Chapter 7: Energy and Chemical Change

1
Chapter 7 Energy andChemical Change

• Chemistry The Molecular Nature of Matter, 6E
• Jespersen/Brady/Hyslop

2
Thermochemistry

• Study of energies given off by or absorbed by
  reactions.
• Thermodynamics
• Study of energy transfer (flow)
• Energy (E)
• Ability to do work or to transfer heat.
• Kinetic Energy (KE)
• Energy of motion
• KE ½mv2

3
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 energy

4
Your Turn!

• Which of the following is not a form of kinetic
  energy?
• A pencil rolls across a desk
• A pencil is sharpened
• A pencil is heated
• A pencil rests on a desk
• A pencil falls to the floor

5
Factors Affecting Potential Energy

• Increase Potential Energy
• Pull apart objects that attract each other
• Book/gravity
• N and S poles of magnets
• Positive and negative charges
• Push together objects that repel each other
• Spring compressed
• N poles on two magnets
• 2 like charges

6
Factors Affecting Potential Energy

• Decrease Potential Energy
• Objects that attract each other come together
• Book falls
• N and S poles of 2 magnets
• Positive and negative charges
• Objects that repel each other move apart
• Spring released
• N poles on 2 magnets
• 2 like charges

7
Your Turn!

• Which of the following represents a decrease in
  the potential energy of the system?
• A book is raised 6 feet above the floor.
• A ball rolls downhill.
• Two electrons come close together.
• A spring is stretched completely.
• Two atomic nuclei approach each other.

8
Law of Conservation of Energy

• Energy can neither be created nor destroyed
• Can only be converted fromone form to another
• Total Energy of universe is constant

9
Temperature vs. Heat

• Temperature
• Proportional to average kinetic energy of
  objects particles
• Higher average kinetic energy means
• Higher temperature
• Faster moving molecules
• Heat
• Energy transferred between objects
• Caused by temperature difference
• Always passes spontaneously from warmer objects
  to colder objects
• Transfers until both are the same temperature

10
Heat Transfer
hot
cold

• Hot and cold objects placed in contact
• Molecules in hot object moving faster
• KE transfers from hotter to colder object
• ? average KE of hotter object
• ? average KE of colder object
• Over time
• Average KEs of both objects becomes the same
• Temperature of both becomes the same

11
Units of Energy

• Joule (J)
• KE possessed by 2 kg object moving at speed of 1
  m/s.
• If calculated value is greater than 1000 J, use
  kJ
• 1 kJ 1000 J

12
Units of Energy

• calorie (cal)
• Energy needed to raise T of 1 g H2O by 1 C
• 1 cal 4.184 J (exactly)
• 1 kcal 1000 cal
• 1 kcal 4.184 kJ
• 1 nutritional Calorie (Cal)
• (note capital C)
• 1 Cal 1000 cal 1 kcal
• 1 kcal 4.184 kJ

13
Your Turn!

• Which is a unit of energy?
• pascal
• newton
• joule
• watt
• ampere

14
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 reach same temperature
• Thermal Equilibrium
• When both cup and coffee reach same average
  Kinetic Energy and same temperature
• Energy transferred through heat comes from
  objects internal energy

15
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

16
Molecular Kinetic Energy (MKE)

• Energy associated with molecular motion
• All particles within object are in constant
  motion
• Molecules moving through space/solution
• Atoms jiggle and vibrate within molecules
• Electrons move around atoms
• Each particle has certain value of MKE at any
  given time
• Particles continually exchanging energy during
  collisions
• In isolated sample, total kinetic energy of all
  molecules is constant

17
?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 is positive ()
• Final energy gt initial energy
• Ex. Photosynthesis or charging battery
• As system absorbs energy
• Increase potential energy
• Available for later use

18
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

19
Temperature and Average Kinetic Energy

• Large collection of molecules (gas)
• Wide distribution of kinetic energy (KE)
• Small number with KE 0
• Collisions momentarily stopped molecules motion
• Very small number with very high KE
• Unbalanced collisions give high velocity
• Most molecules intermediate KEs
• Result distribution of energies

20
Distribution of Kinetic Energy
1 lower T 2 higher T
At higher T, distribution shifts to higher KE
21
Kinetic Energy Distribution

• Temperature
• Average KE of all atoms and molecules in object
• Average speed of particles
• Kelvin Temperature of sample
• T(K)? Avg MKE ½ mvavg2
• At higher temperature
• Most molecules moving at higher average speed
• Cold object Small average MKE
• Hot object Large average MKE
• Note At 0 K KE 0 so v 0

22
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 same at same T.
• At same T gas, liquid, and solid have
• Same average KE
• But very different PE

23
Your Turn!

• Which statement about kinetic energy (KE) is
  true?
• Atoms and molecules in gases, liquids and solids
  possess KE since they are in constant motion.
• At the same temperature, gases, liquids and
  solids all have different KE distributions.
• Molecules in gases are in constant motion, while
  molecules in liquids and solids are not.
• Molecules in gases and liquids are in constant
  motion, while molecules in solids are not.
• As the temperature increases, molecules move more
  slowly.

24
?E, Change in Internal Energy

• ?E Eproducts Ereactants
• Energy change can appear entirely as heat
• Can measure heat
• Cant measure Eproduct or Ereactant
• Fortunately, we are more interested in ?E
• Energy of system depends only on its current
  condition
• DOES NOT depend on
• How system got it
• What E for system might be sometime in future

25
State of Object or System

• Complete list of properties that specify objects
  current condition
• For Chemistry
• Defined by physical properties
• Chemical composition
• Substances
• Number of moles
• Pressure
• Temperature
• Volume

26
State Functions

• Any property that only depends on objects
  current state or condition
• Independence from method, path or mechanism by
  which change occurs is important feature of all
  state functions
• Some State functions
• Internal energy ?E Ef Ei
• Pressure ?P Pf Pi
• Temperature ?t tf ti
• Volume ?V Vf Vi

27
State of an object

• If T 25C, tells us all we need to know
• Dont need to know how system got to that T,
  just that this is where it currently is
• If T ? to 35C, then change in T is simply
• ?t tfinal tinitial
• Dont need to know how this occurred, just need
  to know initial and final values
• What does ?t tell us?
• Change in average MKE of particles in object
• Change in objects total KE
• Heat energy

28
Defining the System

• System
• 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 Ex. Walls of beaker
• Invisible Ex. Line separating warm and cold
  fronts

29
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
30
Three Types of Systems

• Isolated System
• No energy or matter can cross boundary
• Energy and mass are constant
• Ex. Thermos bottle
• Adiabatic Process
• Process that occurs in isolated system
• Process where neither energy nor matter crosses
  the system/surrounding boundary

Isolated system
31
Your Turn!

• A closed system can __________
• include the surroundings.
• absorb energy and mass.
• not change its temperature.
• not absorb or lose energy and mass.
• absorb or lose energy, but not mass.

32
Heat (q)

• Cant 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

33
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 J/C or J?C 1
• Extensive property
• Depends on two factors
• Sample size or amount (mass)
• Doubling amount doubles heat capacity
• Identity of substance
• Water vs. iron

34
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 3.1 103 J
35
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.

4.18 J/C
36
Your Turn!

• What is the heat capacity of 300 g of water if it
  requires 2510 J to raise the temperature of the
  water by 2.00 C?
• 4.18 J/C
• 418 J/C
• 837 J/C
• 1.26 103 J/C
• 2.51 103 J/C

37
Specific Heat (s)

• Amount of Heat Energy needed to raise T of 1 g
  substance by 1 C
• C s m or
• Intensive property
• Ratio of two extensive properties
• Units
• J/(gC) or J g?1C?1
• Unique to each substance
• Large specific heat means substance releases
  large amount of heat as it cools

38
Learning Check

• Calculate the specific heat of water if it 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

4.18 J/(g?C)
4.18 J/(g?C)
39
Your Turn!

• The specific heat of silver 0.235 J g1 C1.
  What is the heat capacity of a 100. g sample of
  silver?
• 0.235 J/C
• 2.35 J/C
• 23.5 J/C
• 235 J/C
• 2.35 103 J/C

40
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41
Using Specific Heat

• Heat Exchanged (Specific Heat ? mass) ? ?t
• q s ? m ? ?t
• Units J/(g ? C) ? g ? C J
• Substances with high specific heats resist ?T
  changes
• Makes it difficult to change temperature widely
• Water has unusually high specific heat
• Important to body (60 water)
• Used to cushion temperature changes
• Why coastal temperatures are different from
  inland temperatures

42
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 2.53C.

43
Your Turn!

• The specific heat of copper metal is 0.385
  J/(gC). 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?
• 547 J
• 1.37 104 J
• 3.47 104 J
• 34.7 J
• 4.74 104 J

44
Direction of Heat Flow

• Heat energy transferred between 2 objects
• Heat lost by one object 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
• Ex. 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

45
Your Turn!

• A cast iron skillet is moved from a hot oven to a
  sink full of water. Which of the following is
  false?
• The water heats
• The skillet cools
• The heat transfer for the skillet has a negative
  () sign
• The heat transfer for the skillet is the same as
  the heat transfer for the water
• None of these are false

46
Ex. 1 Using Heat Capacity

• A ball bearing at 260 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 1046J /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
2. Determine how much heat gained by water
qwater Cwater??twater 1046 J/C ? 12.3 C
12,866 J
47
Ex. 1 Using Heat Capacity (cont)

• A ball bearing at 260 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
  1046J /C

3. Determine how much heat ball bearing lost
qball bearing qwater
12,866 J
4. Determine T change of ball bearing
?t ball bearing (37.3 C 260 C)
222.7 C
5. Calculate C of ball bearing
57.8 J/C
48
Ex. 2 Specific Heat Calculation

• How much energy must you lose from a 250. mL cup
  of coffee for the temperature to fall from 65.0C
  to a 37.0 C? Assume density of coffee 1.00
  g/mL, scoffee swater 4.18 J g?1 C?1

q s ? m ? ?t
?t 37.0 65.0 C 28.0 C
q 4.18 J g?1 C?1?250. mL?1.00 g/mL?( 28.0 C)
q (29.3 ? 103 J)
29.3 kJ
49
Ex. 3 Using Specific Heat

• If a 38.6 g piece of silver 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 g1 C1.
• 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

59.6 C
84.1 C
50
Your Turn!

• What is the heat capacity of a container if 100.g
  of water (s 4.18 J/gC) at 100.C are added to
  100.g of water at 25C in the container and the
  final temperature is 61C?
• 35 J/C
• 4.12 103 J/C
• 21 J/C
• 4.53 103 J/C
• 50. J/C

51
Your Turn! - Solution

• What is the heat capacity of a container if 100.g
  of water (s 4.18 J/gC) at 100.C are added to
  100.g of water at 25C in the container and the
  final temperature is 61C?
• qlost by hot water m?ts
• (100.g)(61C 100.C)(4.18 J/gC) 16,302
  J
• qgained by cold water (100.g)(61C
  25C)(4.18J/gC)
• 15,048 J
• qlost by system 15,048 J (16,302 J) 1254
  J
• qcontainer qlost by system 1254 J

35 J/C
52
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 compounds potential energy
• Chemical energy
• Potential energy stored in chemical bonds
• Chemical reactions
• Generally involve both breaking and making
  chemical bonds

53
Chemical Reactions

• Forming Bonds
• Things that attract each other move closer
  together
• Decrease PE of reacting system
• Releases energy
• Profits of reaction
• Breaking Bonds
• Things that are attracted to each other are
  forced apart
• Increase PE of reacting system
• Requires energy
• Costs of reaction

54
Exothermic Reaction

• Reaction where products have less chemical energy
  than reactants
• Some chemical energy converted to kinetic energy
• Reaction releases heat to surroundings
• Heat leaves the system q negative ( )
• Heat/energy is product
• Reaction gets warmer (?T)
• Ex.
• CH4(g) 2O2(g) ? CO2(g) 2H2O(g) heat

55
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 positive ()
• Heat/energy is reactant
• Reaction becomes colder (T ?)
• Ex. Photosynthesis
• 6CO2(g) 6H2O(g) solar energy ?
  C6H12O6(s) 6O2(g)

56
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 stronger bond
• Weak bonds require less energy to break than
  strong bonds
• Key to understanding reaction energies
• Ex. If reaction has
• Weak bonds in reactants and
• Stronger bonds in products
• Heat released

57
Why Fuels Release Heat

• Methane and oxygen have weaker bonds
• Water and carbon dioxide have stronger bonds

58
Your Turn!

• Chemical energy is
• the kinetic energy resulting from violent
  decomposition of energetic chemicals.
• the heat energy associated with combustion
  reactions.
• the electrical energy produced by fuel cells.
• the potential energy which resides in chemical
  bonds.
• the energy living plants receive from solar
  radiation.

59
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

60
Heats of Reaction

• Calorimeter design not standard
• Depends on
• Type of reaction
• Precision desired
• Usually measure heat of reaction under 1 of 2
  sets of conditions
• Constant volume, qV
• Closed, rigid container
• Constant pressure, qP
• Open to atmosphere

61
What is Pressure?

• Amount of force acting on unit area
• Atmospheric Pressure
• Pressure exerted by Earths 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

62
Comparing qV and qP

• Reactions involving large volume changes,
• Consumption or production of gas
• Difference between qV and qP can be significant
• 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 piston, cylinder, bucket, and water

63
Comparing qV and qP

• Heat capacity of calorimeter 8.101 kJ/C
• Reaction run twice, identical amounts of
  reactants
• Run 1 qV
• Run at constant volume
• Calorimeter absorbs all heat produced in
  reaction
• Pin locked
• ti 24.00 ?C tf 28.91 ?C
• qCal C?t
• 8.101J/?C ? (28.91 24.00)?C 39.8 kJ
• qV qCal 39.8 kJ

64
Calculating difference in 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 C?t
• 8.101J/?C ? (31.54 ? 27.32)?C
• 34.2 kJ
• qP qCal 34.2 kJ

65
Why are qv and qp different in reactions with
significant volume change?

• 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

66
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 gt Vinitial
• ?V must be positive
• So expansion work is negative
• Work done by system

67
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.
• 629 L atm
• 629 L atm
• 315 L atm
• 171 L atm
• 315 L atm

Work P?V
?V 189.0 L 152.0 L
w 17.0 atm 37.0 L
68
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?
• Work was performed on the system.
• Work was performed by the system.
• The internal energy of the system increased.
• The internal energy of the system decreased.
• The internal energy of the system remained
  unchanged.

69
First Law of Thermodynamics

• In an isolated system, the change in internal
  energy (?E) is constant
• ?E Ef Ei 0
• Cant measure internal energy of anything
• Can measure changes in energy
• ?E is state function
• ?E work heat
• ?E q w
• ?E heat input work input

70
First Law of Thermodynamics

• 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 materials gain (have ) energy
• Others lose (have ) energy
• By monitoring surroundings, we can predict what
  is happening to system

71
First Law of Thermodynamics

• ?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

72
?E is Independent of Path

• q and w
• NOT path independent
• NOT state functions
• Depend on how change takes place

73
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 gtgt q)
• ?E is same for each path
• Partitioning between 2 paths differs

74
Heat and Work

• Two ways system can exchange internal energy with
  surroundings
• Heat
• Heat absorbed, Systems E ?
• Heat lost, Systems E ?
• Work
• Is exchanged when pushing force moves something
  through distance
• Ex. Compression or expansion of systems gas

75
Heat and Work

• 2. Work (cont)
• System does work on surroundings
• Pushes back an opposing force
• Systems E ?
• Ex. Gases produced by combustion push piston
• System has work done on it by surroundings
• Surroundings push on system
• Systems E ?
• Ex. Compression of systems gas

76
Heats of Reaction

• Measure ?E of system by measuring amount of heat
  and work exchanged in system
• ?E q w
• Amount of work done by or on system depends on
• Pressure
• Volume
• Heat of reaction measured under conditions of
  either constant volume or constant pressure

77
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?
• 15.2 J
• 15.2 J
• 9.2 J
• 9.2 J
• 3.0 J

E q w
E 3.0 J (12.2 J)
78
Your Turn!

• Which of the following is not an expression for
  the First Law of Thermodynamics?
• Energy is conserved
• Energy is neither created nor destroyed
• The energy of the universe is constant
• Energy can be converted from work to heat
• The energy of the universe is increasing

79
Heat at Constant Volume, qV

• Carry out reaction in rigid walled container
• ?V 0
• No work done
• All energy comes out as heat
• ?E qv 0 qv
• Heat of reaction at constant value
• Constant Volume calorimetry
• Used for combustion reactions
• Determine heat of combustion
• Bomb calorimeter

80
Bomb Calorimeter (Constant V)

• Apparatus for measuring ?E in reactions at
  constant volume
• Vessel in center with rigid walls
• Heavily insulated vat
• Water bath
• No heat escapes
• ?E qv

81
Ex. 3 Calorimeter Problem

• When 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.

?t 26.049 C 22.000 C 4.049 C
qabsorbed by calorimeter C?t 9.032kJ/C
4.049 C
36.57 kJ
qreleased by oil qcalorimeter 36.57 kJ
8.740 Cal/g oil
82
Ex. 3 Calorimeter Problem (cont)

• Olive oil is almost pure glyceryl trioleate,
  C57H104O6. The equation for its combustion is
• C57H104O6(l) 80 O2(g) ? 57 CO2(g) 52 H2O(l)
• 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.

?E qV 3.238 104 kJ/mol oil
83
Your Turn!

• A bomb calorimeter has a heat capacity of 2.47
  kJ/K. When a 3.74103 mol sample of ethylene
  (C2H4, MM 28.04 g/mol) was burned in this
  calorimeter, the temperature increased by 2.14 K.
  Calculate the energy of combustion for one mole
  of ethylene.
• 5.29 kJ/mol
• 5.29 kJ/mol
• 148 kJ/mol
• 1410 kJ/mol
• 1410 Kj/mol

qcal C?t
2.47 kJ/K 2.14 K 5.286 kJ
qethylene qcal 5.286 kJ
84
Heat at Constant Pressure (qP)

• Chemists usually do NOT run reactions at constant
  V
• Usually do reactions in open containers
• Open to atmosphere constant P
• Heat of reaction at constant Pressure (qP)
• ?E qP w
• Inconvenient
• Must measure volume changes to determine ?E
• Define corrected internal energy for constant
  pressure conditions

85
Enthalpy (H)

• Heat of reaction at constant Pressure (qP)
• H E PV
• Similar to E, but for systems at constant P
• Now have P?V work heat transfer
• H state function
• At constant Pressure
• ?H ?E P?V (qP w) P?V
• If only work is PV work, w P ?V
• ?H (qP w) w qP

86
Enthalpy Change (?H)

• ?H 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

87
Enthalpy vs. Internal Energy

• ?H ?E P?V
• Rearranging gives
• ?H ?E P?V
• Difference between ?H and ?E is P?V
• Reactions where form or consume gases
• P?V can be large
• Reactions involving only liquids and solids
• ?V negligible
• So ?H ?E

88
Measuring Energy Changes

• All forms of E can be converted quantitatively
  into heat.
• Thus, can completely determine amount of any form
  of E by converting it to heat.
• How?
• 1 way Let heat flow from exothermic reaction
  into mass of cooler water and measure ? T.
• To do this we need physical property of H2O
• Capacity of H2O to absorb heat

89
Coffee Cup Calorimeter

• Simple
• Measures qP
• Open to atmosphere
• Constant P
• Let heat flow from exothermic reaction into mass
  of cooler water and measure ?T
• Very little heat lost
• Calculate heat of reaction
• qP C?t

90
Ex. 4 Coffee Cup Calorimetry
NaOH 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
g1C1. Density 1.00 M HCl 1.02 g mL1 1.00
M NaOH 1.04 g mL1.
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
91
Ex. 4 Coffee Cup Calorimetry
qcal 103.0 g ? 4.184 J g1 C1 ? 6.7 C 2887
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
58 kJ/mol
92
Ex. 5 Coffee Cup Calorimetry

• When 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/gC, density is 1.00 g/mL, and the
  calorimeter absorbs a negligible amount of heat.

Write balanced net ionic equation
2NaOH(aq) H2SO4(aq) ? Na2SO4(aq) 2H2O(aq)
Determine heat absorbed by calorimeter
masssoln (25.0 mL 50.0 mL) 1.00 g/mL 75.0
g
qsoln 75.0g(33.9 25.0)C4.184J/gC
2.8103J
93
Ex 5 Determine Limiting Reagent
0.04935 mol H2SO4 present
0.0987 mol NaOH needed
0.0500 mol NaOH present
NaOH is limiting
56 kJ/mol
94
Your Turn!

• A 43.29 g sample of solid is transferred from
  boiling water (T 99.8C) to 152 g water at
  22.5C in a coffee cup. The temperature of the
  water rose to 24.3C. Calculate the specific
  heat of the solid.
• 1.1 103 J g1 C1
• 1.1 103 J g1 C1
• 1.0 J g1 C1
• 0.35 J g1 C1
• 0.25 J g1 C1

q m s ?t
1.1 103 J
qsample qwater 1.1 103 J
95
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

96
?H in Chemical Reactions

• Standard Conditions for ?H 's
• 25 C and 1 atm
• Standard Heat of Reaction (?H)
• Enthalpy change for reaction at 1 atm and 25 C
• Ex. 5
• 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

97
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
• ?Hrxn different for different states
• CH4(g) 2O2(g) ? CO2(g) 2H2O(l) ?H 890.5
  kJ
• CH4(g) 2O2(g) ? CO2(g) 2H2O(g) ?H 802.3
  kJ
• Difference energy to vaporize water

98
Thermochemical Equation

• Write ?H immediately after equation
• N2(g) 3H2(g) ? 2NH3(g) ?H 92.38 kJ
• Assumes coefficients moles
• 92.38 kJ released ? 2 moles of NH3 formed
• If 10 mole of NH3 formed
• 5N2(g) 15H2(g) ? 10NH3(g) ?H 461.9 kJ
• ?Hrxn (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 46.19 kJ

99
State Matters!

• C3H8(g) 5O2(g) ? 3 CO2(g) 4 H2O(g)
• ?H 2043 kJ
• C3H8(g) 5O2(g) ? 3 CO2(g) 4 H2O(l)
• ?H 2219 kJ
• Note there is difference in energy because
  states do not match
• If H2O(l) ? H2O(g) ?H 44 kJ/mol
• 4H2O(l) ? 4H2O(g) ?H 176 kJ/mol
• Or 2219 kJ 176 kJ 2043 kJ

100
Learning Check

• Consider the following reaction
• 2C2H2(g) 5O2(g) ? 4CO2(g) 2H2O(g)
• ?E 2511 kJ
• The reactants (acetylene and oxygen) have 2511 kJ
  more energy than products. How many kJ are
  released for 1 mol C2H2?

1,256 kJ
101
Learning Check

• Given the equation below, how many kJ are
  required for 44 g CO2 (MM 44.01 g/mol)?
• 6CO2(g) 6H2O ? C6H12O6(s) 6O2(g) ?H 2816
  kJ
• If 100. kJ are provided, what mass of CO2 can be
  converted to glucose?

470 kJ
9.38 g
102
Your Turn!

• Based on the reaction
• CH4(g) 4Cl2(g) ? CCl4(g) 4HCl(g)
• ?H 434 kJ/mol CH4
• What energy change occurs when 1.2 moles of
  methane reacts?
• 3.6 102 kJ
• 5.2 102 kJ
• 4.3 102 kJ
• 3.6 102 kJ
• 5.2 102 kJ

?H 434 kJ/mol 1.2 mol
?H 520.8 kJ
103
Running Thermochemical Equations in Reverse

• Consider
• CH4(g) 2O2(g) ? CO2(g) 2H2O(g)
• ?H 802.3 kJ
• Reverse thermochemical equation
• Must change sign of ?H
• CO2(g) 2H2O(g) ? CH4(g) 2O2(g)
• ?H 802.3 kJ

104
Reverse Thermochemical equation, 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 reverse reaction

105
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
• Lets see if this is true

106
Ex. 7 Multiple Paths Same ?H

• Path a Single step
• C(s) O2(g) ? CO2(g) ?H 393.5 kJ
• Path b Two step
• Step 1 C(s) ½O2(g) ? CO(g) ?H 110.5 kJ
• Step 2 CO(g) ½O2(g) ? CO2(g) ?H 283.0 kJ
• Net Rxn C(s) O2(g) ? CO2(g) ?H 393.5 kJ
• Chemically and thermochemically, identical
  results
• True for Exothermic reaction or for Endothermic
  reaction

107
Ex.8 Multiple Paths Same ?H

• Path a N2(g) 2O2(g) ? 2NO2(g) ?H 68 kJ
• Path b
• Step 1 N2(g) O2(g) ? 2NO(g) ?H 180. kJ
• Step 2 2NO(g) O2(g) ? 2NO2(g) ?H 112 kJ
• Net rxn N2(g) 2O2(g) ? 2NO2(g) ?H 68 kJ
• Hesss Law of Heat Summation
• For any reaction that can be written into steps,
  value of ?H for reactions sum of ?H values of
  each individual step

108
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

109
Enthalpy Diagrams

• Use to measure ?H
• Arrow down ?H
• Arrow up ?H
• Calculate cycle
• 1 step process sum of 2 step process

Ex. H2O2(l) ? H2O(l) ½O2(g) 286kJ 188kJ
?Hrxn ?Hrxn 286 kJ (188 kJ ) ?Hrxn 98
kJ
110
Hesss Law

• Hesss 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 for reaction that cant be
  measured directly
• Thermochemical equations for individual steps of
  reaction sequence may be combined to obtain
  thermochemical equation of overall reaction

111
Rules for Manipulating Thermochemical Equations

1. When equation is reversed, sign of ?H must also
   be reversed.
2. If all coefficients of equation are multiplied or
   divided by same factor, value of ?H 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

112
Strategy for Adding Reactions Together

• Choose most complex compound in equation for
  one-step path
• Choose equation in multi-step path that contains
  that compound
• Write equation down so that compound
• Is on appropriate side of equation
• has appropriate coefficient for our reaction
• Repeat steps 1 3 for next most complex
  compound, etc.

113
Strategy for Adding Reactions (Cont.)

• Choose equation that allows you to
• cancel intermediates
• multiply by appropriate coefficient
• Add reactions together and cancel like terms
• Add energies together, modifying enthalpy values
  in same way equation modified
• If reversed equation, change sign on enthalpy
• If doubled equation, double energy

114
Ex.9 Calculate ?H for

• Cgraphite(s) ?? Cdiamond(s)
• Given Cgr(s) O2(g) ? CO2(g) ?H 394 kJ
• Cdia(s) O2(g) ? CO2(g) ?H 396 kJ
• To get desired equation, must reverse 2nd
  equation and add resulting equations
• Cgr(s) O2(g) ? CO2(g) ?H 394 kJ
• CO2(g) ? Cdia(s) O2(g) ?H (396 kJ)
• Cgr(s) O2(g) CO2(g) ? Cdia(s) O2(g)
  CO2(g)
• ?H 394 kJ 396 kJ 2 kJ

1?

115
Learning Check Ex.10

• Calculate ?H for 2 Cgr(s) H2(g) ?? C2H2(g)
• Given the following
• C2H2(g) 5/2O2(g) ? 2CO2(g) H2O(l)
  ?H 1299.6 kJ
• Cgr(s) O2(g) ? CO2(g) ?H 393.5 kJ
• H2(g) ½O2(g) ? H2O(l) ?H 285.8 kJ

116
Ex.10 Calculate for 2Cgr (s) H2(g) ??
C2H2(g)

• a 2CO2(g) H2O(l) ? C2H2(g) 5/2O2(g)
  ?H (1299.6 kJ) 1299.6 kJ
• 2b 2Cgr(s) 2O2(g) ? 2CO2(g) ?H
  (2? 393.5 kJ) 787.0 kJ
• c H2(g) ½O2(g) ? H2O(l) ?H 285.8 kJ
• 2CO2(g) H2O(l) 2Cgr(s) 2O2(g) H2(g)
  ½O2(g) ? C2H2(g) 5/2O2(g) 2CO2(g)
  H2O(l)
• 2Cgr(s) H2(g) ? C2H2(g) ?H 226.8 kJ

117
Your Turn!

• Which of the following is a statement of Hess's
  Law?
• ?H for a reaction in the forward direction is
  equal to ? H for the reaction in the reverse
  direction.
• ?H for a reaction depends on the physical states
  of the reactants and products.
• If a reaction takes place in steps, ?H for the
  reaction will be the sum of ?Hs for the
  individual steps.
• If you multiply a reaction by a number, you
  multiply ?H by the same number.
• ?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.

118
Your Turn!

• Given the following data
• C2H2(g) O2(g) ? 2CO2(g) H2O(l) ?H
  1300. kJ
• C(s) O2(g) ? CO2(g) ?H 394 kJ
• H2(g) O2(g) ? H2O(l) ?H 286 kJ
• Calculate for the reaction
• 2C(s) H2(g) ? C2H2(g)
• 226 kJ
• 1980 kJ
• 620 kJ
• 226 kJ
• 620 kJ

?H 1300. kJ 2(394 kJ) (286 kJ)
119
Tabulating ?H values

• Need to Tabulate ?H values
• Major problem is vast number of reactions
• Define standard reaction and tabulate these
• Use Hesss 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

120
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 Cgr(s)
H H2(g)
Al Al(s)
Ne Ne(g)
Note All ?Hf of elements in their std states
0 Forming element from itself.

• See Appendix C in back of textbook and Table 7.2

121
Uses of Standard Enthalpy (Heat) of Formation,
?Hf

• 1. From definition of ?Hf, can write balanced
  equations directly
• ?HfC2H5OH(l)
• 2C(s, gr) 3H2(g) ½O2(g) ? C2H5OH(l)
  ?Hf 277.03 kJ/mol
• ?HfFe2O3(s)
• 2Fe(s) 3/2O2(g) ? Fe2O3(s) ?Hf 822.2
  kJ/mol

122
Your Turn!

• What is the reaction that corresponds to the
  standard enthalpy of formation of NaHCO3(s), ?Hf
  947.7 kJ/mol?
• Na(s) ½H2(g) 3/2O2(g) C(s, gr) ? NaHCO3(s)
• Na(g) H(g) 3O2(g) C4(g) ? NaHCO3(s)
• Na(aq) H(aq) 3O2(aq) C4(aq) ?
  NaHCO3(s)
• NaHCO3(s) ? Na(s) ½H2(g) 3/2O2(g) C(s, gr)
• Na(aq) HCO3(aq) ? NaHCO3(s)

123
Using ?Hf

• 2. Way to apply Hesss Law without needing to
  manipulate thermochemical equations
• Consider the reaction
• aA bB ? cC dD
• ?Hreaction c?Hf(C) d?Hf(D)
• a?Hf(A) b?Hf(B)

124
Ex.11 Calculate ?Hrxn Using ?Hf

• Calculate ?Hrxn using ?Hf data for the reaction
• SO3(g) ?? SO2(g) ½O2(g)
• Add ?Hf for each product times its coefficient
• Subtract ?Hf for each reactant times its
  coefficient.

?Hrxn 99 kJ/mol
125
Learning Check

• Calculate ?Hrxn using ?Hf for the reaction
  4NH3(g) 7O2(g) ? 4NO2(g) 6H2O(l)
• ?Hrxn 136 1715.4 184 kJ
• ?Hrxn 1395 kJ

126
Check Using Hesss Law

• 4NH3(g) ? ½ N2(g) 3/2 H2(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)

Same as before
127
Other Calculations

• Dont always want to know ?Hrxn
• Can use Hesss Law and ?Hrxn to calculate ?Hf
  for compound where not known
• Ex. Given the following data, what is the value
  of ?Hf(C2H3O2, aq)?
• Na(aq) C2H3O2(aq) 3H2O(l) ?
  NaC2H3O23H2O(s)
• ?Hrxn 19.7 kJ/mol

?Hf(Na, aq) 239.7 kJ/mol
?Hf(NaC2H3O23H2O, s) 710.4 kJ/mol
?Hf(H2O, l) 285.9 kJ/mol
128
Ex. 13 cont.

• ?Hrxn ?Hf(NaC2H3O23H2O, s) ?Hf(Na, aq)
  ?Hf(C2H3O2, aq) 3?Hf(H2O, l)
• Rearranging
• ?Hf(C2H3O2, aq) ?Hf(NaC2H3O23H2O, s)
  ?Hf(Na, aq) ?Hrxn 3?Hf (H2O, l)
• ?Hf(C2H3O2, aq) 710.4kJ/mol
  (239.7kJ/mol) (19.7kJ/mol) 3(285.9kJ/mol)
• 406.7 kJ/mol

129
Learning Check

• Calculate ?H for this reaction using ?Hf data.
• 2Fe(s) 6H2O(l) ? 2Fe(OH)3(s) 3H2(g)
• ?Hf 0 285.8 696.5 0
• ?Hrxn 2?Hf(Fe(OH)3, s) 3?Hf(H2, g)
  2 ?Hf(Fe, s) 6?Hf(H2O, l)
• ?Hrxn 2 mol( 696.5 kJ/mol) 30 20
  6 mol(285.8 kJ/mol)
• ?Hrxn 1393 kJ 1714.8 kJ
• ?Hrxn 321.8 kJ

130
Learning Check

• Calculate ?H for this reaction using ?Hf data.
• CO2(g) 2H2O(l) ? 2O2(g)
  CH4(g)
• ?Hf 393.5 285.8 0 74.8
• ?Hrxn 2?Hf(O2, g) ?Hf(CH4, g)
  ?Hf(CO2, g) 2 ?Hf(H2O, l)
• ?Hrxn 20 1 mol (74.8 kJ/mol) 1 mol
  (393.5 kJ/mol) 2 mol (285.8 kJ/mol)
• ?Hrxn 74.8 kJ 393.5 kJ 571.6 kJ
• ?Hrxn 890.3 kJ