Unit 2: Thermochemistry

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Unit 2 Thermochemistry

• Energy Changes

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ENERGY IN CHEMISTRY

• Energy - What is it?
• Energy is something for which we all have an
  instinctive feel.  We may wake up in the morning
  feeling "energized" and ready to go to work but
  beyond that, the concept of energy is a difficult
  one to grasp for many, because it is a somewhat
  abstract concept. As a result, energy is not
  defined by what it is, but by what it does.
• Energy - The capacity to do work or produce heat
  
• the capacity to move matter.

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1. Potential Energy

• Potential Energy - The ability an object has to
  do work because of its position or condition. 
  (Ep)
• Your lifted object, a compressed spring, a
  stretched elastic band, a drawn longbow, etc. all
  must have work done on them to get them into
  their higher energy condition.  They then, have
  the ability to do work because of their new
  position or condition relative to their prior
  position or condition.  For example,  when work
  is done on an object to lift it above some base
  level, its energy gain is called gravitational
  potential energy and its value can be calculated
  using the formula
• Ep mgh            
• where m mass of object in kilograms
  g acceleration due to gravity (9.81 m/s2 on
  Earth)   h height the object was
  lifted

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2. Kinetic Energy

• Kinetic Energy - The ability an object has to do
  work by virtue of its motion. (Ek)
• If you pick up another object from your desk and
  throw it as hard as you can (please don't really
  do this) you have once again done work on an
  object, but this time that object's ability to do
  work is due to the fact that it is moving. As
  that object moves through the air it has the
  capacity to do work on anything with which it
  collides. The work it can do on that object (its
  Kinetic Energy) is equal to the work done setting
  it in motion in the first place. The formula for
  kinetic energy is
• Ek 1 mv2   Where m mass of the object in
  motion          2                      v
  velocity of the object in motion

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The Laws of Thermodynamics

• There are several basic laws by which the entire
  universe operates
• The Laws of Thermodynamics describe the rules by
  which matter and energy interact in the universe.
  We know that the energy of the universe is
  constant. Whatever quantity of energy the
  universe was born with is what it possesses today
  and is exactly what it will finish with.
• 1. The First Law of Thermodynamics states that
  energy can neither be created nor destroyed, but
  it can be converted from one form to
  another. (Euniverse 0) 
• 2. The Second Law of Thermodynamics states that
  whenever one form of energy is converted to
  another, some of that energy will be donated to
  the universe as heat energy and that heat will
  flow spontaneously from a region of high heat
  content to a region of lower heat content.

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Matter and Kinetic Energy

• Atoms and molecules are little pieces of matter,
  which means that they have mass and occupy space.
  Work is done on atoms and molecules to get them
  moving and once they are in motion they can do
  work, they possess kinetic energy. Here is a
  brief word about how this relates to temperature
  
• Temperature is a relative measure of the average
  kinetic energy of a population of atoms or
  molecules.  Temperature tracks changes in kinetic
  energy by giving us a relative idea of how the
  particles are changing their rate of motion. An
  increasing temperature indicates that particles
  moving faster while a decreasing temperature
  signals that the particles are slowing down.

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Matter and Potential Energy

• A bond is a force of attraction between two
  particles. So, work must be done on a system to
  overcome these attractive forces when
  intermolecular, intramolecular, ionic, or
  metallic bonds are broken. It's really no
  different than stretching an elastic band or a
  spring until they break.  Breaking bonds is an
  energy consuming process. The separated
  components (atoms or ions), are now in a
  different position or condition than they were
  before the bonds were broken, work had to be done
  to get them there. The separated particles then,
  have the potential to return that energy to the
  surroundings if the bonds are allowed to reform.
  In other words, when atoms or ions recombine it
  is an energy giving process.  Atoms and molecules
  in the separated state can possess a greater
  potential energy relative to their combined
  state, potential energy is stored in the bonds of
  chemical systems.

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To conclude

• Matter can possess and store both kinetic and
  potential energy
• a) As matter moves, it can do work on other
  pieces of matter and therefore must possess
  kinetic energy     
• b) As matter rearranges its "bonds" by changing
  its state, or undergoingchemical changes to
  become new and different substances, it can gain
  or lose potential energy

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SYSTEM and SURROUNDINGS

• When energy is transferred, it is important to
  define what is gaining energy and what is losing
  energy.  For the purposes of our study, the
  universe is divided into two parts, the system
  and the surroundings. 
• The system is that part of the universe upon
  which we choose to examine and study in detail.  
  It is usually the reactants and products of a
  chemical reaction, or perhaps a single substance
  as in a phase change like boiling or freezing. It
  is any process that can be represented by a
  balanced chemical equation.
• The surroundings are everything else in the
  universe that is not part of the system.  It
  includes any matter that is in the immediate
  vicinity of the system that does not include the
  reactants and products.  Energy lost by a system
  can be gained by the surroundings or energy lost
  by the surroundings can be gained by a system. It
  works both ways.
• Euniverse Esystem Esurroundings
• 0 Esystem Esurroundings
• Esystem - Esurroundings OR -
  Esystem Esurroundings

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TOTAL INTERNAL ENERGY

• The total internal energy of a system (E) is the
  sum of all possible kinetic and potential
  energies in a system.  Energy is stored in so
  many ways in matter that it is impossible to know
  all of them or their actual values, therefore it
  is impossible to know the absolute internal
  energy of any system. So of what value is this
  concept?  You will see in a moment.

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ENERGY "CHANGE"

• Imagine you are walking down a crowded street.
  How much money does the man walking in front of
  you have in his pocket? Obviously, you don't
  know. But there is a hole in his pocket. Money is
  falling out and you are picking it up. Just
  before you can return the money to him, he jumps
  in a cab and drives away. How much money does he
  have in his pocket now? The answer is the same -
  you don't know. All you can know for certain is
  the amount he has lost, which of course is the
  difference between what he had when you first saw
  him and what he now has as he drives away. You
  can only know by what amount his total cash value
  has changed (?).  Note that " ? " , called a
  delta sign means the change or difference between
  two values.

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• This money analogy is similar to thermochemistry.
  What is important for us is the change in
  internal energy as a system proceeds from
  reactants to products.  If you pick up a piece of
  paper from your desk and consider it to be a
  source of energy for a moment, you would be
  astonished at the magnitude of the total internal
  energy (E) stored within it. But precisely how
  much energy is that? You don't know? No one
  does!  But if you burn the piece of paper, some
  of that energy will emerge in the form of heat. 
  Once the combustion has gone to completion, how
  much energy is now stored in the product
  molecules? Again, you don't know. All you can
  measure is the difference between the energy
  content of the paper before combustion, and the
  energy content of the products after combustion
  is complete.
• By measuring the amount of heat that flows from
  the system to the surroundings, you would know by
  what amount the system has changed its total
  internal energy content (?E).  ?E represents the
  change in the system's total internal energy, and
  it can be measured.

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EXOTHERMIC and ENDOTHERMIC

• The total internal energy of a system is a
  property that can be changed either by a flow of
  energy from the system to the surroundings or by
  a flow of energy from the surroundings to the
  system, as illustrated by the following diagram

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• The amount of energy gained or lost by a system
  is represented by a number which indicates the
  quantity of energy flow, (e.g. 35 kJ) and by a
  sign that indicates the direction of the energy
  flow.
• If energy flows out of a system (an exothermic
  process) the magnitude of that energy is
  accompanied by a negative sign indicating that
  the system's energy is decreasing.
• Similarly, if energy flows into a system (an
  endothermic process), the magnitude of that
  energy is accompanied by a positive sign,
  indicating that the system's energy is
  increasing. 

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ENTHALPY

• Enthalpy (denoted , H) is a property of a
  substance that is related to it's internal
  energy.  Enthalpy is sometimes described as the
  heat content of a substance.  Just as we cannot
  measure the internal energy, we cannot measure
  the absolute value of enthalpy.  However, in
  reactions that take place under constant
  (atmospheric) pressure with little or no change
  in volume, we can make the assumption that the
  change in enthalpy is the same as the change in
  internal energy.  That is,
• ? H (change in heat content)    ? E (change in
  internal energy)
• 1.  If ?H is negative, the system is called
  exothermic 2.  If ?H is positive, the system is
  called endothermic
• ?H is a measure of the actual quantity of heat
  transferred during a reaction.  Change in
  internal energy, change in enthalpy, and heat of
  reaction are terms often used interchangeably. 

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The following diagrams summarize the changes
that occur as exothermic and endothermic systems
proceed

• Reactants ? Products 100 kJ
  Reactants 100 kJ ? Products
• Exothermic (system loses enthalpy)                
          Endothermic (system gains enthalpy)
• Note that the gain or loss of energy is always
  from the system's point of view.

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ENERGY CONVERSIONS

• Exothermic Example
• Imagine filling a container with a potentially
  explosive mixture of methane gas and oxygen gas
  according to the following system
• CH4(g) 2O2(g) ---gt CO2(g) 2H2O(g) 802.3
  kJ
• This system has the potential to do work. The
  reactants can explode and do a considerable
  amount of work on the surroundings. If someone
  lights a match, some of that chemical potential
  energy will be converted to heat which is donated
  to the surroundings in the form of kinetic
  energy, and the surroundings warm up. Chemical
  potential energy of the system has been converted
  to kinetic energy in the surroundings.
• Exothermic systems convert Ep of the system to Ek
  of the surroundings. The surroundings warm up.

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• Endothermic Example
• Now imagine the following system in which an ice
  cube is placed in a glass of water
• H2O(s) 6.03 kJ -------gt H2O(l)
• Notice that in an endothermic reaction, as the
  ice melts, the surrounding water cools down as it
  loses kinetic energy.
• Question Where does the surrounding water's
  kinetic energy go?Answer    It is absorbed by
  the melting ice and used to break the bonds that
  are holding the water molecules together in its
  solid state.
• The water cools down while the ice melts. The
  kinetic energy of the surroundings has been
  converted to potential energy in the melted ice.
• Endothermic systems convert Ek of the
  surroundings to Ep of the system.  The
  surroundings cool down.

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Practice Questions

• 1. Analyze each of the following scenarios, by
  filling in the following table (the first one is
  done for you)
• a) When solid KBr is dissolved in water, the
  solution gets colder.b) Natural gas (CH4(g)) is
  burned in a forced air furnace.c) When
  concentrated H2SO4 is added to water, the
  solution gets very hot.d) A candle burns, while
  the air around it warms up.

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• 2. The equation for the fermentation of glucose
  to alcohol and carbon dioxide is
• C6H12O6(s) -------gt 2C2H5OH(l) 2CO2(g)
• a)  The enthalpy change for the reaction is
  -68.1 kJ. Is the reaction exothermic or
  endothermic?
• b)  Is energy absorbed or released as the
  reaction proceeds?  Explain how you know this.

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Calculating Energy Gain or Loss by Systems

• In this section, you will use all your prior
  knowledge to learn how to mathematically
  calculate and graphically show the ?H values for
  the following changes
•         1.  Kinetic Energy Changes (Warming and
  Cooling)         2.  Potential Energy Changes
  (during phase changes)         3.  Combinations
  of successive Kinetic and Potential
  Energy Changes         4.  Using Heat Loss
  Heat Gained Principles to                        
  a) Calculate Molar Heats of Fusion and
  Vaporization of substances                   
        b) Solve Heat Transfer Problems (kinetic
  energy lost kinetic energy gained)
                          c) Calculate enthalpy
  changes for Chemical Reactions (The art of
  Calorimetry)

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Calculating Kinetic Energy Changes (?Ek) -
Warming and Cooling

• It is possible to calculate the energy gain or
  loss (?H) of any substance as it is warmed or
  cooled through any given temperature change. 
• Remember that any substance that goes through a
  temperature change is changing its kinetic energy
  content only . 
• We will consider the symbols ?H and ?Ek to refer
  the same thing, the enthalpy change (?H) of these
  systems is of the kinetic energy type (?Ek).
• (The textbook also refers to this as q)
• A graph of temperature versus time as heat is
  transferred to a system, called a heating curve,
  can be used to visually show what is occurring
  during a change in enthalpy.

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The enthalpy (or kinetic energy) change, ?H or
(?Ek) of a substance will depend on 3 factors

• 1.  The mass of the substance changing
  temperature (m) or the volume of the substance
  (v)
• 2.  The temperature change through which the
  substance passes. (? t)
• 3.  The ability of that substance to absorb
  heat. (c)  This constant is known as the specific
  heat capacity.
• The enthalpy change of a substance (?H) that
  undergoes a kinetic energy change is calculated
  using
•                              ?Ek ?H mc?t
  vc?t
• This equation is always used when calculating
  the kinetic energy change of a substance as it
  changes temperature.

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A few words about specific heat capacity

• Specific heat capacity is a measure of the amount
  of energy required to raise the temperature of
  1.00 g of a substance by 1.00C.  It is also of
  course, a measure of the amount of heat released
  when 1.00 g of a substance cools by 1.00C. 
  Therefore, the larger the "c" value for any given
  substance, the better it is at storing, retaining
  and later releasing heat.  Liquid water's
  specific heat capacity is 4.19J/gC.  That is,
  it requires 4.19 J of energy to raise the
  temperature of one gram of water by 1.00C.  This
  is one of the higher specific heat capacities in
  nature.
• There is an alternative set of units that can be
  used when expressing specific heat capacity.  For
  example, if you wanted to raise the temperature
  of 1000 g of water by one degree, it would
  require 4190 J of energy.  Written in shorter
  form you could say that the specific heat
  capacity of liquid water is 4190 J/1000gC, or
  more properly, as 4.19kJ/kgC.
• Every substance has its own unique value for "c"
  for each state (solid, liquid, or gas).  See
  page 4 of your data booklet for the specific heat
  capacities of ice and water vapor.  The specific
  heat capacities of all the elements in the state
  at which they exist at room temperature are
  listed on the back cover of your textbook. 
• The values for selected elements and compounds
  are found on pages 4 and 5 of your data booklet. 

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Example 1

• What amount of energy must be applied to 50.0 g
  of solid silver in order to increase its
  temperature from 25.0C to 300C?  This is a
  kinetic energy change so....
•  Solution Heating Curve
• Organizing the given data produces
• ?Ek    ? m     50.0 g Ag(s) ?t    275C
  cAg  0.237 J/gC  (from the back cover of your
  textbook)   
• Substituting into ?Ek mc?t produces
• ?H    50.0 g x 275C  x  0.237 J/gC        
    3.26 x 103 J  of energy absorbed by the silver
• This problem could also have been solved using
  the alternative units, that is kJ/kgC.  But you
  must be sure that all your units are consistent
  with the "c units" you choose to use.

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Example 2

• 4000 J of energy were required to heat a sample
  of solid chromium  from 12.0C to 500.0C.  What
  mass of chromium was heated?  First, you must
  recognize this as a kinetic energy change, which
  requires the use of the formula
• Solution Heating Curve
•     Organizing the given data produces
• ?Ek    4000 J m    ?  ?t    (500.0 -
  12.0)C    488.0C  cCr    0.448 J/gC   
• Transposing  ?Ek mc?t in terms of mass (m)
  produces
• m    ?Ek          c?t   
• Substituting
• m                4000 J                    
  488.0C x  0.448 J/gC
• m    18.3 g of chromium

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Practice Questions

• Include a heating curve for each question.
• How much heat is necessary to raise the
  temperature of 150 mL of water by 2.5C? 
  (Answer  1.6 kJ) Note  Water is generally
  considered to have a density of 1.00 g/mL.
• How much energy is released to the surroundings
  if 140 mL of water at 82.0C is cooled to 21.0C?
      (Answer  -35.8 kJ)
• If 0.500 kg of ethanol is heated from 25.0C to
• 26.3C , calculate the heat absorbed. The
  specific     heat capacity of ethanol is 2.45
  J/gC.  
• (Answer  1.59 kJ)
• Textbook p.288 7-11 (HW check)
• Worksheet 43 (fax in)

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Calculating Potential Energy Changes - During
Phase Changes

• The amount of energy absorbed or released during
  a phase change can also be calculated.  Systems
  that gain or lose energy during a phase change
  are undergoing a potential energy change.  The
  formula to be used for the melting or freezing
  process is
• ?Ep (moles of substance changing phase) x
  (molar heat of fusion for that substance)
• ?Ep ?H nHfusion
• The molar heats of fusion have been determined
  for pure substances and are published in your
  data booklet (some are also on page 293 of your
  textbook).  For example, the molar heat of fusion
  for water is 6.03 kJ/mol, this is the amount of
  heat needed to melt one mole of ice at at 0C.

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Example 1

• How much heat is necessary to melt 25.5 g of ice
  at 0C to liquid water at 0C?
• Since   ?H nHfusion  and n m/M
                       25.5 g H2O   ?       1 mol
  H2O     ?    6.03 kJ                            
                               18.02 g H2O
         1  mol H2O                      8.53
  kJ
• Heating Curve (include the states of the
  compound)

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Similarly, the formula used for the vaporization
or condensation process is

• ?Ep (moles of substance changing phase) x
  (molar heat of vaporization for that substance)
• ?Ep ?H nHvaporization
• The molar heats of vaporization for many
  substances have also been determined and
  published.  For example, the molar heat of
  vaporization for water  is 40.8 kJ/mol.  This is
  the amount of heat needed to vaporize one mole of
  liquid water at at 100C.

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Example 2

• How much energy is released when 56.5 g of water
  vapour at 100C condenses to liquid water at
  100C?
• Since ?H    nHvaporization    
• ?H    56.5 g H2O(g)    ?     1 mol
  H2O(g)      ?       - 40.8 kJ                    
                                
   18.02 g H2O(g)          1 mol H2O(g)
                - 128 kJ
• Heating Curve

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Practice Questions

• Determine the amount of heat lost by the
  surroundings in order to melt 50.0 g of ice at
  0.00C.  (Answer 16.7 kJ lost)
• Textbook p.295 14-17 (HW check)
• Worksheet 44 (fax in)

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Successive Potential and Kinetic Energy Changes 

• Many systems will involve a succession of kinetic
  and potential energy changes as a substance
  warms, changes phase, and warms again, perhaps
  through as many as five different steps.   Using
  a warming/cooling curve as a guide, individual
  calculations can be made for each separate step
  in a warming or cooling process. Using water as
  an example, a brief description of how to make
  these calculations follows
• As water is warmed from below its melting point
  to above its boiling point, it passes through
  three kinetic energy changes and two potential
  energy changes.  So, the energy change for each
  separate step must be calculated using the
  appropriate equations
•                 Kinetic energy changes are
  calculated using ?H mc?t                
  Potential energy changes are calculated using
                      ?H nHfusion for melting
                      ?H nHvaporization for
  vaporizing

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Example 1

• Calculate the heat required to change 90.0 g of
  ice at -30C into steam at 150C.
• Since ?Htotal Ek1 Ep1 Ek2 Ep2 Ek3 we
  can write that
•            ?Htotal mc?tice nHfusion ice
  mc?twater nHvap water mc?tsteam
• Substituting     a) Warming the solid water
  from -30C to 0C. This is a Ek change. (?H
  mc?tice)                                     ?H
  0.0900 kg ? 2.01kJ  ? 30.0C  5.42 kJ
                                                   
                     kgC
•     b) Melting the solid. This is an Ep change.
  (?H nice Hfusion of ice )                      
                 ?H 90.0 g ice  ?   1 mol ice    
  ?     6.03kJ    30.1 kJ                       
                                               
  18.02 g ice          1 mol ice
•     c) Warming the liquid water from 0C to
  100C. This is a Ek change. (?H mc?twater)
                                      ?H 0.0900
  kg ? 4.19 kJ  ? 100C  37.7 kJ
                                                   
                       kgC     d) Boiling the
  liquid. This is an Ep change. (?H nwater
  Hvaporization)                                   
    ?H 90.0 g H2O   ?    1 mol water   ?    40.8
  kJ    203.7 kJ                                 
                                           18.02 g
  water        1 mol water
•     e) Warming the steam from 100.0C to 150.0C.
  This is an Ek change.(?H mc?tsteam)
                                          ?H
  0.0900 kg ?   2.01 kJ  ?    50.0C  9.04 kJ
                                                   
                            kgC
•                                                   
                            Total Heat required
  _____?____kJ

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When solving problems that involve both kinetic
energy changes and phase potential energy changes
you should always follow these steps

• 1. Always sketch a warming/cooling curve of the
  system before you do the calculations, as this
  will allow you to determine the number of steps
  through which the warming/cooling substance will
  pass.  Place the melting and boiling points on
  the graph as well as the phase at each point.
• 2. Use the graph to determine how many, and what
  type of energy changes the substance will pass
  through during the warming/cooling process.
• 3. Look up the necessary constants (e.g. heat
  capacity (c) , and the molar heats of fusion and
  vaporization).
• 4. Set up the formula and perform the
  calculations.  Make sure your units are
  consistent.

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Example 2

• Determine the heat needed to warm a 197 g of Gold
  (Au), from 300C to 2000C.
• Sketch the graph including all the relevant
  temperatures and constants.
• Hfusion 12.55 kJ/mol
• Hvaporization 310.9 kJ/mol
• Specific heat capacity of Au(s)   0.129
  kJ/kgC
• Specific heat capacity of Au(l)     0.256
  kJ/kgC
• The graph should look like this

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• The gold is going through 3 stages, two kinetic
  energy changes and one potential energy change
  so
• Set up the formula and perform the calculations
  ?Htotal mc?tgold(s) nHfusion mc?tgold(l)
• All you need to do now is plug in the appropriate
  numbers and perform the calculations. 
•              You should get an answer of  79.2
  kJ.

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Practice Questions

• Find the heat required to melt 40.0 g of ice at
  0.0?C and raise the temperature of the melted ice
  to 50.0 ?C. (Answer 21.8 kJ)
• Textbook p.297 21, 22(a,b) (HW check)
• Worksheet 45 (fax in)

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Heat Lost Heat Gained

• In the questions you have done so far, you have
  only been concerned with the amount of energy
  gained by a substance or lost by a substance. 
  You have not had to consider where the gained
  energy came from or where the lost energy went. 
  The following pages will ask you to consider what
  happens to both the system and the surroundings
  as a change occurs.  Energy lost by one will be
  gained by the other, they are equal as you will
  soon see.
• If you recall, the first Law of Thermodynamics
  states that energy can neither be created nor
  destroyed.  So energy lost by an exothermic
  system for example, cannot just disappear. It has
  to be absorbed somewhere else in the universe,
  and that "somewhere" is usually the surroundings.
  Equating the amount of energy lost by one part of
  the universe with the amount of energy gained by
  another part of the universe is the key to
  successfully performing a variety of energy
  calculations.
• ?Hlost by system ?Hgained by surroundings
• OR
• ?Hlost by surroundings ?Hgained by system

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Calculating Molar Heats of Fusion or Vaporization

• Although molar heats of fusion and vaporization
  are known for pure substances, it is possible to
  determine them experimentally using basic
  thermochemical principles. If a pure substance
  simply changes phase without any further change
  in temperature, calculations are quite simple.
  However, if a substance changes phase and then
  undergoes a further cooling or warming after the
  phase change is complete, the calculation becomes
  a little more complicated.  Again, it is useful
  to sketch a warming or cooling curve including
  data for both the system and the surroundings.

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Example

• A 70.0 g sample of solid cesium at its melting
  point (28.0C) is sealed in a glass vial and
  lowered into 250 mL of water at 90.00C. Some
  time after the cesium had melted, the final
  temperature of the water and liquid cesium was
  found to be   78.98C. Determine the molar heat
  of melting (fusion) for cesium based on these
  data.  (Note Cesium will react violently with
  water, so it must be in a separate container or
  vial, for this experiment)
• 1.  Draw the graph illustrating the changes to
  both the system (Cs in the vial) and surroundings
  (H2O in the container)

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• 2.  Write the formula    
• The cesium (or system) goes through two changes. 
  First it melts (?Ep) and then it warms (?Ek) to
  achieve thermal equilibrium with the water.  
• The liquid water (surroundings) cools (?Ek).  
• The first step is to equate the system's gain of
  energy to the surrounding's loss of energy
• ?Hlost by water ?Hgained by Cs(s) in melting
  ?Hgained by Cs(l) in warming
• or...  mc?twater n(Cs) Hfusion Cs
  mc?t(liquid Cs)
• Rearranging the formula in terms of Hfusion
  (requires a little algebra skill) gives
• Substituting the given values into the equation
  produces
• ?Hfusion(Cs) 20.3 kJ/mol

43
A few things of which to take note

• Treat temperature change (?t) as an absolute
  value. In other words, temperature change
  shouldn't be negative.
• Keep all your values positive and put a sign in
  only at your final answer. You should be able to
  tell whether the heat change is exothermic (-
  sign) or endothermic ( sign).
• Molar heats should be expressed per mole so
  kJ/mol rather than just kJ.

44
Heat Lost Heat Gained continued... Ek lost
Ek gained

• Heat Transfer
• Sometimes only kinetic energy is involved in an
  energy transfer.  The concept of heat transfer,
  wherein the kinetic energy of a warm body is
  transferred to a cooler body,  (Ek lost Ek
  gained) can be used to solve a number of energy
  related problems.

45
Example 1

• Silver at 150.0C is dropped into 500.0 mL of
  water at 5.00C. The final temperature of the
  silver and water at thermal equilibrium  is
  35.0C. Notice that only kinetic energy is
  exchanged.  What is the mass of the silver?   cAg
  0.237 J/gC.
• 1.  The graph

46

• 2.  The equation  Since the silver loses kinetic
  energy and the water gains it...
• ?Hlost by Ag ?Hgained by water
•   mc?t(Ag(s)) mc?t(liquid water)
• 3.  The calculations
• mAg  ?  0.237 kJ/kgC   ?  (150.0 - 35.0)C 
    0.500 kg  ?  4.19 kJ/kgC   ? (35.0 - 5.00)C
• mAg    62.85 kJ        
• 27.255 kJ/kg
• mAg 2.31 kg

47
Example 2

• This is an example of a problem with which
  students often have difficulty.  It asks for the
  final temperature that two substances will arrive
  at when they reach thermal equilibrium.  The
  problem itself is not that difficult but it is
  important to set the problem up correctly.
• Find the final temperature of the system prepared
  by placing 100.0 g of hot copper metal originally
  at 98.8C into 100.00 g of water at 25.0C. 
  Assume that no heat escapes to the surroundings. 
  Obtain any necessary constants from your data
  booklet.
• 1.  The Graph
• The ?t for both the cooling copper and the
  warming water must be expressed as positive
  values, therefore
• The change in temperature (?t) for the cooling
  copper will be  (98.8 - tf)C. The change in
  temperature for the warming water will be (tf -
  25.0)C.

48

• 2.  The equation  Since the copper loses kinetic
  energy and the water gains it...
• ?Hlost by Cu ?Hgained by water
• mc?t(Cu) mc?t(liquid water)
• 3.  The calculations  Substitute the appropriate
  values into the heat loss heat gain equation
• copper cooling               water warming
• (0.1000 kg)(0.385 kJ/kgC)(98.8C - tf) 
    (0.10000 kg)(4.19 kJ/kgC)(tf - 25.0C)
• Perform the calculations and see if you can get
  31.2C as your final answer.

49
Practice Questions

• Water at 50.0C was cooled to 0C by adding 18.02
  g of ice at 0C. Find the mass of the water
  cooled. (Assume the ice just melts) (Answer
  28.8 g)
• The same quantity of heat that was lost by 13.71
  kg of iron as it cooled from 240.0C to 100.0C 
  was added to 1.00 kg of steam at 110.0C.
• a)  What temperature change was experienced by
  the steam? (Answer ?t(steam) 424C)
•     b)  What was the maximum temperature reached
  by the steam? 
• (Answer 534C)

50
Chemical ChangesThe Art of Calorimetry

• It is possible to experimentally measure the heat
  that flows into or out of any chemical system.
  The process, known as calorimetry, is the science
  of measuring heat. 
• Now here is the tricky part... it is not possible
  to directly measure the amount of potential
  energy lost or gained by a system.  However,
  since the heat lost by a system goes into its
  surroundings, all we need to do is design an
  apparatus that will absorb and trap the heat
  gained by the surroundings and hold it long
  enough for us to measure it.  We can then assume
  that the kinetic energy (heat) gained by the
  surroundings is equal to the potential energy
  lost by the system. 
• To say it another way, by measuring the
  surroundings' gain of energy, we know the
  system's loss. Such an apparatus is known as a
  calorimeter.
• A calorimeter is designed to permit a change to
  occur either in or near liquid water.  Liquid
  water is a superb heat sponge due to its high
  specific heat capacity, it will absorb and retain
  great quantities of energy with little change in
  temperature. It is also readily available,
  nontoxic and economical.  There are basically two
  kinds of calorimeters

51
1. The Conventional Calorimeter

• Although some are very sophisticated in design,
  the one you see here is typical of those used in
  a high school laboratory.  It is simple,
  effective and inexpensive.  It is designed for
  calorimetric studies in systems that take place
  in the calorimeter water.
• The nested styrofoam cups and cork lid prevent
  heat loss or gain from the greater universe and
  the copper wire is used for stirring the contents
  to ensure uniform temperature throughout as the
  reaction proceeds.

52
2. The Bomb Calorimeter 

• Illustrated beside, these are designed to study
  changes that do not take place in water. 
  Combustion reactions for example simply cannot
  take place underwater for obvious reasons.  The
  idea here is to have the reaction (usually an
  explosion of a gas or finely divided sample of a
  solid) take place in a steel container near water
  (usually submerged in it), so that the heat can
  flow from the reaction vessel to the calorimeter
  water and be measured there.
• watch Bomb Calorimeter Video (from Learn Alberta
  site) and complete the handout.

53
Calculating Enthalpy Changes

• For an exothermic system, the potential energy
  change that occurs as the system proceeds can be
  summarized as follows
• Potential Energy lost by a system Kinetic
  energy gained by surroundings     
  ?Ep lost by a system ?Ek (Heat) gained by water
  in a calorimeter
• Since ?Ep nH  and  ?Ek  mc?t,  substitution
  into the equality above produces the equation
• nH mc?t
• Where n moles of substance in the system H
  enthalpy change of the system m mass of
  water in the calorimeter (in g or kg) ?t
  temperature change of the the calorimeter water
  (surroundings) c specific heat capacity of
  calorimeter water, (either 4.19 J/gC or 4.19
  kJ/kgC)

54
Example 1

• When 2.5 g of KOH(s) dissolves in 125 mL of
  water, the temperature increases by 1.8C.  Find
  the molar heat of dissociation (also known as
  molar heat of solution) of KOH(aq).
• The system is   KOH(s) -------gt K(aq) OH-(aq)
  The surroundings are the calorimeter water in
  which the KOH dissociated.  It's now a solution.
  Since the water (surroundings) warmed up, we know
  that the system is exothermic, therefore
• Potential energy lost by the system Kinetic
  energy gained by the calorimeter water
• n(KOH)?H mc?t ?H 
  mc?t              n
• Important!!  When the calorimeter water is turned
  into a solution as a result of the action of a
  chemical system, we assume that the specific heat
  capacity and the density of the resulting
  solution is the same as for pure water, since all
  solutions are mostly water anyway.  So don't go
  looking for the specific heat capacity of KOH or
  any other solute, as these solutions are mostly
  water!  Use 4.19 J/gC, or  4.19 kJ/kgC  for
  "c" and 1.00 g/mL for density as you have in the
  past.  So..
• Substituting
• ?H 0.125 kg  x  4.19 kJ/kgC  x
  1.8C               2.5 g KOH  x  1 mol
  KOH                                      56.11 g
  KOH
• ?H  21 kJ/mol released or - 21 kJ/mol

55
Example 2

• 25.0 g of molten aluminum at its melting point
  were dropped into 200 mL of water in a
  calorimeter. When the aluminum has just
  solidified, the water had increased in
  temperature by 15.0C. Calculate the molar heat
  of solidification of aluminum.
• The system is   Al(l) ---------gt Al(s)The
  surroundings are the water in which the aluminum
  solidified. Since the surroundings increased in
  temperature, it must have gained the heat that
  the reacting system lost so the system is
  exothermic
• Potential energy lost by the system Kinetic
  energy gained by the calorimeter water n(Al)?H
  mc?t ?H  mc?t              n(Al)
• Substituting
• ?H 0.200 kg x 4.19 kJ/kgC x
  15.0C            25.0 g Al x 1 mol Al
                               26.98 g Al
• ?H 13.6 kJ/mol released or - 13.6 kJ/mol

56
Example 3

• If 100 mL of 0.500 mol/L HCl(aq) and 100 mL of
  0.500 mol/L KOH(aq) are mixed in a calorimeter,
  the temperature of the total solution increases
  from 25.0?C to 55.0 ?C.  Calculate the heat of
  neutralization per mole of HCl(aq) for this
  system.
• The system is  HCl(aq) KOH(aq) -------gt
  KCl(aq) HOH(l), a neutralization reaction
• The surroundings are  the combined calorimeter
  water in which this reaction is taking place, 
  (200 mL of combined solution).  Since the
  surroundings increased in temperature, it must
  have gained the heat that the reacting system
  lost, so the system must be exothermic
• Potential energy lost by the system Kinetic
  energy gained by the calorimeter water
• n(HCl) ?H mc?t?H  mc?t              n(HCl)
• Subsituting
• ?H   0.200 kg x 4.19 kJ/kgC x
  (55.0-25.0)C                               0.500
  mol HCl  x  0.100 L                             
                   L
• ?H   - 503 kJ/mol

57
Practice Questions

• 1. To raise the temperature of a calorimeter and
  its contents by 1C requires 5028 J. When 0.50
  mol of fuel is burned in the calorimeter the
  temperature increases by 4.0C. Using these data,
  the molar heat of combustion of the fuel is
• A. - 1.0 x 102 kJ/mol B. - 60 kJ/mol C. - 40
  kJ/mol D. - 20 kJ/mol (Answer C)
• 2.   When one mole of methane is burned 4000.0 g
  of water are heated by 52.7C. Determine the
  amount of heat involved. (Answer - 883 kJ)
• 3.   A new sugar was synthesized for use in space
  travel. Its molar mass is 2.50 x 102 g/mol. When
  1.00 g of the sugar was burned in a calorimeter,
  the temperature of 1.00 x 102 g of water
  increased by 30.6C.  Calculate the molar heat of
  combustion for this new sugar. (Answer - 3.21
  MJ)
• Watch Calorimetry video (from Learn Alberta site)
  and complete the handout.
• Textbook p.303-8 27,29,32,33 (HW check)
• Worksheet 46 (fax in)

58
Energy Changes

• This covers Chapter 10 in the textbook.
• Assignments
• Worksheet 43-46
• Extra Practice Questions
• Ch.10 Review p.310
• Ch.10 Pre-test
• The Key