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First Law of Thermodynamics
• energy cannot be created or destroyed.
• Therefore, the total energy of the
universe is a constant.
• Energy can, however, be converted
from one form to another or transferred
from a system to the surroundings or
vice versa.
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‫אנרגיה חופשית וספונטניות‬-14
Enthalpy/Entropy
• Enthalpy is the heat absorbed by a
system during a constant-pressure
process.
• Entropy is a measure of the
randomness in a system.
• Both play a role in determining whether
a process is spontaneous.
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‫אנרגיה חופשית וספונטניות‬-14
Spontaneous Processes
• Spontaneous processes
proceed without any outside
assistance.
• The gas in vessel A will
spontaneously effuse into
vessel B, but it will not
spontaneously return to
vessel A.
• Processes that are
spontaneous in one direction
are nonspontaneous in the
reverse direction.
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‫אנרגיה חופשית וספונטניות‬-14
Experimental Factors Affect
Spontaneous Processes
• Temperature and pressure can affect spontaneity.
• An example of how temperature affects spontaneity is ice
melting or freezing.
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Reversible and Irreversible
Processes
Reversible process: The system
changes so that the system and
surroundings can be returned to
the original state by exactly
reversing the process. This
maximizes work done by a
system on the surroundings.
5
Irreversible processes cannot
be undone by exactly
reversing the change to the
system or cannot have the
process exactly followed in
reverse. Also, any
spontaneous process is
irreversible!
‫אנרגיה חופשית וספונטניות‬-14
Entropy
• Entropy can be thought of as a measure
of the randomness of a system.
• It is a state function:
• It can be found by heat transfer from
surroundings at a given temperature:
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‫אנרגיה חופשית וספונטניות‬-14
Second Law of Thermodynamics
The entropy of the universe increases in
any spontaneous processes.
This results in the following
relationships:
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‫אנרגיה חופשית וספונטניות‬-14
Entropy on the Molecular Scale
• Boltzmann described entropy on the molecular level.
• Gas molecule expansion: Two molecules are in the
apparatus above; both start in one side. What is the
likelihood they both will end up there? (1/2)2
6.02×1023
• If one mole is used? (1/2)
! (No chance!)
• Gases spontaneously expand to fill the volume given.
• Most probable arrangement of molecules: approximately
equal molecules in each side
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‫אנרגיה חופשית וספונטניות‬-14
Statistical Thermodynamics
• Thermodynamics looks at bulk properties of
substances (the big picture).
• We have seen what happens on the
molecular scale.
• How do they relate?
• We use statistics (probability) to relate them.
The field is called statistical thermodynamics.
• Microstate: A single possible arrangement of
position and kinetic energy of molecules
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‫אנרגיה חופשית וספונטניות‬-14
Boltzmann’s Use of Microstates
• Because there are so many possible
microstates, we can’t look at every picture.
• W represents the number of microstates.
• Entropy is a measure of how many
microstates are associated with a particular
macroscopic state.
• The connection between the number of
microstates and the entropy of the system is:
k = Boltzmann constant = 1.38 × 10−23 J/K
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‫אנרגיה חופשית וספונטניות‬-14
Entropy Change
• Since entropy is a state function, the final
value minus the initial value will give the
overall change.
• In this case, an increase in the number of
microstates results in a positive entropy
change (more disorder).
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‫אנרגיה חופשית וספונטניות‬-14
W
These are energetically
equivalent states for the
expansion of a gas.
It doesn’t matter, in terms
of potential energy,
whether the molecules
are all in one flask or
evenly distributed.
But one of these states is
more probable than the
other two.
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‫אנרגיה חופשית וספונטניות‬-14
Macrostates → Microstates
This macrostate can be achieved through several
different arrangements of the particles.
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‫אנרגיה חופשית וספונטניות‬-14
Macrostates → Microstates
These microstates
all have the same
macrostate.
So there are six
different particle
arrangements that
result in the same
macrostate.
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‫אנרגיה חופשית וספונטניות‬-14
Macrostates and Probability
• There is only one possible
arrangement that gives state A
and one that gives state B.
• There are six possible
arrangements that give state C.
• The macrostate with the highest
entropy also has the greatest
dispersal of energy.
• Therefore, state C has higher
entropy than either state A or
state B.
• There is six times the
probability of having the state C
macrostate than either state A
or state B.
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‫אנרגיה חופשית וספונטניות‬-14
Effect of Volume and Temperature
Change on the System
• If we increase volume, there are more positions
possible for the molecules. This results in more
microstates, so increased entropy.
• If we increase temperature, the average kinetic
energy increases. This results in a greater
distribution of molecular speeds. Therefore, there
are more possible kinetic energy values, resulting
in more microstates, increasing entropy.
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‫אנרגיה חופשית וספונטניות‬-14
Molecular Motions
• Molecules exhibit several types of motion.
 Translational: Movement of the entire molecule
from one place to another
 Vibrational: Periodic motion of atoms within a
molecule
 Rotational: Rotation of the molecule about an axis
• Note: More atoms means more microstates (more
possible molecular motions).
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‫אנרגיה חופשית וספונטניות‬-14
Entropy on the Molecular Scale
• The number of microstates and, therefore,
the entropy tend to increase with increases in
 temperature.
 volume.
 the number of independently moving
molecules.
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‫אנרגיה חופשית וספונטניות‬-14
Entropy and Physical States
•
•
•



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Entropy increases with the freedom of motion of molecules.
S(g) > S(l) > S(s)
Entropy of a system increases for processes where
gases form from either solids or liquids.
liquids or solutions form from solids.
the number of gas molecules increases
during a chemical reaction.
‫אנרגיה חופשית וספונטניות‬-14
Entropy
DS = Sfinal  Sinitial
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‫אנרגיה חופשית וספונטניות‬-14
‫‪Entropy‬‬
‫‪-14‬אנרגיה חופשית וספונטניות‬
‫‪21‬‬
‫‪Entropy‬‬
‫‪-14‬אנרגיה חופשית וספונטניות‬
‫‪22‬‬
Entropy Changes in Surroundings
• Heat that flows into or out of the system
changes the entropy of the surroundings.
• For an isothermal process
• At constant pressure, qsys is simply DH°
for the system.
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‫אנרגיה חופשית וספונטניות‬-14
Measuring Dispersal of E: Entropy
Calculate ΔfusS when 25.0 g of Al(s) melts at its normal
melting point (660.3 °C; 1 atm). ΔfusH° (Al)=10.7 kJ mol-1
qrev
ΔfusH°
qp = ΔH
ΔfusS =
=
Reversible if
T
T
normal melting
T = 660.3 °C = 933.5 K
25.0 g
nAl =
= 0.9266 mol
-1
26.98 g mol
0.9266 mol(10.7 kJ mol-1) = 0.01062 kJ K-1
ΔfusS =
933.5 K
24
ΔfusS = 10.6 J K-1 (system)
‫אנרגיה חופשית וספונטניות‬-14
The Third Law of Thermodynamics: Absolute Entropy
A perfect crystal, at absolute zero (0 K) has:
• Minimum molecular motion (min. E dispersal).
S=0
(perfect crystal at 0 K)
(The 3rd law of thermodynamics)
Measure the heat to change from 0 K  room-T
(reversibly). Gives ΔS and the absolute S at room T.
qrev
ΔS = Sfinal – Sinitial = T
Sroom T = ΔS – Sinitial = ΔS
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‫אנרגיה חופשית וספונטניות‬-14
Absolute Entropy Values
Standard molar entropies (S°), are measured by
heating 1 mol of substance from 0 K to:
• A specified T (often 25 °C = 298.15 K)
• At constant P = 1 bar
S° has J K-1mol-1 units.
S° is always positive, for all materials.
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‫אנרגיה חופשית וספונטניות‬-14
Standard Molar Entropies (298.15 K)
Substance
S° (J K-1 mol-1) Substance
S° (J K-1 mol-1)
C (graphite)
5.740
C(g)
158.096
CH4(g)
186.264
C2H6(g)
229.60
Cl2(g)
Br2(ℓ)
I2(s)
HCl(g)
223.066
152.231
116.135
186.908
C3H8(g)
C2H4(g)
CH3OH(ℓ)
269.9
219.4
126.8
H2O(g)
H2O(ℓ)
NaCl(s)
188.825
69.91
72.13
CO(g)
197.674
KOH(s)
78.9
CO2(g)
F2(g)
213.74
202.78
NaCl(aq)
KOH(aq)
27
115.5
91.6
‫אנרגיה חופשית וספונטניות‬-14
Qualitative Guidelines for Entropy
Sgas >> Sliquid > Ssolid
Gas molecules have few restrictions on motion.
Liquid molecules can slide past each other.
Solid molecules vibrate about fixed lattice points.
‫אנרגיה חופשית וספונטניות‬-14
S[Cl2(g)] = 223
S[Br2(ℓ)] = 152
S[I2(s)] = 116 J K-1 mol-1
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Qualitative Guidelines for Entropy
Larger, more complex, molecules have larger S:
S°complex-molecule > S°simple-molecule
S° (JK-1mol-1)
gas
mass
S° (JK-1mol-1)
CH4
186
Ar
40
155
C2H6
C3H8
230
270
CO2
C3H8
44
44
214
270
Alkane
Larger molecules have more ways to distribute E
(more bonds to hold potential and kinetic E).
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‫אנרגיה חופשית וספונטניות‬-14
Qualitative Guidelines for Entropy
Ionic solids
Weaker ionic forces = larger S
(Less tightly bound = more extensive motion)
S°
(J K-1 mol-1)
MgO
NaF
NaCl
27
52
72
Charges
Q1 Q2
+2
+1
+1
-2
-1
-1
Radii
Mn+ Xn-
d
(pm)
86 126
116 119
116 167
212
235
283
Cl-
Na+
Q1Q2
Ionic force, F = k 2
d
30
d
Na+
F-
‫אנרגיה חופשית וספונטניות‬-14
Qualitative Guidelines for Entropy
Solid or liquid dissolving
Larger matter dispersal. S usually* increases:
molecule
S°(pure)
CH3COOH(ℓ)
160
NH4NO3(s)
151
S°(aq)
179
ΔS°
+19
260
+109
J K-1 mol-1 units
*Strong solvation may increase order. ΔS < 0 is
possible.
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‫אנרגיה חופשית וספונטניות‬-14
Qualitative Guidelines for Entropy
Gas dissolving
Gas-molecule motion becomes restricted. ΔS < 0.
Gas
S°(pure)
CO2
H2
CH3OH
32
214
131
240
S°(aq)
ΔS°
118
58
133
-96
-73
-107
‫אנרגיה חופשית וספונטניות‬-14
Predicting Entropy Changes
CaCO3(s) → CaO(s) + CO2(g) ΔS = 161 J K-1mol-1
Large entropy increase:
• Solid → gas, ΔS > 0.
• 1 molecule → 2 molecules, ΔS > 0.
H2(g) + F2(g) → 2 HF(g)
ΔS = 14 J K-1mol-1
• All gases, S ≈ constant.
• 2 reactant gases, 2 product gases, S ≈ constant.
Actual ΔS shows a small increase.
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‫אנרגיה חופשית וספונטניות‬-14
Predicting Entropy Changes
Predict whether S will increase or decrease for:
2 CO(g) + O2(g) → 2 CO2(g)
Decrease
3 gas molecules → 2 gas molecules.
NaCl(s) → Na+(aq) + Cl-(aq)
Increase
Ions locked in a crystal → ions dispersed
in water.
H2O(ℓ) → H2O(s)
Decrease Liquid molecules less restricted than solid
molecules.
‫אנרגיה חופשית וספונטניות‬-14
34
Calculating Entropy Changes
Entropy is a state function (like H). For a reaction:
ΔrS° = (nproductS°product) – (nreactantS°reactant)
Example
Methanol is a common fuel additive. Calculate ΔrS° at
298 K for its combustion:
2 CH3OH(ℓ) + 3 O2(g) → 2 CO2(g) + 4 H2O(ℓ)
Look up S° values (J K-1 mol-1):
CH3OH(ℓ) = 126.8
O2(g) = 205.138
CO2(g)
= 213.74
H2O(ℓ) = 69.91
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‫אנרגיה חופשית וספונטניות‬-14
Calculating Entropy Changes
Example: ΔrS° at 298 K for: 2 CH3OH(ℓ) + 3 O2(g) → 2 CO2(g) + 4 H2O(ℓ)
ΔrS° = (nproductS°product) – (nreactantS°reactant)
= (2S°CO2 + 4S°H2O) – (2S°CH3OH + 3S°O2)
= [2(213.74) + 4(69.91)] – [2(126.8) + 3(205.14)]
= 707.12 – 869.02
= –161.90 J K-1 mol-1
S decreases as expected: 3 gas + 2 ℓ → 2 gas + 4 ℓ
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‫אנרגיה חופשית וספונטניות‬-14
Entropy & the 2nd Law of Thermodynamics
E dispersal is accompanied by an increase in disorder
of a system:
Increased disorder = increased S
Second Law of Thermodynamics
“The total entropy of the universe is continuously
increasing.”
• The universe is slowly becoming more disordered.
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‫אנרגיה חופשית וספונטניות‬-14
Entropy & the 2nd Law of Thermodynamics
The entropy of a system can increase or decrease.
• Suniverse always goes up.
• Suniverse = Ssystem + Ssurr
(No subscript? S = Ssystem ).
• If Ssystem falls, Ssurr must increase by a larger amount.
qrev
ΔrS°surr =
T
38
=
ΔrHsurr
T
=
-ΔrHsys
T
‫אנרגיה חופשית וספונטניות‬-14
Entropy & the 2nd Law of Thermodynamics
Calculate ΔrS°univ for the combustion of octane in air:
2 C8H18(g) + 25 O2(g)  16 CO2(g) + 18 H2O(ℓ)
For the reaction ΔrH° = -11,024 kJ mol-1 and ΔrS° =
-1383.9 J mol-1K-1 at 298.15 K.
° = ΔrS°sys + ΔrSsurr
°
ΔrSuniv
given
-ΔHsys / T
° = -1383.9 J/K + (+11,024 x 103 J)/298.15 K
ΔrSuniv
= -1383.9 J/K + 36.975 kJ/K
39
= +35,591 J/K
‫אנרגיה חופשית וספונטניות‬-14
Gibbs Free Energy
Neither entropy (S), nor enthalpy (H ), alone can predict
whether a reaction is product favored.
Spontaneous (product favored) reactions can:
• Be exothermic or endothermic.
• Increase or decrease Ssystem.
The Gibbs free energy (G), combines H and S.
ΔrG does predict if a reaction is product favored or not.
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‫אנרגיה חופשית וספונטניות‬-14
Gibbs Free Energy
For a constant T change, ΔG is equal to:
ΔrG = ΔrH - TΔrS
Josiah Willard Gibbs
If G:
Decreases (ΔrG < 0) a reaction is product favored
Increases (ΔrG > 0) a reaction is reactant favored
(at constant P and T).
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‫אנרגיה חופשית וספונטניות‬-14
Effect of T on Reaction Direction
ΔrG = ΔrH – TΔrS
Product Favored?
ΔrH ΔrS
Low T
High T
<0
>0
Yes
Yes
<0
<0
Yes
No
>0
>0
No
Yes
>0
<0
No
No
42
Comments
exothermic; disorder increase
always possible
exothermic; disorder decrease
possible at low T
endothermic; disorder increase
possible at high T
endothermic; disorder decrease
never occurs
‫אנרגיה חופשית וספונטניות‬-14
Gibbs Free Energy
ΔrG° can be calculated from ΔrH° and ΔrS° values…
… or from Gibbs free energies of formation (ΔfG°).
ΔrG° = (nproductΔfG°product) – (nreactantΔfG°reactant)
ΔfG° equals:
• ΔrG to make 1 mol of compound from its elements.
• 0 for an element in its most stable state (like ΔfH°)
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‫אנרגיה חופשית וספונטניות‬-14
Gibbs Free Energy
Calculate ΔrG° for the following reaction at 25°C:
CO(g) + 2 H2(g) → CH3OH(ℓ)
ΔfG° values: CO(g) = -137.2 kJ/mol
CH3OH(ℓ) = -166.3 kJ/mol
ΔG° = ΔfG°CH3OH – (2ΔfG°H2 + 1ΔfG°CO)
element in its
standard state
= -166.3 kJ/mol – [2(0) + 1(-137.2 kJ/mol)]
= -29.1 kJ/mol
Product favored
44
‫אנרגיה חופשית וספונטניות‬-14
Effect of T on Reaction Direction
A reactions will change from reactant to productfavored (or visa versa) when ΔrG° = 0.
Δ rG° = 0 = Δ rH° − TΔ rS°
ΔrH° and ΔrS° typically remain ≈constant over large T
ranges and thus the “switchover” T can be calculated:
0 = Δ rH° − TΔ rS°
Δ rH° = TΔ rS°
ΔrH°
T=
ΔrS°
45
‫אנרגיה חופשית וספונטניות‬-14
Effect of T on Reaction Direction
46
‫אנרגיה חופשית וספונטניות‬-14
Effect of T on Reaction Direction
At what T will the following reaction become product
favored?
CH4(g) + H2O(g)
CO(g) + 3 H2(g)
ΔfH° (kJ/mol)
−74.81
S° (J K-1 mol-1) 186.26
−241.82
188.83
−110.53
197.67
0
130.68
ΔrS°=[197.67 +3(130.68)] − [186.26 + 188.83] J K-1mol-1
= 214.62 J K-1 mol-1
47
ΔrH° = [-110.53 + 3(0)] − [-74.81 + -241.82] kJ mol-1
= 206.10 kJ mol-1
It will switch direction at:
2.0610 x 105 J mol-1
Δ rH
T=
=
= 960.30 K
-1
-1
214.62 J K mol
Δ rS
‫אנרגיה חופשית וספונטניות‬-14
Variation of G During a Reaction
ΔrG° equals the change in G which occurs when pure
reactants are completely converted to pure products.
How does G vary during a reaction?
Consider:
NaOH(s) + CO2(g)  NaHCO3(s)
ΔrG°= −77.2 kJ/mol
The extent of reaction (x) equals the fraction of
reactants converted to products.
48
‫אנרגיה חופשית וספונטניות‬-14
Variation of G During a Reaction
ninitial
nx=0.5
nx=1.0
49
NaOH(s) + CO2(g)  NaHCO3(s)
ΔrG°= −77.2 kJ/mol
1 mol
½ mol
0
ΔrG°= −33.6 kJ
ΔrG°= −77.2 kJ
1 mol
½ mol
0
0
½ mol
1 mol
‫אנרגיה חופשית וספונטניות‬-14
Reactions that Reach Equilibrium
If a reaction reaches equilibrium, G has a minimum:
cis-2-butene(g)
50
trans-2-butene(g)
ΔrG° = -2.08 kJ/mol
‫אנרגיה חופשית וספונטניות‬-14
Free Energy and Equilibrium
Under any conditions, standard or
nonstandard, the free energy change can be
found this way:
DG = DG° + RT ln Q
(Under standard conditions, concentrations are 1 M,
so Q = 1 and ln Q = 0; the last term drops out.)
51
‫אנרגיה חופשית וספונטניות‬-14
Free Energy and Equilibrium
• At equilibrium, Q = K, and DG = 0.
• The equation becomes
0 = DG° + RT ln K
• Rearranging, this becomes
DG° = RT ln K
or
DG/RT
K=e
52
‫אנרגיה חופשית וספונטניות‬-14
Reactions that Reach Equilibrium
Gibbs free energy G and the equilibrium constant K
both show if a reaction is product-favored and are
related.
ΔrG° = −RT ln K°
K° must be unitless.
Each quantity in the K° expression is divided by its
standard state value:
• concentrations by 1 mol L-1 (solids and liquids).
• pressures by 1 bar (gas reactions).
53
‫אנרגיה חופשית וספונטניות‬-14
Reactions that Reach Equilibrium
Evaluate K° (298 K) for the reaction:
2 NO2(g)
N2O4(g)
ΔrG° = ΔfG°(N2O4) - 2ΔfG°(NO2)
= 97.89 – 2(51.31) kJ/mol = -4.73 kJ/mol
3 J/mol
ΔrG°
-4.73
x
10
ln K° = –
=–
= 1.91
-1
-1
RT
(8.314 J K mol )(298 K)
K° = 6.75
54
‫אנרגיה חופשית וספונטניות‬-14
Gibbs Free Energy & Maximum Work
ΔrG = maximum useful work that can be done by a
reaction on its surroundings (constant T and P).
ΔrG = wsystem = - wmax
“free” energy = available energy
For:
2 H2(g) + O2(g)
2 H2O(g)
ΔrG°= -474.3 kJ/mol
Every 2 moles of H2 consumed can do up to 474.3 kJ
of work.
55
‫אנרגיה חופשית וספונטניות‬-14
Free Energy and Reversible Reactions
• The change in free energy is a theoretical
limit as to the amount of work that can
be done.
• If the reaction achieves its theoretical limit, it
is a reversible reaction.
56
‫אנרגיה חופשית וספונטניות‬-14
Gibbs Free Energy & Maximum Work
ΔrG also equals the minimum work required to cause a
reactant-favored process to occur.
2 H2O(g)
2 H2(g) + O2(g)
ΔrG°= 474.3 kJ/mol
A minimum of 474.3 kJ of work must be used to
produce 2 mol of H2 from liquid water.
57
‫אנרגיה חופשית וספונטניות‬-14
Gibbs Free Energy & Biological Systems
Energy released by oxidizing glucose:
C6H12O6(aq) + 6 O2(g) → 6 CO2(g) + 6 H2O(ℓ)
at pH =7
ΔrG°′= -2870 kJ/mol
Is stored in 32 more manageable “chunks”:
32[ADP3-(aq) + H2PO4-(aq) → ATP4-(aq) + H2O(ℓ)
ΔrG°′= 30.5 kJ/mol]
32x = 976 kJ
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‫אנרגיה חופשית וספונטניות‬-14
Gibbs Free Energy & Biological Systems
A three-stage process extracts G from nutrients.
• Digestion: large molecules are broken down into smaller
ones.
• Conversion of smaller molecules to acetyl coenzyme A.
• Citric acid cycle: acetyl groups are oxidized to CO2 and H2O
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‫אנרגיה חופשית וספונטניות‬-14
Photosynthesis & Gibbs Free Energy
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6 CO2(g) + 6 H2O(ℓ) → C6H12O6(aq) + 6 O2(g)
‫אנרגיה חופשית וספונטניות‬-14
Thermodynamics versus Kinetics
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‫אנרגיה חופשית וספונטניות‬-14
Thermodynamic & Kinetic Stability
Thermodynamically stable: process is reactant-favored.
• Thermodynamically stable:
Pd + O2 → PdO2
ΔrG° = +326 kJ/mol
• Thermodynamically unstable:
4 Al + 3 O2 → 2 Al2O3
ΔrG° = -3164.6 kJ/mol
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Kinetically stable: product-favored, but too slow to be
important.
• Diamond is thermodynamically unstable
Cdiamond → Cgraphite ΔrG° = -2.9 kJ/mol
• but it is kinetically stable
• Ea is too large
‫אנרגיה חופשית וספונטניות‬-14