Unit 2: Energy Icons are used to prioritize notes in this section. Note This: There are important items on this page that should be copied into your notebook or highlighted in your printed notes. - PowerPoint PPT Presentation
1 Unit 2: Energy Icons are used to prioritize notes in this section. Note This: There are important items on this page that should be copied into your notebook or highlighted in your printed notes. Important Examples: There are sample problems on this page. You MUST copy the solutions into a notebook—Even if you have printed the notes!!! Look at This: There are diagrams or charts on this page. Look at them and make sure you understand them. (but you don’t need to copy them) Extra Information: This page contains background information that you should read, but you don’t need to copy it. R Review: There is review material on this page. It is up to you to decide if you want to copy or highlight it, depending on how well you remember it.
Transcript
1
Unit 2: EnergyIcons are used to prioritize notes in this section.Note This: There are important items on this page that should be copied into your notebook or highlighted in your printed notes.
Important Examples: There are sample problems on this page. You MUST copy the solutions into a notebook—Even if you have printed the notes!!!
Look at This: There are diagrams or charts on this page. Look at them and make sure you understand them. (but you don’t need to copy them)
Extra Information: This page contains background information that you should read, but you don’t need to copy it.
R Review: There is review material on this page. It is up to you to decide if you want to copy or highlight it, depending on how well you remember it.
2••••••••••••••••••••••••••••••••
Chapter 3
Energy Transfer
••••••••••••••••
••••••••••••••••
3
Heat, Work & Temperature
Overview: Heat, work and temperature are frequently confused by students. The terms are closely related to each other, but they are not the same. Heat and work are a forms of energy transfer, temperature is a measure of molecular agitation. Heat and temperature both involve random movement of particles, while work is an orderly movement. Heat and work measurements depend on the mass of a material, while temperature does not.
3.1
4
Energy• Energy is sometimes defined as the ability to do
work or to produce heat. • This means that energy, work and heat can all be
measured with the same unit: the joule (J)
• Work is a transfer of kinetic energy that results in an orderly movement of particles, such as the motion of an object, or the expansion of a balloon.
Workand
5
Heat• Heat is a transfer of thermal energy that occurs
when two systems of different temperatures come into contact. • Heat is transferred as a result of molecular agitation,
that is random motion of particles.
• Click on the boxes below to link to more on heat
When two systems of different temperatures are placed in contact…Thermal energy (or heat) flows from the hotter one into the colder one…Until their temperature is the same.
This concept will be developed fully in a later lesson.
PREVIEW
7
Temperature
• Temperature is a measurement of the agitation of the particles (atoms or molecules) in a system.• While temperature is related to the kinetic energy of
the individual particles, it does not reflect the energy of the system as a whole. It cannot be measured in joules. It is measured in Kelvins (K) or in degrees (°C)
• If particles were sufficiently cooled, they would (theoretically) reach a temperature where they had no agitation at all. This temperature is called absolute zero, and is –273.15°C.
You remember this from the Gas Laws. We often round this to 3 significant digits: -273°C
8
Law of Conservation of Energy
• Overview:• The law of conservation of energy states that
energy can be transformed or transferred, but that it cannot be created or destroyed in any chemical or physical change*.
• Conservation of energy is the first law of thermodynamics, the study of energy changes.
• In this section we will study thermodynamic systems.
3.2
*It’s a bit more complicated with nuclear changes, since matter can change into energy and vice-versa at the nuclear level, as Einstein calculated E=mc2
9
Thermodynamic Systems
• A system is the location where we are observing an energy transfer or change.• A system can be as simple as a
beaker with a thermometer, or as complex as a Dewar flask or a calorimeter with an array of thermocouples.
• There are three different types of system.
Nothing is lost, Nothing is created,
Everything is transformed.
Antoine Lavoisier
Promoter of the Laws of
Conservation
10
Three Types of System• An Open System• Both matter and energy can easily enter or
leave the system, for example, a beaker.• A Closed System• Energy can enter or leave the system, but
matter cannot, for example, a sealed balloon.
• An Isolated System• Neither energy nor matter can easily enter
or leave the system. For example, an insulated container or calorimeter.
11
Imperfection of Isolated Systems
• In the real world it is impossible to create a perfectly isolated system. • Even the best Dewar flask or
Thermos™ bottle will eventually let some heat in or out.
• However, if experiments are conducted quickly, an insulated container is close enough to an isolated system to give us acceptable results.
A Dewar Flask
12
Calorimeters• A calorimeter is a device used
to measure heat transfer.• An inner chamber where the
system (material or reaction that produces the heat) is placed.
• A thermometer .• An outer chamber that holds
water (or another substance of known specific heat capacity)
• Insulation to prevent loss of heat to the surroundings.
Simplified calorimeter
Typical calorimeter
13
Homemade Calorimeters• For quick experiments a simple,
but effective calorimeter can be made out of Styrofoam cups and a good thermometer.
• The chemicals are mixed in the cup, and the Styrofoam insulates long enough to measure temperature differences.
• Measurements must be made quickly with this type of calorimeter, before heat loss can occur.
When using a homemade calorimeter promptness is important. Every minute you waste will reduce the accuracy of your measurements. Make sure you gather your data quickly but carefully!
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Energy Relationships• Overview:• When a substance changes temperature, thermal
energy is absorbed or released. The thermal energy is related to the mass and specific heat capacity of the substance by the familiar formula: Q = mc
• As a general rule, the amount of thermal energy lost by one system is equal to the energy gained by another.–Q1 = Q2
3.3
15
Specific Heat Capacity (c)• Specific heat capacity is
the amount of energy required to raise temperature of one gram of a substance by 1°C.
• Specific heat is a characteristic property, it is different for each substance, if measured at a standard temperature.
Remember the specific heat capacity of water. Textbooks sometimes use 4.19 or 4.18 if the problem requires no more than 3 significant figures.
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Heat Gained or Lost by a Substance• The heat gained or lost by a substance can be
measured by the calorimeter formulas:
Where: Q = thermal energy (heat), in joulesm=mass of a substance, in gramsc = specific heat capacity of a substance, in J/(g °C)∙ΔT = temperature change (Tf -Ti), in °C
Where: Tf = the final temperature of the substance, in °CTi = the initial temperature of the substance, in °C
17
Sample Question• Calculate the energy that is absorbed by a 3.00 kg block of
aluminum when its temperature changes from 17.1 °C to 35.5 °C.
Data:m= 3.00kg = 3000gc =0.90 J/(g∙°C)Ti =17.1 °C Tf =35.5 °C
To find:ΔT=Q=
Step 1: Calculate ΔT:
ΔT = Tf – Ti
= 35.5 °C – 17.1 °C = 18.4 °C
18.4 °C Step 2: Calculate Q:
Q = mcΔT= 3000 g • 0.90 J/(g∙ °C) • 18.4 °C=
49 680 J
Answer: The aluminum block must absorb 49.7 kilojoules of thermal energy.
49 680 J
Always show equation
Always show equation
Note: large amounts of energy should be converted to kilojoules and answers rounded to a sensible number of significant digits.
= 49.68 kJ
18
• Read pages 134-136• Examine sample questions on page 136.• Do questions 1 to 16 on page 137
19
Calculating Energy Transfer
Overview:Energy transfer occurs when energy moves from one body to another. When two systems at different temperatures come in contact, the thermal energy from the hotter system is transferred into the cooler system until both systems reach the same temperature. The amount of heat lost by the hotter system should exactly equal the amount of heat gained by the cooler system, unless heat is lost to the surroundings.
3.4
20
Heat Content of an Object
• The heat content of an object or system that is available for transfer is calculated by the same formula used in calorimetry, which was described in the previous lesson:
• Now we are going to combine two systems, so we will be calculating
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Heat Transfer
Hot Object Cold ObjectEqual Temperature
Thermal Energy
When two systems of different temperatures are placed in contact…Thermal energy (or heat) flows from the hotter one into the colder one…Until their temperature is the same.
Where: Q1 = Heat lost by the 1st systemQ2 = Heat gained by the 2nd system
The heat lost by the warmer object will equal the heat gained by the cooler one.
22
Heat Transfer Calculations
Combining
Gives us:
Where: m1 = mass of system 1, in gramsc1 = specific heat capacity of system 1, in J/(g.°C)ΔT1 = temperature change in system 1 (Tf -Ti1), in °Cm2 = mass of system 1, in gramsc2 = specific heat capacity of system 1, in J/(g.°C)ΔT1 = temperature change in system 2 (Tf -Ti2), in °C
Special note: Tf is the same for both systems, but Ti is different
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Sample Question• Calculate the mass of cold water at 10 °C that it
would take to cool 10.0 g of hot (95°C) glass to 30°C.
Data:m1 = 10gc1 =0.84 J/(g∙°C)Ti1 =95 °C Tf =30 °C c2 = 4.184 J/(g∙°C)Ti2 = 10 °C
To find:ΔT1
ΔT2
m2
Step 1: Calculate ΔT1 :ΔT1 = Tf – Ti1 = 30 °C – 95 °C = -65 °C
= -65 °C
Step 2: Calculate ΔT2 :ΔT2 = Tf –Ti2 = 30 °C – 10 °C = 20 °C
= +20 °C
Answer: It would take 6.5 g of water to cool the glass.
= 6.5 g
Formula: ̶ m1c1ΔT1=m2c2ΔT2
Step 3: Calculate the mass of water (m2)If , then m2 = =
= 6.5248565965 g
From Table
Round to a sensible number of significant digits.
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Finding the Final Temperature• Quite often questions will ask for the final
temperature of the two systems. • This rearrangement of the previous formula is
sometimes useful:
Where: Ti1 = initial temperature of system 1Ti2 = initial temperature of system 2Tf = final temperature of both systemsEverything else means the same as before.
The derivation of this formula is shown on page 139 of your textbook, so I won’t repeat it here. You can actually use formulas given before to solve this type of problem, but this form is easier to input if you are working with a TI83 calculator or the equivalent.
25
Sample Question• A 500 g package of frozen strawberries at -4.0 °C has a specific heat capacity
of 3.50 J/(g.°C). It is placed in 2.00 kg of warm (40.0°C) water in an insulated cooler to thaw. What is the final temperature of the raspberries and water?
Data:m1 = 500 gc1 = 3.5 J/(g∙°C)Ti1 =-4.0 °C m2 = 2 kg = 2000 gc2 = 4.19 J/(g∙°C)Ti2 = 40 °C
To find:Tf
Step 1: Calculate Tf:
Tf =
=
= 32.398815 °C= 32.4 °C
Answer: The final temperature is 32.4 degrees celsius.
Formula:
Round to a sensible number of significant digits.
I did this in two steps to show the cancellation of units more
clearly. You can do it in a single step if you have a scientific calculator, as shown on the following calculator display.
26
Personal ObservationPersonally, I have never bothered to
memorize the formula on the previous slide. I just use the three simple formulas
earlier in the chapter: Q=mcΔT, –Q1=Q2 and ΔT=Tf-Ti.
Then I break my calculations into many small steps. It takes me a bit longer, but I
don’t have to commit a really complex formula to memory, and I find I make
fewer mistakes.
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• Read Section 3.4, pp. 138-140• Do Questions 1 to 11 on page 141
• Also:• Chapter End Questions on page 145 to 146
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Chapter 4
Enthalpy Change
29
Enthalpy (H)
• Overview:• Enthalpy is defined as the total energy of a system. In
theory, enthalpy is the sum of the kinetic an potential energy that a system contains at a given pressure. In practice, it is impossible to measure the total enthalpy of a system, since there are too many variables involved. Instead, we concern ourselves with the energy changes that occur in a system. This gives us a way of finding how much potential energy changes to kinetic, or vice versa, during a chemical change.
4.1Page 148
30
Enthalpy (H)
• Enthalpy is the total energy of a system• It includes all the potential energy and all the kinetic
energy of the system.
• In practice, it is impossible to find these values exactly, so we never talk about total enthalpy.
Where: H = total enthalpy of a systemEk = the kinetic energy of the systemEp = the potential energy of the system
31
Enthalpy Change (ΔH)AKA: Heat of Reaction
• Enthalpy Change is the energy exchanged between a system and its surroundings during a physical change or chemical change.
• In theory, the enthalpy change should equal the difference between the enthalpy of the products and the enthalpy of the reactants:
Where: ΔH = the heat of reaction (enthalpy change) Hp = enthalpy of the productsHr = enthalpy of the reactants
32
Enthalpy Change in Practice• Although finding the actual total enthalpy (H)
of a system is nearly impossible, finding the enthalpy change (∆H) is quite easy to do by experiment.
• Since nearly all the energy change of most reactions is in the form of heat (thermal energy), a calorimeter can find the amount of heat absorbed or released during a change. This will be your heat of reaction.
33
Endothermic and Exothermic
• Overview:• Chemical and physical changes can absorb or release
energy. Endothermic changes absorb heat from their surroundings. Exothermic changes release heat into their surroundings.
• For chemical reactions its usually easy to tell. If materials become hotter, an exothermic reaction is occurring. If they get colder, it is an endothermic reaction.
• For physical changes, it is not quite so straight forward.
4.2
34
Endothermic and ExothermicDefinitions
• An Endothermic process is a change that absorbs heat or other forms of energy from the surroundings.• Examples: Photosynthesis (absorbs light), evaporation
(absorbs heat), dissolution of NH4NO3 (absorbs heat)
• An Exothermic process is a change that releases energy into the surroundings.• Examples: Respiration (releases chemical energy),
condensation (releases heat), dissolution of H2SO4 (releases heat)
35
Endothermic and Exothermic
Physical Changes• Physical changes, like freezing, melting, boiling
vaporization, condensation and dissolution can absorb or release energy.
Endothermic Changes Endothermic or Exothermic Exothermic Changes
Vaporization (boiling) (depends on the solute) Condensation (liquid)
Fusion (melting) Solidification (freezing)
Sublimation
HeatAbsorbed
HeatReleased
36
LiquidSolid
Gas
Melting (fusion)
Freezing (solidification)
Vaporization
Liquid
Liquid CondensationSubli
mati
on
Solid
Cond
ensa
tion /
Dep
ositi
onTerminology associated with
Change of PhaseAnd the energy exchanges
Rapid vaporization is called “boiling”,
Slow vaporization is “evaporation”
Sublimation occurs when a material
“evaporates” from a solid straight to a gas, like dry ice or iodine.
ExothermicProcess
EndothermicProcess
36
37
The Staircase Analogyanother way to remember physical changes
GasMost
energyLiquidMore
energySolidLow
energy
Imagine the states of matter to be like a staircase. Solids are lowest in energy, so they are on the bottom step. Gases are highest, so they are the top step.To go from solid to liquid or gas means “climbing” the stairs. Using up energy to go up. These changes are endothermic.
Going from gas to liquid, or from liquid to solid gives you back energy, as you bounce down the stairs. These changes are exothermic
If a pure, cold solid is slowly heated, its temperature will increase, until it starts to melt. Then, even though you are still heating it, the temperature stays the same until all the solid is melted. The energy absorbed to melt it is called the heat of fusion
If instead of heating a cold, solid substance, we cool a hot gaseous substance, we get a similar heat curve, but reversed in appearance. Some of the terminology changes when we reverse a heat curve.
Instead of a boiling point, we have a condensation point. Instead of heat of vaporization we have heat of condensation.
Instead of a melting point we have a freezing point, and the heat change is called heat of solidification.
40
Things You Must be Able to Find From the Heat Curve of a Pure Substance
• The melting/freezing point temperature• From the vertical axis (1st plateau level)
• The vaporization/condensation temperature• From the vertical axis (2nd plateau level)
• The heat of fusion (+) or solidification (–)• From the horizontal axis (difference of 2 measurements)
• The heat of vaporization (+) or condensation (–)• From the horizontal axis (difference of 2 measurements)
• The specific heat capacity of the solid or liquid substance.
We can use the different boiling temperatures of the mixtures components to separate them by capturing the vapours separately—a process called fractional distillation.
The heat curve of a mixture of substances is less precise than that of a pure substance. The mixture “melts” over a range of temperatures, and each component “boils” at a different temperature.
The mixture melts more gradually than pure matter.
Use the heat curve to determine:a) The melting point of imaginarium.b) The boiling point of imaginariumc) The specific heat capacity of liquid imaginariumd) the heat of fusion of imaginariume) the heat of vaporization of imaginarium
*Heat of fusion and vaporization are given in J/g. To convert to the more standard form ( j/mol) we could multiply this answer by the molar mass of imaginarium (if we knew it).
43
Endothermic and Exothermic
Chemical Reactions• During a chemical reaction, • energy is absorbed as the old bonds in the reactants
are broken. This phase of the reaction is endothermic• Energy is released as new bonds form in the products.
This phase is exothermic.• If more energy is absorbed than is released, the
overall reaction is endothermic. If more energy is released than is absorbed, then the overall reaction is exothermic.
44
Energy Released
Energy Absorbed
H
HH
Visualization of a Chemical Change
OOHH
HH
Reactants
2H2 + O2
atoms
4H + 2O
O
O
H HHH
Products
2H2O
O OHH
OOHH
HHO
O
HHH
In the chemical reaction visualized here, hydrogen burns with oxygen to form water. This happens in two stages. ① The collisions of hydrogen (H2) and oxygen (O2) molecules break the molecules into atoms, absorbing a little energy. ② The oxygen and hydrogen atoms then form new bonds, making water molecules and releasing large amounts of energy. The overall reaction is exothermic, but the initial stage is endothermic.
① ②Bond breaking Bond forming
Be aware that this visualization is a bit of a simplification, as explained in the next slide
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In reality, the process of reaction is a bit more complicated. Molecules do not instantly jump apart into atoms… they tend to clump into intermediate complexes, as we shall see in the next chapter.
However, the mathematics of energy is the same for the simple visualization shown on the last slide as it is for the actual reaction pathway, so we’ll use the simple version for now.
46
Simple Enthalpy Diagrams• Simple enthalpy diagrams show the difference between the
enthalpy of the reactants and the products.• In an endothermic reaction, the enthalpy increases (ΔH is positive)• In an exothermic reaction, the enthalpy decreases (ΔH is negative)
Endothermic Reaction Exothermic Reaction
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Assignments
• Read sections 4.1 and 4.2• Do questions 1 to 6 on page 155
48
Energy Balance
Overview:Energy Balance is the sum of the energy released when old bonds are broken (a positive number) and the energy absorbed as new bonds are formed (a negative number). The energy balance can tell us what the enthalpy change is during a reaction. Energy balance can also be calculated from bond energies, many of which have been found experimentally and recorded in charts.
4.3
49
Bond EnergiesBond Energy
H—H 346 kJ/mol
H—O 460 kJ/mol
H—F 570 kJ/mol
H—Cl 432 kJ/mol
C—H 413 kJ/mol
C—C 347 kJ/mol
C==C 607 kJ/mol
C—O 358 kJ/mol
O==O 498 kJ/molC==O 745 kJ/molN==O 631 kJ/mol
• The text book contains tables of bond energies, such as the one on page 419, or the small one here.
• Each bond energy can represent the energy required to break a bond, or the energy released if a bond is formed (per mole).• The energy absorbed while breaking bonds
is represented by a positive number, and the energy released by a negative number,
• so breaking the bond between two hydrogen atoms would require +346 kJ of energy per mole of atoms.
• If you formed new H—H bonds, you would release energy: –346 kJ/mol.
50
Energy Balance Formula
• Your textbook gives the following formula for energy balance:
• This is technically correct, but since the second number is always negative, isn’t it easier to treat this as a subtraction?
• Okay, purists won’t like that, since I’m not using the standards of enthalpy notation, but I think its easier to remember.
51
Energy Balance Diagrams• Energy balance diagrams are a graphical way of representing the
calculations of energy balance.• They show the energy absorbed breaking bonds, and the energy released
forming new bonds.
Exothermic Reaction Endothermic Reaction
1
2
3
4
56
52
Energy Balance diagrams• Energy balance diagrams are, by definition, diagrammatic.• They do NOT need to be drawn to exact scale, although the
arrows should approximately represent the energy proportions.
• Actual numbers are usually not shown on the axes.• The y-axis is labelled Enthalpy, and represents total energy
of the system, presumably in kilojoules or kilojoules/mole.• The x-axis is labelled reaction Progress. It is in arbitrary
time units, representing the beginning, mid-point and end of the reaction.• This means that we don’t actually record how long the reaction
takes, just that it proceeds from reactants to products.
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Energy Balance ExampleBond Energy
See p. 419H—H 346 kJ/mol
H—O 460 kJ/mol
H—F 570 kJ/mol
H—Cl 432 kJ/mol
C—H 413 kJ/mol
C—C 347 kJ/mol
C==C 607 kJ/mol
C—O 358 kJ/mol
O==O 498 kJ/mol
C==O 745 kJ/mol
N==O 631 kJ/mol
Methane (CH4) burns with oxygen (O2) to form carbon dioxide (CO2) and water (H2O):CH4 + 2 O2 CO2 + 2 H2OUse the energy balance of the bonds to find the enthalpy change.
Okay, here we come to a major difference between what the textbook says, and the way I have always seen this done before. The textbook says that the previous equation should be:Exothermic Reaction: 4 Fe(s) + 3 O2 2Fe2O3(s) ∆H =–824.1 kJ
Not what I wrote… 4 Fe(s) + 3 O2 2Fe2O3(s) ∆H =–1648.2 kJ
I think the textbook is wrong. While it’s true that if you look up the ΔH value for this equation in a table of heats of formation, you will find the value -824.1 kJ, but you will also see that they have written the equation differently, with fractional coefficients, sort of like this:Heat of Formation: 2 Fe(s) + O2 Fe2O3(s) ∆H°F =–824.1 kJ/mol
But when you use Hess’ Law (see chapter 7) you always multiply the whole equation, including the ΔH value, by whatever multiplier is needed (in this case 2) and write the total ΔH at the end of the equation.
• Read section 4.3, pp. 156 to 160• Do Questions on page 160, #1 to 6
58
Energy Change and Stoichiometry
Overview:Calculating enthalpy change using stoichiometry allows us to find the energy change that accompanies a chemical reaction when we know the mass of the reactants or products.
Although normally we write equations with whole number coefficients, sometimes during calculations with thermoequations it is more convenient to use fractional coefficients.
4.4
59
Stoichiometry and Thermochemical Equations
• Consider the following thermo-equation:H2(g) + O2(g) H2O(g) + 244kJ
• If we burn 1 mole of H2 and produce 1 mole of steam, we will also produce 244 kJ of energy.
• If we doubled the amounts of the reactants…2H2(g) + O2(g) 2 H2O(g) + 2(244) kJ
• We would produce twice as much energy… that’s 488 kJ
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Sample Problem
Carrying this idea one step further, can we calculate how much energy we would get from burning 20g of H2?
2H2(g) + O2(g) 2 H2O(g) + 488 kJ
Data:mH2= 20gMH2= 2g/mol
nH2 =
To Find:Energy=
Step 1: calculate number of moles of H2n = = = 10 Step 2: calculate the energy released. =
= = 2440 kJ10 mol
2440 kJ
Answer: Burning 20g of hydrogen gas produces 2440 kJ of energy
61
• Read Section 4.4• Answer questions 1 to 19 on pages 162 to 163
• End of Chapter Summary on pages 166 - 167• End of Chapter Questions pages 168 to 169• Check for the date of the chapter test.
62
Chapter 5
Graphical Representation of Enthalpy
63
Activated Complexes
OverviewThe molecular collisions that cause chemical reactions happen very fast, but in those fractions of a second that the particles are colliding many things happen. As reactant particles strike, some of their kinetic energy is momentarily converted to potential energy. The particles temporarily form an unstable cluster or “complex”. Then the complex springs apart into product molecules as some of the potential energy turns back into kinetic energy.
5.1
64
Activated ComplexFormation and Breakup
• When two reactant molecules collide
• They can form an unstable cluster, called an activated complex.
• The activated complex will then break apart into the products.
Activated
Complex
HCl
H Cl
65
Activation Energy
• The reactant particles must collide with sufficient kinetic energy to form an “activated complex” if they are to react.
• Without this “activation energy” they will simply bounce apart without changing.
• Usually, an area of high temperature can provide the activation energy to start a reaction.
Reactants
Activation Energy
Reaction
66
Energy Diagrams5.2
• Overview• An energy diagram is a visualization of the
energy relationship between the reactants, the activated complex and the products.
• An energy diagram is a graph of Potential Energy
• An energy diagram resembles an energy balance diagram, but there are some important differences, as we shall see on the next slide.
67
Energy Diagramsvs. Energy Balance Diagrams
↑Energy Balance
Energy Diagram↓
In an Energy Diagram:• The graph is drawn as a smooth
curve, not abrupt steps.• The vertical axis represents potential
energy only, not total enthalpy.• The high point in the graph
represents the energy needed to form an activated complex, not the total energy needed to break all the bonds.
68
Energy Diagramfor an endothermic reaction
Progress of the Reaction
Pote
ntial
Ene
rgy
Activ
ation
ene
rgy
Heat
of R
eacti
onEa ∆H
Products
Activated Complex
Reactants
69
Energy Diagramfor an exothermic reaction
Progress of the Reaction
Pote
ntial
Ene
rgy
Products
Activated Complex
ReactantsEa
Heat
of R
eacti
on ∆H
70
The “Over the Hill” Analogy• Getting a reaction
started is a bit like pushing a boulder over a hill.
• It takes some energy (Ea) to get it to the top. • ie. start the reaction
• Once it’s at the top it rolls down the other side easily.• ie. the reaction gets
goingDiagram from http://www2.ucdsb.on.ca/tiss/stretton/CHEM2/rate03.htm
71
Slow, Faster and Spontaneous Reactions
Progress of the Reaction
Pote
ntial
Ene
rgy
Reaction with high activation energy. It takes a lot of energy to get started, so the reaction will begin slowly.
Reaction with medium activation energy will begin more quickly.
Reaction with no activation energy will begin spontaneously. (that means right away)
Actually reactions with zero activation energy don’t exist, but some have activation energies that are so low that ambient temperature can cause them to begin.
72
Direct and Reverse Reactions
• Some chemical reactions are reversible.• For example:• Hydrogen & oxygen burn to form water 2 H2 + O2 2 H2O• Water can be decomposed by electrolysis2 H2O 2 H2 + O2
• These two reactions are opposites of each other. • The first is exothermic (it gives off heat and flames) • its opposite is endothermic (it absorbs energy, in the form of electricity)
73
Energy Diagramfor a reversible reaction
Progress of the Reaction
Pote
ntial
Ene
rgy
Products of the direct reaction
Activated Complex
Reactants
of the direct reaction
Ea(direct)
–∆H(direct)
Ea(reverse) +∆H(reverse)
Direct Reaction (Exothermic in this example)
Reverse Reaction (Endothermic in this example)
74
• Read Chapter 5, pp. 171 - 181• Do the Chapter end questions on pages 183-
184
75
Chapter 6
Molar Heat of Reaction
76
Molar Heat of Reaction and Heat of Dissolution
Overview:• The molar heat of a reaction is an enthalpy change
involving one mole of a substance.• When a solute dissolves in a solvent, its particles
disperse themselves between the particles of solvent. Water in particular is a good solvent because it is a polar molecule, having a slight residual electric charge at each end.
• During dissolution, heat may be released or absorbed. The amount of this heat is called the heat of dissolution.
6.1
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Molar Heat of Reaction (ΔH)
• Molar heat of reaction is the enthalpy change involved in the transformation of one mole of a substance.
• In most cases, the symbol ΔH is used for the molar enthalpy, an Q for the heat measured in a calorimeter.
• Since the calorimeter formula usually gives Q in joules, and ΔH is usually measured in kilojoules, you may have to convert the joules into kilojoules first.
∆𝐻=𝑄𝑛
Where: ΔH = molar heat of reaction, in kilojoulesQ = total heat exchanged, converted to kilojoulesn = number of moles of substance transformed.
𝑘𝑖𝑙𝑜𝑗𝑜𝑢𝑙𝑒𝑠= 𝑗𝑜𝑢𝑙𝑒𝑠1000
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Dissolution
• Dissolution occurs when a solute dissolves in a solvent.
• During dissolution… • Heat may be absorbed making
the solution colder. • endothermic dissolution, ΔHd+
• Heat may be released making the solution hotter.• exothermic dissolution, ΔHd–
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Molar Heat of Dissolution (ΔHd)(a specific example of heat of reaction)
• The molar heat of dissolution is the amount of heat released (ΔHd –) or absorbed(ΔHd +) during the dissolution of one mole of solute in a solvent.
• As usual for thermochemistry…• If heat is released, ΔHd is negative (exothermic)
• This means that the temperature of the solvent goes up↑.
• If heat is absorbed, ΔHd is positive (endothermic)• This means that the temperature of the solvent goes down↓.
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Calculating Heat of Dissolutiontypical steps and useful formulas
• Dissolve a certain number of moles of solute in sufficient solvent
• Measure the temperature change• Use the calorimeter formula to find
the total heat exchange (Q in joules)• Convert joules to kilojoules, • Divide the heat (Q in kilojoules) by
the number of moles of solute.
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Sample Problem (easy)112.2 grams of potassium hydroxide, KOH(s) are dissolved in 500mL (5.00x102 g) of water in an insulated cup. The temperature of the solution rises from 12.1°C to 64.7°C . Calculate the molar heat of dissolution of potassium hydroxide.
Step 1: Calculate number of moles of KOHn = = = 2.00 mol
Step 2: Calculate Q = mc ΔT note: use mwater= 500 g x 4.184 J/g°C x52.6°C = 110 039 joules
Step 4: Calculate ΔHΔH = = = 55.02 kJ/mol
Step 3: Convert to kilojoules: 110039 J ≈ 110.0 kJ
Answer: The molar heat of dissolution of KOH is 55.0 kJ/mol.
Step 2: Find ΔT = Tf – Ti = 64.7–12.1 = 52.6°C
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Sample Problem (tough)In an insulated cup, 4.25 g of sodium nitrate (NaNO3) were dissolved in 100 mL of water at 23.4C. The molar heat of dissolution of sodium nitrate is known to be 21.0 kJ/mol. What will the final temperature of the water be?
• Overview:• When an acid and base are mixed, some of the H+
ions from the acid will combine with some OH- ions from the base to produce H2O, water. This process will be discussed in more detail in future chapters.
• Heat is often released during this neutralization, and can be measured by calorimetry.
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Neutralization• A neutralization reaction occurs when an
acidic solution and a basic solution are mixed. The general reaction is:
Molar Heat of Neutralization (ΔHn)(another specific example of heat of reaction)
• The molar heat of neutralization is the quantity of energy that is absorbed or released in the neutralization of one mole of an acid or base.
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Calorimeters and Neutralization
• One way to find the molar heat of neutralization is to mix the two solutions inside a calorimeter – often a Styrofoam cup calorimeter.
• When calculating the molar heat of neutralization in a calorimeter we often simplify the problem by assuming that the acid and base solution have the same specific heat capacity and density as water. This is an over simplification, but it will give us accurate results as long as the acid and base are not concentrated.
• The mass of “water” in the calorimeter is the combined mass of the two solutions.
Assumed properties of dilute acids & bases: c=4.184
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Sample ProblemIn a calorimeter 100 mL of a 0.5 mol/L solution of NaOH is neutralized by an equal amount of of 0.5 mol/L solution of HCl, both at 22.5°C. The temperature after they are mixed is 25.9°C.Calculate the molar heat of neutralization of Sodium Hydroxide.
Step 1: Calculate Q, the quantity of heat transferredQ =mcΔT
= 200 g 4.184 J/(g °C) 3.4°C∙ ∙ ∙= 2845.1 J or 2.8451 kJ
= -2.8451 kJ
Reaction is exothermic, so Q is negative. Q= –2.8451 kJ
Step 2: Find the number of moles of NaOH.nNaOH = CNaOH V∙ NaOH
= 0.5 mol/L 0.100 L∙= 0.05 mol
Step 3: Calculate the molar heat of neutralization.ΔHn = = =56.902
Answer: the molar heat of neutralization of NaOH is 56.9 kJ/mol
= 0.05 mol= 56.9 kJ/mol
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Other Heats of Reaction• Any type of chemical reaction or physical change that
absorbs or releases energy can have a heat of reaction (ΔH). Some common examples are…
Name Symbol Explanation
Heat of vaporization ΔHvap, or ΔHv Heat absorbed when a material evaporates
Heat of fusion ΔHfus Heat absorbed when a material melts
Heat of solidification ΔHsol Heat released by freezing (-ΔHfus)
Heat of condensation ΔHcond Head released by condensation(-ΔHvap)
Heat of sublimation ΔHsub Heat absorbed by sublimation of
Heat of combustion ΔHcomb Heat released by burning of a fuel
ΔHvap
Standard Heat of Formation
Special heat of reaction, representing the enthalpy change in creating one mole of a substance from its elements.
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• Read Chapter 6• Answer the Chapter End Questions on pages
195 and 196
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Chapter 7
Hess’s Law
∑∆𝐻
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Reaction Mechanisms
Overview:Sometimes a reaction occurs in a series of steps, rather than all in one instant. A reaction mechanism is a series of simple reactions that combine into a more complex reaction. Germain Henri Hess discovered that the total enthalpy change of a complex reaction is equal to the sum of the enthalpy changes of the simple steps that make up the more complex mechanism.
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Reaction MechanismReaction Mechanism: A series of simple reactions that add up to more a complex reaction.
For example, the reaction: 2 NO +O2 2 NO2
Can actually occur in 2 steps:1) 2 NO N2O2
2) N2O2 + O2 2 NO2.
This two step process is a reaction mechanism. It produces the same result as the single step version, but is more likely to occur in reality.
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Mechanism Energy Graphs
When a reaction has a multi-step mechanism, its energy graph will show an activation energy for each step.
Because the activation energy of several small steps is usually lower than a single activation energy would be, it is easier for a reaction to occur by means of a mechanism.
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Energy Graph for a Complex Reaction Step 1 Step 2 Step 3
ΔH1
ΔH2
ΔH3ΔHTotal = ΔH1 + ΔH2 + ΔH3
ΔHTotal
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Summation of Enthalpies
• Overview:• According to Hess’ Law, also called the law of heat
summation, if a reaction can be broken down into several simple reactions, its enthalpy change is equal to the sum of the enthalpy changes of each of the simple reactions.
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Hess’ Law
The total enthalpy change of a reaction is the sum of the individual enthalpy changes.
Where: ΔH = Total enthalpy change of the reactionΔH1 = Enthalpy change of 1st simple reactionΔH2 = Enthalpy change of 2nd simple reaction
etc…
Germain Henri Hess1802-1850
Born: August 8, 1802 Geneva, SwitzerlandDied: November 30, 1850 St. Petersburg, RussiaFields: Chemistry, MedicineKnown for: Law of constant heat summation (Hess’ Law)Special Note: When working with Hess’ Law you are
sometimes allowed to break the rule about having
only whole number coefficients in chemical equations.
Using fraction coefficients may be necessary!
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Standard Heat of Formation ()
• The most common use of Hess’ Law is to calculate the enthalpy change of a reaction using data from a table of standard heats of formation (AKA. Standard Molar Enthalpy of Formation)
• These tables list the enthalpy associated with the formation of one mole of a substance from its elements, and these values have been worked out for many substances.
• Your textbook has a list of these on page 418, unfortunately it does not show the formation equations, which renders it awkward to use.
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Sample Table of Standard Heats of Formation
Substance Name Formula Formation Equation (balanced for 1 mole) ΔH°f (kJ/mol)
Water H2O(l) H2(g) + ½ O2(g) H2O(l)-285.8 kJ/mol
Water vapour H2O(g) H2(g) + ½ O2(g) H2O(g)-242.8 kJ/mol
Oxygen gas* O2(g) It’s the common element form at 25℃ 0 kJ/mol
*The ΔH°f value for the most common form of any element, in its most common state at 25°C is always zero. Therefore, all diatomic gases (H2, N2, O2, Cl2, F2) have a heat of formation of zero. Same with the diatomic liquid (Br2) and solid (I2).
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Working Out Standard Formation Equations
• Since the table on page 418 only gives the Standard Enthalpy of Formation values, and not the entire formation equation, we may have to work out the formation equations ourselves.
• Method:• Write out a balanced equation showing the formation of the
required compound from its components.• If the balanced equation produces more than one mole of product,
divide the whole equation to reduce the product to one mole.• Add the given ΔH°f value, from the table on p. 418, to the end of
the equation.
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Sample ProblemFind the standard enthalpy of formation equation for barium oxide:
Formula: BaO(s)Elements: Ba(s)
O2(g)
Step 1: Find the balanced equation. Ba(s) + O2(g) BaO ΔH= ?2 2
Step 2: Divide all coefficients by 2 to get a single mole of BaO Ba(s) + ½ O2(g) BaO ΔH= ?Step 3: Add ΔH°f info from the table on p. 418. Ba(s) + ½ O2(g) BaO ΔH= –944.7 kJ/mol
The equation is: Ba(s) + ½ O2(g) BaO ΔH= –944.7 kJ/mol
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Hess’s Law
• If two or more thermo-chemical equations are added together to give a final equation, then the enthalpy changes can be added to give the enthalpy change for the final equation.
ΔHT = ΔH1 + ΔH2 +...
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Adding Equations
Reactants Products Enthalpy (ΔH)
N2 +O2 2 NO +180.6kJ
2 NO+O2 2 NO2 - 122.2 kJ
N2+2NO+2O2
2 NO + 2NO2 + 66.4 kJ
N2 + 2O2 2 NO2 + 66.4 kJ
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• In the example, we cancelled the like substances at the end of the problem. • You may also cancel them earlier, as long as you are
careful to only cancel a substance in the reactant column with an identical substance in the product column.
• You may multiply an entire equation, including the ΔH value by a simple number to make the coefficients match. • You can only cancel if coefficents they match exactly!
• You may switch the substances in the reactant column with the substances in the product column, but you then have to change the sign of the ΔH value.
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Example 2: (see study guide page 47 question 2)
• Find the heat of combustion of methanol (CH3OH) from the following reactions:• C(s) + O2(g) CO2(g) + 394 kJ• H2(g) + ½ O2(g) H2O(l) + 242 kJ• C(s) + 2H2(g) + ½ O2(g) CH3OH(l)+239 kJ
• Note: for simplicity, the “phase markers” ie.(s) (l) (g) (aq) will be left out of some steps. Be aware that they are important, especially if a material changes state before or after it reacts.
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• The Reactions we know:• C(s) + O2(g) CO2(g) + 394 kJ• H2(g) + ½ O2(g) H2O(l) + 242 kJ• C(s) + 2H2(g) + ½ O2(g) CH3OH(l)+239 kJ
• The reaction we are trying to get will contain the following:• ___CH3OH(l)+ ___O2(g) ___CO2 + ___H2O
• We can balance it to find the coefficients, but I’m not going to do that yet, since all I need to know for now is what side of the arrow each substance is on.
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Solution, as you should show it in your notebooks.
