Thermochemistry – Heat and Chemical Change

1. Thermochemistry – Heat and Chemical Change • Thermochemistryis the study of heat transfer in chemical and physical processes.

2. Objectives • Explain the relationship between energy and heat • Explain the difference between heat and temperature • Construct equations that show the heat changes for chemical and physical processes • Calculate heat changes in chemical and physical processes

3. Energy Transformations • Energyis the capacity to do work or transfer heat.

4. Energy • Potential energy(PE) is possessed by an object by virtue of its position.(Chemical potential energy is the energy stored in chemical bonds.) • Kinetic energy(KE) is the energy of motion. This form of energy depends on the mass (m) and velocity (v) of the moving object. KE = ½ mv2

5. A. The Nature of Energy Potential energy (PE): Kinetic energy (KE): the energy of position the energy of motion

6. Law of conservation of energy • Energy is neither created nor destroyed in a chemical or physical process. (Energy can be converted from one form to another.)

7. B. Temperature and Heat • Temperature is a measure of the average kinetic energy in matter. Hot water (90oC) Cold water (10oC)

8. B. Heat (q) is…. • the energy that transfers between two objects due to a temperature difference between them. • Heat is transferred from a hot object to a colder object (never in the other direction). Hot water (90oC) Cold water (10oC) Water (50oC) Water (50oC)

9. Heat vs. Temperature -Question: • Which represents more heat? • A bathtub full of 100o C water or a teacup full of 100o C water? • Answer: A bathtub full! The amount of heat depends partly on the mass of the substance.

10. In studying heat changes…. System– part of the universe on which we focus attention Surroundings– everything else in the universe • Example: Burning a match

11. Exothermic and Endothermic Processes Exothermic – energy flows out of the system Endothermic – energy flows into the system

12. Units of energy: • The joule (J) is the SI unit of energy. • Another common unit of energy is the calorie (cal). 1 calorie = 4.18 joules

13. Question… • How many joules are in 375 calories? • Answer: • 375 cal x 4.18 J/cal = 1567.5 joules

14. In food, aCalorieis the same as a kilocalorie. If an apple provides 120 Calories, that is the same as… • 120 000 calories, or 501 600 joules of energy. (120 Cal)(1000 cal/1 Cal)(4.18 J/cal) = 501 600 J

15. Specific heat capacity (or specific heat) is… • The amount of energy needed to change the temperature of 1.0 gram of a substance by 1.0 degree Celsius (or 1.0 kelvin).

16. The unit of specific heat is… • Expressed in either joules or calories. • The units are either J/(g .oC) • or cal/(g . oC).

17. Table of Specific Heats:

18. The specific heat of water • The specific heat of water is 4.18 J/g . K • The specific heat of aluminum is 0.902 J/g . K. • It takes over 4 times as much heat to raise the temperature of a gram of water by a certain amount than a gram of aluminum.

19. B. Calculating Heat Transfer • To calculate the energy required for a reaction: Heat transferred = (specific heat) (mass) (change in temp) q = C m T

20. In solving heat problems: • q = C m DT where the quantity of heat =q , and the change in temp (DT ) = (final temp – initial temp) or DT = (Tfinal - Tinitial).

21. Problem: • How many joules must be transferred from a cup of coffee to your body if the temperature of the coffee drops from 60.0oC to 37.0oC (normal body temp)? Assume the cup holds 250. mL (with a density of 1.0 g/mL, the coffee has a mass of 250. g), and that the specific heat of coffee is the same as that of water.

22. Answer: • q = C m DT • q = (4.18 J/g . K) (250. g) (37.0oC – 60.0oC) • = -24100 J = -24.1 kJ • Note: the negative value denotes the direction of heat transfer; it shows that energy is transferred from the coffee as the temperature declines.

23. Problem: • In the previous problem, the cup of coffee lost 24.1 kJ when cooled from 60.0oC to 37.0oC. If this same amount of heat is used to warm a piece of aluminum weighing 250. g. what would be the final temperature of the aluminum if its initial temperature is 37.0oC? (The specific heat of aluminum is 0.902 J/g . K.)

24. Answer: • q = C m DT 24100 J = • (0.902 J/g . K) (250. g) (Tfinal - 37.0oC) • Tfinal = 144oC

25. A. Thermochemistry • Calorimetry • Heat transfer is measured using a calorimeter.

26. Problem: • Suppose you heat a 55.0 g piece of iron in the flame of a Bunsen burner to 425oC and then you plunge it into a beaker of water. The beaker holds 600. mL water (density = 1.00 g/mL), and its temperature before you drop in the hot iron is 25.0oC. What is the final temperature of the water and the piece of iron?

27. Phases of Matter and Intermolecular Forces Adapted from: Wilbraham, Anthony. Chemistry, Addison-Wesley. Upper Saddle River, NJ: Prentice Hall, Inc.,2002.

28. Objectives • Review the organization of particles in the three phases of matter: gases, liquids and solids. • Name and describe the weak attractive forces that hold groups of molecules together. • Determine which type of intermolecular force is important to overcome in converting a substance from a liquid to a gas.

29. Intermolecular Forces • Intermolecular forces – occur between molecules • Intramolecular forces – occur inside the molecules

30. Three types of Intermolecular Forces (IMF) in order of increasing strength: • 1. Dispersion forces • 2. Dipole Interactions • 3. Hydrogen bonding • Dispersion forces and dipole interactions are the weakest and are sometimes referred to a van der Waals forces.

31. Dispersion Forces • Dispersion forces are the weakest of all molecular interactions. • Important innonpolar substances, such as F2, N2, and CO2 • They are also called London forces and induced dipole forces.

32. About dispersion forces… • Dispersion forces are caused by the motion of electrons. When two nonpolar molecules encounter one another, attractions and repulsions lead to distortions in their electron clouds, inducing “momentary dipoles”. • The strength of dispersion forcesincreases as the number of electrons in a molecule increases.

33. Question…… • Why are F2 and Cl2 gases at room temperature , while Br2 is a liquid and I2 is a solid? • Answer: F2 has 18 electrons and Cl2 has 34 electrons, while Br2 has 70 electrons and I2 has 106 electrons. More electrons causes a greater force of attraction.

34. Dipole Interactions • Dipole interactions occur when polar molecules are attracted to one another. They are stronger than dispersion forces. • The oppositely charged ends of polar molecules create electrostatic attractions, and these hold polar molecules in the liquid or solid state. They are much weaker than ionic bonds.

35. Hydrogen Bonding • Hydrogen bonds are attractive forces in which a hydrogen atom that is covalently bonded to a very electronegative atom is attracted to an unshared pair of another very electronegative atom. • Hydrogen bonding is an extreme case of dipole interaction. • This occurs only when H is directly bonded to either N, O, or F.

36. Question: • Which has the higher boiling point? (…consider which has the stronger IMF) • 1. O2 or N2? • 2. SO2 or CO2? • 3. HF or HI? • 4. SiH4 or GeH4?

37. Answers: • 1. O2 – both N2 and O2 are nonpolar, both have dispersion forces, but O2 has more electrons • 2. SO2– CO2 is nonpolar (has dispersion forces), while SO2 is polar (has stronger dipole interactions) • 3. HF – HI has dipole interactions, HF has stronger hydrogen bonding • 4. GeH4 – both SiH4 and GeH4 are nonpolar, but GeH4 has more electrons

38. Challenge: • Place these in order of increasing strength of intermolecular forces: • SCl2, NH3, CH4, Cl2

39. Answer: • CH4 (nonpolar, dispersion, few electrons), • Cl2 (nonpolar, dispersion, more electrons), • SCl2 (polar, dipole interactions), • NH3 (H bonded to N, O, or F, hydrogen bonding)

40. Question: • Which has the higher boiling point? • dimethyl ether (H3C-O-CH3) or • ethyl alcohol (H3C-CH2-OH)? • (both have the general formula C2H6O.)

41. Answer: • Ethyl alcohol. • (Dimethyl ether is slightly polar and has dipole interactions, while ethyl alcohol has hydrogen bonding.)

42. The three states of matter • Solids • Liquids • Gases

43. States of Matter - Gases • Gases can be compressed because the particles are spaced far apart. • Gases fill their container uniformly and completely, and they flow because there are only weak forces attracting molecules to one another. • The particles move randomly in a gas.

44. States of Matter - Liquids • Liquids are condensed states, their particles are in close contact with each other. • Liquids take the shape of the container and they can flow because the particles are not rigidly packed and the attraction between particles is weak. • The particles of a liquid move randomly.

45. States of Matter – Solids • Solids are condensed because the particles are packed tightly together. • Solids have a definite shape and volume.

46. A Model for Liquids • Liquid molecules are free to slide past each other, so liquids, like gases, can flow. • Liquids take the shape of their container. • Liquids are condensed, not compressible.

47. Vaporization • Vaporization is the conversion of a liquid to a gas or vapor. • If the conversion occurs at the surface of a liquid that is not boiling, the process is called evaporation. • Evaporation is a cooling process. The particle with the highest kinetic energy tend to escape first, leaving cooler particles behind.

48. Vapor Pressure • Vapor pressure that results from the vapor particles colliding with the walls of the container. • It is a force due to a gas above the surface of a liquid. • (Vapor is the gaseous state of a substance that is a liquid or solid at room temperature).

49. Phase changes with liquids • Evaporation… Liquid  Vapor (gas) • Condensation… Vapor (gas) Liquid

50. Equilibrium vapor pressure… • When the rate of vaporization is equal to the rate of condensation, no net change in vapor pressure occurs. • This is a dynamic equilibrium.