THERMOCHEMISTRY

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THERMOCHEMISTRY
HEAT CAPACITY, SPECIFIC HEAT, ENDOTHERMIC/EXOTHERM
IC, ENTHALPY, STANDARD ENTHALPIES, CALORIMETERY
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INTRO TO THERMOCHEMISTRY

• Chemical rxns involve changes in energy
• Breaking bonds requires energy
• Forming bonds releases energy
• The study of the changes in energy in chem rxns
  is called thermochemistry.
• The energy involved in chemistry is real and
  generally a measurable value
• Energy units are numerous, but we will
  concentrate on the Joule (SI base unit) and the
  calorie (little c, big C is the food Calorie or a
  kilocalorie)

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WHAT IS HEAT?

• Hot cold, are automatically associated with the
  words heat and temperature
• Heat temperature are NOT synonyms
• The temperature of a substance is directly
  related to the energy of its particles,
  specifically its

• The Kinetic Energy defines the temperature
• Particles vibrating fast hot
• Particles vibrating slow cold

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• Thermal energy is the total energy of all the
  particles that make up a substance
• Kinetic energy from vibration of particles
• Potential energy from molecular attraction
  (within or between the particles)
• Thermal energy is dependent upon the amount or
  mass of
  material present
  (KE ½mv2)

• Thermal energy is also related to the type of
  material

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• Different type of materials
• May have the same temp, same mass, but different
  connectivity
• Affected by the potential energy stored in
  chemical bonds or the IMFs holding molecules
  together
• So it is possible to be at same temp (same KE)
  but have very different thermal
  energies
• The different abilities to hold
  onto or release energy is
  referred to as the
  substances heat capacity

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• Thermal energy can be transferred from object to
  object through direct contact
• Molecules collide, transferring energy from
  molecule to molecule

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• Thermal energy can be transferred from object to
  object through direct contact
• Molecules collide, transferring energy from
  molecule to molecule

AKA HEAT
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HEAT CAPACITY

• The measure of how well a material absorbs or
  releases heat energy is its heat capacity
• It can be thought of as a reservoir to hold heat,
  how much it holds before it overflows is its
  capacity
• Heat capacity is a physical property unique to a
  particular material
• Water takes 1 calorie of energy
  to raise temp 1 C
• Steel takes only 0.1 calorie of
  energy to raise temp 1 C

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SPECIFIC HEAT CAPACITY

• The amount of energy it takes to raise the temp
  of a standard amount of an object 1C is that
  objects specific heat capacity (Cp)
• The standard amount 1 gram
• Specific heats can be listed on data tables
• Smaller the specific heat ? the less energy it
  takes the substance to feel hot
• Larger the specific heat ? the more energy it
  takes to heat a substance up (bigger the heat
  reservoir)

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• Specific heats and heat capacities work for gains
  in heat and in losses in heat
• Smaller the specific heat ? the less time it
  takes the substance to cool off
• Larger the specific heat ? the longer time it
  takes the substance to cool off
• Specific heat capacity values are used to
  calculate changes in energy for chemical rxns
• Its important for chemists to know how much
  energy is needed or produced in chemical rxns

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CHANGE IN HEAT ENERGY (ENTHALPY)

• The energy used or produced in a chem rxn is
  called the enthalpy of the rxn
• Burning a 15 gram piece of paper produces a
  particular amount of heat energy or a particular
  amount of enthalpy
• Enthalpy is a value that also contains a
  component of direction (energy in or energy out)
• Heat gained is the out-of
  direction ie exo-

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CHANGE IN HEAT ENERGY (ENTHALPY)

• The energy used or produced in a chem rxn is
  called the enthalpy of the rxn
• Burning a 15 gram piece of paper produces a
  particular amount of heat energy or a particular
  amount of enthalpy
• Enthalpy is a value that also contains a
  component of direction (energy in or energy out)
• Heat released is the out-of
  direction ie exo-

• Heat absorbed is the into
  direction ie endo-

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SURROUNDINGS
HEAT
HEAT
HEAT
HEAT
SYSTEM
SYSTEM
EXOTHERMIC
ENDOTHERMIC
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• Chemical rxns can be classified as either
• Exothermic ? a reaction in which heat energy is
  generated (a product)
• Endothermic ? reaction in which heat energy is
  absorbed (a reactant)
• Exothermic rxns typically feel warm as the rxn
  proceeds
• Give off heat energy, sometimes quite alot
• Endothermic rxns typically feel cooler the longer
  the rxn proceeds
• Absorb heat energy, sometimes enough to get very
  cold

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• Exothermic rxn

• To a cold camper, the important product here is
  the heat energy

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In an exothermic process the amount of energy
given off is more than the initial energy
invested. So the products are less in energy
than the reactants.
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• Endothermic rxn

NH4NO3H2O 752kJ ?NH4OHHNO3

• Similar system as what is found in cold packs

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ENDOTHERMIC RXN
In an endothermic process more energy is required
to cause the rxn to proceed than obtained in
return. So the products are less in energy than
the reactants.
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CHANGE IN ENTHALPY

• Most common measurement of the energy or enthalpy
  in a reaction is actually a change in enthalpy
  (?H)
• The enthalpy absorbed or gained (changed) in a
  rxn is dependent on the amount (mols) of material
  reacting
• We can use the coefficient ratios to energy
  ratios to calculate how much energy a reaction
  used or produced
• DHs can be data provided in a chemical reaction
  and has a magnitude and direction ( or -)

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USING ?H IN CALCULATIONS

• Chemical reaction equations are very powerful
  tools.
• Given a rxn equation with an energy value, We can
  calculate the amount of energy produced or used
  for any given amount of reactants.

(For Example) How much heat will be absorbed if
1.0g of (H2O2) decomposes in a bombardier
beetle to produce a defensive spray of steam
2H2O2? 2H2O O2 ?Hº 190kJ
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2H2O2? 2H2O O2 ?Hº 190kJ
Analyze we know that if we had 2 mols of H2O2
decomposing we would use 190kJ of heat, but how
much would it be for 1.0 g of H2O2
Therefore we have to convert our given
1.0 g of H2O2 to moles of H2O2
1mol H2O2
1.0g H2O2
.02941 mol
34g H2O2
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2H2O2? 2H2O O2 ?Hº 190kJ
Therefore with 2 moles of H2O2 we would use 190
kJ of energy, but we dont have 2 moles we only
have .02941 mols of H2O2, so how much energy
would the bug use?
190kJ
2.8kJ
.02941 mol
2molH2O2
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Example 2
How much heat will be released when 4.77 g of
ethanol (C2H5OH) react with excess O2 according
to the following equation C2H5OH3O2?
2CO23H2O ?Hº -1366.7kJ
analyze we know that if we had 1 mol of ethanol
(assuming coefficient of 1 in rxn equation)
burning we would produce 1366.7kJ of heat, but
how much would it be if only we only had 4.77 g
of ethanol?
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C2H5OH3O2?2CO23H2O ?Hº -1366.7kJ
1mol C2H5OH
-1366.7kJ
4.77g C2H5OH
1mol C2H5OH
46g C2H5OH
-142 kJ
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• We can also track energy changes due to temp
  changes, using ?HmCp?T

?H

• If the temp difference is positive
• The final temperature is higher than the initial
  temperature
• End up with a positive enthalpy

• if the temp change is negative
• the enthalpy ends up negative
• The initial temperature is higher than the final
  temperature

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if you drink 4 glasses of ice water at 0C, how
much heat energy is transferred as this water is
brought to body temp? each glass contains 250 g
of water body temp is 37C.

• mass of 4 glasses of water
• m 4 x 250g 1000g H2O
• change in water temp
• Tf Ti 37C - 0C
• specific heat of water
• Cp 4.18 J/gC (from previous slide)

?HmC?T
?H(1000g)(4.18J/gC)(37C)
?H 160,000J
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Example 2 500 g of a liquid is heated from
25C to 100C. The liquid absorbs 156,900 J of
energy. What is the specific heat of the liquid
and identify it.
DH mCDT
C DH/mDT
C 156,900J/(500g)(75C)
C 4.184 J/gC
H2O
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• Enthalpy is dependent on the conditions of the
  rxn
• Its important to have a standard set of
  conditions, which allows us to compare the affect
  of temps, pressures, etc. On different substances
• Chemists have defined a standard set of
  conditions
• Stand. Temp 298K or 25C
• Stand. Press 1atm or 760mmHg
• Enthalpy produced in a rxn under standard
  conditions is the standard enthalpy (?H)

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• Standard enthalpies can be found on tables of
  data measured as standard enthalpies of
  formations, enthalpies of combustion, enthalpies
  of solution, enthalpies of fusion, and enthalpies
  of vaporization
• Enthalpy of formation is the amount of energy
  involved in the formation of a compound from its
  component elements.
• Enthalpy of combustion is the amount of energy
  produced in a combustion rxn.
• Enthalpy of solution is the amount of energy
  involved in the dissolving of a compound

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• Enthalpy of fusion is the amount of energy
  necessary to melt a substance.
• Enthalpy of vaporization is the amount of energy
  necessary to convert a substance from a liquid to
  a gas.
• All of these energies are measured very carefully
  in a laboratory setting under specific conditions
• At 25 C and 1atm of pressure
• These measured energies are reported in tables to
  be used in calculations all over the world.

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• Calorimetry is the process of measuring heat
  energy
• Measured using a device called a calorimeter
• Uses the heat absorbed by H2O to meas-ure the
  heat given off by a rxn or an object
• The amount of heat soaked up by the water is
  equal to the amount of heat released by the rxn

?Hsys is the system or what is taking place
in the main chamber (rxn etc.) And ?Hsur is
the surroundings which is generally
water.
?HSYS-?HSUR
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A COFFEE CUP CALORIMETER
USED FOR A REACTION IN WATER, OR JUST A
TRANSFER OF HEAT.
A BOMB CALORIMETER
USED WHEN TRYING TO FIND THE AMOUNT OF HEAT
PRODUCED BY BURNING SOMETHING.
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CALORIMETRY

• With calorimetry we use the sign of what happens
  to the water
• When the water loses heat into the system it
  obtains a (-) sign

SIGN MEANS HEAT WAS ABSORBED BY THE RXN
- SIGN MEANS HEAT WAS RELEASED BY WATER
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WATER SURROUNDINGS
SYSTEM
ENDOTHERMIC
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CALORIMETRY

• With calorimetry we use the sign of what happens
  to the water
• When the water gains heat from the system it
  obtains a () sign

- SIGN MEANS HEAT WAS RELEASED BY THE RXN
SIGN MEANS HEAT WAS ABSORBED BY WATER
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WATER SURROUNDINGS
SYSTEM
EXOTHERMIC
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CALORIMETRY

• You calculate the amount of heat absor-bed by the
  water (using ?H mC?T)
• Which leads to the amount of heat given off by
  the rxn
• you know the mass of the water (by weighing it)
• you know the specific heat for water (found on a
  table)
• and you can measure the change in the temp of
  water (using a thermometer)

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A chunk of Al that weighs 72.0g is heated to
100C is dropped in a calorimeter containing
120ml of water at 16.6C. the H2Os temp rises
to 27C.

• mass of Al 72g
• Tinitial of Al 100C
• Tfinal of Al 27C
• CAl .992J/gC (from table)

??HSYS
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• We can do the same calc with the water info

• Mass of H2O 120g
• Tinitial of H2O 16.6C
• Tfinal of H2O 27C
• CH2O 4.18J/gC (from table)

?HSUR
Equal but opposite, means that the Al decreased
in temp, it released its stored heat into the
H2O, causing the temp of the H2O to increase.
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When a 4.25 g sample of solid NH4NO3 dissolves in
60.0 g of water in a calori-meter, the
temperature drops from 21.0C to 16.9C.
Calculate the energy involved in the dissolving
of the NH4NO3.
DHwater (mwater)(Cwater)(DTwater)
DHwater (60g)(4.18J/gC)(16.9C-21.0C)
DHwater -1.03 x 103 J
- DHwater DHNH4NO3
DHNH4NO3 1.03 x 103 J
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THERMOCHEMISTRY
HESSS LAW, AND HEAT VS. TEMPERATURE
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HESSS LAW

• Suppose you wanted to determine the magnitude of
  the heat flow for a rxn, but you are unable to
  perform the rxn.
• The reactants are unstable, and you might start a
  fire in the lab. What can you do?
• You might find other rxns that are easier to
  perform and when added together equal your
  original desired rxn. AKA Hesss Law
• According to Hess's Law, the sum of the heat
  flows for these rxns is equal to the heat flow
  for your original net rxn.

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• Lets use Hesss Law to determine the enthalpy of
  formation of N2O5. The formation rxn is given by
  the following eqn

2 N2(g) 5 O2(g) ? 2 N2O5(g) DH ?

• We dont have any tables including heats of
  formation handy. Fortunately, we do have the
  following standard heats of rxn

2NO(g) O2(g)? 2NO2(g) DHrxn -114kJ/mol
4NO2(g) O2(g)? 2N2O5(g) DHrxn -110kJ/mol
N2(g) O2(g)? 2NO(g) DHrxn 181kJ/mol
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• You may be wondering why its fortunate that we
  have this information. After all, what do all
  of these rxns have to do with the one were
  interested in?
• Im sure glad you asked! By combining the
  preceding equations, we can come up with a whole
  new eqn that describes the process were actually
  interested in.
• The only rules we need to follow are

• If we reverse the rxn, the sign of the DHrxn is
  reversed. For example, if

If 2NO(g) O2(g)? 2NO2(g) DHrxn -114kJ/mol
Then 2NO2(g)?2NO(g) O2(g) DHrxn114kJ/mol
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• If we need to perform a rxn more than once, the
  heat of rxn is multiplied by the number of times
  we do the rxn. For example, if we do the
  following rxn twice

2NO2(g)?2NO(g) O2(g) we end up with
4NO2(g)? 4NO(g) 2O2(g) DHrxn 228kJ/mol

• Lets use Hesss Law to determine the enthalpy of
  formation of N2O5

2 N2(g) 5 O2(g) ? 2 N2O5(g) DH ?
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STEP 1

• There are 2 moles of N2 in the reactants. The
  only eqn that contains nitrogen is

N2(g) O2(g)? 2 NO(g) DHrxn 181kJ/mol
so well multiply this eqn by 2 to give us
2 N2(g) 2 O2(g)? 4 NO(g) DHrxn 362kJ/mol
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STEP 2

• The eqn were trying to solve has 2 moles of
  N2O5(g) as the products. The only eqn that
  contains N2O5 is

4NO2(g) O2(g)? 2N2O5(g) DHrxn -110kJ/mol
Because there are already two mols of N2O5 in the
products, we can leave the heat of rxn the way it
is.
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STEP 3

• Now that we have two eqns, lets add them
  together to see what we still need to do

2 N2(g) 2 O2(g)? 4 NO(g) DHrxn 362kJ/mol
4NO2(g) O2(g)? 2N2O5(g) DHrxn -110kJ/mol
Overall
2 N2(g) 3 O2(g) 4 NO2? 4 NO 2 N2O5(g)
DHrxn 252 kJ/mol
Remember our goal 2 N2(g) 5 O2(g) ? 2
N2O5(g)
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STEP 4

• We need to get rid of the NO2 on the reactant
  side of the eqn and the NO on the product side.
  Fortunately, we have an eqn to help us do this

2NO(g) O2(g)? 2NO2(g) DHrxn -114kJ/mol
By multiplying this eqn by 2, the NO on the
product side and the NO2 on the reactant side
will cancel out.
4NO(g) 2 O2(g)? 4NO2(g) DHrxn -228kJ/mol
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STEP 5

• Adding this eqn to the others, we get

2N2(g) 3O2(g) 4NO2 4NO 2O2 ? 4NO
2N2O5(g) 4NO2(g)
2 N2(g) 5 O2(g) ? 2 N2O5(g)
Or
By adding the heats of rxn, we can find the total
standard heat for the process
DHf 252 kJ 228 kJ 24 kJ/mol
These problems take time and practice to perfect.
However, with a bit of trial and error, you
should be able to figure out the heat of rxn for
just about any process.
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• Example 2
• Given the following information

C2H6?C2H4 H2 137kJ/mol
2H2O?2H2O2 484kJ/mol
2H2O2CO2?C2H43O2 1323kJ/mol
Find the value of ?H for the reaction
2C2H6 7O2 ? 4CO2 6H2O
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• Example 2
• Rearranging and multiplying

2 C2H6 ? 2 C2H4 2 H2 274kJ/mol
2H2O?2H2O2 484kJ/mol
2C2H46O2 ? 4H2O4CO2 -2646kJ/mol
Find the value of ?H for the reaction
2C2H6 7O2 ? 4CO2 6H2O
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2C2H6 7O2 ? 4CO2 6H2O

• Example 2
• Adding eqn 1 with eqn 3

2C2H62C2H46O2 ? 2C2H42H24H2O4CO2 DH
-2372kJ/mol
2C2H66O2 ? 2H24H2O4CO2 DH -2372kJ/mol
We still need to cancel out the H2 in the
products, add an additional O2 to the reactants,
and add 2 H2Os to the prod.
2H2O?2H2O2 484kJ/mol
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2C2H6 7O2 ? 4CO2 6H2O

• Example 2
• Rearranging the eqn and the energy

2C2H66O2 ? 2H24H2O4CO2 DH -2372kJ/mol
2 H2 O2 ? 2 H2O DH -484kJ/mol
2 C2H6 6 O2 2 H2 O2 ? 2 H2 4 H2O 4 CO2
2 H2O
2C2H6 7O2 ? 4CO2 6H2O
DH -2856kJ/mol
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Rule 1 if an equation is reversed,
the sign of ?H changes too.
Rule 2 if the coefficients of an eqn are
multiplied by a factor, the ?H is the multiplied
by the same factor.
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PHASE CHANGES HEAT

• Energy is required to change the phase of a
  substance
• The amount of heat necessary to melt 1 mole of
  substance
• Heat of fusion (?Hfus)
• It takes 6.00 kJ of energy to melt 18 grams of
  ice into liquid water.
• The amount of heat necessary to boil 1 mole of
  substance
• Heat of vaporization (?Hvap)
• It takes 40.6 kJ of energy to boil away 18 grams
  of water.

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• There are 5 distinct sections we can divide the
  curve into
• Ice
• Water ice
• Water only
• Water steam
• Steam only
• We can calculate the amount of energy involved in
  each stage
• There are two types of calculations
• Temperature changes use ?HmC?T
• Phase changes use (mols)?Hfus or (mols)?Hvap

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SOLID ICE

• Lets say we have 18.0g of ice at 10C, we
  begin heating it on a hot plate with sustained
  continuous heat.
• Heat energy absorbs into the ice increas-ing the
  vibrational or kinetic energy of the ice
  molecules
• The temp will increase will continue to
  increase until just before the ice has enough
  energy to change from solid to liquid

• We can calc the energy absorbed by the ice to
  this point
• Use the eqn DHmCDT (cice2.09J/gC)

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DHicemCiceDT
DHice (18g)(2.09J/gC)(0C-(-10C))
DHice376.2J
0
-10
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WATER ICE (MELTING)

• Any additional heat absorbed by the ice goes into
  partially breaking the connections between the
  ice molecules.
• There is no change in the kinetic energy of the
  molecules (curve flattens out)
• No change in temp
• All of the energy goes into breaking the
  connections

• As long there is solid ice present, the temp
  cannot increase.
• The solid liquid are in equilibrium if they are
  both present

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• The ?H required to change from the solid to the
  liquid phase is called the heat of fusion
  depends on the amnt of the substance (DHfus of
  H2O6000J/mol or 6kJ/mol)

• Using the formula DHmelting (mol) DHfus

18g H2O
6000J
0
-10
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ALL WATER

• Now all of the particles are free to flow,
• The heat energy gained now goes into the
  vibrational energy of the molecules.
• The temp of the water increases
• The rate of temp increase now depends on the heat
  capacity of liquid water
• Cwater4.180J/gC

• The temp continues to increase until it just
  reaches boiling stage (for water 100ºC)
• again, ?HwatermCwater?T

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DHwater(18g)(4.18J/gC)(100C-0C)
DHwater 7524J
100
0
7524J
6000J
-10
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STEAM WATER (VAPORIZING)

• Any additional heat absorbed by the water goes
  into completely breaking the connections between
  the water molecules.
• Again the heat does not increase the KE of the
  molecules so the temp does not change,
• the energy is used to vaporize the water

• If there are still connections to break or there
  is liquid present, the temp cannot increase.
• The energy required to change from the liquid to
  the vapor phase is called the heat of
  vaporization using ?Hboiling(mol)?Hvap
• ?Hvap of H2O40,790J/mol

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18g H2O
100
?Hboiling(mol)?HVAP
0
6000J
-10
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STEAM ONLY (VAPOR PHASE)

• Again the heat energy goes into the vibrational
  energy of the molecule.
• Rate of temp increase depends on CH2O
  vapor1.84J/gC
• The temp can increase indefinitely, or until the
  substance decomposes (plasma)
• Well stop at 125C.

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125
DHSTEAM(m)(CWATER VAPOR)(?T)
DH(18g)(1.84J/gC)(125-100)
100
0
6000J
-10
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• To figure out how much energy we need would need
  all together to heat up the water this much, we
  just need to add up the energy of each step.

DHtotal (376J 6000J 7524J 40,790J
829J) DHtotal 55,500J

• Notice, the majority of the energy is needed for
  the vaporization step.
• The connections between molecules of H2O must be
  broken completely to vaporize

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• What volume of propane gas would we need in order
  to perform the process just explored?

C3H8 5O2? 3CO2 4H2O DH -2043kJ

• To vaporize 18 grams of water, it took a total of
  55,519J so we need to convert our total energy
  to mols then to volume.

55.500kJ
0.609 L