Spontaneous Processes and
Thermodynamic Equilibrium
박준원 교수(포항공과대학교 화학과)
General Chemistry
• The nature of spontaneous processes
• Entropy and spontaneity: A molecular statistical interpretation
• Entropy and heat: Macroscopic basis of the second law of
thermodynamics
• Entropy changes in reversible processes
• Entropy changes and spontaneity
• The third law of thermodynamics
• The Gibbs free energy
이번 시간에는!
Spontaneous Processes and
Thermodynamic Equilibrium (I-1)
박준원 교수(포항공과대학교 화학과)
General Chemistry
Chemical reactions are carried out by mixing the
reactants and regulating external conditions such as temperature and pressure. Questions arise immediately:
1. Is it possible for the reaction to occur at the selected conditions?
2. If the reaction is possible, what determines the ratio of products and reactants at equilibrium?
The nature of spontaneous processes
One of the most striking features of spontaneous change is that
it follows a specific direction when starting from a particular initial
condition (Fig 13.1).
We never observe a pile of metal fragments spontaneously
assemble themselves into a speeding bullet. The first law of
thermodynamics provides no guidance in predicting or explaining
directionality (Energy is conserved both in a forward process and
in its reverse.).
1
Spontaneous process familiar in chemical laboratories also follow specific directions:
1. We measure heat transfer from a hot body to a cold one.
2. We observe a gas expanding into a region of lower pressure.
3. We place a drop of red ink into a beaker of water and the ink particles diffuse until the water is uniformly pink.
4. We place 10 g sucrose in a beaker and add 100 mL water at 80℃.The sucrose dissolves to form a uniform solution.
5. We open a container of acetone on the laboratory bench. Some of the molecules evaporate from the liquid and then diffuse through the atmosphere.
All the previous examples are physical processes that are both
spontaneous and rapid.
Likewise spontaneous chemical reactions are inherently directional.
1. H2(g) + 1
2 O2(g) →
2. Na 𝑠 +1
2 Cl2(g) →
3. Cu(𝑠) + 1
2 O2(g) →
Entropy and spontaneity: A molecular statistical interpretation
<Spontaneity and molecular motions>
Consider the free adiabatic expansion
of 1 𝐦𝐨𝐥 of an ideal gas into a vacuum
(Fig. 13.2). The gas is initially held in the
left bulb of volume 𝑉/2 , whereas the
right bulb is evacuated. After the
stopcock is opened, the gas expands to
fill the entire volume, 𝑉.
[ F I G U R E 1 3 . 2 ]
2
Oxtoby, D. W.; Gillis, H. P.; Campion, A., Principles of modern chemistry, 7th ed.; Cengage Learning: Boston, 2012; p 576.
Now examine the same free expansion from a microscopic point of view.
The 16 possible microstates of a system of 4 molecules that may occupy either side of a container. In only one of these are all four molecules on the left side (probability = 1/16).
Just how unlikely is a spontaneous compression of 1 mol gas from the combined volume back to the left side?
1
2×1
2× ⋯×
1
2=1
2
6.0×1023
Oxtoby, D. W.; Gillis, H. P.; Campion, A., Principles of modern chemistry, 7th ed.; Cengage Learning: Boston, 2012; p 576.
[ F I G U R E 1 3 . 3 ]
The probability that all the molecules will be on the left side is 1
in 101.8×1023
. Written out, such a number would more than fill all
the books in the world. Nothing in the laws of mechanics
prevents a gas from compressing spontaneously, but this event is
never seen because it is so improbable.
The directionality of spontaneous change is a consequence of the
large numbers of molecules in the macroscopic systems treated
by thermodynamics.
<Entropy and molecular motions>
In preparation for defining entropy, we need a precise way to describe the “range of possible motions” of the molecules. This is accomplished by counting the number of microscopic, mechanical states, or microstates, available to molecules of the system. This number, denoted by , counts all the possible combinations of positions and momenta available to the 𝑁 molecules in the system when the system has internal energy 𝑈 and volume 𝑉.
The internal energy consists of the total kinetic energy of the atoms, given as
𝑈 = 𝜀𝑖 =
𝑁
𝑖=1
[𝑝𝑥𝑖2 + 𝑝𝑦𝑖
2 + 𝑝𝑧𝑖2 ]
2𝑚
𝑁
𝑖=1
2𝑚𝑈 = [𝑝𝑥𝑖2 + 𝑝𝑦𝑖
2 + 𝑝𝑧𝑖2 ]
𝑁
𝑖=1
Although we do not provide the details of the calculation, the value of for a monatomic ideal gas is given by
Where g is a collection of constants.
The equation connecting entropy 𝑆 and the number of available microstates is
where 𝑘B is Boltzmann’s constant.
The constant is identified as 𝑅/𝑁A. Thus, entropy has the physical dimensions J K−1.
𝑈, 𝑉, 𝑁 = g 𝑉𝑁𝑈(3𝑁 2 )
𝑆 = 𝑘B ln [13.1]
Example 13.3
Consider the free expansion of 1 𝐦𝐨𝐥 gas from 𝐕/𝟐 to 𝑽. Use Boltzmann’s relation to estimate the change in entropy for this process.
number of states available per molecule = 𝑐𝑉
microscopic states available = = (𝑐𝑉)𝑁
∆𝑆 microscopic = 𝑁A𝑘B ln 𝑐𝑉 − 𝑁A𝑘B ln 𝑐𝑉/2
= 𝑁A𝑘B lncVcV/2
= 𝑁A𝑘B ln 2
∆𝑆 (thermodynamic) = 𝑅 ln 2
Spontaneous Processes and
Thermodynamic Equilibrium (I-2)
박준원 교수(포항공과대학교 화학과)
General Chemistry
Entropy and heat: Macroscopic basis of the second law of thermodynamics
The second law of thermodynamics, which is stated as an abstract
generalization of engineering observations on the efficiency of
heat engines, has two important consequences for chemistry. It
defines the entropy function in terms of measurable macroscopic
quantities. And it defines the criterion for predicting whether a
particular process will be spontaneous.
3
<Background of the second law of thermodynamics>
- Hea t eng ine -
The second law of thermodynamics
originated in practical concerns
over the efficiency of steam
engines at the dawn of the
Industrial Age, late in the 18th
century, and required a century for
its complete development as an
engineering tool.
Oxtoby, D. W.; Gillis, H. P.; Campion, A., Principles of modern chemistry, 7th ed.; Cengage Learning: Boston, 2012; p 580.
In essence, the engine extracts thermal energy from a hot reservoir, uses
some of this energy to accomplish useful work, then discards the
remainder into a cooler reservoir (the environment).
The efficiency (that is, the ratio of work
accomplished by the engine in a cycle to the
heat invested to drive that cycle) can be
improved by reducing the unrecoverable losses
to the environment in each cycle. Seeking to
maximize, Sadi Carnot, an officer in Napoleon’s
French Army Corps of Engineers, modeled
operation of the engine with an idealized cycle
process now known as the Carnot cycle. Oxtoby, D. W.; Gillis, H. P.; Campion, A., Principles of modern chemistry, 7th ed.; Cengage Learning: Boston, 2012; p 581.
He concluded that unrecoverable losses of energy to environment cannot
be completely eliminated, no matter how carefully the engine is designed.
Even if the engine operated as a reversible process, its efficiency cannot
be exceed a fundamental limit known as thermodynamic efficiency. Thus,
an engine with 100% efficiency cannot be constructed.
In more general terms, “there is no device that can transform heat
withdrawn from a reservoir completely into work with no other effect.”
“There is no device that can transfer heat from a colder to a warmer
reservoir without net expenditure of work.” <- the second law of
thermodynamics
<Definition of entropy>
Carnot’s analysis of efficiency for a heat engine operating reversibly
showed that in each cycle 𝑞/𝑇 at the high temperature reservoir and
𝑞/𝑇 at the low temperature reservoir summed to zero:
This result suggests that 𝑞/𝑇 is a state function in a reversible process
because the sum of its changes in a cyclic process is zero. Clausius
defined the entropy change ∆𝑆 = 𝑆f − 𝑆i of a system by equation
𝑞h𝑇h+𝑞𝑙𝑇𝑙= 0
∆𝑆 = 𝑆f − 𝑆i = dq𝑟𝑒𝑣T
f
i [13.2]
Because entropy is a state function, its change depends only on the
initial and not at all on the path. “Therefore, we are free to choose any
reversible path that connects the initial and final states for calculating ∆𝑆.”
To date, no disagreements have been found.
Entropy changes in reversible processes
<∆𝑺𝐬𝐲𝐬for Isothermal processes>
Because 𝑻 is constant,
Compression/expansion of an ideal gas
Consider an ideal gas enclosed in a piston-cylinder arrangement that is
maintained at constant temperature in a heat bath.
∆𝑆 = dq𝑟𝑒𝑣T
f
i=1
𝑇 𝑑𝑞rev =
qrevT
f
i [13.3]
𝑞rev = 𝑛𝑅𝑇 ln𝑉2𝑉1
∆𝑆 = 𝑛𝑅 ln𝑉2𝑉1constant 𝑇 [13.4]
4
Phase transitions
Another type of constant-temperature process is a phase
transition such as the melting of a solid at constant temperature.
The reversible heat absorbed by the system when 1 mol of
substance melts is 𝒒𝐫𝐞𝐯 = ∆𝑯𝐟𝐮𝐬,
so
∆𝑆fus =qrevTf=∆HfusTf
[13.5]
<∆𝑺𝐬𝐲𝐬for processes with changing temperature>
For a reversible adiabatic process, 𝒒 = 𝟎, therefore ∆𝑺 = 𝟎.
For a reversible isochoric process, the volume is held constant, the heat transferred is
∆𝑆 = 1
𝑇𝑑𝑞rev
f
i
𝑑𝑞rev = 𝑛𝑐V 𝑑𝑇
∆𝑆 = 1
𝑇
𝑇2
𝑇1
𝑑𝑞rev = 𝑛𝑐V𝑇
𝑇2
𝑇1
𝑑𝑇
∆𝑆 = 𝑛𝑐V 1
𝑇
𝑇2𝑇1𝑑𝑇 = 𝑛𝑐V ln
𝑇2
𝑇1 (𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 𝑉) [13.7]
For a reversible isobaric process,
∆𝑆 = 𝑛𝑐P
𝑇
𝑇2𝑇1𝑑𝑇 = 𝑛𝑐P ln
𝑇2
𝑇1 (𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 𝑃) [13.8]
<∆𝑺 for surroundings>
If the process occurs at constant 𝑷, then
The entropy change of the surroundings is
The heat capacity of the surroundings is assumed so large that the heat
transferred during the process does not change its temperature.
𝑞surr = −∆𝐻sys
∆𝑆surr =−∆HsysTsurr
[13.9]
Spontaneous Processes and
Thermodynamic Equilibrium (II-1)
박준원 교수(포항공과대학교 화학과)
General Chemistry
Entropy changes and spontaneity
Part Two of the second law states that a process can occur spontaneously if the total entropy change for the thermodynamic universe of the process is positive:
<Spontaneous cooling of a hot body>
Example 13.6
A well-insulated ice-water bath at 0.0 oC contains 20 g ice. Throughout the experiment, the bath is maintained at the constant pressure of 1 atm. When a piece of nickel at 100 oC is dropped into the bath, 10.0 g of the ice melts.
∆𝑆tot = ∆𝑆sys + ∆𝑆surr > 0
∆𝑆Ni = −10 J K−1, ∆𝑆bath = 12 J K
−1, Now ∆𝑆tot = +2 J K−1
5
Irreversible expansion of an ideal gas
Consider a gas confined within a piston-cylinder arrangement and held at constant temperature in a heat bath. Suppose the external pressure is abruptly reduced and held constant at the new lower value.
The definition of work done on a system:
[ F I G U R E 13.5 ]
𝑤 = − 𝑃ext 𝑑𝑉
−𝑤irrev = 𝑃ext 𝑑𝑉 < 𝑃 𝑑𝑉 = −𝑤rev
The work performed by the system,
−𝑤irrev is less than −𝑤rev .
Oxtoby, D. W.; Gillis, H. P.; Campion, A., Principles of modern chemistry, 7th ed.; Cengage Learning: Boston, 2012; p 587.
If the reversible and irreversible processes have the same initial and final states, then ∆𝑈 is the same for both.
But because
We must have
and
The heat absorbed is a maximum when the process is conducted reversibly.
∆𝑈 = 𝑤irrev + 𝑞irrev = 𝑤rev + 𝑞rev
−𝑤irrev < −𝑤rev
𝑤irrev > 𝑤rev
𝑞irrev < 𝑞rev
Dividing this expression by 𝑇 , the temperature at which the heat is transferred, gives
The left side is the entropy
so
The last two equations can be
In any spontaneous process the heat absorbed by the system from surroundings at the same temperature is always less than 𝑇∆𝑆sys. In a
reversible process, the heat absorbed is equal to 𝑇∆𝑆sys.
𝑞rev𝑇>𝑞irrev𝑇
∆𝑆sys =𝑞rev𝑇
∆𝑆sys >𝑞rev𝑇
∆𝑆sys ≥ 𝑞
𝑇 (inequality of Clausius) combined as
change ∆𝑆sys,
Now let’s apply Clausius’s inequality to processes occurring
within an isolated system.
In this system, there is no transfer of heat into or out of the
system, and 𝑞 = 0. Therefore, for spontaneous processes within
an isolated system, ∆𝑆 > 0. Thermodynamics universe of a
process is clearly an isolated system. It follows that
1. In a reversible process the total entropy of a system plus its
surroundings is unchanged.
2. In an irreversible process the total entropy of a system plus its
surroundings must increase.
3. A process for which ∆𝑆univ < 0 is not spontaneous.
The third law of thermodynamics
In thermodynamics processes, only changes in entropy, ∆𝑆 , are
measured. It is nevertheless useful to define absolute values of
entropy relative to some reference state. An important
experimental observation that simplifies the choice of reference
state is:
In any thermodynamic process involving only pure phases in
their equilibrium states, the entropy change ∆𝑆 approaches
zero as 𝑇 approaches 0 K.
(Nernst heat theorem)
6
The third law of thermodynamics
The entropy of any pure element in its equilibrium state is
defined to approach zero as 𝑇 approaches 0 K. The most general
form is the third law of thermodynamics.
The entropy of any pure substance (element or compound) in
its equilibrium state approaches zero at the absolute zero of
temperature.
6
<Standard-state entropies>
Because the entropy of any substance in equilibrium state is zero at absolute zero, its entropy at any temperature 𝑇 is given by the entropy increase as it is heated from 0 K to 𝑇. If heat is added at constant temperature, Thus, 𝑆𝑇, the absolute entropy of 1 mol of substance at temperature 𝑇, is given by We define the standard molar entropy to be the absolute molar entropy 𝑆○ at 298.15 K and 1 atm pressure. Standard molar entropies 𝑆○ are tabulated for a number of elements and compounds in Appendix D.
∆𝑆 = 𝑛 𝑐p
𝑇
𝑇2
𝑇1
𝑑𝑇
𝑆𝑇 = 𝑐p
𝑇
𝑇
0
𝑑𝑇
𝑆○ = 𝑐p
𝑇
298.15
0𝑑𝑇 + ∆𝑆 (phase changes between 0 and 298.15 K) [13.10]
Spontaneous Processes and
Thermodynamic Equilibrium (II-2)
박준원 교수(포항공과대학교 화학과)
General Chemistry
The Gibbs free energy
It would be much more convenient to have a state function that predicts
the feasibility of a process in the system without explicit calculations for
the surroundings.
For process that occur at constant temperature and pressure, which is
the most important set of conditions for chemical applications, such a
state function exists. It is called Gibbs free energy and is denoted by 𝐺.
∆𝑆tot > 0 spontaneous [13.11a]
∆𝑆tot = 0 reversible [13.11b]
∆𝑆tot < 0 nonspontaneous [13.11c]
7
<The nature of spontaneous processes at fixed 𝑻 and 𝑷>
During any process conducted at constant 𝑇 and 𝑃, the heat gained by
the system is ∆𝑯𝐬𝐲𝐬 = 𝒒𝐩 and the heat transferred to the surroundings is
− 𝒒𝐩 = −∆𝑯𝐬𝐲𝐬.
Their entropy change is then
The total entropy change is
∆𝑆surr =−∆𝐻sys
𝑇surr
∆𝑆tot = ∆𝑆sys + ∆𝑆s𝑢𝑟𝑟 = ∆𝑆sys −∆𝐻sys
𝑇surr
=−(∆Hsys − Tsurr ∆Ssys)
Tsurr
Because the temperature, 𝑻, is the same for both the system
and the surroundings, we can rewrite this as
We define the Gibbs free energy G as
Therefore, Equation 13.12 becomes
∆𝑆tot =−∆(Hsys −TSsys)
T [13.12]
𝐺 = 𝐻 − 𝑇𝑆 [13.13]
∆𝑆tot =−∆GsysT [13.14]
Because the absolute temperature 𝑻 is always positive, ∆𝑺𝐭𝐨𝐭 and
∆𝑮𝐬𝐲𝐬, must have the opposite sign for processes occurring at
constant 𝑇 and 𝑃. If follows that
∆𝐺sys < 0 spontaneous processes [13.15a]
∆𝐺sys = 0 reversible processes [13.15b]
∆𝐺sys > 0 nonspontaneous processes [13.15c]
<Gibbs free energy and chemical reactions>
The change in the Gibbs free energy provides a criterion for the
spontaneity of any process occurring a constant temperature and
pressure. Because we cannot know the absolute value of the
Gibbs free energy of a substance, it is convenient to define a
standard molar Gibbs free energy of formation, ∆𝐺f○, analogous to
the standard molar enthalpy of formation ∆𝐻f○ (Section 12.6).
From tables of ∆𝐺f○ we can calculate ∆𝐺○ for a wide range of
chemical reactions.
Example 13.10
Calculate ∆𝐺○ for the following reaction, using tabulated values
for ∆𝐺f○ from Appendix D.
Solution
3 NO(g) → N2O(g) + NO2(g)
∆𝐺○ = ∆𝐺f○ N2O + ∆𝐺f
○ NO2 − 3 ∆𝐺f○(NO)
= 1 mol 104.18 kJ mol−1 + 1mol 51.29 kJ mol−1 − 3 mol 86.55 kJ mol−1
= −104.18 kJ
Effects of temperature on ∆𝑮
Values of ∆𝐺○ can be estimated for reactions at other
temperatures and at
𝑃 = 1 atm using the equation in below
The estimates will be close to the true value
if ∆𝐻○ and ∆𝑆○ are not strongly dependent on 𝑇.
∆𝐺○ = ∆𝐻○ − 𝑇 ∆𝑆○ [13.17]
Oxtoby, D. W.; Gillis, H. P.; Campion, A., Principles of modern chemistry, 7th ed.; Cengage Learning: Boston, 2012; p 596.
[ F I G U R E 13.10 ]