1

Unit 2: Energy Icons are used to prioritize notes in this section.

Make some notes: There are SOME important items on this page that should be copied into your notebook or highlighted in your printed notes.

Copy as we go: There are sample problems on this page. I EXPECT YOU to copy the solutions into a notebook— Even if you have downloaded or 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 make notes 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 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.

Work and

5

• 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

6

Heat Transfer

Hot Object Cold Object Equal 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.

This concept will be developed fully in a later lesson.

7

Temperature

• 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 .

You remember this from the Gas Laws. We often round this to 3 significant digits: -273 C

8

Assignments

• Exercises page 130 # 1 to 4

9

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 , 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

10

Thermodynamic Systems

• 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

11

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.

12

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

13

Calorimeters

• An inner chamber (AKA “bomb”) 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

14

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!

15

Energy Relationships 3.3

16

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.

Substance Specific Heat Capacity

Liquid Water 4.184 J/(g∙ C)

Water vapour 1.41 J/(g∙ C)

Ice (solid water) 2.05 J/(g∙ C)

Ethylene glycol 2.20 J/(g∙ C)

Aluminum 0.90 J/(g∙ C)

Copper 0.39 J/(g∙ C)

Glass 0.84J/(g∙ C)

Air (dry) 1.02 J/(g∙ C)

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.

17

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 joules m=mass of a substance, in grams c = 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 C Ti = the initial temperature of the substance, in C

18

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. (From the internet, the specific heat capacity of aluminum is found to be 0.900 J/g°C)

Data: m= 3.00kg = 3000g c =0.900 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 • 0.900 J/(g∙ C) • 18.4 C =

49 680 J

Answer: The aluminum block must absorb 49.7 kilojoules of thermal energy. (3 significant figures were used to match m and c precision)

49 680 J

Note: large amounts of energy should be converted to kilojoules and answers rounded to a sensible number of significant digits.

= 49.68 kJ

19

• Read pages 134-136

• Examine sample questions on page 136.

• Do questions 1 to 16 on page 137

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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

21

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

22

Heat Transfer

Hot Object Cold Object Equal 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 system Q2 = Heat gained by the 2nd system

The heat lost by the warmer object will equal the heat gained by the cooler one.

23

Heat Transfer Calculations

Combining

Gives us:

Where: m1 = mass of system 1, in grams c1 = specific heat capacity of system 1, in J/(g. C) ΔT1 = temperature change in system 1 (Tf -Ti1), in C m2 = mass of system 1, in grams c2 = 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

24

)20)(184.4(

)65)(84.0)(10(

22

1112

CCg

J

CCg

Jg

Tc

Tcmm

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 = 10g c1 =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

= 6.5248565965 g

From Table

Round to a sensible number of significant digits.

Step 3: calculate the mass of water (m2)

25

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 1 Ti2 = initial temperature of system 2 Tf = final temperature of both systems Everything 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.

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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 strawberries and water?

Data: m1 = 500 g c1 = 3.5 J/(g∙ C) Ti1 =-4.0 C m2 = 2 kg = 2000 g c2 = 4.19 J/(g∙ C) Ti2 = 40 C To find: Tf

= 32.4 C

Answer: The final temperature is 32.4 degrees celsius.

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.

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Personal Observation

Personally, 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

29

Chapter 4

Enthalpy Change

30

Enthalpy (H)

• Overview: • Enthalpy is defined as the total energy of a system.

In theory, enthalpy is the sum of the kinetic and 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.

4.1 Page 148

31

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 system Ek = the kinetic energy of the system Ep = the potential energy of the system

32

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 products Hr = enthalpy of the reactants

33

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, from which we get the enthalpy change (∆H)

34

Endothermic and Exothermic

• Overview: • Chemical and physical changes can absorb or

release energy. changes absorb heat from their surroundings. 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

35

Endothermic and Exothermic Definitions

• An 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 process is a change that releases energy into the surroundings.

• Examples: Respiration (releases chemical energy), condensation (releases heat), dissolution of H2SO4 (releases heat)

36

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 (evaporation) Dissolution Condensation (deposition)

Vaporization (boiling) (depends on the solute) Condensation (liquid)

Fusion (melting) Solidification (freezing)

Sublimation

Heat Absorbed

Heat Released

37

Liquid

Solid

Gas

Melting (fusion)

Freezing (solidification) Liquid

Terminology associated with

Change of Phase And 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.

Exothermic Process

Endothermic Process

37

38

The Staircase Analogy another way to remember physical changes

Gas Most

energy Liquid More

energy Solid Low

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

39

Heat Curve of a Pure Substance

150 140 130 120 110 100

90 80 70 60 50 40 30 20 10

0 -10 -20 -30 -40 -50

100 200 300 400 500 600 700 800 900 1000 1100 1200 1300 1400 1500 1600

Tem

per

atu

re (

C)

Energy absorbed by substance (joules / gram)

liquid & gas

solid & liquid

Heat of fusion Heat of vaporization

ΔHfus

ΔH(l)

ΔT(l)

ΔHvap

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

Specific heat capacity of the liquid can be

determined from the inverse slope of the curve

40

Reversing a Heat Curve

150 140 130 120 110 100

90 80 70 60 50 40 30 20 10

0 -10 -20 -30 -40 -50

100 200 300 400 500 600 700 800 900 1000 1100 1200 1300 1400 1500 1600

Tem

per

atu

re (

C)

Energy (joules / gram) released

liquid & gas

solid & liquid

Heat of solidification Heat of condensation

ΔHs ΔHcond

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.

41

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.

• Inverse slope of a section of the curve.

42

Heat Curve of a Mixture

150 140 130 120 110 100

90 80 70 60 50 40 30 20 10

0 -10 -20 -30 -40 -50

100 200 300 400 500 600 700 800 900 1000 1100 1200 1300 1400 1500 1600

Tem

per

atu

re (

C)

Energy absorbed by substance (joules / gram)

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.

43

Sample Problem Heat curve of “Imaginarium”

150 140 130 120 110 100

90 80 70 60 50 40 30 20 10

0 -10 -20 -30 -40 -50

100 200 300 400 500 600 700 800 900 1000 1100 1200 1300 1400 1500 1600

Tem

per

atu

re (

C)

Energy absorbed by substance (joules / gram)

Heat of vaporization

Use the heat curve to determine: a) The melting point of imaginarium. b) The boiling point of imaginarium c) The specific heat capacity of liquid imaginarium d) the heat of fusion of imaginarium e) the heat of vaporization of imaginarium

ΔH(l) = (900-500) =400J/g

ΔT(l) = (70-(-20)) =90℃

Heat of fusion

ΔHfus= (500–300)= 200 J/g

Answers: a) –20 C b) +70 C c) 4.44 J/g C d) 200 J/g* e) 400 J/g*

ΔHvap=(1300-900)=400 J/g

*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).

44

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.

45

Energy Released

Energy Absorbed

H

H H

Visualization of a Chemical Change

O O H

H

H H

Reactants

2H2 + O2

atoms

4H + 2O

O

O

H H

H

H

Products

2H2O

O O H

H

O O H

H

H H O

O

H

H

H

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 ② Bond breaking Bond forming

46

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.

47

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)

48

Assignments

• Read sections 4.1 and 4.2

• Do questions 1 to 6 on page 155

49

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

50

Bond Energies Bond Energy

H—H 436 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

Cl--Cl 243 kJ/mol

N==N 418 kJ/mol

N==O 631 kJ/mol

• The text book contains tables of bond energies, such as the one on page 419, or the small one here (from page 156).

• 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 +436 kJ of energy per mole of atoms.

• If you formed new H—H bonds, you would release energy. (–436 kJ/mol)

p. 156

Also see p. 419

51

Error in Textbook Table on page 156

• The textbook gives the value of 418 kJ/mol for the N N triple bond. This is incorrect.

• That is the value for the N=N double bond.

• Please make a note in the margin of your textbook, showing N=N

• The table on page 419 is correct for this bond.

498 kJ/mol

C==O 745 kJ/mol

Cl--Cl 243 kJ/mol

N N 418 kJ/mol

N==O 631 kJ/mol

52

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.

53

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.

1

2

3

4

5 6

54

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.

55

Energy Balance Example Bond Energy

See p. 419

H—H 436 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

negative

positive

4x

2x

4x

2x

56

Draw an Energy Balance Diagram for the previous sample problem

Energy balance diagram should show:

• Enthalpy on y-axis (kJ/mol, but usually not numbered)

• Reaction progress on x axis (in arbitrary time units)

• Formulas of reactants

• Energy from broken bonds (+ ΔH)

• Separate atoms

• Energy from bonds formed(-ΔH)

• Formulas of products

• The heat of reaction (ΔH)

Enth

alp

y (k

J/m

ol)

Reaction Progress

CH4 + 2 O2

+ΔH broken

= 2648 kJ

C + 4 H + 4 O

-ΔH formed

= -3330 kJ

CO2 + 2 H2O ΔH= -682 kJ

1

2

3

4

5

6

57

Thermochemical Equations

• When we add energy information to a chemical equation, it is called a thermo-chemical equation

• There are two main techniques for doing this:

• We can treat the energy like a reactant or product and put it directly into the equation:

• Or, we can record the ΔH value separately at the end of the equation:

Endothermic Reaction: CaCO3(s) CaO(s) + CO2(g)

Exothermic Reaction: 4 Fe(s) + 3 O2 2Fe2O3(s)

Exothermic Reaction: 4 Fe(s) + 3 O2 2Fe2O3(s) Endothermic Reaction: CaCO3(s) CaO(s) + CO2(g)

58

Disagreement with the Textbook’s method.

Okay, here we come to a place where I disagree with the textbook’s method of writing thermo-chemical equations. The textbook wrote the thermo-chemical equation on the previous slide this way:

Exothermic reaction: 4 Fe(s) + 3 O2(g) 2Fe2O3(s) ΔH=-824 kJ/mol Not what I wrote:

Exothermic reaction: 4 Fe(s) + 3 O2(g) 2Fe2O3(s) ΔH=-1648 kJ Now, there is nothing wrong with what they put in the textbook. They have divided

the total enthalpy by the number of moles, but they should only do this IF they state which substance they are referring to.. Its ΔH=-824 kJ/mol of Fe2O3 ! Without stating which substance, how are we to know they aren’t giving it with reference to the number of moles of iron (in which case it would be ΔH=-412 kJ/mol of Fe) , or of oxygen (ΔH=-549.3 kJ/mol of O2)

Overall, I prefer to write my thermo-chemical equations with the total enthalpy change (just kJ) rather than the molar enthalpy change (kJ/mol), because otherwise its just too confusing.

59

Assignments

• Read section 4.3, pp. 156 to 160

• Do Questions on page 160, #1 to 6

60

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

61

Stoichiometry and Thermochemical Equations

62

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= 20g MH2= 2g/mol nH2 = To Find: Energy=

10 mol

2440 kJ

Answer: Burning 20g of hydrogen gas produces 2440 kJ of energy

63

Chapter 5

Graphical Representation of Enthalpy

64

Activated Complexes

Overview

The 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

65

Activated Complex Formation 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

H Cl

H Cl

66

Activation Energy

• The reactant particles must collide with sufficient kinetic energy to form an “ ” if they are to react.

• Without this “ ” 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

67

Energy Diagrams 5.2

• 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.

68

Energy Diagrams vs. Energy Balance Diagrams

In an Energy Diagram:

• The graph is drawn as a , not .

• The vertical axis represents only, not .

• The high point in the graph represents the energy needed to form an , not the total energy needed to

.

69

Energy Diagram for an endothermic reaction

Progress of the Reaction

Pote

nti

al E

ner

gy

Act

ivat

ion

en

ergy

Hea

t o

f R

eact

ion

Ea

∆H

Products

Activated Complex

Reactants

70

Energy Diagram for an exothermic reaction

Progress of the Reaction

Pote

nti

al E

ner

gy

Products

Activated Complex

Reactants Ea

Hea

t o

f R

eact

ion

∆H

71

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

going Diagram from http://www2.ucdsb.on.ca/tiss/stretton/CHEM2/rate03.htm

Sisyphus

72

Slow, Faster and Spontaneous Reactions

Progress of the Reaction

Pote

nti

al E

ner

gy

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.

73

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 electrolysis

2 H2O 2 H2 + O2

• These two reactions are opposites of each other. • The first is exothermic

(it gives off energy as heat and flames)

• its opposite is endothermic

(it absorbs energy, in the form of electricity)

74

Energy Diagram for a reversible reaction

Progress of the Reaction

Pote

nti

al E

ner

gy

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)

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• Read Chapter 5, pp. 171 - 181

• Do the Chapter end questions on pages 183-184

76

Chapter 6

Molar Heat of Reaction

77

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

78

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, and 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.

Where: ΔH = molar heat of reaction, in kilojoules Q = total heat exchanged, converted to kilojoules n = number of moles of substance transformed.

Q

ΔH n

79

Dissolution

• Dissolution occurs when a solute dissolves in a solvent.

• During dissolution…

•

80

Dissolution in Detail

A substance dissolves in two steps. 1. The solute particles are

separated from each other. For ionic compounds, they may actually dissociate. This phase is endothermic. Energy is absorbed from the solvent to break up solute molecules.

2. The solute particles disperse and rearrange themselves in the solvent. This step is exothermic.

H2O H2O

H2O

H2O H2O H2O

H2O

H2O

S

S

S

S

81

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 (exothermic) • This means that the temperature of the solution goes up↑.

• If heat is absorbed, ΔHd is (endothermic) • This means that the temperature of the solution goes down↓.

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Dissociation of Ions

• When ionic compounds dissolve, many of them will also dissociate into individual ions:

• Examples:

NaCl(s) + H2O Na+(aq) + Cl–(aq) + H2O ΔHd= +4.3 kJ/mol

Since the water doesn’t change, we can leave it out of the equation.

CaBr2(s) Ca2+(aq) + 2Br-

(aq) ΔHd= -81 kJ/mol

H2SO4(s) 2H+ (aq) + SO4

2-(aq) ΔHd= -74 “

NaNO3(l) Na+(aq) + NO3

–(aq) ΔHd= +21 “

exothermic

exothermic

endothermic

83

Calculating Heat of Dissolution typical steps and useful formulas

• Dissolve a certain number of moles of solute in sufficient solvent in a calorimeter (if solute is in grams, change it to moles)

• Measure the temperature change • Use the calorimeter formula to find

the total heat exchange (Q in joules). Now mw= the mass of the solvent (ie water) in the calorimeter.

• Convert joules to kilojoules, • Divide the heat (Q in kilojoules) by the

number of moles of solute.

s

ss

M

mn

TcmQ ww

84

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.

Data mKOH = 112.2 g MKOH = 56.1 g/mol mwater = 500 g cwater= 4.184 J/g C Tf= 64.7 C Ti= 12.1 C ΔT= 52.6 C To Find: nKOH Q ΔH

Preliminary Step: MKOH = 39.1 + 16.0 + 1.0 = 56.1 g/mol

= 110.0 kJ = 55.0 kJ/mol

= 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 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 23.7C. 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?

Data: mwater = 100g 100mL@1g/mL

cwater = 4.184 J/(g∙ C) Ti = 23.7 C ΔHd = +21.0 kJ/mol mNaNO3 = 4.25g To find: MNaNO3 n Q Tf

Step 1: Calculate the number of moles of NaNO3

= 85.0 g/mol

Step 2: Calculate the quantity of heat (Q)

Answer: The final Temperature is 21.2 C

= 0.0500 mol

= –1.05 kJ = –1050 J

But, since ΔHd is positive, the reaction is endothermic, which means the water loses heat. Make Q negative!

=21.2 C

86

• Page 195, Questions #1 to 10

87

Heat of Neutralization

• 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.

6.2

88

Neutralization

• A neutralization reaction occurs when an acidic solution and a basic solution are mixed. The general reaction is:

• A more specific example:

+ + a +

+ + +

Nitric Acid Potassium hydroxide

Water Potassium nitrate

89

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.

90

Calorimeters and Neutralization

• One way to find the molar heat of neutralization is to mix the two solutions inside a calorimeter – or a Styrofoam cup calorimeter.

• If the two solutions (acid and base) start out at the same temperature, any increase in temperature must come from the heat of neutralization.

91

Simplifying the Problem

• Simplification #1: For dilute solutions, assume the solution has the same properties as water • Density = 1 g / mL Specific heat = 4.184 J/g°C

• While this is not strictly true, it is close enough to give us a good answer, so long as the amount of water in each solution is much greater than the mass of solute.

• Simplification #2: Assume that the mass of water in the calorimeter is the sum of the two individual solutions: • Mass in calorimeter= mass of acid plus mass of base

• mw = mA + mB

92

Molar Concentration Formula

• Another useful formula that we learned long ago is sometimes needed in molar problems:

• This may also be rearranged:

)(

)/(

L

sLmol

V

nC

C = molar concentration ns = moles of solute V =volume (in litres)

n

C V

R

93

Sample Problem In a calorimeter 100 mL (1.00x102) of a 0.500 mol/L solution of NaOH is neutralized by an equal amount of of 0.500 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.

Data: mw = 100g + 100g = 200g cw = 4.184 J/(g∙ C) Tf = 25.9 C Ti = 22.5 C ΔT = 25.9-22.5 = 3.4 C CNaOH= 0.5 mol/L VNaOH = 0.100 L To find: Q n ΔHn

Step 1: Calculate Q, the quantity of heat transferred Q =mwcwΔT = 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 ∙ VNaOH

= 0.5 mol/L ∙ 0.100 L = 0.05 mol

Answer: the molar heat of neutralization of NaOH is -56.9 kJ/mol

= 0.05 mol

= -56.9 kJ/mol

n C V

94

Trickier Problems

What do you do if the temperature of the two initial solutions is not the same?

Find the combined inital temperature of the two solutions, using the formula:

and then assume they both have this combined initial temperatre

2211

111222

cmcm

TcmTcmT ii

ic

95

Trickier Problems

What if you are asked to find the final temperature instead of the ΔH?

Use the ΔH value given (or looked up in a table) to find Q, then solve for ΔT and use that to find the final temperature.

96

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

Heat of dissolution ΔHd Heat aborbed/released during dissolving

Standard Heat of Formation

Special heat of reaction, representing the enthalpy change in creating one mole of a substance from its elements.

97

• Read Chapter 6

• Page 196, questions 11 to 14

98

Chapter 7

Hess’s Law

99

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.

7.1

100

Reaction Mechanism

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, since the mechanism will have a lower activation energy than the single reaction.

101

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.

102

Comparison of Overall Reaction and Reaction mechanism

• Overall Reaction 2NO + 2 H2 N2 + 2H2O

• Reaction Mechanism 2NO + 2 H2 N2O2 + 2H2 N2O + H2O + 2H2O N2 + 2H2O

Photsynthesis,An extreme example

6CO2 + 6H2O C6H12O6 + 6O2

Overall Reaction Mechanism

2x

2x

1 2 3

103

Photosyntheis: An extreme example of reaction mechanisms

Overall Reaction:

6CO2 + 6H2O C6H12O6 + 6O2 Mechanism:

104

Energy Graph for a Complex Reaction

Step 1 Step 2 Step 3

ΔH1

ΔH2

ΔH3

ΔHTotal

105

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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106

Hess’ Law

The total enthalpy change of a reaction is the sum of the individual enthalpy changes that make up its mechanism.

Where: ΔH = Total enthalpy change of the reaction ΔH1 = Enthalpy change of 1st simple reaction ΔH2 = Enthalpy change of 2nd simple reaction etc…

Germain Henri Hess 1802-1850

Born: August 8, 1802 Geneva, Switzerland Died: November 30, 1850 St. Petersburg, Russia Fields: Chemistry, Medicine Known for: Law of constant heat summation (Hess’ Law)

107

• 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.

108

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

Carbon dioxide CO2(g) C(s) + O2(g) CO2(g) -393.5 kJ/mol

Carbon monoxide CO(g) C(s) + ½ O2(g) CO(g) -110.5 kJ/mol

Butane C4H10(g) 4C(s) + 5 H2(g) C4H10(g) -125.6 kJ/mol

Propane C3H8(g) 3C(s) + 4 H2(g) C3H8(g) -104.7 kJ/mol

Methane CH4(g) C(s) + 2 H2(g) CH4(g) -74.4 kJ/mol

Ozone O3(g) 3/2 O2(g) O3(g) +142.7 kJ/mol

Nitrogen dioxide NO2(g) ½ N2(g) + O2(g) NO2(g) +33.2 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).

109

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.

110

Sample Problem

Find 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

111

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 +...

112

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 coefficients they match exactly!

• In the final step, you may subtract the same amount of a reactant from both sides (like solving an algebra equation)

• 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.

114

Example 2: (see study guide page 47 question 2)

• Find the heat of combustion of methanol (CH3OH) and write its thermo-chemical equation using the following heat of formation data from p.418: • CO2(g) ΔH°f = - 394 kJ/mol

• H2O(l) ΔH°f = - 286 kJ/mol

• CH3OH(l) ΔH°f = - 239 kJ/mol

• Note: sometimes the “phase markers” ie.(s) (l) (g) (aq) are left

out, but be aware that they are important, especially if a material changes state before or after it reacts. Note also, I have rounded all ΔH values to 3 sig.figs. for simplicity. Remember, combustion is a form of rapid oxidation!

115

• The Reactions we know:

• C(s) + O2(g) CO2(g) ΔH= -394 kJ

• H2(g) + ½ O2(g) H2O(l) + ΔH= -286 kJ

• C(s) + 2H2(g) + ½ O2(g) CH3OH(l) ΔH= -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.

116

Solution, as you should show it in your notebooks.

flip

x2

Reactants Products ΔH

C(s) + O2(g) CO2(g) -394 kJ

H2(g) + ½ O2(g) H2O(l) -286 kJ

C(s) + 2H2(g) + ½ O2(g) CH3OH(l) -239 kJ

2

2H2(g) + O2(g)

2 H2O(l)

C(s) + 2H2(g) + ½ O2(g) CH3OH(l)

-572 kJ

CH3OH(l) + __O2 __CO2 + __H2O

CH3OH(l) + 2 O2(g) CO2 + 2 H2O + ½O2

1½

+239 kJ

Target:

-727 kJ

ΔH=-727 kJ

Leave Blank Lines

Leave Blank Lines

Leave Blank Lines

117

• Hess’ Law Assignment Sheet • do the sheet and check your answers before starting

the textbook questions.

• Textbook Page 208, Questions #1 to 6 plus 9 and 11

• Question 1 has the additional complication of not telling you what the “target” equation is. You must find it as you add the equations together.

118