Chapter 19: Chemical Thermodynamics

1
Chapter 19 Chemical Thermodynamics

• Tyler Brown
• Hailey Messenger
• Shiv Patel
• Agil Jose

2
19.1 Spontaneous Processes
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19.1 Spontaneous Processes

• Spontaneous Process
• A process that proceeds on its own without any
  outside assistance
• Occurs in a definite direction

Processes that are spontaneous in one direction
are nonspontaneous in the other (Pg 805
Brown and LeMay)
4
19.1 Spontaneous Processes

• Reversible Process
• A process in which the system is changed in a way
  that the system and surroundings can be restored
  to their original state by exactly reversing the
  change (can be completely restored to original
  condition)
• Reversible processes reverse direction whenever
  an infinitesimal change is made in some property
  of the system

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19.1 Spontaneous Processes

• Irreversible Process
• Cannot be reversed to return system and
  surroundings to original state

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19.1 Spontaneous Processes

• Isothermal Process
• A process with a constant temperature

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19.1 Spontaneous Processes

• First Law of Thermodynamics Energy is conserved
• Physical and chemical processes have a
  directional character (Pg 804)
• I.e. Sodium and Chlorine come together on their
  own to make NaCl, but it does not decompose of
  its own accord.

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19.1 Spontaneous Processes

• Experimental conditions help to determine if a
  process is spontaneous
• I.e. When the temperature gt 0C in ordinary
  atmospheric pressure, ice melting is spontaneous.
  
• In these conditions, liquid water turning in to
  ice is not spontaneous

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19.1 Spontaneous Processes

• Entropy The ratio of heat delivered to the
  temperature at which it is delivered
• A reversible change produces the maximum amount
  of work that can be achieved by the system on the
  surroundings (wrev wmax) (Page 807)
• All spontaneous processes are IRREVERSIBLE

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19.1 Spontaneous Processes

• Tell whether or not the following process is
  spontaneous
• Ice melts at -1 degree Celsius

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19.1 Spontaneous Processes

• Tell whether or not the following process is
  spontaneous
• Ice melts at -1 degree Celsius
• This would not be spontaneous because Ice does
  not melt under 0 degrees Celsius. This reaction
  would be spontaneous in the reverse reaction

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19.1 Spontaneous Processes

• 19.1 Problems 7-18

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19.2 Entropy and the Second Law of Thermodynamics
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19.3 The Molecular Interpretation of Entropy
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Molecular motion

• Translational Motion entire molecule moves in
  one direction (like throwing a baseball)
• Vibrational Motion atoms in the molecule
  periodically move toward and away from one
  another
• Rotational Motion molecules spin like a top

16
What is Statistical Thermodynamics?

• Its the use of tools of statistics and
  probability to provide the link between
  macroscopic and microscopic worlds
• We molecules in bulk in a microstate a single
  possible arrangement of the positions and kinetic
  energies of gas molecules in a specific
  thermodynamic state

17
So what equation do we use?

• Boltzmann created this beautiful equation
• S k lnW
• K is Boltzmanns constant which is 1.38 x
  10-23J/K
• -This means entropy is the measure of how many
  microstates are associated with a particular
  macroscopic state

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Relationship between microstates and entropy

• The more microstates, the more entropy
• Increasing volume, temperature, and number of
  molecules increases entropy due to larger number
  of microstates

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GENERALLY SPEAKING

• Entropy increases when
• Gases are formed from either solids or liquids
• Liquids or solutions are formed from solids
• The number of gas molecules increases during a
  chemical reaction

20
The Third Law of Thermodynamics

• The entropy of a pure crystalline substance at
  absolute zero is zero
• S(0 K) 0
• - Makes sense because when there is no molecular
  motion (temp is 0 K), there is only ONE microstate

21
Trends

• Sharp increase of entropy at melting points
• Sharp increase of entropy at boiling points
• Entropy of solids lt Entropy of liquids lt Entropy
  of gas

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Example

• What will the sign of change of entropy be for
  the following reactions?
• 2Na(s) Cl2 (g) 2NaCl (s)
• 2H2 (g) O2 (g) 2H2O (l)

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Answers

• Because of Boltzmanns equation, we know that the
  more particles there are in a system, the more
  microstates there are, so the more entropy there
  is, so
• The reaction to form NaCl has a negative sign for
  change in entropy, since there are less particles
  after the reaction
• The reaction to form water also has a negative
  sign for change in entropy since there are less
  particles

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Example

• In a chemical reaction two gases combine to form
  a solid. What do you expect for the sign of
  change in S? For which of the processes does the
  entropy of the system increase? A)The melting of
  ice cubes at -5 degrees Celsius and 1 atm
  pressure B) dissolution of sugar in a cup of hot
  coffee C) the reaction of nitrogen atoms to form
  N2

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Answer

• Entropy decreases when two gases form a solid, so
  negative sign
• A) Positive because the ice is melting from a
  structured solid to unstructured liquid
• B) Positive because the sugar went from being
  structured solids to numerous separated particles
• C) Negative, there is a huge decrease in the
  number of gas particles (by 1/2 to be exact)

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Practice, practice, practice!

• Page 838 s 19.27 19.40
• A closer look on page 810

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19.4 Entropy Changes in Chemical Reactions
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How do we do it?

• We cant measure change in entropy for a
  reaction, but we CAN find the absolute value of
  the entropy for many substances at any
  temperature
• Standard molar entropies are denoted by S
  (pronounced S knot) for any pure substance in
  their standard state at 1 atm

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Observations on Standard Entropy

• Standard molar entropies (SME) of elements at
  reference temperature 298 K are not zero
• SME of gases greater than liquids
• SME generally increase with molar mass
• SME generally increase with number of atoms in
  molecular formula

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So whats the equation we use?

• ?S SnS(products) SnS(reactants)
• In english
• Change in entropy equals sum of entropies of
  products minus the sum of entropies of reactants

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Entropy Changes in the Surroundings

• The change in entropy of the surroundings will
  depend on how much heat is absorbed or given off
  by the system
• Isothermal process equation
• ?Ssurr-qsys/T
• - At constant pressure, qsys is the enthalpy
  change of the system, ?H

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Example

• Calculate change in standard entropy for the
  following reaction
• 2Na(s) Cl2 (g) 2NaCl (s)

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Answer

• 1. The standard entropy change is equal to the
  standard entropy of products minus the reactants
• 2. So we must first find the standard entropy of
  the products(using a standard entropy table), and
  multiply them by the number of moles in the
  reaction
• Standard entropy of NaCl 72.1 J/K-mol, so
  2(72.1) 144.2
• 3. Then we find the standard entropy of reactants
• Standard entropy of Na 51.2 J/K-mol, so
  2(51.2) 102.4 J/K-mol
• Standard entropy of Cl2 223.1 J/K-mol, 223.1
  J/K-mol
• 4. Now we subtract product and reactants
• 144.2 J/K-mol (102.4 J/K-mol 223.1 J/K-mol)
  - 181.3 J/K-mol

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Example

• Cyclopropane and propylene isomers both have the
  formula C3H6. Based on the molecular structures
  shown, which of these isomers would you expect to
  have the higher standard molar entropy at 25
  degrees Celsius?

Cyclopropane
Propylene
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Answer

• Propylene because it is much less organized
  compared to Cyclopropane

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We love to practice!

• Page 839 s 19.41 19.48
• Example on page 821, figure 19.5

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19.5 Gibbs Free Energy
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19.5 Gibbs Free Energy

• Standard Free Energies of Formation Used to
  calculate standard free-energy change for
  chemical processes

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19.5 Gibbs Free Energy

• Equation for Standard Free-Energy Change
• ?G Sn?Gf (products) - Sm?Gf (reactants)

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19.5 Gibbs Free Energy

• G H TS
• T is the absolute temperature.
• The change in free energy of a system (?G) is
  given by ?G ?H T?S

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19.5 Gibbs Free Energy

• If ?G is negative, the reaction is spontaneous
• If ?G is zero, then reaction is at equilibrium
• If ?G is positive, the forward reaction is not
  spontaneous, but the reverse reaction is.

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19.5 Gibbs Free Energy

• For a certain chemical reaction, ?H -35.4 kJ
  and ?S - 85.5 J/K. Calculate ?G for the
  reaction at 298 K.

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19.5 Gibbs Free Energy

• For a certain chemical reaction, ?H -35.4 kJ
  and ?S - 85.5 J/K. Calculate ?G for the
  reaction at 298 K (19.51 C)
• ?G ?H T?S
• ?G - 35.4 kJ (298K)(-85.5 J/K)
• ?G - 35.4 kJ 25500 J
• ?G -35.4 kJ 25.5 kJ
• ?G -9.9 kJ

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19.5 Gibbs Free Energy

• 19.5 Problems 49 -70

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19.6 Free Energy and Temperature
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19.6 Free Energy and Temperature

• The equation ?G?H(T?S) in which ?H is the
  enthalpy term and T?S is the entropy term, can
  be used to demonstrate how free energy is
  affected by the change in temperature.
• The value of T?S depends directly on the
  absolute temperature, T, ?G will vary with the
  temperature.

47
19.6 Free Energy and Temperature

• T is a positive number at all temperatures other
  than absolute zero.
• Both the enthalpy and entropy can be either
  positive or negative.
• When ?S is positive, which means greater
  randomness than the original state, then the T?S
  is negative. When the ?S is negative, then the
  T?S is positive.

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19.6 Free Energy and Temperature

• ?G will always be negative when the process is
  spontaneous, and the opposite is true when the
  process is not spontaneous.

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19.6 Free Energy and Temperature

• H2O(s) ? H2O(l) ?Hgt0 ?Sgt0
• This process is endothermic, meaning the ?H is
  positive. Entropy also increases with this
  process, meaning that ?S is positive as well,
  which makes T?S a negative value. This means
  that the T?S term dominates, making ?G negative.
  A negative ?G means that this process is
  spontaneous at Tgt0ºC

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19.6 Free Energy and Temperature

• Calculate ?G for 2NO2(g) ? N2O4(g) at 298K using
  appendix C.

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19.6 Free Energy and Temperature

• ?H 2NO2(g) ? N2O4(g) -58.02 kJ/mol
• ?S 2NO2(g) ? N2O4(g) -175.79 J/K

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19.6 Free Energy and Temperature

• ?Gº
• -58.02kJ (298K)(-175.79J/K)(1kJ/1000J)
• -58.02kJ (298K)(-0.176kJ/K)
• -58.02kJ (-52.45kJ)
• ?Gº -5.57kJ
• This means that at 298K,
• 2NO2(g) ? N2O4(g) is a spontaneous reaction.

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19.6 Free Energy and Temperature

• Textbook problems chapter 19, 65,66,79. 80, and
  87

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19.7 Free Energy and the Equilibrium Constant
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Most reactions occur in nonstandard conditions

• To find the free energy in nonstandard reactions,
  we use the standard free energy change to
  calculate
• ?G ?G RT ln(Q)
• R is ideal gas constant 8.314 J/mol-K
• T is absolute temperature
• Q is reaction quotient for that particular
  mixture of interest

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Some more info

• At equilibrium ?G 0
• Q equilibrium constant K

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Trends

• The more negative ?G, is the larger K is
• The more positive ?G is, the smaller K is

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Example

• 2CO(g) O2 (g) 2CO2 (g)
• Calculate the change in free energy for the above
  reaction at 298K, for a reaction mixture that
  consists of 1.0 atm O2 , 2.0 atm CO , and 0.75
  atm CO2

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Answer

• First we must find Q in order to use the equation
  provided earlier. So we solve for the reaction
  quotient
• Q(PCO2 )2 / (PCO)2(PO2 )(.75 atm)2 / (2.0 atm)2
  (1.0 atm) .14atm
• 2. Now calculate standard change in free energy
  for reaction using tables
• Products Reactants, so 2(-394.4kJ/mol)
  (2(-137.2kJ/mol) 0)
• Change in standard free energy is -514.4 kJ/mol
• 3. Now we can finally use ?G ?G RT ln(Q)
• ?G (-514.4 kJ/mol) (8.314 J/mol-K)(298 K) (1
  kJ/1000J)ln(.14)
• (-514.4 kJ/mol) 4.871 kJ/mol -519.271 -
  520 kJ/mol

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Practice Problems

• Page 841
• Problems 19.71 19.95
• You can find answers on pages A-24 thru A-25

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