1. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu How to Use This Presentation To View the presentation as a slideshow with effects select “View” on the menu bar and click on “Slide Show.” To advance through the presentation, click the right-arrow key or the space bar. From the resources slide, click on any resource to see a presentation for that resource. From the Chapter menu screen click on any lesson to go directly to that lesson’s presentation. You may exit the slide show at any time by pressing the Esc key.

2. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Chapter Presentation Transparencies Bellringer Standardized Test PrepVisual Concepts Sample Problems Resources

3. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Table of Contents Chapter 10 Causes of Change Section 1 Energy Transfer Section 2 Using Enthalpy Section 3 Changes in Enthalpy During Chemical Reactions Section 4 Order and Spontaneity

4. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Section 1 Energy Transfer Bellringer Look at the words below. heattemperature Write a definition for each of these words and describe how they are related. Chapter 10

5. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Objectives Define enthalpy. Distinguish between heat and temperature. Perform calculations using molar heat capacity. Section 1 Energy Transfer Chapter 10

6. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Energy as Heat A sample can transfer energy to another sample. One of the simplest ways energy is transferred is as heat. Heat is the energy transferred between objects that are at different temperatures Though energy has many different forms, all energy is measured in units called joules (J). Section 1 Energy Transfer Chapter 10

7. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Energy as Heat, continued Energy is never created or destroyed. The amount of energy transferred from one sample must be equal to the amount of energy received by a second sample. The total energy of the two samples remains exactly the same. Section 1 Energy Transfer Chapter 10

8. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Visual Concepts Heat Chapter 10

9. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Energy as Heat, continued Temperature Temperature is a measure of how hot (or cold) something is; specifically, a measure of the average kinetic energy of the particles in an object. When samples of different temperatures are in contact, energy is transferred from the sample that has the higher temperature to the sample that has the lower temperature. Section 1 Energy Transfer Chapter 10

10. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Energy as Heat, continued Temperature, continued If no other process occurs, the temperature of a sample increases as the sample absorbs energy. The temperature of a sample depends on the average kinetic energy of the sample’s particles. The higher the temperature of a sample is, the faster the sample’s particles move. The temperature increase of a sample also depends on the mass of the sample. Section 1 Energy Transfer Chapter 10

11. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Energy as Heat, continued Heat and Temperature are Different Temperature is an intensive property, which means that the temperature of a sample does not depend on the amount of the sample. Heat is an extensive property, which means that the amount of energy transferred as heat by a sample depends on the amount of the sample. Section 1 Energy Transfer Chapter 10

12. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Visual Concepts Comparing Extensive and Intensive Properties Chapter 10

13. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Visual Concepts Temperature and the Temperature Scale Chapter 10

14. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Energy as Heat, continued A Substance’s Energy Can Be Measure by Enthalpy Measuring the total amount of energy present in a sample of matter is impossible, but changes in energy content can be determined. These changes are determined by measuring the energy that enters or leaves the sample of matter. If 73 J of energy enter a piece of silver and no change in pressure occurs, we know that the enthalpy of the silver has increased by 73 J. Enthalpy, which is represented by the symbol H, is the total energy content of a sample. Section 1 Energy Transfer Chapter 10

15. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Energy as Heat, continued A Substance’s Energy Can Be Measure by Enthalpy, continued Enthalpy is the sum of the internal energy of a system plus the product of the system’s volume multiplied by the pressure that the system exerts on its surroundings If pressure remains constant, the enthalpy increase of a sample of matter equals the energy as heat that is received. This relationship remains true even when a chemical reaction or a change of state occurs. Section 1 Energy Transfer Chapter 10

16. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Energy as Heat, continued A Sample’s Enthalpy Includes the Kinetic Energy of Its Particles The particles in a sample are in constant motion. These particles have kinetic energy. The enthalpy of a sample includes the total kinetic energy of its particles. Both the total and average kinetic energies of a substance’s particles are important to chemistry, because these quantities account for every particle’s kinetic energy. Section 1 Energy Transfer Chapter 10

17. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Molar Heat Capacity Molar heat capacity can be used to determine the enthalpy change of a sample. The molar heat capacity of a pure substance is the energy as heat needed to increase the temperature of 1 mol of the substance by 1 K. Molar heat capacity has the symbol C and the unit J/K  mol. Molar heat capacity is accurately measured only if no other process, such as a chemical reaction, occurs. Section 1 Energy Transfer Chapter 10

18. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Molar Heat Capacity, continued The following equation shows the relationship between heat and molar heat capacity, where q is the heat needed to increase the temperature of n moles of a substance by  T. q = nC  T heat = (amount in moles)  (molar heat capacity)  (change in temperature) Section 1 Energy Transfer Chapter 10

19. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Calculating Molar Heat Capacity Sample Problem A Determine the energy as heat needed to increase the temperature of 10.0 mol of mercury by 7.5 K. The value of C for mercury is 27.8 J/Kmol. Section 1 Energy Transfer Chapter 10

20. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Calculating Molar Heat Capacity, continued Sample Problem A Solution q = nC  T q = (10.0 mol )(27.8 J/Kmol )(7.5 K) q = 2085 J The answer should only have two significant figures, so it is reported as 2100 J or 2.1  10 3 J. Section 1 Energy Transfer Chapter 10

21. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Molar Heat Capacity, continued Molar Heat Capacity Depends on the Number of Atoms One mole of tungsten has a mass of 184 g, while one mole of aluminum has a mass of only about 27 g. You might expect that much more heat is needed to change the temperature of 1 mol W than is needed to change the temperature of 1 mol Al. The molar heat capacities of all of the metals are nearly the same. The temperature of 1 mol of any solid metal is raised 1 K when the metal absorbs about 25 J of heat. Section 1 Energy Transfer Chapter 10

22. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Molar Heat Capacity, continued Molar Heat Capacity Depends on the Number of Atoms Section 1 Energy Transfer Chapter 10

23. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Molar Heat Capacity, continued Molar Heat Capacity Is Related to Specific Heat The specific heat of a substance is represented by c p and is the energy as heat needed to raise the temperature of one gram of substance by one Kelvin. The molar heat capacity of a substance, C, is related to moles of a substance not to the mass of a substance. Because the molar mass is the mass of 1 mol of a substance, the following equation is true. M (g/mol)  c p (J/K  g) = C (J/K  mol) (molar mass)(specific heat) = (molar heat capacity) Section 1 Energy Transfer Chapter 10

24. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Visual Concepts Specific Heat Chapter 10

25. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Molar Heat Capacity, continued Heat Results in Disorderly Particle Motion When a substance receives energy in the form of heat, its enthalpy increases and the kinetic energy of the particles that make up the substance increases. The direction in which any particle moves is not related to the direction in which its neighboring particles move. The motions of these particles are random. Other types of energy can produce orderly motion or orderly positioning of particles. Section 1 Energy Transfer Chapter 10

26. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Section 2 Using Enthalpy Bellringer Make a list of objects that have absorbed energy and evidence that the objects have gained energy. For example, a burner on an electric stove gains energy when it is turned on. Evidence of this is that the burner glows red and it is hot. Chapter 10

27. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Objectives Define thermodynamics. Calculate the enthalpy change for a given amount of substance for a given change in temperature. Section 2 Using Enthalpy Chapter 10

28. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Molar Enthalpy Change Because enthalpy is the total energy of a system, it is an important quantity. The only way to measure energy is through a change. There’s no way to determine the true value of H.  H can be measured as a change occurs. The enthalpy change for one mole of a pure substance is called molar enthalpy change. Section 2 Using Enthalpy Chapter 10

29. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Molar Enthalpy Change, continued Molar Heat Capacity Governs the Changes When a pure substance is only heated or cooled, the amount of heat involved is the same as the enthalpy change.  H = q for the heating or cooling of substances Section 2 Using Enthalpy Chapter 10

30. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Molar Enthalpy Change, continued Molar Heat Capacity Governs the Changes, continued The molar enthalpy change is related to the molar heat capacity by the following equation. molar enthalpy change = C  T molar enthalpy change = (molar heat capacity)(temperature change) This equation does not apply to chemical reactions or changes of state. Section 2 Using Enthalpy Chapter 10

31. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Visual Concepts Enthalpy Change Chapter 10

32. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Calculating Molar Enthalpy Change for Heating Sample Problem B How much does the molar enthalpy change when ice warms from  5.4  C to  0.2  C? Section 2 Using Enthalpy Chapter 10

33. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Calculating Molar Enthalpy Change for Heating Sample Problem B Solution T initial = -5.4  C = 267.8 K and T final = -0.2  C = 273.0 K For H 2 O(s), C = 37.4 J/K  mol  T = T final  T initial = 5.2 K  H = C  T Section 2 Using Enthalpy Chapter 10

34. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Calculating Molar Enthalpy Change for Cooling Sample Problem C Calculate the molar enthalpy change when an aluminum can that has a temperature of 19.2  C is cooled to a temperature of 4.00  C. Section 2 Using Enthalpy Chapter 10

35. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Calculating Molar Enthalpy Change for Cooling Sample Problem C Solution T initial = -19.2  C = 292 K and T final = 4.00  C = 277 K For Al(s), C = 24.2 J/K  mol.  T = T final  T initial = 277K  292 K =  15 K  H = C  T  H = (24.2 J/K  mol)(  15 K ) =  360 J/K  mol Section 2 Using Enthalpy Chapter 10

36. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Molar Enthalpy Change, continued Enthalpy Changes of Endothermic or Exothermic Processes Enthalpy changes can be used to determine if a process is endothermic or exothermic. Processes that have positive enthalpy changes are endothermic. Processes that have negative enthalpy changes are exothermic. Section 2 Using Enthalpy Chapter 10

37. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Enthalpy of a System of Several Substances Thermodynamics is the branch of science concerned with the energy changes that accompany chemical and physical changes. Enthalpy is one of three thermodynamic properties. Section 2 Using Enthalpy Chapter 10

38. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Enthalpy of a System of Several Substances, continued Writing Equations for Enthalpy Changes An equation can be written for the enthalpy change that occurs during a change of state or a chemical reaction. The following equation is for the hydrogen and bromine reaction. H 2 (g, 298K) + Br 2 (l, 298K)  2HBr(g, 298K)  H = -72.8 kJ The enthalpy change for this reaction and other chemical reactions are written using the symbol  H. The negative enthalpy change indicates the reaction is exothermic. Section 2 Using Enthalpy Chapter 10

39. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Section 3 Changes In Enthalpy During Chemical Reactions Bellringer Write a short paragrpah describing what you know about food calories and how they relate to nutrition, metabolism, weight gain, and weight loss. Chapter 10

40. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Objectives Explain the principles of calorimetry. Use Hess’s law and standard enthalpies of formation to calculate  H. Section 3 Changes In Enthalpy During Chemical Reactions Chapter 10

41. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Changes in Enthalpy Accompany Reactions A change in enthalpy during a reaction depends on many variables. Temperature is one of the most important variables. To standardize the enthalpies of reactions, data are often presented for reactions in which both reactants and products have the standard thermodynamic temperature of 25.00  C or 298.15 K. Section 3 Changes In Enthalpy During Chemical Reactions Chapter 10

42. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Changes in Enthalpy Accompany Reactions, continued Chemists usually present a thermodynamic value for a chemical reaction by using the chemical equation. Section 3 Changes In Enthalpy During Chemical Reactions Chapter 10 This equation shows that when 0.5 mol of H 2 reacts with 0.5 mol of Br 2 to produce 1 mol HBr and all have a temperature of 298.15 K, the enthalpy decreases by 36.4 kJ.

43. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Chemical Calorimetry For the H 2 and Br 2 reaction, in which  H is negative, the total energy of the reaction decreases. The energy is released as heat by the system. If the reaction was endothermic, energy in the form of heat would be absorbed by the system and the enthalpy would increase. Section 3 Changes In Enthalpy During Chemical Reactions Chapter 10

44. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Chemical Calorimetry, continued The experimental measurement of an enthalpy change for a reaction is called calorimetry. Calorimetry is the measurement of heat-related constants, such as specific heat or latent heat Combustion reactions are always exothermic. The enthalpy changes of combustion reactions are determined using a bomb calorimeter. A calorimeter Is a device used to measure the heat absorbed or released in a chemical or physical change Section 3 Changes In Enthalpy During Chemical Reactions Chapter 10

45. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Chemical Calorimetry, continued A calorimeter is a sturdy, steel vessel in which the sample is ignited electrically in the presence of high- pressure oxygen. The energy from combustion is absorbed by a surrounding water bath and by the calorimeter. The water and the other parts of the calorimeter have known specific heats. A measured temperature increase can be used to calculate the energy released in the combustion reaction and then the enthalpy change. Section 3 Changes In Enthalpy During Chemical Reactions Chapter 10

46. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Bomb Calorimeter Chapter 10 Section 3 Changes In Enthalpy During Chemical Reactions

47. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Visual Concepts Calorimeter Chapter 10

48. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Chemical Calorimetry, continued Nutritionists Use Bomb Calorimetry Inside the pressurized oxygen atmosphere of a bomb calorimeter, most organic matter, including food, fabrics, and plastics, will ignite easily and burn rapidly. Sample sizes are chosen so that there is excess oxygen during the combustion reactions. Under these conditions, the reactions go to completion and produce carbon dioxide, water, and possibly other compounds. Section 3 Changes In Enthalpy During Chemical Reactions Chapter 10

49. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Chemical Calorimetry, continued Adiabatic Calorimetry Is Another Strategy Instead of using a water bath to absorb the energy generated by a chemical reaction, adiabatic calorimetry uses an insulating vessel. The word adiabatic means “not allowing energy to pass through.” No energy can enter or escape this type of vessel. The reaction mixture increases in temperature if the reaction is exothermic or decreases in temperature if the reaction is endothermic. Section 3 Changes In Enthalpy During Chemical Reactions Chapter 10

50. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Chemical Calorimetry, continued Adiabatic Calorimetry Is Another Strategy, continued If the system’s specific heat is known, the reaction enthalpy can be calculated. Adiabatic calorimetry is used for reactions that are not ignited, such as for reactions in aqueous solution. Section 3 Changes In Enthalpy During Chemical Reactions Chapter 10

51. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Hess’s Law Any two processes that both start with the same reactants in the same state and finish with the same products in the same state will have the same enthalpy change. Hess’s law states that the overall enthalpy change in a reaction is equal to the sum of the enthalpy changes for the individual steps in the process. Section 3 Changes In Enthalpy During Chemical Reactions Chapter 10

52. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Hess’s Law, continued When phosphorus is burned in excess chlorine 4 mol of phosphorus pentachloride, PCl 5, is synthesized. P 4 (s) + 10Cl 2 (g)  4PCl 5 (g)  H = -1596 kJ Phosphorus pentachloride may also be prepared in a two-step process. Step 1: P 4 (s) + 6Cl 2 (g)  4PCl 3 (g)  H = -1224 kJ Step 2: PCl 3 (g) + Cl 2 (g)  PCl 5 (g)  H = -93 kJ Section 3 Changes In Enthalpy During Chemical Reactions Chapter 10

53. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Hess’s Law, continued The second reaction must take place four times for each occurrence of the first reaction in the two-step process. This two-step process is more accurately described by the following equations. P 4 (s) + 6Cl 2 (g)  4PCl 3 (g)  H = -1224 kJ 4PCl 3 (g) + 4Cl 2 (g)  4PCl 5 (g)  H = 4(-93 kJ) = -372 kJ Section 3 Changes In Enthalpy During Chemical Reactions Chapter 10

54. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Hess’s Law, continued So, the total change in enthalpy by the two-step process is as follows: (-1224 kJ) + (-372 kJ) = -1596 kJ This enthalpy change,  H, for the two-step process is the same as the enthalpy change for the direct route of the formation of PCl 5. This example is in agreement with Hess’s law. Section 3 Changes In Enthalpy During Chemical Reactions Chapter 10

55. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Visual Concepts Hess’s Law Chapter 10

56. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Hess’s Law, continued Using Hess’s Law and Algebra Chemical equations can be manipulated using rules of algebra to get a desired equation. When equations are added or subtracted, enthalpy changes must be added or subtracted. When equations are multiplied by a constant, the enthalpy changes must also be multiplied by that constant. Section 3 Changes In Enthalpy During Chemical Reactions Chapter 10

57. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Hess’s Law, continued Using Hess’s Law and Algebra The enthalpy of the formation of CO, when CO 2 and solid carbon are reactants, is found using the equations below. 2C(s) + O 2 (g)  2CO(g)  H = -221 kJ C(s) + O 2 (g)  CO 2 (g)  H = -393 kJ You cannot simply add these equations because CO 2 would not be a reactant. Section 3 Changes In Enthalpy During Chemical Reactions Chapter 10

58. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Hess’s Law, continued Using Hess’s Law and Algebra, continued You subtract or reverse the second equation, carbon dioxide will be on the correct side of the equation. –C(s) – O 2 (g)  – CO 2 (g)  H = –(– 393 kJ) CO 2 (g)  C(s) + O 2 (g)  H = 393 kJ Reversing an equation causes the enthalpy of the new reaction to be the negative of the enthalpy of the original reaction. Section 3 Changes In Enthalpy During Chemical Reactions Chapter 10

59. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Hess’s Law, continued Using Hess’s Law and Algebra, continued Adding the two equations gives the equation for the formation of CO by using CO 2 and C. 2C(s) + O 2 (g)  2CO(g)  H = – 221 kJ CO 2 (g)  C(s) + O 2 (g)  H = 393 kJ 2C(s) + O 2 (g) + CO 2 (g)  2CO(g) + C(s) + O 2 (g)  H = 172 kJ Oxygen and carbon that appear on both sides of the equation can be canceled. C(s) + CO2(g)  2CO(g)  H = 172 kJ Section 3 Changes In Enthalpy During Chemical Reactions Chapter 10

60. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Hess’s Law, continued Standard Enthalpies of Formation The enthalpy change in forming 1 mol of a substance from elements in their standard states is called the standard enthalpy of formation of the substance,  Section 3 Changes In Enthalpy During Chemical Reactions Chapter 10 The values of the standard enthalpies of formation for elements are 0. The following equation is used to determine the enthalpy change of a chemical reaction from the standard enthalpies of formation.  H reaction =  H products -  H reactants

61. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Hess’s Law, continued Standard Enthalpies of Formation, continued Standard Enthalpies of Formation Section 3 Changes In Enthalpy During Chemical Reactions Chapter 10

62. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Calculating a Standard Enthalpies of Formation Sample Problem D Calculate the standard enthalpy of formation of pentane, C 5 H 12, using the given information. (1) C(s) + O 2 (g)  CO 2 (g)  = -393.5 kJ/mol (2) H 2 (g) + ½O 2 (g)  H 2 O(l)  = -285.8 kJ/mol (3) C 5 H 12 (g) + 8O 2 (g)  5CO 2 (g) + 6H 2 O(l)  H = -3535.6 kJ/mol Section 3 Changes In Enthalpy During Chemical Reactions Chapter 10

63. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Calculating a Standard Enthalpies of Formation Sample Problem D Solution Section 3 Changes In Enthalpy During Chemical Reactions Chapter 10

64. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Calculating a Reaction’s Change in Enthalpy Sample Problem E Calculate the change in enthalpy for this reaction. 2H 2 (g) + 2CO 2 (g)  2H 2 O(g) + 2CO(g) State whether the reaction is exothermic or endothermic. Section 3 Changes In Enthalpy During Chemical Reactions Chapter 10

65. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu CO 2 (g),  = -393.5 kJ/mol. H 2 (g),  = 0 kJ/mol CO(g),  = -110.5 kJ/mol Calculating a Reaction’s Change in Enthalpy, continued Sample Problem E Solution Standard enthalpies of formation as follows: H 2 O(g),  = -241.8 kJ/mol Section 3 Changes In Enthalpy During Chemical Reactions Chapter 10  H = (2 mol)(-241.8 kJ/mol) + (2 mol)(-110.5 kJ/mol) - (2 mol)(0 kJ/mol) - (2 mol)(-393.5 kJ/mol) = 82.4 kJ

66. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Section 4 Order and Spontaneity Bellringer Make a list of examples that show both increasing and decreasing order. For example, making your bed or baking a cake are examples of increasing order. Breaking a dish or shredding paper are examples of decreasing order. Chapter 10

67. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Objectives Define entropy, and discuss the factors that influence the sign and magnitude of  S for a chemical reaction. Describe Gibbs energy, and discuss the factors that influence the sign and magnitude of  G. Indicate whether  G values describe spontaneous or nonspontaneous reactions. Section 4 Order and Spontaneity Chapter 10

68. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Entropy Some reactions happen easily, but others do not. Sodium and chlorine react when they are brought together. Nitrogen and oxygen coexist in the air you breathe without forming poisonous nitrogen monoxide, NO. One factor you can use to predict whether reactions will occur is enthalpy. A reaction is more likely to occur if it is accompanied by a decrease in enthalpy or if  H is negative. Section 4 Order and Spontaneity Chapter 10

69. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Entropy, continued Another factor known as entropy can determine if a process will occur. Entropy,  S, is a measure of the disorder in a system and is a thermodynamic property. Entropy is not a form of energy and has the units joules per kelvin, J/K. A process is more likely to occur if it is accompanied by an increase in entropy.  S is positive. Section 4 Order and Spontaneity Chapter 10

70. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Entropy, continued Factors That Affect Entropy Ions in a solution disperse throughout the solution. This process of dispersion is called diffusion and causes the increase in entropy. Entropy also increases as solutions become more dilute or when the pressure of a gas is reduced. In both cases, the molecules fill larger spaces and so become more disordered. Entropies also increase with temperature, but this effect is not great unless a phase change occurs. Section 4 Order and Spontaneity Chapter 10

71. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Entropy, continued Factors That Affect Entropy, continued The entropy can change during a reaction. The entropy of a system can increase when the total number of moles of product is greater than the total number of moles of reactant. Entropy can increase in a system when the total number of particles in the system increases. Entropy also increases when a reaction produces more gas particles, because gases are more disordered than liquids or solids. Section 4 Order and Spontaneity Chapter 10

72. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Entropy, continued Factors That Affect Entropy, continued Entropy decreases as sodium chloride forms: 2 mol of sodium combine with 1 mol of chlorine to form 2 mol of sodium chloride. 2Na(s) + Cl 2 (g)  2NaCl(s)  S = -181 J/K This decrease in entropy is because of the order present in crystalline sodium chloride. Section 4 Order and Spontaneity Chapter 10

73. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Entropy, continued Factors That Affect Entropy, continued Entropy increases when 1 mol of sodium chloride dissolves in water to form 1 mol of aqueous sodium ions and 1 mol of aqueous chlorine ions. NaCl(s)  Na + (aq) + Cl – (aq)  S = 43 J/K This increase in entropy is because of the order lost when a crystalline solid dissociates to form ions. Section 4 Order and Spontaneity Chapter 10

74. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Entropy, continued Factors That Affect Entropy, continued Standard Entropy Changes for Some Reactions Section 4 Order and Spontaneity Chapter 10

75. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Visual Concepts Relating Enthalpy, Entropy, and Free Energy Changes Chapter 10

76. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Entropy, continued Hess’s Law Also Applies to Entropy The decomposition of nitrogen triiodide to form nitrogen and iodine creates 4 mol of gas from 2 mol of a solid. 2NI 3 (s)  N 2 (g) + 3I 2 (g) This reaction has such a large entropy increase that the reaction proceeds once the reaction is initiated by a mechanical shock. Section 4 Order and Spontaneity Chapter 10

77. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Entropy, continued Hess’s Law Also Applies to Entropy, continued Molar entropy has the same unit, J/Kmol, as molar heat capacity. Molar entropies can be calculated from molar heat capacity data. Entropies can also be calculated by using Hess’s law and entropy data for other reactions. You can manipulate chemical equations using rules of algebra to get a desired equation. Section 4 Order and Spontaneity Chapter 10

78. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Entropy, continued Hess’s Law Also Applies to Entropy, continued When equations are added or subtracted, entropy changes must be added or subtracted. When equations are multiplied by a constant, the entropy changes must also be multiplied by that constant. Atoms and molecules that appear on both sides of the equation can be canceled. Section 4 Order and Spontaneity Chapter 10

79. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Entropy, continued Hess’s Law Also Applies to Entropy, continued The standard entropy is represented by the symbol S 0. The standard entropy of the substance is the entropy of 1 mol of a substance at a standard temperature, 298.15 K. Elements can have standard entropies that have values other than zero. Most standard entropies are positive; this is not true of standard enthalpies of formation. The entropy change of a reaction can be calculated by using the following equation.  S reaction = S products - S reactants Section 4 Order and Spontaneity Chapter 10

80. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Entropy, continued Hess’s Law Also Applies to Entropy, continued Standard Entropies of Some Substances Section 4 Order and Spontaneity Chapter 10

81. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Hess’s Law and Entropy Sample Problem F Calculate the entropy change that accompanies the following reaction. Section 4 Order and Spontaneity Chapter 10

82. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Hess’s Law and Entropy, continued Sample Problem F Solution Section 4 Order and Spontaneity Chapter 10 The standard entropies are as follows. For H 2 O, S 0 = 188.7 J/Kmol. For CO, S 0 = 197.6 J/Kmol. For H 2, S 0 = 130.7 J/Kmol. For CO 2, S 0 = 213.8 J/Kmol.

83. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Hess’s Law and Entropy, continued Sample Problem F Solution, continued Section 4 Order and Spontaneity Chapter 10

84. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Gibbs Energy The tendency for a reaction to occur depends on both  H and  S. If  H is negative and  S is positive for a reaction, the reaction will likely occur. If  H is positive and  S is negative for a reaction, the reaction will not occur. How can you predict what will happen if  H and  S are both positive or both negative? Section 4 Order and Spontaneity Chapter 10

85. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Gibbs Energy, continued Gibbs energy is the energy in a system that is available for work represented by the symbol G. G = H – TS Another name for Gibbs energy is free energy. Section 4 Order and Spontaneity Chapter 10

86. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Gibbs Energy, continued Gibbs Energy Determines Spontaneity A spontaneous reaction is one that does occur or is likely to occur without continuous outside assistance, such as input of energy. A reaction is spontaneous if the Gibbs energy change is negative. A nonspontaneous reaction will never occur without assistance. If a reaction has a  G greater than 0, the reaction is nonspontaneous. If a reaction has a  G of exactly zero, the reaction is at equilibrium. Section 4 Order and Spontaneity Chapter 10

87. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Gibbs Energy, continued Entropy and Enthalpy Determine Gibbs Energy Reactions that have large negative  G values often release energy and increase disorder. The vigorous reaction of potassium metal and water is an example of this type of reaction. 2K(s) + 2H 2 O(l)  2K + (aq) + 2OH – (aq) + H 2 (g)  H = -392 kJ  S = 0.047 kJ/K The change in Gibbs energy for the reaction is  G =  H – T  S = -392 kJ - (298.15 K)(0.047 kJ/K) = -406 kJ Section 4 Order and Spontaneity Chapter 10

88. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Gibbs Energy, continued Entropy and Enthalpy Determine Gibbs Energy, continued You can calculate  G in another way because lists of standard Gibbs energies of formation exist. The standard Gibbs energy of formation,  G 0 f, of a substance is the change in energy that accompanies the formation of 1 mol of the substance from its elements at 298.15 K. Section 4 Order and Spontaneity Chapter 10

89. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Gibbs Energy, continued Entropy and Enthalpy Determine Gibbs Energy, continued The standard Gibbs energies of formation can be used to find the  G for any reaction. Hess’s law also applies when calculating  G.  G reaction =  G products –  G reactants Section 4 Order and Spontaneity Chapter 10

90. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Gibbs Energy, continued Standard Gibbs Energies of Formation Section 4 Order and Spontaneity Chapter 10

91. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Visual Concepts Equation for Free Energy Change Chapter 10

92. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Calculating a Change in Gibbs Energy from  H and  S Sample Problem G Given that the changes in enthalpy and entropy are – 139 kJ and 277 J/K respectively for the reaction given below, calculate the change in Gibbs energy. Then, state whether the reaction is spontaneous at 25  C. C 6 H 12 O 6 (aq)  2C 2 H 5 OH(aq) + 2CO 2 (g) Section 4 Order and Spontaneity Chapter 10

93. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Calculating a Change in Gibbs Energy from  H and  S, continued Sample Problem G Solution Section 4 Order and Spontaneity Chapter 10  H = –139 kJ  S = 277 J/K T = 25  C = (25 + 273.15) K = 298 K  G =  H - T  S  G = (-139 kJ) - (298 K)(277 J/K)  G = (-139 kJ) - (83 kJ)  G = -222 kJ

94. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Calculating a Gibbs Energy Change using  Values Sample Problem H Calculate  G for the following water-gas reaction. C(s) + H 2 O(g)  CO(g) + H 2 (g) Is this reaction spontaneous? Section 4 Order and Spontaneity Chapter 10

95. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu For H 2 O(g),  = -228.6 kJ/mol. For C(s) (graphite),  = 0 kJ/mol. For H 2 (g),  = 0 kJ/mol. For CO(g),  = -137.2 kJ/mol. Gibbs Energy, continued Sample Problem H Solution Section 4 Order and Spontaneity Chapter 10  G =  G(products) -  G(reactants)

96. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Sample Problem H Solution, continued  G =  G(products) -  G(reactants) = Calculating a Gibbs Energy Change using  Values, continued Section 4 Order and Spontaneity Chapter 10 The reaction is nonspontaneous under standard conditions. [(mol CO(g))(  for CO(g)) + (mol H 2 (g))(  for H 2 (g))] – [(mol C(s))(  for C(s)) + (mol H 2 O(g))(  for H 2 O(g))] = [(1 mol)(-137.2 kJ/mol) + (1 mol)(0 kJ/mol)] – [(1 mol)(0 kJ/mol) – (1 mol)(-228.6 kJ/mol)] = (-137.2 + 228.6) kJ = 91.4 kJ

97. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Gibbs Energy, continued Predicting Spontaneity Does temperature affect spontaneity? Consider the equation for  G.  G =  H - T  S The terms  H and  S change very little as temperature changes. The presence of T in the equation for  G indicates that temperature may greatly affect  G. Section 4 Order and Spontaneity Chapter 10

98. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Gibbs Energy, continued Predicting Spontaneity, continued Suppose a reaction has both a positive  H value and a positive  S value. If the reaction occurs at a low temperature, the value for T  S will be small and will have little impact on the value of  G. The value of  G will be similar to the value of  H and will have a positive value. Section 4 Order and Spontaneity Chapter 10

99. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Gibbs Energy, continued Predicting Spontaneity When the same reaction proceeds at a high enough temperature, T  S will be larger than  H and  G will be negative. Increasing the temperature of a reaction can make a nonspontaneous reaction spontaneous. A nonspontaneous reaction cannot occur unless some form of energy is added to the system. Section 4 Order and Spontaneity Chapter 10

100. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu 1. Which of these thermodynamic values can be determined using an adiabatic calorimeter? A.∆G B.∆H C.∆S D.∆T Standardized Test Preparation Understanding Concepts Chapter 10

101. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu 1. Which of these thermodynamic values can be determined using an adiabatic calorimeter? A.∆G B.∆H C.∆S D.∆T Standardized Test Preparation Understanding Concepts Chapter 10

102. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu 2. Which of these statements about the temperature of a substance is true? F.Temperature is a measure of the entropy of the substance. G.Temperature is a measure of the total kinetic energy of its atoms. H.Temperature is a measure of the average kinetic energy of its atoms. I.Temperature is a measure of the molar heat capacity of the substance. Standardized Test Preparation Understanding Concepts Chapter 10

103. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu 2. Which of these statements about the temperature of a substance is true? F.Temperature is a measure of the entropy of the substance. G.Temperature is a measure of the total kinetic energy of its atoms. H.Temperature is a measure of the average kinetic energy of its atoms. I.Temperature is a measure of the molar heat capacity of the substance. Standardized Test Preparation Understanding Concepts Chapter 10

104. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu 3. Which of the following pairs of conditions will favor a spontaneous reaction? A.a decrease in entropy and a decrease in enthalpy B.a decrease in entropy and an increase in enthalpy C.an increase in entropy and a decrease in enthalpy D.an increase in entropy and an increase in enthalpy Standardized Test Preparation Understanding Concepts Chapter 10

105. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu 3. Which of the following pairs of conditions will favor a spontaneous reaction? A.a decrease in entropy and a decrease in enthalpy B.a decrease in entropy and an increase in enthalpy C.an increase in entropy and a decrease in enthalpy D.an increase in entropy and an increase in enthalpy Standardized Test Preparation Understanding Concepts Chapter 10

106. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu 4. What is the standard enthalpy of formation of N 2 ? Standardized Test Preparation Understanding Concepts Chapter 10

107. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu 4. What is the standard enthalpy of formation of N 2 ? Answer: Because the nitrogen molecule is the normal form of the element nitrogen, its standard enthalpy of formation is defined as zero. Standardized Test Preparation Understanding Concepts Chapter 10

108. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu 5. What are the circumstances that cause a nonspontaneous reaction to occur? Standardized Test Preparation Understanding Concepts Chapter 10

109. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu 5.What are the circumstances that cause a nonspontaneous reaction to occur? Answer: Nonspontaneous reactions require an input of energy, such as heat or light. Standardized Test Preparation Understanding Concepts Chapter 10

110. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Standardized Test Preparation Understanding Concepts Chapter 10 6.Based on changes in entropy and enthalpy, predict whether this reaction is spontaneous or nonspontaneous: 2AB(s)  A 2 (g) + B 2 (g) + 799 kJ

111. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Standardized Test Preparation Understanding Concepts Chapter 10 6.Based on changes in entropy and enthalpy, predict whether this reaction is spontaneous or nonspontaneous: 2AB(s)  A 2 (g) + B 2 (g) + 799 kJ Answer: ∆H is negative and ∆S is positive. Therefore ∆G is negative and the reaction is spontaneous.

112. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Read the passage below. Then answer the questions. Almost all of the organisms on Earth rely on the process of photosynthesis to provide the energy needed for the functions of living. During photosynthesis carbon dioxide and water combine to form glucose and oxygen. The photosynthesis reaction is represented by the chemical equation: 6CO 2 (g) + 6H 2 O(l) + energy  C 6 H 12 O 6 (s) + 6O 2 (g). The source of energy for the reaction is light from the sun. Standardized Test Preparation Reading Skills Chapter 10

113. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Standardized Test Preparation Reading Skills Chapter 10 7. If the absolute value of T∆S is smaller than the absolute value of ∆H, is photosynthesis a spontaneous reaction? Explain your answer.

114. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu 7. If the absolute value of T∆S is smaller than the absolute value of ∆H, is photosynthesis a spontaneous reaction? Explain your answer. Answer: The reaction is endothermic, requiring an input of energy. If the absolute value of T∆S is smaller than that of ∆H, then the Gibbs free energy is positive, whether ∆S is positive or negative, so the reaction is nonspontaneous. Standardized Test Preparation Reading Skills Chapter 10

115. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Standardized Test Preparation Understanding Concepts Chapter 10 8. Based on the nature of the reactants and products, what can be deduced about the entropy change during photosynthesis? F.Entropy increases because one of the products is an element. G.Entropy decreases because there are substantially fewer molecules of product than of reactants. H.Entropy increases because one of the products is a solid while one of the reactants is a liquid. I.Entropy does not change during the reaction because both sides of the reaction include the same number of atoms.

116. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu 8. Based on the nature of the reactants and products, what can be deduced about the entropy change during photosynthesis? F.Entropy increases because one of the products is an element. G.Entropy decreases because there are substantially fewer molecules of product than of reactants. H.Entropy increases because one of the products is a solid while one of the reactants is a liquid. I.Entropy does not change during the reaction because both sides of the reaction include the same number of atoms. Standardized Test Preparation Understanding Concepts Chapter 10

117. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Standardized Test Preparation Reading Skills Chapter 10 9. If a manufacturing process was developed to make glucose from carbon dioxide and water, using heat as the energy source, how would the amount of energy required compare to that of the process that plants use?

118. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu 9. If a manufacturing process was developed to make glucose from carbon dioxide and water, using heat as the energy source, how would the amount of energy required compare to that of the process that plants use? Answer: The same amount of energy would be needed. Standardized Test Preparation Reading Skills Chapter 10

119. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu The table below shows molar heat capacities (joules per kelvins  mole) of elements and compounds. Use it to answer questions 10 through 13. Standardized Test Preparation Interpreting Graphics Chapter 10

120. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu 10.What is the specific heat of aluminum chloride, which has a molar mass of 133.4? A.-1.45 J/Kg B.-0.69 J/Kg C.0.69 J/Kg D.1.45 J/Kg Standardized Test Preparation Interpreting Graphics Chapter 10

121. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu 10.What is the specific heat of aluminum chloride, which has a molar mass of 133.4? A.-1.45 J/Kg B.-0.69 J/Kg C.0.69 J/Kg D.1.45 J/Kg Standardized Test Preparation Interpreting Graphics Chapter 10

122. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu 11.What is the relationship between the number of atoms per unit of an ionic compound and its molar heat capacity? Standardized Test Preparation Interpreting Graphics Chapter 10

123. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu 11.What is the relationship between the number of atoms per unit of an ionic compound and its molar heat capacity? Answer: The molar heat capacity is about 25 J/Kmol for each atom in a mole of the ionic compound. Standardized Test Preparation Interpreting Graphics Chapter 10

124. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Standardized Test Preparation Interpreting Graphics Chapter 10 12. Which of these statements best describes the general relationship of molar heat capacity of two different metals? F.The molar heat capacity of the two metals is about the same. G.The difference in molar heat capacity of two metals depends on the temperature. H.The molar heat capacity of the metal with the lower atomic mass is generally smaller. I.The molar heat capacity of the metal with the higher atomic mass is generally smaller.

125. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu 12. Which of these statements best describes the general relationship of molar heat capacity of two different metals? F.The molar heat capacity of the two metals is about the same. G.The difference in molar heat capacity of two metals depends on the temperature. H.The molar heat capacity of the metal with the lower atomic mass is generally smaller. I.The molar heat capacity of the metal with the higher atomic mass is generally smaller. Standardized Test Preparation Interpreting Graphics Chapter 10

126. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu 13.How many joules of heat are required to raise the temperature of one mole of liquid water by 2.00°C? Standardized Test Preparation Interpreting Graphics Chapter 10

127. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu 13.How many joules of heat are required to raise the temperature of one mole of liquid water by 2.00°C? Answer: 150.6 J Standardized Test Preparation Interpreting Graphics Chapter 10