spontaneous processes and thermodynamic equilibrium · 2017. 1. 7. · 80℃.the sucrose dissolves...

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!


Thermodynamic Equilibrium (I-1)


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


[ 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


Thermodynamic Equilibrium (I-2)


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.


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]


Thermodynamic Equilibrium (II-1)


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 .


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]


Thermodynamic Equilibrium (II-2)


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]


[ F I G U R E 13.10 ]

spontaneous processes and thermodynamic equilibrium · 2017. 1. 7. · 80℃.the sucrose dissolves...

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