Chapter 8. THERMODYNAMICS: THE
SECOND AND THIRD LAW
ENTROPY
8.1 Spontaneous Change
8.2 Entropy and Disorder
8.3 Changes in Entropy
8.4 Entropy Changes Accompanying Changes in Physical
State
8.5 A Molecular Interpretation of Entropy
8.6 The Equivalence of Statistical and Thermodynamic
Entropies
8.7 Standard Molar Entropies
8.8 Standard Reaction Entropies
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ENTROPY (Sections 8.1-8.8)
- The 1st law of thermodynamics says if a reaction
takes place, then the total energy of the universe
remains unchanged. It cannot be used to predict the
directionality of a process.
- The natural progression of a system and its
surroundings (or “the universe”) is from order to
disorder, from organized to random.
- A new thermodynamic state function is needed to
predict both directionality and extent of disorder.
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8.1 Spontaneous Change
 Spontaneous
change is a
change that
has a tendency
to occur
without
needing to be
driven by an
external
influence.
- Spontaneous changes need
not be fast: e.g. C(diamond) 
Heat flow
Mixing of gases
C(graphite); H2(g) + 1/2O2(g)  H2O(l)
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8.2 Entropy and Disorder
- Energy and matter tend to disperse in a disorderly fashion.
 Entropy, S is defined as a measure of disorder.
 The second law of thermodynamics:
The entropy of an isolated system increases in any
spontaneous change.
unit: J·K-1
At constant temperature
- Entropy is a state function; the change in entropy of
a system is independent of the path between its initial
and final states.
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Self-Test 8.1A
Calculate the change in entropy of a large block of ice
when 50. J of energy is removed from it as heat at 0 oC
in the freezer.
Solution
_ 50. J
S =
273.15 K
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=
_
0.18 J K-1
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8.3 Changes in Entropy
- Thermal disorder: arising from the thermal motion
of the molecules
- Positional disorder: related to the locations of the
molecules
 S for a process with changing temperature:
→
(CV if V is constant, CP if P is constant)
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 S for a reversible, isothermal expansion of an ideal gas
S =
qrev
T
=
( _ wrev)
T
=
nRT ln(V2/V1)
T
Hence for the isothermal expansion of an ideal gas,
S = nR ln V2
V1
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or
S = nR ln P1
P2
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Self-Test 8.2A
The temperature of 1.00 mol He(g) is increased from
25.0 oC to 300. oC at constant volume. What is the
change in entropy of the helium? Assume ideal
behavior and use Cv,m = 3/2R.
Solution
25.0 oC = 298 K; 300. o C = 573 K
 S = nCv,m ln(T2/T1)
= (1.00 mol) x 3 x (8.3145 J mol -1 K-1) x ln 573 K
298 K
2
= +8.15 J K-1
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Self-Test 8.3A
Calculate the change in molar entropy of an ideal gas
when it is compressed isothermally to one-third its
initial volume.
Solution
S = nR ln(V2/V1)
-1
-1
= (1.00 mol) x (8.3145 J K mol ) x ln 0.33
= _ 9.13 J K-1 mol-1
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Self-Test 8.4A
Calculate the change in entropy when the pressure
of 1.50 mol Ne(g) is decreased isothermally from
20.00 bar to 5.00 bar. Assume ideal behavior.
Solution
 S = nR ln(P1/P2)
-1
-1
= (1.50 mol) x (8.3145 J K mol ) x ln
= + 17.3 J K-1
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20.00 bar
5.00 bar
EXAMPLE 8.5
In an experiment, 1.00 mol Ar(g) was compressed
suddenly (and irreversibly) from 5.00 L to 1.00 L by driving in a piston.
and in the process its temperature was increased from 20.0 oC to 25.2 oC.
What is the change in entropy of the gas?
To solve this problem, we consider two
reversible stages between initial and
final states. Then S(irrev) = S(rev 1) +
S(rev 2).
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8.4 Entropy Changes Accompanying Changes
in Physical State
- Phase transition: solid → liquid, Tf (fusion or melting point)
liquid → solid, Tb (boiling point)
- At the transition temperature (such as Tb),
The temperature remains constant as heat is supplied.
The transfer of heat is reversible.
The heat supplied is equal to the enthalpy change due to the
constant pressure (at 1 atm).
 Entropy of vaporization, Svap
qrev = Hvap
> 0 in all cases
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- Standard entropy of vaporization, Svapo :
Svap at 1 bar
Some Standard entropies of vaporization at Tb
(Table 8.1)
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 Trouton’s rule: Svapo = ~85 J·K-1mol-1
There is approximately the same increase in positional
disorder for most liquids when evaporating.
- Exceptions; water, methanol, ethanol, ··· due to
extensive hydrogen bonding in liquid phases
- Standard entropy of fusion, Sfuso
> 0 in all cases
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 Temperature dependence of Svapo
-To determine the entropy of vaporization of water at 25 oC
(not at Tb), we can use an entropy change cycle:
Svap (Tb) = Hvap /Tb
S1 = nCp,m(liq.) ln
Liquid (Tb = 100 oC)
373.15 K
298.15 K
Vapor (Tb)
S2 = nCp,m(vap.) ln 298.15 K
373.15 K
Liquid (25 oC)
Vapor (25 oC)
o (25 oC)
Svap
Svap (25 oC) = S1(liq. heating; 25 oC
Tb)
+ Svap(Tb)
+ S2(vap. condensing: Tb 25 oC)
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297s
Exercise 8.43
Calculate the standard entropy of vaporization of water at 85 oC, given
that its standard entropy of vaporization at 100 oC is 109.0 J·K-1·mol-1 and
the molar heat capacities at constant pressure of liquid water and water
vapor are 75.3 J·K-1·mol-1 and 33.6 J·K-1·mol-1, respectively, in this range.
Solution
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8.5 A Molecular Interpretation of Entropy
 The third law of thermodynamics
S → 0 as T → 0
 Boltzmann formula
- statistical entropy:
S = kB ln W
kB = 1.381 × 10-23 J·K-1 = R/NA
- W : the number of microstates
the number of ways that the atoms or molecules in the sample
can be arranged and yet still give rise to the same total energy
- When we measure the bulk properties of a system, we are measuring
an average taken over the many microstates (ensemble) that the
system has occupied during the measurement.
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EXAMPLE 8.7
Calculate the entropy of a tiny solid made up of four diatomic molecules
of a compound such as carbon monoxide, CO, at T = 0 when (a) the four
molecules have formed a perfectly ordered crystal in which all molecules
are aligned with their C atoms on the left and (b) the four molecules lie
in random orientations, but parallel.
(a) 4 CO molecules perfectly ordered:
(b) 4 CO in random, but parallel:
(c) 1 mol CO in random, but parallel:
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- Residual entropy at T = 0, arising from positional disorder
4.6 J·K-1 for the entropy of 1 mol CO < 5.76 J·K-1
Nearly random arrangement due to a small electric
dipole moment
- Solid HCl; S ~ 0 at T = 0 due to the bigger dipole moment
leading strict head-to-tail arrangement
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EXAMPLE 8.8
The entropy of 1.00 mol FClO3(s) at T = 0 is 10.1 J·K-1. Interpret it.
4 orientations possible
nearly random arrangement
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8.6 The Equivalence of Statistical and
Thermodynamic Entropies
Thermodynamic entropy
S = qrev/T
behavior of bulk matter
=
Statistical entropy
S = k ln W
behavior of molecules
- Consider a one-dimensional box, for the
statistical entropy,
• At T = 0, only the lowest energy level occupied
→ W = 1 and S = 0
• At T > 0, W > 1 and S > 0
• When the box length is increased at constant T,
the molecules are distributed across more levels.
→ W and S increase.
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- W = constant × V
- For N molecules,
- The change when a sample expands isothermally from V1 to V2 is,
V2
= nR lnV
1
- By raising the temperature,
The molecules have access to larger number of
energy levels → W and S increase.
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- The equations used to calculate changes in the
statistical entropy and the thermal entropy lead to the
same result.
1. Both are state functions.
Number of microstates depends only on its current state.
2. Both are extensive (dependent on “extent”) properties.
2 × no. of molecules = entropy changes from k ln W
to 2k ln W
3. Both increase in a spontaneous change.
In any irreversible change, the overall disorder increases
→ no. of microstates increases.
4. Both increase with temperature.
When T increases, more microstates become accessible.
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302s
Exercise 8.27
If SO2F2 adopts a disordered arrangement in its crystal form, what
would its residual molar entropy be?
Solution
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302s
Exercise 8.29
8.29 Which substance in each of the following pairs has the
higher molar entropy at 298 K: (a) HBr(g) or HF(g); (b) NH3(g)
or Ne(g); (c) I2(s) or I2(l); (d) 1.0 mol Ar(g) at 1.00 atm or 1.0
mol Ar(g) at 2.00 atm?
Solution
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302s
Exercise 8.31
List the following substances in order of increasing molar
entropy at 298 K: H2O(l), H2O(g), H2O(s), C(s, diamond). Explain
your reasoning.
Solution
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302s
Exercise 8.33
Which substance in each of the following pairs would you expect to
have the higher standard molar entropy at 298 K? Explain your
reasoning: (a) Iodine vapor or bromine vapor;
(b) the two liquids cyclopentane and 1-pentene;
(c) ethene (ethylene) or an equivalent mass of polyethylene, a
substance formed by the polymerization of ethylene.
Solution
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302s
Exercise 8.35
Without performing any calculations, predict whether there is an
Increase or a decrease in entropy for each of the following
processes:
(a)Cl2(g) + H2O(l) → HCl(aq) + HClO(aq);
(b)Cu3(PO4)2(s) → 3 Cu2+(aq) + 2 PO43-(aq);
(c)SO2(g) + Br2(g) + 2 H2O(l) → H2SO4(aq) + 2 HBr(aq).
Solution
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8.7 Standard Molar Entropies
Molar entropy, S(T), can be determined from measurement
of Cp at different temperatures.
For heating at constant P,
CP and CP/T → 0
as T → 0
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 Standard molar entropy, Smo is the molar entropy of the
pure substance at 1 bar.
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 Standard molar entropy, Smo :
light heavy
- Diamond (2.4 J·K-1) vs. lead (64.8 J·K-1):
rigid bonds vs. vibrational energy levels
- H2 (130.7 J·K-1) vs. N2 (191.6 J·K-1):
the greater the mass, the closer energy
levels
- CaCO3 (92.9 J·K-1) vs. CaO (39.8 J·K-1):
large, complex vs. smaller, simpler
- In general, Smo: gases >> liquids > solids
Related to freedom of movement and disordered state
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8.8 Standard Reaction Entropies
 Standard reaction entropy, So, is the difference
between the standard molar entropies of the products
and those of the reactants, taking into account their
stoichiometric coefficients.
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EXAMPLE 8.9
Calculate So for N2(g) + 3H2(g) → 2NH3(g) at
25 oC.
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307s
Exercise 8.35
Use data in Table 8.3 or Appendix 2A to calculate the standard reaction
entropy for each of the following reactions at 25 C. Interpret the sign and
magnitude of the reaction entropy. (a) The formation of 1.00 mol H2O(l)
from the elements in their most stable state at 298 K. (b) The oxidation of
1.00 mol CO(g) to carbon dioxide. (c) The decomposition of 1.00 mol
calcite, CaCO3(s), to carbon dioxide gas and solid calcium oxide. (d) The
decomposition of potassium chlorate: 4 KClO3 (s)  3 KClO4 (s) + KCl (s)
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307s
Solutions
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307s
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Chapter 8. THERMODYNAMICS: THE
SECOND AND THIRD LAW
GLOBAL CHANGES IN ENTROPY
8.9 The Surroundings
8.10 The Overall Change in Entropy
8.11 Equilibrium
GIBBS-FREE ENERGY
8.12 Focusing on the System
8.13 Gibbs Free Energy of Reaction
8.14 The Gibbs Free Energy and Nonexpansion Work
8.15 The Effect of Temperature
8.16 Impact on Biology: Gibbs Free Energy Changes
in Biological Systems
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GLOBAL CHANGES IN ENTROPY
(Sections 8.9-8.11)
8.9 The Surroundings
- The second law refers to an isolated
system (system + surroundings =
universe).
- Only if the total entropy
change is positive will the
process be spontaneous.
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Sometimes Ssurr can be difficult to compute, but in general
it can be obtained from the enthalpy change for the process
(that is, of the system).
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Self-Test 8.14A
Calculate the entropy change of the surroundings
when 1.00 mol H2O(l) vaporizes at 90 o C and 1 bar.
Take the enthalpy of vaporization of water as +40.7
kJ mol -1.
Solution
90 o C = 363 K
Ssurr = _ H/T
-1
(
+40
700
J
mol
)
_
=
363 K
_ 112 J K-1
=
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8.10 The Overall Change in Entropy
- To use the entropy to
judge the direction of
spontaneous change,
we must consider the
change in the entropy of
the system plus the
entropy change in the
surroundings:
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-Spontaneous
exothermic
(H<0)
reactions:
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EXAMPLE
8.11
Is the reaction spontaneous at 298 K?
2 Mg(s) + O2(g) → 2 MgO(s) So = -217 J·K-1 Ho = -1202 kJ
The reaction is spontaneous
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-Spontaneous endothermic
(H>0) reactions:
There can still be an
overall increase in
entropy
if the disorder of the
system increases
enough.
Summary
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- A process produces maximum work if it takes place reversibly.
 Clausius inequality
S =
>
S
- For an isolated system (universe), q = 0
The entropy of an isolated system cannot decrease.
- For two given states of the system,ΔS is a state
function (path-independent) but ΔStot is not.
(See EXAMPLE 8.12)
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EXAMPLE
8.12
Calculate S, Ssurr, and Stot for (a) the isothermal, reversible
expansion and (b) the isothermal, free expansion of 1.00 mol of
ideal gas molecules from 8.00 L to 20.00 L at 292 K. Explain any
differences between the two paths.
(a) Isothermal reversible expansion at 292 K
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(b) Isothermal free expansion 292 K
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312s
Exercise 8.49
Initially a sample of ideal gas at 323 K occupies 1.67 L at 4.95 atm.
The gas is allowed to expand to 7.33 L by two pathways:
(a) isothermal, reversible expansion (b) isothermal, irreversible free
expansion.
Calculate ΔStot, ΔS, ΔSsurr for these pathways.
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8.11 Equilibrium
 Dynamic equilibrium is one where there is no net
tendency to change but microscopic forward and reverse
processes occur at matching rates.
Thermal equilibrium: no net flow of energy as heat
Mechanical equilibrium: no tendency to expand or contract
Chemical equilibrium: no net change in composition
at thermodynamic equilibrium
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GIBBS FREE ENERGY
(Sections 8.12-8.16)
8.12 Focusing on the System
-
at constant T and P
 Gibbs free energy, G
G = H - TS
- Change in Gibbs free energy
-
at constant T and P
The direction of spontaneous change is the direction of
decreasing Gibbs free energy.
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at constant T and P
Gsys < 0
Spontaneous, irreversible
Gsys = 0
Reversible
Gsys > 0
Nonspontaneous
- the condition for equilibrium, Stot = 0, and
G = 0
at constant T and P
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EXAMPLE
8.13
Calculate the change in molar Gibbs free energy, Gm, for the process
H2O(s) → H2O(l) at 1 atm and (a) 10 oC; (b) 0 oC. Decide for each
temperature whether melting is spontaneous or not. Treat Hfus and
Sfus as independent of temperature.
(a) At 10 oC,
= -0.22 kJ·mol-1 < 0; spontaneous melting
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(b) At 0 oC,
equilibrium (reversible)
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Variation of G with temperature: phase
transitions
- G decreases as its T is raised
at constant P.
G↓ = H – T↑S
; H and S vary
little with T, S > 0
- Decreasing rate of Gm: vapor >>
liquid > solid
Sm(vapor) >> Sm(liquid) > Sm(solid)
In most cases (opposite), heating
leads to melting, then boiling.
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In some cases, at certain pressures,
G for the liquid may never be lower
than those of the other two phases.
The liquid phase is unstable and the
phase transition is solid to vapor
(sublimation).
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8.13 Gibbs Free Energy of Reaction
 Gibbs free energy of reaction
 Standard Gibbs free energy of reaction
(standard state: pure form at 1 bar)
- Go is fixed for a given reaction and temperature.
- G depends on the composition of the reaction
mixture and so it varies – and might even change sign
as the reaction proceeds.
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 Standard Gibbs free energy of formation, Gfo , is the
standard Gibbs free energy of reaction per mole for
the formation of a compound from its elements in
their most stable form.
- For the most stable forms of elements, Gfo = 0
E.g. Gfo(I2, s) = 0; Gfo(I2, g) > 0
Examples of
most stable
forms of
elements
(Table 8.6)
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Some Standard Gibbs free Energies of Formation at
25 oC (kJ mol-1) (Table 8.7)
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EXAMPLE
8.14
Calculate the standard Gibbs free energy of formation of HI(g) at 25 oC
from its standard molar entropy and standard enthalpy of formation.
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Thermodynamic stability and reactivity
- Thermodynamically stable compound;
- Thermodynamically unstable compound;
but nonlabile or even inert
• Stable and unstable: thermodynamic tendency to
decompose into its elements
• Labile, nonlabile, and inert: the rate at which a
thermodynamic tendency to react is realized
- Another solution for standard Gibbs free energies of reaction:
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EXAMPLE
8.15
Calculate the standard Gibbs free energy of the reaction
4 NH3(g) + 5 O2(g) → 4 NO(g) + 6 H2O(g)
and decide whether the reaction is spontaneous under standard
conditions at 25 oC.
The reaction is
spontaneous
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8.14 The Gibbs Free Energy and Nonexpansion
Work
- The Gibbs free energy is a measure of the energy free to
do nonexpansion work.
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
- If we know the change in Gibbs free energy of a process taking
place at constant T and P, then we immediately know how
much nonexpansion work it can do.
E.g. Bioenergetics of glucose oxidation
- The maximum nonexpansion work obtainable from 1
mol of glucose is +2879 kJ at 1 bar.
- 180 g of glucose can be used to build 170 (= 2879/17)
mole of peptide links.
In practice, only about 10 moles of peptide links can be built.
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8.15 The Effect of Temperature
- The values of Ho and So do not change much with temperature.
- However, Go does depend much on temperature.
Go = Ho - TSo
1) For an exothermic reaction (Ho<0)
with So<0 (disorder decrease L-R),
Go<0 at low T
but it may become >0 at high T.
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2) For an endothermic reaction (Ho>0)
with So>0 (disorder increase L-R),
Go>0 at low T
but it may become <0 at high T.
3) For an endothermic reaction (Ho>0)
with So<0,
Go>0 at all T
and the reaction is never
spontaneous.
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4) For an exothermic reaction (Ho<0)
with So>0,
Go<0 at all T
and the reaction is always spontaneous.
The Gibbs free energy increases with T for
reactions with a negative So and
decreases with T for reactions with a
positive So.
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EXAMPLE
8.16
Estimate T at which it is thermodynamically possible for carbon to
reduce iron(III) oxide to iron under standard conditions by the
endothermic reaction, 2Fe2O3(s) + 3C(s)  4Fe(s) + 3CO2(g)
, above 565 oC
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8.16 Impact on Biology: Gibbs Free Energy
Changes in Biological Systems
- A reaction that produces a lot of entropy can drive
another nonspontaneous reaction forward.
- A process may be driven uphill in
Gibbs free energy by another
reaction that rolls downhill.
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 Hydrolysis of adenosine triphosphate (ATP) to adenosine
diphosphate (ADP): the reaction used for driving
nonspontaneous biochemical reactions
- The nonspontaneous reaction restoring of ATP from
ADP is driven by the food we eat.
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The End!
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Thank you for listening to this lecture,
and please continue to General
Chemistry II
to know a variety of Green World!
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