Transcript Slide 1

The Second Law Of Thermodynamics
All natural reactions obey the first law of thermodynamics –
energy cannot be destroyed..
Nevertheless , not all process (which obeys first law)
occur spontaneously in nature.
Example:
1)Heat does not flow from low temperature to high temperature.
2)Water will not change to ice by itself.
The first law of thermodynamics provides no universally
applicable criterion as to whether a particular reaction will
occur or not occur.
The second law is an empirical law – there is no prove but then there is no contradiction being reported.
There several ways of defining the second law of
thermodynamics.
A more precise statement on second law was given by
Lord Kelvin (1851) and independently by Rudolph
Claussius (1854)
According to Claussius’s Statement
It is impossible to construct a machine that, operating in a
cycle , will take heat from a reservoir at constant temperature
and convert it into work without accompanying changes in the
reservoir or its surroundings.
Claussius proved that the second law lead to the
existence of a state function known as entropy (S)
For a reversible process
q rev ------------------(1)
dS 
T
Entropy is a measure of degree of chaos.
Consider two processes ( reversible and irreversible
process) that occur along the same initial and final state.
From first law :
Reversible : ΔU = qrev + wrev
Irreversible : ΔU = qirr + wirr
Since ΔU for both reactions are the same , so
-wrev – (-wirr) = qrev - qirr
Since–wrev > -wirr
Therefore qrev >qirr
For an infinitesimal process, the equation can be written as
δqrev > δqirr
-----------------------(1)
For a reversible process
dS 
q irr
T
------------------(2)
By combining equation (1) and (2)
dS 
q
T
The > sign is for an irreversible process and the = sign is for
a reversible process.
For an isolated system/adiabatic (δq = 0 ), the second
law can be written as
dS > 0
Irreversible
dS = 0
Reversible
dS < 0
Impossible to take place
For a finite process at constant temperature
Reversible :
TΔS = q
Irreversible :
TΔS > q
Isothermal Expansion Of Ideal Gas
A) Reversible Isothermal Expansion
a) Work done by gas (-wrev)
nRT
 w   PdV  
dV
V
dV
V
 nRT 
 nRT lnV Vfi
V
= (0.5)(8.314)(298.15) ln 2
- wrev = 859 J
b) The change of the internal energy(ΔU)
Since U for ideal gas only depends on T and Tf = Ti
, hence Uf = Ui or ΔU=0
c) The heat provided to the system ,q
q + w = ΔU = 0, hence q = -w = 859 J
d) Heat provided to the surrounding
qsurrounding = qsystem = -859
e) The entropy change of the system
qrev = ∫ T dS
(T=constant)
859 J = TΔSsys= 298.15 ΔSsys
ΔSsys = 2.88 J.K-1
f) The entropy change for surrounding
ΔSsurr = qsurr/ Tsurr = -859/298.15 = -2.88 JK-1
g) The change for universe
= ΔSsis + ΔSsurr
=0
B) Irreversible Isothermal Expansion
a) Work done by the system.
- w = ∫Pext dV = 0
[Pext =0] FREE EXPANSION!
b) ΔU system
ΔU = 0
[becauseTf = Ti ]
c) Heat provided to the system , q = 0
d) Heat provided to the surrounding , q = 0
e) The entropy change of a system
ΔSsisirr = ΔSsisrev
because S is a state function
ΔSsisirr = 2.88 JK-1
f) ΔSsurr = 0
because qsurr = 0
g) ΔSuniv = ΔSsis + ΔSsurr = 2.88 JK-1
>0
Variation of entropy change with temperature
To calculate the entropy change involving the change of
temperature from T1 to T2 .
T2
T2
T1
T1
 dS 

q rev
T
ΔS  ST 2  ST1 
CP 
T2

T1
q rev
T
dq rev
dT
T2
CP
ST 2  ST1  
T1 T
ST 2
T2
CP
 ST1  
T1 T
----------------------------(3)
Usually ST1 is the entropy at 298K
For a chemical reaction involving reactants and
product entropy and Cp need to be calculated,
If there is a phase transformation, changes in the
entropy change must be considered
ΔST 2
T2
ΔCP1
ΔCP2
ΔH
 ΔST1  
dT  
dt 
T x --------------------(4)
T1 T
Tx T
Tx
Free Energy
Consider a process where the heat been
absorbed by the system and the work is done on
the system .
According to first law ,
dU = δq + δw
For reversible process,
dU = TdS + δw
Δw is the maximum work
Maximum work, δwmax = [dU – TdS] ------------ (1)
= mechanical work + non mechanical
work
** in thermodynamics as well as materials it is the
non-mechanical work that is important.
The term (dU – TdS) are called a work function or
Helmhotz free energy which are defined as
A = U - TS
A is a state function.
At constant temperature , T
dA = dU - TdS
Compare with equation (1)
δw = dA
The function A is the total work done by the
system.
If we only consider non mechanical work δw’
δw’ = δw - PdV
δw’ = - [dU – TdS] - PdV
δw’ = - [dU + PdV – TdS]
δw’ = [dH – TdS]
The term [ dH – TdS] is known as Gibbs free
energy or the energy for non mechanical work.
Gibbs free energy are defined as
G = H - TS
dG = dH - TdS At constant temperature , T and pressure , P
By using free energy the following criteria
have been established
ΔG = 0
the system is in equilibrium
ΔG = - ve
the reaction tends to proceed spontaneously.
ΔG = +ve
the reaction will occur spontaneously
in the opposite direction.
Calculation of free energy
The Gibbs free energy change for a reaction
ΔG = ∑nGproduct - ∑nGreactant
For reaction at temperature T , the Gibbs free
energy change is given as
ΔGT =ΔHT - TΔST
T


ΔGT  ΔHT   ΔCP dT  T
298


T


ΔCP
ΔST   T dT 
298


This equation allowed us to calculate ΔG for
all reaction at any temperature T from ΔH298 ,
ΔS298 and ΔCP
Gibbs free energy and thermodynamics
functions
G =H - TS
G =(U + PV) - TS
dG =dU + PdV + VdP – TdS - SdT
Since dU = δq – PdV and dS = δq/T for reversible
process
Since ,
dU = TdS - PdV
Hence, dG= VdP - SdT
At constant pressure,
 G 

  S
 T  P
At constant temperature,
 G 

 V
 P T
Gibbs – Helmhotz Equation
ΔG =ΔH - TΔS
 δΔG 
ΔG  ΔH T 
 ------------Gibbs – Helmhotz Equation
 δT 
The Gibbs – Helmhotz Equation could also be
written in the following form
  ΔG  
δ  T  
ΔH



  2
T
 δT 


P
or
  ΔG  
δ  T  
   ΔH
 
 δ1  
  T  
P
Gibbs – Helmhotz equation permits the calculation of
ΔG for reaction at temperature from its value at another
temperature.
The relationship betweenΔG and K
Gibbs free energy are defined as
G = H - TS
= U + PV - TS
dG = dU + PdV +VdP – TdS -SdT
dG = TdS – PdV + PdV + VdP –TdS -SdT
dG = VdP - SdT
dG = VdP
= RT/P dP
G2
P2
RT
 dG   P dP
G1
P1
 P2
G2 G1  RTln
 P1

 G1=Go1 and P1=1 atm

G2 = Go1 + RTlnP2
Usually activity,a is considered
G2 = Go1 + RTlna2
aA + bB ═ cC + dD
Consider the above reactions,
ΔG = Gproduct – Greactant
= (cGC + dGD ) – (aGA + bGB )
For 1 mole forward reaction
c d
a
o
C aD
ΔG  ΔG  RTln a b
a A aB
At equilibrium ,
ΔG = 0
aCc .aDd
K  a b
a A .aB
K=thermodynamic
equilibrium
constant
Hence,
ΔGo = -RTlnK
vant’Hoff isotherm
This equation is only applicable at equilibrium
condition.
Dissociation Pressure or Equilibrium Partial
Pressure
2M (p) + O2 (g) ═ 2MO (p)
2
aMO
K  2
aM Po 2
1
K 
Po 2
aMO= aM = 1
PO2 = partial
pressure
1
ΔG  RTlnK  RTln
 RTlnPO2
PO2
o
T
Rumusan nilai K
i) The higher the value of K,the more stable the oxide
ii) PO2 (eq) which is the reciprocal of K, represent the so
called dissociation pressure or equilibrium pressure
if PO2 < PO2 (eq) oxide will decompose
PO2 > PO2 (eq) metal will oxidise
Effect of temperature on K values
From van’t Hoff isotherm
ΔGo = -RTlnK
Differentiate this equation,
δ ΔG 
d(lnK)
 RlnK  RT
δT
δT
multiply both sides by T
Tδ ΔG 
2 d(lnK)
 RTlnK  RT
δT
δT
d(lnK)
ΔG  ΔH  ΔG  RT
dT
o
o
d lnK  ΔH o

2
dT
RT
o
2
………………………………
ΔH o
lnK  
 const
RT
Van’t Hoff Equation
Summary
i) Plot ln KP versus 1/T
ii) ΔHo +ve (endothermic KP
gradient= -ΔHo/R
with increasing T
iii) ΔHo -ve (exothermic) KP decrease as T increase
K2
ΔH o T2 1
 d(lnK)  R  T 2 dT
K1
T1
K2
ln
 K1

ΔH  1 1 
  
  
R T2 T1 

o
This equation enable
calculation of K at another
temperature if K at certain
temperature is known.
Vapor Pressure
Vapor pressure of a condensed phases at T is the
pressure of the vapour in equilibrium with the
liquid/solid when the two form a system
Vapour A
Vapour
Pressure
PA
Temp, T
Liquid/Solid
Consider two phases A and B in equilibrium
A═B
Since the system is in equilibrium ,
ΔG = GB – GA = 0
GB = GA
We know that for G to be a function of T and P the total
differential.
dG = VdP – S dT
For A,
dGA = VAdP - SAdT
For B,
dGB = VBdP - SBdT
To maintain the equilibrium ,
dGB = dGA
VBdP – SBdT = VAdP – SAdT
S B  S A ΔS
 dP 


 
 dT e q V B V A ΔV
From ΔG =ΔH – TΔS
at equilibrium
ΔH = TΔS
ΔS = ΔH/T
Hence,
ΔH
 dP 

 
 dT e q TΔ V
------------------ Clapeyron Equation
This equation give the relationship between T and P
that is required to maintain the equilibrium between
two phase.
If Claperyon equation is applied to vapor condensed
phase system , ΔV is the molar volume accompanying
the vaporization or sublimation process. Similarly with
ΔH.
ΔV = V vapor – V condensed phase
ΔV = V vapor – V condensed phase
But ,
V vapor >>> V condensed phase
ΔV = V vapor
Hence
ΔH
 dP 

 
 dT e q TVwap
Vwap = RT/P
estimation according to ideal gas
ΔH
ΔHP
 dP 


 
2
RT
dT
RT



e q T


 P 
Rearrange the equation ,
1
P
ΔH
 dP 

 
2
 dT e q RT
d(lnP) ΔH

2
dT
RT
For most constituents in the small temperature range ,
latent heat can be assumed to be constant.
ΔH dT
ΔH  1 
d(lnP) 

d 
2
R T
R T 
 P2
ln
 P1

ΔH
  
R

1
1
  
T2 T1 
ΔH 1
logP  
 const
2.303R T
3 equations above are known as Clausius – Claperyon Equation.
A few notes about equilibrium of vapour
condensed phase.
In equilibrium:
i) The same number of molecules will condensed and
vapourised at a given time.
ii) The liquid/solid atoms have the same free energy as
the atoms in the vapour phase.
iii) If PA is the vapor pressure of the liquid at equilibrium ,
then the liquid will vapourise when P < PA .
iv) The boiling point of the liquid is the temperature
where the vapour pressure is 1 atm.
V ) Vapor in equilibrium does not behave like an
ideal gas , hence vapour pressure is obtained
from the Clausius-Claperyon equation.
vi) For most metals, the vapour pressure is stated
in the formed ,
log10 P= A – B/T
in standard table.
vii) As estimation for the heat of sublimation if the
data are not available , can be calculated from
ΔHsub = ΔHf + ΔHV
ΔHf = heat of fusion; ΔHV =heat of vapourisation
 Pressure is the most important characteristic
especially in metallurgy.
i) Metal with low melting point will have the highest
pressure compared to metal with high melting
point at given temperature .
ii) Duhring Law
“ The ratio of absolute temperature for two material
at the same pressure is constant” .
Example: Tb ,Zn = 1180K and Tb ,Hg =634K
If PZn =0.1atm at 990K , hence PHg = 0.1 at ?
634/1180 = T/990 T=634/1180 x 990 = 532K (524K)
iii) The vapor pressure will be reduced when a pure
metal become a solution as in alloy.
iv) For ideal solution containing 2 components , the vapor
pressure above the alloy are from component 1 (P1 ) and
component 2 (P2 ).
v) Raolt Law : Partial pressure proportional to mole
fraction
vi) If Po1 and Po2 are the vapor pressure in pure
state hence ,
P1 = Po1N1
N1=mole fraction
component 1
P2 = Po2N2
N2=mole fraction
component 2
END OF MODULE 3