Chapt04 Lecture

27
1 Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Chapter 4 Lecture Outline Prepared by Jennifer N. Robertson- Honecker West Virginia University

Transcript of Chapt04 Lecture

Page 1: Chapt04 Lecture

1Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

Chapter 4LectureOutline

Prepared byJennifer N. Robertson-Honecker

West Virginia University

Page 2: Chapt04 Lecture

2

Energy

• Energy is the capacity to do work.

• Potential energy is stored energy.• Kinetic energy is the energy of motion.

• The law of conservation of energy states that the total energy in a system does not change. Energy cannot be created or destroyed.

Page 3: Chapt04 Lecture

3

EnergyThe Units of Energy

• A calorie (cal) is the amount of energy needed to raise the temperature of 1 g of water by 1 oC.

• A joule (J) is another unit of energy.

1 cal = 4.184 J

• Both joules and calories can be reported in the larger units kilojoules (kJ) and kilocalories (kcal).

1,000 J = 1 kJ 1,000 cal = 1 kcal

1 kcal = 4.184 kJ

Page 4: Chapt04 Lecture

4

Cal/g cal/g

Protein 4 4,000

Carbohydrate 4 4,000

Fat 9 9,000

Focus on The Human BodyEnergy and Nutrition

• The amount of stored energy in food is measured using nutritional Calories (upper case C), where 1 Cal = 1,000 cal.

• Upon metabolism, proteins, carbohydrates, and fat each release a predictable amount of energy, the caloric value of the substance.

Page 5: Chapt04 Lecture

5

Identify the original quantity and the desired quantity.3 g protein23 g carbohydrates ? Caloriginal quantities desired quantity

Step [1]

Step [2] Write out conversion factors.4 Cal 4 Cal .1 g protein 1 g carbohydrate

Focus on The Human BodyEnergy and Nutrition

Sample Problem 4.2

If a baked potato contains 3 g of protein, a trace of fat, and 23 g of carbohydrates, estimate its number of Calories.

Page 6: Chapt04 Lecture

6

Multiply the original quantity by the conversion factor for both protein and carbohydrates and add up the results.

Total Cal = Cal due to protein + Cal due to carbohydrate

= 3 g × 4 Cal + 23 g × 4 Cal . 1 g protein 1 g carbohydrate

Step [3]

grams cancel

Total Cal = 12 Cal + 92 Cal= 104 Cal, rounded to 100 Cal

Focus on The Human BodyEnergy and Nutrition

Page 7: Chapt04 Lecture

7

The Three States of Matter

Whether a substance exists as a gas, liquid, or solid depends on the balance between the kinetic energy of its particles and the strength of the interactions between the particles.

Gas: kinetic energy is high and particles are far apart. The attractive forces between molecules are negligible allowing them to move freely.

Liquid: attractive forces hold the molecules much more closely together. The distance between molecules and the kinetic energy is much less.

Solid: attractive forces between molecules are even stronger. The distance between particles is small and there is little freedom of motion.

Page 8: Chapt04 Lecture

The Three States of Matter

Page 9: Chapt04 Lecture

9

Intermolecular Forces, Boiling Point,and Melting Point

Intermolecular forces are the attractive forces thatexist between molecules.

In order of increasing strength, these are:

• London dispersion forces

• Dipole–dipole interactions

• Hydrogen bonding

The strength of the intermolecular forces determineswhether a compound has a high or low melting pointand boiling point, and thus whether it is a solid,liquid, or gas at a given temperature.

Page 10: Chapt04 Lecture

10

Intermolecular ForcesLondon Dispersion Forces

London dispersion forces are very weak interactionsdue to the momentary changes in electron densityin a molecule.

• The change in electron density creates a temporary dipole.

• All covalent compounds exhibit London dispersion forces.

• The weak interaction between these temporary dipoles constitutes London dispersion forces.

• The larger the molecule, the larger the attractive force, and the stronger the intermolecular forces.

Page 11: Chapt04 Lecture

11

Intermolecular ForcesLondon Dispersion Forces

More e− densityin one region creates a partialnegative charge (δ−).

Less e− densityin one regioncreates a partialpositive charge (δ+).

Page 12: Chapt04 Lecture

12

Intermolecular ForcesDipole–Dipole Interactions

Dipole–dipole interactions are the attractive forcesbetween the permanent dipoles of two polar molecules.

Page 13: Chapt04 Lecture

13

Intermolecular ForcesHydrogen Bonding

Hydrogen bonding occurs when a hydrogen atombonded to O, N, or F is electrostatically attracted to an O, N, or F atom in another molecule.

Hydrogen bonds are the strongest of the three typesof intermolecular forces.

Page 14: Chapt04 Lecture

14

Intermolecular ForcesHydrogen Bonding in DNA

Page 15: Chapt04 Lecture

15

Intermolecular Forces

Page 16: Chapt04 Lecture

16

Intermolecular ForcesBoiling Point and Melting Point

• The boiling point is the temperature at which a liquid is converted to the gas phase.

• The melting point is the temperature at which a solid is converted to the liquid phase.

• The stronger the intermolecular forces, the higher the boiling point and melting point.

Page 17: Chapt04 Lecture

17

Intermolecular ForcesBoiling Point and Melting Point

Page 18: Chapt04 Lecture

18

Intermolecular ForcesBoiling Point and Melting Point

• Both propane and butane have London dispersion forces and nonpolar bonds.

• In this case, the larger molecule will have stronger attractive forces.

Page 19: Chapt04 Lecture

19

Energy and Phase Changes

• When energy is absorbed, a process is said to be endothermic.• When energy is released, a process is said to be exothermic.• In a phase change, the physical state of a substance is altered without changing its composition.

Page 20: Chapt04 Lecture

20

Converting a Solid to a Liquid

• Converting a solid to a liquid is called melting.

• Melting is endothermic—it absorbs heat from the surroundings.

• Freezing converts a liquid to a solid.

• Freezing is exothermic—it gives off heat to the surroundings.

Page 21: Chapt04 Lecture

21

Energy and Phase ChangesConverting a Solid to a Liquid

solid water liquid water

The amount of energy needed to melt 1 gram of a substance is called its heat of fusion.

Page 22: Chapt04 Lecture

22

Converting a Liquid to a Gas

• Vaporization is the conversion of liquids into the gas phase.

• Vaporization is endothermic—it absorbs heat from the surroundings.

• Condensation is the conversion of gases into the liquid phase.

• Condensation is exothermic—it gives off heat to the surroundings.

Page 23: Chapt04 Lecture

23

Energy and Phase ChangesConverting a Liquid to a Gas

liquid water gaseous water

The amount of energy needed to vaporize 1 gram of a substance is called its heat of vaporization.

Page 24: Chapt04 Lecture

24

Converting a Solid to a Gas

• Sublimation is the conversion of solids directly into the gas phase.

• Sublimation is endothermic—it absorbs heat from the surroundings.

• Deposition is the conversion of gases into the solid phase.

• Deposition is exothermic—it gives off heat to the surroundings.

Page 25: Chapt04 Lecture

25

Energy and Phase ChangesConverting a Solid to a Gas

solid CO2gaseous CO2

Page 26: Chapt04 Lecture

26

Heating and Cooling Curves

A heating curve shows how the temperature of a substance (plotted on the vertical axis) changes as heat is added.

The plateau B C occurs at the melting point, while the plateau D E occurs at the boiling point.

A

B

C

D

E

melting

boiling

solid

liquid

gas

Page 27: Chapt04 Lecture

condensation

27

Heating and Cooling Curves

A cooling curve illustrates how the temperature of a substance (plotted on the vertical axis)changes as heat is removed.

The plateau W X occurs at the boiling point, while the plateau Y Z occurs at the freezing point.

V

XW

freezingsolid

liquid

gas

Y

Z