Chapter%2019%20Principles%20of%20Reactivity:%20Entropy%20and%20Free%20Energy

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Chapter 19Principles of Reactivity Entropy and
Free Energy
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Entropy and Free Energy

• How to predict if a reaction can occur, given
  enough time?
• THERMODYNAMICS

PLAY MOVIE
How to predict if a reaction can occur at a
reasonable rate? KINETICS
PLAY MOVIE
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Thermodynamics

• If the state of a chemical system is such that a
  rearrangement of its atoms and molecules would
  decrease the energy of the system---
• AND the K is greater than 1,
• then this is a product-favored system.
• Most product-favored reactions are
  exothermicbut this is not the only criterion

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Thermodynamics

• Both product- and reactant-favored reactions can
  proceed to equilibrium in a spontaneous process.
• AgCl(s) e Ag(aq) Cl(aq) K 1.8 x 10-10
• Reaction is not product-favored, but it moves
  spontaneously toward equilibrium.
• Spontaneous does not imply anything about time
  for reaction to occur.

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Thermodynamics and Kinetics

• Diamond is thermodynamically favored to convert
  to graphite, but not kinetically favored.

Paper burns a product-favored reaction. Also
kinetically favored once reaction is begun.
PLAY MOVIE
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Spontaneous Reactions

• In general, spontaneous reactions are exothermic.
• Fe2O3(s) 2 Al(s) f 2 Fe(s) Al2O3(s)
• ?rH - 848 kJ

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Spontaneous Reactions

• But many spontaneous reactions or processes are
  endothermic or even have ?H 0.

PLAY MOVIE
?H 0
NH4NO3(s) heat f NH4NO3(aq)
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Entropy, S

• One property common to spontaneous processes is
  that the energy of the final state is more
  dispersed.
• In a spontaneous process energy goes from being
  more concentrated to being more dispersed.
• The thermodynamic property related to energy
  dispersal is ENTROPY, S.
• 2nd Law of Thermo a spontaneous process results
  in an increase in the entropy of the universe.

Reaction of K with water
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Directionality of Reactions

• Probability suggests that a spontaneous reaction
  will result in the dispersal of energy.
• Energy Dispersal

PLAY MOVIE
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Directionality of ReactionsEnergy Dispersal

• Exothermic reactions involve a release of stored
  chemical potential energy to the surroundings.
• The stored potential energy starts out in a few
  molecules but is finally dispersed over a great
  many molecules.
• The final statewith energy dispersedis more
  probable and makes a reaction spontaneous.

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Energy Dispersal
To begin, particle 1 has 2 packets of energy and
2-4 have none (upper left). With time it is more
probable energy is dispersed over two particles.
Each of these ways to distribute energy is called
a microstate.
See Figure 19.4
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Directionality of Reactions

• Matter energy dispersal

As the size of the container increases, the
number of microstates accessible to the system
increases, and the density of states increases.
Entropy increases.
PLAY MOVIE
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• The entropy of liquid water is greater than the
  entropy of solid water (ice) at 0 C.
• Energy is more dispersed in liquid water than in
  solid water.

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Entropy, S
So (J/Kmol) H2O(liq) 69.95 H2O(gas) 188.8
PLAY MOVIE

• S (solids) lt S (liquids) lt S (gases)

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Entropy and States of Matter
S(Br2 liq) lt S(Br2 gas)
S(H2O sol) lt S(H2O liq)
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Entropy, S

• Entropy of a substance increases with temperature.

Molecular motions of heptane at different temps.
Molecular motions of heptane, C7H16
PLAY MOVIE
PLAY MOVIE
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Entropy, S

• Increase in molecular complexity generally leads
  to increase in S.

PLAY MOVIE
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Entropy, S

• Entropies of ionic solids depend on coulombic
  attractions.

So (J/Kmol) MgO 26.9 NaF 51.5
PLAY MOVIE
Mg2 O2-
Na F-
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Entropy, S

• Liquids or solids dissolve in a solvent in a
  spontaneous process owing to the increase in
  entropy. Matter (and energy) are more dispersed.

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Standard Molar Entropies
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Entropy Changes for Phase Changes

• For a phase change, ?S q/T
• where q heat transferred in phase change
• For H2O (liq) f H2O(g)
• ?H q 40,700 J/mol

PLAY MOVIE
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Entropy and Temperature
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Calculating ?S for a Reaction
?So ? So (products) - ? So (reactants)

• Consider 2 H2(g) O2(g) f 2 H2O(liq)
• ?So 2 So (H2O) - 2 So (H2) So (O2)
• ?So 2 mol (69.9 J/Kmol) - 2 mol (130.7
  J/Kmol) 1 mol (205.3 J/Kmol)
• ?So -326.9 J/K
• Note that there is a decrease in S because 3 mol
  of gas give 2 mol of liquid.

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2nd Law of Thermodynamics

• A reaction is spontaneous if ?S for the universe
  is positive.
• ?Suniverse ?Ssystem ?Ssurroundings
• ?Suniverse gt 0 for spontaneous process
• Calculate the entropy created by energy dispersal
  in the system and surroundings.

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• Dissolving NH4NO3 in wateran entropy driven
  process.

PLAY MOVIE
?Suniverse ?Ssystem ?Ssurroundings
PLAY MOVIE
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• 2 H2(g) O2(g) f 2 H2O(liq)
• ?Sosystem -326.9 J/K

Can calc. that ?rHo ?Hosystem -571.7 kJ
?Sosurroundings 1917 J/K
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• 2 H2(g) O2(g) f 2 H2O(liq)
• ?Sosystem -326.9 J/K
• ?Sosurroundings 1917 J/K
• ?Souniverse 1590. J/K

• The entropy of the universe is increasing, so the
  reaction is product-favored.

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Spontaneous or Not?
Remember that ?Hsys is proportional to
?Ssurr An exothermic process has ?Ssurr gt 0.
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Gibbs Free Energy, G
?Suniv ?Ssurr ?Ssys

• Multiply through by -T
• -T?Suniv ?Hsys - T?Ssys
• -T?Suniv change in Gibbs free energy for the
  system ?Gsystem
• Under standard conditions
• ?Gosys ?Hosys - T?Sosys

J. Willard Gibbs1839-1903
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?Go ?Ho - T?So

• Gibbs free energy change
• total energy change for system
• - energy lost in energy dispersal
• If reaction is
• exothermic (negative ?Ho)
• and entropy increases (positive ?So)
• then ?Go must be NEGATIVE
• reaction is spontaneous (and product-favored).

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?Go ?Ho - T?So

• Gibbs free energy change
• total energy change for system
• - energy lost in energy dispersal
• If reaction is
• endothermic (positive ?Ho)
• and entropy decreases (negative ?So)
• then ?Go must be POSITIVE
• reaction is not spontaneous (and is
  reactant-favored).

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Gibbs Free Energy, G

• ?Go ?Ho - T?So
• ?Ho ?So ?Go Reaction
• exo() increase() Prod-favored
• endo() decrease(-) React-favored
• exo() decrease(-) ? T dependent
• endo() increase() ? T dependent

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Gibbs Free Energy, G

• ?Go ?Ho - T?So
• Two methods of calculating ?Go
• a) Determine ?rHo and ?rSo and use Gibbs
  equation.
• b) Use tabulated values of free energies of
  formation, ?fGo.

?rGo ? ?fGo (products) - ? ?fGo (reactants)
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Free Energies of Formation
Note that ?fG for an element 0
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Calculating ?rGo

• Combustion of acetylene
• C2H2(g) 5/2 O2(g) f 2 CO2(g) H2O(g)
• Use enthalpies of formation to calculate
• ?rHo -1238 kJ
• Use standard molar entropies to calculate
• ?rSo -97.4 J/K or -0.0974 kJ/K
• ?rGo -1238 kJ - (298 K)(-0.0974 kJ/K)
• -1209 kJ
• Reaction is product-favored in spite of negative
  ?rSo.
• Reaction is enthalpy driven

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Calculating ?rGo
PLAY MOVIE
NH4NO3(s) heat f NH4NO3(aq)

• Is the dissolution of ammonium nitrate
  product-favored?
• If so, is it enthalpy- or entropy-driven?

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Calculating ?rGo
NH4NO3(s) heat f NH4NO3(aq)

• From tables of thermodynamic data we find
• ?rHo 25.7 kJ
• ?rSo 108.7 J/K or 0.1087 kJ/K
• ?rGo 25.7 kJ - (298 K)(0.1087 J/K)
• -6.7 kJ
• Reaction is product-favored in spite of negative
  ?rHo.
• Reaction is entropy driven

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Gibbs Free Energy, G

• ?Go ?Ho - T?So
• Two methods of calculating ?Go
• a) Determine ?rHo and ?rSo and use Gibbs
  equation.
• b) Use tabulated values of free energies of
  formation, ?fGo.

?rGo ? ?fGo (products) - ? ?fGo (reactants)
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Calculating ?Gorxn
?rGo ? ?Gfo (products) - ? ?Gfo (reactants)

• Combustion of carbon
• C(graphite) O2(g) f CO2(g)
• ?rGo ?fGo(CO2) - ?fGo(graph) ?fGo(O2)
• ?rGo -394.4 kJ - 0 0
• Note that free energy of formation of an element
  in its standard state is 0.
• ?rGo -394.4 kJ
• Reaction is product-favored as expected.

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Free Energy and Temperature

• 2 Fe2O3(s) 3 C(s) f 4 Fe(s) 3 CO2(g)
• ?rHo 467.9 kJ ?rSo 560.3 J/K
• ?rGo 300.8 kJ
• Reaction is reactant-favored at 298 K
• At what T does ?rGo just change from being ()
  to being (-)?
• When ?rGo 0 ?rHo - T?rSo

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Thermodynamics and Keq

• FACT ?rGo is the change in free energy when pure
  reactants convert COMPLETELY to pure products.
• FACT Product-favored systems have Keq gt 1.
• Therefore, both ?rG and Keq are related to
  reaction favorability.

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Thermodynamics and Keq

• Keq is related to reaction favorability and so to
  ?rGo.
• The larger the value of K the more negative the
  value of ?rGo
• ?rGo - RT lnK
• where R 8.31 J/Kmol

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?rGo - RT lnK

• Calculate K for the reaction
• N2O4 f 2 NO2 ?rGo 4.8 kJ
• ?rGo 4800 J - (8.31 J/K)(298 K) ln K

K 0.14 When ?rGo gt 0, then K lt 1
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?G, ?G, and Keq

• ?G is change in free energy at non-standard
  conditions.
• ?G is related to ?G
• ?G ?G RT ln Q where Q reaction
  quotient
• When Q lt K or Q gt K, reaction is spontaneous.
• When Q K reaction is at equilibrium
• When ?G 0 reaction is at equilibrium
• Therefore, ?G - RT ln K

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?G, ?G, and Keq
Product Favored, ?G negative, K gt 1
See Active Figure 19.13
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• Product-favored
• 2 NO2 e N2O4
• ?rGo 4.8 kJ
• State with both reactants and products present is
  more stable than complete conversion.
• K gt 1, more products than reactants.

PLAY MOVIE
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?G, ?G, and Keq
Reactant Favored, ?G positive, K lt 1
See Active Figure 19.13
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?G, ?G, and Keq

• Reactant-favored
• N2O4 e 2 NO2 ?rGo 4.8 kJ
• State with both reactants and products present is
  more stable than complete conversion.
• K lt 1, more reactants than products

PLAY MOVIE
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Thermodynamics and Keq

• Keq is related to reaction favorability.
• When ?rGo lt 0, reaction moves energetically
  downhill
• ?rGo is the change in free energy when reactants
  convert COMPLETELY to products.

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A Summary
The relation of ?rG, ?rG, Q, K, reaction
spontaneity, and product- or reactant
favorability.