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Gibbs Free Energy

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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
О”Suniverse= О”Ssurroundings+ О”Ssystem = - О”Hsystem / T + О”Ssystem
-T О”Suniverse = О”Hsystem - T О”Ssystem = О”Gsystem
О”G is the change in free energy for the system.
О”Gosystem = О”Hosystem - T О”Sosystem (standard state)
О”Gorxn from О”Gof and from О”Horxn and О”Sorxn
О”Gorxn= S О”Gfo(products) - S О”Gfo(reactants
О”Gorxn = О”Horxn - T О”Sorxn
(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 О”Suniv = О”Hsys - T О”Ssys
-T О”Suniv = change in Gibbs free energy for the
system = О”Gsystem
Under standard conditions —
О”Go = О”Ho - T О”So
3
Gibbs Free Energy, G
4
О”Go = О”Ho - T О”So
Gibbs free energy change =
total energy change for system - energy lost
in disordering the system
If reaction is exothermic (О”Ho negative) and
entropy increases (О”So is +), then О”Go must be
negative and reaction product-favored in the
standard state.
If reaction is endothermic (О”Ho is +), and entropy
decreases (О”So is -), then О”Go 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 О”S 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 О”S favoring
spontaneity at high temperature, thus
overcoming the positive О”H, and a
negative О”H favoring spontaneity at a low
temperature when О”S is negative.
• Table next slide and Figure 20.8
7
Gibbs Free Energy, G
О”Go = О”Ho - T О”So
О”Ho
О”So
О”Go Reaction
exo(-)
increase(+)
-
Prod-favored
endo(+)
decrease(-)
+
React-favored
exo(-)
decrease(-)
?
T dependent
endo(+)
increase(+)
?
T dependent
8
Figure 20.8
9
Gibbs Free Energy, G
О”Go = О”Ho - T О”So
Two methods of calculating О”Go
a)
Determine О”Horxn and О”Sorxn and use
Gibbs equation.
b)
Use tabulated values of free energies of
formation, О”Gfo.
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
О”Horxn = -1238 kJ
Use standard molar entropies to calculate
О”Sorxn = -97.4 J/K or -0.0974 kJ/K
О”Gorxn = -1238 kJ - (298 K)(-0.0974 kJ/K)
= -1209 kJ
Reaction is product-favored (in the standard state) in
spite of negative О”Sorxn.
Reaction is “enthalpy driven”.
11
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:
О”Horxn = +25.7 kJ
О”Sorxn = +108.7 J/K or +0.1087 kJ/K
О”Gorxn = +25.7 kJ - (298 K)(+0.1087 kJ/K)
= -6.7 kJ
Reaction is product-favored (in the standard state) in spite
of negative О”Horxn.
Reaction is “entropy driven”.
Calculating DGorxn
DGorxn = S DGfo (products) - S DGfo (reactants)
13
Combustion of carbon
C(graphite) + O2(g) --> CO2(g)
О”Gorxn= О”Gfo(CO2) - [О”Gfo(graph) + О”Gfo(O2)]
О”Gorxn = -394.4 kJ - [ 0 + 0 ]
Note that free energy of formation of an element in
its standard state is 0.
О”Gorxn = -394.4 kJ
Reaction is product-favored as expected.
14
Free energy and Temperature
• If we assume that ΔHo and ΔSo are relatively independent of
temperature, we can calculate О”Go at any particular
temperature we choose.
О”GTorxn = О”Horxn - T О”Sorxn
15
Free Energy and Temperature
2 Fe2O3(s) + 3 C(s) ---> 4 Fe(s) + 3 CO2(g)
О”Horxn = +467.9 kJ О”Sorxn = +560.3 J/K
О”Gorxn = +300.8 kJ
Reaction is reactant-favored at 298 K
At what T does О”Gorxn just change from being (+) to
being (-)?
When О”Gorxn = 0 = О”Horxn - T О”Sorxn.
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 О”So, 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, О”GT = 0, and О”GTo = -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 О”GTo is a function of temperature.
О”GTorxn = О”Horxn - TО”Sorxn
• A useful combined form of these equations is:
О”GTo = О”Ho - TО”So = -RT
ln KT.
18
Figure 20.11
Thermodynamics and Keq
Keq is related to reaction favorability.
When О”Gorxn < 0, reaction moves
energetically “downhill”
О”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.
In this case О”Grxn is < О”Gorxn , 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
О”Gorxn = -4.8 kJ
Here О”Grxn is less than
О”Gorxn , 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
О”Gorxn = +4.8 kJ
Here О”Gorxn is greater
than О”Grxn , 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 О”Gorxn.
The larger the value of О”Gorxn the larger the
value of K.
o
О”G rxn
= - RT lnK
where R = 8.31 J/K•mol
23
Thermodynamics and Keq
DGorxn = - RT lnK
Calculate K for the reaction
N2O4 --->2 NO2
О”Gorxn = +4.8 kJ
О”Gorxn = +4800 J = - (8.31 J/K)(298 K) ln K
lnK = -
4800 J
= - 1.94
(8.31 J/K)(298K)
K = 0.14
When О”Gorxn > 0, then K < 1
24
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,
О”G = nRT О”P/P.
25
THERMODYNAMICS AND THE
EQUILIBRIUM CONSTANT
G = Go + nRT ln a, where a, is the activity,
unitless concentration.
• For a reaction, we arrive at:
О”GT = О”GTo + RT
• Where Q = K at equilibrium.
ln QT
26
DERIVATIONS
• Derive the relationships between
О”GTo and KT.
• Derive the relationships between
О”Ho , KT , and T.
27
THE EQUILIBRIUM CONSTANT
• The temperature dependence of K can be calculate
using the equation:
О”GTo = О”Ho - T О”So = - RT ln KT.
• This equation can be written at two temperatures,
T1 and T2, and combined to eliminate О”So.
• This produces the very useful equation:
пѓ© KT2 пѓ№
ln пѓЄ K пѓє
пѓЄпѓ« T1 пѓєпѓ»
- D H пѓ©1 - 1 пѓ№
=
R
пѓЄT2 T1пѓє
пѓ«
пѓ»
28
THE EQUILIBRIUM CONSTANT
A plot of ln K vs. 1/T yields a straight
line with a slope of - О” Ho/R.
29
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".
30
Real World Examples of:
Chapter 20: Second Law of Thermodynamics
“The total entropy of the universe is always increasing”
liq u id
va p o riza tio n
co n d en sa tio n
gas
-1
H 2 O (l)
40.7 kJm ol
- 40.7 kJm ol
-1
H 2O ( g )
31
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.
пѓ¦ D H vap пѓ¶
ln Pvap = - пѓ§
пѓ·пЂ«C
пѓЁ RT пѓё
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 2O ( g )
PH
2O
Pw a ter va p o r
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
2O
Pw a ter va p o r
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) п‚® H 2 O ( g )
D G = G rxn пЂ« R T ln Q
п‚°
D G rxn = - R T ln K p
п‚°
D G rxn = D H rxn - T D S rxn пѓћ 44.0 kJ - 35.5 kJ = 8.58 kJ
п‚°
K p = .0 3 1 3
п‚°
п‚°
40
Gibbs free energy
D G rxn = 8 .5 8 kJ
п‚°
K p = .0 3 1 3b ar
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?
D G = G rxn пЂ« R T ln Q
п‚°
Q =
PH
2O
Pw a ter va p o r
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
Kp =
PH
2O
•Evidence that
P
under normal conditions
is easily manipulated
w a ter va p o r
•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?
пѓј The concentration of the reactants. The more concentrated
the faster the rate;
пѓ� Temperature. Usually reactions speed up with increasing
temperature;
пЃ± Physical state of reactants;
пЃ± Catalyst (or inhibitor). A catalyst speeds up a reaction, an
inhibitor slows it down.
44
,H
E ster + H 2 O п‚ѕ О”п‚ѕ
п‚® carboxylic acid + alcohol
+
Exp
Temp K
k [L/molп‚ґs]
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 пѓ¦ 1 пѓ¶
ln k = ln A пѓ§ пѓ·
R пѓЁT пѓё
46
Testing the “Rule of Thumb”
пѓ¦ K 2 пѓ¶ пѓ¦ Ea пѓ¶ пѓ¦ 1
1 пѓ¶
ln пѓ§
пѓ·пѓ§ пѓ·=пѓ§
пѓ·
пѓЁ K 1 пѓё пѓЁ R пѓё пѓЁ T1 T 2 пѓё
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 ) пЂ« C l2 ( g ) п‚® 2 H C l
heat пЂ« C l 2 ( g ) п‚® 2 C l ( g )
Cl(g ) пЂ« H 2 (g ) п‚® HCl(g ) пЂ« H (g )
H ( g ) пЂ« C l2 ( g ) п‚® H C l ( g ) пЂ« C l ( g )
Once the rxn is initiated, no further heat is needed.
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