Bio on

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Lecture 51
Bio on Iditarod Champion Susan Butcher
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Susan Butcher
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I think I had to learn at about fifteen that I
was going to have to set my own path."
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• 1,161 miles

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• In 1985, I was traveling alone at night in the
  lead of the race and ran into an obviously crazed
  moose. She was
• starving to death. There was something wrong
  with her. She was just skin and bones. And rather
  than run
• away, she turned to charge the team. I thought
  she would just run through me. I stopped the
  team, threw the
• sled over. She had plenty of room to pass us
  along the trail. She came into the team and
  stopped. She just
• started stomping and kicking the dogs. She
  charged at me. for twenty minutes, I held her off
  with my ax

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Susan Butcher broke by an amazing 31 hours
Gentle but strong
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By Eddie P.
thermodynamics
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Read abstract from Lesson 31 written notes now
and read the complete document for homework

• Do handout 31

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Quiz on limiting reagents

• Get a piece of paper out

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Explosion.. The release of energy
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Last time we showed howthe energy of a system is
equal to the heat (q) work (w).E q w
We also learned that the amount of energy E can
never be know only the change in energy ?E
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Lastly we learned that because work pressure x
volume (wp?v)and because we deal in an open
system where you can not measure pressure or
volume we really can not measure work as Eq
wbut instead soE q
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E q so the energy of the systems we will deal
with in this class is not Classic E but a
watered down special Energy that doesn't
involve work (typically a very a small factor)
called Enthalpy ?H
What do you mean, because it is a open system
you are going to calculate the heat of a system
and not take into account work
We use enthalpy ?H and not energy because we are
in an open system and are not involving work
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Some reactions give energy out of a system and
some actually suck energy into a system

• Remember, we dont know how much energy is
  actually in the system but we can calculate the
  change in enthalpy or ?H

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The change in enthalpy is equal to ?H Hfinal
- Hinitial
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?H HProducts H reactants

• Makes sense

Walk into the restaurant with 100 dollars walk
out with 70 dollars What is your ?
Temperature went from 27 degrees C to 24 degrees
what is ?T
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Obviously the ?H for the entire cosmos will never
change because of the law of conservation of
energy. However in a particular chemical
reaction or system, energy will be lost or gained
depending on the change of energy of the
products vs the reactants.
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The change in enthalpy that accompanies a
particular chemical reaction is called the
enthalpy of reaction (? Hrxn).
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Energy is measured in joules sometimes calories

• one Calorie is equal to 4.1 joules. the amount
  of energy to raise one gram on degree centigrade
  is 4.1 joules
• One calorie is the amount of heat necessary to
  raise one gram one degree

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• A joule J is a very small number
• The energy to bring a cheeseburger to your mouth
  is a joule

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Joules are typically written in kJ or kilojoules

• 1 Calorie 4.18 joule

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How do we measure heat?

• With a thermometer invented by a guy named
  Fahrenheit

How does a thermometer work
Gases expand, so does mercury
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Fahrenheit used a scale that went from 32 degrees
freezing to 212 boiling

• A different scale was proposed by a guy named
  Celsius that was labeled with 0 being freezing
  and 100 being boiling.

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Because there was 100 increments between
freezing and boiling it is refereed to the
centigrade scale.
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Different substances have different capacities
to ABSORB HEAT

• Some things get hot fast
• Others things take for ever to warm up

The tendency for a material to absorb heat is
known as a materials heat capacity.
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Heat capacityis the amount of heat required to
change a substances temperature by one degree.
The heat capacity is an extensive variable and
so the quantity of the substance must be stated
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Heat capacity for grams is specific heat
capacityfor Moles is molar heat capacity
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Specific heat of water is 4.18 J/Co g Specific
heat of aluminum is .89 J/Co g This means that
it takes 4.18 joules to raise one gram of water
1Co and .89 joules to raise on gram of
aluminum 1Co
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Things with low specific heats get hotter faster
Wood or Metal
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The reason why water takes so much energy to get
the molecules moving is those wild hydrogen bonds

• Hydrogen bonds act like shock absorbers and make
  it so it takes more energy for the water molecule
  to start moving.
• But once they start moving they keep going
• Hot water bottle

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Mass vs Number of particles

• Molar heat capacity mole
• Specific Heat capacity gram

What is the difference???
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We are going to use specific heat capacity that
deals in grams

• Molar heat does enable us to make one interesting
  observation

Which break would send the balls flying
faster The full break or The two ball break?
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Dulong-Petit Law is approximately true it says
that it takes 25 Joules to raise 1 mole of atoms
1degree K
Specific Heat x molar mass 25jmol C
Molar heat capacity for Ag is 25.3 Molar
heat capacity for Al is 24.2 Molar heat
capacity for NaCl is 50.5 Molar heat
capacity for BaCl2 is 75.1
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Heat transference is dependent on three factors

• a)     THE CAPACITY OF A SUBSTANCE TO ABSORB HEAT
• b)    THE MASS OF THE SUBSTANCE
• c)     THE CHANGE OF THE TEMPERATURE OF THE MASS

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To see how the temperature of an object will
change with applied heat you use the equation

• q s m ?T

q is the heat transferred in calories  s is the
specific heat in CALORIES  m is the mass in
grams  ?T is the temperature in Co    
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How much energy is required to heat a nail with a
mass of 7 g from 25 C until it becomes red hot
at 750 degrees the specific heat of iron is .45
J/gCo
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If 5750 J of energy are added to a 455g piece of
24Co granite (sg of .8) what is the final
temperature of the granite
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30 g sample of a unknown metal is heated form 22
C to 59.2 during the process 1000 joules of
heat is absorbed by the metal. What is the
specific heat of the metal
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• Using our nifty equation no matter what the
  substance just insert the number of joules of
  heat that enters a system plug in the mass of the
  substance and its specific gravity whippo sappo
  we find the new and altered temperature.

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Tanks hidden in the sand during the Iraq war
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Heat of reaction

• Heat of Reaction is the general term denoting the
  amount of energy that is gained or lost during a
  chemical process. It is expressed by the symbol
  ?H rxn.
• If the sign of the term is negative , then the
  process is exothermic. If the sign of the term is
  positive, then the process is endothermic.

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These are called a bomb calorimeter because they
have been known to explode
calorimeter
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You calculate how much the water changes temp ?T
you know the mass of water and you know the
specific heat of water is and calculate the q
q s (m) ?T
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Lavoisier wrote how he, in a special chamber,
ignited Hydrogen and oxygen and 259 pounds of ice
melted from the heat.

• One pound of Charcoal melts 96pounds 8oz of
  ice
• One pound of phosphorus melts 100 pounds of ice

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A typical reaction that is exothermic and puts
out heat is C(s) O2 (g) ?CO2 (g) ? H 394 kJ

Here we see that the product has lost 394 kJ in
the formation of CO2. It is said that the heat
of formation of CO2 is 394 kJ or ? Hf 394 kJ
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exothermic
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endothermic
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What about the heat of formation of water where
the equation is

• 2H2 O2 ? 2H2O ? Hf -483.6 kJ

This gives us the heat of formation of How many
H2O?
2
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Heat of Reaction is for one mole of product

• 2H2(g) O2(g) ? 2 H2O(l) H? - 572 kJ
  (Exothermic)
• H2(g) ½ O2(g) ? H2O(l) H? - 286 kJ
  (Exothermic

Always write a thermochemical reaction in
regards to 1 mole
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The Heat of Reaction is broken down into two
different classes of reactions

• Heat of Formation ?Hf going from most basic
  into 1 mole of a compound
• 2. Heat of Combustion ? Hc going from most basic
  plus oxygen gas into one mole of compound

Hf
?Hf
?Hc
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Example 1When 1 mole of CH4 is burned,
(enthalpy, H) 890 kJ of energy is released as
heat. Calculate ?H for a process in which a 5.8
g sample of CH4 is burned.
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Example 2
C O2 ? CO2 ?H -393.5 kJ

• The above equation indicates how many kJ of
  enthalpy (heat) is released for every mole of
  carbon dioxide produced from Carbon and oxygen
  gas. If we need to heat a fire alarm in order to
  warn the world that Lex Luthor (arch enemy of
  Superman) is on the loose, how many grams of
  carbon do we need to burn to activate the 1000 kJ
  alarm.

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There are Tables that have the heats of formation
on them. They are

• Compound Heat of Formation, kJ/mole  
• LiF -612
• LiCl -409
• LiBr -350
• LiI -271
• NaF -569
• NaCl -411
• NaBr -360
• NaI -288

2Li F2 ? 2LiF Li ½ F2 ?LiF ?H -612
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Reversing a chemical reaction results in the same
magnitude of enthalpy but of the opposite sign.
C O2 ? CO2 ?H -393.5 kJ
CO2 ? C O2 ?H 393.5 kJ
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It is important to be able to calculate the
amount of heat that is available in a chemical
reaction so you can predict how much energy will
be released during the experiment you can also
get an idea of how much energy is being locked in
an compound ie food
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Books and catalogues have heats of formation for
many standard chemical equationsbut not all

• C ½ O2 ? CO ? H -100.5 kJ
• Elements ? compounds
• S O2 ? SO2 ? H -297kJ

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These books and charts in addition to giving
heats of formation also give heats of combustion.
These are written ?Hcomb

• The ?Hcomb is the heat of combustion of one mole
  of a reactant burned with oxygen

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What if you wanted to calculate the heat of
formation for a specific equation that was not in
a chart or table?

• How could you calculate a enthalpy from known
  values????????/

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• All you know is that
• The price of a coke and fries is 1.25
• The price of a hot dog and fries is 1.50
• The price of a coke and hot dog is 1.75
• Can you determine the price of fries????

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• C F 1.25
• H F 1.50
• C F H F 2.75
• We know that C H 1.75 so if we take this out
  of the equation then and 2F 1.00

So we see that by combining known facts we can
by adding and subtracting factors determine a
unknown value.
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Can we really add chemical equations the same way
you would find the cost of French fries from the
known value of coke and fries and the value of
hot dog and fries

• Hesss Law

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This is the essence and importance of the state
function

• It doesn't matter what turns and shifts you take
  its where you end up.

I was up then down now I am down 10 dollars
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We know that when buying a hot dog the math works
but what about adding chemical equations does the
math hold true?

• Victor Franz Hess (1883-1964), Austrian-American
  Nobel laureate, studied this field

Hess determined this was also the case for
chemical reactions. It is Hesss law.
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Hesss law states that the overall heat of a
reaction is equal to the sum of the overall steps
in the reaction.
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To find the value of the unknown heat of
formation you calculate the heat of the known
reactions either formation or combustion
In order to do this problem we need to build the
unknown equation using the known!!!!

• Sn 2Cl2 ? SnCl4 ? Hf ?

We Know Sn Cl2 ? SnCl2 ? Hf -324
kJ SnCl2 Cl2 ? SnCl4 ? Hf -186 kJ
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Note N2 O2 Cl2 H2 C S Naare
considered elements And do not have a enthalpy
change ? Hf
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To find Sn 2Cl2 ? SnCl4 ? Hf ?We add the
two known parts

• Sn Cl2 ? SnCl2 ? Hf -324 kJ
• SnCl2 Cl2 ? SnCl4 ? Hf -186 kJ

Sn Cl2 SnCl2 Cl2 ? SnCl2 SnCl4
Sn 2Cl2 ? SnCl4 ? Hf -510 kJ
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The idea that you can add or subtract chemical
equations to calculate unknown values is called
Hesss Law

• When calculating the unknown value you must
  account for each element or compound

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Methane Oxygen gas yields how much energy

• CH4 O2 ? ? Hf ?

Balance the equation
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CH4 2O2 ? CO2 2H2O

• You must consider each item
• CH4 formation
• 2O2 no change
• CO2 formation
• 2H2O formation

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according to chart
The equation we are trying to build CH4 2O2 ?
CO2 2H2O

• C 2H2 ?CH4 ? Hf -74.8 kJ
• Flip the equation change the charge
• CH4 ? C 2H2 ? H 74.8

2O2 ? 2O2 ? Hf 0 you cant burn O2
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C O2 ? CO2 ? Hf -393.5kJ
The equation we are trying to build CH4 2O2 ?
CO2 2H2O

• We need 2 H2O
• H2 ½ O2? H2O ? Hf -285.8kJ
• 2H2 O2?2H2O? Hf 2(-285.8) -571.6 kJ

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So to react CH4 and 2O2to create CO2 and 2H2O we

• CH4 ? C 2H2 ? H 74.8
• 2O2 ? 2O2 ? Hf 0
• C O2 ? CO2 ? Hf -393.5kJ
• 2H2 O2?2H2O ? Hf -571.6 kJ

CH4 2O2 C O2 2H2 O2 ? C 2H2 2O2 CO2
2H2O
CH4 2O2? CO2 2 H2O ? Hf -890.36 kJ
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C 2H2 ? CH4 ?Hfo ?

• We know from the book that
• C O2 ? CO2 ?Hcomb 393.5 kJ
• H2 ½O2 ? H2O ?Hcomb -285.8 kJ
• CH4 2O2?CO2 2H2O ?Hcomb 890.8

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We need the CH4 on the other side so just switch
it and change the charge on the ?Hcomb kJ

• CH42O2?CO22H2O ?Hcomb 890.8

Switched sides switched kJ CO22H2O ? CH4 2O2
?Hcomb 890.8
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C 2H2 ? CH4

• C O2 ? CO2 ? Hcomb -393.5kJ
• 2H2 O2 ? 2H2O ?Hcomb-571.6 kJ
• CO2 2H2O ? CH4 2O2 ?Hcomb 890.8
• CO2 2H2 O2 CO2 2H2O?CO2 2H2O CH42O2

C 2H2? CH4 ?Hf -74.3
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Calculate the enthalpy change for

• 2H2O2(l) ? O2 (g) H2O (l)

2CO (g) O2(g) ? 2CO2(g)
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