unit 10 thermodynamics – chapter 17 organic chemistry – chapters 22 & 24

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UNIT 10 Thermodynamics – chapter 17 Organic Chemistry – chapters 22 & 24

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Page 1: UNIT 10 Thermodynamics – chapter 17 Organic Chemistry – chapters 22 & 24

UNIT 10Thermodynamics –

chapter 17

Organic Chemistry –chapters 22 & 24

Page 2: UNIT 10 Thermodynamics – chapter 17 Organic Chemistry – chapters 22 & 24

Chapter 17 -Thermochemistry – Heat and Chemical Change

Page 3: UNIT 10 Thermodynamics – chapter 17 Organic Chemistry – chapters 22 & 24

Observing Heat Flow

1. Hook index fingers through each end of rubber band. Without stretching, place it against your upper lip or forehead. Note the temperature of the rubber band.

2. Move rubber band away from your skin, quickly stretch and hold it, and then place it back against your skin. Note any temp. change.

3. Fully stretch the band, and then allow it to return to its original shape. Place it against your skin and note any temp. changes.

Page 4: UNIT 10 Thermodynamics – chapter 17 Organic Chemistry – chapters 22 & 24

Think About It

1. Did the rubber band feel cooler or warmer after it was stretched in Step 2?

2. Did the rubber band feel cooler or warmer after it returned to its original shape in Step 3?

3. Think about temperature changes you observed and form some initial answers to the following questions. What is heat? In what direction does heat flow?

Page 5: UNIT 10 Thermodynamics – chapter 17 Organic Chemistry – chapters 22 & 24

THERMOCHEMISTRY

study of heat changes that occur during chemical reactions and changes of state

Page 6: UNIT 10 Thermodynamics – chapter 17 Organic Chemistry – chapters 22 & 24

Some important concepts…

• Energy = capacity to do work.

• Temperature = measure of Kinetic Energy

• Chemical potential energy =Energy stored in chemical bonds of a

substance.

Page 7: UNIT 10 Thermodynamics – chapter 17 Organic Chemistry – chapters 22 & 24

HEAT

• Energy that transfers from one object to another because of a temperature difference.

• Always flows from warmer object to a cooler object.

• if the objects remain in contact, they will become equal in temperature

• represented by “q”

Page 8: UNIT 10 Thermodynamics – chapter 17 Organic Chemistry – chapters 22 & 24

EXO vs. ENDO

Exothermic =

• the system loses heat as the surroundings heat up (heat is given off)

• “q” is negative (-)

Endothermic =

• the system gains heat as the surroundings cool down (heat is absorbed)

• “q” is positive (+)

Page 9: UNIT 10 Thermodynamics – chapter 17 Organic Chemistry – chapters 22 & 24

How to measure Heat Flow

• Heat flow is measured in two common units (the calorie and the joule).

• calorie (cal) = the amount of heat needed to raise the temperature of 1 g of pure water 1oC.

• Important Relationships:

• 1 food Calorie = 1000 calories

• 4.184 J = 1 cal

Page 10: UNIT 10 Thermodynamics – chapter 17 Organic Chemistry – chapters 22 & 24

The accurate and precise measurement of heat change for chemical and physical processes.

The insulated device used to measure the absorption or release of heat in chemical or physical processes is called a calorimeter.

Page 11: UNIT 10 Thermodynamics – chapter 17 Organic Chemistry – chapters 22 & 24

• What happens to the HEAT in lava once it flows out of a volcano?

• Heat is released• Why does lava cool more quickly in

water than on land?• Water has greater capacity to absorb

heat than does air; temp. difference between the lava and water may be greater than between lava and air

Page 12: UNIT 10 Thermodynamics – chapter 17 Organic Chemistry – chapters 22 & 24

Heat Capacity and Specific Heat

• Heat Capacity

–Amount of heat needed to increase the temperature of an object exactly 1oC.

–Depends on both the mass and chemical composition.

• Specific Heat

–Amount of heat needed to raise the temperature of 1 g of the substance 1oC.

Page 13: UNIT 10 Thermodynamics – chapter 17 Organic Chemistry – chapters 22 & 24

• What types of things have High Specific Heat?

• Water (1 calorie raises 1g 1oC)

• What types of things have low specific heat?

• Metals (1 calorie raises 1g Fe 9oC)

• What is the relationship between specific heat and heat capacity?

Page 14: UNIT 10 Thermodynamics – chapter 17 Organic Chemistry – chapters 22 & 24

Specific Heat Formula

Q = m T Cp

• Q = Heat (J or cal)• m = mass (grams)T = change in Temp. (Tf – Ti)• Cp = Specific Heat (J / goC) or

(cal / goC)

Page 15: UNIT 10 Thermodynamics – chapter 17 Organic Chemistry – chapters 22 & 24

Sample ProblemFind the specific heat of copper when its weight is 95.4 grams. The temperature went from 50oC to 27oC and the heat capacity is 849 J.

q = 849 J (heat)

m = 95.4 (mass)

T = 50oC - 27oC = 23oC

849 J = .387 J/goC

95.4g x 23oC

Page 16: UNIT 10 Thermodynamics – chapter 17 Organic Chemistry – chapters 22 & 24

Heat Gained• How much heat is required when

49.0 grams of water is heated from 4oC to 43oC? Calculate in Joules.

Specific Heat of water = 1 cal/g x oC or 4.18 J/g x oC

mass = 49.0 g T = 43oC – 4 oC = 39oC

Heat Gained = Specific Heat x m x TCalories = 1.00 cal x 49.0 g x 39oC = 1911 cal

g oC

Page 17: UNIT 10 Thermodynamics – chapter 17 Organic Chemistry – chapters 22 & 24

Heat Gained (Problem continued)

Joules = 4.18 J x 49g x g oC

39 oC = 7988 J

Page 18: UNIT 10 Thermodynamics – chapter 17 Organic Chemistry – chapters 22 & 24

More Practice

• When 435 J of heat is added to 3.4 g of olive oil at 21oC, the temperature increases to 85oC. What is the specific heat of the oil?

• How much heat is required to raise the temperature of 250.0 g of mercury 52oC? The specific heat of mercury is 0.14 J/goC.

1.99 J/goC

1820 J

Page 19: UNIT 10 Thermodynamics – chapter 17 Organic Chemistry – chapters 22 & 24

Thermochemical Equation

• An equation that includes the amount of heat gained or released in the chemical equation

Page 20: UNIT 10 Thermodynamics – chapter 17 Organic Chemistry – chapters 22 & 24

CaCO3 + 176 kJ CaO + CO2

Endothermic Equations:

(Heat used)

C + O2 CO2 + 393.5 kJ

Exothermic Equations:

(Heat released)

Page 21: UNIT 10 Thermodynamics – chapter 17 Organic Chemistry – chapters 22 & 24

Calculate the energy required to decompose 5.20 moles CaCO3.

5.20 moles CaCO3 x 176 kJ = 915 kJ

1 mole

Page 22: UNIT 10 Thermodynamics – chapter 17 Organic Chemistry – chapters 22 & 24

Sample Problem

How much heat is released when 3.84 g of SO3 reacts with H2O?

SO3 + H2O H2SO4 + 129.6 KJ

3.84 g SO3 x 1 mole x 129.6 KJ = 6.22 KJ

80 g SO3 1 mole

Page 23: UNIT 10 Thermodynamics – chapter 17 Organic Chemistry – chapters 22 & 24

• Copy picture of Heating Curve for Water

1. In which region(s) of the graph is temperature constant?

2. Describe how the amount of energy required to melt a given mass of ice compares to the energy required to vaporize the same mass of water? Explain.

3. Which region of the graph represents the coexistence of solid and liquid? Liquid and Vapor?

HEAT IN CHANGES OF STATE

Page 24: UNIT 10 Thermodynamics – chapter 17 Organic Chemistry – chapters 22 & 24

4. When a material condenses or freezes is energy released or absorbed? Explain your reasoning.

5. When a material evaporates or melts, is energy released or absorbed? Explain your reasoning.

Page 25: UNIT 10 Thermodynamics – chapter 17 Organic Chemistry – chapters 22 & 24

Energy (as heat) during Changes of State

• Compounds absorb energy (heat) when breaking bonds.

• Solids absorb energy to melt into liquids. (ice cubes melting)

• Liquids absorb energy to evaporate into gases.

• Energy goes to changing state (breaking bonds), there is no temperature change!

Page 26: UNIT 10 Thermodynamics – chapter 17 Organic Chemistry – chapters 22 & 24

• Compounds release energy when forming bonds.

• Gases release energy to condense into a liquid.

• Liquids release energy to freeze into a solid.

• Energy goes to changing state (making bonds), there is no temperature change!

Page 27: UNIT 10 Thermodynamics – chapter 17 Organic Chemistry – chapters 22 & 24

Equations for Heating Curve

• No Phase Change:

Q = m(▲T)Cp

• Latent Heat of Fusion: (Solid Liquid)

Q = m ▲Hfus

• Latent Heat of Vaporization: (Liquid Gas)

Q= m ▲Hvap

Page 28: UNIT 10 Thermodynamics – chapter 17 Organic Chemistry – chapters 22 & 24

Practice Problem

• How much heat (in kJ) is absorbed when 24.8 g water at 100oC is converted to steam at 100oC?

• 56,048 J 56.05 kJ

• How much heat is released when 10g of water at 0oC is converted to 10g of ice at 0oC?

Page 29: UNIT 10 Thermodynamics – chapter 17 Organic Chemistry – chapters 22 & 24

• How much heat (in kJ) is absorbed when 5 g of water at 50oC is converted to steam and heated to 110oC?

• How much heat (in KJ) is released when 10g of steam at 115oC is converted to 10g of ice at -5oC ?

Page 30: UNIT 10 Thermodynamics – chapter 17 Organic Chemistry – chapters 22 & 24

• molar heat of fusion = heat absorbed by one mole of a substance in melting from a solid to a liquid at constant temperature.

• molar heat of solidification = heat lost when one mole of a liquid solidifies at a constant temperature

Hfus = - Hsolid

Page 31: UNIT 10 Thermodynamics – chapter 17 Organic Chemistry – chapters 22 & 24

Enthalpy Amount of heat that a substance has at

a given temperature and pressure = H

H = Change in enthalpy

Heat of reaction - heat that is released or absorbed during a chemical reaction

H = negative for exothermic(-)

H = positive for endothermic(+)

Page 32: UNIT 10 Thermodynamics – chapter 17 Organic Chemistry – chapters 22 & 24

E

ReactantsProducts

Exothermic

H is negative

Endothermic

ReactantsProducts

H is positive

E

Page 33: UNIT 10 Thermodynamics – chapter 17 Organic Chemistry – chapters 22 & 24

If you add two or more thermochemical equations to give a final equation, then you can also add the heats of reaction to give the final heat of reaction.

Page 34: UNIT 10 Thermodynamics – chapter 17 Organic Chemistry – chapters 22 & 24

Enthalpy changes occur when a compound is formed from its elementsDiatomic molecules = 0

Use table 11-6 pg 316

H is the difference between standard enthalpies of formation of all reactants and products

Page 35: UNIT 10 Thermodynamics – chapter 17 Organic Chemistry – chapters 22 & 24

Formula for Enthalpy Problems

H = H f (products) - H f ( reactants)

= change H = enthalpy = standard conditions (25oC and 101.3

kpa)

f = formation of the compound