Thermodynamics

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Chemical Thermodynamics and Energetics

Prof.Jadhav Swapnil. S.

Thermo = Heat dynamics = science of motion or transfer

Introduction-

Energy – is defined as its capacity to do work. A system with higher energy can do more work.

Different forms of energy

Kinetic Energy Energy due to motion.

Potential Energy Energy due to position and state of matter.

Heat Energy Thermal energy

Mechanical Energy Heat energy converted into mechanical energy in engine.

Electrical Energy Energy produced in the galvanic cell.

Chemical Energy Energy produced in the chemical reaction.

All these forms of energy are intercovertible.

Thermodynamics:-

“Thermodynamics is branch of science that deals with the different forms of energy, qualitative relation between them and energy changes occur in physical and chemical processes.”

Scope of Thermodynamics:-

1) It provides information regarding the feasibility of a chemical reaction.

2) It tells us whether a particular reaction is possible under the given set of condition. (T, V, C)

3) It tells us the direction in which the reaction tends to occurs.

4) Important laws of physical chemistry (ie. lowering of vapour pressure, phase rule, distribution law) can be derived from the law of thermodynamic.

Limitations of Thermodynamics:-

1) Thermodynamics does not gives information regarding the rates of physical or chemical process. Ie. It does not bother about time factor.

2) It is concerned only with the initial and final state of system and not the mechanism of process.

3) Thermodynamics is applicable to macroscopic system consisting matter in bulk.

4) It does not applicable to microscopic system of individual atoms or molecule.

5) It ignores the internal structure of atoms and molecules.

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Basic concepts in Thermodynamics:-

Types of system:-

1) Open system:-

2) Closed system:-

3) Isolated system:-

1) System:- The part of universe which is under thermodynamic study is called system.

2) Surrounding:- The remaining part of universe excluding system is surroundings.

Open system Closed system Isolated system

3) Boundery:- The real or imaginary surface separating the system from the surroundings.

Boundary

A system that can exchange both the matter and energy with its surroundings is called an open system.

A system that can exchange only energy but not matter with its surroundings is called a closed system.

A system that cannot exchange either energy or matter with its surroundings is called an isolated system.

Eg. Hot tea in an open cup.

Eg. Hot tea in a cup covered with saucer.

Eg. Hot tea in thermos flask.

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Properties of system:-

Note. (1) Extensive Properties are additives. (2) Intensive Properties are not additives. (3) The ratio of two Extensive Properties is an intensive property.

State and state function:-

When these variables are changes, then system is said to be in different state.

i.e

State function:-

Extensive Property:-

The property which depends upon the quantity of matter is known as Extensive Property.

Eg. Mass, Volume, Work, Entropy, Enthalpy,

Surface area, Internal energy etc.

Intensive Property:-

The property which depends upon the quantity of matter is known as intensive Property.

Eg. Temperature, Pressure, Concentration, Specific

gravity etc.

P1 , V1 , T1 P2 , V2 , T2 Change in state

State- I State- II

A known set of measurable properties like temperature, pressure, volume define a state of system.

Any property of the system whose value depends only on the current state of system and does not depend on the path fallowed by system is called state function.

∆P

∆V

A (P1 ,V1 )

B(P2 ,V2 )

Consider a system at point ‘A’ P = P1 , V = V1

After condition pressure and volume are changed, at point ‘B’ P = P2 , V = V2

The change in pressure ∆P = P2 — P1 and The change in volume ∆V = V2 — V1 does not depend upon the path fallowed by system from point ‘A’ to ‘B’.

Hence ∆P and ∆V are state function.

Thermodynamic Equilibrium:- A system is said to be in thermodynamic equilibrium when the state functions of the system do not change with time.

1) Thermal Equilibrium 2) Chemical Equilibrium 3) Mechanical Equilibrium

Temperature constant Chemical composition constant Mechanical properties (or Pressure) constant

P

V

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Types of processes:-

Process: - The transition from one state to another state is called process.

Isothermal process Adiabatic process A process in which temperature of the system remain constant throughout the process is called an isothermal process.

A process in which no heat can exchange between system and surroundings is called an adiabatic process.

Temperature of the system remains constant. Ie ∆T = 0

Temperature of the system remains changes. Ie ∆T ≠ 0

Exchange of heat between system and surroundings.

No exchange of heat between system and surroundings.

Internal energy of the system remains constant. Ie ∆U = 0

Internal energy of the system remains changes. Ie ∆U ≠ 0

System is not thermally isolated. System is thermally isolated.

Eg. Fusion of ice. Eg. Expansion of gas in vacuum.

Isothermal process:-

A process in which temperature of the system remain constant throughout the process is called an isothermal process.

For isothermal process, ∆T = 0 , ∆U = 0 U = internal energy

Isobaric process:-

A process in which pressure of the system remain constant throughout the process is called an isobaric process.

For isobaric process, ∆P = 0 This process is carried out in open vessel.

Isochoric process:-

A process in which volume of the system remain constant throughout the process is called an isochoric process.

For isochoric process, ∆V= 0 This process is carried out in closed vessel.

Adiabatic process:-

A process in which no heat can exchange between system and surroundings is called an adiabatic process.

For adiabatic process, q = 0

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Reversible process:-

A process in which the driving force is infinitesimally greater than the opposing force and direction of the process can be reversed by slight increase in the opposing force is called Reversible process.

Features of Reversible process:-

(1) Driving force and opposing force are slight different from each other. (2) It takes infinite time.

(3) It takes place in infinite number of steps. (4) The process can be reversed at any point.

(5) Reversible process is occurs at a very slow rate. (6) Maximum work is obtained.

Nature of heat and work:-

Work (W):- In mechanics, work is product of force and displacement, W = f × d

where f = force in the direction of displacement d = displacement

Pressure – volume work

Work

Maximum work

Pressure – volume work:-

The work is done due to expansion or compression of a gas against an external opposing pressure P is called Pressure – volume work.

Expression:-

Consider a gas having pressure( P), Volume (V1 ) and

Temperature (T) enclosed in a cylinder fitted with frictionless

Piston having area (A) as shown in fig.

As gas expand against external pressure Pext , the

Piston moves in upward direction through a distance ‘d’

Then work done (W) by gas is,

W = Opposing force × displacement W = — f × d …... (1)

Change of energy in closed system occur (1) by transfer of energy as work and (2) by transfer of energy as heat

Pext

P

A

Ideal Gas

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Pressure is force per unit area

Pext = f / A ===> f = Pext × A …….…. (2)

From eq. (1) and (2), W = — Pext × A × d ………… (3)

But, A × d = ∆V = V2 — V1 V1 = initial volume V2 = final volume

Eq. (3) becomes,

Or ……….. (4)

Maximum Work:- When work is done there are two forces [1] acting force (driving force)(P)

and [2]Opposing force ( Pext )

When opposing force is zero, no work is done and the opposing force is increased more, more work will be obtained. Ie. W = — Pext × ∆V W = 0 × ∆V W = 0

If opposing force = driving force, Pext = P then ∆V = 0.

Ie. W = — Pext × ∆V W = — Pext × 0 W = 0

Maximum work done is obtained, when driving force is slightly greater than opposing force.

Ie. P > — Pext

If P— Pext = ∆P Pext = P—∆P

W = — Pext × ∆V

W = — (P—∆P) × ∆V = (∆P—P) × ∆V Work is maximum when ∆P is smallest.

Condition for maximum work:

Maximum work is obtained when-

(1) Process is thermodynamic ally reversible. (2) Change takes place in infinite number of steps. (3) Driving force is slightly greater than opposing force. (4) System is in mechanical equilibrium with its surrounding.

W = — Pext × ∆V W = — Pext × (V2 — V1 )

Free Expansion:- Expansion against zero opposing force is called Free Expansion.

When gas expand in vacuum, there is no opposing force i.e Pext = 0

W = — Pext × ∆V W = 0 × ∆V W = 0

Hence, no work is done when gas expand in vacuum.

Note= For getting work done opposing force is necessary.

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Expression for Maximum Work:- ( Isothermal reversible expansion )

Consider ‘n’ moles of an ideal gas enclosed in a Cylinder fitted with frictionless Piston having area (A). Let, P = pressure, V = volume and T = temperature of the gas. Let the gas expand isothermally and reversibly. The opposing pressure ( Pext ) is slightly less than pressure of gas (P) hence, Pext = P — dP ………(1)

Suppose gas expand by a small volume dV, then work done dW Is given by

dW = — Pext × dV

= — (P — dP) × dV

= — P.dV + dP . dV

dP . dV isvery small hence neglected

dW = — P.dV ………….. (3)

Now, total work done when gas expand from initial volume V1 to final volume V2 is obtained by integrating eq. (3) between limits V1 and V2

Wmax = ∫ 푑푊 = ∫ — P. dV

…………….(4)

For ‘n’ moles ideal gas eq. is PV = nRT ===> P = nRT / V

Wmax = ∫ — 푑푉

= — nRT ∫ — 푑푉

= — nRT [loge V]

V + dV

dv

V

Cylinder

P—dP

P n

Fig.

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= — nRT [loge V2 — loge V1 ] = — nRT loge

According to Boyle’s law, P1 V1 = P2 V2 =

Path function:-

Work (W)and heat (q) are path functions

A) Path dependence nature of work

1. When gas expand in vacuum work done is zero

2. When gas expand from V1 to V2 work done is W = — Pext × (V2 — V1 )

3. When gas expand isothermally and reversibly, work done is Wmax = — 2.303 nRT log10

B) Path dependence nature of heat

1. For adiabatic process q = 0

2. For isothermal process q ≠ 0

Unit of energy and work:- Joule (J)

1 J = 1 kgm3 /s2 1 atm = 101.3 × 10—3 kgm3 /s2

1 J = 1 pa.m3 1 L.atm = 101.3 kgm3 /s = 101.3 J

1 J = 107 ergs 1 J = 0.239 cal. 1 cal. = 4.184 J 1 kcal. = 4.184 J

Sign convention of work done, heat and energy:

+W work done on system heat absorbed +q

—w work done by system heat released —q

+ q : heat is absorbed by system from surrounding.

— q: heat is released by system from surrounding.

+ w: work is done on system by surrounding. — w: work is done by system on surrounding.

Wmax = — 2.303 nRT log10 푷ퟏ푷ퟐ

Wmax = — 2.303 nRT log10 푽ퟐ푽ퟏ

System

A

B

Adiabatic q = 0

Isothermal q ≠ 0

A (V1 )

B (V2 )

1

3

2

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Internal Energy (U):-

(1) Internal energy is a sum of kinetic energy and potential energy.

(2) Internal energy is a state function. (3) Internal energy is an extensive property.

(4) Internal energy depends on temperature hence for isothermal process, (∆T = 0) ∆U = 0

(5) Change in Internal energy ∆U = U2 — U1

(6) —∆E = energy is evolved by system, +∆E = energy is absorbed by system.

Note:- Kinetic energy = Translation + Rotational + Vibrational + Electronic

Potential energy = Intramolecular + Intermolecular

First law of thermodynamics:- It is simply law of conservation of energy.

Statements

1. Energy can neither be created nor destroyed, although it can be converted from one form to another

2. The total internal energy of an isolated system is constant.

3. The total quantity of energy of the universe is constant.

4. Whenever a one form of energy disappears, exactly an equivalent amount of other form must appear.

5. It is impossible to construct perpetual motion machine which produce work continuously without

consuming energy.

Mathematical equation of First law of thermodynamics:-

Consider U1 = initial internal energy. Suppose ‘q’ amount of energy supplied to system which produce ‘W’ amount of work done.

Thus, final internal energy U2 is given by, U2 = U1 + q + W ==> U2 — U1 = q + w

∆U = q + w ………… (A)

∆U = increase in internal energy of system.

Eq. (A) represents equation of First law of thermodynamics.

For infinitesimal change eq. (A) becomes dU = dq + dW

U = Utran. + Urot. + Uvibr. + Uele. + Uintra. + Uinter.

“The energy associated with a substance or system is called internal energy.”

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Modified First law of thermodynamics:-

According to Einstein theory, mass can be covered into energy. (ie. Energy produced due to loss of mass.)

Hence modified First law of thermodynamics is stated as, “total mass and energy of an isolated system (or universe) remain unchanged.”

First law of thermodynamics for various processes:-

We have ∆U = q + w …….. (A)

1) For isothermal process:- ∆T = 0 and ∆U = 0

Eq. (A) becomes, 0 = q + W

Ie. q = — W or — q = W

+ W : work done by system on surroundings. — W : work done by surroundings on system.

2) For isobaric process:- ∆P = 0

∆U = qp + W

= qp + ( — Pext × ∆V )

∆U = qp — Pext × ∆V or qp = ∆U + Pext × ∆V

Heat used to increase internal energy of system.

3) For isochoric process:- ∆V = 0

∆U = qV + W

= qV + ( — Pext × ∆V )

∆U = qV + (0) or ∆U = qV

Change in internal energy of system is due to heat transfer.

4) For adiabatic process:- q = 0

Eq. (A) becomes, ∆U = 0 + W

Ie. ∆U = W or — ∆U = — W

1. work done by system, internal energy increases.

2. Work done by surroundings internal energy decreases.

Einstein eq. is E = mC2

Where E = energy produced by loss of mass ‘m’. C = velocity of light = 3 × 108 m/s

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Second law of thermodynamics:-

Statements

1) “The spontaneous flow of heat is always unidirectional from higher temperature to lower temperature.”

2) “Heat cannot be completely converted into an equivalent amount of work without producing permanent changes either in the system or its surroundings.”

3) “No machine has yet been made that has an unit efficiency.”

4) “The entropy of the system is constantly increasing.”

Enthalpy (H):-

(PV energy is energy related with the pressure and volume.)

Mathematically, the enthalpy is defined as H = U + PV ………… (1)

Since P, V and U are state functions, H is also state function.

Hence change in enthalpy (∆H) is given by ∆H = H2 — H1 …………(2)

If H1 = U1 + P1 V1 and H2 = U2 + P2 V2

Then from eq. (2) we have ∆H = U2 + P2 V2 — ( U1 + P1 V1 )

= ( U2 — U1 ) + ( P2 V2 — P1 V1 )

= ∆U + ( P2 V2 — P1 V1 )

At constant pressure i.e P1 = P2 = P then

∆H = ∆U + ( P V2 — P V1 ) = ∆U + P ( V2 — V1 )

………… (3)

Hence, for isobaric process enthalpy of system is the sum of increase in internal energy the work of expansion.

Also we have, qp = ∆U + Pext × ∆V

If Pext = P then qp = ∆U + P∆V ……….. (4)

From eq. (3) & (4) we write since, qP is state function.

Hence, for isobaric process enthalpy change is the heat absorbed by system.

Note:- Enthalpy is Heat content of the system. Heat evolved or absorbed at constant pressure is expressed in terms of enthalpy.

∆H = ∆U + P∆ V

∆H = qp

It is defined as “the sum of internal energy of the system and PV energy.”

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Relation between ∆H and ∆U for chemical reaction:-

Que. Obtain relationship betn ∆H and ∆U for chemical reaction.

Ans. We have, ∆H = ∆U + P∆ V …………. (1)

For solid and liquid, ∆V = 0 then ∆H = ∆U + P(0)

∆H = ∆U ………………… (2)

For gas, ∆V is not neglected.

∆H = ∆U + P ( V2 — V1 )

= ∆U + PV2 — PV1 …………… (3)

V1= initial volume (volume of gaseous reactant) , V2 = final volume (volume of gaseous product)

According to ideal gas eq. PV = nRT

Suppose, n1 = moles of gaseous reactant. n2 = moles of gaseous product.

PV1 = n1 RT and PV2 = n2 RT

From eq. (3) ∆H = ∆U + ( n2 RT — n1 RT )

= ∆U + RT ( n2 — n1 )

Work done in chemical reaction:-

We have W = — Pext . ∆V ……… (1)

Suppose Pext = P then W = — P . ∆V = — P ( V2 — V1 )

= — PV2 + PV1 …………….. (2)

According to ideal gas eq. PV = nRT

If n1 = moles of gaseous reactant. n2 = moles of gaseous product.

Then PV1 = n1 RT and PV2 = n2 RT

Eq. (2) becomes, W = — n2 RT + n1 RT = — RT ( n2 — n1 )

W = — RT . ∆n

∆H = ∆U + RT.∆n

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Condition under which ∆H = ∆U

1) When reaction is carried out in a closed vessel ∆U = 0, ∆H = ∆U + P∆V ∆H = ∆U

2) When reaction involves solid and liquid ∆V = 0, ∆H = ∆U + P∆V ∆H = ∆U

3) When reaction in which n1 = n2 , ∆n = 0 ∆H = ∆U + RT∆n ∆H = ∆U

Sign conventions of ∆H

1) When heat is absorbed by system (ie. for endothermic reaction) ∆H is positive.

2) When heat is evolved by system (ie. for exothermic reaction) ∆H is negative.

Enthalpies of physical changes:-

A) Enthalpy of phase transition:-

Whenever there is a change in the state of matter, the process is called phase transition or change.

It is accompanied by the change in enthalpy (Heat content) of the system.

Commonly observed phase transitions are:-

Solid liquid fusion (melting)

Liquid vapour vaporization

Solid vapour sublimation

1) Enthalpy of fusion (Heat of fusion):- ( ∆fus. H)

The enthalpy change when one mole of a solid substance is converted into liquid at its melting point at constant pressure is called its enthalpy of fusion.

¤ The reverse of fusion is freezing of liquid.

2) Enthalpy of vaporization:- ( ∆vap. H)

The enthalpy change when one mole of a liquid is converted into vapour at its boiling point at constant pressure is called its enthalpy of vaporization.

¤ The reverse of vaporization is condensation of vapour.

H2 O (s) H2 O (l) ∆fus. H = + 6.01 kJ/mol at 00 C

H2 O (l) H2 O (g) ∆vap. H = + 40.7 kJ/mol at 1000 C

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3) Enthalpy of sublimation:- ( ∆sub. H)

The enthalpy change when one mole of a solid is directly converted into vapour at constant temperature and pressure is called its enthalpy of sublimation.

Relationship between ∆fus. H , ∆vap. H and ∆Sub. H

Eg for water,

B) Enthalpy of atomic or molecular changes:-

1) Enthalpy of ionization:- (∆ion. H)

The enthalpy change when removal of an electron from each atom or ion in one mole of a gaseous atom or ion called enthalpy of ionization.

# 494 kJ / mol of heat is required to ionize one mole of Na atom.

Note:-

2) Enthalpy of atomization:- (∆atm. H)

The enthalpy change when the dissociation of all the molecule in one mole of a gas phase substance into gaseous atoms is called enthalpy of atomization.

OR The enthalpy change when one mole of a gas phase substance dissociates into its gaseous state constituent atoms is called enthalpy of atomization.

H2 O (s) H2 O (g) ∆sub. H = + 51.08 kJ/mol at 00 C

∆sub. H = ∆fus. H + ∆vap. H

51.08 = 6.01 + 45.07

Vaporization

Fusion

Sublimation

Gas

Liquid

Solid

Na (g) Na+ (g) + e— ∆ion. H = 494 kJ/mol

Ca (g) Ca+ (g) + e— ∆ion. H = 590 kJ/mol …… first ionization enthalpy

Ca+ (g) Ca++ (g) + e— ∆ion. H = 590 kJ/mol …… second ionization enthalpy

The reverse of ionization is electron gain process. Hence electron gain enthalpy is defined as,

“The enthalpy change when one mole of a gas phase atom of an element accept electron so as to form gaseous anion.”

Cl (g) + e— Cl— (g) ∆eg. H = — 349 kJ/mol

Cl2 (g) Cl (g) + Cl (g) ∆atm. H = 242 kJ/mol

CH4 (g) C (g) + 4 H (g) ∆atm. H = 1660 kJ/mol

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3) Enthalpy of solution:- ( ∆sol. H )

The enthalpy change when one mole of a substance is dissolved in a specified quantity of solvent at a given temperature is called enthalpy of solution.

Note:- The ∆sol. H is the sum of three kinds of interactions.

i) Solute – solute interaction (ie crystal lattice energy) ∆L H

ii) Solvent – solvent interaction

ii) Solute – solvent interaction (ie enthalpy of hydration) ∆hyd. H

1) Lattice enthalpy:- ( ∆L H )

The enthalpy change when one mole of a ionic solid breaks down into its ions is called Lattice enthalpy.

2) Enthalpy of hydration :-( ∆hyd. H )

The enthalpy change involved in the hydration of one mole of a gaseous ions is called enthalpy of hydration .

So,

NaCl (s) + H2 O (l) 1 dm3 NaCl (aq.) ∆sol. H = + 4 kJ/mol

XY (s) X+ (g) + Y— (g) or X+ (g) + Y— (g) XY (s) ∆L. H = + ve kJ/mol

X+ (g) + Y— (g) X+ (aq.) + Y— (aq.)

XY X+ (aq.) + Y— (aq.)

X+ (g) + Y— (g)

∆sol. H

∆L H ∆hyd. H

Ie. ∆sol. H = ∆L H + ∆hyd. H

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4) Enthalpy of dilution:-

The enthalpy change that occurs when a solution of one concentration is diluted to form the solution of another concentration.

Thermochemistry:-

Enthalpy of chemical reactions (Heat of reaction):-

The enthalpy change occur by chemical reaction is called Enthalpy reaction.

Consider a rean aA + bB cC + dD

The enthalpy change for rean is ∆H = (c HC + d HD ) — (a HA + b HB )

Where HA , HB , HC and HD are the enthalpies of A, B, C and D respectively.

Hence enthalpy of chemical reaction is the difference between the sum of enthalpies of product and sum of enthalpies of reactant.

Born – Haber cycle:-

Maw Born & Fritz Haber proposed the method to calculate lattice enthalpy.

Eg. Formation of NaCl. Na (s) + Cl2 NaCl (s) ∆f H

1) Sublimation of Na metal:- Na (s) Na (g) ∆sub H

2) Ionization of Na atom:- Na (g) Na+ (g) + e— (g) ∆ion H

3) Dissociation of mole Cl2 :- Cl2 (g) Cl (g) ∆a H

4) Electron gain by Cl atom:- Cl (g) + e— Cl— (g) ∆eg. H

5) Combination of Na+ and Cl— atom:- Na+ (g) + Cl— (g) NaCl (s) ∆L. H

∆f H = ∆sub H + ∆ion H + ∆a H + ∆eg. H + ∆L H

HCl (g) + 50 H2 O (finite amount) HCl (50 H2 O ) ∆s. H1 = — 73.26 kJ

HCl (g) + H2 O (infinite amount) HCl (aq. ) ∆s. H2 = — 75.14 kJ

∆H = ∆ H2 — ∆ H1 = (— 75.14 )— (— 73.26 ) = — 1.88 kJ enthalpy of dilution.

Thermochemistry deals with the study of enthalpy changes in chemical reactions.

∆H = ∑ HProduct — ∑ HReactant

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Exothermic and Endothermic reactions:-

Exothermic reactions Endothermic reactions

1. The reaction in which heat released in the surrounding is called Exothermic reaction.

1. The reaction in which heat absorbed from surrounding is called Endothermic reaction.

2. Exothermic reactions are occurs by evolution of heat.

2. Endothermic reactions are occurs by absorption of heat.

3. ∑ HProduct < ∑ HReactant 3. ∑ HProduct > ∑ HReactant

4. ∆H is negative 4. ∆H is positive

5. Products are more stable than reactants. 5. Reactants are more stable than Products.

6. i) C (s) + O2 CO2 ∆H = — 393.5 kJ/mol ii)Making ice cubes, iii) condensation of water vapour

6. i) N2 + 2O2 2NO2 ∆H = + 66.4 kJ/mol ii) Melting of ice, iii) Evaporation of water.

Standard Enthalpy of chemical reactions:- (∆H0 )

“The standard Enthalpy of chemical reactions is the enthalpy change for a reaction when all the substances involved are in their standard states.”

Note: The standard state of substance is the pure form in which it is most stable at pressure 1 atm & 25 0C temperature.

Standard states of element = H2 (g) , Hg (l) , Na (s) , C (S)

Standard states of substance = C2 H5 OH (l), H2 O (l) , CaCO3 (S) , CO2

Thermochemical equation:-

1) The equation must be balanced

2) The physical state of reactants and products are indicated by placing the letters (s) for solid, (l) for liquid and (g) for gas.

3) The numerical value and appropriate sign of enthalpy change must be shown on right hand side of equation.

4) A thermodynamic eq. can be reversed, if the sign of ∆H is changed.

C (s) + O2 CO2 ∆H = — 393.5 kJ/mol CO2 C (s) + O2 ∆H = + 393.5 kJ/mol

5) The enthalpy of excess of water shown by (aq) is taken as zero.

NH4 Cl (s) + aq. NH4 Cl (aq) ∆H0 = + 15.1 kJ

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Standard Enthalpy of formation:- (∆fH0 ) (Standard heat of formation ):-

The Standard Enthalpy of formation of a compound is defined as, the enthalpy change that take place when one mole of pure compound under standard condition is formed from its constituent elements which is in their standard states.

Eg. Formation of one mole of liquid water, H2 (g) + ½ O2 (g) H2 O (l) ∆fH0 = —286 kJ

Calculation of Standard Enthalpy of reaction from Standard Enthalpy of formation:-

Consider the reaction aA + bB cC + dD

Standard Enthalpy of reaction is given by ∆H0 = ∑ ∆f 퐻 — ∑ ∆f 퐻

= {c ∆fH0 (C)+ d H0∆fH

0 (D)} — {a ∆fH0 (A) + b ∆fH

0 (B)}

Where a, b, c and d are stiochiometric coefficient.

Eg. Formation of CO2 (g) C (s) + O2 CO2 ∆fH = — 393.5 kJ/mol

∆H0 = ∑ ∆f 퐻 — ∑ ∆f 퐻 = { ∆fH0 (CO2 )} — { ∆fH

0 (C) + ∆fH0 (O2 )}

By convention enthalpies of C (s) and O2 (g) is zero

∆H0 = ∆fH0 (CO2 ) — {0 + 0} = ∆fH

0 (CO2 )

Hence enthalpy of CO2 is equal to its enthalpy.

Standard Enthalpy of combustion:- (∆CH0 )

The Standard Enthalpy of combustion of a substance is the enthalpy change that takes place when one mole of pure substance in the standard state reacts completely with oxygen.

Eg. Complete combustion of 1 mole of CO is,

CO (g) + ½ O2 (g) CO2 (g) ∆cH0 = —283 kJ

Combustion of ethyne C2 H2 (g) + 5/2 O2 (g) 2CO2 (g) + H2O (l) ∆cH0 = — 1300 kJ

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Bond Enthalpy (Bond energy):- (D)

“The amount of energy required to break a particular covalent bond in 1 mole of gaseous molecule to produce gaseous atom or radical is called bond enthalpy.”

# Bond energy is always positive because energy is always needed to break the bond.

Eg. 1) H2 (g) H (g) + H (g) ∆H0 = + 436.4 kJ

Ie. to break H—H bond in H2 , 436.4 kJ energy is required. Ie. DH—H = 436.4 kJ or ∆H0 (H—H)= + 436.4 kJ

2) HCl (g) H (g) + Cl (g) ∆H0 = + 431.9 kJ Ie ∆H0 (H—Cl) = + 431.9 kJ

# The bond enthalpy for diatomic molecule is the same as enthalpy of atomization.

# In polyatomic molecule, every covalent bond has its own specific bond enthalpy. Hence average bond enthalpy has to be considered.

Eg. Dissociation of water Step i) H2 O (g) OH (g) + H (g) ∆H0 = 499 kJ

Step ii) OH (g) O (g) + H (g) ∆H0 = 428 kJ

Overall Rean H2 O (g) 2H (g) + O (g) ∆H0 = 927 kJ

Average bond enthalpy of O—H bond = = 463.5 kJ hence ∆H0 (O—H) = 463.5 kJ

Reaction enthalpy from bond enthalpy:-

Que. Define bond enthalpy. How is it useful to calculate reaction enthalpy?

Ans.

In a chemical reaction, in reactants bonds broke while in product bond formed. For enthalpy calculation we assume all the bonds are identical and bond enthalpy independent of nature of substance.

The enthalpy of chemical reaction is the difference between the sum of bond enthalpies of reactants and the sum of bond enthalpies of product.

∆H0 (Reaction) = ∑ ∆H0 (Reactants broken bonds) — ∑ ∆H0 (Product formed bonds)

Eg. H2 (g) + I2 HI (g) ∆H0 ={ [ ∆H0 (H—H) + ∆H0 (I—I) ] — 2∆H0 (H—I) }

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Hess’s Law of constant heat summation:-

The law states that, “the enthalpy change for a reaction is the same whether the reaction takes place in one step or a series of steps.” OR

“The enthalpy change for a reaction is same regardless of the path of the reaction.”

Proof / illustration:- Consider, a conversion of A to C

i) Conversion of A to C in single step

A C ∆H0 = ∆H1 ………. (1)

ii) Conversion of A to C in two steps

A B ∆H0 = ∆H2

B C ∆H0 = ∆H3

Overall A C ∆H0 = ∆H2 + ∆H3 ……. (2)

According to Hess’s Law,

Eg. Formation of CO2 from C and O2

1) Formation of CO2 in single step C (s) + O2 (g) CO2 ∆H1 = — 394 kJ

2) Formation of CO2 in two step C (s) + O2 (g) CO ∆H2 = — 83 kJ

CO (g) + O2 (g) CO2 ∆H3 = — 311 kJ

According to Hess’s Law, ∆H1 = ∆H2 + ∆H3 — 394 =— 83 — 311 — 394 = — 394

Thus heat evolved in both methods is same. Hence Hess’s Law is proved experimentally.

Application of Hess’s Law:-

1) The thermodynamic equations can be added, subtracted or multiplied by numeral like simple algebraic equation.

2) It can be used to calculate enthalpy of reaction which does not occurs directly.

3) The enthalpy of formation of any compound from its elements can be calculated.

4) It can be used to calculate enthalpy of formation, combustion, neutralization etc.

5) Enthalpy of formation of slow reaction can be calculated.

6) Enthalpy of reaction which is not possible to determine experimentally can be calculated.

B

A C

∆H2 ∆H3

∆H1

∆H1 = ∆H2 + ∆H3

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Spontaneous Process (Irreversible process):-

The process that takes place on its own accord without any external influence is called spontaneous process. It is also called natural process.

Eg. (1) All natural process are spontaneous i) cliff jumping, ii)

(2) Flow of gas from higher pressure to lower pressure. (3) Flow of heat from hot body to cold body

(4) Flow of water from higher level to lower level

(5) Dissolution of KCl in water

(6) Acid – Base neutralization reaction

Note:- Spontaneity:- The tendency of process to occur naturally is called spontaneity.

Criteria for spontaneity:-

(1) A spontaneous change is one-way or unidirectional. Ie. for reverse change work is required.

(2) A spontaneous change is time is no factor. Ie. Spontaneous reaction may be slow or rapid.

(3) A spontaneous process is continuing till the system attains the equilibrium state.

(4) Once system in equilibrium, it does not undergo any further spontaneous change.

Energy and Spontaneity:-

Generally Spontaneous process are exothermic (∆H0 = — ve) ie. Exothermility favors Spontaneity. But there are many processes that are endothermic but Spontaneous.

Eg. 1) Melting of ice Ice H2O ∆H = + 4 kJ

2) Dissolution of NaCl in water NaCl (s) + H2O (aq) Na+ (aq) + Cl— (aq) ∆H = + 3.9 kJ

3) Conversion of N2O4 to NO2 above 400 k N2O4 (g) NO2 (aq) ∆H = + 57 kJ

From above example it is evident that energy is not only criteria for spontaneity. Hence with energy there must be another factor which governs spontaneity. That factor is entropy. 2nd law of thermodynamics introduces entropy (S).

KCl (s) + H2O KCl (aq.)

NaOH (s) + HCl NaCl + H2O

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Entropy (S):-

Greater the disorder of a system, higher is the entropy and vice versa. Hence solid state (ice) has lowest entropy while gaseous state (steam) has highest entropy.

Entropy is a state function. Hence change in entropy ∆S is ∆S = Sfinal — Sinitial

1) I2 (s) (ordered state) I2 (aq) (disordered state ) ∆S is +ve

2) 2 H2 (g) + O2 2 H2 O (l) ∆S is —ve

Quantitative definition of entropy:-

The entropy change (∆S) of a system in a process is the ratio of amount heat transferred in a reversible manner to the temperature at that heat transfer (T).

∆S = (qrev /T)

Entropy and spontaneity:- (2nd law of thermodynamics)

A process occur by an increase in entropy tend to be spontaneous.

Consider two spontaneous process 1) 2H2O2 (l) 2 H2O (l) + O2 (g) ∆S = +ve

2) 2H2 (g) + O2 (g) 2 H2O (l) ∆S = —ve

From above two examples it is clear that in spontaneous process the entropy may increase or decrease. This discrepancy is explained by second law of thermodynamics.

The thermodynamic property which measures the molecular disorder or randomness is called entropy.

H2 O

More disordered state disordered state

Ordered State Disordered State

Highly disordered State

Unit of ∆S SI unit = J/k or J/mol.k CGS unit = cal/k or cal/mol.k

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2nd law of thermodynamics:-

“The total entropy of the system and surroundings increases in a spontaneous process.”

Thus,

OR The total entropy of the system is constantly increasing.

Gibbs Energy (Gibbs Function):- (G)

For most of the chemical reactions there are changes in both enthalpy and entropy. Hence to decide the spontaneity of process in terms of both enthalpy and entropy change we define new thermodynamic function, called Gibbs energy (G).

Gibbs free energy is defined as , G = H — TS

Where, H is enthalpy and S is entropy of system. T is absolute temperature.

H, S and T are state functions, G is also state function. Hence change in Gibbs Energy (∆G)is defined as , ∆G = ∆H — T∆S

Gibbs Energy and spontaneity:-

We have ∆STotal = ∆Ssys. + ∆ Ssurr. ∆STotal = ∆S + ∆ Ssurr. ……… (1)

If + ∆H is the enthalpy change of system, then enthalpy change of surrounding is —∆H.

∆S = qrev. / T = qsur. / T = — ∆H / T …….. (2)

From eq. (1) & (2), total entropy ∆STotal is, ∆STotal = ∆S — ∆

Multiplying above eq. by — T — T∆STotal = — T∆S + ∆H

Ie — T∆STotal = ∆H— T∆S ……… (3)

The change in Gibbs Energy (∆G)is ∆G = ∆H — T∆S .……… (4)

From eq. (3) & (4) ∆G = — T∆STotal

Ie. ∆G = T ( —∆STotal ) because T is always positive.

Hence if ∆STotal > 0 then ∆G < 0

∆Suniverse = ∆STotal = ∆Ssys. + ∆ Ssurr. > 0 , for spontaneous process

(1) ∆G < 0, the process is spontaneous (2) ∆G > 0, the process is non-spontaneous

(3) ∆G = 0, the process is at equilibrium.

(1) ∆STotal > 0 , the process is spontaneous. (2) ∆STotal < 0 , the process is non-spontaneous.

(3) ∆STotal = 0 , the process is at equilibrium.

Unit of ∆G SI unit = J or kJ or kJ/mol CGS unit = cal. Or k.cal or k.cal/mol

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¤ The spontaneity of reaction with respect to sign dependence of ∆H, ∆S and ∆G

∆H ∆S ∆G Spontaneity of reaction — (Exothermic)

+ — Reactions are spontaneous at all the temperatures.

— (Exothermic)

— — Or + Reactions become spontaneous at low temperatures. When ІT.∆SІ<І∆HІ

+ (Endothermic)

+ — Or + Reactions become spontaneous at high temperatures. When T.∆S > ∆H

+ (Endothermic)

— + Reactions are non-spontaneous at all the temperatures.

Temperature of equilibrium:- We have ∆G = ∆H — T∆S ………….. (1)

At equilibrium, ∆G = 0 Ie. process is neither spontaneous nor non-spontaneous.

Eq. (1) becomes, 0 = ∆H — T∆S ie. ∆H = T∆S T = ∆∆

∆G and equilibrium constant:-

The change in Gibbs Energy (∆G) of a reaction is related to standard Gibbs Energy change (∆G0 ) by the relationship

∆G = ∆G0 + RT lnQ = ∆G0 + 2.303 RT log10 Q ………..(A)

Where, ∆G0 = Standard Gibbs energy change R = Gas constant (8.314 kJ/mol)

Q = Reaction quotient T = Temperature

For Q consider a reaction aA + bB cC + dD

[ C ]c [ D ]d OR 푃 × 푃

[ A ]a [ B ]b 푃 × 푃

Where [] indicate concentration and P indicate partial pressure

At equilibrium Q = k and ∆G = 0 K = equilibrium constant

Eq. (A) becomes 0 = ∆G0 + 2.303 RT log10 k Ie. ∆G0 = — 2.303 RT log10 k

QC = Q P =

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Third law of thermodynamics:-

It states that, “at absolute zero temperature, the entropy of perfectly ordered crystalline substance is zero.”

Thus, S = 0 at T = 0 for perfectly ordered crystalline substance. Substance contain impurities S >0

# The entropy of solid greater than zero (S > 0) at T = 0 & is called residual entropy of substance.

Importance of 3rd law of thermodynamics:-

(1) It gives absolute starting place from which to measure entropy.

(2) It allows determining the absolute entropy of any substance at any desired temperature.

(3) The standard molar entropy (S0 ) of pure substance can be measured.

(4) Standard entropy change ∆S0 for a reaction can be calculated.

(5) Spontaneity of reaction can be predicted.

Que. How is absolute entropy measured for a pure crystalline substance at high temperature? Or why is it impossible for any substance to have an absolute entropy zero at temp. greater than 0 k?

Ans:- For perfectly ordered crystalline substance at absolute zero temperature, the entropy of is zero S0 = 0. When this substance is heated to the temp. T its entropy will increase to ST then increase in entropy is given by

∆S = ST — S0 since S0 = 0 then ∆S = ST

Thus the value of absolute entropy at different temperature can be calculated. Hence at temp. greater than absolute zero, it is impossible for any substance to have an absolute entropy zero.

Note: ST is determined by ST = ∫ . Where CP is heat capacity of solid.

Standard molar entropy (S0 ):-

The absolute entropy of one mole of pure substance at 1 atm. Pressure and 25 0C temperature is called Standard molar entropy of the substance.

Usefulness of Standard molar entropy:-

(1) To determine the ∆ S0 for a reaction. (2) It is used to compare entropies of different substance under same condition of temperature and pressure. (3) Spontaneity of the process can be predicted.

Thermodynamics

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Transcript of Thermodynamics