thermodynamics ii entropy

8/18/2019 Thermodynamics II Entropy

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•Tells whether something can orcannot not happen (spontaneous)

•Provides equilibrium concentrationsor maximum yield

•Says nothing about rate(speed)— kinetics gives rate

THERMODYNAMICS

Zumdahl


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EXAMPLES OF SPONTANEOUS PROCESSES

•A ball rolls down a hill but never spontaneously rolls back up the hill.

•If exposed to air and moisture, steel rusts spontaneously. However,the iron oxide in rust does not spontaneously change back to ironmetal and oxygen gas.

•A gas fills its container uniformly. It never spontaneously collects atone end of the container.

•Heat flow always occurs from a hot object to a cooler one. The reverseprocess never occurs spontaneously.

•Wood burns spontaneously in an exothermic reaction to form carbondioxide and water, but wood is not formed when carbon dioxide andwater are heated together.

•Diamond is converted to graphite at 1 atm.

Spontaneous processesare irreversible—withoutthe input of energy.


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TWO PARTS TO SPONTANEITY(1) ∆ H system < 0

Mcquarrie


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TWO PARTS TO SPONTANEITY (2) ∆ S system > 0 (S = ENTROPY)

∆ H ≈ 0

∆ S > 0

What is entropy?

Entropy is a measure ofthe amount of disorder orrandomness in a system

FE6

Zumdahl


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SPONTANEOUS ENDOTHERMIC ( ∆ H > 0) PROCESS

∆ S > 0

Whitten


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INTRODUCTON TO ENTROPY (S)

There are two types of entropy or disorder

Thermal:

Positional:

Low T High T

(In each example, the greater disorder and entropy are on the right)


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GENERAL TRENDS IN ENTROPY

0. Entropy always increases when heat is added to a system.

1. Entropy increases in going from the solid to the liquid to thegaseous state.

(Not to scale; volume gas about

103

bigger than volume liquid at 1atm)

Reger


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2. Entropy generally increases when a condensedphase (solid or liquid) dissolves in a solvent,

Reger


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3. Entropy decreases when a gas dissolves in a solvent, mainly because thedissolved molecules are confined to a smaller volume.

4. Entropy increases as the temperature increases. The kinetic energy of theparticles in a sample increases as the temperature increases. The disorderincreases as the motion of the particles increases.

5. Frequently, an increase in the moles of gas in a reaction indicate an

increase in Entropy.

6 moles 3 moles

∆ S < 0


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6. Entropy is greater for more complicated and more massivemolecules, since there are more ways to take up energy (morethermal disorder).

M.W.4

204084

131

M.W.3871

160254

Mcquarrie


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201 J/(mol K) 230 J/(mol K)220 J/(mol K)

EXAMPLE: Arrange C 2H2, C 2H4, and C 2H6 in order of increasingentropy.

Mcquarrie


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H2O(g) (0.02 atm, 25ºC) H2O(g) (0.002 atm, 25ºC)

Mcquarrie


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THE MOLAR ENTROPY OF OXYGEN AT 1 ATM

Increasing thermal disorder Increasing positional disorder

The entropy of a perfect crystal at 0K is 0 (Third law of thermodynamics).

Mcquarrie


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ENTROPY INCREASES IN PHASE TRANSITIONFROM SOLID TO LIQUID TO GAS

When solid melts, molecules

(Not to scale; volume gas about10 3 bigger than volume liquid at 1atm)

free to move throughout liquidincreased positional

disorder and hence entropy.

Mcquarrie


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REVERSIBLE AND IRREVERSIBLE PROCESSES

+

Ice and water at equilibrium at 0ºC.

Add δ q

Subtract δ q

Irreversible process (system not at equilibrium)

Add δ q

Subtract δ q

Reversible process (system at equilibrium)


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REVERSIBLE AND NONREVERSIBLE PROCESSES

Gas at1 atm

Weight provides 2 atm pressure

Irreversible process (system not at equilibrium)

Add δ m

Subtract δ m

Gas at1 atm

Weight provides 1 atm pressure

Add δ m

Subtractδ

m

(e.g., grains of sand)

Reversible process (system at equilibrium)


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ENTROPY DEFINED

2

1

revdqS T

∆ ≡

∫ at constant Trev

qS

T ∆ ≡

EXAMPLE 1 : Calculate the increase of entropy when 1 mole of ice melts at0ºC and 1 atm.

ENTROPY DEFINED


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ENTROPY DEFINED2

1

revdqS T

∆ ≡ ∫ at constant Trevq

S T

∆ ≡

EXAMPLE 1 : Calculate the increase of entropy when 1 mole of ice melts at0ºC and 1 atm. ∆H fus = 6.00 kJ/mol .

1

6,000 11 22.0

273 fusrev

fus

J mol x H q molS J K

T T K −∆∆ ≡ = = =

ENTROPY DEFINED


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ENTROPY DEFINED2

1

revdqS T


S T

∆ ≡


1

6,000 11 22.0

273 fusrev

fus

J mol x H q molS J K

T T K −∆∆ ≡ = = =

EXAMPLE 2 : Calculate the increase of entropy when 3 moles of watervaporizes at 100 0ºC and 1 atm. ∆H vap = 44 kJ/mol .

ENTROPY DEFINED


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ENTROPY DEFINED2

1

revdqS T


S T

∆ ≡


1

6,000 1 J mol1 22.0 fusrev

fus 273 x H q molS J K

T T K −∆∆ ≡ = = =

EXAMPLE 2 : Calculate the increase of entropy when 3 moles of watervaporizes at 100 0ºC and 1 atm. ∆H vap = 44 kJ/mol .

1

44,000 3 J mol x H 1

350373vaprev

b

q molS J K T T K

−∆

∆ ≡ = = =

Notice that the units of entropy are the

same as the units of heat capacity.

TEMPERATURE DEPENDENCE OF ENTROPY


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U C O O

2

1

revdqS T

∆ ≡ ∫ q rev = C P dT at constant P

q rev = C V dT at constant V

2

1

T

P

T

C dT S

T ∆ = ∫ 2

1

ln for sufficiently smallPT

C T T

= ∆

2

1

T V

T

C dT

S T ∆ =

∫2

1ln for sufficiently smallV T

C T T = ∆

Here it is even more apparent that entropyand heat capacity have the same units

EXAMPLE 3 : Calculate the change in entropy when a 50.0 g Cu block isheated from 25ºC to 82ºC at 1 atm . (The specific heat is 0.38 J/ (g K ).)

12

1

0.38 355ln 50 ln 3.3298P

T J S C g J K T g K

−⎛ ⎞∆ = = ⋅ = ⋅⎜ ⎟⋅⎝ ⎠

ABSOLUTE ENTROPIES AND THE THIRD LAW


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2

00

T P

T

C dT S S S

T ∆ = − =∫

2T P

T

C dT S

Entropy at 0 K = S o = 0 for perfect crystal

T = ∫ Tables: ∆ G, ∆ H , but S

0

Cu metal

Suppose want standard (298K and 1 atm) entropy for substance that meltsat T

f < 298:

McQuarrie


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EXAMPLE 4 : Calculate the change in entropy when a 35 g water at 40ºC iscooled to ice at -15ºC at 1 atm . (The specific heat of ice and water are 2.06and 4.18 J/ (g K ), respectively.)

f 2

1

Hln ln

P

f water iceP

f f

T T S C C

T T T

∆∆ = − +

( ) ( )273 11,700 258146.3 / ln 72.1 / ln313 273K 273

J S J K J K ∆ = − +

( )4.1835 146 /water water P S J C mC g J K g K = = =⋅

( )2.0635 72.1 /ice iceP S J

C mC g J K

g K

= = =⋅

20.0 42.7 4.07 67 /S J K ∆ = − − − = −

35 600011,700

1 18.0 mol f g mol J

H J g

⎛ ⎞∆ = ⋅ =⎜ ⎟⎝ ⎠

For cooling T 1 > T f > T 2

thermodynamics ii entropy

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