Chapter 6

Energy

and

Chemical Reactions

Macroscale

Kinetic Energy• energy that something

has because it is moving

Potential Energy• energy that something

has because of its position

Molecular Kinetic Energy• energy of motion of

molecules = temperature

Chemical Potential Energy• reactive molecules have the

potential to form stable ones and release energy

Nanoscale

Energy Units

1J = 1 kg m2/sec2

1 cal = 4.184 J

1kcal = 1 Cal

thus

1 Cal = 1 kcal = 1000 cal = 4.184 kJ = 4184 J

First Law of Thermodynamics

• energy can neither be created nor destroyed

• the total amount of energy in the universe is a constant

• energy can be transformed from one form to another

Measuring Temperature

K.E. of motion Temperature

Energy Transfer

Energy is always transferred from the hotter to the cooler sample

Heat Transfer

Energy is always transferred from the hotter to the cooler sample

Internal Energy E

• the sum of the individual energies of all nanoscale particles (atoms, ions, or molecules) in that sample

• including all chemical bonds and intermolecular interactions

• we cannot measure the internal energy accurately

• but chemistry depends only on changes in internal energy E

Thermochemistry

Divide universe into two partssystem that part of the universe under

investigation

surroundings the rest of the universe

universe = system + surroundings

consider everything from the system’s viewpoint

Internal Energy

First Law of Thermodynamics

System can exchange energy with surroundings via heat or work

heat q

work w

internal energy change E

E = q + w

Combustion of gasoline in your car engine results in the production of 124 kJ of energy. If the engine does 31 kJ of work to move the car how much heat must be released

by the radiator?

1. 31 kJ

2. 124 kJ

3. 93 kJ

4. 155 kJ

Internal Energy, Heat, and Work

Which of these changes always results in a decrease in the internal energy of the system?

1. The system absorbs heat and does work on the surroundings.

2. The system releases heat and does work on the surroundings.

3. The system absorbs heat and work is done on it by the surroundings.

4. The system releases heat and work is done on it by the surroundings.

Specific Heat c

• the amount of heat necessary to raise the temperature of 1 gram of the substance 1°C

• independent of mass

• substance dependent

• Specific Heat of Water = 4.184 J/g°C

Specific Heat Capacity

Heat

q = m c T

where q heat, J

m mass, g

c specific heat, J/g.CT = change in temperature, C

Molar Heat Capacity

• the heat necessary to raise the temperature of one mole of substance by 1C

• substance dependent

• cm

• metals: cm ≈ 25 J/mol.C

Hot and Cold Iron

Freezing and Melting

Heat Transfer

Two substances in an isolated system – no energy exchange with the surroundings

E = 0, no work

qlost + qgained = E = 0, qlost = -qgained

(m c T)lost = - (m c T)gained

T = final temperature – initial temperature

T = Tf – Ti Tf is same for both substances

EXAMPLE If 100. g of iron at 100.0oC is placed in 200. g of water at 20.0oC in an insulated container, what will the temperature, oC, of the iron and water when both are at the same temperature? The specific heat of iron is 0.106 cal/goC.

(100.g 0.106cal/goC (Tf - 100.)oC) = qlost

- qgained = - (200.g 1.00cal/goC (Tf - 20.0)oC)

10.6(Tf - 100.oC) = - 200.(Tf - 20.0oC)

10.6Tf - 1060oC = - 200.Tf + 4000oC

(10.6 + 200.)Tf = (1060 + 4000)oC

Tf = (5060/211.)oC = 24.0oC

At 25ºC, the same quantity of heat is added to 1.00-g samples of gold and iron.

Metal Specific Heat Capacity (J g-1 ºC-1) Gold 0.128Iron 0.451

Identify the correct statement.

1. The temperatures of both will remain the same.

2. The temperatures of both will increase by the same amount.

3. The temperatures of both will decrease by the same amount.

4. The temperature of the gold sample will rise higher than that of iron.

5. The temperature of the iron sample will rise higher than that of gold.

Heating, Temperature Change, and Phase Change

Vaporization and Condensation

Which of the following is an exothermic process?

1. Water boiling

2. Ice melting

3. Heating soup

4. Steam condensing

EXAMPLE: How much heat is required to heat 10.0 g of ice at -15.0oC to steam at 127.0oC?

q ice = m c Tice cice = 2.09J/goC

qwater = m c Twater cwater = 4.18J/goC

qsteam = m c tsteam csteam = 2.03J/goC

qfusion = m heat of fusion333J/g

qboil. = m heat of vaporization 2260J/g

qoverall = qice + qfusion + qwater + qboil. + qsteam

EXAMPLE: How much heat is required to heat 10.0 g of ice at -15.0oC to steam at 127.0oC? qoverall = qice + qfusion + qwater + qboil. + qsteam

q = (10.0g 2.09J/goC 15.0oC)

+ (10.0g 333J/g)

+ (10.0g 4.18J/goC 100.0oC)

+ (10.0g 2260J/g)

+ (10.0g 2.03J/goC 27.0oC)

q = (314 + 3.33×103 + 4.18×103 + 2.26×104 + 548)J

= 23.3 kJ

Heat Flow in Reactions

exothermic –reaction that gives off energy

q < 0

isolated system E=0 heat released by reaction raises the temperature of the system

constant T, heat is released to the surroundings

endothermic – reaction that absorbs energy

q > 0

Expansion Type Work

w = -PV system does work

P

P

Vinitial

V

V = Vfinal - Vinitial

qp = +2kJ

Do 250 J of work to compress a gas, 180 J of heat are released by the gas

What is E for the gas?

1. 430 J

2. 70 J

3. -70 J

4. -180 J

5. -250 J

Enthalpy H

E = q + w

at constant V, wexpansion = 0

E = qv

at constant P, wexpansion = -PV

E = qp - PV

DefinePV) = PV

H = qp

EnthalpyEnthalpy heat at constant pressure

qp = H = Hproducts - Hreactants

Exothermic ReactionH = (Hproducts - Hreactants) < 0

H2O(l) H2O(s) H < 0

Endothermic ReactionH = (Hproducts - Hreactants) > 0

H2O(l) H2O(g) H > 0

State FunctionsH and E along with P, T, V are state

functions. They are the same no matter what path we take for the change.

• q and w are not state functions, they depend on which path we take between two points.

initial

final

Eq

w

q

wE=Efinal-Einitial

q and w can be anything