cy 6151- unit ii chemical thermodynamics presentation pdf/unit ii...“the heat content of the...

CY 6151- UNIT IICHEMICAL 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 equationMaxwell relations Van’t Hoff isotherm and isochore .


• The term thermodynamics means flow of heat.

• In general it deals with the inter conversion of various kinds ofenergy in physical and chemical systems.

• Thermodynamics –

– Predicts the feasibility of a physical process or chemicalreaction under given condition of temperature and pressure.

– Predicts whether a chemical reaction would occurspontaneously or not under a given set of conditions

– Helps to determine the extent to which a reaction would takesplace.


Terms used in thermodynamics


Types of System• Homogeneous system

• Heterogeneous system

• Isolated system

• Open system

• Closed system


Properties of a system

• Extensive properties: These are thermodynamicproperties which depend on the quantity of matterspecified in the system e.g. mass, volume energy etc.

• Intensive properties: These are thermodynamicproperties which depend on characteristics of matter butindependent of its amount e.g. pressure, temperature,viscosity,density m.p, b.p etc.


Thermodynamic process:

• Isothermal process• Adiabatic process• Isobaric process • Isochoric process• Reversible process• Irreversible process


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.


• 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

Internal Energy: U or E


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


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)


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.


T

qS

rev

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.


Significance of entropy• Measure of disorder of the system: All spontaneous process are

accompanied by increase in entropy as well as increases in thedisorder .Increase in entropy implies increase in disorder.

• Measure of probability: An irreversible process tend to proceedfrom less probable state to more probable state. Since entropyincreases in a spontaneous process, entropy may be defined as afunction of probability of thermodynamic state.

• Entropy and unavailable energy: When heat is supplied to thesystem, some portion of heat is used to do some work. This portionof heat is available energy. The remaining portion is calledunavailable energy. Hence entropy is defined as unavailableenergy per unit temperature.


Entropy change for a reversible ( non spontaneous)process

If the system absorbs q amount of heat from the surroundings at temperature T , the

increase in entropy of the system is given by

∆𝑆𝑠𝑦𝑠𝑡𝑒𝑚 = 𝑞

𝑇

But the entropy of the surroundings decrease , because the surroundings loose the same

of heat q i.e ∆𝑆 = −𝑞

𝑇

Hence, the net change in the entropy is given by

∆𝑆𝑇𝑜𝑡𝑎𝑙 = ∆𝑆𝑆𝑦𝑠𝑡𝑒𝑚 + ∆𝑆𝑠𝑢𝑟𝑟𝑜𝑢𝑛𝑑𝑖𝑛𝑔𝑠

∆𝑆𝑇𝑜𝑡𝑎𝑙 =𝑞

𝑇 +

−𝑞

𝑇

∆𝑆𝑇𝑜𝑡𝑎𝑙 = 0

i.e in a reversible isothermal process, there is no net change entropy.


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

𝑇1𝑇2

Since T1 > 𝑇2 : 𝑇1 − 𝑇2 is positive

∆𝑆𝑇𝑜𝑡𝑎𝑙 > 0

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.


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.



CLAUSIUSINEQUALITY OR THEOREM

CLAUSIUS INEQUALITY OR THEOREMClaussius theorem is a mathematical explanation for II law of

thermodynamics.

It states that the cyclic integral of is always less than or equal to zero

Wh ere = 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 irreversibleprocess.

q


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


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


Gibbs free energy.

W useful


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

The I law of thermodynamics equation for an infinitesimal change may be written

as.

If work done .dw is only due to expansion then

Gibbs –Helmholtz equation (in terms of free energy and enthalpy)

SdTTdSVdPPdVdEdG

dwdEdq

PdVdEdq

PdVdEdqTdS


Combining equation 1 and 2

At constant Pressure dP = O and the above equation becomes

Or

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

SdTVdPdG

ST

G

P

SdTdG


d(∆G)= -∆S dT

∆G = ∆H -T∆S

ST

G

)(

GT

GTH P

)(

GT

GTH P

)(Gibbs- Helmoltz Equation


Gibbs –Helmholtz equation Applications

Calculation of enthalpy change for the cell reactionIf 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

=

- =

- =


𝑛𝐹𝐸=∆𝐻 − 𝑛𝐹𝑇 𝜕𝐸

𝜕𝑇 𝑃

Or

∆𝐻 = −𝑛𝐹𝐸 + 𝑛𝐹𝑇 𝜕𝐸

𝜕𝑇 𝑃

∆𝐻 = −𝑛𝐹 𝐸 − 𝑇 𝜕𝐸

𝜕𝑇 𝑃

Calculation of emf of the cell

-𝑛𝐹𝐸=∆𝐻 − 𝑛𝐹𝑇 𝜕𝐸

𝜕𝑇 𝑃

𝐸=∆𝐻

𝑛𝐹+ 𝑇

𝜕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


Spontaneous and Non spontaneous process• The physical or chemical changes which proceed by

themselves without the intervention of any external agents areknown as spontaneous process.

• All natural process are spontaneous.

• All spontaneous process proceed in one direction and arethermodynamically 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.


Spontaneous process(Click to see)

Free energy and spontaneity ∆G = ∆H – T∆S

The Sign of ∆G depends on magnitude of ∆H and ∆SFor a spontaneus process

Thus spontaneous process involve a decrease in free energy.

VeG

dG

TSPVEd

OTdSPdVdE

PdVdETdS

PdVdETdS

PdVdEdq

dqorTdST

dqdS

0

0)(

0)(or

law I


• 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


Conditions for spontaneityΔH 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


MAXWELL RELATIONSHIP


=


𝐝𝐄 = 𝐪 − 𝐏𝐝𝐕

𝛛𝐓

𝛛𝐕 𝐒

= − 𝛛𝐏

𝛛𝐒 𝐕

𝐇 = 𝐄 + 𝐏𝐕

𝛛𝐓

𝛛𝐏 𝐒

= 𝛛𝐕

𝛛𝐒 𝐏

𝐆 = 𝐇 − 𝐓𝐒

𝛛𝐒

𝛛𝐏 𝐓

= − 𝛛𝐕

𝛛𝐓 𝐏

𝐀 = 𝐄 − 𝐓𝐒

𝛛𝐒

𝛛𝐕 𝐓

= 𝛛𝐏

𝛛𝐓 𝐕


Clausius-Clapeyron Equation

Benoit Paul Emile Clapeyron1799-1864French Engineer / Physicist

Expanded on Carnot’s work

Rudolf Clausius1822-1888German Mathematician / Physicist

“Discovered” the Second LawIntroduced the concept of entropy

CLAUSIUS –CLAYPERON EQUATIONConsider 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


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


Let us consider the following equilibrium

Solid ↔ vapour Vv ›› Vs

Liquid↔ vapour Vv ›› Vl

PV = RT ; V= RT /P


Integrating between the limits. P1 and P2 coressponding to T1 to

This is Claussius –Clapeyron equation.


Van’t Hoff isotherm

It gives the quantitative relation ship between the free energy ∆ G

and equilibrium constant K

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.

Consider a general reaction aA + bB ⇄ cC +dD


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


∆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


Van’t Hoff isochoreThe 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


PROBLEMS



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

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


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


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


Zeroth LawIf 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)


A B

C

cy 6151- unit ii chemical thermodynamics presentation pdf/unit ii...“the heat content of the...

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