hot object to the cold object. Energy: Some Basics - this flow continues until the two objects are at the same temperature (thermal equilibrium).

9 Chapter 69 Directionality of Heat Exothermic – Heat is transferred from the system to the surroundings (object will feel hot). Endothermic – Heat is transferred to the system from the surroundings (object will fell cold). Energy: Some Basics

10 Chapter 610 Energy Units SI Unit for energy is the joule, J: A more traditional unit is the Calorie Calorie (cal) – amount of energy required to raise 1.0 g of water 1 o C. 1cal = 4.184J Energy: Some Basics

11 Chapter 611 Specific Heat Capacity The amount of heat transferred is dependant on three quantities: –Quantity of material –Size of temperature change –Identity of the material

12 Chapter 612 Specific Heat Capacity q = energy c = specific heat capacity T = temperature change

13 Chapter 613 Specific Heat Capacity exothermic- T-q endothermic+ T+q

14 Chapter 614 Specific Heat Capacity Specific heat capacity can be either per gram (J/g( o C) or per mole (J/mol( o C). The smaller a substances specific heat capacity, the better a thermal conductor it is.

15 Chapter 615 Energy and Changes of State

16 Chapter 616 Energy and Changes of State In the previous slide there is a continuous, steady application of energy. The sections that show increasing temperature are the result of the particular phase being warmed. q = cm( T) The flat sections occur when all the applied energy is used to change the phase of the substance. Fusion – solid liquid Vaporization – liquid gas

17 Chapter 617 Energy and Changes of State The energy required to change the phase of a substance is unique and is described in a physical constant. Solid Liquid Heat of Fusion (water, 333J/g) Liquid Gas Heat of Vaporization (water, 2256J/g) These constants can be used to determine the energy used in melting or vaporizing a substance. q = (Heat of Fusion)(mass of sample) q = (Heat of Vapor.)(mass of sample)

18 Chapter 618 Energy and Changes of State q = cm( T) q = (Heat of Vapor.)(mass)

19 Chapter 619 First Law of Thermodynamics Internal Energy Internal Energy – sum of all kinetic and potential energy in an object. It is very hard to determine an objects internal energy, but it is possible to determine the change in energy ( E). Change in internal energy, E = E final - E initial –A positive E means E final > E initial or the system gained energy from the surroundings (endothermic) –A negative E means E final < E initial or the system lost energy to the surroundings (exothermic)

20 Chapter 620 Relating E to Heat and Work E = q + w E = q + w q = heatw = work Both heat energy and work can change a systems internal energy. First Law of Thermodynamics

21 Chapter 621 State Functions State function – a process that is determined by its initial and final conditions. First Law of Thermodynamics

22 Chapter 622 State Functions State function – a process that is determined by its initial and final conditions. A process that is not path dependant. Work (w) and heat (q) are not state functions. Energy change ( E) is a state function. First Law of Thermodynamics

23 Chapter 623 Enthalpy (H) - Heat transferred between the system and surroundings carried out under constant pressure. E = q + w E = q + w Most reactions occur under constant pressure, so E = q + (-P( V)) E = q + (-P( V)) V = 0 If volume is also constant, V = 0 E = q p E = q p So, Energy change is due to heat transfer, E = H = q p E = H = q p First Law of Thermodynamics

24 Chapter 624 Enthalpy Change ( H) – The heat evolved or absorbed in a reaction at constant pressure H = H final - H initial = q P Enthalpy

25 Chapter 625 Enthalpy Change ( H) – The heat evolved or absorbed in a reaction at constant pressure H and H are state functions, depending only on the initial and final states. Enthalpy

26 Chapter 626 Enthalpies of Reaction 2 H 2 (g) + O 2 (g) 2 H 2 O(g) H = J

27 Chapter 627 For a reaction 1.Enthalpy is an extensive property (magnitude H is directly proportional to amount): CH 4 (g) + 2O 2 (g) CO 2 (g) + 2H 2 O(g) H = -802 kJ 2CH 4 (g) + 4O 2 (g) 2CO 2 (g) + 4H 2 O(g) H = kJ Enthalpies of Reaction

28 Chapter 628 For a reaction 1.Enthalpy is an extensive property (magnitude H is directly proportional to amount): 2.When we reverse a reaction, we change the sign of H: CO 2 (g) + 2H 2 O(g) CH 4 (g) + 2O 2 (g) H = +802 kJ CH 4 (g) + 2O 2 (g) CO 2 (g) + 2H 2 O(g) H = -802 kJ Enthalpies of Reaction

29 Chapter 629 For a reaction 1.Enthalpy is an extensive property (magnitude H is directly proportional to amount): 2.When we reverse a reaction, we change the sign of H: 3.Change in enthalpy depends on state: CH 4 (g) + 2 O 2 (g) CO 2 (g) + 2 H 2 O(g) H = -802 kJ CH 4 (g) + 2 O 2 (g) CO 2 (g) + 2 H 2 O(l) H = -890 kJ Enthalpies of Reaction

30 Chapter Mg(s) + O 2 (g) 2 MgO(s) H = kJ a) Is this reaction endothermic or exothermic? Enthalpies of Reaction

31 Chapter Mg(s) + O 2 (g) 2 MgO(s) H = kJ a) Is this reaction endothermic or exothermic? Exothermic, this is indicated by the negative H. Enthalpies of Reaction

32 Chapter Mg(s) + O 2 (g) 2 MgO(s) H = kJ b)Calculate the amount of heat transferred when 2.4g of Mg reacts at constant pressure. Enthalpies of Reaction

33 Chapter Mg(s) + O 2 (g) 2 MgO(s) H = kJ b)Calculate the amount of heat transferred when 2.4g of Mg reacts at constant pressure. Enthalpies of Reaction

34 Chapter Mg(s) + O 2 (g) 2 MgO(s) H = kJ b)Calculate the amount of heat transferred when 2.4g of Mg reacts at constant pressure. Enthalpies of Reaction

35 Chapter Mg(s) + O 2 (g) 2 MgO(s) H = kJ b)Calculate the amount of heat transferred when 2.4g of Mg reacts at constant pressure. Enthalpies of Reaction

36 Chapter Mg(s) + O 2 (g) 2 MgO(s) H = kJ c) How many grams of MgO are produced during an enthalpy change of 96.0 kJ? Enthalpies of Reaction

37 Chapter Mg(s) + O 2 (g) 2 MgO(s) H = kJ c) How many grams of MgO are produced during an enthalpy change of 96.0 kJ? Enthalpies of Reaction

38 Chapter Mg(s) + O 2 (g) 2 MgO(s) H = kJ c) How many grams of MgO are produced during an enthalpy change of 96.0 kJ? Enthalpies of Reaction

39 Chapter Mg(s) + O 2 (g) 2 MgO(s) H = kJ c) How many grams of MgO are produced during an enthalpy change of 96.0 kJ? Enthalpies of Reaction

40 Chapter Mg(s) + O 2 (g) 2 MgO(s) H = kJ d) How many kilojoules of heat are absorbed when 7.50g of MgO is decomposed into Mg and O 2 at constant pressure? Enthalpies of Reaction

41 Chapter MgO(s) 2 Mg(s) + O 2 (g) H = 1205 kJ d) How many kilojoules of heat are absorbed when 7.50g of MgO is decomposed into Mg and O 2 at constant pressure? Enthalpies of Reaction

42 Chapter MgO(s) 2 Mg(s) + O 2 (g) H = 1205 kJ d) How many kilojoules of heat are absorbed when 7.50g of MgO is decomposed into Mg and O 2 at constant pressure? Enthalpies of Reaction

43 Chapter MgO(s) 2 Mg(s) + O 2 (g) H = 1205 kJ d) How many kilojoules of heat are absorbed when 7.50g of MgO is decomposed into Mg and O 2 at constant pressure? Enthalpies of Reaction

44 Chapter 644 Constant-Pressure Calorimetry Calorimetry

45 Chapter 645 Constant-Pressure Calorimetry Atmospheric pressure is constant! H = q P q system = -q surroundings -The surroundings are composed of the water in the calorimeter and the calorimeter. q system = -(q water + q calorimeter ) Calorimetry

46 Chapter 646 Constant-Pressure Calorimetry Atmospheric pressure is constant! H = q P q system = -q surroundings -The surroundings are composed of the water in the calorimeter and the calorimeter. -For most calculations, the q calorimeter can be ignored. q system = - q water c system m system T system = - c water m water T water Calorimetry

47 Chapter 647 Bomb Calorimetry (Constant-Volume Calorimetry) Calorimetry

48 Chapter 648 Bomb Calorimetry (Constant-Volume Calorimetry) -Special calorimetry for combustion reactions -Substance of interest is placed in a bomb and filled to a high pressure of oxygen -The sealed bomb is ignited and the heat from the reaction is transferred to the water -This calculation must take into account the heat capacity of the calorimeter (this is grouped together with the heat capacity of water). q rxn = -C calorimeter ( T) Calorimetry

49 Chapter 649 NH 4 NO 3 (s) NH 4 + (aq) + NO 3 - (aq) T water = 16.9 o C – 22.0 o C = -5.1 o C m water = 60.0g c water = 4.184J/g o C m sample = 4.25g q sample = -q water q sample = -c water m water T water q sample = -(4.184J/g o C)(60.0g)(-5.1 o C) q sample = J - Now calculate H in kJ/mol Calorimetry

50 Chapter 650 NH 4 NO 3 (s) NH 4 + (aq) + NO 3 - (aq) T water = 16.9 o C – 22.0 o C = -5.1 o C m water = 60.0g c water = 4.184J/g o C m sample = 4.25g q sample = J moles NH 4 NO 3 = 4.25g/80.032g/mol = mol H = q sample /moles H = J/0.0529mol H = 24.2 kJ/mol Calorimetry

51 Chapter C 8 H O 2 16 CO H 2 O T water = o C – o C = 7.42 o C C cal = 11.66kJ/ o C m sample = 1.80g q rxn = -C cal ( T water ) q rxn = kJ/ o C(7.42 o C) q rxn = kJ Calorimetry

52 Chapter C 8 H O 2 16 CO H 2 O T water = o C – o C = 7.42 o C C cal = 11.66kJ/ o C m sample = 1.80g q rxn = kJ H combustion (in kJ/g) H combustion = kJ/1.80g = Calorimetry

53 Chapter C 8 H O 2 16 CO H 2 O T water = o C – o C = 7.42 o C C cal = 11.66kJ/ o C m sample = 1.80g q rxn = kJ H combustion (in kJ/g) H combustion = kJ/1.80g = kJ/g H combustion (in kJ/mol) H combustion = kJ/ mol = Calorimetry

54 Chapter C 8 H O 2 16 CO H 2 O T water = o C – o C = 7.42 o C C cal = 11.66kJ/ o C m sample = 1.80g q rxn = kJ H combustion (in kJ/g) H combustion = kJ/1.80g = kJ/g H combustion (in kJ/mol) H combustion = kJ/ mol = kJ/mol Calorimetry

55 Chapter 655 Hesss law - if a reaction is carried out in a series of steps, H for the overall reaction is the sum of Hs for each individual step. For example: CH 4 (g) + 2O 2 (g) CO 2 (g) + 2H 2 O(g) H = -802 kJ 2H 2 O(g) 2H 2 O(l) H = -88 kJ CH 4 (g) + 2O 2 (g) CO 2 (g) + 2H 2 O(l) H = -890 kJ Hesss Law

56 Chapter 656 Enthalpies of Formation (Heat of Formation) -There are many type of H, depending on what you want to know H vapor – enthalpy of vaporization (liquid gas) H fusion – enthalpy of fusion (solid liquid) H combustion – enthalpy of combustion (energy from burning a substance)

57 Chapter 657 Enthalpies of Formation (Heat of Formation) -A fundamental H is the Standard Enthalpy of Formation ( ) Standard Enthalpy of Formation ( ) – The enthalpy change that accompanies the formation of one mole of a substance from the most stable forms of its component elements at 298 Kelvin and 1 atmosphere pressure. The standard enthalpy of formation of the most stable form on any element is zero

58 Chapter 658 Enthalpies of Formation

59 Chapter 659 Enthalpies of Formation Using Enthalpies of Formation to Calculate Enthalpies of Reaction For a reaction:

60 Chapter 660 Homework Problems 4, 14, 20, 24, 28, 36, 40, 44, 46, 52, 54, 56a


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