entropy and free energy how to predict if a reaction can occur, given enough time? thermodynamics...
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Entropy and Free EnergyEntropy and Free EnergyEntropy and Free EnergyEntropy and Free EnergyHow to predict if a How to predict if a
reaction can occur, reaction can occur, given enough time?given enough time?
THERMODYNAMICSTHERMODYNAMICS
How to predict if a How to predict if a reaction can occur at reaction can occur at a reasonable rate?a reasonable rate?
KINETICSKINETICSCopyright © 1999 by Harcourt Brace & CompanyAll rights reserved.Requests for permission to make copies of any part of the work should be mailed to: Permissions Department, Harcourt Brace & Company, 6277 Sea Harbor Drive, Orlando, Florida
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ThermodynamicsThermodynamicsThermodynamicsThermodynamics• Is the state of a chemical system such that a Is the state of a chemical system such that a
rearrangement of its atoms and molecules would rearrangement of its atoms and molecules would decrease the energy of the system? decrease the energy of the system?
• If yes, system is favored to react — a If yes, system is favored to react — a product-product-favoredfavored system.system.
• Most product-favored reactions are Most product-favored reactions are exothermic.exothermic.
• Often referred to as Often referred to as spontaneousspontaneous reactions.reactions.
• Spontaneous does not imply anything about time Spontaneous does not imply anything about time for reaction to occur.for reaction to occur.
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Diamond is Diamond is thermodynamically thermodynamically favored to convert to favored to convert to graphite, but not graphite, but not kinetically favored.kinetically favored.
Thermodynamics and KineticsThermodynamics and KineticsThermodynamics and KineticsThermodynamics and Kinetics
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Thermodynamics and KineticsThermodynamics and KineticsThermodynamics and KineticsThermodynamics and KineticsDiamond is Diamond is
thermodynamically thermodynamically favored to convert to favored to convert to graphite, but not graphite, but not kinetically favored.kinetically favored.
Paper burns — a Paper burns — a product-favored product-favored reaction. Also reaction. Also kinetically favored kinetically favored once reaction is once reaction is begun.begun.
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Product-Favored ReactionsProduct-Favored ReactionsIn general, product-In general, product-
favored reactions are favored reactions are exothermicexothermic..
FeFe22OO33(s) + 2 Al(s) (s) + 2 Al(s) ------
> 2 Fe(s) + Al> 2 Fe(s) + Al22OO33(s)(s)
H = - 848 kJH = - 848 kJ
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Product-Favored ReactionsProduct-Favored ReactionsBut many spontaneous reactions or But many spontaneous reactions or
processes are endothermic or even processes are endothermic or even have have H = 0.H = 0.
NHNH44NONO33(s) + heat ---> NH(s) + heat ---> NH44NONO33(aq)(aq)
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Entropy, SEntropy, SEntropy, SEntropy, SOne property common to One property common to
product-favored processes is product-favored processes is that the final state is more that the final state is more DISORDEREDDISORDERED or or RANDOMRANDOM than the original.than the original.
Spontaneity is related to an Spontaneity is related to an increase in randomness.increase in randomness.
The thermodynamic property The thermodynamic property related to randomness is related to randomness is ENTROPY, SENTROPY, S..
Reaction of K Reaction of K with waterwith water
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The entropy of liquid water is greater than the entropy of solid water (ice) at 0 C.
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How probable is it that reactant How probable is it that reactant molecules will react? molecules will react?
PROBABILITYPROBABILITY suggests that a suggests that a product-favored reaction will product-favored reaction will result in the result in the dispersal of energy dispersal of energy or of matter or of matter or both.or both.
Directionality of ReactionsDirectionality of Reactions
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Probability suggests that a product-Probability suggests that a product-favored reaction will result in the favored reaction will result in the dispersal of energy or of matter or dispersal of energy or of matter or both.both.
Matter DispersalMatter Dispersal
Directionality of ReactionsDirectionality of Reactions
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Probability suggests that a product-Probability suggests that a product-favored reaction will result in the favored reaction will result in the dispersal of energy or of matter or dispersal of energy or of matter or both.both.
Matter DispersalMatter Dispersal
Directionality of ReactionsDirectionality of Reactions
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Probability suggests that a product-Probability suggests that a product-favored reaction will result in the favored reaction will result in the dispersal of energy or of matter or dispersal of energy or of matter or both.both.
Energy DispersalEnergy Dispersal
Directionality of ReactionsDirectionality of Reactions
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Probability suggests that a product-Probability suggests that a product-favored reaction will result in the favored reaction will result in the dispersal of energy or of matter or dispersal of energy or of matter or both.both.
Energy DispersalEnergy Dispersal
Directionality of ReactionsDirectionality of Reactions
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Directionality of Directionality of Reactions —Reactions —
Energy DispersalEnergy DispersalExothermic reactions involve a release of Exothermic reactions involve a release of
stored chemical potential energy to the stored chemical potential energy to the surroundings. surroundings.
The stored potential energy starts out in a few The stored potential energy starts out in a few molecules but is finally dispersed over a molecules but is finally dispersed over a great many molecules. great many molecules.
The final state—with energy dispersed—is The final state—with energy dispersed—is more probable and makes a reaction more probable and makes a reaction product-favored.product-favored.
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S (gases) > S (liquids) > S (solids)S (gases) > S (liquids) > S (solids)
SSoo (J/K•mol) (J/K•mol)
HH22O(liq)O(liq) 69.9169.91
HH22O(gas)O(gas) 188.8 188.8
SSoo (J/K•mol) (J/K•mol)
HH22O(liq)O(liq) 69.9169.91
HH22O(gas)O(gas) 188.8 188.8
Entropy, SEntropy, SEntropy, SEntropy, S
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Entropy of a substance increases Entropy of a substance increases with temperature.with temperature.
Molecular motions Molecular motions of heptane, Cof heptane, C77HH1616
Molecular motions of Molecular motions of heptane at different temps.heptane at different temps.
Entropy, SEntropy, SEntropy, SEntropy, S
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Increase in molecular complexity Increase in molecular complexity generally leads to increase in S.generally leads to increase in S.
SSoo (J/K•mol) (J/K•mol)
CHCH44 248.2248.2
CC22HH66 336.1 336.1
CC33HH88 419.4419.4
SSoo (J/K•mol) (J/K•mol)
CHCH44 248.2248.2
CC22HH66 336.1 336.1
CC33HH88 419.4419.4
Entropy, SEntropy, SEntropy, SEntropy, S
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Entropies of ionic solids depend on Entropies of ionic solids depend on coulombic attractions.coulombic attractions.
SSoo (J/K•mol) (J/K•mol)
MgOMgO 26.926.9
NaFNaF 51.551.5
SSoo (J/K•mol) (J/K•mol)
MgOMgO 26.926.9
NaFNaF 51.551.5
Entropy, SEntropy, SEntropy, SEntropy, S
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Entropy usually increases when a Entropy usually increases when a pure liquid or solid dissolves in a pure liquid or solid dissolves in a solvent.solvent.
Entropy, SEntropy, SEntropy, SEntropy, S
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Entropy Changes for Phase ChangesEntropy Changes for Phase Changes
For a phase change, For a phase change, S = q/TS = q/T
where q = heat transferred in where q = heat transferred in phase changephase change
For HFor H22O (liq) ---> HO (liq) ---> H22O(g)O(g)
H = q = +40,700 J/molH = q = +40,700 J/mol
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Entropy Changes for Phase ChangesEntropy Changes for Phase Changes
For a phase change, For a phase change, S = q/TS = q/T
where q = heat transferred in where q = heat transferred in phase changephase change
For HFor H22O (liq) ---> HO (liq) ---> H22O(g)O(g)
H = q = +40,700 J/molH = q = +40,700 J/mol
S = qT
= 40, 700 J/mol
373.15 K = + 109 J/K • molS =
qT
= 40, 700 J/mol
373.15 K = + 109 J/K • mol
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Consider 2 HConsider 2 H22(g) + O(g) + O22(g) ---> 2 H(g) ---> 2 H22O(liq)O(liq)
SSoo = 2 S = 2 Soo (H (H22O) - [2 SO) - [2 Soo (H (H22) + S) + Soo (O (O22)])]
SSoo = 2 mol (69.9 J/K•mol) - = 2 mol (69.9 J/K•mol) - [2 mol (130.7 J/K•mol) + [2 mol (130.7 J/K•mol) +
1 mol (205.3 J/K•mol)]1 mol (205.3 J/K•mol)]
SSoo = -326.9 J/K = -326.9 J/K
Note that there is a Note that there is a decrease in S decrease in S because 3 because 3 mol of gas give 2 mol of liquid.mol of gas give 2 mol of liquid.
Calculating Calculating S for a ReactionS for a Reaction
SSoo = = S Soo (products) - (products) - S Soo (reactants) (reactants)SSoo = = S Soo (products) - (products) - S Soo (reactants) (reactants)
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2nd Law of Thermodynamics2nd Law of ThermodynamicsA reaction is spontaneous (product-favored) if ²S A reaction is spontaneous (product-favored) if ²S
for the universe is positive.for the universe is positive.
SSuniverseuniverse = = SSsystemsystem + + SSsurroundingssurroundings
SSuniverseuniverse > 0 for product-favored process > 0 for product-favored process
First calc. entropy created by matter dispersal First calc. entropy created by matter dispersal ((SSsystemsystem))
Next, calc. entropy created by energy dispersal Next, calc. entropy created by energy dispersal ((SSsurroundsurround))
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Dissolving NH4NO3 in water—an entropy driven process.
2nd Law of Thermodynamics2nd Law of Thermodynamics
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2 H2 H22(g) + O(g) + O22(g) ---> 2 H(g) ---> 2 H22O(liq)O(liq)
SSoosystemsystem = -326.9 J/K = -326.9 J/K
2nd Law of Thermodynamics2nd Law of Thermodynamics
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2 H2 H22(g) + O(g) + O22(g) ---> 2 H(g) ---> 2 H22O(liq)O(liq)
SSoosystemsystem = -326.9 J/K = -326.9 J/K
Sosurroundings =
qsurroundings
T =
-Hsystem
TSo
surroundings = qsurroundings
T =
-Hsystem
T
2nd Law of Thermodynamics2nd Law of Thermodynamics
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2 H2 H22(g) + O(g) + O22(g) ---> 2 H(g) ---> 2 H22O(liq)O(liq)
SSoosystemsystem = -326.9 J/K = -326.9 J/K
Can calc. that Can calc. that HHoorxnrxn = = HHoo
systemsystem = -571.7 kJ = -571.7 kJ
Sosurroundings =
qsurroundings
T =
-Hsystem
TSo
surroundings = qsurroundings
T =
-Hsystem
T
2nd Law of Thermodynamics2nd Law of Thermodynamics
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2 H2 H22(g) + O(g) + O22(g) ---> 2 H(g) ---> 2 H22O(liq)O(liq)
SSoosystemsystem = -326.9 J/K = -326.9 J/K
Can calc. that Can calc. that HHoorxnrxn = = HHoo
systemsystem = -571.7 kJ = -571.7 kJ
Sosurroundings =
qsurroundings
T =
-Hsystem
TSo
surroundings = qsurroundings
T =
-Hsystem
T
Sosurroundings =
- (-571.7 kJ)(1000 J/kJ)
298.15 KSo
surroundings = - (-571.7 kJ)(1000 J/kJ)
298.15 K
2nd Law of Thermodynamics2nd Law of Thermodynamics
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2 H2 H22(g) + O(g) + O22(g) ---> 2 H(g) ---> 2 H22O(liq)O(liq)
SSoosystemsystem = -326.9 J/K = -326.9 J/K
Can calc. that Can calc. that HHoorxnrxn = = HHoo
systemsystem = -571.7 kJ = -571.7 kJ
SSoosurroundingssurroundings = +1917 J/K = +1917 J/K
Sosurroundings =
qsurroundings
T =
-Hsystem
TSo
surroundings = qsurroundings
T =
-Hsystem
T
Sosurroundings =
- (-571.7 kJ)(1000 J/kJ)
298.15 KSo
surroundings = - (-571.7 kJ)(1000 J/kJ)
298.15 K
2nd Law of Thermodynamics2nd Law of Thermodynamics
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2 H2 H22(g) + O(g) + O22(g) ---> 2 H(g) ---> 2 H22O(liq)O(liq)
SSoosystemsystem = -326.9 J/K = -326.9 J/K
SSoosurroundingssurroundings = +1917 J/K = +1917 J/K
SSoouniverse universe = +1590. J/K= +1590. J/K
The entropy of the universe The entropy of the universe is increasing, so the is increasing, so the reaction is product-favored. reaction is product-favored.
2nd Law of Thermodynamics2nd Law of Thermodynamics
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2 H2 H22(g) + O(g) + O22(g) ---> 2 H(g) ---> 2 H22O(liq)O(liq)
SSoosystemsystem = -326.9 J/K = -326.9 J/K
SSoosurroundingssurroundings = +1917 J/K = +1917 J/K
SSoouniverse universe = +1590. J/K= +1590. J/K
The entropy of the universe is increasing, so the The entropy of the universe is increasing, so the reaction is product-favored.reaction is product-favored.
2nd Law of Thermodynamics2nd Law of Thermodynamics
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Gibbs Free Energy, GGibbs Free Energy, GGibbs Free Energy, GGibbs Free Energy, G
²S²Sunivuniv = ²S = ²Ssurrsurr + ²S + ²Ssyssys
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Gibbs Free Energy, GGibbs Free Energy, GGibbs Free Energy, GGibbs Free Energy, G
²S²Sunivuniv = ²S = ²Ssurrsurr + ²S + ²Ssyssys
Suniv = Hsys
T + Ssys
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Gibbs Free Energy, GGibbs Free Energy, GGibbs Free Energy, GGibbs Free Energy, G
²S²Sunivuniv = ²S = ²Ssurrsurr + ²S + ²Ssyssys
Multiply through by -TMultiply through by -T
Suniv = Hsys
T + Ssys
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Gibbs Free Energy, GGibbs Free Energy, GGibbs Free Energy, GGibbs Free Energy, G
SSunivuniv = = S Ssurrsurr + + S Ssyssys
Multiply through by -TMultiply through by -T
-T -T S Sunivuniv = = H Hsyssys - T - T S Ssyssys
Suniv = Hsys
T + Ssys
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Gibbs Free Energy, GGibbs Free Energy, GGibbs Free Energy, GGibbs Free Energy, G
SSunivuniv = = S Ssurrsurr + + S Ssyssys
Multiply through by -TMultiply through by -T
-T -T S Sunivuniv = = H Hsyssys - T - T S Ssyssys
-T -T S Sunivuniv = change in Gibbs free energy = change in Gibbs free energy for the system = for the system = G Gsystemsystem
Suniv = Hsys
T + Ssys
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Gibbs Free Energy, GGibbs Free Energy, GGibbs Free Energy, GGibbs Free Energy, GSSunivuniv = = SSsurrsurr + + SSsyssys
Multiply through by -TMultiply through by -T
-T-TSSunivuniv = = HHsyssys - T - TSSsyssys
-T-TSSunivuniv = change in Gibbs free energy for the = change in Gibbs free energy for the system = system = GGsystemsystem
Under standard conditions —Under standard conditions —
GGoo = = HHoo - T - TSSoo
Suniv = Hsys
T + Ssys
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Gibbs Free Energy, GGibbs Free Energy, GGibbs Free Energy, GGibbs Free Energy, G
GGoo = = HHoo - T - T SSoo
Gibbs free energy change = Gibbs free energy change =
total energy change for system total energy change for system - - energy lost in disordering the systemenergy lost in disordering the system
If reaction is exothermic (If reaction is exothermic (HHoo negative) and negative) and entropy increases (entropy increases (SSoo is +), then is +), then GGoo must be must be negative and reaction product-favored.negative and reaction product-favored.
If reaction is endothermic (If reaction is endothermic (HHoo is +), and is +), and entropy decreases (entropy decreases (SSoo is -), then is -), then GGoo must be must be + and reaction is reactant-favored.+ and reaction is reactant-favored.
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Gibbs Free Energy, GGibbs Free Energy, GGibbs Free Energy, GGibbs Free Energy, G
GGoo = = HHoo - T - TSSoo
HHoo SSoo GGoo ReactionReaction
exo(-)exo(-) increase(+)increase(+) -- Prod-favoredProd-favored
endo(+)endo(+) decrease(-)decrease(-) ++ React-favoredReact-favored
exo(-)exo(-) decrease(-)decrease(-) ?? T dependentT dependent
endo(+)endo(+) increase(+)increase(+) ?? T dependentT dependent
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Gibbs Free Energy, GGibbs Free Energy, GGibbs Free Energy, GGibbs Free Energy, G
GGoo = = HHoo - T - TSSoo
Two methods of calculating Two methods of calculating GGoo
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Gibbs Free Energy, GGibbs Free Energy, GGibbs Free Energy, GGibbs Free Energy, G
GGoo = = HHoo - T - TSSoo
Two methods of calculating Two methods of calculating GGoo
a)a) Determine Determine HHoorxnrxn and and SSoo
rxnrxn and use and use
GIbbs equation.GIbbs equation.
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Gibbs Free Energy, GGibbs Free Energy, GGibbs Free Energy, GGibbs Free Energy, G
GGoo = = HHoo - T - TSSoo
Two methods of calculating Two methods of calculating GGoo
a)a) Determine Determine HHoorxnrxn and and SSoo
rxnrxn and use and use
GIbbs equation.GIbbs equation.
b)b) Use tabulated values of free energies Use tabulated values of free energies
of formation, of formation, GGffoo..
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Gibbs Free Energy, GGibbs Free Energy, GGibbs Free Energy, GGibbs Free Energy, G
GGoo = = HHoo - T - TSSoo
Two methods of calculating Two methods of calculating GGoo
a)a) Determine Determine HHoorxnrxn and and SSoo
rxnrxn and use and use
GIbbs equation.GIbbs equation.
b)b) Use tabulated values of free energies Use tabulated values of free energies
of formation, of formation, GGffoo..
²G²Goorxnrxn = = ²G ²Gff
oo (products) - (products) - ²G ²Gffoo (reactants) (reactants)²G²Goo
rxnrxn = = ²G ²Gffoo (products) - (products) - ²G ²Gff
oo (reactants) (reactants)
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Calculating Calculating GGoorxnrxn
Calculating Calculating GGoorxnrxn
Combustion of acetyleneCombustion of acetylene
CC22HH22(g) + 5/2 O(g) + 5/2 O22(g) --> 2 CO(g) --> 2 CO22(g) + H(g) + H22O(g)O(g)
Use enthalpies of formation to calculateUse enthalpies of formation to calculate
HHoorxnrxn = -1238 kJ = -1238 kJ
Use standard molar entropies to calculateUse standard molar entropies to calculate
SSoorxnrxn = -97.4 J/K or -0.0974 kJ/K = -97.4 J/K or -0.0974 kJ/K
GGoorxnrxn = -1238 kJ - (298 K)(-0.0974 J/K) = -1238 kJ - (298 K)(-0.0974 J/K)
= -1209 kJ= -1209 kJ
Reaction is Reaction is product-favoredproduct-favored in spite of negative in spite of negative SSoorxnrxn. .
Reaction is Reaction is “enthalpy driven”“enthalpy driven”
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Calculating Calculating GGoorxnrxn
Calculating Calculating GGoorxnrxn
Is the dissolution of ammonium nitrate product-Is the dissolution of ammonium nitrate product-favored? favored?
If so, is it enthalpy- or entropy-driven?If so, is it enthalpy- or entropy-driven?
NHNH44NONO33(s) + heat ---> NH(s) + heat ---> NH44NONO33(aq)(aq)
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Calculating Calculating GGoorxnrxn
Calculating Calculating GGoorxnrxn
From tables of thermodynamic data we findFrom tables of thermodynamic data we find
HHoorxnrxn = +25.7 kJ = +25.7 kJ
SSoorxnrxn = +108.7 J/K or +0.1087 kJ/K = +108.7 J/K or +0.1087 kJ/K
GGoorxnrxn = +25.7 kJ - (298 K)(+0.1087 J/K) = +25.7 kJ - (298 K)(+0.1087 J/K)
= -6.7 kJ= -6.7 kJ
Reaction is Reaction is product-favoredproduct-favored in spite of negative in spite of negative HHoorxnrxn. .
Reaction is Reaction is “entropy driven”“entropy driven”
NHNH44NONO33(s) + heat ---> NH(s) + heat ---> NH44NONO33(aq)(aq)
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Calculating Calculating GGoorxnrxn
Calculating Calculating GGoorxnrxn
Combustion of carbonCombustion of carbon
C(graphite) + OC(graphite) + O22(g) --> CO(g) --> CO22(g) (g)
GGoorxnrxn = = GGff
oo(CO(CO22) - [) - [GGffoo(graph) + (graph) + GGff
oo(O(O22)])]
GGoorxnrxn = -394.4 kJ - [ 0 + 0] = -394.4 kJ - [ 0 + 0]
Note that free energy of formation of an element Note that free energy of formation of an element in its standard state is 0.in its standard state is 0.
GGoorxnrxn = -394.4 kJ = -394.4 kJ
Reaction is Reaction is product-favoredproduct-favored as expected. as expected.
GGoorxnrxn = = ²G ²Gff
oo (products) - (products) - GGffoo (reactants) (reactants)GGoo
rxnrxn = = ²G ²Gffoo (products) - (products) - GGff
oo (reactants) (reactants)
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Free Energy and TemperatureFree Energy and Temperature
2 Fe2 Fe22OO33(s) + 3 C(s) ---> 4 Fe(s) + 3 CO(s) + 3 C(s) ---> 4 Fe(s) + 3 CO22(g)(g)
HHoorxnrxn = +467.9 kJ = +467.9 kJ SSoo
rxnrxn = +560.3 J/K = +560.3 J/K
GGoorxnrxn = +300.8 kJ = +300.8 kJ
Reaction is Reaction is reactant-favoredreactant-favored at 298 K at 298 K
At what T does At what T does GGoorxnrxn just change from being just change from being
(+) to being (-)? (+) to being (-)?
When When GGoorxnrxn = 0 = = 0 = HHoo
rxnrxn - T - TSSoorxnrxn
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Free Energy and TemperatureFree Energy and Temperature
2 Fe2 Fe22OO33(s) + 3 C(s) ---> 4 Fe(s) + 3 CO(s) + 3 C(s) ---> 4 Fe(s) + 3 CO22(g)(g)
HHoorxnrxn = +467.9 kJ = +467.9 kJ SSoo
rxnrxn = +560.3 J/K = +560.3 J/K
GGoorxnrxn = +300.8 kJ = +300.8 kJ
Reaction is Reaction is reactant-favoredreactant-favored at 298 K at 298 K
At what T does At what T does GGoorxnrxn just change from being just change from being
(+) to being (-)? (+) to being (-)?
When When GGoorxnrxn = 0 = = 0 = HHoo
rxnrxn - T - TSSoorxnrxn
T = HrxnSrxn
= 467.9 kJ
0.5603 kJ/K = 835.1 K
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KKeqeq is related to reaction favorability. is related to reaction favorability.
When When GGoorxnrxn < 0, reaction moves energetically < 0, reaction moves energetically
“downhill”“downhill”
²G²Goorxnrxn is the change in free energy as reactants is the change in free energy as reactants
convert completely to products.convert completely to products.
But systems often reach a state of equilibrium But systems often reach a state of equilibrium in which reactants have not converted in which reactants have not converted completely to products.completely to products.
In this case In this case GGrxnrxn is < is < GGoorxnrxn , so state with both , so state with both
reactants and products present is more reactants and products present is more stable than complete conversion.stable than complete conversion.
Thermodynamics and KThermodynamics and Keqeq
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Product-favored Product-favored reaction. reaction.
2 NO2 NO22 ---> N ---> N22OO44
GGoorxnrxn = -4.8 kJ = -4.8 kJ
Here Here GGrxnrxn is less than is less than GGoo
rxnrxn , so the state , so the state with both reactants with both reactants and products and products present is more present is more stable than complete stable than complete conversion.conversion.
Thermodynamics and KThermodynamics and Keqeq
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Reactant-favored Reactant-favored reaction. reaction.
NN22OO44 --->2 NO --->2 NO22 GGoo
rxnrxn = +4.8 kJ = +4.8 kJ
Here Here GGoorxnrxn is greater is greater
than than GGrxnrxn , so the , so the state with both state with both reactants and reactants and products present is products present is more stable than more stable than complete conversion.complete conversion.
Thermodynamics and KThermodynamics and Keqeq
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KKeqeq is related to reaction favorability and so is related to reaction favorability and so to to GGoo
rxnrxn..
The larger the value of The larger the value of GGoorxnrxn the larger the the larger the
value of K.value of K.
GGoorxnrxn = - RT lnK = - RT lnK
where R = 8.31 J/K•molwhere R = 8.31 J/K•mol
Thermodynamics and KThermodynamics and Keqeq
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Calculate K for the reactionCalculate K for the reaction
NN22OO44 --->2 NO --->2 NO22 GGoorxnrxn = +4.8 kJ = +4.8 kJ
GGoorxnrxn = - RT lnK = - RT lnK
Thermodynamics and KThermodynamics and Keqeq
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Calculate K for the reactionCalculate K for the reaction
NN22OO44 --->2 NO --->2 NO22 GGoorxnrxn = +4.8 kJ = +4.8 kJ
GGoorxnrxn = +4800 J = - (8.31 J/K)(298 K) ln K = +4800 J = - (8.31 J/K)(298 K) ln K
GGoorxnrxn = - RT lnK = - RT lnK
Thermodynamics and KThermodynamics and Keqeq
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Calculate K for the reactionCalculate K for the reaction
NN22OO44 --->2 NO --->2 NO22 GGoorxnrxn = +4.8 kJ = +4.8 kJ
GGoorxnrxn = +4800 J = - (8.31 J/K)(298 K) ln K = +4800 J = - (8.31 J/K)(298 K) ln K
GGoorxnrxn = - RT lnK = - RT lnK
lnK = -4800 J
(8.31 J/K)(298K) = - 1.94
Thermodynamics and KThermodynamics and Keqeq
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Calculate K for the reactionCalculate K for the reaction
NN22OO44 --->2 NO --->2 NO22 GGoorxnrxn = +4.8 kJ = +4.8 kJ
GGoorxnrxn = +4800 J = - (8.31 J/K)(298 K) ln K = +4800 J = - (8.31 J/K)(298 K) ln K
K = 0.14K = 0.14
GGoorxnrxn = - RT lnK = - RT lnK
lnK = -4800 J
(8.31 J/K)(298K) = - 1.94
Thermodynamics and KThermodynamics and Keqeq
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Calculate K for the reactionCalculate K for the reaction
NN22OO44 --->2 NO --->2 NO22 GGoorxnrxn = +4.8 kJ = +4.8 kJ
GGoorxnrxn = +4800 J = - (8.31 J/K)(298 K) ln K = +4800 J = - (8.31 J/K)(298 K) ln K
K = 0.14K = 0.14
When When GGoorxnrxn > 0, then K < 1 > 0, then K < 1
GGoorxnrxn = - RT lnK = - RT lnK
lnK = -4800 J
(8.31 J/K)(298K) = - 1.94
Thermodynamics and KThermodynamics and Keqeq