Transcript Spontaneity, Entropy, & Free Energy Chapter 16 1st Law of Thermodynamics The first law of thermodynamics is a statement of the law of conservation.

Spontaneity, Entropy, & Free
Energy
Chapter 16
1st Law of Thermodynamics
The first law of thermodynamics is a
statement of the law of conservation of
energy: energy can neither be created nor
destroyed. The energy of the universe is
constant, but the various forms of energy
can be interchanged in physical and
chemical processes.
Spontaneous Processes
and Entropy
Thermodynamics lets us predict whether a
process will occur but gives no information
about the amount of time required for the
process.
A spontaneous process is one that occurs
without outside intervention.
Kinetics & Thermodynamics
Chemical kinetics focuses on the pathway
between reactants and products--the
kinetics of a reaction depends upon
activation energy, temperature,
concentration, and catalysts.
Thermodynamics only considers the initial
and final states.
To describe a reaction fully, both
kinetics and thermodynamics are
necessary.
Domain of
thermodynamics
(the initial and
final states)
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Energy
Domain of kinetics
(the reaction pathway)
Reactants
Products
Reaction progress
The rate of a reaction depends on the pathway from
reactants to products. Thermodynamics tells whether
the reaction is spontaneous and depends upon initial
& final states only.
Entropy
The driving force for a spontaneous
process is an increase in the entropy of the
universe.
Entropy, S, can be viewed as a measure of
randomness, or disorder.
Nature spontaneously proceeds toward the
states that have the highest probabilities of
existing.
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The expansion of an ideal gas into an evacuated bulb.
Positional Entropy
A gas expands into a vacuum because the
expanded state has the highest positional
probability of states available to the
system.
Therefore,
Ssolid < Sliquid << Sgas
Positional Entropy
Which of the following has higher positional
entropy?
a) Solid CO2 or gaseous CO2?
b) N2 gas at 1 atm or N2 gas at 1.0 x 10-2 atm?
Entropy
What is the sign of the entropy change for
the following?
a) Solid sugar is added to water to form a
solution?
S is positive
b) Iodine vapor condenses on a cold
surface to form crystals?
S is negative
The Second Law of
Thermodynamics
. . . in any spontaneous process there is
always an increase in the entropy of the
universe.
Suniv > 0
for a spontaneous process.
SUniverse
Suniverse is positive -- reaction is spontaneous.
Suniverse is negative -- reaction is spontaneous
in the reverse direction.
Suniverse = 0 -- reaction is at equilibrium.
G -- Free Energy
Two tendencies exist in nature:
•
tendency toward higher entropy -- S
•
tendency toward lower energy -- H
If the two processes oppose each other (e.g.
melting ice cube), then the direction is
decided by the Free Energy, G, and
depends upon the temperature.
Free Energy
G = H  TS
(from the standpoint of the system)
A process (at constant T, P) is spontaneous in
the direction in which free energy decreases:
Gsys means +Suniv
Entropy changes in the surroundings are
primarily determined by the heat flow. An
exothermic process in the system increases
the entropy of the surroundings.
G, H, & S
Spontaneous reactions are indicated by the
following signs:
G = negative
H = negative
S = positive
Temperature Dependence
Ho & So are not temperature dependent.
Go is temperature dependent.
G = H  TS
Ssurroundings
Ssurr is positive -- heat flows into the
surroundings out of the system.
Ssurr is negative -- heat flows out of the
surroundings and into the system.
Ssurr = - Hsystem
T
Ssurroundings Calculations
Sb2S3(s) + 3Fe(s) ---> 2Sb(s) + 3FeS(s) H = -125 kJ
Sb4O6(s) + 6C(s) ---> 4Sb(s) + 6CO(g) H = 778 kJ
What is Ssurr for these reactions at 250C & 1 atm.
Ssurr = - Hsystem
T
Ssurr = -(-125kJ/298K)
Ssurr = 419 J/K
Ssurr = - Hsystem
T
Ssurr = -(778kJ/298K)
Ssurr = -2.61 x 103 J/K
Effect of H and S on
Spontaneity
H
S
Result

+
spontaneous at all temps
+
+
spontaneous at high temps


spontaneous at low temps
+

not spontaneous at any temp
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Table 16.3 Interplay of Ssys and Ssurr in Determining the Sign of Suniv
Signs of Entropy Changes
Ssys
Ssurr
Suniv



Yes



No (reaction will occur
in opposite direction)



Yes, if Ssys has a
larger magnitude
than Ssurr



Yes, if Ssurr has a
larger magnitude
than Ssys
Process Spontaneous?
Calculations showing that the melting of ice is
temperature dependent. The process is spontaneous
above 0oC.
16_05T
Table 16.5 Various Possible Combinations of H and S for a Process and the
Resulting Dependence of Spontaneity on Temperature
Case
S positive, H negative
S positive, H positive
S negative, H negative
S negative, H positive
Result
Spontaneous at all temperatures
Spontaneous at high temperatures
(where exothermicity is relatively unimportant)
Spontaneous at low temperatures
(where exothermicity is dominant)
Process not spontaneous at any temperatures
(reverse process is spontaneous at all temperatures)
Free Energy G
G = H  TS
G = negative -- spontaneous
G = positive -- spontaneous in
opposite direction
G = 0 -- at equilibrium
Boiling Point Calculations
What is the normal boiling point for liquid Br2?
Br2(l) ---> Br2(g)
Ho = 31.0 kJ/mol & So = 93.0 J/Kmol
At equilibrium, Go = 0
Go = Ho  TS0 = 0
Ho = TS0
T = Ho/S0
T = 3.10 x 104 J/mol/(93.0J/Kmol)
T = 333K
The Third Law of
Thermodynamics
. . . the entropy of a perfect crystal at 0 K is zero.
Because S is explicitly known (= 0) at 0 K,
S values at other temps can be calculated.
See Appendix 4 for values of S0.
Soreaction
Calculate S at 25 oC for the reaction
2NiS(s) + 3O2(g) ---> 2SO2(g) +2NiO(s)
S = npS(products)  nrS(reactants)
S = [(2 mol SO2)(248 J/Kmol) + (2 mol NiO)(38
J/Kmol)] - [(2 mol NiS)(53 J/Kmol) + (3 mol
O2)(205 J/Kmol)]
S = 496 J/K + 76 J/K - 106 J/K - 615 J/K
S = -149 J/K # gaseous molecules decreases!
So
So increases with:
•
solid ---> liquid ---> gas
•
greater complexity of molecules (have a
greater number of rotations and
vibrations)
•
greater temperature (if volume
increases)
•
lower pressure (if volume increases)
Free Energy Change and
Chemical Reactions
G = standard free energy change that
occurs if reactants in their standard
state are converted to products in
their standard state.
G = npGf(products)  nrGf(reactants)
The more negative the value of G, the
further a reaction will go to the
right to reach equilibrium.
G Calculations
Calculate H, S, & G for the reaction
2 SO2(g) + O2(g) ----> 2 SO3(g)
H = npHf(products)  nrHf(reactants)
H = [(2 mol SO3)(-396 kJ/mol)]-[(2 mol
SO2)(-297 kJ/mol) + (0 kJ/mol)]
H = - 792 kJ + 594 kJ
H = -198 kJ
G Calculations
Continued
S = npS(products)  nrS(reactants)
S = [(2 mol SO3)(257 J/Kmol)]-[(2 mol
SO2)(248 J/Kmol) + (1 mol O2)(205
J/Kmol)]
S = 514 J/K - 496 J/K - 205 J/K
S = -187 J/K
G Calculations
Continued
Go = Ho  TSo
Go = - 198 kJ - (298 K)(-187 J/K)(1kJ/1000J)
Go = - 198 kJ + 55.7 kJ
Go = - 142 kJ
The reaction is spontaneous at 25 oC and 1 atm.
Hess’s Law & Go
Cdiamond(s) + O2(g) ---> CO2(g) Go = -397 kJ
Cgraphite(s) + O2(g) ---> CO2(g) Go = -394 kJ
Calculate Go for the reaction
Cdiamond(s) ---> Cgraphite(s)
Cdiamond(s) + O2(g) ---> CO2(g) Go = -397 kJ
CO2(g) ---> Cgraphite(s) + O2(g) Go = +394 kJ
Cdiamond(s) ---> Cgraphite(s) Go = -3 kJ
Diamond is kinetically stable, but thermodynamically unstable.
Go & Temperature
Go depends upon temperature. If a
reaction must be carried out at
temperatures higher than 25 oC,
then Go must be recalculated from
the Ho & So values for the
reaction.
Free Energy & Pressure
The equilibrium position represents the
lowest free energy value available to a
particular system (reaction).
G is pressure dependent
S is pressure dependent
H is not pressure dependent
Free Energy and Pressure
G = G + RT ln(Q)
Q = reaction quotient from the law
of mass action.
Free Energy Calculations
. CO(g) + 2H2(g) ---> CH3OH(l)
Calculate Go for this reaction where CO(g)
is 5.0 atm and H2(g) is 3.0 atm are
converted to liquid methanol.
G = npGf(products)  nrGf(reactants)
G = [(1 mol CH3OH)(- 166 kJ/mol)]-[(1
mol CO)(-137 kJ/mol) + (0 kJ)]
G =  166 kJ + 137 kJ
G =  2.9 x 104 J
Free Energy Calculations
Continued
1
Q
(PCO )( PH2 )
1
Q
2
(5.0)(3.0)
Q
= 2.2 x 10-2
Free Energy Calculations
Continued
G = G + RT ln(Q)
G = (-2.9 x 104 J/mol rxn) + (8.3145
J/Kmol)(298 K) ln(2.2 x 10-2)
G = ( 2.9 x 104 J/mol rxn) - (9.4 x 103
J/mol rxn)
G = - 38 kJ/ mol rxn
Note: G is significantly more negative
than G, implying that the reaction is
more spontaneous at reactant pressures
greater than 1 atm. Why?
16_352
A
A
B
B
C
(a)
(b)
A system can achieve the lowest possible free energy
by going to equilibrium, not by going to completion.
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GA
G
GB
(a)
GA (PA decreasing)
G
GB (PB increasing)
(b)
G
GA
GB
(c)
As A is changed into B, the pressure and free energy of A
decreases, while the pressure and free energy of B increases
until they become equal at equilibrium.
16_354
Equilibrium
occurs here
G
0
(a)
Equilibrium
occurs here
G
0.5
1.0
Fraction of A reacted
G
0
(b)
Equilibrium
occurs here
0.5
1.0
Fraction of A reacted
0
0.5
1.0
Fraction of A reacted
(c)
Graph a) represents equilibrium starting from only reactants,
while Graph b) starts from products only. Graph c) represents
the graph for the total system.
Free Energy and
Equilibrium
G = RT ln(K)
K = equilibrium constant
This is so because G = 0 and Q = K at
equilibrium.
o
G &
K
Go = 0
K=1
G < 0
K > 1 (favored)
G > 0
K < 1 (not favored)
Equilibrium Calculations
4Fe(s) + 3O2(g) <---> 2Fe2O3(s)
Calculate K for this reaction at 25 oC.
Go = - 1.490 x 106 J
Ho = - 1.652 x 106 J
So = -543 J/K
G = RT ln(K)
K = e - G/RT
K = e 601 or 10261
K is very large because G is very negative.
Temperature Dependence of K
H
S
ln( K )  
(1 / T ) 
R
R
o
y = mx + b
(H and S  independent of temperature
over a small temperature range)
If the temperature increases, K decreases
for exothermic reactions, but increases for
endothermic reactions.
Free Energy & Work
The maximum possible useful work
obtainable from a process at constant
temperature and pressure is equal to the
change in free energy:
wmax = G
Reversible vs. Irreversible
Processes
Reversible: The universe is exactly the same
as it was before the cyclic process.
Irreversible: The universe is different after
the cyclic process.
All real processes are irreversible -- (some
work is changed to heat).  w < G
Work is changed to heat in the surroundings
and the entropy of the universe increases.
Laws of Thermodynamics
First Law: You can’t win, you can only
break even.
Second Law: You can’t break even.