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Brady/Jespersen/Hyslop, Chemistry7E, Copyright © 2015 John Wiley & Sons, Inc. All Rights Reserved.

Chapter 6 Energy and

Chemical Change

Chemistry, 7th Edition

International Student Version

Brady/Jespersen/Hyslop


.

▪ Explore the difference between potential and kinetic energy

▪ Apply the principle of the conservation of energy

▪ Examine the connection between energy, heat, and temperature

▪ Learn about state functions

▪ Determine the amount of heat exchanged from the temperature change of an object

▪ Understand exo- an endothermic reactions

2

Chapter in Context


.

▪ Learn and apply the first law of thermodynamics

▪ Understand the difference between the heat of reaction under constant pressure versus constant volume

▪ Explore the utility of and assumptions in thermochemical equations

▪ Uses Hess’s law to predict enthalpies of reaction

▪ Determine and use standard heats of reaction to solve problems

3

Chapter in Context, cont’d


.4

Thermochemistry

▪ Study of energies given off by or absorbed by reactions

Thermodynamics

▪ Study of heat transfer or heat flow

Energy (E )

▪ Ability to do work or to transfer heat

Kinetic Energy (KE)

▪ Energy of motion

▪ KE = ½mv 2

6.1


.

Potential Energy (PE)▪ Stored energy

▪ Exists in natural attractions

and repulsions

▪ Gravity

▪ Positive and negative charges

▪ Springs

Chemical Energy

▪ PE possessed by chemicals

▪ Stored in chemical bonds

▪ Breaking bonds requires energy

▪ Forming bonds releases energy5

H2 + O2 → H2O

6.1


.6

Your Turn!Which of the following is not a form of kinetic energy?

A. A pencil rolls across a desk

B. A pencil is sharpened

C. A pencil is heated

D. A pencil rests on a desk

E. A pencil falls to the floor

6.1


.

Factors Affecting Potential Energy

Increase Potential Energy

▪ Pull apart objects that attract each other

▪ A book is attracted to the earth by gravity

▪ North and south poles of magnets


▪ Push together objects that repel each other

▪ Spring compressed

▪ Same poles on two magnets

▪ Two like charges7

6.1


.

Decrease Potential Energy

▪ Objects that attract each other come together▪ Book falls

▪ North and south poles of two magnets


▪ Objects that repel each other move apart▪ North poles on two

magnets

▪ Spring released

▪ Two like charges

8

Factors Affecting Potential Energy6.1


.

Your Turn!Which of the following represents a decrease in the potential energy of the system?

A. A book is raised six feet above the floor

B. A ball rolls downhill

C. Two electrons come close together

D. A spring is stretched completely

E. Two atomic nuclei approach each other

9

6.1


.

Your Turn!Which of the following represents an increase in the potential energy of the system?

A. Na → Na+ + e–

B. A periodic table falls off the wall

C. Ca2+ + 2e–→ Ca

D. A firecracker explodes

E. Gasoline is burned and creates CO2 and H2O

10

6.1


.

Law of Conservation of Energy

▪ Energy can neither be created nor destroyed

▪ Can only be converted from one form to another

▪ Total energy of universe is constant

11

Total Energy

Potential Energy

Kinetic Energy

= +

6.1


.12

Units of EnergyJoule (J)

▪ KE possessed by 2 kg object moving at speed of 1 m/s.

▪ If calculated value is greater than 1000 J, use kilojoules (kJ)

▪ 1 kJ = 1000 J

( )2

s 1

m 1 kg 2

2

1 J1

=

2

2

s

mkg 1 J1

=

6.1


.13

Units of Energy

A calorie (cal)

▪ Energy needed to raise the temperature of 1 g H2O by 1 °C

▪ 1 cal = 4.184 J (exactly)

▪ 1 kcal = 1000 cal

▪ 1 kcal = 4.184 kJ

A nutritional Calorie (Cal)

▪ note capital C

▪ 1 Cal = 1000 cal = 1 kcal

▪ 1 kcal = 4.184 kJ

6.1


.

Your Turn!Which is a unit of energy?

A. Pascal

B. Newton

C. Joule

D. Watt

E. Ampere

14

6.1


.

Your Turn!Convert 175.2 kJ into nutritional Calories

A. 41.87 103 Cal

B. 41.87 cal

C. 733.0 Cal

D. 41.87 Cal

E. 733.0 103 Cal

kcal 1

Cal 1

kJ 184.4

kcal 1kJ 2.175 = 41.87 Cal

15

6.1


.

Your Turn!Convert 57.9 kJ into calories

A. 13800 Cal

B. 13.8 cal

C. 242 kcal

D. 242 Cal

E. 13.8 kcal

=J 4.184

cal 1

kJ 1

J 1000kJ 9.57

13.8 kcal=cal 1000

kcalcal13800

16

6.1


.17

Temperature vs. Heat Temperature▪ Proportional to average kinetic energy of object’s

particles

▪ Higher average kinetic energy means▪ Higher temperature

▪ Faster moving molecules

Heat▪ Total amount of energy transferred between

objects

▪ Heat transfer is caused by a temperature difference

▪ Always passes spontaneously from warmer objects to colder objects

▪ Transfers until both are the same temperature

6.2


.18

Heat Transfer▪ Hot and cold objects placed in

contact▪ Molecules in hot object moving

faster

▪ KE transfers from hotter to colder object▪ A decrease in average KE of hotter

object

▪ An increase in average KE of colder

object

▪ Over time▪ Average KEs of both objects becomes the same

▪ Temperature of both becomes the same

hot cold

6.2


.19

Heat

▪ Pour hot coffee into cold cup

▪ Heat flows from hot coffee to cold cup

▪ Faster coffee molecules bump into wall of cup

▪ Transfer kinetic energy

▪ Eventually, the cup and the coffee reach the same temperature

Thermal Equilibrium

▪ When both cup and coffee reach same average kinetic energy and same temperature

▪ Energy transferred through heat comes from object’s internal energy

6.2


.

Internal Energy (E )

▪ Sum of energies of all particles in system

E = total energy of system

E = potential + kinetic = PE + KE

Change in Internal Energy

E = Efinal – Einitial

▪ means change

▪ final – initial

▪ What we can actually measure

▪ Want to know change in E associated with given process

20

6.2


.21

E, Change in Internal Energy

▪ For reaction: reactants ⎯→ products

▪ E = Eproducts – Ereactants

▪ Can use to do something useful

▪ Work

▪ Heat

▪ If system absorbs energy during reaction

▪ Energy coming into system has a positive sign (+)

▪ Final energy > initial energy

Example: Photosynthesis or charging battery

▪ As system absorbs energy

▪ Increase potential energy

▪ Available for later use

6.2


.

Kinetic Molecular Theory▪ Kinetic Molecular Theory tells us

Temperature

▪ Related to average kinetic energy of particles in object

Internal energy

▪ Related to average total molecular kinetic energy

▪ Includes molecular potential energy

Average kinetic energy

▪ Implies distribution of kinetic energies among molecules in object

22

6.2


.23

Temperature and Average Kinetic Energy

In a large collection of gas molecules

▪ Wide distribution of kinetic energy (KE)

▪ Small number with KE = 0

▪ Collisions can momentarily stop a molecule’s motion

▪ Very small number with very high KE

▪ Most molecules intermediate KEs

▪ Collisions tend to average kinetic energies

▪ Result is a distribution of energies

6.2


.24

Distribution of Kinetic Energy

1: lower temperature2: higher temperature

At higher temperature, distribution shifts to higher kinetic energy

6.2


.25

Distribution of Kinetic EnergyTemperature

▪ Average KE of all atoms and molecules in object

▪ Average speed of particles

▪ Kelvin temperature of sample

▪ T (K) Avg KE = ½ mvavg2

▪ At higher temperature

▪ Most molecules moving at higher average speed

▪ Cold object = Small average KE

▪ Hot object = Large average KE

Note: At 0 K KE = 0 so v = 0

6.2


.26

Kinetic Theory: Liquids and Solids

▪ Atoms and molecules in liquids and solids also constantly moving

▪ Particles of solids jiggle and vibrate in place

▪ Distributions of KEs of particles in gas, liquid and solid are the same at same temperatures

▪ At same temperature, gas, liquid, and solid have

▪ Same average kinetic energy

▪ But very different potential energy

6.2


.

Your Turn!

Which statement about kinetic energy (KE) is true?

A. Atoms and molecules in gases, liquids and solids possess KE since they are in constant motion.

B. At the same temperature, gases, liquids and solids all have different KE distributions.

C. Molecules in gases are in constant motion, while molecules in liquids and solids are not.

D. Molecules in gases and liquids are in constant motion, while molecules in solids are not.

E. As the temperature increases, molecules move more slowly.

27

6.2


.

E, Change in Internal Energy

▪ E = Eproducts – Ereactants

▪ Energy change can appear entirely as heat

▪ Can measure heat

▪ Can’t measure Eproduct or Ereactant

▪ Importantly, we can measure E

▪ Energy of system depends only on its current condition

▪ DOES NOT depend on:

▪ How system got it

▪ What energy the system might have sometime in future

28

6.2


.

State of Object or System

▪ Complete list of properties that specify object’s current condition

For Chemistry

▪ Defined by physical properties

▪ Chemical composition

▪ Substances

▪ Number of moles

▪ Pressure

▪ Temperature

▪ Volume

29

6.2


.30

State Functions

▪ Any property that only depends on object’s current state or condition

▪ Independence from method, path or mechanism by which change occurs is important feature of all state functions

▪ Some State functions, E, P, t, and V :

▪ Internal energy E

▪ Pressure P

▪ Temperature t

▪ Volume V

6.2


.31

State of an object▪ If tc = 25 °C, tells us all we need to know

▪ Don’t need to know how system got to that temperature, just that this is where it currently is

▪ If temperature increases to 35 °C, then change in temperature is simply:

▪ t = tfinal – tinitial

▪ Don’t need to know how this occurred, just need to know initial and final values

▪ What does t tell us?

▪ Change in average KE of particles in object

▪ Change in object’s total KE

▪ Heat energy

6.2


.32

Defining the SystemSystem

▪ What we are interested in studying

▪ Reaction in beaker

Surroundings

▪ Everything else

▪ Room in which reaction is run

Boundary

▪ Separation between system and surroundings

▪ Visible Example: Walls of beaker

▪ Invisible Example: Line separating warm and cold fronts

6.3


.33

Three Types of Systems

Open System

▪ Open to atmosphere

▪ Gain or lose mass and energy across boundary

▪ Most reactions done in open systems

Closed System

▪ Not open to atmosphere

▪ Energy can cross boundary, but mass cannot

Open system

Closed system

6.3


.34

Three Types of Systems

Isolated System

▪ No energy or matter can cross boundary

▪ Energy and mass are constant

Example: Thermos bottle

Isolated system

6.3


.

Adiabatic Process

Adiabatic Process

▪ Process that occurs in isolated system

▪ Process where neither energy nor matter crosses the system/surrounding boundary

35

6.3


.

Your Turn!A closed system can __________

A. include the surroundings

B. absorb energy and mass

C. not change its temperature

D. not absorb or lose energy and mass

E. absorb or lose energy, but not mass

36

6.3


.37

Heat (q)

▪ Cannot measure heat directly

▪ Heat (q) gained or lost by an object

▪ Directly proportional to temperature change (t) it undergoes

▪ Adding heat, increases temperature

▪ Removing heat, decreases temperature

▪ Measure changes in temperature to quantify amount of heat transferred

q = C × t

▪ C = heat capacity

6.3


.38

Heat Capacity (C )▪ Amount of heat (q) required to raise

temperature of object by 1 °C

Heat Exchanged = Heat Capacity × t

q = C × t

▪ Units for C = J/°C or J°C –1

▪ Extensive property

▪ Depends on two factors

1. Sample size or amount (mass)

▪ Doubling amount doubles heat capacity

2. Identity of substance

▪ Water vs. iron

6.3


.39

Learning Check: Heat Capacity

A cup of water is used in an experiment. Its heat capacity is known to be 720 J/˚C. How much heat will it absorb if the experimental temperature changed from 19.2 ˚C to 23.5 ˚C?

q = C ´Dt

q = 720 J

°C´ 23.5 -19.2 °C( )

q = 720 J

°C´ 4.3 °C( )

q = 3.1 × 103 J

6.3


.

Learning Check: Heat Capacity

If it requires 4.184 J to raise the temperature of 1.00 g of water by 1.00 C, calculate the heat capacity of 1.00 g of water.

40

t

qC

=

C1.00 g water

=4.184 J

1.00 C= 4.18 J/°C

6.3


.

Your Turn!

What is the heat capacity of 300. g of an object if it requires 2510. J to raise the temperature of the object by 2.00˚ C?

A. 4.18 J/˚C

B. 418 J/˚C

C. 837 J/˚C

D. 1.26 × 103 J/˚C

E. 2.51 × 103 J/°C

41

Cobject

=2510 J

2.00 C= 1255 J/°C

6.3


.

Your Turn!A copper mug has a heat capacity of 77.5 J/˚C. After adding hot water to the mug, the temperature of the mug changed from 77.0 ˚F to 185 ˚F. How much heat did the mug absorb from the water?

A. 4.65 × 103 J

B. 1.29 J

C. 8.37 × 103 J

D. 5.97 × 103 J

E. 1.43 × 104 J

tCJ/ 77.5 t Cq o ==

Note units: temps. must be in oC

C 0.60C

J 77.5q o

o=

ti = 25.0 oC, tf = 85.0 oC, So t = 85.0 oC – 25.0 oC = 60.0 oC

= 4.65 × 103 J42

6.3


.43

Specific Heat (s)▪ Amount of heat energy needed to raise

temperature of 1 g substance by 1 °C

C = s× m or

▪ Intensive property

▪ Ratio of two extensive properties

▪ Units

▪ J/(g °C) or J g−1 °C−1

▪ Unique to each substance

▪ Large specific heat means substance releases large amount of heat as it cools

m

Cs =

6.3


.

Learning Check

▪ Calculate the specific heat of water if the heat capacity of 100. g of water is 418 J/°C.

▪ What is the specific heat of water if heat capacity of 1.00 g of water is 4.18 J/°C?

▪ Thus, heat capacity is independent of amount of substance

44

s =C

m==

g .100

C J/418

s

==g 00.1

C J/18.4

s

4.18 J/g°C

4.18 J/g°C

6.3


.

Your Turn!

The specific heat of silver 0.235 J g–1 °C–1. What is the heat capacity of a 100. g sample of silver?

A. 0.235 J/°C

B. 2.35 J/°C

C. 23.5 J/°C

D. 235 J/°C

E. 2.35 × 103 J/°C

45

C = s ´m

C = 0.235 J

g C´100. g

C

J C

523.=

6.3


.46

Specific Heats of Some Substances6.3


.47

Using Specific HeatHeat Exchanged = (Specific Heat mass) t

q = s m t

Units = J/(g °C) g °C = J

▪ Substances with high specific heats resist changes in temperature when heat is applied

▪ Water has unusually high specific heat

▪ Important to body (~60% water)

▪ Used to cushion temperature changes

▪ Why coastal temperatures are different from inland temperatures

6.3


.48

Learning Check: Specific Heat

Calculate the specific heat of a metal if it takes 235 J to raise the temperature of a 32.91 g sample by 2.53 °C.

s =q

m ´ Dt=

235 J

32.91 g´2.53 °C

q = m ´ s ´Dt

s = 2.82J

g °C

6.3


.

Your Turn!The specific heat of copper metal is 0.385 J/(g ˚C). How many J of heat are necessary to raise the temperature of a 1.42 kg block of copper from 25.0 ˚C to 88.5 ˚C?

A. 547 J

B. 1.37 × 104 J

C. 3.47 × 104 J

D. 34.7 J

E. 4.74 × 104 J

49

q = m ´ s ´Dt

q = 1420 g´ 0.385J

g °C´ 63.5 °C

Dt = (88.5 -25.0) °C

Jq 410473 = .

6.3


.

Direction of Heat Flow▪ Heat is the energy transferred between two

objects

▪ Heat lost by one object has the same magnitude as heat gained by other object

▪ Sign of q indicates direction of heat flow

▪ Heat is gained, q is positive (+)

▪ Heat is lost, q is negative (–)

q1 = –q2

Example: A piece of warm iron is placed into beaker of cool water. Iron loses 10.0 J of heat, water gains 10.0 J of heat

qiron = –10.0 J qwater = +10.0 J 50

6.3


.51

Your Turn!A cast iron skillet is moved from a hot oven to a sink full of water. Which of the following is false?

A. The water heats

B. The skillet cools

C. The heat transfer for the skillet has a negative (–) sign

D. The heat transfer for the skillet has the same sign as the heat transfer for the water

6.3


.52

Example 1: Using Heat CapacityA ball bearing at 260.0 ˚C is dropped into a cup containing 250. g of water. The water warms from 25.0 to 37.3 ˚C. What is the heat capacity of the ball bearing in J/˚C?

Heat capacity of the cup of water = 1046 J / ˚C

qlost by ball bearing = –qgained by water

1. Determine temperature change of water

t water = (37.3 ˚ C – 25.0 ˚C) = 12.3 ˚C

qwater = Cwater twater = 1046 J/˚C 12.3 ˚C

2. Determine how much heat gained by water

= 12.87 ×103 J

6.3


.

Ex. 1: Using Heat Capacity (cont)

53

3. Determine how much heat ball bearing lost

qball bearing = – qwater

4. Determine T change of ball bearing

t ball bearing = (37.3 ˚C – 260.0 ˚C)

C 7.222

J 10 12.87–

t

qC

3

−

=

=

5. Calculate C of ball bearing

= –12.87 × 103 J

= –222.7 ˚ C

= 57.8 J/˚ C

A ball bearing at 260.0 ˚C is dropped into a cup containing 250. g of water. The water warms from 25.0 to 37.3 ˚C. What is the heat capacity of the ball bearing in J/˚C? C of the cup of water = 1046 J /˚ C

6.3


.54

Ex. 2: Specific Heat CalculationHow much heat energy must you lose from a 250. mL cup of coffee for the temperature to fall from 65.0 ˚C to 37.0 ˚C? (Assume density of coffee = 1.00 g/mL, scoffee = swater = 4.18 J/g˚C)

q = s m t

t = 37.0 – 65.0 ˚ C = – 28.0 ˚C

q = 4.18 J/g˚C 250. mL 1.00 g/mL (– 28.0 ˚C)

q = (–29.3 103 J) = –29.3 kJ

6.3


.

Ex. 3: Using Specific HeatIf a 38.6 g piece of gold absorbs 297 J of heat, what will the final temperature of the gold be if the initial temperature is 24.5 ˚C? The specific heat of gold is 0.129 J/g˚C.

Need to find tfinal t = tf – ti

First use q = s m t to calculate t

Next calculate tfinal

59.6 °C = tf – 24.5 °C

tf = 59.6 °C + 24.5 °C

55

Dt =q

s ´m g C J/g

J

6381290

297

.. =

= 59.6 ˚ C

= 84.1 °C

6.3


.

Your Turn!What is the heat capacity of a container if 100. g of water (s = 4.18 J/g °C) at 100. ˚C are added to

100. g of water at 25.0 ˚C in the container and the final temperature is 61.0 ˚C?

A. 35 J/˚C

B. 4.12 × 103 J/˚C

C. 21 J/˚C

D. 4.53 × 103 J/˚C

E. 50. J/˚C

56

6.3


.

Your Turn! - Solution What is the heat capacity of a container if 100. g of water (s = 4.18 J/g ˚C) at 100. ˚C are added to 100. g of water at 25.0 ˚C in the container and the final temperature is 61.0 ˚C?

qlost by hot water = m × t× s

=(100. g)(61.0 °C – 100.˚C)(4.18 J/g˚C) = –1.63 104 J

qgained by cold water = (100. g)(61.0°C – 25.0˚C)(4.18J/g ˚C)

= 1.50 104 J

qlost by system = 1.50104 J + (–1.63104 J) = –1.3103 J

qcontainer = –q lost by system = +1.3 103 J

C25.0) - (61.0

J10 1.3

Δt

qC

3

== = 36 J/°C

57

6.3


.58

Chemical Bonds and Energy

Chemical bond

▪ Attractive forces that bind

▪ Atoms to each other in molecules, or

▪ Ions to each other in ionic compounds

▪ Give rise to compound’s potential energy

Chemical energy

▪ Potential energy stored in chemical bonds

Chemical reactions

▪ Generally involve both breaking and making chemical bonds

6.4


.

Chemical ReactionsForming Bonds

▪ Atoms that are attracted to each other are moved closer together

▪ Decrease the potential energy of reacting system

▪ Releases energy

Breaking Bonds

▪ Atoms that are attracted to each other are forced apart

▪ Increase the potential energy of reacting system

▪ Requires energy

59

6.4


.60

Exothermic Reaction▪ Reaction where products have less chemical

energy than reactants

▪ Some chemical heat energy converted to kinetic energy

▪ Reaction releases heat energy to surroundings

▪ Heat leaves the system; q is negative ( – )

▪ Heat energy is a product

▪ Reaction gets warmer, temperature increases

Example:

CH4(g) + 2O2(g) → CO2(g) + 2H2O(g) + heat

6.4


.61

Endothermic Reaction

▪ Reaction where products have more chemical energy than reactants

▪ Some kinetic energy converted to chemical energy

▪ Reaction absorbs heat from surroundings

▪ Heat added to system; q is positive (+)

▪ Heat energy is a reactant

▪ Reaction becomes colder,

temperature decreases

Example: Photosynthesis

6CO2(g) + 6H2O(g) + solar energy →C6H12O6(s) + 6O2(g)

6.4


.

Bond Strength▪ Measure of how much energy is needed to

break bond or how much energy is released when bond is formed.

▪ Larger amount of energy equals a stronger bond

▪ Weak bonds require less energy to break than strong bonds

▪ Key to understanding reaction energies

Example: If reaction has

▪ Weak bonds in reactants and

▪ Stronger bonds in products

▪ Heat released62

6.4


.

Why Fuels Release Heat

▪ Methane and oxygen have weaker bonds

▪ Water and carbon dioxide have stronger bonds

63

6.4


.

Your Turn!Chemical energy is

A. the kinetic energy resulting from violent decomposition of energetic chemicals

B. the heat energy associated with combustion reactions

C. the electrical energy produced by fuel cells

D. the potential energy which resides in chemical bonds

E. the energy living plants receive from solar radiation

64

6.4


.

Your Turn!Which of the following is an endothermic process?

A. Na+ + e–→ Na

B. wood burning

C. CH4(g) → C(g) + 4H(g)

D. a bomb exploding

E. water condensing

Attraction of charges releases heat

Produces heat (i.e., heat released)

Breaking bonds requires heat

Produces heat (i.e., heat released)

Heat must be released

65

6.4


.

Heat of Reaction ▪ Amount of heat absorbed or released in chemical

reaction

▪ Determined by measuring temperature change they cause in surroundings

Calorimeter▪ Instrument used to measure temperature changes

▪ Container of known heat capacity

▪ Use results to calculate heat of reaction

Calorimetry▪ Science of using calorimeter to determine heats of

reaction

66

6.5


.67

Heats of Reaction▪ Calorimeter design not standard

▪ Depends on

▪ Type of reaction

▪ Precision desired

▪ Usually measure heat of reaction under one of two sets of conditions

▪ Constant volume, qV

▪ Closed, rigid container

▪ Constant pressure, qP


6.5


.

What is Pressure?

▪ Amount of force acting on unit area

Atmospheric Pressure

▪ Pressure exerted by Earth’s atmosphere by virtue of its weight.

▪ ~14.7 lb/in2

▪ Container open to atmosphere

▪ Under constant P conditions

▪ P ~ 14.7 lb/in2 ~ 1 atm ~ 1 bar

area

forcePressure =

68

6.5


.

Comparing qV and qP

▪ Difference between qV and qP can be significant

▪ Reactions involving large volume changes,

▪ Consumption or production of gas

▪ Consider gas phase reaction in cylinder immersed in bucket of water

▪ Reaction vessel is cylinder topped by piston

▪ Piston can be locked in place with pin

▪ Cylinder immersed in insulated bucket containing weighed amount of water

▪ Calorimeter consists of piston, cylinder, bucket, and water

69

6.5


.

Comparing qV and qP

70

▪ Heat capacity of calorimeter = 8.101 kJ/°C

▪ Reaction run twice, identical amounts of reactants

Run 1: qV - Constant Volume

▪ Same reaction run once at constant

volume and once at constant pressure

▪ Pin locked

▪ ti = 24.00 C; tf = 28.91 C

qCal = Ct

= 8.101 J/C (28.91 – 24.00)C = 39.8 kJ

qV = – qCal = –39.8 kJ

6.5


.71

Comparing qV and qP

Run 2: qP

▪ Run at atmospheric pressure

▪ Pin unlocked

▪ ti = 27.32 C; tf = 31.54 C

▪ Heat absorbed by calorimeter is

qCal = Ct

= 8.101 J/C (31.54 − 27.32)C

= 34.2 kJ

qP = – qCal = –34.2 kJ

6.5


.

Comparing qV and qP

▪ qV = -39.8 kJ

▪ qP = -34.2 kJ

▪ System (reacting mixture) expands, pushes against atmosphere, does work

▪ Uses up some energy that would otherwise be heat

▪ Work = (–39.8 kJ) – (–34.2 kJ) = –5.6 kJ

▪ Expansion work or pressure volume work

▪ Minus sign means energy leaving system

72

6.5


.

Work Convention

Work = –P× V

▪ P = opposing pressure against which piston pushes

▪ V = change in volume of gas during expansion

▪ V = Vfinal – Vinitial

▪ For expansion

▪ Since Vfinal > Vinitial

▪ V must be positive

▪ So expansion work is negative

▪ Work done by system

73

6.5


.

Your Turn!Calculate the work associated with the expansion of a gas from 152.0 L to 189.0 L at a constant pressure of 17.0 atm.

A. 629 L atm

B. –629 L atm

C. –315 L atm

D. 171 L atm

E. 315 L atm

74

Work = –P × V

w = –17.0 atm × 37.0 L

V = 189.0 L – 152.0 L = 37.0 L

6.5


.

Your Turn!A chemical reaction took place in a 6 liter cylindrical enclosure fitted with a piston. Over the course of the reaction, the system underwent a volume change from 0.400 liters to 3.20 liters. Which statement below is always true?

A. Work was performed on the system.

B. Work was performed by the system.

C. The internal energy of the system increased.

D. The internal energy of the system decreased.

E. The internal energy of the system remained

unchanged.

75

6.5


.76

First Law of Thermodynamics

▪ In an isolated system, the change in internal energy (E) is constant:

E = Ef – Ei = 0

▪ Can’t measure internal energy of anything

▪ Can measure changes in energy

E is state function

E = heat + work

E = q + w

E = heat input + work input

6.5


.77


▪ Energy of system may be transferred as heat or work, but not lost or gained

▪ If we monitor heat transfers (q) of all materials involved and all work processes, can predict that their sum will be zero

▪ Some energy transfers will be positive, gain in energy

▪ Some energy transfers will be negative, a loss in energy

▪ By monitoring surroundings, we can predict what is happening to system

6.5


.78


▪ E = q + w

q is (+) Heat absorbed by system (IN)

q is (–) Heat released by system (OUT)

w is (+) Work done on system (IN)

w is (–) Work done by system (OUT)

Endothermic reaction

▪ E = +

Exothermic reaction

▪ E = –

6.5


.79

E is Independent of Path

q and w

▪ NOT path independent

▪ NOT state functions

▪ Depend on how change takes place

6.5


.80

Discharge of Car Battery

Path a

▪ Short out with wrench

▪ All energy converted to heat, no work

▪ E = q (w = 0)

Path b

▪ Run motor

▪ Energy converted to work and little heat

▪ E = w + q (w >> q)

▪ E is same for each path

▪ Partitioning between two paths differs

6.5


.

Your Turn!

A gas releases 3.0 J of heat and then performs 12.2 J of work. What is the change in internal energy of the gas?

A. –15.2 J

B. 15.2 J

C. –9.2 J

D. 9.2 J

E. 3.0 J

81

E = q + w

E = – 3.0 J + (–12.2 J)

6.5


.

Your Turn!Which of the following is not an expression for the First Law of Thermodynamics?

A. Energy is conserved

B. Energy is neither created nor destroyed

C. The energy of the universe is constant

D. Energy can be converted from work to heat

E. The energy of the universe is increasing

82

6.5


.

Your Turn!

A reaction contracts by 1.534 L under a constant pressure of 2.134 atm while releasing 200.7 J of heat to the surrounding. What is the change in internal energy of the system? 1.000 Latm = 101.3 J

A. 131.0 J

B. 532.4 J

C. -131.0 J

D. -532.4 J

E. 331.6 J

Work = –P ×V = -2.134 atm ×1.534 L = 3.274 Latm

E = q + w = -200.7 J + w

J 7.331Latm 000.1

J 101.3Latm 274.3 ==

E = q + w = -200.7 J + 331.7= 131.0 J

83

6.5


.84

Bomb Calorimeter (Constant V)▪ Apparatus for measuring

E in reactions at constant volume

▪ Vessel in center with rigid walls

▪ No change in volume, so V = 0 so PV = 0

▪ Heavily insulated vat

▪ Water bath

▪ No heat escapes

▪ E = q - PV

▪ E = q + 0 = qv

▪ subscript ‘v’ emphasizes constant volume

6.6


.85

Example 4: Calorimeter ProblemWhen 1.000 g of olive oil is completely burned in pure oxygen in a bomb calorimeter, the temperature of the

water bath increases from 22.000 ˚C to 26.049 ˚C.

a) How many Calories are in olive oil, per gram? The

heat capacity of the calorimeter is 9.032 kJ/˚C.

= 36.57 kJ

qreleased by oil = – qcalorimeter = – 36.57 kJ

qabsorbed by calorimeter = Ct = 9.032 kJ/°C × 4.049 ˚C

t = 26.049 ˚C – 22.000 ˚C = 4.049 ˚C

kcal 1

Cal 1

kJ 4.184

kcal 1

g 0001

kJ 5736cal/g) (in

−=

.

.oilq

–8.740 Cal/g oil

6.6


.86

Ex. 4 Calorimeter Problem (cont)b) Olive oil is almost pure glyceryl trioleate,

C57H104O6. The equation for its combustion is

C57H104O6(l) + 80O2(g) → 57CO2(g) + 52H2O

What is E for the combustion of one mole of glyceryl trioleate (MM = 885.4 g/mol)? Assume the olive oil burned in part a) was pure glyceryl trioleate.

610457

610457

610457 OHC mol 1

OHC g 4.885

OHC g 000.1

kJ 57.36

−

E = qV = –3.238 × 104 kJ/mol oil

6.6


.

Your Turn!A bomb calorimeter has a heat capacity of 2.47 kJ/K. When a 3.74×10–3 mol sample of ethylene was burned in this calorimeter, the temperature increased by 2.14 K. Calculate the energy of combustion for one mole of ethylene.

A. –5.29 kJ/mol

B. 5.29 kJ/mol

C. –148 kJ/mol

D. –1410 kJ/mol

E. 1410 kJ/mol

87

= 2.47 kJ/K × 2.14 K = 5.286 kJ

qethylene = – qcal = – 5.286 kJ

qcal = Ct

DEethylene

=-5.286 kJ

3.74 ´10-3mol

kJ/mol Eethylene 1410−=

6.6


.88

Enthalpy (H)

▪ Heat of reaction at constant pressure (qP)

H = E + PV

▪ Similar to E, but for systems at constant P

▪ Now have PV work + heat transfer

▪ H = state function

▪ At constant pressure

H = E + PV = (qP + w) + PV

If only work is P–V work, w = – P V

H = (qP + w) – w = qP

6.6


.89

Enthalpy Change (H)H is a state function

▪ H = Hfinal – Hinitial

▪ H = Hproducts – Hreactants

▪ Significance of sign of H

Endothermic reaction

▪ System absorbs energy from surroundings

▪ H positive

Exothermic reaction

▪ System loses energy to surroundings

▪ H negative

6.6


.90

Coffee Cup Calorimeter▪ Simple

▪ Measures qP


▪ Constant P

▪ Let heat be exchanged between reaction and water, and measure change in temperature

▪ Very little heat lost

▪ Calculate heat of reaction

▪ qP = Ct

6.6


.91

Ex. 5: Coffee Cup CalorimetryNaOH and HCl undergo rapid and exothermic reaction when you mix 50.0 mL of 1.00 M HCl and 50.0 mL of 1.00 M NaOH. The initial t = 25.5 °C and final t = 32.2 °C. What is H in kJ/mole of HCl? Assume for these solutions s = 4.184 J g–1°C–1. Density: 1.00 M HCl = 1.02 g mL–1; 1.00 M NaOH = 1.04 g mL–1.

NaOH(aq) + HCl(aq) → NaCl(aq) + H2O(aq)

qabsorbed by solution = mass s t

massHCl = 50.0 mL 1.02 g/mL = 51.0 g

massNaOH = 50.0 mL 1.04 g/mL = 52.0 g

massfinal solution = 51.0 g + 52.0 g = 103.0 g

t = (32.2 – 25.5) °C = 6.7 °C

6.6


.92

Ex. 5: Coffee Cup Calorimetryqcal = 103.0 g 4.184 J g–1 °C–1 6.7 °C = 2890 J

Rounds to qcal = 2.9 103 J = 2.9 kJ

qrxn = –qcalorimeter = –2.9 kJ

= 0.0500 mol HCl

Heat evolved per mol HCl =

DH =-2.9 kJ

0.0500 mol HCl= -58 kJ/mol

soln HClL 1

HCl mol 1soln HClL 0500.0

6.6


.

Ex. 6: Coffee Cup CalorimetryWhen 50.0 mL of 0.987 M H2SO4 is added to 25.0 mL of 2.00 M NaOH at 25.0 °C in a calorimeter, the temperature of the aqueous solution increases to 33.9 °C. Calculate H in kJ/mole of limiting reactant. Assume: specific heat of the solution is 4.184 J/g°C, density is 1.00 g/mL, and the calorimeter absorbs a negligible amount of heat.

93

2NaOH(aq) + H2SO4(aq) → Na2SO4(aq) + 2H2O(aq)

Write balanced equation

qsoln = 75.0 g × (33.9 – 25.0)°C × 4.184 J/g°C = 2.8×103 J

Determine heat absorbed by calorimeter

masssoln = (25.0 mL + 50.0 mL) × 1.00 g/mL = 75.0 g

6.6


.

Ex. 6: Determine Limiting Reagent

= 0.04935 mol H2SO4 present

= 0.0500 mol NaOH present NaOH is limiting

= 0.0987 mol NaOHneeded

SOHL 1

SOH mol .9870

SOHmL 1000

SOHL 1 SOHmL 50.0

42

42

42

4242

0.04935 mol H2SO

4´

2 mol NaOH

1 mol H2SO

4

NaOHL 1

NaOH mol 2

NaOHmL 1000

NaOHL 1NaOHmL 25.0

= –56 kJ/mol J1000

kJ 1

NaOH mol 0.0500

J102.8 3

−

=H

94

6.6


.95

Your Turn!A 43.29 g sample of solid is transferred from boiling water (t = 99.8 ˚C) to 152 g water at 22.5 ˚C in a coffee cup. The temperature of the water rose to 24.3 ˚C. Calculate the specific heat of the solid.

A. –1.1 × 103 J g–1 ˚C–1

B. 1.1 × 103 J g–1 ˚C–1

C. 1.0 J g–1 ˚C–1

D. 0.35 J g–1 ˚C–1

E. 0.25 J g–1 ˚C–1

q = m × s × t

= 1.1 × 103 J

qwater = 152 g´4.184 J

g °C´ 24.3 - 22.5( ) °C

qsample = – qwater = – 1.1 × 103 J

C99.8) – (24.3g 43.29

J1011 3

−=

.s

6.6


.96

Enthalpy Changes in Chemical Reactions

▪ Focus on systems

▪ Endothermic

▪ Reactants + heat ⎯→ products

▪ Exothermic

▪ Reactants ⎯→ products + heat

▪ Want convenient way to use enthalpies to calculate reaction enthalpies

▪ Need way to tabulate enthalpies of reactions

6.6


.

The Standard State▪ A standard state specifies all the necessary

parameters to describe a system. Generally this includes the pressure, temperature, and amount and state of the substances involved

▪ Standard state in thermochemistry

▪ Pressure = 1 atmosphere

▪ Temperature = 25 °C = 298 K

▪ Amount of substance = 1 mol (for formation reactions and phase transitions)

▪ Amount of substance = moles in an equation (balanced with the smallest whole number coefficients)

97

6.7


.

Thermodynamic Quantities▪ E and H are state functions and are also

extensive properties

▪ E and H are measurable changes but still

extensive properties

▪ Often used where n is not standard, or specified

▪ E ° and H ° are standard changes and

intensive properties

▪ Units of kJ /mol for formation reactions and phase

changes (e.g., H°f or H°vap)

▪ Units of kJ for balanced chemical equations

(H°reaction)

98

6.7


.99

H in Chemical Reactions

Standard Conditions for H 's

▪ 25 °C and 1 atm and 1 mole

Standard Heat of Reaction (H° )

▪ Enthalpy change for reaction at 1 atm and 25 °C

Example:

N2(g) + 3H2(g) ⎯→ 2 NH3(g)

1.000 mol 3.000 mol 2.000 mol

▪ When N2 and H2 react to form NH3 at 25 °C and 1 atm 92.38 kJ released

▪ H = –92.38 kJ

6.7


.100

Thermochemical Equation▪ Write H immediately after equation

N2(g) + 3H2(g) → 2NH3(g) H = –92.38 kJ

▪ Must give physical states of products and reactants

▪ H different for different states

CH4(g) + 2O2(g) → CO2(g) + 2H2O(l ) H ° rxn = –890.5 kJ

CH4(g) + 2O2(g) → CO2(g) + 2H2O(g) H ° rxn = –802.3 kJ

▪ Difference is equal to the energy to vaporize water

6.7


.101

Thermochemical Equation

▪ Write H immediately after equation

N2(g) + 3H2(g) → 2NH3(g) H= –92.38 kJ

▪ Assumes coefficients is the number of moles

▪ 92.38 kJ released when 2 moles of NH3 formed

▪ If 10 mole of NH3 formed

5N2(g) + 15H2(g) → 10NH3(g) H= –461.9 kJ

▪ H° = (5 × –92.38 kJ) = – 461.9 kJ

▪ Can have fractional coefficients

▪ Fraction of mole, NOT fraction of molecule

½N2(g) + 3/2H2(g) → NH3(g) H°rxn = –46.19 kJ

6.7


.102

State Matters!

C3H8(g) + 5O2(g) → 3 CO2(g) + 4 H2O(g)

ΔH °rxn= –2043 kJ

C3H8(g) + 5O2(g) → 3 CO2(g) + 4 H2O(l )

ΔH °rxn = –2219 kJ

Note: there is difference in energy because states do not match

If H2O(l ) → H2O(g) ΔH °vap = 44 kJ/mol

4H2O(l ) → 4H2O(g) ΔH °vap = 176 kJ/mol

Or –2219 kJ + 176 kJ = –2043 kJ

6.7


.103

Consider the following reaction:

2C2H2(g) + 5O2(g) → 4CO2(g) + 2H2O(g)

ΔE ° = –2511 kJ

The reactants (acetylene and oxygen) have 2511 kJ more energy than products. How many kJ are released for 1 mol C2H2?

Learning Check:

–1,256 kJ-2511 kJ

2 mol C2H

2

´1 mol C2H

2=

6.7


.104

Learning Check:

Given the equation below, how many kJ are required for 44 g CO2 (MM = 44.01 g/mol) to react with H2O?

6CO2(g) + 6H2O → C6H12O6(s) + 6O2(g) ΔH˚reaction = 2816 kJ

▪ If 100. kJ are provided, what mass of CO2 can be converted to glucose?

9.4 g

470 kJ=22

22

CO mol 6

kJ 2816

CO g 44.01

CO mol 1CO g 44

=2

22

CO mol 1

CO g44

kJ 2816

CO mol 6kJ 100

6.7


.

Your Turn!Based on the reaction

CH4(g) + 4Cl2(g) → CCl4(g) + 4HCl(g)

H˚reaction = – 434 kJ/mol CH4

What energy change occurs when 1.2 moles of methane reacts?

A. –3.6 × 102 kJ

B. 5.2 × 102 kJ

C. –4.3 × 102 kJ

D. 3.6 × 102 kJ

E. –5.2 × 102 kJ

105

H = –434 kJ/mol × 1.2 mol

H = –520.8 kJ = 5.2 × 102 kJ

6.7


.

Rxn. 1: C3H8(g) + 5O2(g) → 3 CO2(g) + 4 H2O(g)

ΔH °rxn= –2043 kJ

Rxn. 2: C3H8(g) + 5O2(g) → 3 CO2(g) + 4 H2O(l )

ΔH °rxn = –2219 kJ

Why does rxn. 2 release more heat than rxn. 1?

A. It shouldn’t, and the H values should be equal

B. More reactants must have been used in rxn. 2

C. In rxn. 1 some of the heat of the reaction is used up converting liquid water to gas, so less heat can be given off.

D. Liquids always have lower temperatures than gases

Your Turn!

106

6.7


.107

Reversing Thermochemical Equations

Consider

CH4(g) + 2O2(g) → CO2(g) + 2H2O(g)

H°reaction = – 802.3 kJ

▪ Reverse thermochemical equation

▪ Must change sign of H

CO2(g) + 2H2O(g) → CH4(g) + 2O2(g)

H°reaction = 802.3 kJ

6.7


.108

Reversing Thermochemical Equations Changes sign of H▪ Makes sense:

▪ Get energy out when form products

▪ Must put energy in to go back to reactants

▪ Consequence of Law of Conservation of Energy

▪ Like mathematical equation

▪ If you know H ° for reaction, you also know H °for the reverse

6.7


.109

Multiple Paths; Same H°▪ Can often get from reactants to products by

several different paths

▪ Should get same H °

▪ Enthalpy is state function and path independent

▪ Let’s see if this is true

Intermediate A

Products Reactants

Intermediate B

6.8


.110

Ex. 7: Multiple Paths; Same H°Path a: Single step

C(s) + O2(g) → CO2(g) H°rxn = –393.5 kJ

Path b: Two step

Step 1: C(s) + ½O2(g) → CO(g) H°rxn = –110.5 kJ

Step 2: CO(g) + ½O2(g) → CO2(g) H°rxn = –283.0 kJ

Net Rxn: C(s) + O2(g) → CO2(g) H°rxn = –393.5 kJ

▪ Chemically and thermochemically, identical results

▪ True for exothermic and endothermic reactions

6.8


.111

Ex. 8: Multiple Paths; Same H rxn

Path a: N2(g) + 2O2(g) → 2NO2(g) H°rxn = 68 kJ

Path b:

Step 1: N2(g) + O2(g) → 2NO(g) H°rxn = 180. kJ

Step 2: 2NO(g) + O2(g) → 2NO2(g) H°rxn = –112 kJ

Net rxn: N2(g) + 2O2(g) → 2NO2(g) H°rxn = 68 kJ

Hess’s Law of Heat Summation

▪ For any reaction that can be written into steps, value

of H°rxn for reactions = sum of H°rxn values of each individual step

6.8


.

Enthalpy Diagrams▪ Graphical description of Hess’ Law

▪ Vertical axis = enthalpy scale

▪ Horizontal line =various states of reactions

▪ Higher up = larger enthalpy

▪ Lower down = smaller enthalpy

112

6.8


.

Enthalpy Diagrams▪ Use to measure Hrxn

▪ Arrow down Hrxn = negative

▪ Arrow up Hrxn = positive

▪ Calculate cycle▪ One step process =

sum of two step process

113

Example: H2O2(l ) → H2O(l ) + ½O2(g)

–286 kJ = –188 kJ + Hrxn

Hrxn = –286 kJ – (–188 kJ )

Hrxn = –98 kJ

6.8


.

What is your change in altitude if you climb up a 7,005 ft. mountain, drive a car from that spot down 2,003 ft, then parachute down another 11,508 ft. into a canyon?

Draw a diagram that reflects your changes and your calculation.

A. -6506 ft.

B. 20516 ft.

C. 16310 ft.

D. -20516 ft.

E. No change

Your Turn!

Start

7,005 ft.

5,002 ft.-2,003 ft.

-11,508 ft.

Alt. = -6,506 ft.

Note: Change is negative because altitude decreasedFinal state is lower than initial state

-6,506 ft.

+7,005 ft.

114

6.8


.115

Hess’s Law

Hess’s Law of Heat Summation

▪ Going from reactants to products

▪ Enthalpy change is same whether reaction takes place in one step or many

▪ Chief Use

▪ Calculation of H °rxn for reaction that can’t be measured directly

▪ Thermochemical equations for individual steps of reaction sequence may be combined to obtain thermochemical equation of overall reaction

6.8


.116

Rules for Manipulating Thermochemical Equations

1. When equation is reversed, sign of H°rxn must also be reversed.

2. If all coefficients of equation are multiplied or divided by same factor, value of H°rxn must likewise be multiplied or divided by that factor

3. Formulas canceled from both sides of equation must be for substance in same physical states

6.8


.117

Strategy for Adding Reactions Together:

1. Choose most complex compound in equation for one-step path

2. Choose equation in multi-step path that contains that compound

3. Write equation down so that compound

▪ is on appropriate side of equation

▪ has appropriate coefficient for our reaction

4. Repeat steps 1 – 3 for next most complex compound, etc.

6.8


.118

Strategy for Adding Reactions (Cont.)

5. Choose equation that allows you to

▪ cancel intermediates

▪ multiply by appropriate coefficient

6. Add reactions together and cancel like terms

7. Add energies together, modifying enthalpy values in same way equation modified

▪ If reversed equation, change sign on enthalpy

▪ If doubled equation, double energy

6.8


.119

Example 9: Calculate Horxn for

C (s, graphite) ⎯→ C (s, diamond)

Given C (s, gr) + O2(g) → CO2(g) H°rxn = –394 kJ

C (s, dia) + O2(g) → CO2(g) H°rxn = –396 kJ

▪ To get desired equation, must reverse second equation and add resulting equations

C(s, gr) + O2(g) → CO2(g) H°rxn = –394 kJ

CO2(g) → C(s, dia) + O2(g) H°rxn = –(–396 kJ)

C(s, gr) + O2(g) + CO2(g) → C(s, dia) + O2(g) + CO2(g)

H° = –394 kJ + 396 kJ = + 2 kJ

–1[ ]

6.8


.120

Learning Check: Example 10

Calculate H°rxn for 2 C (s, gr) + H2(g) ⎯→ C2H2(g)

Given the following:

a. C2H2(g) + 5/2O2(g) → 2CO2(g) + H2O(l )

H °rxn = –1299.6 kJ

b. C(s, gr) + O2(g) → CO2(g) H °rxn = –393.5 kJ

c. H2(g) + ½O2(g) → H2O(l ) H °rxn = –285.8 kJ

6.8


.121

Example 10: Calculate Horxn for

2C(s, gr) + H2(g) ⎯→ C2H2(g)

2CO2(g) + H2O(l ) → C2H2(g) + 5/2O2(g) H°rxn = – (–1299.6 kJ) = +1299.6 kJ

2C(s, gr) + 2O2(g) → 2CO2(g) H°rxn =(2–393.5 kJ) = –787.0 kJ

H2(g) + ½O2(g) → H2O(l ) H°rxn = –285.8 kJ

2CO2(g) + H2O(l ) + 2C(s, gr) + 2O2(g) + H2(g) + ½O2(g) → C2H2(g) + 5/2O2(g) + 2CO2(g) + H2O(l )

2C(s, gr) + H2(g) → C2H2(g) H°rxn = +226.8 kJ

6.8


.

Your Turn!Which of the following is a statement of Hess's Law?

A. H for a reaction in the forward direction is equal to Hfor the reaction in the reverse direction.

B. H for a reaction depends on the physical states of the reactants and products.

C. If a reaction takes place in steps, H for the reaction will be the sum of Hs for the individual steps.

D. If you multiply a reaction by a number, you multiply Hby the same number.

E. H for a reaction in the forward direction is equal in magnitude and opposite in sign to H for the reaction in the reverse direction.

122

6.8


.

Your Turn!Given the following data:

C2H2(g) + O2(g) → 2CO2(g) + H2O(l ) H rxn= –1300. kJ

C(s) + O2(g) → CO2(g) Hrxn = –394 kJ

H2(g) + O2(g) → H2O(l ) Hrxn = –286 kJ

Calculate for the reaction

2C(s) + H2(g) → C2H2(g)

A. 226 kJ

B. –1980 kJ

C. –620 kJ

D. –226 kJ

E. 620 kJ

123

Hrxn = +1300. kJ + 2(–394 kJ) + (–286 kJ)

6.8


.124

Tabulating H°values▪ Need to Tabulate H° values

▪ Major problem is vast number of reactions

▪ Define standard reaction and tabulate these

▪ Use Hess’s Law to calculate H° for any other reaction

Standard Enthalpy of Formation, Hf°

▪ Amount of heat absorbed or evolved when one mole of substance is formed at 1 atm (1 bar) and 25 °C (298 K) from elements in their standard states

▪ Standard heat of formation

6.9


.125

Standard State▪ Most stable form and physical state of

element at 1 atm (1 bar) and 25 °C (298 K)

Element Standard

state

O O2(g)

C C (s, gr)

H H2(g)

Al Al(s)

Ne Ne(g)

▪ See Appendix in back of textbook and Table 6.2

Note: All Hf° of elements in their standard states = 0

Forming element from itself.

6.9


.126

Uses of Standard Enthalpy (Heat) of Formation, Hf°

1. From definition of Hf°, can write balanced equations directly

Hf° of C2H5OH(l )

2C(s, gr) + 3H2(g) + ½O2(g) → C2H5OH(l )

Hf° = –277.03 kJ/mol

Hf° of Fe2O3(s)

2Fe(s) + 3/2O2(g) → Fe2O3(s) Hf° = –822.2 kJ/mol

6.9


.

Your Turn!Which reaction corresponds to the standard enthalpy of formation of NaHCO3(s), Hf ° = – 947.7 kJ/mol?

A.Na(s) + ½H2(g) + 3/2O2(g) + C(s, gr) → NaHCO3(s)

B. Na+(g) + H+(g) + 3O2–(g) + C4+(g) → NaHCO3(s)

C. Na+(aq) + H+(aq) + 3O2–(aq) + C4+(aq) → NaHCO3(s)

D. NaHCO3(s) → Na(s) + ½H2(g) + 3/2O2(g) + C(s, gr)

E. Na+(aq) + HCO3–(aq) → NaHCO3(s)

127

6.9


.

Your Turn!Which reaction corresponds to the standard enthalpy of formation of C4H3Br2NO2(l)?

A. 4C(s, gr) + 3H(g) + 2Br(s) + N(g) + 2O(g) →C4H3Br2NO2(l)

B. 8C(s, gr) + 3H2(g) + 2Br2(g) + N2(g) + 2O2 (g) →

2C4H3Br2NO2(l)

C. 4C(s, gr) + 3/2H2(g) + Br2(l) + ½N2(g) + O2(g) →C4H3Br2NO2(l)

D. 4C(s, gr) + 3/2H2(g) + Br2(s) + ½N2(g) + ½O2(g) →C4H3Br2NO2(l)

128

6.9


.

Your Turn!

The standard enthalpy of formation of sulfur dioxide is-296.9 kJ. What is H for the formation of 16.03 g ofsulfur dioxide in its standard state from its elements intheir standard states?

A. 148.4 kJ

B. -296.9 kJ

C. -4,759 kJ

D. -148.4 kJ

E. 593.6 kJ

22

22

SO mol

kJ 9.296

SO g07.32

SO molSO g03.16

−

= ̶148.4 kJ

129

6.9


.130

Using Hf°2. Way to apply Hess’s Law without needing to manipulate

thermochemical equations

Consider the reaction:

aA + bB → cC + dD

H°reaction = c × H°f(C) + d × H°f(D) – {a×H°f(A) + b×H°f(B)}

▪ H°rxn has units of kJ because

▪ Coefficients heats of formation have units of mol kJ/mol

H°reaction = –Sum of all H°f

of all of the products

Sum of all H°f of all of the reactants

DHrxn

o = DHf

o products( ) ´ moles of product( )éë

ùûå -

DHf

o reactants( ) ´ moles of reactant( )éë

ùûå

H°rxn has units of kJ H°f has units of kJ/mol

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Ex. 11: Calculate Horxn Using Hf°

Calculate H°rxn using Hf° data for the reactionSO3(g) ⎯→ SO2(g) + ½O2(g)

1. Multiply each Hf° (in kJ/mol) by the number of moles in the equation2. Add the Hf° (in kJ/mol) multiplied by the number of moles in the equation of each product3. Subtract the Hf° (in kJ/mol) multiplied by the number of moles in the equation of each reactant

)(SO)(O)(SO )(3f)(2f21)(2frxn ggg HHHH −+=

DHrxn

= -297 kJ/mol+ 12(0 kJ/mol) - ( - 396 kJ/mol)

H°rxn = 99 kJ

DHrxn

o = DHf


ùûå -

DHf


ùûå

H°rxn has units of kJ

H°f has units of kJ/mol

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.132

Learning CheckCalculate H°rxn using Hf° for the reaction

4NH3(g) + 7O2(g) → 4NO2(g) + 6H2O(l )

H°rxn = [136 – 1715.4 + 184] kJ

H°rxn = –1395 kJ

DHrxn

= 4DHf NO

2(g )

+ 6DHf H

2O(l )

- 4DHf NH

3(g )

- 7DHf O

2(g )

DHrxn

= 4 mol(34 kJ/mol) + 6 mol( - 285.9 kJ/mol)

- 4 mol( - 46.0 kJ/mol) - 7 mol(0 kJ/mol)

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Check Using Hess’s Law

4 ×[NH3(g) → ½N2(g) + 3/2H2(g)] – 4 × Hf°(NH3, g)

7 ×[ O2(g) → O2(g) ] –7 × Hf°(O2, g)

4 ×[ O2(g) + ½ N2(g) → NO2(g)] 4 × Hf°(NO2, g)

6 ×[ H2(g) + ½ O2(g) → H2O(l ) ] 6 × Hf°(H2O, l)

4NH3(g) + 7O2(g) ⎯→ 4NO2(g) + 6H2O(l )

DHrxn

= 4DHf NO

2(g )

+ 6DHf H

2O(l )

- 4DHf NH

3(g )

- 7DHf O

2(g )

Same as before

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Example 12: Other Calculations▪ Don’t always want to know H°rxn

▪ Can use Hess’s Law and H°rxn to calculate Hf° for compound where not known

Example: Given the following data, what is the value of Hf°(C2H3O2

–, aq)?

Na+(aq) + C2H3O2–(aq) + 3H2O(l ) → NaC2H3O2·3H2O(s)

H°rxn = –19.7 kJ/mol

Na+(aq) Hof = –239.7 kJ/mol

NaC2H3O2•3H2O(s) Hof = 710.4 kJ/mol

H2O(l) Hof = 285.9 kJ/mol

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Example 12 cont.

H°rxn = Hf° (NaC2H3O2·3H2O, s) – Hf° (Na+,

aq) – Hf° (C2H3O2–, aq) – 3Hf° (H2O, l )

Rearranging

Hf°(C2H3O2–, aq) = Hf°(NaC2H3O2·3H2O, s) –

Hf°(Na+, aq) – H°rxn – 3Hf° (H2O, l)

Hf°(C2H3O2–, aq) =

–710.4 kJ/mol – (–239.7kJ/mol) – (–19.7 kJ/mol) – 3(–285.9 kJ/mol)

= +406.7 kJ/mol

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Learning CheckCalculate H for this reaction using Hf° data.

2Fe(s) + 6H2O(l) → 2Fe(OH)3(s) + 3H2(g)

Hf° 0 –285.8 –696.5 0

H°rxn = 2×Hf°(Fe(OH)3, s) + 3×Hf°(H2, g) – 2× Hf°(Fe, s) – 6×Hf°(H2O, l )

H°rxn = 2 mol× (– 696.5 kJ/mol) + 3×0 – 2×0 – 6 mol× (–285.8 kJ/mol)

H°rxn = –1393 kJ + 1714.8 kJ

H°rxn = 321.8 kJ

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.137

Learning CheckCalculate H°rxn for this reaction using Hf° data.

CO2(g) + 2H2O(l ) → 2O2(g) + CH4(g)

Hf° –393.5 –285.8 0 – 74.8

H°rxn = 2×Hf°(O2, g) + Hf°(CH4, g) –Hf°(CO2, g) – 2× Hf°(H2O, l )

H°rxn = 2 × 0 + 1 mol × (–74.8 kJ/mol) – 1 mol × (–393.5 kJ/mol) – 2 mol × (–285.8 kJ/mol)

H°rxn = –74.8 kJ + 393.5 kJ + 571.6 kJ

H°rxn = 890.3 kJ

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.

Your Turn!

Calculate Hf° for FeO(s) using the information below. Hf° values are shown below each substance.

Fe3O4(s) + CO(g) → 3FeO(s) + CO2(g) Ho=21.9 kJ

-1120.9 kJ -110.5 kJ ?? – 393.5 kJ

A. 272.0 kJ B. -816.0 kJ C. -272.0 kJ D. 26.00 JE. -38.60 kJ

138

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.

H°rxn = [3Hf° (FeO, s) + Hf°(O2, g)] – [Hf°(Fe3O4, s) + Hf°(CO, g)]

Your Turn! sol’nDH

rxn

o = DHf


ùûå -

DHf


ùûå

+21.9 kJ = [3Hf° (FeO, s) + -393.5 kJ)] – [-1120.9 kJ + -110.5 kJ)]

+21.9 kJ = [3Hf° (FeO, s) + 837.9 kJ]

-816.0 kJ = 3Hf° (FeO, s)

Important

-272.0 kJ = Hf° (FeO, s)

139

6.9

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