Transcript Entropy and Free Energy

Entropy and Free Energy
(Kotz Ch 20) - Lecture #2
• Spontaneous vs. non-spontaneous
• thermodynamics vs. kinetics
• entropy = randomness (So)
• Gibbs free energy (Go)
• Go for reactions - predicting spontaneous direction
• thermodynamics of coupled reactions
• Grxn versus Gorxn
• predicting equilibrium constants from Gorxn
10 Nov 97
Entropy & Free Energy (Ch 20) - lect. 2
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Entropy and Free Energy
( Kotz Ch 20 )
• some processes are spontaneous;
others never occur. WHY ?
• How can we predict if a reaction
can occur, given enough time?
THERMODYNAMICS
9-paper.mov
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• Note: Thermodynamics DOES NOT say how
quickly (or slowly) a reaction will occur.
• To predict if a reaction can occur at a
reasonable rate, one needs to consider: KINETICS
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Product-Favored Reactions
In general, productfavored reactions
are exothermic.
E.g. thermite reaction
Fe2O3(s) + 2 Al(s)
 2 Fe(s) + Al2O3(s)
H = - 848 kJ
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Non-exothermic spontaneous reactions
But many spontaneous reactions or
processes are endothermic . . .
NH4NO3(s) + heat  NH4+ (aq) + NO3- (aq)
Hsol = +25.7 kJ/mol
or have H = 0 . . .
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PROBABILITY - predictor of most stable state
WHY DO PROCESSES with  H = 0 occur ?
Consider expansion of gases to equal pressure:
This is spontaneous because the final state,
with equal # molecules in each flask,
is much more probable than the initial state,
with all molecules in flask 1, none in flask 2
SYSTEM CHANGES to state of HIGHER PROBABILITY
For entropy-driven reactions - the more RANDOM state.
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Standard Entropies, So
• Every substance at a given temperature and in
a specific phase has a well-defined Entropy
• At 298o the entropy of a substance is called
So - with UNITS of J.K-1.mol-1
• The larger the value of So, the greater the
degree of disorder or randomness
e.g. So (in J.K-1mol-1) :
Br2 (liq) = 152.2
Br2 (gas) = 245.5
For any process:
So =  So(final) -  So(initial)
So(vap., Br2) = (245.5-152.2) = 93.3 J.K-1mol-1
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Entropy and Phase
Vapour
So (J/K•mol)
H2O(gas) 188.8
Ice
Water
H2O(liq)
69.9
H2O (s)
47.9
S (gases) > S (liquids) > S (solids)
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Entropy and Temperature
The entropy of a substance increases
with temperature.
Molecular motions of
heptane at different
temps.
Higher T means :
• more randomness
• larger S
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9_heptane.mov
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Entropy and complexity
Increase in molecular complexity
generally leads to increase in S.
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So (J/K•mol)
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CH4
248.2
C2H6
336.1
C3H8
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Entropy of Ionic Substances
• Ionic Solids : Entropy depends on extent
of motion of ions. This depends on the
strength of coulombic attraction.
ion pairs
So (J/K•mol)
MgO
Mg2+ / O2-
26.9
NaF
Na+ / F-
51.5
• Entropy increases when a pure liquid or
solid dissolves in a solvent.
NH4NO3(s)  NH4+ (aq) + NO3- (aq)
Ssol = So(aq. ions) - So(s) = 259.8 - 151.1
= 108.7 J.K-1mol-1
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Entropy Change in a Phase Changes
For a phase change,
S
= q/T
where q = heat transferred in phase change
H2O (liq)  H2O(g)
For vaporization of water:
H = q = +40,700 J/mol
q
40,700 J/mol
S =
=
= + 109 J/K• mol
T
373.15 K
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Calculating S for a Reaction
 So =
 So (products) -  So (reactants)
Consider 2 H2(g) + O2(g)  2 H2O(liq)
So = 2 So (H2O) - [2 So (H2) + So (O2)]
So = 2 mol (69.9 J/K•mol) [2 mol (130.7 J/K•mol) + 1 mol (205.3 J/K•mol)]
So = -326.9 J/K
Note that there is a decrease in S because 3 mol
of gas give 2 mol of liquid.
If S DECREASES,
why is this a SPONTANEOUS REACTION??
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2nd Law of Thermodynamics
A reaction is spontaneous (product-favored) if
S for the universe is positive.
Suniverse
=
Ssystem
+ Ssurroundings
Suniverse
> 0 for product-favored process
First calc. entropy created by matter dispersal
(Ssystem)
Next, calc. entropy created by energy dispersal
(Ssurround)
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Calculating S(universe)
2 H2(g) + O2(g)  2 H2O(liq)
Sosystem = -326.9 J/K
S o surroundin gs =
q surroundin gs
T
- H o system
=
T
Can calculate that Horxn = Hosystem = -571.7 kJ
S
o
surroundin gs
- (-571.7 kJ)(1000 J/kJ)
=
298.15 K
Sosurroundings = +1917 J/K
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Calculating S(universe) (2)
2 H2(g) + O2(g)  2 H2O(liq)
Sosystem
= -326.9 J/K (less matter dispersal)
Sosurroundings
= +1917 J/K (more energy dispersal)
Souniverse
= +1590 J/K
The entropy of the universe increases
so the reaction is spontaneous.
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The Laws of Thermodynamics
0. Two bodies in thermal equilibrium are at same T
1. Energy can never be created or destroyed.
E =q+w
2. The total entropy of the UNIVERSE
( = system plus surroundings) MUST INCREASE
in every spontaneous process.
 STOTAL =  Ssystem +  Ssurroundings > 0
3. The entropy (S) of a pure, perfectly crystalline
compound at T = 0 K is ZERO. (no disorder)
ST=0 = 0 (perfect xll)
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Gibbs Free Energy, G
Suniv = Ssurr + Ssys
Suniv =
Hsys
+ Ssys
T
Multiply through by -T
-TSuniv = Hsys - TSsys
-TSuniv = change in Gibbs free energy
for the system = Gsystem
Under standard conditions —
Go
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=
Ho
-
TSo
Entropy & Free Energy (Ch 20) - lect. 2
The Gibbs
Equation
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Standard Gibbs Free Energies, Gof
• Every substance in a specific state has a
Gibbs Free Energy, G = H - TS
• recall: only H can be measured. Therefore:
there is no absolute scale for G
• only G values can be determined
• Gof the Gibbs Free Energy of formation (from
elements) is used as the “standard value”
• We set the scale of G to be consistent with that
for H 10 Nov 97
Gof for elements in standard states = 0
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Sign of G for Spontaneous processes
2nd LAW requirement for SPONTANEITY is :
STOTAL = Ssystem + Ssurroundings > 0
Multiply by T
and
Thus
TSsystem + TSsurroundings > 0
Ssurroundings = Hosystem/T
TSsystem - Hosystem > 0
Multiply by -1 (->reverse > to <), drop subscript “system”
Ho -TS < 0
Go < 0
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and Go = Ho -TS
for all SPONTANEOUS processes
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Sign of Gibbs Free Energy, G
Go = Ho - TSo
• change in Gibbs free energy =
(total free energy change for system - free energy lost in
disordering the system)
• If reaction is exothermic (Ho is -ve) and
entropy increases (So is +ve), then
Go must be -ve and reaction CAN proceed.
• If reaction is endothermic (Ho is +ve), and
entropy decreases (So is -ve), then
Go must be +ve; reaction CANNOT proceed.
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Gibbs Free Energy changes for reactions
o
G
=
Ho
So
exo (-)
endo(+)
exo (-)
endo(+)
increase(+)
decrease(-)
decrease(-)
increase(+)
o
H
-
o
TS
Go
+
?
?
Reaction
Product-favored
Reactant-favored
T dependent
T dependent
Spontaneous in last 2 cases only if
Temperature is such that Go < 0
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Methods of calculating G
Go = Ho - TSo
Two methods of calculating Go
a) Determine Horxn and Sorxn and use Gibbs
equation.
b) Use tabulated values of free energies of
formation, Gfo.
Gorxn =  Gfo (products) -   Gfo (reactants)
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Calculating Gorxn
EXAMPLE: Combustion of acetylene
C2H2(g) + 5/2 O2(g)  2 CO2(g) + H2O(g)
From standard enthalpies of formation: Horxn = -1238 kJ
From standard molar entropies: Sorxn = - 0.0974 kJ/K
Calculate Gorxn from Go = Ho - TSo
Gorxn = -1238 kJ - (298 K)(-0.0974 kJ/K)
= -1209 kJ
Reaction is product-favored in spite of negative
Sorxn. Reaction is “enthalpy driven”
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Calculating Gorxn for NH4NO3(s)
EXAMPLE 2:
NH4NO3(s)  NH4NO3(aq)
Is the dissolution of ammonium nitrate
product-favored?
If so, is it enthalpy- or entropy-driven?
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Gorxn for NH4NO3(s)
 NH4NO3(aq)
From tables of thermodynamic data we find
Horxn = +25.7 kJ
Sorxn = +108.7 J/K or +0.1087 kJ/K
Gorxn = +25.7 kJ - (298 K)(+0.1087 kJ/K)
= -6.7 kJ
Reaction is product-favored
. . . in spite of positive Horxn.
Reaction is “entropy driven”
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Calculating Gorxn
Gorxn =  Gfo (products) -  Gfo (reactants)
EXAMPLE 3: Combustion of carbon
C(graphite) + O2(g)  CO2(g)
Gorxn = Gfo(CO2) - [Gfo(graph) + Gfo(O2)]
Gorxn = -394.4 kJ - [ 0 + 0]
Note that free energy of formation of an element
in its standard state is 0.
Gorxn = -394.4 kJ
Reaction is product-favored as expected.
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Free Energy and Temperature
2 Fe2O3(s) + 3 C(s)  4 Fe(s) + 3 CO2(g)
Horxn = +467.9 kJ
Sorxn = +560.3 J/K
Gorxn = 467.9 kJ - (298K)(0.560kJ/K) = +300.8 kJ
Reaction is reactant-favored at 298 K
At what T does Gorxn just change from (+) to (-)?
i.e. what is T for Gorxn = 0 = Horxn - TSorxn
If Gorxn = 0
then
Horxn = TSorxn
so T = Ho/So ~ 468kJ/0.56kJ/K = 836 K or 563oC
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Go for COUPLED CHEMICAL REACTIONS
Reduction of iron oxide by CO is an example of
using TWO reactions coupled to each other in order
to drive a thermodynamically forbidden reaction:
Fe2O3(s)  4 Fe(s) + 3/2 O2(g)
Gorxn = +742 kJ
with a thermodynamically allowed reaction:
3/2 C(s) + 3/2 O2 (g)  3/2 CO2(g)
Overall :
Gorxn = -592 kJ
Fe2O3(s) + 3/2 C(s)  2 Fe(s) + 3/2 CO2(g)
Gorxn= +301 kJ @ 25oC
BUT
Gorxn < 0 kJ for T > 563oC
See Kotz, pp933-935 for analysis of the thermite reaction
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Other examples of coupled reactions:
Copper smelting
Cu2S (s)  2 Cu (s) + S (s)
Couple this with:
S (s) + O2 (g)  SO2 (s)
Overall:
Gorxn= +86.2 kJ
(FORBIDDEN)
Gorxn= -300.1 kJ
Cu2S (s) + O2 (g)  2 Cu (s) + SO2 (s)
Gorxn= +86.2 kJ + -300.1 kJ = -213.9 kJ (ALLOWED)
Coupled reactions VERY COMMON in Biochemistry :
e.g. all bio-synthesis driven by
ATP  ADP for which Horxn = -20 kJ
Sorxn = +34 J/K
Gorxn = -30 kJ @ 37oC
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Thermodynamics and Keq
• Keq is related to reaction favorability.
• If Gorxn < 0, reaction is product-favored.
• Gorxn is the change in free energy as
reactants convert completely to products.
• But systems often reach a state of
equilibrium in which reactants have not
converted completely to products.
• How to describe thermodynamically ?
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Grxn versus Gorxn
Under any condition of a reacting system,
we can define Grxn in terms of the
REACTION QUOTIENT, Q
Grxn = Gorxn + RT ln Q
If Grxn < 0 then reaction proceeds to right
If Grxn > 0 then reaction proceeds to left
At equilibrium, Grxn = 0. Also, Q = K. Thus
Gorxn = - RT lnK
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Thermodynamics and Keq (2)
9_G_NO2.mov
20m09an1.mov
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2 NO2  N2O4
Gorxn = -4.8 kJ
• pure NO2 has Grxn < 0.
• Reaction proceeds until
Grxn = 0 - the minimum in
G(reaction) - see graph.
• At this point, both N2O4 and
NO2 are present, with more
N2O4.
• This is a product-favored
reaction.
Entropy & Free Energy (Ch 20) - lect. 2
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Thermodynamics and Keq (3)
N2O4 2 NO2
Gorxn = +4.8 kJ
9_G_N2O4.mov
20m09an2.mov
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• pure N2O4 has Grxn < 0.
• Reaction proceeds until
Grxn = 0 - the minimum in
G(reaction) - see graph.
• At this point, both N2O4
and NO2 are present, with
more NO2.
• This is a reactant-favored
reaction.
Entropy & Free Energy (Ch 20) - lect. 2
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Thermodynamics and Keq (4)
Keq is related to reaction favorability and so
to Gorxn.
The larger the value of Gorxn the larger the
value of K.
o
G
=
RT
lnK
rxn
where R = 8.31 J/K•mol
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Thermodynamics and Keq (5)
Gorxn = - RT lnK
Calculate K for the reaction
N2O4  2 NO2
Gorxn = +4.8 kJ
Gorxn = +4800 J = - (8.31 J/K)(298 K) ln K
4800 J
lnK = = - 1.94
(8.31 J/K)(298K)
K = 0.14
When Gorxn > 0, then K < 1 - reactant favoured
When Gorxn < 0, then K >1 - product favoured
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Entropy and Free Energy
(Kotz Ch 20)
• Spontaneous vs. non-spontaneous
• thermodynamics vs. kinetics
• entropy = randomness (So)
• Gibbs free energy (Go)
• Go for reactions - predicting spontaneous direction
• thermodynamics of coupled reactions
• Grxn versus Gorxn
• predicting equilibrium constants from Gorxn
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