spontaneity of chemical and physical processes: the second and third laws of thermodynamics 1
TRANSCRIPT
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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01
15
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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
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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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