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Chapter 7: Energy and
Chemical Change
Chemistry: The Molecular Nature of Matter, 6E
Jespersen/Brady/Hyslop
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
2
Thermochemistry
Study of energies given off by or absorbed by reactions.
Thermodynamics Study of heat transfer or heat flowEnergy (E )
Ability to do work or to transfer heat.
Kinetic Energy (KE) Energy of motion KE = ½mv 2
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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
3
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
4
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
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Factors Affecting Potential EnergyIncrease Potential Energy
Pull apart objects that attract each other A book is attracted to the earth
by gravity North and south poles of magnets Positive and negative charges
Push together objects that repel each other Spring compressed Same poles on two magnets Two like charges
5
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Decrease Potential Energy Objects that attract each other come
together Book falls North and south poles of two magnets Positive and negative charges
Objects that repel each other move apart Spring released North poles on two
magnets Two like charges
6
Factors Affecting Potential Energy
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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.
7
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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
8
Total Energy
Potential Energy
Kinetic Energy
= +
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
9
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
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
10
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
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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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 1m 1
kg 221
J1
2
2
smkg 1
J1
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
12
Units of EnergyA 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
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Your Turn!Which is a unit of energy?
A. Pascal
B. Newton
C. Joule
D. Watt
E. Ampere
13
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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, 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
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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
15
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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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 energyEx. Photosynthesis or charging battery As system absorbs energy
Increase potential energy Available for later use
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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 object17
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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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
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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Distribution of Kinetic Energy
1: lower temperature2: higher temperature
At higher temperature, distribution shifts to higher kinetic energy
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
20
Kinetic Energy DistributionTemperature
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
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
21
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
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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.
22
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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
23
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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
24
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
25
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
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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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
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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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 Ex. Walls of beaker Invisible Ex. Line separating warm and cold
fronts
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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Three Types of SystemsOpen 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
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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Three Types of SystemsIsolated 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
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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.
30
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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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
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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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
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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Learning Check: Heat CapacityA 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
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Learning Check: Heat CapacityIf 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.
34
tq
C
4.18 J/°C
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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
35
1255 J/°C
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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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 g1 °C1
Unique to each substance Large specific heat means substance
releases large amount of heat as it cools
mC
s
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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 of substance
37
g .100C J/418
s
g 00.1
C J/18.4 s
4.18 J/g °C
4.18 J/g °C
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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
38
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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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
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
41
Learning Check: Specific HeatCalculate 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.
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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
42
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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
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 heatqiron = –10.0 J qwater = +10.0 J
43
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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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 heatsB. The skillet coolsC. The heat transfer for the skillet has a
negative (–) signD. The heat transfer for the skillet is the same as the heat transfer for the water
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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Ex. 1 Using Heat CapacityA 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 = 1046 J / ˚Cqlost by ball bearing = –qgained by water
1. Determine temperature change of watert water = (37.3 ˚ C – 25.0 ˚C) = 12.3 ˚C
qwater = Cwater twater = 1046 J/˚C 12.3 ˚C2. Determine how much heat gained by water
= 12.87 ×103 J
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Ex. 1 Using Heat Capacity (cont)
46
3. Determine how much heat ball bearing lostqball bearing = – qwater
4. Determine T change of ball bearing
t ball bearing = (37.3 ˚C – 260 ˚C)5. Calculate C of ball bearing
= –12.87 × 103 J
= –222.7 ˚ C
= 57.8 J/˚ C
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 = 1046 J /˚ C
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
47
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 a 37.0 ˚C? (Assume density of coffee = 1.00 g/mL, scoffee = swater = 4.18 J g1 ˚C1)q = s m t
t = 37.0 – 65.0 ˚ C = – 28.0 ˚C
q = 4.18 J g1 ˚ C1250. mL1.00 g/mL(– 28.0 ˚ C)q = (–29.3 103 J) = –29.3 kJ
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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–1 ˚C–1.
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 48
= 59.6 ˚ C
= 84.1 °C
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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 ˚C in the container and the final temperature is 61˚ C?§ 35 J/˚C§ 4.12 × 103 J/˚ C§ 21 J/˚C§ 4.53 × 103 J/˚C§ 50. J/˚C
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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 ˚C in the container and the final temperature is 61˚C?qlost by hot water = m × t × s
=(100. g)(61 °C – 100.˚C)(4.18 J/g ˚C) = –16,302 J
qgained by cold water = (100. g)(61 °C – 25 ˚C)(4.18J/g ˚C)
= 15,048 J qlost by system = 15,048 J + (–16,302 J) = –1254 J
qcontainer = –q lost by system = +1254 J= 35 J/°C
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
51
Chemical Bonds and EnergyChemical 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
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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
52
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
53
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
Ex. CH4(g) + 2O2(g) CO2(g) + 2H2O(g) + heat
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
54
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
Ex. Photosynthesis
6CO2(g) + 6H2O(g) + solar energy C6H12O6(s) + 6O2(g)
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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
Ex. If reaction has Weak bonds in reactants and Stronger bonds in products Heat released
55
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Why Fuels Release Heat
Methane and oxygen have weaker bonds Water and carbon dioxide have stronger bonds
56
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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.
57
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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 reaction58
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
59
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
Open to atmosphere
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
60
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
areaforce
Pressure
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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 water61
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Comparing qV and qP 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 kJqV = – qCal = –39.8 kJ
62
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
63
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
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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
64
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Work ConventionWork = –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
65
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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
66
Work = –P × V
w = –17.0 atm × 37.0 L
V = 189.0 L – 152.0 L
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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.
67
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
68
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
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
69
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 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
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
70
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 = –
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
71
E is Independent of Pathq and w
NOT path independent
NOT state functions
Depend on how change takes place
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
72
Discharge of Car BatteryPath 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
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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
73
E = q + w
E = – 3.0 J + (–12.2 J)
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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
74
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
75
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
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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Ex. 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 kJqreleased 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 1Cal 1
kJ 4.184kcal 1
g 0001kJ 5736
cal/g) (in
.
.oilq
–8.740 Cal/g oil
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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 1OHC g 4.885
OHC g 000.1kJ 57.36
E = qV = –3.238 × 104 kJ/mol oil
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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
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= 2.47 kJ/K × 2.14 K = 5.286 kJqethylene = – qcal = – 5.286 kJ
qcal = Ct
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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
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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
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Coffee Cup Calorimeter Simple Measures qP
Open to atmosphere 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
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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
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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 kJqrxn = –qcalorimeter = –2.9 kJ
= 0.0500 mol HCl
Heat evolved per mol HCl =
= -58 kJ/mol
soln HClL 1 HCl mol 1
soln HClL 0500.0
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Ex. 5: 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.
84
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
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Ex. 6 Determine Limiting Reagent
= 0.04935 mol H2SO4 present
= 0.0500 mol NaOH present NaOH is
limiting
= 0.0987 mol NaOH needed
SOHL 1 SOH mol .9870
SOHmL 1000 SOHL 1
SOHmL 50.042
42
42
4242
NaOHL 1 NaOH mol 2
NaOHmL 1000NaOHL 1
NaOHmL 25.0
= –56 kJ/mol J1000kJ 1
NaOH mol 0.0500 J102.8 3
H
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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
qsample = – qwater = – 1.1 × 103 J
C99.8) – (24.3g 43.29
J1011 3
.s
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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
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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)
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Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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)
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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
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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
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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
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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
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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
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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) ΔH˚reaction = 2816 kJ
If 100. kJ are provided, what mass of CO2 can be converted to glucose?
9.38 g
470 kJ22
22 CO mol 6
kJ 2816CO g 44.01
CO mol 1CO g 44
2
22
CO mol 1 CO g 44.0
kJ 2816CO mol 6
kJ 100
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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
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H = –434 kJ/mol × 1.2 mol
H = –520.8 kJ = 5.2 × 102 kJ
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Running Thermochemical Equations in Reverse
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
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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 the reverse
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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
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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 reaction or for endothermic reaction
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Ex. 8 Multiple Paths; Same H
°rxnPath 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
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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
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Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Enthalpy Diagrams Use to measure Hrxn
Arrow down Hrxn = negative
Arrow up Hrxn = positive
Calculate cycle One step process =
sum of two step process
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Ex. H2O2(l ) H2O(l ) + ½O2(g)
–286 kJ = –188 kJ + Hrxn
Hrxn = –286 kJ – (–188 kJ )
Hrxn = –98 kJ
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Hess’s LawHess’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
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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
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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.
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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
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Ex. 9 Calculate H°rxn 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[ ]
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Learning Check: Ex. 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
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Ex.10 Calculate 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
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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 H for 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 H by 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.
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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 reaction2C(s) + H2(g) C2H2(g)
A. 226 kJ
B. –1980 kJ
C. –620 kJ
D. –226 kJ
E. 620 kJ112
Hrxn = +1300. kJ + 2(–394 kJ) + (–286 kJ)
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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
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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 C in back of textbook and
Table 7.2
Note: All Hf° of elements in their standard states = 0
Forming element from itself.
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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
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Your Turn!What is the reaction that 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)
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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
H°rxn has units of kJ H°f has units of kJ/mol
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Ex. 11 Calculate H°rxn 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
H°rxn = 99 kJ
H°rxn has units of kJH°f has units of kJ/mol
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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
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Check Using Hess’s Law4 ×[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 )Same as before
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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
Ex. 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
Naaq) H f= –239.7 kJ/mol
NaC2H3O2•3H2O(s) H f= 710.4 kJ/mol
H2O(l) H f= 285.9 kJ/mol
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Ex. 13 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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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