Transcript Spontaneity, Entropy, and Free Energy

Spontaneity, Entropy, and 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.
 1st law does not give a hint as to WHY a process
occurs in a given direction.
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.
SPONTANEOUS DOES NOT MEAN FAST!
Tells
us the information about the direction of the reaction
AP exam wants you to use the phrase “thermodynamically
favorable”
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)
16_343
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 thermodynamically favorable 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.
Think
of your bedroom…it naturally tends to get
messy. An ordered room requires everything to be in
its place…There are many more ways for things to be
out of place than in place
Entropy
 Entropy is a thermodynamic function that describes
the number of arrangements (positions and/or
energy levels) that are available to a system existing
in a system
 Is closely linked to probability
 Think of the expansion of an ideal gas, more
possibility of it being spread out than concentrated
in one area
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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
SUniverse
 Suniverse is positive -- reaction is spontaneous.
 Suniverse is negative -- reaction is spontaneous in
the reverse direction.
 Suniverse = 0 -- reaction is at equilibrium.
Concept Check
14
Predict the sign of S for each of the following, and
explain:
+ a) The evaporation of alcohol
– b) The freezing of water
– c) Compressing an ideal gas at constant
temperature
+ d) Heating an ideal gas at constant pressure
+ e) Dissolving NaCl in water
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.
Second Law of Thermodynamics
 In any spontaneous process there is always an
increase in the entropy of the universe.
 The entropy of the universe is increasing.
 The total energy of the universe is constant, but
the entropy is increasing.
Suniverse = ΔSsystem + ΔSsurroundings
16
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rights reserved
ΔSsurr
 ∆Ssys depends on what we have already talked about

Increases when
Going from solid to liquid to gas
 Solution formation occurs
 Pressure descreases (increases in volume)
 # of molecules going from reactants to products increases

 ∆Ssurr sign depends on direction of heat flow &
temperature

Higher temp-lower ∆S surr; Lower temp- higher ∆S Surr
 Consider:
H2O(l)  H2O (g)
 ∆S sys: +
 ∆S surr: - (endothermic process)
Ssurroundings
 Ssurr is positive -- heat flows into the surroundings
out of the system. (EXOTHERMIC)
 Ssurr is negative -- heat flows out of the
surroundings and into the system.
(ENDOTHERMIC)
Ssurr = - Hsystem
T
Ex. Is this rxn thermodynamically favored at
25˚C?
 N2(g) + 3 H2(g)  2NH3 (g)
∆H= -92.6 kJ,
∆Ssys= -199 J/K
Ssurr = - Hsystem
T
Ssurr = - (-92.6)
(298)
Ssurr = .31074 kJ/K = 310.74 J/K
Suniverse = ΔSsystem + Δssurroundings = -199 + 310.74 = 111.74 J/K
YES is THERMODYNAMICALLY FAVORED because +
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 = - Hsystem
T
 Ssurr = 419 J/K
Ssurr = -(778kJ/298K)
Ssurr = -2.61 x 103 J/K
16_03T
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?
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
system)
(from the standpoint of the
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
 Thermodynamically favored, 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
Free Energy G
 G = H  TS
 G = negative -- spontaneous
 G = positive -- spontaneous in
opposite direction
 G = 0 -- at equilibrium
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
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)
Concept Check
A liquid is vaporized at its boiling point. Predict the signs
of:
w
– (gas expanding)
+ (endothermic process)
q
+ (endothermic process)
H
S
+ (going from liquid to gas)
Ssurr – ( -∆H/T, ∆H is positive)
0 (∆H favors reverse process, ∆S favors forward process; these
G
tendencies will exactly balance out at boiling point, and
system will be at equilbrium)
Explain your answers.
Concept Check
Gas A2 reacts with gas B2 to form gas AB at constant
temperature and pressure. The bond energy of AB is
much greater than that of either reactant.
Predict the signs of:
H
Ssurr
–
+
Explain.
S
0
Suniv
+
 Since the average bond energy of the products is
greater than the average bond energies of the
reactants, the reaction is exothermic as written.
 Thus, the sign of H is negative; Ssurr is positive; S
is close to zero (cannot tell for sure); and Suniv is
positive.
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
Go = Ho  TS0
 At temperatures above 333 K, TS has a larger
magnitude than H, and G would be negative
(thermodynamically favored)
 At temperatures below 333 K, TS is smaller than
H, and G would be positive (reverse reaction
favored)
H
S
G
Thermodynamically
favored?
-
+
-
YES
+
-
+
YES
-
-
- high temps
+low temps
+
+
-low temps
+high temp
YES @ high
temps
YES @ low
temps
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.
Standard Entropy Values (S°)
 Represent the increase in entropy that occurs
when a substance is heated from 0 K to 298 K at
1 atm pressure.
ΔS°reaction = ΣnpS°products – ΣnrS°reactants
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)
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.
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.
o
G &
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.
Thermodynamics & Kinetics
 A reaction may be favorable thermodynamically, but
scientists need to study the kinetics of the reaction to
see if it will proceed at a fast enough rate to be useful
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(P)
or
G = G + RT ln(Q)
Q = reaction quotient from the
law of mass action.
Free Energy Calculations
CO(g) + 2H2(g) ---> CH3OH(l)
 Calculate G for this reaction where CO(g) is
5.0 atm and H2(g) is 3.0 atm are converted to
liquid methanol.
G = G + RT ln(Q)
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? (Le Chatelier’s Princ)
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.
16_353
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.
Go & 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.
 Ho = - 1.652 x 106 J (from values in table)
 So = -543 J/K
(from values in table)
 Go = - 1.490 x 106 J (from ∆ G= ∆ H-T∆S)
 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.
Thermodynamically Unfavorable Processes that
readily occur
 Light or electricity may be used to cause a process to
occur that is thermodynamically unfavorable
 Examples:


Charging a battery, process of electrolysis (next chapter)
Photosynthesis

Rxn has a ∆G0 of +2880 kJ/mol but light helps the multistep
process occur through the absorption of several photons in the
Visible Spectrum range (400-700 nm)
 Coupling Reactions may also lead to reactions that
are not thermodynamically favorable occur
Thermodynamically unfavorable reaction can be driven
by a favorable reaction
Review
Free Energy Change
 The free energy change (ΔG) of a chemical process is a measure of the energy
available to do work.
 Standard free energy change ΔGo is when reaction conditions are standard: T is
250 C, P = 1 atm, and conc of all reactants is 1M.
 For biochemical reactions, the pH has to be 7. The standard free energy change
at pH 7 is designated as ΔGo’
 ΔG = Gproducts – Greactants
• If ΔG <0 , the forward reaction is favored relative to the reverse reaction. The
products are more stable than the reactants. As reaction proceeds, energy is
released, it can be used to do work
• If ΔG = 0, reactants and products are at equilibrium
• If ΔG >0, reactants are at lower energy than products; energy needs to be
supplied for the reaction to proceed.
Thermodynamically unfavorable reaction can be
driven by a favorable reaction
 ΔGo’ is directly dependent on the equilibrium constant K’eq.
 For a reaction:
•
•
A + B
C +D
ΔG = ΔGo’ + RT. ln [C] [D]/[A] [B]
 ΔG depends on the nature of the reactants (expressed in the ΔGo’ term) as well as
the concentration of the reactant and products
 At equilibrium, ΔG = 0, and K’eq = [C] [D]/[A] [B]
•
•
•
ΔGo’ = - RT.ln K’eq or ΔGo’ = - 2.303 RT log K’eq
since R= 1.987 x 10-3 kcal.mol-1.deg-1 and T = 298K,
ΔGo’ = - 1.36 log K’eq or K’eq = 10 –ΔGo’ /1.36

Thermodynamically unfavorable reaction can be
driven
by
a
favorable
reaction
Two or more reactions can be coupled when one or more products of one reaction are
reactants of the next reaction.
 When multiple reactions are coupled, it becomes a pathway.
 A pathway must satisfy minimally two criteria:
 The individual reactions must be specific, yielding only one particular product or
set of products.
 The entire set of reactions in a pathway must be thermodynamically favored

An important thermodynamic fact: the overall free energy change for a chemically
coupled series of reactions is equal to the sum of the free-energy changes of the
individual steps
AB+C
G0’ = + 5 kcal mol-1
BD
G0’ = - 8 kcal mol-1
*******************************
AC+D
G0’ = - 3 kcal mol-1

Thus, a thermodynamically unfavorable reaction can be driven by a
thermodynamically favorable reaction to which it is coupled.
ATP is the universal currency of free energy
• Metabolism is facilitated by the use of a common energy currency, ATP.
• Part of the free energy derived from the oxidation of foodstuffs and from
light is transformed into ATP - the energy currency.
• ATP: The role of ATP in energy metabolism is paramount!
• Nucleotide consisting of an adenine, a ribose, and a triphosphate unit
• Is an energy rich molecule
• triphosphate unit contains 2 phosphoanhydride bonds.
•A large amount of free energy is liberated when ATP is hydrolyzed to
ADP & Pi, or ATP to AMP & PPi
•Under typical cellular conditions, the actual G for these hydrolysis is
approximately -12 kcal mol-1
ATP + H2O  ADP + Pi
G0’ = -7.3 kcal mol-1
ATP + H2O  AMP + PPi
G0’ = -10.9 kcal mol-1
Adenine
Ribose
Phosphate units
ATP hydrolysis drives metabolism by shifting the equilibrium of
coupled reactions
 Thermodynamically unfavorable reactions can proceed when coupled with ATP hydrolysis. For
example:
• A
B
ΔGo’ = 4 kcal.mol-1,
o
• K’eq = 10 –ΔG ’ /1.36
• then K’eq = 1.15 x 10-3 and equilibrium is reached at very low [B].
 If this reaction is coupled to another reaction that has a very low (negative) ΔGo’, then overall
ΔGo’ of coupled reaction is lowered such that Keq is increased significantly.
 Biochemical reactions are usually coupled to the hydrolysis of ATP:
•
•
•
•
•
ATP
ADP + Pi
ΔGo’= -7.3 kcal.mol-1
Overall ΔGo’ = 4 + (-7.3) = -3.3 kcal.mol-1
Overall Keq’ = [B] /[A] x [ADP][Pi]/[ATP] = 267
In the cell, [ATP]/[ADP][Pi] is maintained at 500 M-1, therefore
Keq’ = [B] /[A] x (1/500) =267 or [B]/[A] = 267 x 500 = 1.34 x 105
 Coupling with 1 ATP changed equilibrium ratio by a factor of about 108. For nATPs it changes by
108n
Example
 Glycolysis (breaking down glucose)
 Protein Synthesis
 Contraction / Relaxation of Muscles
Relaxed muscle + energy  contracted muscle
ATP + H2O  ADP + P + energy
YOUTUBE video on ATP Coupling
 http://www.youtube.com/watch?v=4O6DFv2p5js&f
eature=player_detailpage