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 this Unit, 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 contributionsin chemical systems;

••• explain state functions: U, H.

••• correlate ΔU and ΔH;

••• measure experimentally ΔUand ΔH;

••• define standard states for ΔH;

••• calculate enthalpy changes forvarious types of reactions;

••• state and apply Hess’s law ofconstant heat summation;

••• differentiate between extensiveand intensive properties;

••• define spontaneous and non-spontaneous processes;

••• explain entropy as athermodynamic state functionand apply it for spontaneity;

••• explain Gibbs energy change(ΔG);

••• establish relationship betweenΔG and spontaneity, ΔG andequilibrium constant.

Chemical energy stored by molecules can be released as heatduring chemical reactions when a fuel like methane, cookinggas 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 likedry cell. Thus, various forms of energy are interrelated andunder certain conditions, these may be transformed fromone form into another. The study of these energytransformations forms the subject matter of thermodynamics.The laws of thermodynamics deal with energy changes ofmacroscopic systems involving a large number of moleculesrather than microscopic systems containing a few molecules.Thermodynamics is not concerned about how and at whatrate these energy transformations are carried out, but isbased 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 toanother equilibrium state. Macroscopic properties likepressure 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 achemical reaction/process? Will it occur or not?

What drives a chemical reaction/process?

To what extent do the chemical reactions proceed?

UNIT 6

THERMODYNAMICS 155

6.1 THERMODYNAMIC STATE

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

A system 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 theuniverse .

The universe = The system + The surroundingsHowever, the entire universe other than the

system 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 abeaker, 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 calledboundary. This is designed to allow us tocontrol 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 SystemIn an open system, there is exchange of energyand matter between system and surroundings[Fig. 6.2 (a)]. The presence of reactants in anopen beaker is an example of an open system*.Here the boundary is an imaginary surfaceenclosing the beaker and reactants.2. Closed SystemIn a closed system, there is no exchange ofmatter, but exchange of energy is possiblebetween 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 SystemIn an isolated system, there is no exchange ofenergy or matter between the system and thesurroundings [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 SystemThe 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 recallfrom 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 by state 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 likep, 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 youmay think of, the sum of all these is the energyof the system. In thermodynamics, we call itthe internal energy, U of 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 thesurroundings is called the adiabatic wall(Fig 6.3).

Fig. 6.3 An adiabatic system which does notpermit 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 TA. Let the

THERMODYNAMICS 157

internal energy of the system in state A becalled UA. We can change the state of the systemin 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 TB. It isfound that TB > TA and the change intemperature, ΔT = TB–TA. Let the internalenergy of the system in state B be UB and thechange in internal energy, ΔU =UB– UA.

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 intemperature is same as in the earlier case, say,TB – TA.

In fact, the experiments in the abovemanner were done by J. P. Joule between1840–50 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 energy U, 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 of U in one stateand that in another state, ΔU i.e.,

2 1= =Δ −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 done by the system,wad

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 25°C to 35°C, the change in temperatureis 35°C–25°C = +10°C, whether we go straightup to 35°C or we cool the system for a fewdegrees, then take the system to the finaltemperature. Thus, T is 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, becausechange 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) HeatWe 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 insection 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 acontainer having thermally conducting walls,say made up of copper and enclose it in a hugeheat reservoir at temperature, TB. The heatabsorbed by the system (water), q can bemeasured in terms of temperature difference ,TB – TA. In this case change in internal energy,ΔU= q, when no work is done at constantvolume.

The q is positive, when heat is transferredfrom the surroundings to the system and q isnegative when heat is transferred fromsystem to the surroundings.

(c) The general case

Let us consider the general case in which achange of state is brought about both by

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doing work and by transfer of heat. We writechange in internal energy for this case as:

ΔU = q + w (6.1)

For a given change in state, q and w canvary depending on how the change is carriedout. However, q +w = ΔU will depend only oninitial and final state. It will be independent ofthe way the change is carried out. If there isno transfer of energy as heat or as work(isolated system) i.e., if w = 0 and q = 0, thenΔ U = 0.

The equation 6.1 i.e., ΔU = q + w ismathematical statement of the first law ofthermodynamics, which states that

The energy of an isolated system isconstant.

It is commonly stated as the law ofconservation of energy i.e., energy can neitherbe created nor be destroyed.

Note: There is considerable difference betweenthe character of the thermodynamic propertyenergy and that of a mechanical property suchas volume. We can specify an unambiguous(absolute) value for volume of a system in aparticular state, but not the absolute value ofthe internal energy. However, we can measureonly the changes in the internal energy, ΔU ofthe system.

Problem 6.1Express the change in internal energy ofa system when

(i) No heat is absorbed by the systemfrom the surroundings, but work (w)is done on the system. What type ofwall does the system have ?

(ii) No work is done on the system, butq amount of heat is taken out fromthe system and given to thesurroundings. What type of wall doesthe system have?

(iii) w amount of work is done by thesystem and q amount of heat issupplied to the system. What type ofsystem would it be?

Solution

(i) Δ U = w ad, wall is adiabatic

(ii) Δ U = – q, thermally conducting walls

(iii)Δ 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-volumework, let us consider a cylinder whichcontains one mole of an ideal gas fitted with africtionless piston. Total volume of the gas isV

i and pressure of the gas inside is p. If

external pressure is pex which is greater thanp, piston is moved inward till the pressureinside becomes equal to pex. 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.

THERMODYNAMICS 159

be achieved in a single step and the finalvolume be Vf . During this compression,suppose piston moves a distance, l and iscross-sectional area of the piston is A[Fig. 6.5(a)].

then, volume change = l × A = ΔV = (Vf – V

i )

We also know, for

pressurear

=

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 . (–ΔV) = – pex ΔV = – pex (Vf

– Vi ) (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

f – V

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−∑ Δ [Fig. 6.5 (b)]

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

in + dp) in

case 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 = (p

in + 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) duringcompression from initial volume, Vi tofinal volume, Vf . Work done on the gasis represented by the shaded area.

Fig. 6.5 (b) pV-plot when pressure is not constantand changes in finite steps duringcompression from initial volume, Vi tofinal 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 canbe 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 d

Since dp × dV is very small we can write

w = − ∫f

i

V

rev in

V

p dV (6.4)

Now, the pressure of the gas (pin

which wecan write as p now) can be expressed in termsof its volume through gas equation. For n molof an ideal gas i.e., pV =nRT

R⇒ = n T

pV

Therefore, at constant temperature (isothermalprocess),

revw R= − ∫f

i

V

V

n T

= – 2.303 nRT log f

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

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

Δ = − ΔexU q p V

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

ΔU = qV

the subscript V in qV

denotes that heat issupplied at constant volume.

Isothermal and free expansion of anideal gasFor isothermal (T = constant) expansion of anideal gas into vacuum ; w = 0 since p

ex = 0.

Also, Joule determined experimentally thatq = 0; therefore, ΔU = 0

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

1. For isothermal irreversible change

q = – w = pex

(Vf – V

i )

2. For isothermal reversible change

q = – w = nRT ln f

i

V

V

= 2.303 nRT log f

i

V

V

3. For adiabatic change, q = 0,

ΔU = 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 howmuch 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 6.4

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

Solution

We have q = – w = 2.303 × 10 10

log2

= 16.1 litre-atm

THERMODYNAMICS 161

6.2.2 Enthalpy, H(a) A useful new state functionWe know that the heat absorbed at constantvolume is equal to change in the internalenergy i.e., ΔU = q

V. But most of chemical

reactions are carried out not at constantvolume, 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) asΔ = − ΔpU q p V at constant pressure, whereq

p is heat absorbed by the system and –pΔV

represent 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

U2–U1 = qp – p (V2 – V1)

On rearranging, we get

qp = (U2 + pV2) – (U1 + pV1) (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= H2 – H1 = ΔH

Although q is a path dependent function,H is a state function because it depends on U,p and V, all of which are state functions.Therefore, ΔH is independent of path. Hence,q

p is also independent of path.

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

ΔH = ΔU + ΔpV

Since p is constant, we can write

ΔH = ΔU + pΔV (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 ΔH = qp, heat absorbed by the

system at constant pressure.

ΔΔΔH is negative for exothermic reactionswhich evolve heat during the reaction and

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

At constant volume (ΔV = 0), ΔU = qV,

therefore equation 6.8 becomes

ΔH = ΔU = qV

The difference between ΔH and ΔU is 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. IfVA is the total volume of the gaseous reactants,VB is the total volume of the gaseous products,nA is the number of moles of gaseous reactantsand nB is the number of moles of gaseousproducts, all at constant pressure andtemperature, then using the ideal gas law, wewrite,

pVA = nART

and pVB = nBRT

Thus, pVB – pVA = nBRT – nART = (nB–nA)RT

or p (VB – VA) = (nB – nA) RT

or p ΔV = ΔngRT (6.9)

Here, Δng refers to the number of moles of

gaseous products minus the number of molesof gaseous reactants.

Substituting the value of pΔV fromequation 6.9 in equation 6.8, we get

ΔH = ΔU + ΔngRT (6.10)

The equation 6.10 is useful for calculatingΔH from ΔU and 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 100°C is 41kJ mol–1. Calculate theinternal energy change, when

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

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

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Solution

(i) The change ( ) →2 2H O Hl

Δ = Δ + ΔH U n

or ΔU = ΔH – RΔ gn T , substituting the

values, we get

41.00 kJ

8.3 J

Δ =

×

U

= 41.00 kJ m= 37.904 kJ mol–1

(ii) The change ( )2 2H O H O→l

There is negligible change in volume,

So, we can put RΔ = Δ gp V n T in this

case,

Δ ≅ ΔH U

so, 41.Δ =U

(b) Extensive and Intensive PropertiesIn 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. areintensive properties. A molar property, χm, isthe value of an extensive property χ of thesystem for 1 mol of the substance. If n is the

amount of matter, m

χχ =n

is independent of

the amount of matter. Other examples aremolar volume, Vm and molar heat capacity, Cm.Let us understand the distinction betweenextensive and intensive properties byconsidering a gas enclosed in a container ofvolume V and 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, 2V

, but thetemperature 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 CapacityIn 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

= × Δq coeff T

The magnitude of the coefficient dependson the size, composition and nature of thesystem. We can also write it as q = C ΔT

The coefficient, C is called the heat capacity.

Thus, we can measure the heat suppliedby monitoring the temperature rise, providedwe know the heat capacity.

When C is large, a given amount of heatresults in only a small temperature rise. Waterhas a large heat capacity i.e., a lot of energy isneeded to raise its temperature.

C is directly proportional to amount ofsubstance. 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 substanceby 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, ΔT as

= × × Δq c m T (6.11)

(d) The relationship between Cp and CV foran ideal gas

At constant volume, the heat capacity, C isdenoted by C

V and at constant pressure, this

is denoted by Cp . Let us find the relationship

between the two.

We can write equation for heat, q

at constant volume as qV = Δ = ΔVC T U

at constant pressure as qp = Δ = ΔpC T H

The difference between Cp and C

V can be

derived for an ideal gas as:

For a mole of an ideal gas, ΔH = ΔU + Δ(pV )

= ΔU + Δ(RT )

= ΔU + RΔT

R∴ Δ = Δ +H U (6.12)

On putting the values of ΔH and ΔU,we have

Δ = Δp VC T C T

R= +p VC C

R− =p VC C (6.13)

6.3 MEASUREMENT OF ΔΔΔU AND ΔΔΔH:CALORIMETRY

We can measure energy changes associatedwith 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 heatevolved in the process by measuringtemperature changes. Measurements aremade under two different conditions:

i) at constant volume, qV

ii) at constant pressure, qp

(a) ΔΔΔU measurementsFor 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 heatis 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 associatedwith reactions are measured at constantvolume. Under these conditions, no work is

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done as the reaction is carried out at constantvolume in the bomb calorimeter. Even forreactions involving gases, there is no workdone as ΔV = 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) ΔΔΔH measurementsMeasurement of heat change at constantpressure (generally under atmosphericpressure) can be done in a calorimeter shownin Fig. 6.8. We know that Δ = pH q (atconstant p) and, therefore, heat absorbed orevolved, q

p at constant pressure is also called

the heat of reaction or enthalpy of reaction, ΔrH.

In an exothermic reaction, heat is evolved,and system loses heat to the surroundings.Therefore, qp will be negative and ΔrH will alsobe negative. Similarly in an endothermicreaction, heat is absorbed, q

p is positive and

ΔrH will 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

V is the heat

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

q = CV × ΔT

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− × Δ =

= −

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

Thus, ΔU for the combustion of the 1g ofgraphite = – 20.7 kJK–1

For combustion of 1 mol of graphite,

112.0 g mol=

1

−

= – 2.48 ×102 kJ mol–1

6.4 ENTHALPY CHANGE, ΔΔΔrH OF AREACTION – 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 ΔrH

Fig. 6.8 Calorimeter for measuring heat changesat constant pressure (atmosphericpressure).

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ΔrH = (sum of enthalpies of products) – (sum

of enthalpies of reactants)

ai productsi

H= ∑ (6.14)

(Here symbol ∑ (sigma) is used for

summation and ai and b

i are the stoichiometric

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

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

ar i pi

H HΔ = ∑

= [Hm (CO2 ,g) + 2Hm (H2O, l)]– [Hm (CH4 , 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 reactionsEnthalpy of a reaction depends on theconditions 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 at1 bar; standard state of solid iron at 500 K ispure iron at 1 bar. Usually data are taken at298 K.

Standard conditions are denoted by adding

the superscriptV to the symbol ΔH, e.g., HΔ V

(b) Enthalpy changes during phasetransformations

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 remainsconstant (at 273 K).

2 2H O(s) H O→

Here Δfus

H 0 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, ΔΔΔfusH

0.

Melting of a solid is endothermic, so allenthalpies of fusion are positive. Water requiresheat for evaporation. At constant temperatureof its boiling point Tb and at constant pressure:

2 2H O(l) H O→

Δvap

H 0 is the standard enthalpy of vaporization.

Amount of heat required to vaporizeone mole of a liquid at constanttemperature and under standard pressure(1bar) is called its standard enthalpy ofvaporization or molar enthalpy ofvaporization, ΔΔΔvapH

0.

Sublimation is direct conversion of a solidinto its vapour. Solid CO2 or ‘dry ice’ sublimesat 195K with Δ

subH 0=25.2 kJ mol–1;

naphthalene sublimes slowly and for this73.0ΔsubH =V .

Standard enthalpy of sublimation,Δ

subH 0 is the change in enthalpy when one

mole of a solid substance sublimes at aconstant temperature and under standardpressure (1bar).

The magnitude of the enthalpy changedepends on the strength of the intermolecularinteractions in the substance undergoing thephase transfomations. For example, the stronghydrogen bonds between water molecules holdthem tightly in liquid phase. For an organicliquid, such as acetone, the intermolecular

CHEMISTRY166

dipole-dipole interactions are significantlyweaker. Thus, it requires less heat to vaporise1 mol of acetone than it does to vaporize 1 molof water. Table 6.1 gives values of standardenthalpy changes of fusion and vaporisationfor 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 100°C.

ΔvapHV for water

at 373K = 40.66 kJ mol–1

Solution

We can represent the process ofevaporation as

v218gH O(l) ⎯⎯

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

1

18g18g mol−= =

Δ = Δvap vapU HV

(assuming steam behaving as an idealgas).

vap gn R

(1)(8.314 J

Δ − Δ

−

HV

40.6

37.56

Δ =

=vapU

V

(c) Standard enthalpy of formation

The standard enthalpy change for theformation 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 ΔΔΔfH

0, wherethe subscript ‘ f ’ indicates that one mole ofthe compound in question has been formed inits standard state from its elements in theirmost stable states of aggregation. The referencestate of an element is its most stable state ofaggregation at 25°C and 1 bar pressure.For example, the reference state of dihydrogenis H2 gas and those of dioxygen, carbon andsulphur are O2 gas, Cgraphite and Srhombic

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

2 2H (g) ½O (g

285

+

Δ = −f HV

C (graphite, s) 22H (g) CH+ →

Table 6.1 Standard Enthalpy Changes of Fusion and Vaporisation

(Tf and Tb are melting and boiling points, respectively)

THERMODYNAMICS 167

74.8f HΔ = −V

( )2C graphite,s

It is important to understand that astandard molar enthalpy of formation, Δ

fH0,

is just a special case of ΔrH0, where one mole

of 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 (+ 178f HΔ = −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, Δ

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 HΔ = ΔV .

Therefore, by dividing all coefficients in thebalanced equation by 2, expression for

Table 6.2 Standard Molar Enthalpies of Formation (ΔΔΔf H�) at 298K of a

Few Selected Substances

CHEMISTRY168

enthalpy of formation of HBr (g) is written as

2 2½H (g) ½Br+

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

By convention, standard enthalpy forformation, Δ

fH 0, of an element in reference

state, 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) C→

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 fi

H HΔ = Δ∑V V

(6.15)where a and b represent the coefficients of theproducts and reactants in the balancedequation. Let us apply the above equation fordecomposition of calcium carbonate. Here,coefficients ‘a’ and ‘b’ are 1 each.

Therefore,

= [Δ Δr fH HV V

=1( 635.1 kJ −

= 178.3 kJ mol–1

Thus, the decomposition of CaCO3 (s) is anendothermic process and you have to heat itfor getting the desired products.

(d) Thermochemical equationsA balanced chemical equation together withthe value of its Δ

rH is 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 thecombustion of liquid ethanol at constanttemperature and pressure. The negative signof enthalpy change indicates that this is anexothermic reaction.

It would be necessary to remember thefollowing conventions regarding thermo-chemical equations.

1. The coefficients in a balanced thermo-chemical equation refer to the number ofmoles (never molecules) of reactants andproducts involved in the reaction.

2. The numerical value of ΔrH0 refers to the

number of moles of substances specifiedby an equation. Standard enthalpy changeΔ

rH 0 will have units as kJ mol–1.

To illustrate the concept, let us considerthe calculation of heat of reaction for thefollowing reaction :

( )2 3Fe O s 3H+

From the Table (6.2) of standard enthalpy offormation (Δ

f H0), we find :

( )2H O,f H lΔ V = –285.83 kJ mol–1;

( 2 3Fe O ,sf HΔ V = – 824.2 kJ mol–1;

Also (Fe

(H

f

f

H

H

Δ

Δ

V

V

Then,

1r HΔ V = 3(–285.83 kJ mol–1)

– 1(– 824.2 kJ mol–1)

= (–857.5 + 824.2) kJ mol–1

= –33.3 kJ mol–1

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

rH 0 is

THERMODYNAMICS 169

kJ mol–1, which means per mole of reaction.Once we balance the chemical equation in aparticular way, as above, this defines the moleof reaction. If we had balanced the equationdifferently, for example,

( )2 3

1 3Fe O s

2 2+

then this amount of reaction would be onemole of reaction and Δ

rH 0 would be

(2

3= 28

2Δ −r H

V

(1824.2kJ

2− −

= (– 428.7 + 412.1) kJ mol–1

= –16.6 kJ mol–1 = ½ 1r HΔ V

It shows that enthalpy is an extensive quantity.

3. When a chemical equation is reversed, thevalue of Δ

rH 0 is reversed in sign. For

example

(2 2N (g) 3H g+

3 22NH (g) N→

(e) Hess’s 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 Hess’s 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, someCO2 gas is always produced in this reaction.Therefore, we cannot measure enthalpy changefor the above reaction directly. However, if wecan find some other reactions involving relatedspecies, it is possible to calculate the enthalpychange for the above reaction.

Let us consider the following reactions:

( )C graphite,s(i)

( ) (2

1CO g O

2+

(ii)

We can combine the above two reactionsin such a way so as to obtain the desiredreaction. To get one mole of CO(g) on the right,we reverse equation (ii). In this, heat isabsorbed instead of being released, so wechange sign of Δ

rH 0 value

( )2CO g CO→

283.r HΔ = +V (iii)

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

( )C graphite,s

for which (= 39r HΔ −V

= – 110.5 kJ mol–1

In general, if enthalpy of an overall reactionA→B along one route is Δ

rH and Δ

rH1, Δ

rH2,

ΔrH3..... representing enthalpies of reactions

leading to same product, B along anotherroute,then we have

ΔrH = Δ

rH1 + Δ

rH2 + Δ

rH3 ... (6.16)

CHEMISTRY170

4 10

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 : ΔΔΔcH

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 (C4H10). During complete combustionof one mole of butane, 2658 kJ of heat isreleased. We can write the thermochemicalreactions for this as:

C13

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) 6+

ΔOur body also generates energy from food

by 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 H2O (1) areproduced and 3267.0 kJ of heat isliberated. Calculate the standard enthalpyof formation, Δ

f H 0 of benzene. Standard

enthalpies of formation of CO2(g) and

2

( )

H O(l) are –393.5 kJ mol–1 and – 285.83kJ mol–1 respectively.

Solution

The formation reaction of benezene isgiven by :

6C graphite

The enthalpy of combustion of 1 mol ofbenzene is :

( )6 6

15C H l O

2

cH

+

Δ

( )

The enthalpy of formation of 1 mol ofCO2(g) :

C graphite

f H

+

Δ V

The enthalpy of formation of 1 mol ofH2O(l) is :

( ) (2 2

1H g O

2

f H

+

Δ = −V

( )

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

6C graphite

Δ

( )2 2

33H g O

2+

Δ

( )

Summing up the above two equations :

6C graphite +

It can be represented as:

A B

C D

ΔrH1

ΔrH2

ΔrH3

ΔrH

THERMODYNAMICS 171

(i) Bond dissociation enthalpy

(ii) Mean bond enthalpy

Let us discuss these terms with referenceto diatomic and polyatomic molecules.

Diatomic Molecules: Consider the followingprocess in which the bonds in one mole ofdihydrogen gas (H2) are broken:

H2(g) → 2H(g) ; ΔH–HH0 = 435.0 kJ mol–1

The enthalpy change involved in this processis the bond dissociation enthalpy of H–H 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) ; ΔC–ClH0 = 242 kJ mol–1

O2(g) → 2O(g) ; ΔO=OH 0 = 428 kJ mol–1

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, CH4. Theoverall thermochemical equation for itsatomization reaction is given below:

4CH (g) C(g→

Δ

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 (g→

3 2CH (g) CH (g→

2CH (g) CH(g→

CH(g) C(g)→ +

Therefore,

4CH (g) C(g)→

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

3218Δ = −f HV

Reversing equation (ii);

( )2 26CO g 3H

f H

+

Δ V

( )

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

6C graphite

(b) Enthalpy of atomization(symbol: ΔΔΔaH

0)

Consider the following example of atomizationof dihydrogen

H2(g) → 2H(g); ΔaH 0 = 435.0 kJ mol–1

You can see that H atoms are formed bybreaking H–H bonds in dihydrogen. Theenthalpy change in this process is known asenthalpy of atomization, Δ

aH 0. It is the

enthalpy change on breaking one mole ofbonds completely to obtain atoms in the gasphase.

In case of diatomic molecules, likedihydrogen (given above), the enthalpy ofatomization is also the bond dissociationenthalpy. The other examples of enthalpy ofatomization can be

CH4(g) → C(g) + 4H(g); ΔaH 0 = 1665 kJ mol–1

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

Na(s) → Na(g) ; ΔaH0 = 108.4 kJ mol–1

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

(c) Bond Enthalpy (symbol: ΔΔΔbondH0)

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.

CHEMISTRY172

For example in CH4, ΔC–HH 0 is calculated as:

C H = ¼ (−Δ ΔHV

= 416 kJ mol–1

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

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

in a great deal*. Using Hess’s 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 ofold 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, Δ

rH0 is related

to bond enthalpies of the reactants and

products in gas phase reactions as:

bon

b

r HΔ =

−∑∑

V

(6.17)**This relationship is particularly more

useful when the required values of Δf H0 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 ingaseous state.

(d) Enthalpy of Solution (symbol : ΔΔΔsolH0 )

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

** If we use enthalpy of bond formation, (Δf H

bond ), which is the enthalpy change when one mole of a particular type of

bond is formed from gaseous atom, then r f bondsH HΔ = Δ∑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 H347 293 351 439 289 264 259 330 276 238 C

159 201 272 - 209 - 201 243 - N138 184 368 351 - 205 - 201 O

159 540 490 327 255 197 - F176 213 226 360 289 213 Si

213 230 331 272 213 P213 251 213 - S

243 218 209 Cl192 180 Br

151 I

Table 6.3(a) Some Mean Single Bond Enthalpies in kJ mol–1

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

N ≡ N 946 C ≡ C 837

C = N 615 C = O 741

C ≡ N 891 C ≡ O 1070

Table 6.3(b) Some Mean Multiple Bond Enthalpies in kJ mol–1

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

THERMODYNAMICS 173

in a specified amount of solvent. The enthalpyof solution at infinite dilution is the enthalpychange observed on dissolving the substancein 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 N+ − →

ΔSince it is impossible to determine lattice

enthalpies directly by experiment, we use anindirect method where we construct anenthalpy diagram called a Born-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, 108.4Δ =subHV

2. Na(g) Na (g+→ , the ionization of

sodium atoms, ionization enthalpyΔ

iH0 = 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), Δsol

H0, inwater is, therefore, determined by the selectivevalues of the lattice enthalpy,Δ

latticeH0 and

enthalpy of hydration of ions, Δhyd

H0 as

=Δ Δsol latticeH HV

For most of the ionic compounds, Δsol

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 ofenthalpy 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.

CHEMISTRY174

bond dissociation enthalpy.

112

2Δ =bondHV

.

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

chlorine atoms. The electron gain enthalpy,Δ

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 ( becauseΔn

g = 2) and is equal to + 783 kJ mol–1.

Now we use the value of lattice enthalpy tocalculate enthalpy of solution from theexpression:

sol latticHΔ = ΔV

For one mole of NaCl(s),lattice enthalpy = + 788 kJ mol–1

and Δhyd

H 0 = – 784 kJ mol–1( from the literature)Δ

sol H 0 = + 788 kJ mol–1 – 784 kJ mol–1

= + 4 kJ mol–1

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 processeswhether chemical or physical will tend toproceed spontaneously in one direction only.For example, a gas expanding to fill theavailable volume, burning carbon in dioxygengiving 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 processeswhether these will take place or not.

Let us first understand what do we meanby 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 Affinity

Ionization 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

ΔrH0(T ) = Δ

rH0(0) +

0

dT

r pC TΔ∫V

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

rC

p

0 = + 5/2 R (for ionization) Δ

rC

p

0 = – 5/2 R (for electron gain)Therefore,Δ

rH0 (ionization enthalpy)

= E0 (ionization energy) + 5/2 RTΔ

rH0 (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 Hess’s law, we get,

411Δ =latticeHV

7latticeHΔ = +V

for NaCl(s) Na→

THERMODYNAMICS 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 toproceed without the assistance of externalagency’. 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 forspontaneity ?

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:

12

N2(g) + 32

H2(g) = NH3(g) ;

Δr H0 = – 46.1 kJ mol–1

12

H2(g) + 12

Cl2(g) = HCl (g) ;

Δr H0 = – 92.32 kJ mol–1

H2(g) + 12

O2(g) → H2O(l) ;

Δr H0 = –285.8 kJ mol–1

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:

12

N2(g) + O2(g) → NO2(g);

Δr H0 = +33.2 kJ mol–1

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

0 = +128.5 kJ mol–1

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 spontaneityThen, what drives the spontaneous process ina given direction ? Let us examine such a casein which ΔH = 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

CHEMISTRY176

isolated from the surroundings as shown inFig. 6.11.

At this point, we introduce anotherthermodynamic function, entropy denoted asS. The above mentioned disorder is themanifestation 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 wouldbe to relate this process to the heat involved ina process which would make entropy athermodynamic concept. Entropy, like anyother thermodynamic property such asinternal energy U and enthalpy H is a statefunction and ΔS is 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 ΔS with q ? Wait !Experience suggests us that the distributionof 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 movablepartition [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 containerwhen 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 !

THERMODYNAMICS 177

temperature. This suggests that the entropychange is inversely proportional to thetemperature. ΔS is related with q and T for areversible reaction as :

ΔS = revq

T(6.18)

The total entropy change ( ΔStotal

) for thesystem and surroundings of a spontaneousprocess is given by

total systeS SΔ = Δ (6.19)

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

We can say that entropy for a spontaneousprocess 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

ΔSsys = ,sys revq

T

We find that both for reversible andirreversible expansion for an ideal gas, underisothermal conditions, ΔU = 0, but ΔS

total i.e.,

sys surrS SΔ + Δ is not zero for irreversible

process. Thus, ΔU does not discriminatebetween reversible and irreversible process,whereas ΔS does.

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 2NaHCO

(iv) ( ) (2H g 2H g→

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, thesebegin 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 increasesleading 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 JK–1mol–1at298 K. Inspite of negative entropy changeof this reaction, why is the reactionspontaneous?

(ΔrH 0for this reaction is

–1648 × 103 J mol–1)

Solution

One decides the spontaneity of a reactionby considering

(total sysS SΔ Δ + . For calculating

ΔSsurr

, we have to consider the heatabsorbed by the surroundings which isequal to – Δ

rH0... At temperature T, entropy

change of the surroundings is

ΔΔ = − rsurr

HS

T

( 1648 10

298

− ×= −

15530 JK m−=Thus, total entropy change for thisreaction

CHEMISTRY178

5530 JKr totalSΔ =

4980.6 JK−=This shows that the above reaction isspontaneous.

(c) Gibbs energy and spontaneityWe have seen that for a system, it is the totalentropy change, ΔS

total which 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 inprevious 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, G is an extensive propertyand a state function.

The change in Gibbs energy for the system,ΔG

sys can be written as

=sys sysG HΔ Δ −

At constant temperature, 0TΔ =

=sys syG H∴ Δ Δ

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

G H T SΔ = Δ − Δ (6.21)

Thus, Gibbs energy change = enthalpychange – temperature × entropy change, andis referred to as the Gibbs equation, one of themost important equations in chemistry. Here,we have considered both terms together forspontaneity: energy (in terms of ΔH) andentropy (ΔS, a measure of disorder) asindicated earlier. Dimensionally if we analyse,we find that ΔG has units of energy because,both ΔH and the ΔT S are energy terms, sinceTΔS = (K) (J/K) = J.

Now let us consider how GΔ is related toreaction spontaneity.

We know,

ΔStotal

= ΔSsys

+ ΔSsurr

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,

=ΔΔ surr

surr

HS

T

Δ = Δtotal sysS S

Rearranging the above equation:

TΔStotal

= TΔSsys

– ΔHsys

For spontanious process, 0Δ >totalS , so

TΔSsys

– ΔHsys 0>

(⇒ − Δ −sysH T

Using equation 6.21, the above equation canbe written as

0−Δ >G

Δ = Δ − ΔG H T (6.22)

ΔHsys

is the enthalpy change of a reaction,TΔS

sys is the energy which is not available to

do useful work. So ΔG 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.

ΔG gives a criteria for spontaneity atconstant pressure and temperature.

(i) If ΔG is negative (< 0), the process isspontaneous.

(ii) If ΔG is positive (> 0), the process is nonspontaneous.

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

THERMODYNAMICS 179

(a) The positive entropy change of the systemcan be ‘small’ in which case T must be large.(b) The positive entropy change of the systemcan be ’large’, in which case T may 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 oflower free energy.

So, the criterion for equilibrium

A B C D+ +ƒ ; is

ΔrG = 0

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

rG

0 is related to the equilibrium constant of

the reaction as follows:

0 = ΔrG0 + RT ln K

or ΔrG0 = – RT ln K

or ΔrG0 = – 2.303 RT log K (6.23)

We also know that

= r rG HΔ Δ −V V (6.24)

For strongly endothermic reactions, thevalue of Δ

rH0 may be large and positive. In

such a case, value of K will be much smallerthan 1 and the reaction is unlikely to formmuch product. In case of exothermic reactions,Δ

rH0 is large and negative, and Δ

rG0 is likely to

be large and negative too. In such cases, K willbe much larger than 1. We may expect stronglyexothermic reactions to have a large K, andhence can go to near completion. Δ

rG0 also

depends upon ΔrS0, if the changes in the

entropy of reaction is also taken into account,the value of K or extent of chemical reactionwill also be affected, depending upon whetherΔ

rS0 is positive or negative.

ΔrH0 Δ

rS0 Δ

rG0 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

CHEMISTRY180

Using equation (6.24),

(i) It is possible to obtain an estimate of ΔGV

from the measurement of ΔHV and ΔSV,and then calculate K at any temperaturefor economic yields of the products.

(ii) If K is measured directly in the laboratory,value of ΔG0 at any other temperature canbe calculated.

Problem 6.11

Calculate ΔrG0 for conversion of oxygen

to ozone, 3/2 O2(g) → O3(g) at 298 K. if Kp

for this conversion is 2.47 × 10–29.

Solution

We know ΔrG0 = – 2.303 RT log K

p and

R = 8.314 JK–1 mol–1

Therefore, ΔrG0 =

– 2.303 (8.314 J K–1 mol–1) × (298 K) (log 2.47 × 10–29)

= 163000 J mol–1

= 163 kJ mol–1.

Problem 6.12

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

( )32NH g CO+ .

Standard Gibbs energy change, ΔrG0 at

the given temperature is –13.6 kJ mol–1.

Solution

We know, log K =–

2.303RrG

T

Δ V

= ((–13.6

2.303 8.314

×

= 2.38Hence K = antilog 2.38 = 2.4 × 102.

Problem 6.13

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

SolutionN2O4(g) ˆ ˆ ˆ†‡ ˆ ˆˆ 2NO2(g)If N2O4 is 50% dissociated, the molefraction of both the substances is givenby

2 4N Ox =

1 0.51 0.5

−+

; 2NOx =

2 0.51 0.5

×+

2 4N Op =

0.51 atm,

1.5×

2NOp =

1

1 atm.1.5

×

The equilibrium constant Kp is given by

Kp =

( )2

2 4

2

NO

N O (1.5=

p

p

= 1.33 atm.SinceΔ

rG0 = –RT ln K

p

ΔrG0 = (– 8.314 JK–1 mol–1) × (333 K)

× (2.303) × (0.1239)= – 763.8 kJ mol–1

THERMODYNAMICS 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 via ΔU = q + w. ΔU, change in internal energy, depends on initial andfinal states only and is a state function, whereas q and w depend on the path and arenot the state functions. We follow sign conventions of q and 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 = CΔT. Work can be measured by w = –p

exΔV, in case of expansion

of gases. Under reversible process, we can put pex = p for infinitesimal changes in the

volume making wrev = – p dV. In this condition, we can use gas equation, pV = nRT.

At constant volume, w = 0, then ΔU = qV , heat transfer at constant volume. But in

study of chemical reactions, we usually have constant pressure. We define anotherstate function enthalpy. Enthalpy change, ΔH = ΔU + Δn

gRT, can be found directly from

the heat changes at constant pressure, ΔH = 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 Hess’s law. Enthalpychange for chemical reactions can be determined by

(Δ = Δ∑r i ff

H a H

and in gaseous state by

ΔrH0 = Σ bond enthalpies of the reactants – Σ bond enthalpies of the products

First law of thermodynamics does not guide us about the direction of chemicalreactions i.e., what is the driving force of a chemical reaction. For isolated systems,ΔU = 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, ΔU = 0, ΔS > 0, so entropy change distinguishesa spontaneous change, while energy change does not. Entropy changes can be measured

by the equation ΔS = revq

T for a reversible process. revq

T is 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:

ΔrG = Δ

rH – T Δ

rS

For a spontaneous change, ΔGsys < 0 and at equilibrium, ΔGsys

= 0.

Standard Gibbs energy change is related to equilibrium constant by

ΔrG0 = – RT ln K.

K can be calculated from this equation, if we know ΔrG0 which can be found from

r rG H TΔ = Δ −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.

CHEMISTRY182

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) ΔT = 0

(ii) Δp = 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 ΔU 0of combustion of methane is – X kJ mol–1. The value of ΔH0 is

(i) = ΔU 0

(ii) > ΔU 0

(iii) < ΔU 0

(iv) = 0

6.5 The enthalpy of combustion of methane, graphite and dihydrogenat 298 K are, –890.3 kJ mol–1 –393.5 kJ mol–1, and –285.8 kJ mol–1

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

(i) –74.8 kJ mol–1 (ii) –52.27 kJ mol–1

(iii) +74.8 kJ mol–1 (iv) +52.26 kJ mol–1.

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 carriedout in a bomb calorimeter, and ΔU was found to be –742.7 kJ mol–1

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

NH2CN(g) + 32

O2(g) → N2(g) + CO2(g) + H2O(l)

THERMODYNAMICS 183

6.9 Calculate the number of kJ of heat necessary to raise the temperatureof 60.0 g of aluminium from 35°C to 55°C. Molar heat capacity of Alis 24 J mol–1 K–1.

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

fusH = 6.03 kJ mol–1 at 0°C.

Cp [H2O(l)] = 75.3 J mol–1 K–1

Cp [H2O(s)] = 36.8 J mol–1 K–1

6.11 Enthalpy of combustion of carbon to CO2 is –393.5 kJ mol–1. Calculatethe heat released upon formation of 35.2 g of CO2 from carbon anddioxygen gas.

6.12 Enthalpies of formation of CO(g), CO2(g), N2O(g) and N2O4(g) are –110,– 393, 81 and 9.7 kJ mol–1 respectively. Find the value of Δ

rH for the

reaction:

N2O4(g) + 3CO(g) → N2O(g) + 3CO2(g)

6.13 Given

N2(g) + 3H2(g) → 2NH3(g) ; ΔrH0 = –92.4 kJ mol–1

What is the standard enthalpy of formation of NH3 gas?

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

CH3OH (l) + 32

O2(g) → CO2(g) + 2H2O(l) ; ΔrH0 = –726 kJ mol–1

C(g) + O2(g) → CO2(g) ; ΔcH0 = –393 kJ mol–1

H2(g) + 12

O2(g) → H2O(l) ; Δf H0 = –286 kJ mol–1.

6.15 Calculate the enthalpy change for the process

CCl4(g) → C(g) + 4 Cl(g)

and calculate bond enthalpy of C – Cl in CCl4(g).

Δvap

H0(CCl4) = 30.5 kJ mol–1.

ΔfH0 (CCl4) = –135.5 kJ mol–1.

ΔaH0 (C) = 715.0 kJ mol–1 , where Δ

aH0 is enthalpy of atomisation

ΔaH0 (Cl2) = 242 kJ mol–1

6.16 For an isolated system, ΔU = 0, what will be ΔS ?

6.17 For the reaction at 298 K,

2A + B → C

ΔH = 400 kJ mol–1 and ΔS = 0.2 kJ K–1 mol–1

At what temperature will the reaction become spontaneousconsidering ΔH and ΔS to be constant over the temperature range.

6.18 For the reaction,

2 Cl(g) → Cl2(g), what are the signs of ΔH and ΔS ?

6.19 For the reaction

2 A(g) + B(g) → 2D(g)

ΔU 0 = –10.5 kJ and ΔS0 = –44.1 JK–1.

Calculate ΔG0 for the reaction, and predict whether the reactionmay occur spontaneously.

CHEMISTRY184

6.20 The equilibrium constant for a reaction is 10. What will be the valueof ΔG0 ? R = 8.314 JK–1 mol–1, T = 300 K.

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

12

N2(g) + 12

O2(g) → NO(g) ; ΔrH0 = 90 kJ mol–1

NO(g) + 12

O2(g) → NO2(g) : ΔrH0= –74 kJ mol–1

6.22 Calculate the entropy change in surroundings when 1.00 mol ofH2O(l) is formed under standard conditions. Δ

f H0 = –286 kJ mol–1.