INTRODUCTION

The word ‘thermodynamics’ implies flow of heat. It deals with energy changes accompanying all

types of physical and chemical processes.

It helps to lay down the criteria for predicting feasibility or spontaneity of a process, including a

chemical reaction, under a given set of conditions. It also helps to determine the extent to which a

process, including a chemical reaction, can proceed before attainment of equilibrium.

Thermodynamics is based on two generalizations called the first and second law of

thermodynamics. These are based on human experience.

SOME BASIC TERMS

System

A system is defined as any specified portion of matter under study which is separated from the

rest of the universe with a bounding surface. A system may consist of one or more substances.

Surroundings

The rest of the universe which might be in a position to exchange energy and matter with the

system is called the surroundings.

Types of system

(i) Isolated system

A system which can exchange neither energy nor matter with its surrounding is called an isolated

system.

(ii) Open system

A system which can exchange matter as well as energy with its surroundings is said to be an

open system.

(iii) Closed system

A system which can exchange energy but not matter with its surroundings is called a closed

system.

Macroscopic properties

The properties associated with a macroscopic system (i.e. consisting of large number of particles)

are called macroscopic properties. These properties are pressure, volume, temperature,

composition, density etc.

Extensive and Intensive properties

An extensive property of a system is that which depends upon the amount of the substance

present in the system like mass, volume and energy.

An intensive property of a system is that which is independent of the amount of the substance

present in the system like temperature, pressure, density, concentration, viscosity, surface

tension, refractive index etc.

State of a system

When macroscopic properties of a system have definite values, the system is said to be in definite state. Whenever there is a change in any one of the macroscopic properties, the system is said to change into a different state. Thus, the state of a system is fixed by its macroscopic properties.

State variables

Since the state of a system changed with change in any of the macroscopic properties, these properties are called state variables or the thermodynamics parameters which depends only upon the initial and final states of the system and independent of the manner as to how the change is brought are called state functions. Some common state functions are internal energy, enthalpy, entropy, free energy, pressure, temperature, volume etc.

Thermodynamic equilibrium

A system in which the macroscopic properties do not undergo any change with time is said to be in thermodynamic equilibrium.

Thermodynamic process and their types

The operation by which a system changes form one state to another is called a process. Whenever a system changes from one state to another it is accompanied by change in energy. In case of open systems, there may be change of matter as well.

The following types of process are known

Isothermal process

A process is said to be isothermal if the temperature of the system remains constant during each stage of the process.

Adiabatic process

A process is said to be adiabatic if the heat enters or leaves the system during any step of the process.

Isobaric process

A process is said to be isobaric if the pressure of the system remains constant during each step of the process.

Illustration 1. Thermodynamics is concerned with (A) total energy in a system (B) energy changes in a system (C) rate of a chemical change (D) mass changes in nuclear reactions

Solution: (B)

Isochoric Process

A process is said to be isochoric if the volume of the system

remains constant during each step of the process.

Reversible and Irreversible process

A process which is carried out infinitesimally slowly in such a

manner that the system remains almost in a state of

equilibrium at every stage or a process carried out

infinitesimally slowly so that the driving force is only

infinitesimally greater than the opposing force is called a

reversible process.

Isobaric

Isothermal

Adiabatic

Volume

Pre

ssu

re

Iso

cho

ric

Any process which does not take place in the above manner i.e. a process which does not take

place infinitesimally slowly, is said to be an irreversible process.

In fact, all the natural processes are irreversible processes.

INTERNAL ENERGY

Every substance is associated with a definite amount of energy which depends upon its chemical nature as well as upon its temperature, pressure and volume. This energy is known as internal energy. Internal energy of the system is the energy possessed by all its constituent molecules. Internal energy is a state property i.e. its value depends only upon the state of the substance but does not depend upon how that state is achieved. The absolute value of internal energy of a substance can not be determined. However determining the absolute values of internal energies is neither necessary nor required. It is the change in internal energy accompanying a chemical or a physical process that is of interest and this is a measurable quantity.

The first law of thermodynamics

The first law of thermodynamics states that energy can neither be created nor destroyed,

although it can be transformed from one form to another. This is also known as the law of

conservation of energy.

MATHEMATICAL EXPRESSION OF FIRST LAW

Let UA be the energy of a system in its state A and UB be the energy in its state B. Suppose the

system while undergoing change from state A to state B absorbs heat q from the surroundings

and also performs some work (mechanical or electrical), equal to w. The absorption of heat by

the system tends to raise the energy of the system. The performance of work by the system, on

the other hand, tends to lower the energy of the system because performance of work requires

expenditure of energy. Hence the change of internal energy, ∆U, accompanying the above

process will be given by

B AU U U q w∆ = − = −

In general, if in a given process the quantity of heat transferred from the surrounding to the

system is q and work done in the process is w, then the change in internal energy,

∆U = q + w

This is the mathematical statement of the first law of thermodynamics.

If work is done by the surroundings on the system (as during the compression of a gas), w is

taken as positive so that ∆U = q + w. if however work is done by the system on the surroundings

(as during the expansion of a gas), w is taken as negative so that ∆U = q – w.

Illustration 2. 1 mole of ideal monoatomic gas at 27°C expands adiabatically against a constant

external pressure of 1.5 atm from a volume of 4dm3 to 16 dm3.

Calculate (i) q (ii) w and (iii) ∆U

Solution: (i) Since process is adiabatic ∴ q = 0

(ii) As the gas expands against the constant external pressure.

W = ( )2 1P V 1.5 V V− ∆ = − −

= ( ) 31.5 16 4 18 atm dm− − = −

(iii) ∆U = q + w = ( ) 30 18 18 atm dm+ − = −

Exercise 1.

Calculate the internal energy change, when a system absorbs 5 KJ of heat and does

1 KJ of work.

ENTHALPY OF A SYSTEM

The quantity U + PV is known as the enthalpy of the system and is denoted by H. It represents

the total energy stored in the system. Thus

H = U + PV

It may be noted that like internal energy, enthalpy is also an extensive property as well as a state

function. The absolute value of enthalpy can not be determined, however the change in enthalpy

can be experimentally determined.

∆H = ∆U + ∆(PV)

Various kinds of processes:

(i) Isothermal reversible expansion of an Ideal gas: Since internal energy of an Ideal gas

is a function of temperature and it remains constant throughout the process hence

∆E = 0 and ∆H = ∆E + ∆PVQ ∆E = 0

and P1V1 = P2V2 at constant temperature for a given amount of the gas

∴∆H= 0

Calculation of q and w:Q ∆E = q + w

For an Isothermal process, w = -q

This shows that in an Isothermal expansion, the work done by the gas is equal to amount

of heat absorbed.

and w = - n RT ln(V2/V1) = - n RT ln(P1/P2).

Illustration 3. 10 gm of Helium at 127°C is expanded isothermally from 100 atm to 1 atm

Calculate the work done when the expansion is carried out (i) in single step (ii) in

three steps the intermediate pressure being 60 and 30 atm respectively and (iii)

reversibly.

Solution: (i) Work done = V.∆P

V = 5

5

10 8.314 40083.14 10

4 100 10−× × = × ×

m3

So W = 5

83.14

10 (100-1)× 105 = 8230.86 J.

(ii) In three steps

VI = 83.14× 10-5 m3

WI = (83.14× 10-5)× (100-60)× 105

= 3325.6 Jules

V II = 5 35

2.5 8.314 400138.56 10 m

60 10−× × = ×

×WII = V. ∆P

WII = 138.56× 10-5 (60-30)× 105

= 4156.99 ≈ 4157 J.

VIII = 5 35

2.5 8.314 400277.13 10 m

30 10−× × = ×

×WIII = 277.13× 10-5 (30-1)× 105

WIII = 8036.86 J.

W total = WI + WII + WIII

= 3325.6+4156.909+8036.86 = 15519.45 J.

(iii) For reversible process

W = 2.303 nRT log 1

2

P

P

= 2.303 × 10 100

8.314 400 log4 1

× × × W = 38294.28 Jules

Exercise 2.

Calculate the final volume of one mole of an ideal gas initially at 0°C and 1 atm

pressure, if it absorbs 1000 cal of heat during a reversible isothermal expansion.

Exercise 3.

Carbon monoxide is allowed to expand isothermally and reversibly from 10m3 to 20

m3 at 300 K and work obtained is 4.754 KJ. Calculate the number of moles of carbon

monoxide.

(ii) Adiabatic Reversible Expansion of an Ideal gas:

Q q = 0

∴∆E= -w.

Total change in the internal energy is equal to external work done by the system.

∴ Work done by the system = ∆E= Cv∆T.

and Cp-Cv = R

On dividing all the terms by Cv.

p v

v v v

C C R

C C C− =

p

v

C

C= γ

v

R( 1)

C∴ γ − =

and Cv

R

( 1)=

γ −

2

1

T

T

Rw dT

( 1)∴ =

γ − ∫

2 1

Rw (T T )

( 1)= −

γ −and ∆H = ∆E + P∆V.

Thus if T2>T1, w = +ve i.e. work is done on the system.

Thus if T2<T1, w = -ve i.e. work is done by the system.

Limitations of the first law. Need for the second law

A major limitation of the first law of thermodynamics is that its merely indicates that in any

process there is an exact equivalence between the various forms of energies involved, but it

provides no information concerning the spontaneity or feasibility of the process. For example, the

first law does not indicate whether heat can flow from a cold end to a hot end or not.

The answers to the above questions are provided by the second law of thermodynamics.

Spontaneous and non – spontaneous process

If in the expansion of a gas the opposing pressure is infinitesimally smaller than the pressure of the gas, the expansion takes place infinitesimally slowly i.e. reversible. If however, the opposing pressure is much smaller than the pressure of the gas the expansion takes place rapidly i.e. irreversibly. Natural processes are spontaneous and irreversible.

SECOND LAW OF THERMODYNAMICS

The second law of Thermodynamics helps us to determine the direction in which energy can be

transformed. It also helps us to predict whether a given process or chemical reaction can occur

spontaneously or not.

According to Kelvin: “It is impossible to use a cyclic process to extract heat from a reservoir and

to convert it into work without transferring at the same time a certain amount of heat from a hotter

to colder part of the body”.

Entropy Change: Entropy change is the state function and it is the ratio of heat change in a

reversible process by the temperature.

∆S = revq

T

Thermodynamically irreversible process is always accompanied by an increase in the entropy of

the system and its surroundings taken together while in a thermodynamically reversible process,

the entropy of the system and its surroundings taken together remains unaltered.

Illustration 4. Calculate entropy change for vaporization of 1 mole of liquid water to steam at

100°C if ∆HV = 40.8 kJmol−1.

Solution: For entropy change of vaporization

VV

HS

T

∆∆ =

31 1

V

40.8 10S 109.38 Jk mol

373− −×∆ = =

Illustration 5. A system changes its state irreversibly at 300 K in which it absorbs 300 cals of

heat. When the same change is carried out reversibly the amount of heat

absorbed is 900 cals. The change in entropy of the system is equal to

(A) 1 cal K–1 (B) 3 cals K–1

(C) 2 cal K–1 (D) 1.5 cals K–1

Solution: (B)

Physical Significance of Entropy: Entropy is the measure of disorderness because

spontaneous processes are accompanied by increase in entropy as well as increase in the

disorder of the system. Thus, increase in entropy implies increase in disorder.

Illustration 6. Which of the following statement is true?

(i) A closed system shows exchange of mass and not energy with surroundings.

(ii) Entropy change for fusion reaction is positive.

(iii) Heat is a measure of quantity of energy whereas temperature is a measure of

intensity of energy.

Solution: (i) False (ii) True (iii) True

Some Other State Function: For a spontaneous process entropy change is positive and if it is

zero, the system remains in a state of equilibrium. Two other functions are also there to decide

the feasibility of the reactions like work function A and free energy change G.

A = E – TS…….(i)

G = H – TS…….(ii)

And ∆A = ∆E - T∆S……(iii)

∆G = ∆H - T∆S………...(iv) (for a finite change at constant temperature)

Since, ∆S = qrev./T Hence from eq. (i)

∴∆A = ∆E – qrev………………..(v)

and according to first law of Thermodynamics

∆E - qrev = wrev. …………….(vi)

If during the change, work is done by the system, it would carry a negative sign,

-wrev = ∆E – qrev…………….(vii)

Comparing the equation (v) and (vii)

-∆A = wrev

Since the process is carried out reversibly where w represents the maximum work. It is thus clear

that decrease in function A gives maximum work done that can be done by the system during the

given change. The work function A is also called as Helmholtz function.

From equation (iv)

∆G = ∆H - T∆S

and ∆H = ∆E + P∆V

∴∆G = ∆E + P∆V - T∆S

Comparing it with eq. (iii)

∆G = ∆A + P∆V

Since, ∆A is equal to – w, hence.

∆G = - w + P∆V.

- ∆G = w- P∆V

Hence decrease in free energy gives maximum work obtainable from a system other than that

due to change of volume at constant temperature and pressure. This is called as Net Work.

Net Work = w-P∆V = -∆G

The Net Work may be electrical work or chemical work.

Criterion of spontaneity: For a spontaneous process ∆G should be -ve

GIBBS FREE ENERGY

This is another thermodynamic quantity that helps in predicting the spontaneity of a process, is

called Gibbs energy (G).

It is defined mathematically by the equation.

G = H − TS

Where H = heat content, S = entropy of the system, T = absolute temperature

Illustration 7. Which of the following will fit into the blank?

When two phases of the same single substance remain in equilibrium with one

another at a constant P and T, their molar _________ must be equal.

(A) Internal energy (B) Enthalpy

(C) Entropy (D) Free energy

Solution: (D)

Free energy change

For isothermal process.

2 1G G G H T S∆ = − = ∆ − ∆

∆G = change in Gibbs free energy of the system.

It is that thermodynamic quantity of a system the decrease in whose value during a process is

equal to the maximum possible useful work that can be obtained from the system.

Illustration 8. Calculate free energy change when one mole of NaCl is dissolved in water at 25°C. Lattice energy = 700 kJ/mol. S∆ at 25°C = 26.5 1Jmol− , Hydration energy

of NaCl = −696 kJ/mol.

(A) −3.9 kJ (B) −8kJ

(C) −12kJ (D) −16kJ

Solution: (A)

RELATIONSHIP BETWEEN FREE ENERGY AND EQUILIBRIUM CONSTANT

The free energy change of the reaction in any state, ∆G (when equilibrium has not been attained)

is related to the standard free energy change of the reaction, ∆G0 (which is equal to the difference

in free energies of formation of the products and reactants both in their standard states)

according to the equation. 0G G RTlnQ∆ = ∆ +

Where Q is the reaction quotient

When equilibrium is attained, there is no further free energy change i.e. ∆G = 0 and Q becomes

equal to equilibrium constant. Hence the above equation becomes.

( )0

eq.G RTlnK∆ = −

or ( )0

eq.G 2.303 RT logK∆ = −

In case of galvanic cells. Gibbs energy change ∆G is related to the electrical work done by the

cell.

∆G = −nFE(cell) where n = no. of moles of electrons involved

F = the Faraday constant

E = emf of the cell If reactants and products are in their standard states 0 0

cellG nFE∆ = −

Illustration 9. Calculate ∆G0 for conversion of oxygen to ozone ( ) ( )→2 3

3O g O g

2at 298

K, if Kp for this conversion is 2.47 × 10−29.

Solution:0

pG 2.303RTlogK∆ = −

Where R = 8.314 J/K mol, Kp = 2.47 × 10−29, T = 298K

∆G0 = 16300 J/mol = 163 KJ/mol

THIRD LAW OF THERMODYNAMICS

The entropy of a pure crystalline substance increases with increase of temperature, because molecular motion increases with increase of temperature and vice - versa. Or the entropy of a perfectly crystalline solid approaches zero as the absolute zero of temperature is approached. This is third law of thermodynamics.

THERMOCHEMISTRY

The branch of chemistry which deals with energy changes involved in chemical reactions is called

thermochemistry. The energy change that occurs in a chemical reactions is largely due to change

of bond energy.

Change of internal energy in a chemical reaction

Let us consider a chemical reaction taking place at constant temperature and at constant volume.

In such a case, w = 0 and hence from the first law

∆U = qv

Where qv is the heat exchanged at constant volume, or heat or enthalpy of reaction at constant

volume.

Change of Enthalpy in a chemical reaction

Let qP be the heat exchanged in the chemical reaction taking place at constant pressure, Then

evidently,

∆H = qP = Heat or Enthalpy of reaction at constant pressure.

Exothermic and Endothermic reaction

Reaction that give out heat, i.e. which are accompanied by evolution of heat, are called exothermic reaction. In such reactions ∆H is negative. On the other hand, reaction that intake heat, i.e. which are accompanied by absorption of heat are called endothermic reactions. In these reactions ∆H is positive.

Illustration 10. Fill in the blanks with appropriate word in following:

(i) Combustion of reactions are usually ……………………..

(ii) Combustion of F2 in oxygen is ……………………..

Solution: (ii) Exothermic

(iii) Endothermic

Enthalpy of reaction

It is the enthalpy change taking place during the reaction when the number of moles of reactants and products are same as the stoichiometric coefficient indicates in the balanced chemical equation. The enthalpy change of the reaction depends upon the conditions like temperature, pressure etc under which the chemical reaction is carried out. Therefore, it is necessary to select the standard state conditions. According to thermodynamics conventions, the standard state refers to 1 bar pressure and 298 K temperature. The enthalpy change of a reaction at this

standard state conditions is called standard enthalpy of the reaction ( )0H∆ .

Different types of enthalpy

(i) Enthalpy of formation: Enthalpy change when one mole of a given compound is formed

from its elements.

H2(g) + 1/2O2(g) → 2H2O(l), ∆H = –890.36 kJ / mol

Exercise 4.Calculate 0

fHΔ for chloride ion from the following data:

( ) ( ) ( )→2 2

1 1H g + Cl g HCl g H =-92.4 KJΔ

2 2( ) ( ) ( )→ + -

2HCl g +nH O H aq +Cl aq H =-74.8 KJΔ

( )( )0 +fH H aq =0.0 KJΔ

(ii) Enthalpy of combustion: Enthalpy change when one mole of a substance is burnt in

oxygen.

CH4 + 2O2(g) → CO2 + 2H2O(l), ∆H = –890.36 kJ / mol

Exercise 5.

The heat liberated on complete combustion of 7.8 g benzene is 327 KJ. This heat has

been measured at constant volume and at 27°C. Calculate heat of combustion of

benzene at constant pressure at 27°C.

(iii) Enthalpy of Neutralization: Enthalpy change when one equivalent of an acid is neutralized

by a base or vice – versa in dilute solution. This is constant and its value is –13.7 kcal for

neutralization of any strong acid by a base since in dilute solutions they completely dissociate

into ions.

H+ (aq) + OH– (aq) → H2O(l), ∆H = –13.7 kcal

For weak acids and bases, heat of neutralization is different because they are not dissociated

completely and during dissociation some heat is absorbed. So total heat evolved during

neutralization will be less.

e.g. HCN + NaOH → NaCN + H2O, ∆H = –2.9 kcal

Heat of ionization in this reaction is equal to (–2.9 + 13.7) kcal = 10.8 kcal

Illustration 11. Heat of neutralization of a strong acid by a strong base is equal to ∆H of

(A) H+ + OH− → H2O

(B) H2O + H+ → H3O+

(C) 2H2 + O2 = 2H2O

(D) CH3COOH+ NaOH = CH3COONa + H2O

Solution: (A) Since heat of neutralization of strong acid and strong base is equal to the

heat of formation of water.

i.e., NaOH + HCl → NaCl + H2O + Q

Were Q = heat of neutralization

⇒ Na+ + OH– + H+ + Cl– → Na++Cl– + H2O + Q

⇒ H+ + OH– → H2O + Q

(iv) Enthalpy of hydration: Enthalpy of hydration of a given anhydrous or partially hydrated salt

is the enthalpy change when it combines with the requisite no.of mole of water to form a

specific hydrate. For example, the hydration of anhydrous copper sulphate is represented by

CuSO4(s) + 5H2O (l) → CuSO45H2O(s), ∆H° = –18.69 kcal

Illustration 12. Ionisation energy of Al = 5137 kJ mole–1 (∆H) hydration of Al3+ = – 4665 kJ

mole–1. (∆H)hydration for Cl– = – 381 kJ mole–1. Which of the following statement is

correct

(A) AlCl3 would remain covalent in aqueous solution

(B) Only at infinite dilution AlCl3 undergoes ionisation

(C) In aqueous solution AlCl3 becomes ionic

(D) None of these

Solution: If AlCl3 is present in ionic state in aqueous solution, therefore it has Al3+ & 3Cl–

ions

Standard heat of hydration of Al3+ & 3Cl- ions

= – 4665 + 3 × (–381) kJ mole–1 = -5808 kJ/mole

Required energy of ionisation of Al = 5137 kJ mole–1

∴ Hydration energy overcomes ionisation energy

∴ AlCl3 would be ionic in aqueous solution

Hence (C) is the correct answer.

(v) Enthalpy of Transition: Enthalpy change when one mole of a substance is transformed

from one allotropic form to another allotropic form.

C (graphite) → C(diamond), ∆H° = 1.9 kJ/mol

Illustration 13. The heat of transition for carbon from the following is

CDiamond + O2(g) → CO2(g) ∆H = – 94.3 kcal

CAmorphous + O2(g) → CO2(g) ∆H = – 97.6 kcal

(A) 3.3 kJ / mol (B) 3.3 kcal / mol

(C) –3.3 kJ / mol (D) – 3.3 kcal / mol

Solution: Given

CD + O2(g) → CO2(g) ∆H = –94.3 kcal/mole …(1)

CA + O2(g) → CO2(g) ∆H = – 97.6 kcal/mole …(2)

———————————————————————————

Subtracting equation (2) from equation (1):

CD – CA → 0; ∆H = +3.3 kcal/mole

CD → CA ∆H = +3.3 kcal/mole

∴ (B)

Illustration 14. From the reaction P(white) → P (Red): ∆H = - 18.4 kJ, It follows that

(A) Red P is readily formed from white P

(B) White P is readily formed from red P

(C) White P can not be converted to red P

(D) White P can be converted into red P and red P is more stable

Solution: (D)

HESS’S LAW

This law states that the amount of heat evolved or absorbed in a process, including a chemical

change is the same whether the process takes place in one or several steps.

Suppose in a process the system changes from state A to state B in one step and the heat

exchanged in this change is q. Now suppose the system changes from state A to state B in three

steps involving a change from A to C, C to D and finally from D to B. If q1, q2 and q3 are the heats

exchanged in the first, second and third step, respectively then according to Hess’s law

q1 + q2 + q3 = q

Hess’s law is simply a corollary of the first law of thermodynamics. It implies that enthalpy change

of a reaction depends on the initial and final state and is independent of the manner by which the

change is brought about.

Illustration 15.

A ∆H D

B ∆H2

∆H1

C

∆H3

In this case express ∆H in terms of ∆H1, ∆H2, ∆H3 .

Solution: ∆H = ∆H1 + ∆H2 + ∆H3

Illustration 16. H2O (l) → H2(g) + 1

2O2(g) ∆H = + 890.36 kJ / mole

What is ∆H for H2O (l) from its constituent elements

Solution: H2O(l) → H2(g) + 2

1O

2(g) ∆H = + 890.36 kJ / mole

H2(s) + 1

2O2(g) → H2O(l) ∆H = – 890.36 kJ / mole

∴ ( )2

f H O( H∆ = –890.36 kJ / mole

Exercise 6.

(i) C + O2 → CO2(g) ∆H = –94 Kcals

C + 1

2O2 → CO(s) ∆H = –26.4 Kcals

CO + 1

2O2 → CO2(g) ∆H =?

(ii) What is heat evolved using neutralisation of HCN by a strong base? Heat of ionization

of HCN is 10.8 Kcal.

APPLICATION OF HESS’S LAW

1. Calculation of enthalpies of formation

There are large number of compounds such as C6H6, CO, C2H6 etc whose direct synthesis from

their constituent element is not possible. Their ∆H0f values can be determined indirectly by Hess’s

law. e.g. let us consider Hess’s law cycle for CO2 (g) to calculate the ∆H0f of CO(g) which can not

determined otherwise.

( ) ( ) ( ) ( )2 2 1

1C s O g CO g O g H

2+ → + ∆ = ?

( ) ( ) ( )2 2 2

1CO g O g CO g H 283KJ/mole

2+ → ∆ = −

( ) ( ) ( )2 2 3C s O g CO g H 0393KJ/mole+ → ∆ = −

According to Hess’s law,

3 1 2H H H∆ = ∆ + ∆ or 1 3 2H H H∆ = ∆ − ∆

= −393 − (−283) ⇒ −110 KJ/mole

2. Calculation of standard Enthalpies of reactions

From the knowledge of the standard enthalpies of formation of reactants and products the

standard enthalpy of reaction can be calculated using Hess’s law.

According to Hess’s law ( ) ( )0 0 0

f fH P H R H∆ = ∆ + ∆∑ ∑

( ) ( )0 0 0f fH H P H R∴ ∆ = ∆ − ∆∑ ∑

0 sum of s tandard enthalpies of sum of s tandard enthalpiesH

formation of products of formation of reac tants

∴∆ = −

3. In the calculation of bond energies

BOND ENERGY

Bond energy for any particular type of bond in a compound may be defined as the average

amount of energy required to dissociate one mole, viz Avogadro’s number of bonds of that type

present in the compound. Bond energy is also called the enthalpy of formation of the bond.

Calculation:

For diatomic molecules like H2, O2, N2, HCl, HF etc, the bond energies are equal to their

dissociation energies. For polyatomic molecules, the bond energy of a particular bond is found

from the values of the enthalpies of formation. Similarly the bond energies of heteronuclear

diatomic molecules like HCl, HF etc can be obtained directly from experiments or may be

calculated from the bond energies of homonuclear diatomic molecules.

Illustration 17. Calculate the bond energy of HCl. Given that the bond energies of H2 and Cl2 are 430 KJmol−1 and 242 KJ mol−1 respectively and ∆H0

f for HCl is −91 KJ mol−1.

Solution: ( ) ( ) 12H g 2H g H 430 KJmol ...(i)−→ ∆ = +

( ) ( ) 12Cl g 2Cl g H 242 KJmol ...(ii)−→ ∆ = +

( ) ( )HCl H g Cl g H ? ...(iii)→ + ∆ =

For the reaction (iii)

∆H = ( ) ( )0 0f fH product H reac tant∆ − ∆∑ ∑

= ( ) ( ) ( )0 0 0f f fH H H Cl H HCl ∆ + ∆ − ∆ = ( )1 1

430 242 912 2

× + × − −

∆H = 427 KJ mol−1

Illustration 18. Given that 2H2(g) + O2(g) → H2O(g) , ∆H = –115.4 kcal the bond energy of H–H and O = O bond respectively is 104 kcal and 119 kcal, then the O–H bond energy in water vapour is (A) 110.6 kcal / mol (B) –110.6 kcal(C) 105 kcal / mol (D) None

Solution: We know that heat of reaction ∆H = ΣB.E. (reactant) – ΣB.E (product)For the reaction,2H–H(g) + O = O (g) → 2H – O–H(g)∆H = –115.4 kcal, B.E. of H–H = 104 kcalB.E. of O=O = 119 kcalSince one H2O molecule contains two O–H bonds –115.4 = (2 × 104) + 119 – 4 (O–H) bond energy ∴ 4 (O–H) bond energy = (2× 104) +119+115.4

i.e., O–H bond energy =( )2 104 119 115.4

4

× + += 110.6 kcal mol–1

Hence, (A) is correct.

Illustration 19. Given the bond energies of N ≡ N, H – H and N – H bonds are 945, 436 and 391

kJ/mol respectively, the enthalpy of the reaction.

N2(g) + 3H2(g) → 2NH3(g) is

(A) – 93kJ (B) 102kJ

(C) 90kJ (D) 105kJ

Solution: (A)

Exercise 7. Estimate the average S – F bond energy in 6SF . The values of standard enthalpy of

formation of ( ) ( ) ( )6 g g gSF ,S and F are 1100, 275 and 80 KJ/mole respectively.

LATTICE ENERGY OF AN IONIC CRYSTAL (BORN–HABER CYCLE)

M(s) + 2

1X2(g) → MX(s)

Step 1 Step 2

M(g) X(g) Step5 Step 3 Step 4

M+(g) + X–(g)

Born – Haber cycle

The change in enthalpy that occurs when 1 mole of a solid crystalline substance is formed from

its gaseous ions, is known as Lattice energy.

Step 1: Conversion of metal to gaseous atoms

M(s) → M(g) , ∆H1 = sublimation

Step 2: Dissociation of X2 molecules to X atoms

X2(g) → 2X (g), ∆H2 = Dissociation energy

Step 3: Conversion of gaseous metal atom to metal ions by losing electron

M(g) → M+ (g) + e–, ∆H3 = (Ionization energy)

Step 4: X(g) atoms gain an electron to form X– ions

X(g) + e– → X–(g), ∆H4 = Electron affinity

Step 5: M+ (g) and X– (g) get together and form the crystal lattice

M+ (g) + X– (g) → MX(s) ∆H5 = lattice energy

Applying Hess’s law we get

∆H1 + 1/2 ∆H2 + ∆H3 + ∆H4 + ∆H5 = ∆Hf (MX)

On putting the various known values, we can calculate the lattice energy.

Illustration 20. What is the expression of lattice energy (U) of CaBr2 using BornHaber cycle?

Solution: Ca(s) ∆Hf

S

Ca(g)

IE1 + IE2

Ca2+

+ Br2

D

2Br

2E.A

2Br–

CaBr2(s)

2CaBrU

+

= S + IE1 + IE2 + D - 2E.A – 2CaBrU

Illustration 21. What is the relation between ∆H and ∆E in this reaction?

CH4(g) + 2O2(g) CO2(g) + 2H2O(l)

Solution: ∆H = ∆E + ∆nRT

∆n = no. of mole of products - no. of moles of reactants = 1– 3 = –2

∆H = ∆E – 2RT

Illustration 22. What is the expression of lattice energy (U) of CaBr2?

Using Born Haber cycle?

Solution: Ca(s) ∆Hf

S

Ca(g)

IE1 + IE2

Ca2+(g)

+ Br2(g)

D

2Br (g)

2E.A

2Br–(g)

CaBr2(s)

2CaBrU

+

fH∆ = S + IE1 + IE2 + D - 2E.A – 2CaBrU

Illustration 23. The lattice energy of solid NaCl is 180 kcal/mole. The dissolution of the solid in

water in the forms of ions is endothermic to the extent of 1 kcal/mol. If the solution energies of Na+ and Cl− are in the ratio 6:5, what is the enthalpy of

hydration of Na+ ion?

(A) − 85.6 kcal/mol (B) −97.5 kcal/mol

(C) 82.6 kacl/mol (D) +100 kcal/mol

Solution: (B)

Exercise 8.

(i) For a system C(s) + O2(g) = CO2(g) which of the following is correct:

(a) ∆H = ∆E (b) ∆H > ∆E (c) ∆H < ∆E (d) ∆H = 0

(ii) C6H6(l) + 15

2O2(g) → 3H2O + 6CO2(g) ∆H = –3264.4 kJ mole–1.

What is the energy evolved when 7.8 gm of benzene is burnt in air?

BOMB CALORIMETER

The bomb calorimeter used for determining change in internal energy at constant volume if

reaction for the combustion is known than enthalpy of combustion can be estimated by using

formula: ∆H = ∆E + ∆nRT.

– – – – – – – – – – – – – – – –

+ – ← O2

IGNITION WIRES

INSULATING CONTAINER

WATER

STEEL BOMB

SAMPLE

Bomb Calorimeter

This apparatus was devised by Berthelot (1881) to measure the heat of combustion of organic compounds. A modified form of the apparatus shown in Figure consists of a sealed combustion chamber, called a bomb, containing a weighed quantity of the substance in a dish alongwith oxygen under about 20 atm pressure. The bomb is lowered in water contained in an insulated copper vessel. This vessel is provided with a stirrer and a thermometer reading up to 1/100th of a degree. It is also surrounded by an outer jacket to ensure complete insulation from the atmosphere. The temperature of water is noted before the substance is ignited by an electric current. After combustion, the rise in temperature of the system is noted on the thermometer and heat of combustion can be calculated from the heat gained by water and the calorimeter. By knowing the heat capacity of calorimeter and also the rise in temperature, the heat of combustion can be calculated by using the expression

Heat exchange = Z × ∆T

Z–Heat capacity of calorimeter system ∆T– rise in tempHeat changes at constant volumes are expressed in ∆E and Heat changes at constant pressure are expressed in ∆H. Also, ∆H = ∆E + ∆nRT∆n = Moles of gaseous product– Moles of gaseous reactant.

Illustration 24. A sample of 0.16 g CH4 was subjected to combustion at 27°C in a bomb calorimeter. The temperature of the calorimeter system (including water) was found to rise by 0.5°C. Calculate the heart of combustion of methane at (i) constant volume and (ii) constant pressure. The thermal capacity of calorimeter system is 17.7 KJ K−1 & R = 8.314 JK−1mol−1.

Solution: Heat of combustion at constant volume, ∆E = Heat capacity of calorimeter system × rise in temperature

Mol. mass of compound

mass of compound×

= 16

17.7 0.5 8850.16

× × =

∴ ∆E = −885 kJ mol−1

( ) ( ) ( ) ( )4 2 2 2CH g 2O g CO g 2H O+ → + l

∆n = 1 – 3 = −2, T = 300 K, R = 8.314 × 10−3 kJ K−1mol−1

∆H = ∆E + ∆nRT

( ) 3 1 1885 2 8.314 10 kJk mol− − −= − + − × ×

∆H = ∆E + ∆nRT

( ) 3885 2 8.314 10 300−= − + − × × ×

1885 4.988 889.988 kJmol−= − − = −

Application of bond energies

(i) Determination of enthalpies of reactions

Suppose we want to determine the enthalpy of the reaction.

C C

H

H H

H

(g) H H(g) C CH

H H

H H

H (g) ΔH ?=

If bond energies given for C C, C = C, CH, and H H are 347.3, 615.0, 416.2 and 435.1KJ mol−1 respectively.

( )C=C H-H C-H C-C C-HH= H + H +4 H - H +6 HΔ Δ Δ Δ Δ Δ

= (615.0 + 435.1) − (347.3 + 832.4) ⇒ −129.6 KJ

(ii) Determination of enthalpies of formation of compounds

Consider the formation of acetone.

3C(g) 6H(g) O(g) H C C C H

H O H

H H

ΔH ?=

( ) ( )[ ]→

− s gf H-H O-O C-C C-H C=OC C

1H = 3 H + H +3 H 2 H +6 H + HΔ Δ Δ Δ Δ Δ Δ

2

by putting the value of different bond energies you can determine the ∆Hf.

(iii) Determination of resonance energy

If a compound exhibits resonance, there is a considerable difference between the enthalpies of

formation as calculated from bond energies and those determined experimentally. As an example

we may consider the dissociation of benzene. ( ) ( ) ( )6 6C H g 6C g 6H g→ +

Assuming that benzene ring consists of three single and three double bonds (Kekule’s structure)

the calculated dissociation energy comes out to be 5384.1 KJ from bond energies data.

d C C C C C HH 3 H 3 H 6 H− = −∆ = ∆ + ∆ + ∆

The experimental value is known to be 5535.1 KJ/mol. Evidently, the energy required for the

dissociation of benzene is 151 KJ more that the calculated value. The difference of 151 KJ gives

the resonance energy of benzene.

Exercise 9.

Calculate the enthapy of combustion of benzene (l) on the basis of the following.

(i) Resonance energy of benzene (l) = – 152 kJ mole–1

(ii) Enthalpy of hydrogenation of cyclohexene (l) = – 119 kJ mole–1

(iii) (∆Hf0)C6H12 = – 156 kJ mole–1

(iv) (∆H0f)H2O = – 285.8 kJ mole–1

(v) (∆Hf0)CO2 = – 393.5 kJ mole–1

HEAT CAPACITY AND SPECIFIC HEAT

The heat capacity (C) of a sample of substance is the quantity of heat needed to raise the

temperature of the sample of substance by one degree Celsius (or Kelvin).

q = c∆t

Heat capacity is directly proportional to the amount of substance.

The specific heat capacity is the quantity of heat required to raise the temperature of one gram of

a substance by one degree Celsius at constant pressure.

q = s × m × ∆t

where q is the heat required to raise temperature

m = mass in grams

s = specific heat of the substance

∆t = temperature difference

VARIATION OF HEAT OF REACTION WITH TEMPERATURE

The heat of reaction depends on the temperature. The relation between the two is known as

Kirchoff’s equation.

(i) 2 1

2 1

H H

T T

∆ − ∆− = ∆CP (ii)

2 1

2 1

E E

T T

∆ − ∆− = ∆CV

∆CP = molar heat capacity of products – molar heat capacity of reactants (at constant pressure)

∆Cv = molar heat capacity of products – molar heat capacity of reactants (at constant volume)

Illustration 25. The standard heat of formation listed for gaseous NH3 is −11.02 kcal/mol at

298 K. Given that at 298 K, the constant pressure heat capacities of gaseous N2,

H2 and NH3 are respectively 6.96, 6.89, 8.38 cal/mol. Determine ∆H0298K and

∆H773 K for the reactions,

( ) ( ) ( )2 2 3

1 3N g H g NH g

2 2+ →

Solution: ( ) ( ) ( )2 2 3

1 3N g H g NH g

2 2+ →

( ) ( ) ( )298K f FP RH H H 11.02 0∆ = − = − −∑ ∑

= −11.02 kcal mol−1

2 1p

2 1

H HC

T T

∆ − ∆= ∆

−

( )2 3H 11.02 1 3

8.38 6.96 6.89 10773 298 2 2

−∆ − − = − × − × × − ∆H2 = −13.6 kcal mol−1

Exercise 10.

The specific heats of iodine vapour and solid are 0.031 and 0.055 cals/g respectively.

If heat of sublimation of iodine is 24 cals/g at 200°C, Calculate its value at 250 °C.

ANSWERS TO EXERCISES

Exercise 1:4 kJ

Exercise 2:139.7 litre

Exercise 3:2.75

Exercise 4:167.2 KJ−

Exercise 5:3263.76 KJ

Exercise 6:(i) 67.6 kcal (ii) 2.9 kcal

Exercise 8:(i) (a) (ii) 326.44 KJ

Exercise 9:

+ H2 → ∆H = – 119 kJ mole–1

+ H2 → ∆H= 3 × (–119) kJ mole–1

= -357 kJ mole–1

– 357 = (∆Hf0)Product (∆Hf

0)Reactant

= – 156 – (∆Hf0)Benzeene

(∆Hf0)Benzene = (– 156 + 357)KJ mole–1 =201 KJ mole–1

Resonance energy [(∆Hf0)Benzene]Actual – [(∆Hf

0)Benzene]Theoretical

or, – 152 = [(∆Hf0)Benzene]Actual -201

or [(∆Hf0)Benzene]Actual = 49 kJ mole–1

C6H6(l) + 2

15O

2 (g) → 6CO2(g) + 3H2O(l)

(∆H0)reaction = ( ) ( )2 2

0 0f fCO H O

6 H 3 H∆ + ∆ – (∆Hf0)Benzene

= 6 (– 393.5) + 3(–285.8) – 49= 3267.4 kJ mole–1

Exercise 10:Solid I2 is converted in I2 vapourI2(s) = I2(g) ∆H = 24 cal/g at 200°C∆Cp = (CP)I2(g) − (Cp)I2(s) = 0.031 – 0.055 = – 0.24 cal g–1

From Kirchoff’s equation

( )p250 C 200 C 250 C 200 CH H C T T∆ − ∆ = ∆ −o o o o

250 CH∆ o = 24 – 0.024 (523 – 473) = 24 – 1.2 = 22.8 cals g–1

MISCELLANEOUS EXERCISES

Exercise 1: State, whether the following statements are True or False?(i) fusion sub vapH H H∆ = ∆ − ∆(ii) Bond formation is always exothermic(iii) Enthalpy of neutralisation of 4NH OH with HCl is higher than enthalpy of

NaOH with HCl.(iv) Heat of reaction is independent of temperature.(v) Heat of combustion of a fuel = caloric value of fuel

Exercise 2: A solution of 5 gm of Haemoglobin (molecular weight = 64000) in 100 cc of solution shows a temperature raise of o0.031 C for complete oxygenation. Each mole of haemoglobin binds 4 mole of oxygen. If the heat capacity of the solution is 1 34.18 k cm ,− − calculate H∆ per gm mole of oxygen bond.

Exercise 3: At o25 C, the enthalpy change for the reaction

2 4 2 2 4 2H SO 5H O H SO .5H O (all liquids)+ = is -580.32 kJ / mole. Calculate the

temperature change if 1 mole of 2 4H SO is dropped into 5 mole of o

2H O at 25 C.

Assume no heat loss to the surroundings and that the specific heat capacity of solution is 1 14.184 Jk g .− −

Exercise 4: The ofH 41.84 kJ/mol∆ = − for the neutralisation reaction

3 2 3HCO (aq) OH (aq) H O( ) CO (aq)− − −+ → +l

Compute ( )2 6 qC H . for the reaction 2CO

Exercise 5: The of 2H (CaBr (s)) 675 kJ/mole∆ = − . The first and second ionization energies of

Ca are 590 and 1145 kJ / mole. The enthalpy of formation of Ca is 178 kJ / mol. The bond enthalpy of 2Br is 193 kJ / mole and enthalpy of vapourisation of 2Br is 31 kJ / mole. The electron affinity of Br(g) is 325 kJ / mole. Calculate the lattice energy of 2CaBr (s).

Exercise 6: The heats of combustion of yellow phosphorus and red phosphorous are –9.19 kJ and – 8.78 kJ respectively, and then calculate the heat of transition of yellow phosphorus to red phosphorous.

Exercise 7: Calculate the enthalpy change when infinitely dilute solutions of CaCl2 and

Na2CO3 mixed oΔHf for Ca2+ (aq), −2CO3 (aq) and CaCO3(s) are – 129.80,

– 161.65, – 288.5 kcal mole–1 respectively.

Exercise 8: Calculate the heat of formation of ethyl acetate from ethyl alcohol and acetic acid. Given that heat of combustion of ethyl alcohol is 34 kcal and of acetic acid, it is 21 kcal and of ethyl acetate, it is 55.4 kcal.

Exercise 9: For the given heat of reaction,(i) C(s) + O2(g) = CO2(g) + 97 kcal(ii) CO2(g) + C(s) = 2CO(g) – 39 kcalCalculate the heat of combustion of CO(g).

Exercise 10: 1 mole of an ideal gas undergoes reversible isothermal expansion from an initial volume V1 to a final volume 10V1 and does 10 KJ of work. The initial pressure was 1 × 107 Pa.(i) Calculate V1 (i) If there were 2 mole of gas, what must its temperature have been?

ANSWER TO MISCELLANEOUS EXERCISES

Exercise 1: (i) True(ii) True(iii) False(iv) False(v) False

Exercise 2: -41.47 kJ

Exercise 3: o73.7 C

Exercise 4: 14.0 kJ / mol

Exercise 5: -2162 kJ / mole

Exercise 6: (i) P4 (yellow) +5O2(g) → P4O10 + 9.19 kJ(ii) P4(red) + 5O2(g) → P4O10 + 8.78 kJ ——————————————— subtracting, P4(yellow) – P4 (red) = 1.13 kJ

⇒ P4(yellow) = P4(red) + 1.13 kJSo, heat of transition of yellow to red phosphorus is – 1.13 kJ

Exercise 7: On mixing CaCl2 (aq) and Na2CO3

CaCl2 + Na2CO3 → CaCO3↓ + 2NaClSolutions are very dilute and thus 100% dissociation occurs

Ca2+(aq)+2Cl−(aq)+2Na+ (aq) +2

3CO− (aq) → CaCO3↓ + 2Na+(aq) +2Cl−(aq) or Ca2+

(aq)+ + CO32– (aq) → CaCO3(s)

∴ ∆H = ΣH°products – ΣH°reactants

or ∆H = 2 23 3

o o of CaCo f Ca f CO

H [ H H ]+ −∆ − ∆ + ∆

Q ∆H° of a compound = ∆H° formation = –288.5 – (–129.8 – 161.65)= 2.95 kcal/mole

Exercise 8: -400 cals

Exercise 9: Subtracting equation (ii) from equation (i), we get C(s) + O2(g) = CO2(g) + 97 kcalCO2(g) + C(s) = 2CO(g) – 39 kcal————————————————————————or, –CO2(g) + O2(g) = CO2(g) – 2CO(g) + 136 kcalor, 2CO(g) + O2 = 2CO2(g) + 136 kcalor, CO(g) + 1/2 O2(g) = CO2(g) + 68 kcalRequired value = 68 kcal

Exercise 10: (i) 4 34.34 10 m−×(ii) 261.13 K

SOLVED PROBLEMS

Subjective:

Board Type Questions

Prob 1. Why standard entropy of an elementary substance is not zero whereas standard enthalpy of formation is taken as zero?

Sol. A substance has a perfectly ordered arrangement of its constituent particles only at absolute zero. Hence entropy is zero only at absolute zero. Enthalpy of formation is the heat change involved in the formation of one mole of the substance from its elements. An element formed from itself means no heat change.

Prob 2. Out of carbon (diamond) and carbon (graphite), whose enthalpy of formation is taken as zero and why?

Sol. The enthalpy of formation of graphite is taken as zero because it is a more commonly found stable form of carbon.

Prob 3. Justify, an exothermic reaction is always thermodynamically spontaneous.

Sol. Exothermic reactions are generally thermodynamically spontaneous because even if it is accompanied by decrease of randomness, the heat released is absorbed by the surroundings so that the entropy of the surroundings increases to such an extent that

totalS∆ is positive.

Prob 4. Is qp always greater than qv? Explain why or why not?

Sol. qp is not greater than qv always. It depends upon whether ∆ng is +ve or −ve.

Prob 5. Justify, many thermodynamically feasible reactions do not occur under ordinary conditions.

Sol. Under ordinary conditions, the average energy of the reactants may be less than threshold energy. They require some activation energy to initiate the reaction.

IIT Level Questions

Prob 6. The standard heats of formation at 298 K for CCl4 (g), H2O (g), CO2 (g) and HCl (g) are

−25.5, −57.8, −94.1 and −22.1 Kcal/mole. Calculate the 0298H∆ for the reaction.

( ) ( ) ( ) ( )→4 2 2CCl g +2H O g CO g +4HCl g ∆H = ?

Sol. ∆ H = 2 4 2CO HCl CCl H OH 4 H H 2 H ∆ + × ∆ − ∆ + × ∆

= [ ] [ ]94.1 4 22.1 25.5 2 57.8− + × − − + × − ⇒ +135.4 Kcal

Prob 7. The molar heats of combustion of C2H2 (g), C(graphite) and H2 (g) are 310.62 Kcal, 94.05 Kcal and 68.32 Kcal respectively. Calculate the standard heat of formation of C2H2 (g).

Sol. The required equation is

2 2 22C H C H ; H ?+ → ∆ =Writing the thermochemical equation of the given data

2 2 2 2 2

5(i) C H O 2CO H O H 310.62kcal

2+ → + ∆ = −

2 2(ii) C O CO H 94.05kcal+ → ∆ = −

2 2 2

1(iii) H O H O H 68.32kcal

2+ → ∆ = −

(iii) + 2 × (ii) − (i) 2 2 22C H C H H 54.20kcal+ → ∆ =

Prob 8. The molar heat of formation of NH4NO3 (s) is −367.54 KJ and those of N2O (g) and H2O( )l are 81.46 and −285.8 KJ respectively at 25°C and 1 atmospheric pressure.

Calculate ∆H and ∆E of reaction. ( ) ( ) ( )→ l4 3 2 2NH NO s N O g +2H O

Sol. p RH H H∆ = ∆ − ∆

= −122.60 KJ H E nRT∆ = ∆ + ∆

−122600 = ∆E + 1 × 8.314 × 298 ∴ ∆E = −125.07 KJ

Prob 9. Determine the value of ∆H and ∆E for the reversible isothermal evaporation of 90.0 gm of water at 100°C. Assume that water vapour behaves as an ideal gas and heat of evaporation of water is 540 cal/gm.

Sol. ∆H = 90 × 540 = 48.6 Kcal H E P V∆ = ∆ + ∆

Volume of liquid is negligible as compared to volume of vapour So ∆V = Vvapour

H E nRT∆ = ∆ +

∆E = 90

48600 2 37318

− × × = 44.87 kcal

Prob 10. An athelete is given 100 gm glucose of energy equivalent to 15600 kJ. He utilizes 50% of this gained energy in an event. In order to avoid storage of energy in body, Calculate the wt of water he would need to prespire. Enthalpy of 2H O for evaporation is 44 kJ / mole.

Sol. Energy gained by athelete = 1560 kJ

Energy utilized in event 50

1560 780 kJ100

= × =

Energy left = 1560 – 780 = 780 kJSince 44 kJ energy used to evaporate = 18 gm 2H O

∴780 kJ energy used to evaporate 2

18 780319.09 of H O

44

×= =

Prob 11. The specific heat at constant volume for a gas 0.075 cal / g and at constant pressure is 0.125 cal / gm. Calculate:(i) The molecular weight of the gas(ii) Atomicity of gas(iii) Number of atoms in its 1 mole

Sol. (i) Specific heat at constant pressure PC 0.125cal / gm=Specific heat at constant volume, VC 0.075 cal / gm=

As we know, P V

RC C

M− =

or, P V

R 2M 40

C C (0.125 0.075)= = =

− −

(ii) For atomicity, P

V

C 0.125Y 1.66

C 0.075= = =

Hence gas is mono atomic.(iii) Since gas is mono atomic. Hence 1 mole of gas contains 236.023 10 atoms= ×

Prob 12. When 1 mole of ice at o0 C and 4.6 mm of Hg is converted to water vapours at a

constant temperature and pressure. Find H and E,∆ ∆ if the latent heat of fusion of ice is

80 cal / gm and latent heat of vaporisation of liquid water at o0 C is

596 cal / gm. The volume of ice in comparison to volume of vapours may be neglected.

Sol. Ice vapour→

f VH H H∆ = ∆ + ∆80 18 596 18= × + ×

= 12168 cal / moleH E P V∆ = ∆ + ∆V∆ = Volume of vapours at 4.6 mm and o0 C (as iceV neglible)=

Now applying PV = nRT1 8.314 273

P V nRT cal4.18

× ×∆ = = = 543 cal

E H P V∆ = ∆ − ∆ = 12168 – 543 = 11625 cal

Prob 13. Calculate the standard enthalpy of reaction ( ) ( ) ( ) ( )2ZnO s CO g Zn s CO g+ → +

Given ( ) ( )0 1 0 1f f 2H ZnO, s 348.28kJmol ; H CO , g 393.51kJmol− −∆ = − ∆ = −

( )0 1fH CO, g 110.53 kJmol−∆ = −

Sol. We have ( ) ( ) ( ) ( )0 0 0 0 0f f f 2 f fH H Zn, s H CO , g H ZnO,s H CO, g∆ = ∆ + ∆ − ∆ − ∆

= ( ) ( ) ( ){ } 1 10 393.51 348.28 110.53 kJ mol 65.3 kJmol− −+ − − − − − =

Prob 14. Calculate the enthalpy of vaporization for water from the following H2(g) + 1/2 O2(g) → H2O (g) ∆H = – 57.0 kcalH2(g) + 1/2 O2(g) → H2O(l) ∆H = – 68.3 kcalAlso calculate the heat required to change 1 gm H2O (l) to H2O (g).

Sol. H2(g) + 1/2 O2(g) → H2O (g); ∆H = –57.0 kcal ----------------- (1)H2(g) + 1/2 O2(g) → H2O (l); ∆H = – 68.3 kcal------------------(2)Subtracting (2) from (1) H2O (l) → H2O (g) ; ∆H = 11.3 kcal ∴ Enthalpy of vaporization for H2O = 11.3 kcalAlso 18 g H2O requires enthalpy of vaporization = 11. 3 kcal

1 g H2O requires 11.3

18 kcal = 0.628 kcal

Prob 15. The standard enthalpy of combustion of H2, C6H10 and Cyclohexane (C6H12) are – 241, – 3800, – 3920 kJ mole–1 at 25°C respectively. Calculate the heat of hydrogenation of cyclohexene.

Sol. We have to find ∆ H for C6H10 + H2 → C6H12

Given H2 + 1/2 O2 → H2O ∆H = – 241 kJ --------------- (1)

C6H10 + 17

2O2 → 6CO2 + 5H2O ∆H = – 3800 kJ ---------------(2)

C6H12 + 9O2 →6CO2 + 6H2O ∆H = – 3920 kJ ---------------(3)Adding equation (1) and (2) and then subtracting equation 3 C6H10 + H2 → C6H12 ∆H = – 121 kJ∴ Heat of hydrogenation of cylohexene = – 121 kJ

Objective:

Prob 1. A system is taken from state A to state B along two different paths 1 and 2. The heat absorbed and work done by the system along these paths are Q1 and Q2 and W1 and W2 respectively. Then (A) Q1 = Q2 (B) W1 + Q1 = Q2 + W2 (C) W1 = W2 (D) Q1 − W1 = Q2 − W2

Sol. (D)

Prob 2. In which of the following process does the entropy decrease? (A) dissolving of NaCl in water (B) evaporation of water (C) conversion of CO2(g) into dry ice (D) none

Sol. (C)

Prob 3. Calculate the enthalpy change when 50 ml of 0.01 M Ca(OH)2 reacts with 25 ml of 0.01 M HCl. Given that ∆H0

neut of a strong acid and strong base is 140 cal/ equivalent (A) 14.0 cal (B) 35 cal (C) 10.0 cal (D) 7.5 cal

Sol. (B)

Prob 4 In a reversible adiabatic change ∆S is (A) infinity (B) zero (C) equal to CvdT (D) equal to nRln V2/V1

Sol. (B)

Prob 5 At constant temperature and pressure which one of the following statements is correct for the reaction? CO(g) + 1/2O2(g) → CO2(g) (A) ∆H = ∆E(B) ∆H < ∆E(C) ∆H > ∆E(D) ∆H is independent physical state of reactant

Sol. (B)

Prob 6 For the reaction,C7H8(l) + 9O2(g) → 7CO2(g) + 4H2O(l), the calculated heat of reaction is 232 kJ/mol and observed heat of reaction is 50.4 kJ/mol, then the resonance energy is (A) – 182.2 kJ / mol (B) + 182.2 kJ / mol (C) 172 kJ/ mol (D) None

Sol. (A) As we know that,

Resonance energy = ∆H° (observed) – ∆H° (calculated)

= (50.4 – 232.6) kJ / mol

= – 182.2 kJ mol–1

Fill in the Blanks

Prob 7 An isolated system is one which neither shows exchange of ………….. nor ………………. with surroundings.

Sol. (heat, mass)

Prob 8 During fusion, the entropy of the system …………………….

Sol. (increases)

Prob 9 →2 2N + O 2 NO,shows an ………………….. of heat.

Sol. (absorption)

Prob 10 For spontaneous reaction ΔG is ………………………

Sol. (negative)

Prob 11 Bomb calorimeter used for determining change in internal energy at constant……………………

Sol. Volume

True and False

Prob 12. Specific heat is an intensive property.

Sol. True

Prob 13. A thermodynamic equilibrium represents the state when all the three equilibrium (i.e. chemical, thermal and mechanical equilibrium) are attained at a time.

Sol. True

Prob 14. The process is isothermal if temperature of the system remains constant throughout the course of studies.

Sol. True

Prob 15. Rivers flowing from mountain to field shows decrease in entropy.

Sol. False

Prob 16. Enthalpy of combustion at a given temperature is defined as the enthalpy change for the compete combustion of 1 gm of substance.

Sol. False

ASSIGNMENT PROBLEMS

Subjective:

Level- O

1. Why is the enthalpy of sublimation is equal to the sum of enthalpy of fusion and enthalpy of vaporization?

2. Under what conditions of G, H or S,∆ ∆ ∆ a reaction will be spontaneous at all temperatures?

3. Why is the heat of neutralisation of a strong acid by a strong base is constant as 57.0 kJ / mole?

4. When does entropy increases in a reaction.

5. Entropy of the solution is higher than that of pure liquid, why?

6. The G at m. pt.∆ of ice is zero.

7. At temperature T, the endothermic reaction A B→ proceeds almost to completion. Why S is ve?∆ +

8. Why standard heat of formation of diamond is not zero though it is an element?

9. Can the absolute value of internal energy be determined? Why or why not?

10. Same mass of diamond and graphite (both being carbon) are burnt in oxygen. Will the heat produced be same or different? Why?

11. Why in chemical reactions generally heat is either evolved or absorbed?

12. Why is the enthalpy of sublimation equal to the sum of enthalpy of fusion and enthalpy of vaporisation?

13. When an ideal gas expands in vacuum, there is neither absorption nor evolution of heat. Why?

14. Calculate the heat change for the following reaction: ( ) ( ) ( ) ( )4 2 2 2CH g 2O g CO g 2H O+ → + l

0fH∆ for CH4, H2O and CO2 are −17.89, −68.3 and −94.05 kcal/mole

15. The heat of reaction ( ) ( ) ( )2

1C s O g CO g

2+ → at 17°C and at constant volume is −29.29

kcal. Calculate the heat of reaction at constant pressure.

16. Thermochemistry is the study of relationship between heat energy & ………………………..

17. Define Gibbs free energy.

18. Total energy for a reversible isothermal cycle is……………………………

19. Find out the value of equilibrium constant for the following reaction at 298 K.

( ) ( ) ( ) ( )3 2 2 2 22NH g CO g NH CONH aq. H O+ +���� l����

Standard Gibbs energy change ∆r G− at the given temperature is −13.6 kJ/mol.

20. What you can conclude from this graph

Gib

bs

e

ne

rgy

Reactant

Products

Eq

uil

ibri

um

Level – I

1. Which of the following can be determined? Absolute internal energy, absolute enthalpy, absolute entropy

2. Why would you expect a decrease in entropy as a gas condenses into liquid? Compare it with the entropy decrease when a liquid sample is converted into solid.

3. Under what conditions will the reaction occur, if both ∆S and ∆H are positive?

4. Justify, the entropy of a substance increases on going from liquid to vapour state at any temperature.

5. One mole of an ideal gas is heated at constant pressure from 0°C to 100°C. (a) Calculate work done. (b) If the gas were expanded isothermally & reversibly at 0°C from 1 atm to some other pressure Pt, what must be the final pressure if the maximum work is equal to the work involved in (a).

6. Air contains 99% N2 and O2 gases. Then why do not they combine to form NO under the standard conditions? Given that the standard free energy of formation of NO(g) is 86.7 KJ mol−1.

7. Calculate the heat of following reaction

2 2 3N 3H 2NH+ →

Given the bond energies of N ≡ N, H − H and N H bonds are 226, 104 and 93 kcal respectively.

8. When 2 moles of C2H6 are completely burnt 3129 KJ of heat is liberated. Calculate the heat of formation, ∆Hf for C2H6; ∆Hf for CO2 and H2O are −395 and −286 KJ respectively.

9. Calculate the heat of formation of ethane at 25°C. The bond enthalpies for H − H, C − C and C − H bonds are 104.2 kcal, 80 kcal and 99.5 kcal respectively. Heat of vaporization of carbon is 171.7 kcal.

10. Define the following terms:(a) Internal energy (b) Endothermic reaction(c) Hess law (d) Calorific value

11. 5 mole of an ideal gas expand isothermally & reversibly from a pressure of 10 atm to 2 atm at 300 K. What is the largest mass which can be lifted through a height of 1 mitre in this expansion?

12. The equilibrium constant for the reaction:

( ) ( ) ( ) ( )2 2 2CO g H g CO g H O g+ +����� � �� at 298 K is 73. Calculate the value of the standard

free energy change (R = 8.314 JK−1mol−1).

13. An insulated container contains 1 mole of a liquid molar volumes 100 ml at 1 bar. When liquid is steeply passed to 100 bar, volume decrease t0 99 ml, find ∆H & ∆U for the process.

14. AB, A2 & B2 are diatomic molecules. If the bond enthalpies of A2, AB & B2 are in the ratio 1:1:0.5 & the enthalpy of formation of AB from A2 & B2 is −100 KJ mol−1. What is the bond enthalpy of A2?

Level – II

1. Calculate the C H bond energy in methane at 25°C from the data. Heat of formation of methane is −17.9 kcal, heat of vaporization of carbon is 171.1 cal and heat of formation of hydrogen atoms is 52.1 kcal/mol.

2. The heats of combustion of hydrogen, ethane and ethylene are 68.4, 370.4 and 393.5 kcal per molecule respectively. Calculate the energy when ethylene is reduce to ethane.

3. The molar heat of formation of NH4NO3 is 367.54 KJ and those of N2O(g) and H2O ( )l are

81.46 KJ and −285.78 KJ, respectively at 25°C and 1.0 atm pressure. Calculate ∆H and ∆E for the reaction.

( ) ( ) ( )4 3 2 2NH NO s N O g 2H O→ + l

4. Given that ( ) ( ) ( )2 2H g I g 2HI g H 12.46 kcal+ → ∆ = −

( ) ( )2I g 2I g H 35.8 kcal→ ∆ =

( ) ( )2H g 2H g H 103.7 kcal→ ∆ =Calculate the bond energy of H – I

5. Calculate the maximum work done when pressure on 10 g of hydrogen is reduced from 20 to 1 atm at a constant temperature of 273 K. The gas behaves ideally. Calculate Q.

6. Standard heat of formation of HgO(s) at 298 K and at constant pressure is -90.8 kJ / mole. Excess of HgO(s) absorbs 41.84 kJ of heat at constant volume, calculate the amount of Hg that can be obtained at constant volume and 298 K, Atomic weight of Hg = 200.

7. Calculate the heat of formation of anhydrous 2 6Al Cl from

2 6 22Al(s) 6HCl(aq) Al Cl (aq) 3H (g), H 239.76 kcal+ → + ∆ = −

2 2H (g) Cl (g) 2HCl(g), H 44.0 kcal+ → ∆ = −HCl(g) Aq HCl(aq), H 17.32 kcal+ → ∆ = −

2 6 2 6Al Cl (s) Aq Al Cl (aq), H 153.69 kcal+ → ∆ = −

8. Calculate the heat of formation of 2Ag O from following data:

(i) 3 3 2Pb 2AgNO (aq) Pb(NO ) (aq) 2Ag 509 cals+ → + +

(ii) 3 3 2 2PbO 2HNO (aq) Pb(NO ) (aq) H O 178 cals+ → + +

(iii) 2

1Pb O PbO 503 cals

2+ → +

(iv) 2 3 2 3Ag O 2HNO (aq) H O AgNO (aq) 104 cals+ → + +

9. Calculate resonance energy of 3CH COOH from the following data if the observed heat of

formation of 3CH COOH is -439.7 kJ.Bond Energy Heat of atomisation (kJ)C – H = 413 C = 716.7C – C = 348 H = 218.0C = O = 732 O = 249.1C – O = 351O – H = 463

10. For a reaction ( ) ( ) ( )2 2

1M O s 2M s O g

2→ + , ∆H = 30 KJ mol−1 and ∆S = 0.07 KJ K−1 mol−1 at

1 atm. Calculate upto which temperature, the reaction would not be spontaneous.

Objective:

Level – I

1. An intensive property is that property which depends upon

(A) the nature of the substance

(B) the amount of the substance

(C) both the amount and nature of the substance

(D) neither the nature nor the amount of the substance

2. ( ) ( ) ( ) ( )0f 2 2 2H of CO g , CO g , N O g and NO g∆ in KJ/mol are respectively −393, −110, 81 and 34.

Calculate ∆H in KJ of the following reaction. ( ) ( ) ( ) ( )2 2 22NO g 2CO g N O g 3CO g+ → +

(A) 836 (B) 1460

(C) −836 (D) −1460

3. Temperature of 1 mole of a gas is increased by 1° at constant pressure work done is

(A) R (B) 2R

(C) R/2 (D) 3R

4. Which of the following thermodynamic quantities is an outcome of the second law of

thermodynamics?

(A) enthalpy (B) internal energy

(C) work (D) entropy

5. The difference between heats of reaction at constant pressure and constant volume for the

reaction ( ) ( ) ( ) ( )6 6 2 2 22C H 15O g 12CO g 6H O+ → +l l at 25°C in KJ mol−1 is

(A) −7.43 KJmol−1 (B) 7.43 KJmol−1

(C) 2.477 KJmol−1 (D) −2.477 KJmol−1

6. For a reaction at equilibrium (A) 0G G 0∆ = ∆ ≠ (B) 0G 0∆ =(C) 0G G 0∆ = ∆ = (D) 0G 0, G 0∆ = ∆ ≠

7. When 1 mole gas is heated at constant volume, temperature is raised from 298 to 309 K.

Heat supplied to the gas is 500 J. Then which statement is correct?

(A) q = w = 500 J, ∆u = 0 (B) q = ∆u = 500 J, w = 0

(C) q = w = 500 J, ∆u = 0 (D) ∆u = 0, q = w = −500 J

8. In thermodynamics a process is called reversible when

(A) surroundings and system change into each other

(B) there is no boundary between system and surroundings

(C) the surroundings are always in equilibrium with the system

(D) the system changes into the surroundings spontaneously

9. What is true for the reaction? ( ) ( ) ( )5 3 2PCl g PCl g Cl g→ +

(A) H E∆ = ∆ (B) H E∆ > ∆(C) H E∆ < ∆ (D) none

10. Calculate the work done when 1 mol of an ideal gas is compressed reversibly from 1.00 bar

to 5.0 bar at a constant temperature of 300 K

(A) −14.01 KJ (B) 16.02 KJ

(C) 4.01 KJ (D) −8.02 KJ

11. The factor that does not influence the heat of reaction is

(A) the physical state of reactants and products

(B) the temperature

(C) the pressure or volume

(D) the method by which the final products are obtained

12. (∆H − ∆E) for the formation of NH3 from N2 and H2 is

(A) RT (B) 2RT

(C) -RT (D) -2RT

13. ( ) ( )1 atm

vapgA A , H 460.6 cal /mol∆ =l���������� boiling point = 50 k, what is boiling point at 10 atm?

(A) 150 K (B) 75 K

(C) 100 K (D) none

14. Heat of neutralization of H2C2O4 (oxalic acid) is -26 Kcal/mole. The dissociation energy of

(A) H2C2O4 ����� � �� 2H+ + 22 4C O− is

(A) 12.3 Kcal/mole (B) 1.4 Kcal/mole

(C) -13.7 Kcal/mole (D) -1.4 Kcal/mole

15. The heats of combustion of yellow phosphorus and red phosphorous are –9.19 kJ and – 8.78 kJ respectively, then heat of transition of yellow phosphorus to red phosphorous is (A) – 18.69 kJ (B) +1.13 kJ (C) +18.69 kJ (D) 0.41 kJ

16. The enthalpies of formation of organic compounds are conveniently determined from their (A) boiling points (B) melting points (C) enthalpies of neutralization (D) enthalpies of combustion

17. Thermodynamic equilibrium involves(A) chemical equilibrium (B) thermal equation (C) mechanical equation (D) all the three

18. Evaporation of water is (A) a process in which neither heat is evolved nor absorbed (B) a process accompanied by chemical reaction (C) an exothermic change(D) an endothermic change

19. The heat content of the system is called (A) internal energy (B) enthalpy (C) free energy (D) entropy

20. The apparatus used for measuring the heat changes of a reaction is called(B) thermometer (B) a colorimeter (C) a calorimeter (D) none of these

The questions given below consist of statements ‘Assertion’ (A) and ‘Reason’ (R). (a) If both ‘A’ and ‘R’ are correct and ‘R’ is correct reason for ‘A’. (b) If both ‘A’ and ‘R’ are correct but ‘R’ is not the correct explanation for ‘A’. (c) If ‘A’ is true but ‘R’ is false. (d) If both ‘A’ and ‘R’ are false.

21. (A) Enthalpy of graphite is lower than that of diamond. (R) Entropy of graphites lower than that of diamond.

22. (A) When a gas at high pressure expands against vacuum the work done is maximum. (R) Work done in expansion depends upon the pressure inside the gas & increase in volume.

23. (A) Molar entropy of vaporization of water is different form ethanol. (R) Water is more polar than methanol

24. (A) A reaction which is spontaneous & accompanied by decrease of randomness must be exothermic.

(R) All exothermic reaction are accompanied by decrease of randomness.

25. (A) The enthalpy of formation of ( )2H O l is greater than that of H2O(g).

(R) Enthalpy change is negative for the condensation reaction.

( ) ( )2 2H O g H O→ l

Level – II

1. Evaporation of water is a spontaneous process although it(A) Is an exothermic reaction (B) is an endothermic reaction (C) Is a photo chemical reaction (D) proceed without heat loss or heat gain

2. For the two reactions given below

( ) ( ) ( )2 2 2 1

1H g O g H O g X KJ

2+ → +

( ) ( ) ( )2 2 2 2

1H g O g H O X KJ

2+ → +l

Select the correct answer (A) X1 > X2 (B) X1 < X2 (C) X1 = X2 (D) X1 + X2 = 0

3. Which of the following is wrong? (A) change in internal energy of an ideal gas on isothermal expansion is zero (B) in a cyclic process w ≠ Q

(C) for an ideal gas T

H0

P

∂ = ∂ (D) all

4. The heats of neutralization of four acids a, b c and d when neutralized against a common base are 13.7, 9.4, 11.2 and 12.4 Kcal respectively. The weakest among these acids is (A) c (B) b (C) a (D) d

5. The bond energies of C ≡ C, C H, H H and C = C are 198, 98, 103, 145 Kcal respectively. The enthalpy change of the reaction

2 2 2CH CH H CH CH≡ + → = is

(A) −152 Kcal (B) 96 Kcal (C) 48 Kcal (D) −40 Kcal

6. Which plot represents for an exothermic reaction? (A)

H

Time

R

P

(B)

H

Time

R

P

(C)

H

Time

R P

(D)

H

Time

R P

7. The molar heat capacity of water in equilibrium with ice at constant pressure is (A) negative (B) zero (C) infinity (D) 40.45 KJK−1mol−1

8. Enthalpy of 4 2 3

1CH O CH OH

2+ → is negative. If enthalpy of combustion of CH4 and

CH3OH are x and y respectively. Then which relation is correct? (A) x > y (B) x < y (C) x = y (D) x ≥ y

9. When 10 ml of a strong acid are added to 10 ml of an alkali, the temperature rises 5°C. If 100 ml of each liquids are mixed, the temperature rise would be (A) 0.5°C (B) 10°C (C) 7.5°C (D) same

10. X is a metal that forms an oxide X2O

2 2

1 1X O X O ; H 120Kcal

2 4→ + ∆ =

When a sample of metal X reacts with one mole of oxygen, what will be the ∆H in that case? (A) 480 kcal (B) −240 kcal (C) −480 kcal (D) 240 kcal

11. AB, A2 and B2 are diatomic molecules. If the bond enthalpies of A2, AB & B2 are in the ratio 1:1:0.5 and enthalpy of formation of AB from A2 and B2 – 100 kJ/mol–1. What is the bond enthalpy of A2?(A) 400 kJ/mol (B) 200 kJ/mol(C) 100 kJ/mol (D) 300 kJ/mol

12. Which of the following corresponds to the definition of enthalpy of formation at 298 K?(A) C(graphite) + 2H2(g) + 1/2 O2(l) → CH3OH(g)(B) C(diamond) + 2H2(g) + 1/2 O2(g) → CH3OH (l)

(C) 2C(graphite) + 4H2(g) + O2(g) → 2CH3OH (l)(D) C(graphite) + 2H2(g) + 1/2 O2(g) → CH3OH(l)

13. The heat of neutralisation of HCl by NaOH is –12.1kJ/mole, the energy of dissociation of HCl is(A) –43.8 kJ (B) 43.8 kJ(C) 68 kJ (D) –68 kJ

14. The dissociation energy of CH4 and C2H6 are respectively 360 & 620 k. cal/mole. The bond energy of C–C is

(A) 260 kcal/mole (B) 180 kcal/mole(C) 130 kcal/mole (D) 80 kcal/mole

15. Identify the intensive property from the following:(A) Enthalpy (B) Temperature(C) Volume (D) Refractive index

16. Which of the following expression is not correct?

(A) 0 0G nFE∆ = − (B)0 0

eqG RT lnK∆ = −

(C) ( )0 0PE RT / nF lnK= (D) 0

PG G RT lnQ∆ = ∆ +

17. For a reaction ( ) ( )A g B g�������� at equilibrium, the partial pressure of B is found to be one

fourth of the partial pressure of A. The value of ∆G0 of the reaction A → B is (A) RT ln4 (B) −RT ln4 (C) RT log4 (D) −RT log4

18. For an irreversible isothermal expansion of an ideal gas

(A) sys surrS S∆ = ∆ (B) sys surrS S∆ == −∆

(C) sys surrS S∆ > ∆ (D) sys surrS S∆ < ∆

19. ( ) ( )2H g 2H g→

(A) H atom has higher entropy (B) hydrogen molecule has entropy(C) both have same entropy (D) none

20. An endothermic reaction is spontaneous only if (A) the entropy of the surrounding increases (B) entropy of the system increases (C) total entropy decreases (D) none

ANSWERS TO ASSIGNMENT PROBLEMS

Subjective:

Level - O

1. Since sublimation involves the processsHSolid Vapour∆→ ………………………….. (1)

or f vH HSolid Liquid Vapour∆ ∆→ → ……….(2)

By (1) and (2), s f vH H H∆ →∆ + ∆

2. At all temperature conditions,H 0 and S 0, so that G 0.∆ < ∆ > ∆ <

3. It is so because only H+ and OH– react in every case and it is the enthalpy of formation of water.

4. (i) When number of molecules of products are more than number of molecules in the reactants.

(ii) When solid state changes to liquid state.(iii) When liquid state changes to gaseous state.(iv) When solid changes to gaseous state.

5. Solute molecules also set free to move in solution and thus disorder increases.

6. The system: ice water,�������� at m. pt., is in equilibrium and thus G 0.∆ =

7. G H T S;∆ = ∆ − ∆Q since the reaction proceeds to completion that is irreversible or

spontaneous and thus G ve. Therefore, since H ve. H T S ve or S∆ = − ∆ = + ∴ ∆ − ∆ = − ∆ should be +ve.

8. The standard state chosen for carbon is graphite and not diamond.

9. No, because it is the sum of different types of energies and some of which can not be determined?

10. Heat evolved will be different. This is because they have different crystal structure.

11. Because all chemical reactions are accompanied by bond rearrangements. The total bond energy of reactants is not equal to the total bond energy of products hence heat is either evolved or absorbed.

12. Sublimation is a process which is assumed to take place in two step:First a solid converts into liquid and second a liquid converts into vapours hence the sum of enthalpy of fusion and enthalpy of vapourisation is equal to enthalpy of sublimation.

13. In an ideal gas, there are no intermolecular forces of attraction. Hence no energy is required to overcome these forces. Moreover, when a gas expands against vacuum, work done is zero (because Pext = 0). Hence internal energy of the system does not change i.e. there is no absorption or evolution of heat.

14. −212.76 kcal

15. −29.00 kcal

16. Chemical energy

18. Zero

19. 2.4 × 102

20. At equilibrium products are much more in abundance than the reactants (k > > 1).

Level – I

1. Absolute entropy

2. In liquid, the molecules have much less freedom of motion as compared to the molecules of the gas. When a liquid changes into solid the entropy becomes very low because in a solid the molecular motion almost stops (except vibrational motion)

3. G H T S∆ = ∆ − ∆ . For a reaction to occur, G∆ should be negative. If both H∆ and S∆ are positive, G∆ can be negative only if T S H∆ > ∆ in magnitude. Thus either S∆ has large positive value so that even if T is low, T∆S is greater than ∆H or if ∆S is small, T should be high so that T∆S > ∆H.

4. The molecules in the vapour state have greater freedom of movement and hence greater randomness than those in the liquid state. Hence entropy increases in going from liquid to vapour state.

5. (a) Work done during heating of gas form 0°C to 100°C is W = −P∆V = −P(V2 − V1) = −P[(nRT2/P)−(nRT1/P)]

( ) ( )2 1nR T T 1 1.987 373 273= − − = − × × −

= −198.7 cal (b) If work equivalent to 198.7 cal is used for gas at 0°C, causing its isothermal expansion,

from 1 atm to pressure Pt WR = −2.303nRTlog(P1/P2)

( )t198.7 2.303 1.987 273log 1/P− = − × ×

∴ Pt = 0.694 atm

6. Standard free energy of formation (∆G°f) for the reaction

( ) ( ) ( )2 2

1 1N g O g NO g

2 2+ → is positive (= +86.7 KJ mol−1). Hence the reaction is

non – spontaneous under the standard conditions.

7. −20 kcal/mol

8. −83.5 KJ

9. −21 kcal

11. Work done by the system

1 1e 10

2 2

P PnRTlog 2.303nRTlog

P P= − = −

= 3102.303 5 8.314 300 log 20.075 10 J

2− × × × = − ×

Let M be the mass Work done in lifting the mass = Mgh = M × 9.8 × 1 J M × 9.8 = 20.075 × 103 M = 2048.469 kg

12. cG 2.303RTlogK∆ = −o

10G 2.303 8.314 298 log 73∆ = − × ×o

= −10.632 kJ

13. ∆H = 9900 bar ml ∆U = 100 bar ml

14. 400 KJ mol−1

Level – II

1. 99.35 kcal 2. +91.5 kcal

3. −857.64 KJ, −860.1175 KJ 4. 75.98 kcal

5. 8180 cal 6. 93.37 gm

7. −321.99 kcal

8. 68 cals

9. −110.3 kJ / mole

10. T < 428.57 K

Objective:

Level – I

1. D 2. C 3. A

4. D 5. A 6. D

7. B 8. C 9. B

10. C 11. D 12. D

13. C 14. D 15. D

16. D 17. D 18. D

19. B 20. C 21. B

22. D 23. B 24. C

25. A

Level – II

1. B 2. A 3. B

4. B 5. D 6. A

7. C 8. B 9. D

10. C 11. A 12. D

13. C 14. D 15. B & D

16. C 17. A 18. C

19. A 20. B