Chemical Thermodynamics

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Chemical Thermodynamics

• Chapter 19

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First Two Law of Thermodynamics

• 1st Law You cant win, you can only break even
• 2nd Law You cant break even

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First Law Review

• Chapter 5 (thermochemistry) we looked at energy
  accounting
• We saw that energy is transferred by doing work
  (w) or absorbing (or releasing) heat (q)
• This transferred energy is stored in a system
  or surroundings as internal energy (?E)

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How about direction?

• Add heat what happens?
• Common sense piston expands
• Need additional thermodynamic principles to
  predict this

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Spontaneous Processes

• Lets look at some everyday processes for
  principles
• If not hindered (or obstructed) concentrated
  energy will disperse or become more diffuse

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Spontaneous Processes

• Nothing about direction in which a process moves
• But we know from everyday experience that
  processes have direction
• Objects fall down not up
• Hot objects cool down
• Iron nails will rust
• The above occur without outside intervention

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Entropy and the Second Law of Thermodynamics

• The Spontaneous Expansion of a Gas
• Why does the gas expand?

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Spontaneous Processes
Reversible and Irreversible Processes
Reversible Can be returned to original state by
exactly reversing the change
Irreversible Cannot be returned to original
state by exactly reversing the change,
instead it must take a different path
with different values of q and w
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Entropy and the Second Law of Thermodynamics

• Entropy
• Suppose a system changes reversibly between state
  1(Vi) and state 2 (Vf). Then, the change in
  entropy is given by
• at constant T where qrev is the amount of heat
  added reversibly to the system.
• Since ?E 0 for reversible path (state
  function), qrev -w.
• w -?pdV p nRT/V w -nRT ln (Vf/Vi)
• ?Ssys nRlnVf/Vi

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Third Law of Thermo

• The entropy of a pure crystal at 0 K is 0 (S0).
• Gives us a starting point.
• All others must begt0.
• Standard Entropies Sº (at 298 K and 1 atm) of
  substances are listed.
• Products - Reactants to find DSº (a state
  function).
• More complex molecules higher Sº.

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Example Calculation

• Use free expansion as a tool to illustrate
• Meaning of irreversible (spontaneous) and
  reversible process
• Use of state functions and reversible path
• Look at q and w and how theyre different from
  irreversible process
• Calculation of ?S qrev/T
• Compare ?Suniv for irreversible and reversible
  process

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Spontaneous Processes
Reversible and Irreversible Processes
Reversible Can be returned to original state by
exactly reversing the change
Irreversible Cannot be returned to original
state by exactly reversing the change,
instead it must take a different path
with different values of q and w
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Entropy and the Second Law of Thermodynamics

• The Second Law of Thermodynamics
• The second law of thermodynamics explains why
  spontaneous processes have a direction.
• In any spontaneous process, the entropy of the
  universe increases.
• ?Suniv ?Ssys ?Ssurr the change in entropy of
  the universe is the sum of the change in entropy
  of the system and the change in entropy of the
  surroundings.
• Entropy is not conserved ?Suniv is increasing.

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Entropy and the Second Law of Thermodynamics

• The Second Law of Thermodynamics
• For a reversible process ?Suniv 0.
• For a spontaneous process (i.e. irreversible)
  ?Suniv gt 0.
• Note the Second Law states that the entropy of
  the universe must increase in a spontaneous
  process. It is possible for the entropy of a
  system to decrease as long as the entropy of the
  surroundings increases.
• For an isolated system, ?Ssys 0 for a
  reversible process and ?Ssys gt 0 for a
  spontaneous process.

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Spontaneous Processes
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Reversible Ice Melting

• Melting 1 mole of ice
• Put in contact with a heat source at T slightly
  greater than 0o C
• ?Ssys qrev/T
• Melting qrev ?Hfusion 6025 J/mole
• ?Ssys (1mole)(6025 J/mole)/273 K
• ?Ssys 22.1 J/K
• ?Ssurr -22.1 J/K
• ?Suniv 0

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Irreversible Melting

• Melting 1 mole of ice
• Put in contact with a heat source at T10C
• As before ?Ssys qrev/T 22.1 J/K
• But ?Ssurr qrev/T -6025 J/283 K
• ?Ssurr -21.3 J/K
• ?Suniv 22.1 J/K-21.3 J/K 0.8 J/K

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Gases

• Gases completely fill their chamber because there
  are many more ways to do that than to leave half
  empty.
• Ssolid ltSliquid ltltSgas
• there is more dispersal in liquid than a solid.
• Even more dispersal in gases.

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Entropy on the Molecular Scale

• Each thermodynamic state has a specific number of
  arrangements (microstates), W, associated with
  it.
• Entropy is
• S k lnW
• where k is the Boltzmann constant, 1.38 ? 10?23
  J/K.

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The Molecular Interpretation of Entropy

• A gas has more degrees of freedom than a liquid
  which in turn has more freedom than a solid.
• Any process that increases the number of gas
  molecules leads to an increase in entropy.
• When NO(g) reacts with O2(g) to form NO2(g), the
  total number of gas molecules decreases, and the
  entropy decreases.
• 2NO(g) O2 ? 2NO2(g)

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Entropy on the Molecular Scale

• Dispersal of energy, and therefore, entropy tends
  to increase with increases in
• Temperature.
• Volume (gases).
• The number of independently moving molecules.
• Molecule complexity

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The Molecular Interpretation of Entropy

• There are three atomic modes of motion
• translation (the moving of a molecule from one
  point in space to another),
• vibration (the shortening and lengthening of
  bonds, including the change in bond angles),
• rotation (spinning about an axis1 of 3 axis
  shown).

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Standard Entropies

• Larger and more complex molecules have greater
  entropies.

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Entropy on the Molecular Scale

• Implications
• more particles
• -gt more states -gt more entropy
• higher T
• -gt more energy states -gt more entropy
• phase (gas vs solid)
• -gt more degrees of freedom -gt more entropy

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The Molecular Interpretation of Entropy

• Boiling corresponds to a much greater change in
  entropy than melting.
• Entropy will increase when
• liquids or solutions are formed from solids,
• gases are formed from solids or liquids,
• the number of gas molecules increase,
• the temperature is increased.
• Absolute entropy at T 0o K is zero this is the
  third law of thermodynamics

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The Molecular Interpretation of Entropy

• Examples
• Determine whether ?S is positive, negative or
  zero for the following reactions.
• HCl(g) NH3(g) ? NH4Cl(s)
• 2H2O2(l) ? 2H2O(l) O2(g)
• Cooling of nitrogen gas from -20C to -50C
• CaCO3(s) ? CaO(s) CO2(g)

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Entropy Changes in Chemical Reactions

• Absolute entropy can be determined from
  complicated measurements.
• Standard molar entropy, S? entropy of a
  substance in its standard state. Similar in
  concept to ?H?.
• Units J/mol-K. Note units of ?H kJ/mol.
• Standard molar entropies of elements are not
  zero.
• For a chemical reaction which produces n moles of
  products from m moles of reactants

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Entropy Changes in Chemical Reactions

• The coefficients must be integers
• Reaction entropies can be calculated using Hesss
  Law.
• Entropy change in the surroundings can be
  calculated from the defining equation for
  entropy.
• Reaction can be any chemical reaction
• For example 2SO2 O2 ? SO3 ?So
• 2S (s, rhombic) 3O2 ? 2SO3 ?Sfo

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Entropy Changes Example Problem
Calculate ?S for the following reaction Al2O3(s)
3H2(g) ? 2Al(s) 3H2O(g) S 51.00
130.5 28.32 188.83 ?Srxn
3(188.83) 2(28.32) 1(51.00) 3(130.5)
566.49 56.64 51.00 391.5
180.6 J/K


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Gibbs Free Energy

• For a spontaneous reaction -
• The entropy of the universe must increase
• ?H can be endothermic or exothermic
• How do we balance ?S and ?H to predict whether a
  reaction is spontaneous?
• Gibbs free energy, G, of a state is
• For a process occurring at constant temperature
• For a process to occur spontaneously, ?G lt 0

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Derivation of Free Energy Relation from 2nd Law

• ?Suniv ?Ssys ?Ssurr
• ?Suniv ?Ssys qrev/T
• ?Suniv ?Ssys ?H/T
• -T?Suniv ?H - T?Ssys
• ?G ?H - T?Ssys

-T( )
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Lets Check

• For the reaction H2O(s) H2O(l)
• DSº 22.1 J/K mol DHº 6030 J/mol
• Calculate DG at 10ºC and -10ºC
• Look at the equation DGDH-TDS
• Spontaneity can be predicted from the sign of DH
  and DS.
• T 10o C, ?G -224 J
• T -10C, ?G 220 J (not spontaneous)

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Free Energy and Temperature
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• Calculating Gibbs Free Energy Changes at
  Temperature
• Use definition DGo DHo-TDSo
• Assume that DHo and DSo are independent of
  temperature
• ?G can also be calculated like ?H and ?S of rxn.

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• Calculating Gibbs Free Energy Changes
• Example
• 2SO2(g) O2(g) ?2SO3(g) at 227 oC
• DHo -196.6 kJ DSo -189.6 J/K
• DGo -196.6 500 (-0.1896) kJ
• DGo -101.8 kJ