Transcript Gibbs Free Energy

1
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.
2
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
3
Gibbs Free Energy, G
4
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.
5
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.
6
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
7
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
9
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
10
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
12
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
15
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
16
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
17
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.
19
Thermodynamics and Keq
20
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.
21
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.
22
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
32
Kinetic Molecular Theory of Gases
33
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.
34
The Clausius-Clapeyron equation is a method for obtaining
enthalpy of vaporization, at any temperature.
 DH vap 
ln Pvap = - 
C

 RT 
35
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?
36
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
37
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.
38
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.
39
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


40
Gibbs free energy
DGrxn = 8.58kJ

K p = .0313bar
DG > 0 reactant favored
DG = 0 equilibrium
DG < 0 product favored,
reaction proceeds to
products
41
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.
42
Summary
Water will Evaporate even though the process takes
energy
•Kinetic Molecular Theory
PH O
•Evidence that K p = P
water vapor
2
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.
43
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.
44
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.
45
Arrhenius equation
k = Ae
-
Ea
RT
Ea  1 
ln k = ln A -  
R T 
46
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
47
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
48
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.