Chapter Sixteen:
SPONTANEITY, ENTROPY,
AND FREE ENERGY
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Contents
16-1 Spontaneous Process and
Entropy
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Figure 16.1 When methane and oxygen react to form carbon
dioxide and water, the products have lower potential energy
than the reactants. This change in potential energy results in
energy flow (heat) to the surroundings.
Whether a reaction is spontaneous based only p750
the properties of the reactants and products.
Thermodynamics
vs. Kinetics
Figure 16.2
Entropy is thermodynamic function that describes the
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number of arrangement that are available to a system
existing in a given state.
Entropy is closely associated with probability.
Nature spontaneously proceeds toward the state that have
the highest probabilities of existing.
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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.
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The expansion of an ideal gas
into an evacuated bulb.
Figure 16.3
A gas expands into a vacuum because the
expanded state has the highest positional
probability of states available to the system.
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Figure 16.4 Possible arrangements (states) of four
molecules in a two-bulbed flask.
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Table 16.1 The microstates that give a
particular arrangement (state).
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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

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Ex16.1 Position Entropy
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For each of the following pairs, choose the substance with
the higher positional entropy (per mole) at a given
temperature. (a) Solid CO2 and gaseous CO2. (b) N2 gas at 1
atm and N2 gas at 1.0 x 10-2 atm.
Solution:
a. The molecules have many more available positions than in
mole of solid CO2. Thus gaseous CO2 has the higher
positional entropy.
b. A mole of N2 gas at 1 x 10-2 atm has a volume 100 times
that (at a given a. Since a mole of gases CO2 has the greater
volume by far, temperature) of amole of N2 gas at 1 atm.
Thus N2 gas at 1 x 10-2 atm has the higher positional entropy.
16-2 Entropy and the Second Law of p755
Thermodynamics
The first law of thermodynamics: the energy of the
universe is constant. Energy is conserved in the
universe, but entropy is not.
The 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.
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Ex 16.3 The Second Law
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In a living cell, large molecules are assembled from
simple ones. Is this process consistent with the second
law of thermodynamics?
Solution:
To reconcile the operation of an order-producing cell with
the second law of thermodynamic, we remember that
Δsuniv , Δssys not must be positive for a process to be
spontaneous. A process for which Δssys is negative can be
spontaneous if the associated Δssurr is both larger and
positive. The operation of a cell is such a process.
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.
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16-3 The Effect of Temperature
on Spontaneity
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The change of state for one mol of water
from liquid to gas, H2O(l) → H2O(g).
The sign of Δssurr depends on the direction of the
heat flow. At constant temperature, an exothermic
process the random motions and thus the entropy
of the surroundings. For this case, Δssurr is
positive.
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The magnitude of depends on the temperature. The
transfer of a given quantity of energy as heat
produces a much greater percent change in the
randomness of the surroundings at a low
temperature than it does at a high temperature.
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The positive value of Δssurr
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Thus Δssurr depends directly on the quantity of
heat transferred and inversely on temperature.
The tendency for the system to lower its energy
becomes a more important driving force at lower
temperatures.
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To express Δssurr in terms of the change in
enthalpy ΔH for a process occurring at constant
pressure:
Heat flow (constant P) = change in enthalpy = ΔH
S surr
H

T
Recall that ΔH consists of two parts: a sign and a
number. The sign indicates the direction of flow,
where a plus sign means into the system
(endothermic) and a minus sign means out of the
system (exothermic). The number indicates the
quantity of energy.
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Ex16.4 Determining Δssurr
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In the metallurgy of antimony, the pure metal is recovered via
different reactions, depending on the composition of the ore.
For example, iron is used to reduce antimony in sulfide ores:
Sb2S3(s) + 3Fe(s) → 2 Sb(s) + 3FeS(s) △ H = - 125 kJ
Carbon is used as the reducing agent for oxide ores:
Sb4O6(s) + 6C(s) → 4Sb(s) + 6CO(g) △ H = 778 kJ
Calculate △Ssurr for each of these reactions at 25 ℃ and 1atm.
Solution:
For the sulfide ore reaction,
S surr
125 kJ
S surr 
0.419 kJ 419 J
K
K
298 K
For the oxide ore reaction,
778 kJ
S surr 
2.61 kJ 2.61 10 3 J
K
K
298
H

T
Table 16.3 Interplay of Ssys and Ssurr
in Determining the Sign of Suniv
H
S surr  
T
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ct
a
e
R
A liquid is vaporized at its boiling point.
Predict the signs of w, q, H, S, Ssurr
and G.
Explain your answers.
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16-4 Free Energy
The thermodynamic function is symbolized by G is
especially useful in dealing with the temperature dependence
of spontaneity, which is defined by the relationship
G = H - TS
ΔG = ΔH –T ΔS
S surr
H

T
G
H


S
T
T
G
H


S S Ssurr  S  S univ
T
T
G
Suniv 
T
at constant T and P
To predict the spontaneity of
malting of ice
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These data predict that the process is spontaneous at
10℃; that is ice melts at this temperature because
ΔSuniv is positive and ΔG0 is negative. The opposite
is true at -10℃, where water freezes spontaneously.
H2O(s) → H2O(l)
H 0  6.03 103 J
mol
S 0 22.1 J
K
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Table 16.4 Results of the calculations of ΔSuniv and ΔG0
at -10℃, 0 ℃, and 10 ℃.
Ex 16.5 Free Energy and Spontaneity
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At what temperature id the following process spontaneous
at 1 atm? Br2(l) → Br2(g) ΔH0 = 51.0 kJ/mol and
ΔS0 = 93.0 J/K‧ mol
What is the normal boiling of lquid Br2?
Ex16.5
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Spontaneous Reactions
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Effect of H and S on Spontaneity
Table 16.5 Various Possible Combinations of ΔH and ΔS
for a Process and the Resulstsing Dependence of
Spontaneity on Temperature
H
S
Result

+
spontaneous at all temps
+
+
spontaneous at high temps

 spontaneous at low temps
+
 not spontaneous at any temp
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16-5 Entropy Changes in
Chemical Reactions
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N2(g) + 3H2(g) → 2NH3(g)
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Ex 16.6 Predicting the Sign of ΔS0
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Predict the sign of ΔS0 for the following reactions.
a. The thermal decomposition of solid calcium carbonate:
CaCO3(s) → CaO(s) + CO2(g)
b. The oxidation of SO2 in air: 2SO2(g) + O2(g) → 2SO3(g)
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Third law of thermodynamics
A perfect crystal represents the lowest entropy;
that is the entropy of a perfect crystal at 0 K is
zero.
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An increase in entropy for a perfect crystal of
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hydrogen chloride as the temperature rises above
0 K.
Entropy Changes in Chemical Reactions
Ex 16.7 Calculating ΔS0
Calculates ΔS0 at 25 ℃ for the reaction
2NiS(s) + 3O2(g) → 2SO2(g) + 2 NiO(s)
Given the standard entropy values on the right-up hand:
Solution:
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Ex16.8 Calculating △S0
Solution
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16-6 Free Energy and Chemical
Reactions
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Calculate ΔG0
One common method uses the equation
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ΔG 0 = ΔH 0 –T ΔS 0
which applies to a reaction carried out at constant
temperature.
Example:
C(s) + O2(g) → CO2(g)
The values of ΔH0 and ΔS0 are known to be –393 kJ
and 3.05 J/K, respectively.
Ex16.9 Calculating △H0, △S0, and △G0
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Consider the reaction 2SO2(g) + O2(g) → 2SO2 (g)
carried out at 25℃ and1 atm. CalculateΔH0, ΔS0, andΔG0
using the following data:
Solution:
(a)
△H 0
(b) △S0
(c) △G0
Substance
△Hf0 (kJ/mol) S0 (j/K . mol)
Ex 16.11 Calculating △G0
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Substance
Solution
△
Gf0 (kJ/mol)
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Ex16.12 Free Energy and Spontaneity
A chemical engineer wants to determine the
feasibility of making ethanol (C2H5OH) by reacting
water with ethylene (C2H4) according to the reaction
C2 H 4 ( g ) H 2O(l )  C2 H 5OH (l )
Solution
16.7 The Dependence of Free
Energy on Pressure
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Free energy
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Reaction quotient
Ex16.13 Calculating △G0
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One method for synthesizing methanol (C2H3OH) involves
reacting carbon monoxide and hydrogen gases:
CO( g ) 2 H 2 ( g )  CH 3OH (l )
Calculate △G0 at 25℃ for the this reaction where carbon
monoxide gas at 5.0 atm and hydrogen gas at 3.0 atm are
converted to liquid methanol.
Solution:
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The Meaning of ΔG for a Chemical
Reaction
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16.8 Free Energy and Equilibrium
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Table 16.6 Qualitative Relationship Between
the Change in Standard Free Energy and the
Equilibrium Constant for a Given Reaction
Ex 16.14 Free Energy and Equilibrium
Solution:
(a)
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(b)
The Temperature Dependence of K
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Note that this a linear equation of the form y = mx + b,
where y = ln(K), m = - △H0 /R = slop, x = 1/T, and b =
△S0 /R = intercept. This means that if values of K for a
given reaction are determined at various temperature, a
plot of ln(k) versus 1/T will be linear, with slop - △H0 /R
= and intercept △S0 /R.
16-9 Free Energy and Work
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In addition to its qualitative usefulness ( telling us
whether a process is spontaneous). The change in free
energy is important quantitatively because it can tell
us how much work can be done with a given process.
In fact, the maximum possible useful work obtainable
from a process at constant temperature and pressure
is equal to the change in free energy:
Wmax = △G
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