spontaneity of chemical and physical processes: the second and third laws of thermodynamics 1

Spontaneity of Chemical and Physical Processes: The Second and Third Laws

of Thermodynamics

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Study of the energy changes that accompany chemical and physical processes.

Based on a set of laws. A tool to predict the spontaneous

directions of a chemical reaction.

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Spontaneity refers to the ability of a process to occur on its own!

Waterfalls “Though the course may change

sometimes, rivers always reach the sea” Page/Plant ‘Ten Years Gone’.

Ice melts at room temperature!

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Spontaneous Process – the process occurs without outside work being done on the system.

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Kelvin statement• Impossible to construct an engine the sole

purpose of which is to completely convert heat into work

Clausius statement• Impossible for heat to flow spontaneously

from low temperature to high temperature

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The First Law - conservation of energy changes.

U = q + w The First Law tells us nothing about the

spontaneous direction of a process.

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We will look at a new property (the entropy).

Entropy is the reason why salts like NaCl (s), KCl (s), NH4NO3(s) spontaneously dissolve in water.

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For the dissolution of KCl (s) in water

KCl (s) K+(aq) + Cl-(aq)

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Low entropy High entropy

The formation of a solution is always accompanied by an

increase in the entropy of the system!

An imaginary engine

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Hot Reservoir

Cold Reservoir

Working Fluid

Isothermal Expansion

• (P1, V1, Th) (P2, V2, Th)

Adiabatic Expansion

• (P2, V2, Th) (P3, V3, Tc)

Isothermal Compression

• (P3, V3, Tc) (P4, V4, Tc)

Adiabatic Compression• (P4, V4, Tc) (P1, V1, Th)

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Cyclic Process

U = 0

qcycle = -wcycle

qcycle = q1 + q3

wcycle = w1 + w2 + w3 + w4

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The Carnot engine represents the maximum efficiency of a thermal process.

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1q

wcycleC

The thermal efficiency of the Carnot engine is a function of Th and Tc

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h

cC T

T1

Run the Carnot engine in reverse as a heat pump.

Extract heat from the cold temperature reservoir (surroundings) and deliver it to the high temperature reservoir.

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The coefficient of performance of the Carnot heat pump• quantity of heat delivered to the high

temperature reservoir per amount of work required.

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Two definitions

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hhp

h c

T

T T

1

hpcycle

q

w

Use a Carnot cycle as a refrigerator. Extract heat from the cold temperature

reservoir (inside) and deliver it to the high temperature reservoir (outside).

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Again, two definitions

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cr

h C

T

T T

3

rcycle

q

w

The entropy of the system is defined as follows

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Tdq

dS rev

Changes in entropy are state functions

S = Sf – Si

Sf = the entropy of the final state

Si = the entropy of the initial state

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Combine the first law of thermodynamics with the definition of entropy.

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PdVdUTT

dqdS rev

1

In general, we can write S as a function of T and V

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dVVS

dTTS

dSTV

Examine the first partial derivative

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0

dTTS

dSV

Under isochoric conditions, the entropy dependence on temperature is related to CV

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TC

TS V

V

For a system undergoing an isochoric temperature change

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For a macroscopic system

dTTC

dS V

2

1

T

T

V dTTC

S

Examine the second partial derivative

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dVVS

dST

From the first law

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wqU For a reversible, isothermal

process wqU rev

For an isothermal process for an ideal gas, U = 0

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1

2lnVV

nRTwqrev

The entropy change is calculated as follows

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1

2ln2

1VV

nRT

dqS

V

V

rev

For a general gas or a liquid or solid

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dVdVTP

dVVS

dSVT

We will revisit this equality later

In general, we can also write S as a function of T and P

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dPPS

dTTS

dSTP

The entropy of the system can also be rewritten

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VdpPdVdHdU

dVTP

dUT

dS

1

From the definition of enthalpy

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dPTV

TdH

VdPPdVPdVdHT

dS

1

From the mathematical consequences of H

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VdPdPPH

dTCT

dS

dPPH

dTCdH

Tp

TP

1

Examine the first partial derivative

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dTTS

dSP

Under isobaric conditions, the entropy dependence on temperature is related to CP

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TC

TS P

P

For a system undergoing an isobaric temperature change

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For a macroscopic system

dTTC

dS P

2

1

T

T

P dTTC

S

Examine the second partial derivative

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dPPS

dST

Under isothermal conditions

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dPVPH

TdS

P

1

For an isothermal process for an ideal gas, (H/ T)p = 0

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dPTV

dS

The entropy change is calculated as follows

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1

2ln2

1PP

nRPdp

nRSP

P

For a general gas or a liquid or solid

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dVTV

dPPS

dSPT

dPVdS

At the transition (phase-change) temperature only

tr = transition type (melting, vapourization, etc.)

trS = trH / Ttr

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The second law of thermodynamics concerns itself with the entropy of the universe (univS).univS unchanged in a reversible process

univS always increases for an irreversible process

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univS = sysS + surrS

sysS = the entropy change of the system.

surrS = the entropy change of the surroundings.

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We need to obtain estimates for both the sysS and the surrS.

Look at the following chemical reaction.

C(s) + 2H2 (g) CH4(g) The entropy change for the systems is

the reaction entropy change, rS. How do we calculate surrS?

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Note that for an exothermic process, an amount of thermal energy is released to the surroundings!

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Heat

Insulation

surroundings System

A small part of the surroundings is warmed (kinetic energy increases).

The entropy increases!

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Note that for an endothermic process, thermal energy is absorbed from the surroundings!

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Heat

surroundings System

Insulation

A small part of the surroundings is cooled (kinetic energy decreases).

The entropy decreases! For a constant pressure process

qp = H

surrS surrH

surrS -sysH

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The entropy of the surroundings is calculated as follows.

surrS = -sysH / T For a chemical reaction

sysH = rH

surrS = -rH/ T

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For an adiabatic process, q = 0!! There is no exchange of thermal energy

between the system and surroundings!

surrS = -qrev / T = 0

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We can make the following generalizations for an adiabatic processunivS is unchanged for an adiabatic,

reversible processunivS always increases for an adiabatic,

irreversible process The entropy of the system can never

decrease during an adiabatic process!

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Mixing of two gases (and two liquids) is an common example of an irreversible process

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X

Valve

Before

Va

Vb

After the valve is opened!

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Valve

After

V2= Va + Vb

For gas 1

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For gas 2

PRTn

PRTn

Rn

VV

RnS

T

a

1

1

211

ln

ln

PRTn

PRTn

Rn

VV

RnS

T

b

2

2

222

ln

ln

The total change in entropy for the two gases

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J

JJTmix xxRnS ln

Spontaneous mixing process - mixS > 0

The Boltzmann probability

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lnBkSkB – the Boltzmann constant (R/NA)

- the thermodynamic probability

Allow a gas to expand from one small container to an extremely large container

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Before expansionN cells

After

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N’ cells

Calculating the entropy change

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1

2

'

ln

ln'

VV

nR

NN

kNSSS BANN

Entropy is related to the dispersal of energy (degree of randomness) of a substance.

Entropy is directly proportional to the absolute temperature.

Cooling the system decreases the disorder.

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At a very low temperature, the disorder decreases to 0 (i.e., 0 J/(K mole) value for S).

The most ordered arrangement of any substance is a perfect crystal!

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The Third Law - the entropy of any perfect crystal is 0 J /(K mole) at 0 K (absolute 0!)

Due to the Third Law, we are able to calculate absolute entropy values.

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For any system, we can write the following for the entropy change between two temperatures 0 and T1.

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1

0T

p dTT

CS

Assuming a constant pressure heating

The Debye ‘T-cubed’ law

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bTaTCV 3metals

3aTCV nonmetals

This equation is valid to 15 K

Above 15 K, the heat capacity data are usually available

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31

01

15

31

T

K

p dTT

CTaS

For a phase change between 0 – T1, we add in the appropriate entropy change.

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TH

S trtr

The entropy changes of all species in the thermodynamic tables are calculated in this manner

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tr

T

T

P

T

K

P

TH

dTTC

dTTC

TaS

tr

tr

1

31

15

31

Burning ethane! C2H6 (g) + 7/2 O2 (g) 2 CO2 (g) + 3 H2O (l)

The entropy change is calculated in a similar fashion to that of the enthalpies

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JSSJ

mJr

Units for entropy values J / (K mole) Temperature and pressure for the

tabulated values are 298.2 K and 1.00 atm.

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For any gaseous reaction (or a reaction involving gases).

g> 0, rS > 0 J/(K mole).

g < 0, rS < 0 J/(K mole).

g = 0, rS 0 J/(K mole). For reactions involving only solids and

liquids – depends on the entropy values of the substances.

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Note – entropy values are absolute! Note – the elements have NON-ZERO

entropy values!

e.g., for H2 (g)

fH = 0 kJ/mole (by def’n)

S = 130.58 J/(K mole)

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