Chap 12, 23

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Chap 12, 23

• Phase Changes
• Thermodynamics

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Essential Questions

• 1. How do enthalpy and entropy affect reactions
• 2. How do you use Hess Law
• 3. How do you use bond dissociation energies to
  calculate ?H and predict spontaniety

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Energy Changes in Chemical Reactions

• Heat - The transfer of thermal energy between two
  bodies that are at different temperatures.
• Thermochemistry - the study of heat change in
  chemical reactions.
• The heat exchange occurs between
• the system (part of universe of interest) and
• the surroundings (the rest of the universe)

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What is enthalpy

• The heat content of a substance
• The energy stored in the bonds

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What does it measure

• the energy stored in the bonds in the substance

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What symbol us used for enthalpy

• ?H

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What is an exothemic rxn

• One that releases heat

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How is it represented

• -?H

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Lets Look at a ?H

• S8 8O2? 8SO2 ?H -71 kcal
• since heat is neg it tells us the rxn is
  exothermic so heat is produced

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How do you represent this in an equation

• S8 8O2? 8SO2 71 kcal
• heat is a product in an exothermic equation
• we can place it in an equation like any other
  substance
• This is called using heat (enthalpy) as a term in
  an equation

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What does ? mean

• Change in

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How do you calculate ?H

• HProd Hreactant

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What is an endothermic rxn

• One that absorbs heat

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How is it represented

• ?H

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How do you represent this in an equation

• S8 8O2 71 kcal ? 8SO2
• Or
• S8 8O2 ? 8SO2 ?H 71 kcal
• In endo rxns, heat is a reactant

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• The coefficients always refer to the number of
  moles of the substance.
• Reversing an equation, changes the sign for
  DH.H2O(s) ? H2O(l) DH 6.01 kJ. if we are
  melting 1 mol of ice, then it absorbs 6.01 kJ
• H2O(l) ? H2O(s) DH -6.01 if we are freezing
  1 mol of water, then DH -6.01 kJ.
• If we multiply both sides of the equation by n,
  we must also multiply DH by n.
• 2H2O(s) ?2 H2O(l) DH 12.02 kJ.
• Equations must always specify the physical states
  of all the substances in the equation.

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Hesss Law

• Hesss Law - When reactants are converted to
  products, the change in enthalpy is the same
  whether the reaction takes place in one step or
  in a series of steps.
• If an equation can be represented by the sum of a
  series of equations then the heat will be the sum
  of those equations
• To calculate the DH, you will be given several
  thermochemical equations.
• Lets see it at work by doing an example.

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• Lets examine the following thermochemical
  equation
• CH4(g) 2O2(g) ? CO2(g) 2H2O(l)
  DH -890.4 kJ
• What would DH be for this equation? 2CO2(g)
  4H2O(l) ? 2CH4(g) 4O2(g)
• Flipping the equation and doubling it gives
• 2(890.4 kJ) 1780.8 kJ

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From the following data, C(g) O2(g) ? CO2(g)
DHorxn -393.5 kJ H2(g) 1/2O2(g) ? H2O(l)
DHorxn -285.8 kJ 2C2H6(g) 7O2(g) ?4CO2(g)
6H2O(l) DHorxn -3119.6 kJ Calculate the
enthalpy change for the reaction 2C(gr) 3H2(g)
? C2H6(g)
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Remember we want to use formulas like terms in
math

• Think of these terms just like xs, ys, and 2xy
  terms you see in math equations.
• You can combine like terms
• Since these are added you can subtract the same
  term from both sides of the equation to move them
  where you want or to cancel
• The arrow is the equal sign

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Since C2H6(g) is a product in the final equation
and it is a reactant in the given we must flip
it. Change the sign! 4CO2(g) 6H2O(l) ?
2C2H6(g) 7O2(g) DHorxn 3119.6 kJ No CO2
in final equation so multiply by 44 C(gr)
4O2(g) ? 4CO2(g) DHorxn 4(-393.5 kJ) Water
must cancel so multiply by 66H2(g) 3O2(g) ? 6
H2O(l) DHorxn 6(-285.8 kJ) Now add together
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Since C2H6(g) is a product in the final equation
and it is a reactant in the given we must flip
it. Change the sign! 4CO2(g)6H2O(l)
?2C2H6(g)7O2(g) DHorxn 3119.6 kJ 4 C(g)
4O2(g) ? 4CO2(g) DHorxn 4(-393.5 kJ) 6H2(g)
3O2(g) ? 6 H2O(l) DHorxn 6(-285.8 kJ) 4 C(g)
6H2(g) ? 2C2H6(g) DHorxn -169.2 kj but this
equation doesnt match. We need to divide by 2
so 4 C(g) 6H2(g) ? 2C2H6(g) DHorxn -169.2
kj/2 2 C(g) 3H2(g) ? C2H6(g) DHorxn -84.6 kj
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What is the difference between heat and
temperature

• Heat is a flow of energy from high temp to low
• Temp is the average K.E. of a substance

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What is entropy

• disorder

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What symbol us used for entropy

• ?S

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How do you calculate ?S

• SProd Sreactant

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What is the natural order for these terms

• To go from high enthalpy to low
• To go from low entropy to high

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What is a spontaneous rxn

• A reaction that releases free energy.
• It should happen without any outside source of
  energy

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What is free energy

• energy from a system available to do work
• driving force for rxns

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What symbol us used for free energy

• ?G

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How can you calculate ?G

• GProd Greactant
• ?H - T?S (check labels!)

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What is the significance of the sign on ?G

• - spontaneous rxn (exergonic)

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How are ?H,?S, ?G related

• Rxns can be endo or exo
• Rxn can gain or reduce in disorder.
• Exo is good gain in entropy is good

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The possibilities are

• ?H(bad) ?S(good) ?which is bigger
• ?H(bad) -?S(bad) no rxn (nonspont)( ?G)
• -?H(good) ?S(good) rxn (spontaneous)( -?G)
• -?H(good) -?S(bad) which is bigger

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What is heat of formation

• The energy involved when 1 mole of a compound is
  formed from its elements

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What is bond dissociation energy

• The standard molar enthalpy change of bond
  dissociation (?Hd) is the energy change when 1
  mole of bonds is broken, the molecules and
  resulting fragments being in the gaseous state at
  298K and a pressure of 100kPa. (Standard
  Thermodynamic conditions)

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Huh?

• Then energy involved when you break a bond
  between two atoms
• Remember Breaking bonds requires energy
  (Endothermic) while making bonds releases energy
  (Exothermic).

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What is average dissoc energy

• Bond dissociation energy refers to a specific
  bond in a molecule, but if a molecule has more
  than one of the same bond (eg the C-H bonds in
  methane), then different dissociation energies
  can occur
• CH4(g)? CH3(g)H(g) ?Hd427 kJ/ mol
• CH3(g) ? CH2(g)H(g) ?Hd371 kJ/mol-

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What is average dissoc energy contd

• For us, it is OK to just know the average amount
  of energy needed to break a particular bond. In
  this case, the process of breaking all the bonds
  in methane ending up with gaseous atoms.
• So this process could be written as CH4(g) ?
  C(g) 4 H (g) 
• The enthalpy change for this reaction is 1646
  kJ/mol , so the average bond enthalpy is 1646 /
  4 412 kJ/mol .

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What is average dissoc energy contd

• We just look up the values on a data tables.
• I know, what table
• a bond dissociation energy table
•  It is important to realize these average bond
  enthalpies.

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How do I use this info

• Like in our other calculations ? means changes so
  
• ?H ??Hd reactants - ??Hd products
•       This basically means that you add up all
  the energies of the broken bonds add up all the
  energies of the bonds that are reformed and
  subtract one from the other.
• Its another version of Hess' Law.

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Example

• Cal the ?H forC3H8(g) 5 O2(g) ?3 CO2(g)4
  H2O(g)
• We need to determine how many of each bond type
  has been broken.
• For some of these you will have to draw the
  Lewis Structure to see the bond types

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Example

• C3H8(g) H H H
• H--C---C----C--H H
  H H

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Example contd

• We need to determine how many of each bond type
  has been broken.
• 8 x C-H 2 x C-C 5 x OO
• Then how many bond types have been formed
• 6xCO 8xH-O

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Example contd

• So using data tables we can look up the average
  bond enthalpies and calculate the enthalpy change
  of the reaction. (Notice they are all
  endothermic.)
• Then plug into the equation above ?H
  (8x413)(2x348)(5x493)
  (6x799)(8x463)
• - 2033 kJ/mol

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How can I use this for ?S

• Since these are all gas phase rxns, look at the
  total number of moles on each side
• This side that has the most has the most S
• If Products have most then ?S. If reactants
  then - ?S

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How can I use this for ?G

• WellYou know ?H from using the tables, and you
  get the sign of ?S from the mole change,
• ?G ?H - T ?S Now you must use some
  generalizations
• If ?H is neg and ?S is pos, ?G is NEG and
  therefore spont rxn