Transcript Entropy and Free Energy

Entropy and Free Energy
1
Spontaneous Physical and Chemical Processes
•A waterfall runs
downhill
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• A lump of
sugar
dissolves in a
cup of coffee
spontaneous
nonspontaneous
18.2
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Spontaneous for T > 0°C
Spontaneous for T < 0°C
ice
liquid water
•At 1 atm, water freezes below 0 0C and ice melts above 0 0C
4
Iron exposed to oxygen and
water forms rust
but will the rust
spontaneously form iron and
oxygen?
5
•Heat flows from a hotter object to a colder
object
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A gas expands in an evacuated bulb
spontaneous
nonspontaneous
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• All these are spontaneous processes.
A spontaneous process,
Has a definite direction in which it
occurs
Occurs without any ongoing outside
intervention.
can be fast or slow
• A nonspontaneous process is 8
one that cannot take place in a
system left to itself.
• If a process is spontaneous, the
reverse process is
nonspontaneous, and vice
versa.
• However, “spontaneous”
signifies nothing about how
fast a process occurs.
What factors make a
process spontaneous?
Enthalpy is a factor in whether a reaction is
spontaneous; but is not the only factor.
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Enthalpy is a factor in whether a reaction is
spontaneous; but is not the only factor.
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Spontaneous reactions
CH4 (g) + 2O2 (g)
CO2 (g) + 2H2O (l) DH0 = -890.4 kJ
H+ (aq) + OH- (aq)
H2O (l) DH0 = -56.2 kJ
18.2
An exothermic reaction (-DH)
favors a spontaneous reaction
since products are at a lower
potential energy.
Potential Energy
Reactants
-DH
Products
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Enthalpy is a factor in whether a reaction is
spontaneous; but is not the only factor.
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Spontaneous reactions
H2O (s)
NH4NO3 (s)
H2O (l) DH0 = +6.01 kJ
H2O
NH4+(aq) + NO3- (aq) DH0 = +25 kJ
On the basis of enthalpy, both of these processes
are unfavorable. But each shares something in
common: the products are more random and
disordered than the reactants.
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• Processes in which the disorder
of the system increases tend
to occur spontaneously.
• The change in disorder along
with the change in enthalpy
affects the spontaneity of a
process.
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• There are two natural tendencies
behind spontaneous processes:
the tendency to achieve a lower
energy state and the tendency
toward a more disordered state
Entropy, S
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The disorder is expressed by a thermodynamic
quantity called ENTROPY (S) (however, entropy
is not “disorder”).
The more disordered or random a system, the
larger its entropy.
Entropy is a state function, doesn’t depend on
the pathway by which the system changes.
ΔS = Sfinal-Sinitial
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Entropy (S) is a measure of the randomness
or disorder of a system.
order
disorder
S
S
DS = Sf - Si
If the change from initial to final results in an increase in randomness
Sf > Si
DS > 0
+DS; favors a spontaneous reaction for a chemical reaction
18.2
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The products are
more disordered
than the reactants
• ΔS is (+)
• ΔS is (-)
Guidelines for Determining if DS is + or -.
1. For any substance, the solid state is more
ordered than the liquid state and the liquid state
is more ordered than gas state
Ssolid < Sliquid << Sgas
H2O (s)
H2O (l)
DS > 0
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Guidelines for Determining if DS is + or -.
S (gases) > S (liquids) > S (solids)
So (J/K•mol)
H2O(liq)
69.91
H2O(gas) 188.8
Guidelines for Determining if DS is + or -
The entropy of
liquid water is
greater than
the entropy of
solid water
(ice) at 0˚ C.
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Guidelines for Determining if DS is + or -. 21
Increase in Entropy in the
Vaporization of Water
Evaporation is
spontaneous because of
the increase in entropy.
Guidelines for Determining if DS is + or -. 22
2. Entropy often
increases
when one
material
dissolves in
another.
Guidelines for Determining if DS is + or -. 23
Entropy usually increases when a
pure liquid or solid dissolves in a
solvent.
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Guidelines for Determining if DS is + or -.
EOS
Guidelines for Determining if DS is + or -.
3. Entropy of a substance increases
with temperature.
Molecular motions of heptane at different temps.
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Guidelines for Determining if DS is + or -.
4. Increase in molecular complexity
generally leads to increase in S.
So (J/K•mol)
CH4
248.2
C2H6
336.1
C3H8
419.4
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Guidelines for Determining if DS is + or -.
5.Entropies of ionic solids depend on
coulombic(electrostatic) attractions.
So (J/K•mol)
MgO
26.9
NaF
51.5
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Guidelines for Determining if DS is + or -.28
6. When gases are produced (or consumed)
•
If a reaction produces more gas molecules
than it consumes, DS0 > 0.
•
If the total number of gas molecules of the
reactants diminishes, DS0 < 0.
•
If there is no net change in the total number of
gas molecules, then DS0 may be positive or
negative BUT DS0 will be a small number.
What is the sign of the entropy change for the following
reaction? 2Zn (s) + O2 (g)
2ZnO (s)
The total number of gas molecules goes down, DS is
negative.
18.3
Guidelines for Determining if DS is + or -.
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7. Entropy increases as number of moles of
gases increases during chemical reaction.
N2 (g) + 3 H2 (g) -> 2 NH3 (g)
4 moles gas
2 moles gas
- DS
How does the entropy of a system
change for each of the following
processes?
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(a) Condensing water vapor
Randomness decreases
Entropy decreases (DS < 0)
(b) Forming sucrose crystals from a supersaturated solution
Randomness decreases
Entropy decreases (DS < 0)
(c) Heating hydrogen gas from 600C to 800C
Randomness increases
Entropy increases (DS > 0)
(d) Subliming dry ice
Randomness increases
Entropy increases (DS > 0)
18.2
Example
Predict whether each of the following leads to an
increase or decrease in the entropy of a system. If in
doubt, explain why.
(a) The synthesis of ammonia:
N2(g) + 3 H2(g)  2 NH3(g)
(b) Preparation of a sucrose solution:
C12H22O11(s) H2O(l)
C12H22O11(aq)
(c) Evaporation to dryness of a solution of urea,
CO(NH2)2, in water:
CO(NH2)2(aq)  CO(NH2)2(s)
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Entropy,S
• The change in entropy is related to the
heat transferred during the process.
Entropy and Temperature
S increases
slightly with T
S increases a
large amount
with phase
changes
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The Second Law of Thermodynamics34
• There is an inherent direction in which
processes occur is called the second law of
thermodynamics.
• It’s usually expressed in terms of entropy.
DSuniv
Dsuniverse
= DSsys+ DSsurr
: the total change in entropy, called the change in
entropy of the universe.
Dssys: the change in the entropy of the system.
Dssurr:
the change in the entropy of the surroundings.
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The Second Law of Thermodynamics
• In any spontaneous process:

Dsuniv
>0
In any spontaneous
process, the ENTROPY
OF UNIVERSE
INCREASES!!!
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2nd Law of Thermodynamics
Dissolving NH4NO3
in water—an
entropy driven
process.
Unlike energy, entropy
is not conserved, Suniv
is continually
increasing.
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The Second Law of Thermodynamics
Dissolving
NH4NO3 in
water—an
entropy
driven
process.
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2nd Law of Thermodynamics
• The second law demands that the
increase in the entropy of the
surroundings must be greater than
the decrease in the entropy of the
system, so that Δsuniv > 0.
2nd Law of Thermodynamics
• The effect of temperature on spontaneity
– H2O(l) --> H2O(g)
– water is the system, everything else is the
surroundings
 DSsystem increases, i.e. DSsystem is positive,
because there are more positions for the
water molecules in the gas state than in the
liquid state
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2nd Law of Thermodynamics
• What happens to the surrounding?
– Heat leaves the surroundings, entering the
system to cause the liquid molecules to
vaporize
– When heat leaves the surroundings, the
motion of the molecules of the
surroundings decrease, which results in a
decrease in the entropy of the
surroundings.
 DSsurroundings is negative
– (The opposite is true for an exothermic
process at constant temperature.)
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2nd Law of Thermodynamics
• Sign of DS depends on the heat flow
– Exothermic Rxn: DSsurr >0
– Endothermic Rxn: DSsurr< 0
• Magnitude of DS is determined by the
temperature
 DSsurr = - DH
T
T = absolute temperature
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2nd Law of Thermodynamics
2 H2(g) + O2(g) ---> 2 H2O(liq)
DSosystem = -326.9 J/K
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2nd Law of Thermodynamics
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2 H2(g) + O2(g) ---> 2 H2O(liq)
DSosystem = -326.9 J/K
DS
o
surroundings
qsurroundings
-DH system
=
=
T
T
2nd Law of Thermodynamics
2 H2(g) + O2(g) ---> 2 H2O(liq)
DSosystem = -326.9 J/K
DS
o
surroundings
qsurroundings
-DH system
=
=
T
T
Can calc. that DHorxn = DHosystem = -571.7 kJ
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2nd Law of Thermodynamics
2 H2(g) + O2(g) ---> 2 H2O(liq)
DSosystem = -326.9 J/K
DS
o
surroundings
qsurroundings
-DH system
=
=
T
T
Can calc. that DHorxn = DHosystem = -571.7 kJ
DS
o
surroundings
- (-571.7 kJ)(1000 J/kJ)
=
298.15 K
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2nd Law of Thermodynamics
2 H2(g) + O2(g) ---> 2 H2O(liq)
DSosystem = -326.9 J/K
DS
o
surroundings
qsurroundings
-DH system
=
=
T
T
Can calc. that DHorxn = DHosystem = -571.7 kJ
DS
o
surroundings
- (-571.7 kJ)(1000 J/kJ)
=
298.15 K
DSosurroundings = +1917 J/K
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2nd Law of Thermodynamics
2 H2(g) + O2(g) ---> 2 H2O(liq)
DSosystem = -326.9 J/K
DSosurroundings = +1917 J/K
DSouniverse = +1590. J/K
The entropy of the universe is increasing, so
the reaction is product-favored or
spontaneous.
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Summary
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Spontaneous or Not?
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Remember that –∆H˚sys is proportional to ∆S˚surr
An exothermic process has ∆S˚surr > 0.
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• The standard molar entropy, So, is the
entropy of one mole of a substance in its
standard state.
According to the Third Law of
Thermodynamics, the entropy of a pure,
perfect crystal can be taken to be zero at 0 K
EOS
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.
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2nd Law of Thermodynamics
A reaction is spontaneous if ∆S for the
universe is positive.
∆Suniverse = ∆Ssystem + ∆Ssurroundings
∆Suniverse > 0 for spontaneous
process
First calc. entropy created by matter
dispersal (∆Ssystem)
Next, calc. entropy created by energy
dispersal (∆Ssurround)
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Gibbs Free Energy, G
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• We can define a new function whose minimization
combines both S and ΔH. This is the ‘Gibbs free
energy, G’ or just ‘free energy.’
• For a spontaneous process, ΔG is the energy
that’s free to do maximum work at constant P&T.
ΔG is dependent on ;
• 1) temp & pressure
• 2) amount of substances
Gibbs Free Energy, G
DSuniv = DSsurr + DSsys
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Gibbs Free Energy, G
ΔSuniv = ΔSsurr + ΔSsys
DHsys
DSuniv =
+ DSsys
T
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Gibbs Free Energy, G
DSuniv = DSsurr + DSsys
DHsys
DSuniv =
+ DSsys
T
Multiply both sides by (-T)
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Gibbs Free Energy, G
D Suniv = D Ssurr + D Ssys
D Hsys
+ D Ssys
DSuniv =
T
Multiply through by -T
-T D Suniv = D Hsys - T D Ssys
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Gibbs Free Energy, G
D Suniv = D Ssurr + D Ssys
DHsys
DSuniv =
+ DSsys
T
Multiply through by -T
-T D Suniv = D Hsys - T D Ssys
-T D Suniv = change in Gibbs free energy
for the system = D Gsystem
-T
D S=Duniv
-T D S
G = D Gsystem
univ
system
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Gibbs Free Energy, G
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DSuniv = DSsurr + DSsys
D Hsys
+ D Ssys
DSuniv =
T
Multiply through by -T
-TDSuniv = DHsys - TDSsys
-TDSuniv = change in Gibbs free energy
for the system = DGsystem
Under standard conditions —
o
DG
=
o
DH
-
o
TDS
(Under constant
temp&pressure)
Gibbs Free Energy, G
o
DG
DHo
=
o
DH
DSo
o
TDS
-
DGo
Reaction
exo(-)
increase(+)
-
Product-favored
endo(+)
decrease(-)
+
Reactant-favored
exo(-)
decrease(-)
?
T dependent
endo(+)
increase(+)
?
T dependent
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The rxn is
spontaneous in the
forward direction.
The rxn is
nonspontaneous in
the forward
direction.work must
be supplied from the
surroundings to make
the rxn
happen.However, the
reverse rxn will be
spontaneous.
The rxn is at
equilibrium
example
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• Consider the reaction
•
• 2SO2(g) + O2(g)  2SO3(g)
•
• Carried out at 25°C and 1 atm. Calculate ΔH°, ΔS°,
and ΔG° using the following data. State whether
the reaction is spontaneous or nonspontaneous.
Substance
S° (JK-1mol-1)
ΔH°f (kJmol-1)
• SO2(g)
248
-297
• SO3(g)
257
-396
• O2(g)
205
Solution:
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• 2SO2(g) + O2(g)  2SO3(g)
• ΔH°rxn= ƩΔH°f (products) - ƩΔH°f (reactants)
ΔH°rxn= [2xΔH°f(SO3(g)] – [2xΔH°f(SO2(g))+ 1xΔH°f(O2(g))]
ΔH°rxn= [2x(-396)] – [2x (-297) + 1x(0)]
ΔH°rxn= - 198 kJ
Solution:
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• 2SO2(g) + O2(g)  2SO3(g)
• ΔS°rxn= ƩS°(products) - ƩS° (reactants)
ΔS°rxn= [2xS°(SO3(g)] – [2xS° (SO2(g))+ 1xS°(O2(g))]
ΔS°rxn= [2x 257)] – [2x 248) + 1x(205)]
ΔH°rxn= - 187 JK-1mol-1= -0.187 kJK-1mol-1
Solution:
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• 2SO2(g) + O2(g)  2SO3(g)
• ΔG°rxn= DHo - TDSo
ΔG°rxn= (- 198 kJ) – (298 K).(-0.187 kJK-1mol-1)
ΔG°rxn= -142.27 kJ
Since ΔG°rxn < 0, the reaction is spontaneous.
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References
• Chemistry, Zumdahl& Zumdahl. 6th edition
• Chemistry, the central science. Brown,
LeMay, & Bursten. 9th edition
• IBID, 3rd edition
• Petrucci, General Chemistry: An Integrated
Approach.Hill, Petrucci, 4th Edition