Transcript Gibbs Free Energy

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Second Law of Thermodynamics
As the reaction goes to products our system becomes more disordered and
the entropy of our system increases.
One driving forces for a chemical reaction is one
of the products is a gas.
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19.3 GIBBS FREE ENERGY
ΔSuniverse= ΔSsurroundings+ ΔSsystem = - ΔHsystem / T + ΔSsystem
-T ΔSuniverse = ΔHsystem - T ΔSsystem = ΔGsystem
ΔG is the change in free energy for the system.
ΔGosystem = ΔHosystem - T ΔSosystem (standard state)
ΔGorxn from ΔGof and from ΔHorxn and ΔSorxn
ΔGorxn= S ΔGfo(products) - S ΔGfo(reactants
ΔGorxn = ΔHorxn - T ΔSorxn
(Don't forget to change the entropy term from J to kJ)
Gibbs Free Energy, G
DSuniv = DSsurr + DSsys
-D H sys
D Suniv =
+ D Ssys
T
Multiply through by -T
-T ΔSuniv = ΔHsys - T ΔSsys
-T ΔSuniv = change in Gibbs free energy for the
system = ΔGsystem
Under standard conditions —
ΔGo = ΔHo - T ΔSo
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Gibbs Free Energy, G
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ΔGo = ΔHo - T ΔSo
Gibbs free energy change =
total energy change for system - energy lost
in disordering the system
If reaction is exothermic (ΔHo negative) and
entropy increases (ΔSo is +), then ΔGo must be
negative and reaction product-favored in the
standard state.
If reaction is endothermic (ΔHo is +), and entropy
decreases (ΔSo is -), then ΔGo must be + and
reaction is reactant-favored in the standard
state.
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Product-Favored or
Reactant-Favored?
• The reaction is product-favored if ΔG is
negative.
• We can see that this is always true if ΔH is
negative and ΔS is positive.
• If ΔH is positive and ΔS is negative, ΔG is
always positive.
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Product-Favored or
Reactant-Favored?
• The other two cases are temperature
dependent with a positive ΔS favoring
spontaneity at high temperature, thus
overcoming the positive ΔH, and a
negative ΔH favoring spontaneity at a low
temperature when ΔS is negative.
• Table next slide and Figure 20.8
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Gibbs Free Energy, G
ΔGo = ΔHo - T ΔSo
ΔHo
ΔSo
ΔGo Reaction
exo(-)
increase(+)
-
Prod-favored
endo(+)
decrease(-)
+
React-favored
exo(-)
decrease(-)
?
T dependent
endo(+)
increase(+)
?
T dependent
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Figure 20.8
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Gibbs Free Energy, G
ΔGo = ΔHo - T ΔSo
Two methods of calculating ΔGo
a)
Determine ΔHorxn and ΔSorxn and use
Gibbs equation.
b)
Use tabulated values of free energies of
formation, ΔGfo.
DGorxn = S DGfo (products) - S DGfo (reactants)
Calculating D
o
G rxn
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Combustion of acetylene
C2H2(g) + 5/2 O2(g) --> 2 CO2(g) + H2O(g)
Use enthalpies of formation to calculate
ΔHorxn = -1238 kJ
Use standard molar entropies to calculate
ΔSorxn = -97.4 J/K or -0.0974 kJ/K
ΔGorxn = -1238 kJ - (298 K)(-0.0974 kJ/K)
= -1209 kJ
Reaction is product-favored (in the standard state) in
spite of negative ΔSorxn.
Reaction is “enthalpy driven”.
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Calculating D
o
G rxn
Is the dissolution of ammonium nitrate productfavored?
If so, is it enthalpy- or entropy-driven?
NH4NO3(s) + heat
NH4NO3(aq)
Calculating D
NH4NO3(s) + heat
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o
G rxn
NH4NO3(aq)
From tables of thermodynamic data:
ΔHorxn = +25.7 kJ
ΔSorxn = +108.7 J/K or +0.1087 kJ/K
ΔGorxn = +25.7 kJ - (298 K)(+0.1087 kJ/K)
= -6.7 kJ
Reaction is product-favored (in the standard state) in spite
of negative ΔHorxn.
Reaction is “entropy driven”.
Calculating DGorxn
DGorxn = S DGfo (products) - S DGfo (reactants)
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Combustion of carbon
C(graphite) + O2(g) --> CO2(g)
ΔGorxn= ΔGfo(CO2) - [ΔGfo(graph) + ΔGfo(O2)]
ΔGorxn = -394.4 kJ - [ 0 + 0 ]
Note that free energy of formation of an element in
its standard state is 0.
ΔGorxn = -394.4 kJ
Reaction is product-favored as expected.
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Free energy and Temperature
• If we assume that ΔHo and ΔSo are relatively independent of
temperature, we can calculate ΔGo at any particular
temperature we choose.
ΔGTorxn = ΔHorxn - T ΔSorxn
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Free Energy and Temperature
2 Fe2O3(s) + 3 C(s) ---> 4 Fe(s) + 3 CO2(g)
ΔHorxn = +467.9 kJ ΔSorxn = +560.3 J/K
ΔGorxn = +300.8 kJ
Reaction is reactant-favored at 298 K
At what T does ΔGorxn just change from being (+) to
being (-)?
When ΔGorxn = 0 = ΔHorxn - T ΔSorxn.
DHrxn
467.9 kJ
T =
=
= 835.1 K
DSrxn
0.5603 kJ/K
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Free Energy and Temperature
• For the reaction below: Calculate ΔGo at 298.15 K
two ways, explain the sign of ΔSo, determine if the
reaction in the standard state is product-favored at
298.15 K, determine which term or terms favor
spontaneity, and calculate the temperature at which
the reaction would first become reactant favored in
the standard state.
3 H2(g) + CO(g) ====> CH4(g) + H2O(g)
20.4 THERMODYNAMICS AND
THE EQUILIBRIUM CONSTANT
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At equilibrium, ΔGT = 0, and ΔGTo = -RT lnKT
• Figure 20.11, shows the relationship between Q and K,
which comes from the concentration dependence of Free
Energy.
• One cannot calculate a new K by simply changing the T in
the equation since ΔGTo is a function of temperature.
ΔGTorxn = ΔHorxn - TΔSorxn
• A useful combined form of these equations is:
ΔGTo = ΔHo - TΔSo = -RT
ln KT.
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Figure 20.11
Thermodynamics and Keq
Keq is related to reaction favorability.
When ΔGorxn < 0, reaction moves
energetically “downhill”
ΔGorxn 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.
In this case ΔGrxn is < ΔGorxn , so state
with both reactants and products
present is more stable than complete
conversion.
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Thermodynamics and Keq
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Product-favored
reaction.
2 NO2 ---> N2O4
ΔGorxn = -4.8 kJ
Here ΔGrxn is less than
ΔGorxn , so the state
with both reactants
and products present
is more stable than
complete conversion.
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Thermodynamics and Keq
Reactant-favored
reaction.
N2O4 --->2 NO2
ΔGorxn = +4.8 kJ
Here ΔGorxn is greater
than ΔGrxn , so the state
with both reactants and
products present is
more stable than
complete conversion.
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Thermodynamics and Keq
Keq is related to reaction favorability and so
to ΔGorxn.
The larger the value of ΔGorxn the larger the
value of K.
o
ΔG rxn
= - RT lnK
where R = 8.31 J/K•mol
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Thermodynamics and Keq
DGorxn = - RT lnK
Calculate K for the reaction
N2O4 --->2 NO2
ΔGorxn = +4.8 kJ
ΔGorxn = +4800 J = - (8.31 J/K)(298 K) ln K
lnK = -
4800 J
= - 1.94
(8.31 J/K)(298K)
K = 0.14
When ΔGorxn > 0, then K < 1
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THERMODYNAMICS AND THE
EQUILIBRIUM CONSTANT
• Let us consider the derivation of these
equations to further our understanding of them
and their interrelationships.
• This derivation starts with the definition of G,
G = H - TS
• At constant temperature, with H = E + PV, and
wext = 0,and w = -PdV, we end up with,
ΔG = nRT ΔP/P.
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THERMODYNAMICS AND THE
EQUILIBRIUM CONSTANT
G = Go + nRT ln a, where a, is the activity,
unitless concentration.
• For a reaction, we arrive at:
ΔGT = ΔGTo + RT
• Where Q = K at equilibrium.
ln QT
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DERIVATIONS
• Derive the relationships between
ΔGTo and KT.
• Derive the relationships between
ΔHo , KT , and T.
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THE EQUILIBRIUM CONSTANT
• The temperature dependence of K can be calculate
using the equation:
ΔGTo = ΔHo - T ΔSo = - RT ln KT.
• This equation can be written at two temperatures,
T1 and T2, and combined to eliminate ΔSo.
• This produces the very useful equation:
 KT2 
ln  K 
 T1 
- D H 1 - 1 
=
R
T 2 T1


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THE EQUILIBRIUM CONSTANT
A plot of ln K vs. 1/T yields a straight
line with a slope of - Δ Ho/R.
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20.5 THERMODYNAMICS AND TIME
• The first and second Laws of
Thermodynamics cannot be proven,
they are laws of experience and tell us
the direction of time in any given
"picture".
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Real World Examples of:
Chapter 20: Second Law of Thermodynamics
“The total entropy of the universe is always increasing”
liquid
H 2 O(l)
vaporization
condensation
40.7 kJmol
gas
-1
-40.7 kJmol
-1
H 2 O( g )
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Faster molecules can break intermolecular forces
like H-bonding to escape a solution
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Kinetic Molecular Theory of Gases
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Experiment: Two Florence flasks with H2O,
one closed and the other open.
This experiment demonstrates:
1. There is sufficient heat in the surroundings to allow liquid
water to escape to the atmosphere;
2. Air currents and gas diffusion prevent the gaseous water
from making contact with the water surface.
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The Clausius-Clapeyron equation is a method for obtaining
enthalpy of vaporization, at any temperature.
 DH vap 
ln Pvap = - 
C

 RT 
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In a closed system…Questions
1. Can we change the equilibrium in this system?
2. Is there any reason for wanting to change the equilibrium in
this system?
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Closed system equilibrium
H 2 O(l)
Kp =
H 2 O( g )
PH O
2
Pwater vapor
Q < K Reaction favors reactants to products
Q > K Reaction favors products to reactants
Q = K Reaction is at equilibrium
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Frost-free freezers
Kp =
PH O
2
Pwater vapor
Q < K favors
reactants to
products
Air in the freezer is
warmed then dried.
The vapor pressure
of ice is 4.579 torr.
Warm, desiccated
air can remove
water vapor.
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Second Law of Thermodynamics
As the reaction goes to products our system becomes more disordered and
the entropy of our system increases.
One driving forces for a chemical reaction is one of the
products is a gas.
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Gibbs free energy
H 2 O(l)  H 2 O( g )

DG = Grxn
 RT ln Q

DGrxn
= - RT ln K p
DGrxn = DH rxn - T DS rxn  44.0kJ - 35.5kJ = 8.58kJ

K p = .0313


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Gibbs free energy
DGrxn = 8.58kJ

K p = .0313bar
DG > 0 reactant favored
DG = 0 equilibrium
DG < 0 product favored,
reaction proceeds to
products
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If Q < K product favored
What are the conditions necessary for the spontaneous
formation of products?
DG = Grxn  RT ln Q

Q=
PH O
2
Pwater vapor
A Q of .0313 allows water vapor to evaporate. The atmosphere
has a PH2O of .001%-4% water vapor (3%-100% humidity), in
the winter Pvap can be as high as 20 torr.
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Summary
Water will Evaporate even though the process takes
energy
•Kinetic Molecular Theory
PH O
•Evidence that K p = P
water vapor
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is easily manipulated
under normal conditions
•Highly ordered H2O(l) has a large DS, this is the main
driving force for producing a higher partial pressure of
water.
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As a rule of thumb the rate of an rxn
doubles, or triples for every 10 degrees
increase in temperature.
What factors affect the rate of a reaction?
 The concentration of the reactants. The more concentrated
the faster the rate;
 Temperature. Usually reactions speed up with increasing
temperature;
 Physical state of reactants;
 Catalyst (or inhibitor). A catalyst speeds up a reaction, an
inhibitor slows it down.
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Δ,H
Ester + H 2 O 
carboxylic acid + alcohol
+
Exp
Temp K
k [L/mols]
DK
K2/K1
1
288
.0521
2
298
.101
10
1.93
3
308
.184
10
1.82
4
318
.332
10
1.80
Concentrations of the Ester and H2O are held constant,
only the temperature changes.
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Arrhenius equation
k = Ae
-
Ea
RT
Ea  1 
ln k = ln A -  
R T 
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Testing the “Rule of Thumb”
 K 2   Ea   1 1 
ln   =    - 
 K1   R   T1 T2 
Setting K2/K1 = 2, T1 = 273, T2 = 283
Solve for Ea, Ea = 44.5kJmol-1
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Theoretical Results
T1
T2
k2/k1
273
283
2.00
373
383
1.45
473
483
1.26
573
583
1.17
673
683
1.12
773
783
1.09
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When can relying on the rule be
dangerous
H 2 ( g )  Cl2 ( g )  2 HCl
heat  Cl2 ( g )  2Cl ( g )
Cl ( g )  H 2 ( g )  HCl ( g )  H ( g )
H ( g )  Cl2 ( g )  HCl ( g )  Cl ( g )
Once the rxn is initiated, no further heat is needed.
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