Download unit ii chemical thermodynamics

CY 6151- UNIT II
CHEMICAL THERMODYNAMICS
APPLIED CHEMISTRY/SVCE
 Terminology of thermodynamics
 Second law
 Entropy - entropy change for an ideal gas, reversible and
irreversible processes ,entropy of phase transitions.
 Clausius inequality.
 Free energy and work function:
 Helmholtz and Gibbs free energy functions
 Criteria of spontaneity
 Gibbs-Helmholtz equation
 Clausius-Clapeyron equation
 Maxwell relations
 Van’t Hoff isotherm and isochore .
APPLIED CHEMISTRY/SVCE
• The term thermodynamics means flow of heat.
• In general it deals with the inter conversion of various kinds of
energy in physical and chemical systems.
• Thermodynamics –
– Predicts the feasibility of a physical process or chemical
reaction under given condition of temperature and pressure.
– Predicts whether a chemical reaction would occur
spontaneously or not under a given set of conditions
– Helps to determine the extent to which a reaction would takes
place.
APPLIED CHEMISTRY/SVCE
Terms used in thermodynamics
APPLIED CHEMISTRY/SVCE
Types of System
•
•
•
•
•
Homogeneous system
Heterogeneous system
Isolated system
Open system
Closed system
APPLIED CHEMISTRY/SVCE
Properties of a system
• Extensive properties: These are thermodynamic
properties which depend on the quantity of matter
specified in the system e.g. mass, volume energy etc.
• Intensive properties: These are thermodynamic
properties which depend on characteristics of matter but
independent of its amount e.g. pressure, temperature,
viscosity,density m.p, b.p etc.
APPLIED CHEMISTRY/SVCE
Thermodynamic process:
•
•
•
•
•
•
Isothermal process
Adiabatic process
Isobaric process
Isochoric process
Reversible process
Irreversible process
APPLIED CHEMISTRY/SVCE
Reversible Process
• Driving force and opposing force
differ by small amount.
• It is a slow process
• The work obtained is more
• It is am imaginary process
• It consists of many steps
• It occurs in both the directions
• It can be reversed by changing
thermodynamic variables
Irreversible process
• Driving force and opposing force
differ by a large amount
• It is a rapid process
• The work obtained is less
• It is a real process
• It has only two steps i.e initial and
final
• It occurs in only one direction
• It can’t be reversed.
APPLIED CHEMISTRY/SVCE
Internal Energy: U or E
• The energy stored in a substance by virtue of its constituent atoms and
molecules is called Internal energy.
• It is the sum of vibrational energy, rotational energy, electronic energy
etc.
• Internal Energy change ((∆E))
• It is the difference in the internal energies of initial and final states of
the system.
∆ E = E final – E initial
But for the chemical reactions
APPLIED CHEMISTRY/SVCE
Enthalpy or Heat content of a system (H)
“The heat content of the system” or “ sum of internal energy and
pressure volume change work done”
H = E + PV
Unit KJ mol-1
Enthalpy change – It is the difference in the enthalpy of initial and
final stages of the system.
For the chemical reaction ΔH = H final –H initial
ΔH = ΔE + P ΔV
At constant Volume ΔH = ΔE
APPLIED CHEMISTRY/SVCE
First law of thermodynamics:
The law of conservation of energy.
Energy can be neither created nor destroyed, but it can be converted
from one form to another.
The mathematical form of First law of thermodynamics is
ΔE = q – w
where ΔE, q and w represent respectively the change in internal
energy, quantity of heat supplied and work done. For a small
change,
dE = dq – dw ----- (1)
Work done (dw) can be represented as PdV in terms of pressure
volume changes for the gas. i.e dw = PdV ----- (2)
APPLIED CHEMISTRY/SVCE
APPLIED CHEMISTRY/SVCE
Second law of thermodynamics :
• Clausius statement : It states that heat cannot flow itself from a cold body to
a hot body spontaneously without the intervention of an external energy.
• Kelvin statement: It is impossible to take heat from a hot body and convert it
completely into work by a cyclic process without transferring a part of heat to
cold body.
• II law in terms of entropy: A spontaneous process is always accompanied
by an increase in entropy of the universe.
• Other statements:
• All spontaneous process are irreversible
• It is not possible to construct a machine functioning in a cycle which can
convert heat completely into equivalent amount of work without produces
changes elsewhere.
APPLIED CHEMISTRY/SVCE
ENTROPY
• Clausius introduced a new thermodynamic function called entropy.
It is a measure of degree of disorder or randomness in a molecular
system. It is also considered as a measure of unavailable form of
energy.
• The change in entropy of equal to the ratio of heat change to the
temperature (T) of the reversible cyclic process.
qrev
S 
T
APPLIED CHEMISTRY/SVCE
Significance of entropy
• Measure of disorder of the system: All spontaneous process are
accompanied by increase in entropy as well as increases in the
disorder .Increase in entropy implies increase in disorder.
• Measure of probability: An irreversible process tend to proceed
from less probable state to more probable state. Since entropy
increases in a spontaneous process, entropy may be defined as a
function of probability of thermodynamic state.
• Entropy and unavailable energy: When heat is supplied to the
system, some portion of heat is used to do some work. This portion
of heat is available energy. The remaining portion is called
unavailable energy. Hence entropy is defined as unavailable
energy per unit temperature.
APPLIED CHEMISTRY/SVCE
Entropy change for a reversible
( non spontaneous)process
the system absorbs q amount of heat from the surroundings at temperature T , the
increase in entropy of the system is given by
If
𝑞
∆𝑆𝑠𝑦𝑠𝑡𝑒𝑚 = 𝑇
But the entropy of the surroundings decrease , because the surroundings loose the same
−𝑞
of heat q i.e ∆𝑆 = 𝑇
Hence, the net change in the entropy is given by
∆𝑆𝑇𝑜𝑡𝑎𝑙 = ∆𝑆𝑆𝑦𝑠𝑡𝑒𝑚 + ∆𝑆𝑠𝑢𝑟𝑟𝑜𝑢𝑛𝑑𝑖𝑛𝑔𝑠
𝑞
∆𝑆𝑇𝑜𝑡𝑎𝑙 =𝑇 +
−𝑞
𝑇
∆𝑆𝑇𝑜𝑡𝑎𝑙 = 0
i.e in a reversible isothermal process, there is no net change entropy.
APPLIED CHEMISTRY/SVCE
Entropy change for a irreversible ( spontaneous) process:
Consider a system maintained at higher temperature T1 and its surrounding
maintained at a lower temperature T2.If q amount of heat passes irreversibly from the
system to surroundings. then,
Decrease in entropy of the system , ∆𝑆𝑠𝑦𝑠𝑡𝑒𝑚 =
−𝑞
𝑇1
Increase in entropy of the surroundings. ∆𝑆𝑠𝑢𝑟𝑟𝑜𝑢𝑛𝑑𝑖𝑛𝑔𝑠 =
+𝑞
𝑇2
Net change in entropy is given by
∆𝑆𝑇𝑜𝑡𝑎𝑙 = ∆𝑆𝑠𝑦𝑠𝑡𝑒𝑚 + ∆𝑆𝑠𝑢𝑟𝑟𝑜𝑢𝑛𝑑𝑖𝑛𝑔𝑠
∆𝑆𝑇𝑜𝑡𝑎𝑙 =
−𝑞
𝑇1
+
𝑞
𝑇2
=q
𝑇1 − 𝑇2
𝑇1 𝑇2
Since T1 > 𝑇2 : 𝑇1 − 𝑇2 is positive
∆𝑆𝑇𝑜𝑡𝑎𝑙 > 0
APPLIED CHEMISTRY/SVCE
Entropy change accompanying
change of phase
Solid to liquid:
Let us consider the process of melting of 1 mole of substance being carried out
reversibly .It would absorb molar heat of fusion at temperature equal to its
melting point.
Where
∆ Hf -molar heat of fusion
Tf - fusion temperature.
APPLIED CHEMISTRY/SVCE
Liquid to vapour
One mole of a substance changes from liquid to vapour state reversibly at its
boiling point Tb Under constant pressure .
∆H V Molar heat of vapourisation.
APPLIED CHEMISTRY/SVCE
CLAUSIUS
INEQUALITY OR THEOREM
APPLIED CHEMISTRY/SVCE
CLAUSIUS INEQUALITY OR THEOREM
Claussius theorem is a mathematical explanation for II law of
thermodynamics.
It states that the cyclic integral of
Wh ere
is always less than or equal to zero
q = Differential heat transfer at the system
boundary during a cycle.
T = Absolute temperature at the boundary.
 = Integration over the entire cycle.
The Clausius inequality is valid for all cyclic, reversible or irreversible
process.
APPLIED CHEMISTRY/SVCE
Helmholtz free energy or work function A
The work function A is defined as
𝐀 = 𝐄 − 𝐓𝐒
E- Energy content of the system
T- Absolute temperature
S- Entropy
APPLIED CHEMISTRY/SVCE
ΔA= ΔE-TΔS
By definition
Δ S = qrev / T --------(1)
According to I law
ΔE= qrev – W-------(2)
Using 1 and 2
ΔA= -Wrev
- ΔA = W max
APPLIED CHEMISTRY/SVCE
.
Gibbs free energy
W useful
APPLIED CHEMISTRY/SVCE
Gibbs –Helmholtz equation
(in terms of free energy and enthalpy)
Free energy (G) is related with enthalpy (H) as G = H-TS
Enthalpy (H) is related with internal energy (E) as H= E+ PV
∴ G = E+ PV – TS
Upon differenciation
dG  dE  PdV  VdP  TdS  SdT
The I law of thermodynamics equation for an infinitesimal change may be written
as. dq  dE  dw
If work done .dw is only due to expansion then
dq  dE  PdV
TdS  dq  dE  PdV
APPLIED CHEMISTRY/SVCE
Combining equation 1 and 2
dG  VdP  SdT
At constant Pressure dP = O and the above equation becomes
dG  SdT
Or
 G 

  S
 T  P
G1 - Initial - T1 at T+ dT G1+ dG1
G2 - Final - F1 at T+ dT G2+ dG2
dG1 = - S1dT : dG2 = - S2dT
dG2 - dG1 = - S2dT –(- S1dT)
d(∆G)= -∆S dT
APPLIED CHEMISTRY/SVCE
d(∆G)= -∆S dT
  (G ) 
 T   S
∆G = ∆H -T∆S
  (G ) 
H  T 
P  G

 T 
 (G ) 
H  T 
P  G

 T 
Gibbs- Helmoltz Equation
APPLIED CHEMISTRY/SVCE
Gibbs –Helmholtz equation
Applications
Calculation of enthalpy change for the cell reaction
If a Galvanic cell produces nF coulombs of electricity in a reversible
manner, it must be equal to the decrease in the free energy
J
Where n- no of electrons involved in the process. F- Faraday 96500
coulombs. E = EMF in V
Gibbs –Helmoltz equation
=
-
=
-
=
APPLIED CHEMISTRY/SVCE
𝑛𝐹𝐸=∆𝐻 − 𝑛𝐹𝑇
𝜕𝐸
𝜕𝑇 𝑃
Or
𝜕𝐸
∆𝐻 = −𝑛𝐹𝐸 + 𝑛𝐹𝑇
𝜕𝑇
∆𝐻 = −𝑛𝐹
𝜕𝐸
𝜕𝑇
𝐸−𝑇
𝑃
𝑃
Calculation of emf of the cell
-𝑛𝐹𝐸=∆𝐻 − 𝑛𝐹𝑇
∆𝐻
𝐸=𝑛𝐹 + 𝑇
𝜕𝐸
𝜕𝑇 𝑃
𝜕𝐸
𝜕𝑇 𝑃
Calculation of entropy change (∆𝑺)
∆𝐺 = ∆𝐻 − 𝑇∆𝑆
∆𝐺= ∆𝐻 + 𝑇
𝜕(∆𝐺)
𝜕𝑇
𝑃
Comparing the above two equations
−∆𝑆 =
𝜕 (∆𝐺)
𝜕𝑇
𝑃
-----------A
APPLIED CHEMISTRY/SVCE
Spontaneous and Non spontaneous process
• The physical or chemical changes which proceed by
themselves without the intervention of any external agents are
known as spontaneous process.
• All natural process are spontaneous.
• All spontaneous process proceed in one direction and are
thermodynamically irreversible.
• Examples:
 Heat flow from a hotter to a colder body till they attain thermal equilibrium.
 Water flows by itself from a higher level to a lower level.
 The expansion of gas into an evacuated space.
• The process which proceed in both directions are called
spontaneous or reversible process.
APPLIED CHEMISTRY/SVCE
Spontaneous process
(Click to see)
Free energy and spontaneity
∆G = ∆H – T∆S
The Sign of ∆G depends on magnitude of ∆H and ∆S
For a spontaneus process
dq
orTdS  dq
T
I law dq  dE  PdV
 TdS  dE  PdV
or TdS  ( dE  PdV )  0
dE  PdV  TdS  O
d ( E  PV  TS )  0
dG  0
G  Ve
dS 
Thus spontaneous process involve a decrease in free energy.
APPLIED CHEMISTRY/SVCE
• The free energy change (∆G)is the criterion for predicting
spontaneity or feasibility of a reaction.
• The sign of ∆G depends on sign and numerical value of ∆H
and T∆S
APPLIED CHEMISTRY/SVCE
Conditions for spontaneity
ΔH
ΔS
ΔG =
ΔH - TΔS
NATURE OF
PROCESS
-Ve
(EXOTHERMIC)
+ Ve
- Ve
spontaneous
-Ve
(EXOTHERMIC)
- Ve
- Ve (low T)
spontaneous
-Ve
(EXOTHERMIC)
- Ve
+ Ve (High T)
Non spontaneous
+Ve
(ENDOTHERMIC)
+ Ve
+ Ve (low T)
Non spontaneous
+Ve
(ENDOTHERMIC)
+ Ve
- Ve (High T)
spontaneous
+Ve
(ENDOTHERMIC)
- Ve
+ Ve
Non spontaneous
APPLIED CHEMISTRY/SVCE
MAXWELL RELATIONSHIP
APPLIED CHEMISTRY/SVCE
APPLIED CHEMISTRY/SVCE
=
APPLIED CHEMISTRY/SVCE
APPLIED CHEMISTRY/SVCE
𝐝𝐄 = 𝐪 − 𝐏𝐝𝐕
𝐇 = 𝐄 + 𝐏𝐕
𝐆 = 𝐇 − 𝐓𝐒
𝛛𝐓
𝛛𝐕
𝛛𝐓
𝛛𝐏
𝛛𝐏
=−
𝛛𝐒
𝐒
=
𝐒
𝛛𝐒
𝛛𝐏 𝐓
𝐀 = 𝐄 − 𝐓𝐒
𝛛𝐒
𝛛𝐕
=−
=
𝐓
𝛛𝐕
𝛛𝐒
𝐕
𝐏
𝛛𝐕
𝛛𝐓 𝐏
𝛛𝐏
𝛛𝐓
APPLIED CHEMISTRY/SVCE
𝐕
Clausius-Clapeyron Equation
Rudolf Clausius
1822-1888
German
Mathematician / Physicist
Benoit Paul Emile Clapeyron
1799-1864
French
Engineer / Physicist
“Discovered” the Second Law
Introduced the concept of entropy
Expanded on Carnot’s work
CLAUSIUS –CLAYPERON EQUATION
Consider a system consisting of only 1 mole of substance in two phases A
and B
The free energies of the substance in two phases A and B be GA and GB
Let the temperature and pressure of the system be T and P respectively.
The system is in equilibrium ,so there is no change in free energy
GA = GB
If the temperature of the system raised to T + dT and Presure becomes
P + dP
GA + d GA = GB + d GB
G = H – TS
G= E + PV –TS
Diffentiating the above equation.
dG = dE + PdV + Vdp-TdS –SdT
dG = VdP-SdT
APPLIED CHEMISTRY/SVCE
dGA = VAdP –SA dT--------(a)
dGB = VBdP –SB dT---------(b)
Where VAand VB are the molar volume of phases A and B respectively.
SA and SB are molar entropies.
Since GA = GB
dGA = dGB
Substituting in equation (a) and (b)
VAdP –SA dT = VBdP –SB dT
Clapeyron Equation
APPLIED CHEMISTRY/SVCE
Let us consider the following equilibrium
Solid ↔ vapour Vv ›› Vs
Liquid↔ vapour Vv ›› Vl
PV = RT ; V= RT /P
APPLIED CHEMISTRY/SVCE
Integrating between the limits. P1 and P2 coressponding to T1 to
This is Claussius –Clapeyron equation.
APPLIED CHEMISTRY/SVCE
Van’t Hoff isotherm
It gives the quantitative relation ship between the free energy ∆ G
and equilibrium constant K
Consider a general reaction aA + bB ⇄ cC +dD
We know G = H –TS = E+ PV –TS
Or dG = dE +PdV +VdP-TdS-SdT
But dq = dE + PdV and dS = dq/T
∴ dG = VdP-SdT
At Constant temperature dG = VdP
PV = RT or V = RT /P for one mole of a gas.
APPLIED CHEMISTRY/SVCE
On integration
G = G0 + RT ln P --(2)
G0 - standard free energy
Let the free energies of A,B, C and D at their respective pressure PA ,PB ,PC and
PD are GA ,GB,GC and GD respectively .
Then the free energy change for the above reaction is given by
∆G = G Products – G reactants.
∆G = (cGC + dGD ) –(aGA + bGB)
From (2)
aGA = aG0 A+ RT ln PA
bGB = bG0 B+ RT ln Pb
cGA = CG0 C+ RT ln PC
dGA = dG D+ RT ln PD
APPLIED CHEMISTRY/SVCE
∆G = (cGC + dGD ) –(aGA + bGB ) =
( CG0 C+ RT ln PC + dG D+ RT ln PD ) –( aG0 A+ RT ln PA + bG0 B+ RT ln Pb )
Where ∆ G0 = standard free energy change of the reaction. At equilibrium ∆ G =0
Van’t Hoff
isotherm
APPLIED CHEMISTRY/SVCE
Van’t Hoff isochore
The effect of variation of equilibrium constant with temperature
∆G 0 = - RT ln Keq
ln Keq = -∆G0 / RT
= -(∆H0-T∆S0) /RT
= - ∆H0/RT + ∆S0/R
Vant’t Hoff isochore
APPLIED CHEMISTRY/SVCE
PROBLEMS
APPLIED CHEMISTRY/SVCE
Calculate the change in entropy acompanying the isothermal
expansion of 4 moles of an ideal gas at 300K until its
volume has increased three times
∆S = 2.3030 nR log (V2/V1) JK -1
= 2.303 Xx 8.314 log 3
= 36.54 JK-1
APPLIED CHEMISTRY/SVCE
Calculate the entropy change when 100 g of ice converted
into water at 0oC. Latent heat of fusion of ice is 80Cal/g.
For 1 g of ice ∆S= 0.293 cal K-1 g-1
For 100 g of ice = 29.3 cal K-1 g-1
APPLIED CHEMISTRY/SVCE
Gibbs free energy of a reaction at 300K and 310 K are
-29 k Cal and 29.5 k.cal respectively. Determine its ∆H
and ∆S at 300 K
d(∆G) = - 29.5-(-29) = -0.5
= T2-T1 = 10 K
∆H = - 14kcal
= 0.05 k.cal K-1
APPLIED CHEMISTRY/SVCE
The equilibrium constant Kp for a reaction is 3.0 at 673 K
and 4.0 at 773 K. Calculate the value of ∆H0 for the
reaction. (R=8.314 JK-1)
K1= 3.0 K2= 4.0
T1 = 673
T2 =773
∆H= 12.490 kJ
APPLIED CHEMISTRY/SVCE
Zeroth Law
If two systems say A and B are in thermal equilibrium with a third
System C separately
(that is A and C are in thermal equilibrium: B and C are in thermal
equilibrium ) then they are in thermal equilibrium with each other
(that is A and B are in thermal equilibrium)
B
A
C
APPLIED CHEMISTRY/SVCE