Chapter%2019:%20Chemical%20Thermodynamics

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Chapter 19 Chemical Thermodynamics

• Joe DiLosa
• Hannah Robinson
• Justin Whetzel
• Emily Penn
• Casey White

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19.1 Spontaneous Processes Vocabulary

• Spontaneous Process a process that proceeds on
  its own without any outside assistance
• Reversible Process A process that can go back
  and forth between states along exactly the same
  path
• Irreversible Process A process that is not
  reversible
• Isothermal Process A process that occurs at a
  constant temperature

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19.1 Spontaneous Processes

• Occur in a definite direction
• Ex When a brick is dropped and it falls down
  (gravity is always doing work on the brick)
• Opposite direction (reverse) would be
  nonspontaneous (takes work to go against gravity)
• If the surroundings must do work on the system to
  return it to its original state, the process is
  irreversible
• Any spontaneous process is irreversible

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19.1 Spontaneous Processes
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19.1 Spontaneous Processes

• A spontaneous reaction does not mean the reaction
  occurs very fast
• Iron rusting is spontaneous
• Experimental conditions are important in
  determining if a reaction is spontaneous
  (temperature and pressure)
• To determine spontaneity, distinguish between the
  reversible and irreversible paths between states

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19.1 Spontaneous Processes

• Which of the following processes are spontaneous
  and which are nonspontaneous (a) the melting of
  ice cubes at -5C and 1 atm pressure, (b)
  dissolution of sugar in a cup of hot coffee, (c)
  the reaction of nitrogen atoms to form N2
  molecules at 25C (d) alignment of iron filings
  in a magnetic field, (e) formation of CH4 and O2
  molecules from CO2 and H2O at room temperature
  and 1 atm of pressure?

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19.1 Spontaneous Processes

• (a) is nonspontaneous because the ice will not
  change state
• (b) is spontaneous because to get the sugar back,
  the temperature of the system must be altered.
• (c) is spontaneous because work must be done to
  form N2 and it is an irreversible process
• (d) is spontaneous because the system must do
  work against the magnetic field to go back to its
  original state
• (e) is nonspontaneous because a change in
  temperature and work must be done to break the
  products up and form the original reactants

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19.1 Spontaneous Processes

• Practice Problems - 7, 8, and 9

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19.2 Entropy and Second Law of Thermodynamics
Vocabulary

• Entropy A thermodynamic function associated
  with the number of different equivalent energy
  states or spatial arrangements in which a system
  may be found
• Second Law of Thermodynamics in general, any
  irreversible process results in an overall
  increase in entropy whereas a reversible process
  results in no overall change in entropy
• Infinitesimal difference - an extremely small
  difference

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19.2 Entropy and Second Law of Thermodynamics

• Entropy q/T S
• A reversible change produces the maximum amount
  of work that can be achieved by the system on the
  surroundings (wrev wmax)
• Reversible processes are those that reverse
  direction whenever an infinitesimal change is
  made in some property of the system

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19.2 Entropy and Second Law of Thermodynamics

• Example of Reversible Process
• Flow of heat between a system and its
  surroundings
• Example of Irreversible Process
• Gas that has been limited to part of a container
  expands to fill the entire container when the
  barrier is removed

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19.2 Entropy and Second Law of Thermodynamics

• ?S Sfinal Sinitial
• ?S qrev/T (where T constant, isothermal
  process)
• qrev ?Hfusion when calculating ?Sfusion for a
  phase change

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19.2 Entropy and Second Law of Thermodynamics

• ?Ssystem qrev/T
• ?Ssurroundings -qrev/T
• ?Stotal ?Suniv ?Ssystem ?Ssurroundings
• For a reversible process ?Stotal 0
• For an irreversible process ?Stotal gt 0

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19.2 Entropy and Second Law of Thermodynamics

• The normal boiling point of methanol (CH3OH) is
  64.7C and its molar enthalpy of vaporization is
  ?Hvap71.8 kJ/mol. When CH3OH boils at its normal
  boiling point does its entropy increase or
  decrease? Calculate the value of ?S when 1.00 mol
  CH3OH (l) is vaporized at 64.7 C
• Increases, the process is endothermic
• ?S qrev/T ?Hvap/T
• 64.7 C 273 337.7 K
• (1 mol)(71.8 kJ/mol)/(337.7 K) 0.213 kJ/K 213
  J/K
• entropy increases

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19.2 Entropy and Second Law of Thermodynamics

• Practice Problems- 20 and 21
• http//ffden-2.phys.uaf.edu/212_fall2003.web.dir/A
  nca_Bertus/page_7.htm

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19.3 The Molecular Interpretation of Entropy
Vocabulary

• Translational Motion movement in which an
  entire molecule moves in a definite direction
• Vibrational Motion the atoms in the molecule
  move periodically toward and away from one
  another
• Rotational Motion the molecules are
• spinning like a top
• Microstate the state of a system at a
  particular instant, one of many possible states
  of the system
• Third Law of Thermodynamics the entropy of a
  pure crystalline substance at absolute zero is
  zero

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19.3 The Molecular Interpretation of Entropy

• Vibration Vibration
• Vibration Rotation
• 
  
  Img src wps.prenhall.com

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19.3 The Molecular Interpretation of Entropy

• S k ln W (W number of microstates)
• k Boltzmanns constant 1.38 x 10-23 J/K
• Entropy is a measure of how many microstates are
  associated with a particular macroscopic state
• ?S k ln Wfinal - k ln Winitial k ln
  Wfinal/Wintial
• Entropy increase with the number of microstates
  of the system

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19.3 The Molecular Interpretation of Entropy

• The of microstates available to a system
• increases with
• Increase in volume
• Increase in temperature
• Increase in the number of molecules
• Because any of these changes increases the
  possible positions and energies of the molecules
  of the system

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19.3 The Molecular Interpretation of Entropy

• Generally expect the entropy of a system to
• increase for processes in which
• Gases are formed from either solids or liquids
• Liquids or solutions are formed from solids
• The of gas molecules increases during a
  chemical reaction

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19.3 The Molecular Interpretation of Entropy

• At absolute zero there is no thermal motion so
  there is only one microstate
• As the temperature increases the molecules gain
  energy in the form of vibrational motion
• When the molecules increase their motion they
  have a greater number of microstates

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19.3 The Molecular Interpretation of Entropy

• http//wpscms.pearsoncmg.com/au_hss_brown_chemistr
  y_1/57/14649/3750314.cw/content/index.html

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19.3 The Molecular Interpretation of Entropy

• Practice Problems- 28, 30, 38, 40

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19.4 Entropy Changes in Chemical Reactions
Vocabulary

• Standard Molar Entropy The molar entropy value
  of substances in their standard states

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19.4 Entropy Changes in Chemical Reactions

• The standard molar entropies of elements are not
  0 at 298K
• The standard molar entropies of gases are greater
  than those of liquids and solids
• standard molar entropies generally increase with
  increasing molar mass
• Standard molar entropies generally increase with
  an increasing number of atoms in the formula of a
  substance

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19.4 Entropy Changes in Chemical Reactions

• Pictures taken from p820 of the Tenth Edition of
  Chemistry The Natural Science textbook written
  by Brown, LeMay, and Bursten

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19.4 Entropy Changes in Chemical Reactions

• ?S SnS(products) - SmS(reactants)
• ?Ssurr -qsys/T
• For a reaction at constant pressure qsys ?H

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19.4 Entropy Changes in Chemical Reactions

• Calculate ?S values for the following reaction
• C2H4 (g) H2 (g) ? C2H6 (g)
• ?S SnS(products) - SmS(reactants)
• 229.5 J/molK (219.4 J/molK 130.58 J/molK)
• -120.5 J/molK

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19.4 Entropy Changes in Chemical Reactions

• Practice Problems - 48

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19.5 - Gibbs Free Energy Vocabulary

• Gibbs free energy- a thermodynamic state function
  that combines entropy and enthalpy
• Standard free energies of formation- the change
  in free energy associated with the formation of a
  substance from its elements under standard
  conditions

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19.5 - Gibbs Free Energy

• Free energy (G) is defined by
• G H-TS
• Where T is absolute temperature, S is entropy,
  and H is enthalpy

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19.5 - Gibbs Free Energy

• When temperature is constant the equation for the
  change in free energy for the system is
• ?G?H-T?S
• If T and P are constant then the sign of ?G and
  spontaneity of a reaction are related.
• If ?G lt 0, reaction proceeds forward
• If ?G 0, reaction is at equilibrium
• If ?G gt 0, the forward reaction is not
  spontaneous because work must be supplied from
  the surroundings but the reverse reaction is be
  spontaneous.

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19.5 - Gibbs Free Energy

• In any spontaneous process at constant
  temperature and pressure the free energy always
  decreases

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19.5 - Gibbs Free Energy

• Conditions of Standard free energies of formation
  (implied)
• 1 atm pressure for gases
• pure solids
• pure liquids
• 1M solutions
• 25 C (normally)
• The free energy of elements is 0 in their
  standard states.

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19.5 - Gibbs Free Energy

• ?Gf standard free energies of formation
• Calculated the same way as ?H
• ?G Sn?Gf (products) - Sm?Gf (reactants)
• ?G -wmax

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19.5 - Gibbs Free Energy

• http//en.wikibooks.org/wiki/Structural_Biochemist
  ry/Enzyme/Gibbs_free_energy_graph

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19.5 Gibbs Free Energy

• Practice Problems - 49, 51, 53, 54, 56, 58, 63,
  66

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19.5 Gibbs Free Energy

• For a certain reaction ?H -35.4 kJ and ?S
  -85.5 J/K. Calculate ?Go for the reaction at 298
  K. Is the reaction spontaneous at 298 K?
• ?G ?H ?ST
• ?G -35.4KJ (-85.5 J/K 298K)
• ?G -9920 J

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19.6 Free Energy and Temperature Vocabulary

• No new vocabulary

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19.6 Free Energy and Temperature
When ?H and T?S are negative, ?G is negative and
the process is spontaneous When ?H and T?S are
positive, ?G is positive and the process is
non-spontaneous When ?H and T?S are opposite
signs, ?G will depend on the magnitudes
http//www.chem1.com/acad/webtext/thermeq/TE4.html
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19.6 Free Energy and Temperature
Picture taken from p828 of the Tenth Edition of
Chemistry The Natural Science textbook written
by Brown, LeMay, and Bursten
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19.6 Free Energy and Temperature

• For a certain reaction ?H -35.4 kJ and ?S
  -85.5 J/K. is the reaction exothermic or
  endothermic? Does the reaction lead to an
  increase or decrease of disorder in the system?
  Calculate ?Go for the reaction at 298 K. Is the
  reaction spontaneous at 298 K?

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19.6 Free Energy and Temperature

• A) The reaction is exothermic because ?H is
  negative
• B) It leads to a decrease in the disorder of the
  system (increase in the order of the system)
  because ?S is negative
• C) ?G ?H ?ST
• ?G -35.4KJ (-85.5 J/K 298K)
• ?G -9920 J
• D) Yes, if all reactants are in their standards
  states because ?G is negative

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19.6 Free Energy and Temperature

• Practice problems (same as 19.5) - 49, 51, 53,
  54, 56, 58, 63, 66

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19.7 Free Energy and Equilibrium Constant
Vocabulary

• Nonstandard conditions conditions other than
  the standard conditions

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19.7 Free Energy and Equilibrium Constant

• ?G ?Go RT ln Q
• Where R 8.314 J/mol-K, T absolute
  temperature, Q reaction quotient
• ?Go ?G under standard conditions
• Because Q 1 and lnQ 0

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19.7 Free Energy and Equilibrium Constant

• ?Go -RT ln K
• At equilibrium, ?G 0
• K e (delta G standard)/RT
• Smaller (more negative ?G0), the larger the K
  value

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19.7 Free Energy and Equilibrium Constant

• Direction of Reaction Keq
  ?G
• Toward forming more products gt1 negative
• Toward forming more reactants lt1 positive
• Products and reactants equal 1 0
• http//www.wiley.com/legacy/college/boyer/04700037
  90/reviews/thermo/thermo_keq.htm

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19.7 Free Energy and Equilibrium Constant

• 71 Explain Quantitatively how ?G changes for
  each of the following reactions as the partial
  pressure of O2 is increased
• A) 2 CO(g) O2(g) ? 2 CO2 (g)

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19.7 Free Energy and Equilibrium Constant

• 2 CO(g) O2(g) ? 2 CO2 (g)
  
• -137.2 kJ/mol 0 kJ/mol -394.4 kJ/mol
• ?G products reactants
• ?G -394.4 kJ/mol 137.2 kJ/mol
• ?G -257.2 kJ/mol

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19.7 Free Energy and Equilibrium Constant

• Practice Problems 71,76,79

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Additional Chapter Problems

• 82, 87, 88, 89, 90, 92