ncert ch6 chemistry class 11

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C:\Chemistry XI\Unit-6\Unit-6(3)-Lay-6.pmd 9.1.2006 (Final), 10.1.6, 13.1.6, 14.1.6, 17.1.6, 24.1.6, 17.2.6

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THERMODYNAMICS

It is the only physical theory of universal content concerningwhich I am convinced that, within the framework of theapplicability of its basic concepts, it will never be overthrown.

Albert Einstein

After studying thisUnit, you willbe able to

explain the terms : system andsurroundings;

discriminate between close,open and isolated systems;

explain internal energy, workand heat;

state first law ofthermodynamics and expressit mathematically;

calculate energy changes aswork and heat contributions

in chemical systems; explain state functions: U, H.

correlate Uand H;

measure experimentally Uand H;

define standard states forH;

calculate enthalpy changes forvarious types of reactions;

state and apply Hesss law ofconstant heat summation;

differentiate between extensiveand intensive properties;

define spontaneous and non-

spontaneous processes; explain entropy as a

thermodynamic state functionand apply it for spontaneity;

explain Gibbs energy change(G);

establish relationship betweenG and spontaneity, Gandequilibrium constant.

Chemical energy stored by molecules can be released as heat

during chemical reactions when a fuel like methane, cooking

gas or coal burns in air. The chemical energy may also beused to do mechanical work when a fuel burns in an engineor to provide electrical energy through a galvanic cell like

dry cell. Thus, various forms of energy are interrelated and

under certain conditions, these may be transformed fromone form into another. The study of these energy

transformations forms the subject matter of thermodynamics.The laws of thermodynamics deal with energy changes ofmacroscopic systems involving a large number of molecules

rather than microscopic systems containing a few molecules.Thermodynamics is not concerned about how and at whatrate these energy transformations are carried out, but is

based on initial and final states of a system undergoing thechange. Laws of thermodynamics apply only when a systemis in equilibrium or moves from one equilibrium state to

another equilibrium state. Macroscopic properties like

pressure and temperature do not change with time for asystem in equilibrium state. In this unit, we would like toanswer some of the important questions throughthermodynamics, like:

How do we determine the energy changes involved in a

chemical reaction/process? Will it occur or not?

What drives a chemical reaction/process?

To what extent do the chemical reactions proceed?

UNIT 6


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6.1 THERMODYNAMIC STATE

We are interested in chemical reactions and theenergy changes accompanying them. In order

to quantify these energy changes, we need toseparate the system which is underobservations, from remaining part of theuniverse.

6.1.1 The System and the Surroundings

Asystem in thermodynamics refers to thatpart of universe in which observations aremade and remaining universe constitutes thesurroundings. The surroundings includeeverything other than the system. System andthe surroundings together constitute the

universe .The universe = The system + The surroundings

However, the entire universe other than thesystem is not affected by the changes takingplace in the system. Therefore, for all practicalpurposes, the surroundings are that portionof the remaining universe which can interactwith the system. Usually, the region of spacein the neighbourhood of the systemconstitutes its surroundings.

For example, if we are studying the reactionbetween two substances A and B kept in a

beaker, the beaker containing the reactionmixture is the system and the room where thebeaker is kept is the surroundings (Fig. 6.1).

be real or imaginary. The wall that separatesthe system from the surroundings is called

boundary. This is designed to allow us to

control and keep track of all movements ofmatter and energy in or out of the system.

6.1.2 Types of the System

We, further classify the systems according tothe movements of matter and energy in or outof the system.

1. Open System

In an open system, there is exchange of energyand matter between system and surroundings[Fig. 6.2 (a)]. The presence of reactants in an

open beaker is an example of an open system*.

Here the boundary is an imaginary surfaceenclosing the beaker and reactants.

2. Closed System

In a closed system, there is no exchange ofmatter, but exchange of energy is possible between system and the surroundings[Fig. 6.2 (b)]. The presence of reactants in aclosed vessel made of conducting material e.g.,copper or steel is an example of a closedsystem.

Fig. 6.2 Open, closed and isolated systems.

Fig. 6.1 System and the surroundings

Note that the system may be defined byphysical boundaries, like beaker or test tube,or the system may simply be defined by a setof Cartesian coordinates specifying aparticular volume in space. It is necessary tothink of the system as separated from thesurroundings by some sort of wall which may

* We could have chosen only the reactants as system then walls of the beakers will act as boundary.


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3. Isolated System

In an isolated system, there is no exchange ofenergy or matter between the system and the

surroundings [Fig. 6.2 (c)]. The presence ofreactants in a thermos flask or any other closedinsulated vessel is an example of an isolatedsystem.

6.1.3 The State of the System

The system must be described in order to makeany useful calculations by specifyingquantitatively each of the properties such asits pressure (p), volume (V), and temperature(T) as well as the composition of the system.We need to describe the system by specifyingit before and after the change. You would recall

from your Physics course that the state of asystem in mechanics is completely specified ata given instant of time, by the position andvelocity of each mass point of the system. Inthermodynamics, a different and much simplerconcept of the state of a system is introduced.It does not need detailed knowledge of motionof each particle because, we deal with averagemeasurable properties of the system. We specifythe state of the system bystate functions orstate variables.

The state of a thermodynamic system isdescribed by its measurable or macroscopic

(bulk) properties. We can describe the state ofa gas by quoting its pressure (p), volume (V),temperature (T), amount (n) etc. Variables like

p, V, T are called state variables or statefunctions because their values depend onlyon the state of the system and not on how it isreached. In order to completely define the stateof a system it is not necessary to define all theproperties of the system; as only a certainnumber of properties can be variedindependently. This number depends on thenature of the system. Once these minimumnumber of macroscopic properties are fixed,others automatically have definite values.

The state of the surroundings can neverbe completely specified; fortunately it is notnecessary to do so.

6.1.4 The Internal Energy as a StateFunction

When we talk about our chemical systemlosing or gaining energy, we need to introduce

a quantity which represents the total energyof the system. It may be chemical, electrical,mechanical or any other type of energy you

may think of, the sum of all these is the energyof the system. In thermodynamics, we call itthe internal energy, Uof the system, which maychange, when

heat passes into or out of the system,

work is done on or by the system,

matter enters or leaves the system.

These systems are classified accordingly asyou have already studied in section 6.1.2.

(a) Work

Let us first examine a change in internalenergy by doing work. We take a systemcontaining some quantity of water in athermos flask or in an insulated beaker. Thiswould not allow exchange of heat between thesystem and surroundings through itsboundary and we call this type of system asadiabatic. The manner in which the state ofsuch a system may be changed will be calledadiabatic process. Adiabatic process is aprocess in which there is no transfer of heatbetween the system and surroundings. Here,the wall separating the system and the

surroundings is called the adiabatic wall(Fig 6.3).

Fig. 6.3 An adiabatic system which does not permit the transfer of heat through itsboundary.

Let us bring the change in the internalenergy of the system by doing some work onit. Let us call the initial state of the system asstate A and its temperature as T

A. Let the


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internal energy of the system in state A becalled U

A. We can change the state of the system

in two different ways.

One way:We do some mechanical work, say1 kJ, by rotating a set of small paddles andthereby churning water. Let the new state becalled B state and its temperature, as T

B. It is

found that TB

> TA

and the change intemperature, T = T

BT

A. Let the internal

energy of the system in state B be UB

and thechange in internal energy, U=U

BU

A.

Second way:We now do an equal amount(i.e., 1kJ) electrical work with the help of animmersion rod and note down the temperaturechange. We find that the change in

temperature is same as in the earlier case, say,T

BT

A.

In fact, the experiments in the abovemanner were done by J. P. Joule between184050 and he was able to show that a givenamount of work done on the system, no matterhow it was done (irrespective of path) producedthe same change of state, as measured by thechange in the temperature of the system.

So, it seems appropriate to define aquantity, the internal energyU, whose valueis characteristic of the state of a system,

whereby the adiabatic work, wad required tobring about a change of state is equal to thedifference between the value ofUin one stateand that in another state, Ui.e.,

2 1= U U U

Therefore, internal energy, U, of the systemis a state function.

The positive sign expresses that wad

ispositive when work is done on the system.Similarly, if the work is doneby the system,w

ad

will be negative.

Can you name some other familiar statefunctions? Some of other familiar statefunctions are V,p, and T. For example, if webring a change in temperature of the systemfrom 25C to 35C, the change in temperatureis 35C25C = +10C, whether we go straightup to 35C or we cool the system for a fewdegrees, then take the system to the finaltemperature. Thus, Tis a state function and

the change in temperature is independent ofthe route taken. Volume of water in a pond,for example, is a state function, because

change in volume of its water is independentof the route by which water is filled in thepond, either by rain or by tubewell or by both,

(b) Heat

We can also change the internal energy of asystem by transfer of heat from thesurroundings to the system or vice-versawithout expenditure of work. This exchangeof energy, which is a result of temperaturedifference is called heat, q. Let us considerbringing about the same change in temperature(the same initial and final states as before in

section 6.1.4 (a) by transfer of heat throughthermally conducting walls instead ofadiabatic walls (Fig. 6.4).

Fig. 6.4 A system which allows heat transferthrough its boundary.

We take water at temperature, TA

in a

container having thermally conducting walls,

say made up of copper and enclose it in a huge

heat reservoir at temperature, TB. The heat

absorbed by the system (water), q can be

measured in terms of temperature difference ,

TB

TA. In this case change in internal energy,

U= q, when no work is done at constant

volume.The q is positive, when heat is transferred

from the surroundings to the system and q is

negative when heat is transferred from

system to the surroundings.

(c) The general case

Let us consider the general case in which a

change of state is brought about both by


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doing work and by transfer of heat. We write

change in internal energy for this case as:

U= q + w (6.1)

For a given change in state, qand w can

vary depending on how the change is carried

out. However, q+w = Uwill depend only on

initial and final state. It will be independent of

the way the change is carried out. If there is

no transfer of energy as heat or as work

(isolated system) i.e., if w = 0 and q= 0, then

U= 0.

The equation 6.1 i.e., U = q + w is

mathematical statement of the first law ofthermodynamics, which states that

The energy of an isolated system is

constant.

It is commonly stated as the law of

conservation of energy i.e., energy can neither

be created nor be destroyed.

Note:There is considerable difference betweenthe character of the thermodynamic property

energy and that of a mechanical property such

as volume. We can specify an unambiguous

(absolute) value for volume of a system in a

particular state, but not the absolute value of

the internal energy. However, we can measure

only the changes in the internal energy, Uof

the system.

Problem 6.1Express the change in internal energy of

a system when

(i) No heat is absorbed by the system

from the surroundings, but work (w)

is done on the system. What type of

wall does the system have ?

(ii) No work is done on the system, but

q amount of heat is taken out fromthe system and given to the

surroundings. What type of wall does

the system have?

(iii) w amount of work is done by the

system and q amount of heat is

supplied to the system. What type of

system would it be?

Solution

(i) U= wad

, wall is adiabatic

(ii) U= q, thermally conducting walls

(iii) U= q w, closed system.

6.2 APPLICATIONS

Many chemical reactions involve the generationof gases capable of doing mechanical work orthe generation of heat. It is important for us toquantify these changes and relate them to thechanges in the internal energy. Let us see how!

6.2.1 Work

First of all, let us concentrate on the nature ofwork a system can do. We will consider onlymechanical work i.e., pressure-volume work.

For understanding pressure-volume work, let us consider a cylinder whichcontains one mole of an ideal gas fitted with africtionless piston. Total volume of the gas isV

iand pressure of the gas inside is p. If

external pressure ispex

which is greater thanp, piston is moved inward till the pressureinside becomes equal top

ex. Let this change

Fig. 6.5(a) Work done on an ideal gas in acylinder when it is compressed by aconstant external pressure, p

ex

(in single step) is equal to the shadedarea.


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be achieved in a single step and the final volume be Vf . During this compression,suppose piston moves a distance, l and is

cross-sectional area of the piston is A[Fig. 6.5(a)].

then, volume change = l A = V= (VfV

i)

We also know,fo

pressurea

=

Therefore, force on the piston =pex

. AIf w is the work done on the system bymovement of the piston then

w = force distance =pex

. A .l

=pex

. (V) = pexV= p

ex(V

fV

i) (6.2)

The negative sign of this expression isrequired to obtain conventional sign for w,which will be positive. It indicates that in caseof compression work is done on the system.Here (V

fV

i) will be negative and negative

multiplied by negative will be positive. Hencethe sign obtained for the work will be positive.

If the pressure is not constant at everystage of compression, but changes in numberof finite steps, work done on the gas will besummed over all the steps and will be equal

to p V [Fig. 6.5 (b)]

If the pressure is not constant but changesduring the process such that it is alwaysinfinitesimally greater than the pressure of the

gas, then, at each stage of compression, thevolume decreases by an infinitesimal amount,dV. In such a case we can calculate the workdone on the gas by the relation

w= f

i

V

ex

V

p dV ( 6.3)

Here,pex

at each stage is equal to (pin

+ dp) incase of compression [Fig. 6.5(c)]. In anexpansion process under similar conditions,the external pressure is always less than thepressure of the system i.e., p

ex= (p

in dp). In

general case we can write,pex= (pin+ dp). Suchprocesses are called reversible processes.

A process or change is said to bereversible, if a change is brought out insuch a way that the process could, at anymoment, be reversed by an infinitesimalchange. A reversible process proceedsinfinitely slowly by a series of equilibriumstates such that system and thesurroundings are always in nearequilibrium with each other. Processes

Fig. 6.5 (c)pV-plot when pressure is not constantand changes in infinite steps(reversible conditions) during

compression from initial volume, Vitofinal volume, Vf. Work done on the gas

is represented by the shaded area.

Fig. 6.5 (b)pV-plot when pressure is not constantand changes in finite steps during

compression from initial volume, Vito

final volume, Vf. Work done on the gasis represented by the shaded area.


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other than reversible processes are knownas irreversible processes.

In chemistry, we face problems that can

be solved if we relate the work term to theinternal pressure of the system. We canrelate work to internal pressure of the systemunder reversible conditions by writingequation 6.3 as follows:

w = f

i

V

rev ex

V

p

Since dp dVis very small we can write

w = f

i

V

rev in

V

p d (6.4)

Now, the pressure of the gas (pinwhich we

can write aspnow) can be expressed in termsof its volume through gas equation. Fornmolof an ideal gas i.e.,pV=nRT

R =

n Tp

V

Therefore, at constant temperature (isothermalprocess),

revw R=

f

i

V

V

n

= 2.303 nRTlogf

i

V

V(6.5)

Free expansion: Expansion of a gas invacuum (p

ex= 0) is called free expansion. No

work is done during free expansion of an idealgas whether the process is reversible orirreversible (equation 6.2 and 6.3).

Now, we can write equation 6.1 in numberof ways depending on the type of processes.

Let us substitute w = pex

V(eq. 6.2) inequation 6.1, and we get

= exU q p

If a process is carried out at constant volume(V= 0), then

U= qV

the subscript V in qV

denotes that heat issupplied at constant volume.

Isothermal and free expansion of anideal gas

For isothermal (T= constant)expansion of an

ideal gas into vacuum ; w = 0 since pex= 0.Also, Joule determined experimentally thatq= 0; therefore, U= 0

Equation 6.1, w = +U q can beexpressed for isothermal irreversible andreversible changes as follows:

1. For isothermal irreversible change

q= w =pex

(Vf

Vi)

2. For isothermal reversible change

q= w = nRTlnf

i

V

V

= 2.303 nRTlogf

i

V

V

3. For adiabatic change, q= 0,

U= wad

Problem 6.2

Two litres of an ideal gas at a pressure of10 atm expands isothermally into avacuum until its total volume is 10 litres.How much heat is absorbed and how

much work is done in the expansion ?Solution

We have q= w = pex

(10 2) = 0(8) = 0

No work is done; no heat is absorbed.

Problem 6.3

Consider the same expansion, but thistime against a constant external pressureof 1 atm.

Solution

We have q= w = pex

(8) = 8 litre-atm

Problem 6.4Consider the same expansion, to a finalvolume of 10 litres conducted reversibly.

Solution

We have q= w = 2.303 1010

log2

= 16.1 litre-atm


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6.2.2 Enthalpy, H

(a) A useful new state function

We know that the heat absorbed at constant

volume is equal to change in the internalenergy i.e., U = q

V. But most of chemical

reactions are carried out not at constant volume, but in flasks or test tubes underconstant atmospheric pressure. We need todefine another state function which may besuitable under these conditions.

We may write equation (6.1) as

= pU q p at constant pressure, where

qp

is heat absorbed by the system and pVrepresent expansion work done by the system.

Let us represent the initial state bysubscript 1 and final state by 2

We can rewrite the above equation as

U2U

1= q

pp(V

2V

1)

On rearranging, we get

qp

= (U2+pV

2) (U

1+ pV

1) (6.6)

Now we can define another thermodynamicfunction, the enthalpy H [Greek wordenthalpien, to warm or heat content] as :

H= U+pV (6.7)

so, equation (6.6) becomes

qp= H

2H

1= H

Although qis a path dependent function,His a state function because it depends on U,

p and V, all of which are state functions.Therefore, His independent of path. Hence,q

pis also independent of path.

For finite changes at constant pressure, wecan write equation 6.7 as

H= U+ pV

Sincepis constant, we can write

H= U+pV (6.8)

It is important to note that when heat isabsorbed by the system at constant pressure,we are actually measuring changes in theenthalpy.

Remember H= qp, heat absorbed by the

system at constant pressure.

His negative for exothermic reactionswhich evolve heat during the reaction and

H is positive for endothermic reactionswhich absorb heat from the surroundings.

At constant volume (V = 0), U = qV,

therefore equation 6.8 becomesH= U= q

V

The difference between Hand Uis notusually significant for systems consisting ofonly solids and / or liquids. Solids and liquidsdo not suffer any significant volume changesupon heating. The difference, however,becomes significant when gases are involved.Let us consider a reaction involving gases. IfVAis the total volume of the gaseous reactants,VB is the total volume of the gaseous products,n

A

is the number of moles of gaseous reactantsand n

Bis the number of moles of gaseous

products, all at constant pressure andtemperature, then using the ideal gas law, we

write,

pVA

= nART

and pVB

= nBRT

Thus, pVB

pVA

= nBRTn

ART= (n

Bn

A)RT

or p(VB

VA) = (n

Bn

A) RT

or pV= ngRT (6.9)

Here, ngrefers to the number of moles of

gaseous products minus the number of molesof gaseous reactants.

Substituting the value ofpV fromequation 6.9 in equation 6.8, we get

H= U+ ngRT (6.10)

The equation 6.10 is useful for calculatingHfrom Uand vice versa.

Problem 6.5

If water vapour is assumed to be a perfectgas, molar enthalpy change forvapourisation of 1 mol of water at 1barand 100C is 41kJ mol1. Calculate theinternal energy change, when

(i) 1 mol of water is vaporised at 1 barpressure and 100C.

(ii) 1 mol of water is converted into ice.


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Solution

(i) The change ( ) 2 2H O Hl

= + H U

orU= H R gn T , substituting the

values, we get

41.00 kJ

8.3 J

=

U

= 41.00 kJ m

= 37.904 kJ mol1

(ii) The change ( )2 2H O Hl

There is negligible change in volume,So, we can put R = gp V n in this

case,

H U

so, 41. =U

(b) Extensive and Intensive Properties

In thermodynamics, a distinction is madebetween extensive properties and intensiveproperties. An extensive property is aproperty whose value depends on the quantityor size of matter present in the system. Forexample, mass, volume, internal energy,enthalpy, heat capacity, etc. are extensiveproperties.

Those properties which do not depend onthe quantity or size of matter present areknown as intensive properties. For exampletemperature, density, pressure etc. are

intensive properties. A molar property, m, is

the value of an extensive property of the

system for 1 mol of the substance. Ifnis the

amount of matter, m =n

is independent of

the amount of matter. Other examples aremolar volume, V

mand molar heat capacity, C

m.

Let us understand the distinction betweenextensive and intensive properties byconsidering a gas enclosed in a container ofvolume Vand at temperature T[Fig. 6.6(a)].Let us make a partition such that volume is

halved, each part [Fig. 6.6 (b)] now has one

half of the original volume,2

V, but the

temperature will still remain the same i.e., T.It is clear that volume is an extensive propertyand temperature is an intensive property.

Fig. 6.6(a) A gas at volume V and temperature T

Fig. 6.6 (b) Partition, each part having half thevolume of the gas

(c) Heat Capacity

In this sub-section, let us see how to measureheat transferred to a system. This heat appearsas a rise in temperature of the system in caseof heat absorbed by the system.

The increase of temperature is proportionalto the heat transferred

= q coeff T

The magnitude of the coefficient depends

on the size, composition and nature of the

system. We can also write it as q= CT

The coefficient, Cis called the heat capacity.

Thus, we can measure the heat supplied

by monitoring the temperature rise, provided

we know the heat capacity.

When Cis large, a given amount of heatresults in only a small temperature rise. Water

has a large heat capacity i.e., a lot of energy is

needed to raise its temperature.

C is directly proportional to amount of

substance. The molar heat capacity of a

substance, Cm=

C

n, is the heat capacity for


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one mole of the substance and is the quantityof heat needed to raise the temperature of onemole by one degree celsius (or one kelvin).

Specific heat, also called specific heat capacityis the quantity of heat required to raise thetemperature of one unit mass of a substance by one degree celsius (or one kelvin). Forfinding out the heat, q, required to raise thetemperatures of a sample, we multiply thespecific heat of the substance, c, by the massm, and temperatures change, T as

= q c m T (6.11)

(d) The relationship between Cp

and CVfor

an ideal gas

At constant volume, the heat capacity, C isdenoted byC

Vand at constant pressure, this

is denoted byCp

. Let us find the relationshipbetween the two.

We can write equation for heat, q

at constant volume as qV= =

VC T U

at constant pressure as qp

= = pC T H

The difference between Cp

and CV

can bederived for an ideal gas as:

For a mole of an ideal gas, H= U+ (pV)

= U+ (RT)= U+ RT

= +H U (6.12)

On putting the values of H and U,we have

= p VC T C T

R= +p VC C

R =p VC C (6.13)

6.3 MEASUREMENT OF U AND H:CALORIMETRY

We can measure energy changes associated with chemical or physical processes by anexperimental technique called calorimetry. Incalorimetry, the process is carried out in avessel called calorimeter, which is immersedin a known volume of a liquid. Knowing the Fig. 6.7 Bomb calorimeter

heat capacity of the liquid in which calorimeteris immersed and the heat capacity ofcalorimeter, it is possible to determine the heat

evolved in the process by measuringtemperature changes. Measurements aremade under two different conditions:

i) at constant volume, qV

ii) at constant pressure, qp

(a)U measurements

For chemical reactions, heat absorbed atconstant volume, is measured in a bombcalorimeter (Fig. 6.7). Here, a steel vessel (thebomb) is immersed in a water bath. The wholedevice is called calorimeter. The steel vessel isimmersed in water bath to ensure that no heat

is lost to the surroundings. A combustiblesubstance is burnt in pure dioxygen suppliedin the steel bomb. Heat evolved during thereaction is transferred to the water around thebomb and its temperature is monitored. Sincethe bomb calorimeter is sealed, its volume doesnot change i.e., the energy changes associated with reactions are measured at constantvolume. Under these conditions, no work is


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done as the reaction is carried out at constant volume in the bomb calorimeter. Even forreactions involving gases, there is no work

done as V= 0. Temperature change of thecalorimeter produced by the completedreaction is then converted to q

V, by using the

known heat capacity of the calorimeter withthe help of equation 6.11.

(b)H measurements

Measurement of heat change at constantpressure (generally under atmosphericpressure) can be done in a calorimeter shownin Fig. 6.8. We know that = pH q (atconstantp) and, therefore, heat absorbed orevolved, q

pat constant pressure is also called

the heat of reaction or enthalpy of reaction, rH.In an exothermic reaction, heat is evolved,

and system loses heat to the surroundings.Therefore, qpwill be negative and rHwill also be negative. Similarly in an endothermicreaction, heat is absorbed, q

pis positive and

rHwill be positive.

Problem 6.6

1g of graphite is burnt in a bombcalorimeter in excess of oxygen at 298 Kand 1 atmospheric pressure according tothe equation

C (graphite) + O2(g) CO2 (g)

During the reaction, temperature risesfrom 298 K to 299 K. If the heat capacityof the bomb calorimeter is 20.7kJ/K,what is the enthalpy change for the abovereaction at 298 K and 1 atm?

Solution

Suppose q is the quantity of heat fromthe reaction mixture and C

Vis the heat

capacity of the calorimeter, then thequantity of heat absorbed by thecalorimeter.

q= CV T

Quantity of heat from the reaction willhave the same magnitude but oppositesign because the heat lost by the system(reaction mixture) is equal to the heatgained by the calorimeter.

=

Vq C T =

=

(Here, negative sign indicates theexothermic nature of the reaction)

Thus, Ufor the combustion of the 1g ofgraphite = 20.7 kJK1

For combustion of 1 mol of graphite,

112.0 g mol=

1

= 2.48 102 kJ mol1

6.4 ENTHALPY CHANGE, rH OF A

REACTION REACTION ENTHALPY

In a chemical reaction, reactants are convertedinto products and is represented by,

Reactants Products

The enthalpy change accompanying areaction is called the reaction enthalpy. Theenthalpy change of a chemical reaction, isgiven by the symbol rH

Fig. 6.8 Calorimeter for measuring heat changesat constant pressure (atmospheric

pressure).


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rH= (sum of enthalpies of products) (sum

of enthalpies of reactants)

ai p ro du ct s

i H= (6.14)

(Here symbol (sigma) is used forsummation and a

iand b

iare the stoichiometric

coefficients of the products and reactantsrespectively in the balanced chemicalequation. For example, for the reaction

CH4

(g) + 2O2

(g) CO2(g) + 2H

2O (l)

ar i p

i

H H =

= [Hm

(CO2,g) + 2H

m(H

2O, l)] [H

m(CH

4, g)

+ 2Hm

(O2, g)]

where Hm

is the molar enthalpy.

Enthalpy change is a very useful quantity.Knowledge of this quantity is required whenone needs to plan the heating or coolingrequired to maintain an industrial chemicalreaction at constant temperature. It is alsorequired to calculate temperature dependenceof equilibrium constant.

(a) Standard enthalpy of reactions

Enthalpy of a reaction depends on the

conditions under which a reaction is carriedout. It is, therefore, necessary that we mustspecify some standard conditions. Thestandard enthalpy of reaction is theenthalpy change for a reaction when allthe participating substances are in theirstandard states.

The standard state of a substance at aspecified temperature is its pure form at1 bar. For example, the standard state of liquidethanol at 298 K is pure liquid ethanol at

1 bar; standard state of solid iron at 500 K is

pure iron at 1 bar. Usually data are taken at298 K.

Standard conditions are denoted by adding

the superscriptV to the symbol H, e.g., H V

(b) Enthalpy changes during phase

transformations

Phase transformations also involve energychanges. Ice, for example, requires heat for

melting. Normally this melting takes place atconstant pressure (atmospheric pressure) andduring phase change, temperature remains

constant (at 273 K).

2 2H O(s) H

Here fus

H0

is enthalpy of fusion in standardstate. If water freezes, then process is reversedand equal amount of heat is given off to thesurroundings.

The enthalpy change that accompaniesmelting of one mole of a solid substancein standard state is called standardenthalpy of fusion or molar enthalpy offusion,

fusH

0.

Melting of a solid is endothermic, so allenthalpies of fusion are positive. Water requiresheat for evaporation. At constant temperatureof its boiling pointT

band at constant pressure:

2 2H O(l) H O

vap

H0

is the standard enthalpy of vaporization.

Amount of heat required to vaporize

one mole of a liquid at constant

temperature and under standard pressure

(1bar) is called its standard enthalpy of

vaporization or molar enthalpy ofvaporization,

vapH

0.

Sublimation is direct conversion of a solid

into its vapour. Solid CO2 or dry ice sublimes

at 195K with sub

H0

=25.2 kJ mol1 ;

naphthalene sublimes slowly and for this

73.sub

H =V .

Standard enthalpy of sublimation,

sub

H0

is the change in enthalpy when one

mole of a solid substance sublimes at a

constant temperature and under standard

pressure (1bar). The magnitude of the enthalpy change

depends on the strength of the intermolecular

interactions in the substance undergoing the

phase transfomations. For example, the strong

hydrogen bonds between water molecules hold

them tightly in liquid phase. For an organic

liquid, such as acetone, the intermolecular


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166

dipole-dipole interactions are significantly

weaker. Thus, it requires less heat to vaporise

1 mol of acetone than it does to vaporize 1 mol

of water. Table 6.1 gives values of standard

enthalpy changes of fusion and vaporisation

for some substances.

Problem 6.7

A swimmer coming out from a pool iscovered with a film of water weighingabout 18g. How much heat must besupplied to evaporate this water at298 K ? Calculate the internal energy ofvaporisation at 100C.

vapHV

for water

at 373K = 40.66 kJ mol1

Solution

We can represent the process ofevaporation as

2

18g H O(l)

No. of moles in 18 g H2O(l) is

1

18g

18gmol= =

= vap vap U H

(assuming steam behaving as an idealgas).

vap gn

(1)(8.314

HV

40.6

37.5

=

=

vapUV

(c) Standard enthalpy of formation

The standard enthalpy change for the

formation of one mole of a compound fromits elements in their most stable states ofaggregation (also known as referencestates) is called Standard Molar Enthalpyof Formation. Its symbol is

fH0, where

the subscript f indicates that one mole of

the compound in question has been formed inits standard state from its elements in theirmost stable states of aggregation. The reference

state of an element is its most stable state ofaggregation at 25C and 1 bar pressure.For example, the reference state of dihydrogen

is H2 gas and those of dioxygen, carbon andsulphur are O

2gas, C

graphiteand S

rhombic

respectively. Some reactions with standardmolar enthalpies of formation are given below.

2 2H (g ) O (

285

+

= fHV

C (graphite, s) 22H (g) C+

Table 6.1 Standard Enthalpy Changes of Fusion and Vaporisation

(Tfand T

bare melting and boiling points, respectively)


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167

74.fH = V

(2C graphite,s

It is important to understand that a

standard molar enthalpy of formation, fH

0

,is just a special case ofrH

0

, where one moleof a compound is formed from its constituentelements, as in the above three equations, where 1 mol of each, water, methane andethanol is formed. In contrast, the enthalpychange for an exothermic reaction:

2CaO(s) CO (+

178fH = V

is not an enthalpy of formation of calciumcarbonate, since calcium carbonate has beenformed from other compounds, and not fromits constituent elements. Also, for the reactiongiven below, enthalpy change is not standardenthalpy of formation,

fH

0

for HBr(g).

2 2H ( g) Br ( )l+

Here two moles, instead of one mole of theproduct is formed from the elements, i.e.,

2r fH H = V

.

Therefore, by dividing all coefficients in the balanced equation by 2, expression for

Table 6.2 Standard Molar Enthalpies of Formation (f

H) at 298K of a

Few Selected Substances


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168

enthalpy of formation of HBr (g) is written as

2 H ( g ) Br +

Standard enthalpies of formation of somecommon substances are given in Table 6.2.

By convention, standard enthalpy forformation,

fH

0

, of an element in referencestate, i.e., its most stable state of aggregationis taken as zero.

Suppose, you are a chemical engineer andwant to know how much heat is required todecompose calcium carbonate to lime andcarbon dioxide, with all the substances in theirstandard state.

3CaCO (s)

Here, we can make use of standard enthalpyof formation and calculate the enthalpychange for the reaction. The following generalequation can be used for the enthalpy changecalculation.

ar i f

i

H H = V

(6.15)where a and b represent the coefficients of theproducts and reactants in the balanced

equation. Let us apply the above equation fordecomposition of calcium carbonate. Here,coefficients a and b are 1 each.

Therefore,

= [ r fH HV V

=1( 635.1 kJ

= 178.3 kJ mol1

Thus, the decomposition of CaCO3(s) is an

endothermic process and you have to heat itfor getting the desired products.

(d) Thermochemical equations

A balanced chemical equation together withthe value of its

rHis called a thermochemical

equation. We specify the physical state

(alongwith allotropic state) of the substance inan equation. For example:

2 5C H OH( ) 3l

+

The above equation describes the

combustion of liquid ethanol at constant

temperature and pressure. The negative sign

of enthalpy change indicates that this is an

exothermic reaction.

It would be necessary to remember the

following conventions regarding thermo-

chemical equations.

1. The coefficients in a balanced thermo-

chemical equation refer to the number ofmoles (never molecules) of reactants and

products involved in the reaction.

2. The numerical value ofrH

0

refers to the

number of moles of substances specified

by an equation. Standard enthalpy change

rH

0

will have units as kJ mol1.

To illustrate the concept, let us consider

the calculation of heat of reaction for thefollowing reaction :

( )2 3Fe O s 3+

From the Table (6.2) of standard enthalpy offormation (

fH

0

), we find :

( )2H O,fH lV

= 285.83 kJ mol1;

( 2 3F e O ,fHV

= 824.2 kJ mol1;

Also (F

(H

f

f

H

H

V

V

Then,

1

rHV

= 3(285.83 kJ mol1)

1( 824.2 kJ mol1)

= (857.5 + 824.2) kJ mol1

= 33.3 kJ mol1

Note that the coefficients used in thesecalculations are pure numbers, which areequal to the respective stoichiometriccoefficients. The unit for

rH

0

is


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169

kJ mol1, which means per mole of reaction.Once we balance the chemical equation in aparticular way, as above, this defines the mole

of reaction. If we had balanced the equationdifferently, for example,

( )2 31

Fe O s2

+

then this amount of reaction would be onemole of reaction and

rH

0

would be

(23

= 22

rH

V

(1

824.2k2

= ( 428.7 + 412.1) kJ mol1

= 16.6 kJ mol1 = 1rHV

It shows that enthalpy is an extensive quantity.

3. When a chemical equation is reversed, the value of

rH

0

is reversed in sign. Forexample

(2 2N (g) 3H g+

3 22NH (g) N

(e) Hesss Law of Constant HeatSummation

We know that enthalpy is a state function,therefore the change in enthalpy isindependent of the path between initial state(reactants) and final state (products). In otherwords, enthalpy change for a reaction is thesame whether it occurs in one step or in aseries of steps. This may be stated as followsin the form of Hesss Law.

If a reaction takes place in several stepsthen its standard reaction enthalpy is thesum of the standard enthalpies of theintermediate reactions into which theoverall reaction may be divided at the sametemperature.

Let us understand the importance of thislaw with the help of an example.

Consider the enthalpy change for thereaction

(C graphite,s

Although CO(g) is the major product, some

CO2

gas is always produced in this reaction.

Therefore, we cannot measure enthalpy change

for the above reaction directly. However, if we

can find some other reactions involving related

species, it is possible to calculate the enthalpy

change for the above reaction.

Let us consider the following reactions:

(C graphite,s

(i)

( ) 21

CO g O2

+(ii)

We can combine the above two reactions

in such a way so as to obtain the desired

reaction. To get one mole of CO(g) on the right,

we reverse equation (ii). In this, heat is

absorbed instead of being released, so we

change sign ofr

H0

value

( )2CO g CO

283.rH =+V (iii)

Adding equation (i) and (iii), we get thedesired equation,

(C graphite,s

for which (= 3rH V

= 110.5 kJ mol1

In general, if enthalpy of an overall reaction

AB along one route is rHand

rH

1,

rH

2,

rH

3..... representing enthalpies of reactions

leading to same product, B along another

route,then we have

rH=

rH

1+

rH

2+

rH

3... (6.16)


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170

6.5 ENTHALPIES FOR DIFFERENT TYPESOF REACTIONS

It is convenient to give name to enthalpiesspecifying the types of reactions.

(a) Standard enthalpy of combustion(symbol :

cH

0)

Combustion reactions are exothermic innature. These are important in industry,rocketry, and other walks of life. Standardenthalpy of combustion is defined as theenthalpy change per mole (or per unit amount)of a substance, when it undergoes combustionand all the reactants and products being intheir standard states at the specifiedtemperature.

Cooking gas in cylinders contains mostlybutane (C

4H

10). During complete combustion

of one mole of butane, 2658 kJ of heat isreleased. We can write the thermochemical

reactions for this as:

4 10

13C H (g)

2+

Similarly, combustion of glucose gives out2802.0 kJ/mol of heat, for which the overallequation is :

6 12 6C H O (g )+

Our body also generates energy from foodby the same overall process as combustion,although the final products are produced aftera series of complex bio-chemical reactionsinvolving enzymes.

Problem 6.8

The combustion of one mole of benzenetakes place at 298 K and 1 atm. After

combustion, CO2(g) and H

2O (1) are

produced and 3267.0 kJ of heat isliberated. Calculate the standard enthalpy

of formation, fH0

of benzene. Standardenthalpies of formation of CO

2(g) and

2H O(l) are 393.5 kJ mol1 and 285.83

kJ mol1 respectively.

Solution

The formation reaction of benezene isgiven by :

( )6C graphite

The enthalpy of combustion of 1 mol ofbenzene is :

( )6 615

C H l2

cH

+

The enthalpy of formation of 1 mol ofCO

2(g) :

( )C graphite

fH

+

The enthalpy of formation of 1 mol of

H2O(l) is :

( ) (2 21

H g O2

fH

+

= V

multiplying eqn. (iii) by 6 and eqn. (iv)by 3 we get:

( )6C graphite

( )2 23

3H g O2+

Summing up the above two equations :

( )6C graphite +

It can be represented as:

A B

C D

rH

1

rH

2

rH

3

rH


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171

(i) Bond dissociation enthalpy

(ii) Mean bond enthalpy

Let us discuss these terms with reference

to diatomic and polyatomic molecules.

Diatomic Molecules:Consider the followingprocess in which the bonds in one mole ofdihydrogen gas (H

2) are broken:

H2(g) 2H(g) ; HHH0

= 435.0 kJ mol1

The enthalpy change involved in this processis the bond dissociation enthalpy of HH bond.The bond dissociation enthalpy is the changein enthalpy when one mole of covalent bondsof a gaseous covalent compound is broken toform products in the gas phase.

Note that it is the same as the enthalpy ofatomization of dihydrogen. This is true for alldiatomic molecules. For example:

Cl2(g) 2Cl(g) ; CClH0

= 242 kJ mol1

O2(g) 2O(g) ; O=OH0

= 428 kJ mol1

In the case of polyatomic molecules, bonddissociation enthalpy is different for differentbonds within the same molecule.

Polyatomic Molecules:Let us now considera polyatomic molecule like methane, CH

4. The

overall thermochemical equation for its

atomization reaction is given below:

4CH (g) C(g

In methane, all the four C H bonds areidentical in bond length and energy. However,the energies required to break the individualC H bonds in each successive step differ :

4 3CH (g) CH (

3 2CH (g) CH (

2CH (g) CH(

CH(g) C(g)

Therefore,

4CH (g) C(g

In such cases we use mean bond enthalpyof C H bond.

321 = fHV

Reversing equation (ii);

( )2 26CO g 3H

fH

+

V

Adding equations (v) and (vi), we get

( )6C graphite

(b) Enthalpy of atomization(symbol:

aH

0)

Consider the following example of atomizationof dihydrogen

H2(g) 2H(g);

aH

0

= 435.0 kJ mol1

You can see that H atoms are formed by breaking HH bonds in dihydrogen. Theenthalpy change in this process is known asenthalpy of atomization,

aH

0

. It is theenthalpy change on breaking one mole ofbonds completely to obtain atoms in the gasphase.

In case of diatomic molecules, likedihydrogen (given above), the enthalpy of

atomization is also the bond dissociationenthalpy. The other examples of enthalpy ofatomization can be

CH4(g) C(g) + 4H(g); aH0

= 1665 kJ mol1

Note that the products are onlyatoms of Cand H in gaseous phase. Now see the followingreaction:

Na(s) Na(g) ; aH

0

= 108.4 kJ mol1

In this case, the enthalpy of atomization issame as the enthalpy of sublimation.

(c) Bond Enthalpy (symbol:bond

H0)

Chemical reactions involve the breaking andmaking of chemical bonds. Energy is requiredto break a bond and energy is released whena bond is formed. It is possible to relate heatof reaction to changes in energy associatedwith breaking and making of chemical bonds. With reference to the enthalpy changesassociated with chemical bonds, two differentterms are used in thermodynamics.


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172

For example in CH4,

CHH

0

is calculated as:

C H = ( HV

= 416 kJ mol1

We find that mean CH bond enthalpy inmethane is 416 kJ/mol. It has been found thatmean CH bond enthalpies differ slightly fromcompound to compound, as in

3 2CH CH Cl,C etc, but it does not differ

in a great deal*. Using Hesss law, bondenthalpies can be calculated. Bond enthalpyvalues of some single and multiple bonds aregiven in Table 6.3. The reaction enthalpies arevery important quantities as these arise fromthe changes that accompany the breaking of

old bonds and formation of the new bonds. Wecan predict enthalpy of a reaction in gas phase,if we know different bond enthalpies. Thestandard enthalpy of reaction,

rH0 is related

to bond enthalpies of the reactants and

products in gas phase reactions as:

bo

b

rH =

V

(6.17)** This relationship is particularly more

useful when the required values offH0 are

not available. The net enthalpy change of areaction is the amount of energy required tobreak all the bonds in the reactant moleculesminus the amount of energy required to breakall the bonds in the product molecules.Remember that this relationship isapproximate and is valid when all substances(reactants and products) in the reaction are in

gaseous state.

(d) Enthalpy of Solution (symbol :sol

H0

)

Enthalpy of solution of a substance is theenthalpy change when one mole of it dissolves

** If we use enthalpy of bond formation, (fHbond), which is the enthalpy change when one mole of a particular type ofbond is formed from gaseous atom, then r f bonds H H =

V V

0

H C N O F Si P S Cl Br I

436 414 389 464 569 293 318 339 431 368 297 H

347 293 351 439 289 264 259 330 276 238 C

159 201 272 - 209 - 201 243 - N

138 184 368 351 - 205 - 201 O159 540 490 327 255 197 - F

176 213 226 360 289 213 Si

213 230 331 272 213 P

213 251 213 - S

243 218 209 Cl

192 180 Br

151 I

Table 6.3(a) Some Mean Single Bond Enthalpies in kJ mol1

N = N 418 C = C 611 O = O 498

N N 946 C C 837C = N 615 C = O 741

C N 891 C O 1070

Table 6.3(b) Some Mean Multiple Bond Enthalpies in kJ mol1

* Note that symbol used for bond dissociation enthalpy and mean bond enthalpy is the same.


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173

in a specified amount of solvent. The enthalpyof solution at infinite dilution is the enthalpychange observed on dissolving the substance

in an infinite amount of solvent when theinteractions between the ions (or solutemolecules) are negligible.

When an ionic compound dissolves in asolvent, the ions leave their ordered positionson the crystal lattice. These are now more free insolution. But solvation of these ions (hydrationin case solvent is water) also occurs at the sametime. This is shown diagrammatically, for anionic compound, AB (s)

Fig. 6.9 Enthalpy diagram for lattice enthalpyof NaCl

( )Na Cl s+

Since it is impossible to determine latticeenthalpies directly by experiment, we use an

indirect method where we construct an

enthalpy diagram called aBorn-Haber Cycle(Fig. 6.9).

Let us now calculate the lattice enthalpyof Na+Cl(s) by following steps given below :

1. Na(s) Na(g) , sublimation of sodium

metal, 108. =sub

HV

2. Na(g) Na (+ , the ionization of

sodium atoms, ionization enthalpy

iH

0= 496 kJ mol

1

3. 21

Cl (g ) Cl(2

, the dissociation of

chlorine, the reaction enthalpy is half the

The enthalpy of solution of AB(s), sol

H0

, inwater is, therefore, determined by the selective values of the lattice enthalpy,

latticeH

0

and

enthalpy of hydration of ions, hydH0 as

= sol lattice

HV

For most of the ionic compounds, sol

H0

ispositive and the dissociation process isendothermic. Therefore the solubility of mostsalts in water increases with rise oftemperature. If the lattice enthalpy is veryhigh, the dissolution of the compound may nottake place at all. Why do many fluorides tendto be less soluble than the correspondingchlorides? Estimates of the magnitudes of

enthalpy changes may be made by using tablesof bond energies (enthalpies) and latticeenergies (enthalpies).

Lattice Enthalpy

The lattice enthalpy of an ionic compound isthe enthalpy change which occurs when onemole of an ionic compound dissociates into itsions in gaseous state.


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174

bond dissociation enthalpy.

11

2 =bond

HV

.

4. 1Cl( g) e ( g)+ electron gained by

chlorine atoms. The electron gain enthalpy,

egH

0= 348.6 kJ mol

1.

You have learnt about ionization enthalpyand electron gain enthalpy in Unit 3. Infact, these terms have been taken fromthermodynamics. Earlier terms, ionizationenergy and electron affinity were in practicein place of the above terms (see the box forjustification).

Internal energy is smaller by 2RT ( becausen

g= 2) and is equal to + 783 kJ mol1.

Now we use the value of lattice enthalpy to

calculate enthalpy of solution from theexpression:

sol lattic H = V

For one mole of NaCl(s),lattice enthalpy = + 788 kJ mol1

and hyd

H0

= 784 kJ mol1

( from theliterature)

sol

H0

= + 788 kJ mol1

784 kJ mol1

= + 4 kJ mol1

The dissolution of NaCl(s) is accompaniedby very little heat change.

6.6 SPONTANEITY

The first law of thermodynamics tells us aboutthe relationship between the heat absorbedand the work performed on or by a system. Itputs no restrictions on the direction of heatflow. However, the flow of heat is unidirectionalfrom higher temperature to lower temperature.In fact, all naturally occurring processes whether chemical or physical will tend toproceed spontaneously in one direction only.For example, a gas expanding to fill theavailable volume, burning carbon in dioxygen

giving carbon dioxide.

But heat will not flow from colder body towarmer body on its own, the gas in a containerwill not spontaneously contract into one corneror carbon dioxide will not form carbon anddioxygen spontaneously. These and manyother spontaneously occurring changes showunidirectional change. We may ask what is thedriving force of spontaneously occurringchanges ? What determines the direction of aspontaneous change ? In this section, we shallestablish some criterion for these processes

whether these will take place or not.Let us first understand what do we mean

by spontaneous reaction or change ? You maythink by your common observation thatspontaneous reaction is one which occursimmediately when contact is made between thereactants. Take the case of combination ofhydrogen and oxygen. These gases may bemixed at room temperature and left for many

Ionization Energy and Electron AffinityIonization energy and electron affinity aredefined at absolute zero. At any othertemperature, heat capacities for thereactants and the products have to betaken into account. Enthalpies of reactionsfor

M(g) M+(g) + e

(for ionization)

M(g) + e

M(g) (for electron gain)

at temperature, T is

rH

0

(T) = rH

0

(0) +0

d

T

r pC T

V

The value ofCp for each species in theabove reaction is 5/2 R (CV= 3/2R)So,

rC

p

0

= + 5/2 R (for ionization)

rC

p

0

= 5/2 R (for electron gain)Therefore,

rH

0

(ionization enthalpy)= E0 (ionization energy) + 5/2 RT

rH

0

(electron gain enthalpy)= A( electron affinity) 5/2 RT

5. Na ( g) Cl (g+ +

The sequence of steps is shown in Fig. 6.9,

and is known as a Born-Haber cycle. Theimportance of the cycle is that, the sum ofthe enthalpy changes round a cycle is zero.

Applying Hesss law, we get,

41 =lattice

HV

7lattice

H = +V

forNaCl(s) Na


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175

years without observing any perceptiblechange. Although the reaction is taking placebetween them, it is at an extremely slow rate.

It is still called spontaneous reaction. Sospontaneity means having the potential to

proceed without the assistance of external

agency. However, it does not tell about therate of the reaction or process. Another aspectof spontaneous reaction or process, as we seeis that these cannot reverse their direction ontheir own. We may summarise it as follows:

A spontaneous process is anirreversible process and may only bereversed by some external agency.

(a) Is decrease in enthalpy a criterion for

spontaneity ?If we examine the phenomenon like flow ofwater down hill or fall of a stone on to theground, we find that there is a net decrease inpotential energy in the direction of change. Byanalogy, we may be tempted to state that achemical reaction is spontaneous in a givendirection, because decrease in energy hastaken place, as in the case of exothermicreactions. For example:

1

2N

2(g) +

3

2H

2(g) = NH

3(g) ;

rH

0

= 46.1 kJ mol1

1

2H

2(g) +

1

2Cl

2(g) = HCl (g) ;

rH

0

= 92.32 kJ mol1

H2(g) +

1

2O

2(g) H

2O(l) ;

rH

0

= 285.8 kJ mol1

The decrease in enthalpy in passing fromreactants to products may be shown for anyexothermic reaction on an enthalpy diagramas shown in Fig. 6.10(a).

Thus, the postulate that driving force for achemical reaction may be due to decrease inenergy sounds reasonable as the basis ofevidence so far !

Now let us examine the following reactions:

1

2N

2(g) + O

2(g) NO

2(g);

rH

0

= +33.2 kJ mol1

C(graphite, s) + 2 S(l) CS2(l);

rH

0

= +128.5 kJ mol1

These reactions though endothermic, arespontaneous. The increase in enthalpy may berepresented on an enthalpy diagram as shownin Fig. 6.10(b).

Fig. 6.10 (a) Enthalpy diagram for exothermicreactions

Fig. 6.10 (b) Enthalpy diagram for endothermicreactions

Therefore, it becomes obvious that whiledecrease in enthalpy may be a contributoryfactor for spontaneity, but it is not true for allcases.

(b) Entropy and spontaneity

Then, what drives the spontaneous process ina given direction ? Let us examine such a casein which H= 0 i.e., there is no change inenthalpy, but still the process is spontaneous.

Let us consider diffusion of two gases intoeach other in a closed container which is


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176

isolated from the surroundings as shown inFig. 6.11.

At this point, we introduce anotherthermodynamic function, entropy denoted asS. The above mentioned disorder is the

manifestation of entropy. To form a mentalpicture, one can think of entropy as a measureof the degree of randomness or disorder in thesystem. The greater the disorder in an isolatedsystem, the higher is the entropy. As far as achemical reaction is concerned, this entropychange can be attributed to rearrangement ofatoms or ions from one pattern in the reactantsto another (in the products). If the structureof the products is very much disordered thanthat of the reactants, there will be a resultantincrease in entropy. The change in entropyaccompanying a chemical reaction may beestimated qualitatively by a consideration ofthe structures of the species taking part in thereaction. Decrease of regularity in structurewould mean increase in entropy. For a givensubstance, the crystalline solid state is thestate of lowest entropy (most ordered), Thegaseous state is state of highest entropy.

Now let us try to quantify entropy. One wayto calculate the degree of disorder or chaoticdistribution of energy among molecules wouldbe through statistical method which is beyondthe scope of this treatment. Other way would

be to relate this process to the heat involved ina process which would make entropy athermodynamic concept. Entropy, like anyother thermodynamic property such asinternal energyUand enthalpyHis a statefunction and Sis independent of path.

Whenever heat is added to the system, itincreases molecular motions causingincreased randomness in the system. Thusheat (q) has randomising influence on thesystem. Can we then equate Swith q? Wait !Experience suggests us that the distribution

of heat also depends on the temperature atwhich heat is added to the system. A systemat higher temperature has greater randomnessin it than one at lower temperature. Thus,temperature is the measure of averagechaotic motion of particles in the system.Heat added to a system at lower temperaturecauses greater randomness than when thesame quantity of heat is added to it at higher

Fig. 6.11 Diffusion of two gases

The two gases, say, gas A and gas B arerepresented by black dots and white dotsrespectively and separated by a movable

partition [Fig. 6.11 (a)]. When the partition iswithdrawn [Fig.6.11( b)], the gases begin todiffuse into each other and after a period oftime, diffusion will be complete.

Let us examine the process. Beforepartition, if we were to pick up the gasmolecules from left container, we would besure that these will be molecules of gas A andsimilarly if we were to pick up the gasmolecules from right container, we would besure that these will be molecules of gas B. But,if we were to pick up molecules from container

when partition is removed, we are not surewhether the molecules picked are of gas A orgas B. We say that the system has become lesspredictable or more chaotic.

We may now formulate another postulate:in an isolated system, there is always atendency for the systems energy to becomemore disordered or chaotic and this could bea criterion for spontaneous change !


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177

temperature. This suggests that the entropychange is inversely proportional to thetemperature. Sis related with qand Tfor a

reversible reaction as :

S =revq

T(6.18)

The total entropy change ( Stotal

) for thesystem and surroundings of a spontaneousprocess is given by

total syst S S = (6.19)

When a system is in equilibrium, theentropy is maximum, and the change inentropy, S= 0.

We can say that entropy for a spontaneous

process increases till it reaches maximum andat equilibrium the change in entropy is zero.Since entropy is a state property, we cancalculate the change in entropy of a reversibleprocess by

Ssys

=,sys rev

q

T

We find that both for reversible andirreversible expansion for an ideal gas, underisothermal conditions, U= 0, butS

total i.e.,

sys surr S S + is not zero for irreversible

process. Thus, U does not discriminatebetween reversible and irreversible process,whereas Sdoes.

Problem 6.9

Predict in which of the following, entropyincreases/decreases :

(i) A liquid crystallizes into a solid.

(ii) Temperature of a crystalline solid israised from 0 K to 115 K.

( )iii 2NaHC

(iv) ( ) (2H g 2H

Solution

(i) After freezing, the molecules attain anordered state and therefore, entropydecreases.

(ii) At 0 K, the contituent particles arestatic and entropy is minimum. Iftemperature is raised to 115 K, these

begin to move and oscillate abouttheir equilibrium positions in thelattice and system becomes moredisordered, therefore entropyincreases.

(iii) Reactant, NaHCO3

is a solid and ithas low entropy. Among productsthere are one solid and two gases.Therefore, the products represent acondition of higher entropy.

(iv) Here one molecule gives two atomsi.e., number of particles increases

leading to more disordered state.Two moles of H atoms have higherentropy than one mole of dihydrogenmolecule.

Problem 6.10

For oxidation of iron,

( ) 24Fe s 3O+

entropy change is 549.4 JK1mol1at298 K. Inspite of negative entropy changeof this reaction, why is the reactionspontaneous?

(rH0for this reaction is

1648 103 J mol1)

Solution

One decides the spontaneity of a reactionby considering

(total sys S S + . For calculatingS

surr, we have to consider the heat

absorbed by the surroundings which isequal to

rH

0

.....At temperature T, entropychange of the surroundings is

= rsurr

HST

( 1648 129

=

15530JK

=

Thus, total entropy change for thisreaction


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CHEMISTRY178


178

5530Jr t ot al

S =

4980.6JK=

This shows that the above reaction isspontaneous.

(c) Gibbs energy and spontaneity

We have seen that for a system, it is the totalentropy change, S

totalwhich decides the

spontaneity of the process. But most of thechemical reactions fall into the category ofeither closed systems or open systems.Therefore, for most of the chemical reactionsthere are changes in both enthalpy andentropy. It is clear from the discussion in

previous sections that neither decrease inenthalpy nor increase in entropy alone candetermine the direction of spontaneous changefor these systems.

For this purpose, we define a newthermodynamic function the Gibbs energy orGibbs function, G, as

G = H TS (6.20)

Gibbs function, Gis an extensive propertyand a state function.

The change in Gibbs energy for the system,

Gsyscan be written as=sys sys G H

At constant temperature, 0T =

=sys s G H

Usually the subscript system is droppedand we simply write this equation as

G H T = (6.21)

Thus, Gibbs energy change = enthalpy

change temperature entropy change, and

is referred to as the Gibbs equation, one of the

most important equations in chemistry. Here,

we have considered both terms together for

spontaneity: energy (in terms of H) and

entropy (S, a measure of disorder) as

indicated earlier. Dimensionally if we analyse,

we find thatGhas units of energy because,

both Hand the T S are energy terms, since

TS= (K) (J/K) = J.

Now let us consider how G is related to

reaction spontaneity.

We know,

Stotal

= Ssys

+ Ssurr

If the system is in thermal equilibrium withthe surrounding, then the temperature of thesurrounding is same as that of the system.Also, increase in enthalpy of the surroundingis equal to decrease in the enthalpy of thesystem.

Therefore, entropy change ofsurroundings,

=

surr

surr

HS

T

= total sys S S

Rearranging the above equation:

TStotal

= TSsys

Hsys

For spontanious process, 0 >total

S , so

TSsys

Hsys 0>

( sysHUsing equation 6.21, the above equation canbe written as

0 >G

= G H T (6.22)

Hsys

is the enthalpy change of a reaction,TS

sysis the energy which is not available to

do useful work. So G is the net energyavailable to do useful work and is thus ameasure of the free energy. For this reason, itis also known as the free energy of the reaction.

Ggives a criteria for spontaneity at

constant pressure and temperature.

(i) If Gis negative (< 0), the process is

spontaneous.

(ii) IfGis positive (> 0), the process is non

spontaneous.

Note : If a reaction has a positive enthalpychange and positive entropy change, it can bespontaneous when TS is large enough tooutweigh H. This can happen in two ways;


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THERMODYNAMICS


179

(a) The positive entropy change of the systemcan be small in which case Tmust be large.(b) The positive entropy change of the system

can be large, in which case Tmay be small.The former is one of the reasons why reactionsare often carried out at high temperature.Table 6.4 summarises the effect of temperatureon spontaneity of reactions.

6.7 GIBBS ENERGY CHANGE ANDEQUILIBRIUM

We have seen how a knowledge of the sign andmagnitude of the free energy change of achemical reaction allows:

(i) Prediction of the spontaneity of thechemical reaction.

(ii) Prediction of the useful work that could beextracted from it.

So far we have considered free energychanges in irreversible reactions. Let us nowexamine the free energy changes in reversiblereactions.

Reversible under strict thermodynamicsense is a special way of carrying out a processsuch that system is at all times in perfectequilibrium with its surroundings. Whenapplied to a chemical reaction, the term

reversible indicates that a given reaction canproceed in either direction simultaneously, sothat a dynamic equilibrium is set up. Thismeans that the reactions in both the directionsshould proceed with a decrease in free energy,which seems impossible. It is possible only if

at equilibrium the free energy of the system isminimum. If it is not, the system wouldspontaneously change to configuration of

lower free energy.So, the criterion for equilibrium

A B C+ + ; is

rG= 0

Gibbs energy for a reaction in which allreactants and products are in standard state,

rG

0

is related to the equilibrium constant ofthe reaction as follows:

0 = rG

0

+ RTln K

or rG

0

= RTln K

or rG

0

= 2.303 RTlog K (6.23)

We also know that

=r rG H V V (6.24)

For strongly endothermic reactions, thevalue of

rH

0

may be large and positive. Insuch a case, value ofKwill be much smallerthan 1 and the reaction is unlikely to formmuch product. In case of exothermic reactions,

rH

0

is large and negative, and rG

0

is likely tobe large and negative too. In such cases, Kwillbe much larger than 1. We may expect stronglyexothermic reactions to have a large K, and

hence can go to near completion. rG0

alsodepends upon rS0

, if the changes in theentropy of reaction is also taken into account,the value ofKor extent of chemical reactionwill also be affected, depending upon whether

rS0

is positive or negative.

rH

0

rS

0

rG

0

Description*

+ Reaction spontaneous at all temperature

(at low T) Reaction spontaneous at low temperature

+ (at high T) Reaction nonspontaneous at high temperature

+ + + (at low T) Reaction nonspontaneous at low temperature

+ + (at high T ) Reaction spontaneous at high temperature

+ + (at all T) Reaction nonspontaneous at all temperatures

* The term low temperature and high temperature are relative. For a particular reaction, high temperature could evenmean room temperature.

Table 6.4 Effect of Temperature on Spontaneity of Reactions


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CHEMISTRY180


180

Using equation (6.24),

(i) It is possible to obtain an estimate ofGV

from the measurement ofHV

and SV

,

and then calculate Kat any temperaturefor economic yields of the products.

(ii) If K is measured directly in the laboratory,value ofG

0

at any other temperature canbe calculated.

Problem 6.11

Calculate rG

0

for conversion of oxygento ozone, 3/2 O

2(g) O

3(g) at 298 K. ifK

p

for this conversion is 2.47 1029.

Solution

We knowrG

0

= 2.303 RT log Kp

and

R = 8.314 JK1 mol1

Therefore, rG

0

= 2.303 (8.314 J K1 mol1)

(298 K) (log 2.47 1029)

= 163000 J mol1

= 163 kJ mol1.

Problem 6.12

Find out the value of equilibrium constantfor the following reaction at 298 K.

( )32NH g C+ .

Standard Gibbs energy change, rG

0

atthe given temperature is 13.6 kJ mol1.

Solution

We know, log K =

2.303R

rG

T

V

=(

(

13.6

2.303 8.314

= 2.38Hence K= antilog 2.38 = 2.4 102.

Problem 6.13

At 60C, dinitrogen tetroxide is fiftypercent dissociated. Calculate thestandard free energy change at thistemperature and at one atmosphere.

Solution

N2O

4(g) 2NO2(g)

If N2O

4is 50% dissociated, the mole

fraction of both the substances is given

by

2 4N Ox =

1 0.5

1 0.5

+;

2NOx =

2 0.5

1 0.5

+

2 4N Op =

0.51 atm,

1.5

2NOp =

1

1 atm.1.5

The equilibrium constantKp

is given by

Kp

=( )

2

2 4

2

NO

N O (1.

=p

p

= 1.33 atm.Since

rG

0

= RTln Kp

rG

0

= ( 8.314 JK1 mol1) (333 K) (2.303) (0.1239)

= 763.8 kJ mol1


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181

SUMMARY

Thermodynamics deals with energy changes in chemical or physical processes andenables us to study these changes quantitatively and to make useful predictions. Forthese purposes, we divide the universe into the system and the surroundings. Chemicalor physical processes lead to evolution or absorption of heat (q), part of which may beconverted into work (w). These quantities are related through the first law ofthermodynamics viaU= q+ w. U, change in internal energy, depends on initial andfinal states only and is a state function, whereas qand w depend on the path and arenot the state functions. We follow sign conventions ofqand w by giving the positive signto these quantities when these are added to the system. We can measure the transfer ofheat from one system to another which causes the change in temperature. The magnitudeof rise in temperature depends on the heat capacity (C) of a substance. Therefore, heatabsorbed or evolved is q= CT. Work can be measured by w = p

exV, in case of expansion

of gases. Under reversible process, we can putpex

= pfor infinitesimal changes in the

volume making wrev

= pdV. In this condition, we can use gas equation, pV= nRT.

At constant volume, w = 0, then U= qV, heat transfer at constant volume. But instudy of chemical reactions, we usually have constant pressure. We define anotherstate function enthalpy. Enthalpy change, H= U+ n

gRT, can be found directly from

the heat changes at constant pressure, H= qp.

There are varieties of enthalpy changes. Changes of phase such as melting, vaporization and sublimation usually occur at constant temperature and can becharacterized by enthalpy changes which are always positive. Enthalpy of formation,combustion and other enthalpy changes can be calculated using Hesss law. Enthalpychange for chemical reactions can be determined by

( = r i ff

H a

and in gaseous state by

rH

0

= bond enthalpies of the reactantsbond enthalpies of the productsFirst law of thermodynamics does not guide us about the direction of chemical

reactions i.e., what is the driving force of a chemical reaction. For isolated systems,U = 0. We define another state function, S, entropy for this purpose. Entropy is ameasure of disorder or randomness. For a spontaneous change, total entropy change ispositive. Therefore, for an isolated system, U= 0, S> 0, so entropy change distinguishesa spontaneous change, while energy change does not. Entropy changes can be measured

by the equation S=rev

q

Tfor a reversible process.

revq

Tis independent of path.

Chemical reactions are generally carried at constant pressure, so we define anotherstate function Gibbs energy, G, which is related to entropy and enthalpy changes ofthe system by the equation:

rG =

rHT

rS

For a spontaneous change, Gsys

< 0 and at equilibrium, Gsys

= 0.

Standard Gibbs energy change is related to equilibrium constant by

rG

0

= RT ln K.

K can be calculated from this equation, if we know rG

0

which can be found from

r rG H T = V V . Temperature is an important factor in the equation. Many reactions

which are non-spontaneous at low temperature, are made spontaneous at hightemperature for systems having positive entropy of reaction.


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CHEMISTRY182


182

EXERCISES

6.1 Choose the correct answer. A thermodynamic state function is aquantity

(i) used to determine heat changes

(ii) whose value is independent of path

(iii) used to determine pressure volume work

(iv) whose value depends on temperature only.

6.2 For the process to occur under adiabatic conditions, the correctcondition is:

(i) T= 0

(ii) p= 0

(iii) q= 0

(iv) w = 0

6.3 The enthalpies of all elements in their standard states are:

(i) unity

(ii) zero

(iii) < 0

(iv) different for each element

6.4 U0

of combustion of methane is X kJ mol1. The value ofH0

is

(i) = U0

(ii) > U0

(iii) < U0

(iv) = 0

6.5 The enthalpy of combustion of methane, graphite and dihydrogenat 298 K are, 890.3 kJ mol1 393.5 kJ mol1, and 285.8 kJ mol1

respectively. Enthalpy of formation of CH4(g) will be

(i) 74.8 kJ mol1 (ii) 52.27 kJ mol1

(iii) +74.8 kJ mol1 ( iv) +52.26 kJ mol1.

6.6 A reaction, A + B C + D + q is found to have a positive entropychange. The reaction will be

(i) possible at high temperature

(ii) possible only at low temperature

(iii) not possible at any temperature

(v) possible at any temperature

6.7 In a process, 701 J of heat is absorbed by a system and 394 J ofwork is done by the system. What is the change in internal energyfor the process?

6.8 The reaction of cyanamide, NH2CN (s), with dioxygen was carried

out in a bomb calorimeter, and Uwas found to be 742.7 kJ mol1

at 298 K. Calculate enthalpy change for the reaction at 298 K.

NH2CN(g) +

3

2O

2(g) N

2(g) + CO

2(g) + H

2O(l)


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THERMODYNAMICS


183

6.9 Calculate the number of kJ of heat necessary to raise the temperatureof 60.0 g of aluminium from 35C to 55C. Molar heat capacity of Alis 24 J mol1 K1.

6.10 Calculate the enthalpy change on freezing of 1.0 mol of waterat10.0C to ice at 10.0C.

fusH= 6.03 kJ mol1 at 0C.

Cp

[H2O(l)] = 75.3 J mol1 K1

Cp

[H2O(s)] = 36.8 J mol1 K1

6.11 Enthalpy of combustion of carbon to CO2

is 393.5 kJ mol1. Calculatethe heat released upon formation of 35.2 g of CO

2from carbon and

dioxygen gas.

6.12 Enthalpies of formation of CO(g), CO2(g), N

2O(g) and N

2O

4(g) are 110,

393, 81 and 9.7 kJ mol1 respectively. Find the value ofrHfor the

reaction:

N2O

4(g) + 3CO(g) N

2O(g) + 3CO

2(g)

6.13 Given

N2(g) + 3H2(g) 2NH3(g) ; rH0

= 92.4 kJ mol1

What is the standard enthalpy of formation of NH3

gas?

6.14 Calculate the standard enthalpy of formation of CH3OH(l) from the

following data:

CH3OH (l) +

3

2O

2(g) CO

2(g) + 2H

2O(l) ;

rH

0 = 726 kJ mol1

C(g) + O2(g) CO

2(g) ;

cH

0 = 393 kJ mol1

H2(g) +

1

2O

2(g) H

2O(l) ;

fH

0 = 286 kJ mol1.

6.15 Calculate the enthalpy change for the process

CCl4(g) C(g) + 4 Cl(g)and calculate bond enthalpy of C Cl in CCl

4(g).

vap

H0(CCl

4) = 30.5 kJ mol1.

fH

0

(CCl4) = 135.5 kJ mol1.

aH

0 (C) = 715.0 kJ mol1 , where aH

0 is enthalpy of atomisation

aH

0 (Cl2) = 242 kJ mol1

6.16 For an isolated system, U= 0, what will be S?

6.17 For the reaction at 298 K,

2A + B C

H= 400 kJ mol1 and S= 0.2 kJ K1 mol1

At what temperature will the reaction become spontaneousconsidering H and S to be constant over the temperature range.

6.18 For the reaction,

2 Cl(g) Cl2(g), what are the signs ofHand S?

6.19 For the reaction

2 A(g) + B(g) 2D(g)

U0 = 10.5 kJ and S0 = 44.1 JK1.

Calculate G0 for the reaction, and predict whether the reactionmay occur spontaneously.


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CHEMISTRY184

184

6.20 The equilibrium constant for a reaction is 10. What will be the valueofG0? R = 8.314 JK1 mol1, T = 300 K.

6.21 Comment on the thermodynamic stability of NO(g), given

1

2N

2(g) +

1

2O

2(g) NO(g) ;

rH

0 = 90 kJ mol1

NO(g) +1

2O

2(g) NO

2(g) :

rH

0

= 74 kJ mol1

6.22 Calculate the entropy change in surroundings when 1.00 mol ofH

2O(l) is formed under standard conditions.

fH

0

= 286 kJ mol1.

ncert ch6 chemistry class 11

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