Definitions of the following:

.
Heat of reaction : Heat of formation-standard heat of formation, Heat of solution.

Heat of dilution, Heat of neutralization, Heatofco mbustio n.

. cids and strong bases R ,
Constancy m the heat of neutraliz ation: Expenm . . . ·n case of strong a · eason ~
ental venficat,on I or~~
observation-ionic neutralization and the heat evolved.
Definition of Calorific value of fuel.
Stateme nt of Hess' Law and its application. Problems based on Hess' Law.
... . h . E t
(111) Second Law of Thermodynamics and its significane, spontan ·ty of a chemical c ange, n ropy, Free E
ei "ble (recapitualtion), spontaneous ne~,
Inadequacy of First Law and need for Second Law; Ideas about revers,
spontanceous processes
and n0tt.
. . . . . 1
Mean mg of entropy - derived from Second Law- statement of Second Law · t ms of entropy; Physica l s1gn,f1cane ofe t ,
m eri n ropy,
· . 'bl I·sotherinal process and irreversible pro
State function and not path function. Entropy change of the universe , revers, e cess. 1
Meaning of thermal death, Gibb's fr.ee energy of the system and Helmhol
tz free energy. Relation ship between Gibb's free energy!
and Helmholtz's free energy.
'
Relation ship between change in Gibb's free energy and equilibrium constan
t of a chemica l reaction . Defining the criteria for
spontan eity of chemica l change in terms of Gibb's free energy.
Note: Numeric als based on the First Law, Second Law of Thermodynamics and
Hess' Law.
(iv) Third Law of Thermodynamics - statement only.
Self explanatory.

6.1. GENERAL INTRODUCTION (ii) It helps to deduce some generalisations of phy~~

One very importan t application of chemical reactions is to chemistry. Laws of thermod ynamics can be used to deduce solfi
produce usable energy. Power companies burn coal, oil or gas to importan t generali sations of physical chemistr y such as van't Hon
generate electrici ty to power factories or to light our homes . law of dilute solution s, Raoult' s law of vapour pressure lowerin~
Breakdo wn of food in our bodies provides energy for doing work distribution law, phase rule and law of chemica l equilibrium.
and for mainten ance of our body temperature. We use the energy (iii) It helps in predicting the yield of products. Certam
from burning gasoline to move us in a car or bus from home to work mathema tical relation ships of thermod ynamics can be used w
or to school. The study of energy changes during chemica l
determin e the extent to which a process including a chemici
reaction s is an importan t aspect of chemistry.
reaction would proceed till the establish ment of equilibrium.~
The branch of chemis try which deals with energy can be used to predict the yield of products .
change s taking place during various physica l
proces ses and chemic al reactio ns is called (iv) It helps in genera lizing experimental resu\11
chemic al energe tics or thermo dynami cs. Thermod ynamics offers the means by which one can generaW
the results of many diffe~ent experim ents. For example, with ~
Thermo dynamic s is based on three generalisations known
as First, Second and Third law of thermod ynamics . Since help of these laws, equation s relating to effect of temperature a&·
thermod ynamic laws deal with energy, they are applicable to all the pressure on certain physical properti es have been developed
phenom ena of nature. There is no formal proof for these laws. 6 •3 • LIMITATION
Howeve r, nothing contrary to these laws has been known to happen S OF THERMODYNAMICS
in case of macrosc opic systems, i.e., systems comparatively large
and involvin g many molecul es, ions, etc.
. T~e~od ynamic s is a powerf~ l theoretical tool which
~pp~ic~tion m all branche s of science. However, it has cef1t
fi~;
It is importan t to note that the laws of thermodynamics are lurutah~ns also, e.g.,
not concerned at all with the atomic or molecular structure of matter. .fz) The _laws of thermod ynamics apply only to matteri11 ~
and not to ind· 'd l
6.2. OBJE CTIV ES (IMPO RTAN CE) zvz ua atoms or molecul es In ot ber wow .~
OF thermodyn ami d . · I cule1
. cs ea1s with large groups of atoms, mo e
THERMODYNAMICS tons rather than · d. •d
m 1v1 ual atoms molecul es or ions. ~
The primary objectives of chemical thermodynamics are : (ii)Itca nonl
Y pre d'zct the possibil
1
. . ' b111 n °
success. This is be ity of a process e55 v.
h . .
(i) To establish criterio n/criteria for predicting the
b ase d only on th cause t e cntenon of feasib1.!tty . Of proC
feasibil ity (or spontan eity) of a physica l or chemic a l · ·• a 5o~
e mittal and final states of a process. It doe
transformation under a given set of conditions. Thennodynamics concern itself with the hurdles in the path of the change. .
helps to predict whether a given process or a chemical reaclion is ( • • ') /
(IOII•
'· ILL I does nor tell anything about the rare of reac , ~i
feasible or not under given conditions of temperature, pressure (i,v) Thermo d 01
·
and concentration. vi' .
mec I·1an1.sm of a process
_vn.amics does n.ot throw any, /ighl
. .
 ..aJCAL THERM ODYNA MICS 6/3
cttE""
DEFIN ITION S OF SOME COM to the system. Steam remains inside the vessel. Thus, matter is not
6,A, rHERM ODYN AMIC TERMS MON exchange d.
(e) Isolated system. A system which can neither exchange
sefore starting the st~~y of tbcrmody namics, it is somewha t matter nor energy with the surround ings is called an isolated
s of the terms which are emp1oyed
seno·a1 to 1cnow the defirut1on
d .
m the study of thennod • · system. For example, take some water in an insulated _v essel ~d
es coJll11lOD terms use
50rnt ribed below. ynarrucs are put a small piece of sodium metal in it. An exotherm ic reactl~n
briet1Ydesc takes place. Neither the hydrogen gas (matter) nor heat (energy) is
for
1. Syste_m . The_ part of the uni v ers e s elected transferred to the surroundi ngs (Fig. 6.lc).
tberJ11odyna m1c study IS termed as a system. Boiling water in a
3ker is an example of a system. CLOSED
t,e ding Th . . METALLIC
z.Su rroun. b . s. eremaun ngpartof tbeuniverse aroun d VESSEL
the system whi_c : not un~er study is called surround ings.
consider a reaction et'>-'.een zmc and dilute sulphuric acid in a test
HEAT
(Ube-Here the reactants m_ the test tube form a system. Everythin g ~
:-:-:-:-:-: -:-:-
-:-.-_-_-_-_--:-:
HEAT>
~
else around the test tube 1s the surround ings.
~
--- WATER :..-_
-------
___________
---- ----_
___
----------
3. Boundar y. The region separatin g the system from the
surrouod~gs is ~alled t~e ~oundar y. Let us consider a gas (Open system) (Closed system)
contained m a cylinder which 1s placed in a constant temperau re (a) (b)
bath maintained at 298 K. In this case, gas constitute s the system,
INSULATION-..J~:7Z.222Z2Z?.:7Z.2~
the walls of the cylinder constitut e the boundari es and the constant
temperature bath is the surround ings.
H2 gas
rypes of bou ndarie s or walls
SODIUM - - , 1 , ~ - ,
(i) A rigid wall : A wall whose shape and position are fixed . METAL

(ii) An impermeable wall : A wall that prevents the passage of

matter through it.
(iii) A permeab le wall : A wall that allows the passage of matter
and energy through it. (Isolated system)
(c)
(iv) A diatberm ic or diatherm al wall : A wall that prevents
the passage of matter but allows the flow of the energy Fig . 6.1 Types of systems .
through it.
if) Macrosco pic system. A system which consists of a large
(v) An adiabatic wall : A wall or boundary that prevents the
nuniber of atoms, ions or molecules is called a macrosco pic system.
passage of matter or energy provided no external force is The measurable properties associated with such a system are called
present (e.g., thermos bottle) . macrosco pic propertie s. For example, temperat ure, pressure ,
4. Types of Systems. The systems may be classified into the concentration, mass, composit ion, density, refractive index, surface
following types : tension, viscosity, internal energy, enthalpy , fugacity , etc. are
(a) Ideal system. In pencil and paper (Theoreti cal) work,_the examples of macrosco pic propertie s of the system.
system treated is called ideal system. However , i_n actual p~actI~e, ~ Homogen eous and hetergen eous systems . (i) Homo-
no system is ideal. An ideal system is always considere d to sunplify geneous system. A system is said to be homo~en eous if it is ~ onn
throug!]oy t For example, a gure solid, or a liquid, a gas or a mixutre
the thermodynamic problems . of gases or a solution of a solid in a liquid or a mixture of two or
. k the system is called
(b) Real system. In expenme nta1wor , more complete ly miscible liquids are all homogen eous systems. In
real system. other words, a homogen eous system consists of only one phase.*
/ . h change matter as well
(c) Open system. A system whic can ex t ~ terogene ous system. A system is said to b e
as energy with the surroundi ngs is called an open syS em, e.g.' heteroge neous if it is not uniform througho ut. A heterogen eous
b aker represent s an open
. t0 the atmosphe re system consists of two or more phases, which are separated from
(I) Evaporati on of water from a e
system. Here vapours of water (matter) move
10
f m the each other by definite bounding surfaces. However , each phase in
and heat (e~ergy) required is absorbed by water ro itself is uniform througho ut. Some common example s of
heterogen eous systems are (i) a mixture of two or more solids (ii) a
surround·mgs (F'1g. 6. 1a ) • b dioxide,
mixture of two or more immiscib le liquids (~a liquid in contact
\\I (ii) In the photosyn thesis, plants take_ u~hc:~: :sence of with itg, vapours.
1
ch~ter (matte r) and sunlight (energy) ~is rocess, oxygen ~ -Macros copic properties (Thermo dynamic Properti es ).
(,n orophy~l and produce carbohydrates. In . s pHence, plants Thermod ynamics is mainly concerne d with the bulk propertie s of
atter) is transEerred to the surround ing . the materials such as mass , pressure, volume, temperat ure, surface
cons • :J'
tztute an open system. tension and viscosity, etc. The properties associated with the bulk
'(d} hi b may exch an~ rgy
0aI d metallic vessel.
closed--SY"Stem. For
but 'd) Closed system. A sys~ m w • A phase is homogeneous and physicalty distinct part of a system wh'ch
exnot lll~ ter with surrounofdwater . 10 a ~ ~~~er (surroundings)
is bounded by a surface and is mechanically separable from other part~ f
the sy stem. sO
r~~ple, consider boiling
6 transferre d from th
g. -lb). Here, heat is
1tt ((rc;o/-rvQ,, f)J(li)fk<JR/(OJ ·➔ ,/.:""'\
..... ., /,..- [ -
/ 1tO
_
(l~~~-(_li. ~
l.9<.v~c-- 1 < __;
I
_ JCf'
~
.; ;_e ·-; ,·.·-t:!~~~:~"
~(l(;
JI ,.-1i2 ' .- ,/ r

\~:;\ ~r i~f..~ ii~.. ".'l t;: :-r, ,ecl~,, J~).,//5,ttt:; ~;::;ji~M~~

- f / O )~ '<.~

r "6 J,1.,,.,~·' A~~ ;re

614
j ."' - t?r
of ma~r, i.e., with ma&oscopic sysreJ
} -cA..1'1t-~ Y1 ),,,
a
called macr~opi'
7 t . ;_RT (for one mole of the gas) where P is the pr
is pl - and T is the temperature and R is cons,,,-ets~Ure. \, ·
p~ L

vo u.me (P V d T) kn -..u . cv1.

properties. nl two of the variables , an ~e owu, the th ~ ~..
T_he macroscopic properties can be subdivided into two ~Ily calculated. Hence, the state 0 ~ a ~lillple hotnogeneolrd ~ .,
categones as follows : be completely defined by specifymg only two of ,1._ Us s1~.
may · bles ge erall "!C villi '
✓ (a) Extensive properties. T he properties which depend upon out of P, V and T. The two vana . n Yspecified are ~
the amount oft~ s_u~stanc~_!"_!t_!bstances present in the system and temperature. These are calledmdependentvariabJes. Pre~
are calJed ex tensive properties. Examples of such properties are variable, i.e., volume w~ose value depends upon the v}~
~ ' volume, heat capacity, internal energy, enthalpy, entropy, pressure and temperature is call~ dependent Variable. Iles •1
Gibbs free energy etc. Extensive properties of a single substance It is important to note tha~ m a closed system, cons· .
depend upon the number of moles (n) of the substance present one or more components, m ass 1s not a state Variable. IS~ ~
'.1°d also on an~ two of the three variables P, V and T (called Thermod ynamic equilibrium. A system is said
mdependent vanables). The total value of an extensive property is th
8
od amic equilibrium if the macr oscopicprope . 10~,
equal to the sum of the values for the separate parts into which the erm . yn . us phases do not undergo anv chang r:ttesor,
system may, for convenience, be divided. sys1em m 'ano th od . • 'lib . e "'1th ~ -
In . . Actually the term erm ynannc eqlll nu.m im li
Of three kinds of equilib · 10
tensive properties. Th e properties which de,Pend onlv ' • · P I!! ~
u nature of the substance and are independent of the!'llop.nt simuJtane~us eXIS tence n a the syste.:
'<ff' tbe su bstan ce present in the system are called intensive These are · .. . A . .
p roperties. (a) Thermal eq~bnum. system IS said to be in th~
Examples of intensive properties are pressure, temperature, equilibrium if there is no flow of heat from one portion ofr.t
density, viscosity, surface tension, dipole moment, dielectric system to a nother. This is possible only if the temperaturerelllati
constant, concentration, refractive index, electromotive force, th t
the same throughout in all parts of e syS em.
chemicalpotential, molefraction,freeziogpoint,boilingpoiot, (b) M echanjcal equilibrium . A system is said tohe i
etc. Intensive properties are not additive. When a system is di vided mechanical equiUbrium if no mechanical work is done byone 111
into a number of parts, each part will have the same value of of the system on a noth er part of the system. This is possible onJyi
intensive property . For example, temperature is an intens ive the pressure remains constant throughout in all pans ofthe s~
property because if we take one litre of water kept at 298 K and (cl Chemical eq uiJj brium. A sys1em is said to be in chema
then divide it into five parts each containing separately 500 mL, equilibrium iftl1e composition of th e various phases in the~
250 mL, 125 mL, 75 mL and 50 mL, the temperature of each part of docs not change ,\ith time.
the water will be the same. 9. Thermodynamic properties. As already stated, asySliil
It is important to note that if x and y are arbitrary extensive can be defined by measurable properties like pressure, temperatm.
variables then (x+y) is an extensive variable but xly and Ox !Oy are volume and the amount of substance (number of moles of vm
intensi ve variables. For example, mass and volume are extensive species present in th e sys te m ). These properties are calkJ
properties but density (MN ) and specific volume (V/M ) are thennodynamic propertie-s. Two additional quantities, i.e.,work(M I
intensive properties . The product, ratio and sum of in tensive and heat (q), due to their frequent occurrence, are also regarded t
properties are intensive. Thus an extensive property if expressed the thennodynamic properties. However, these two quantitiesatt
per mole or per gram or per unit volume becomes an intensive not the state variables. Some important thennodynamic functi~
property. Hence, molar heat capacity (heat capacity/mole), molar along with their mathematical definitions are listed in Table6.l.
energy (energy/mole), molar volume (volume/mole), specific heat
(he at c apacity/gram ), concen tration (moles/litre) , mole ~ --Ta~fe:~:~}iar~ous Thermodynamic Functions
fraction (n/N), molar U, A , G, H, S denoted by U, A, G, H, S Name of Symbol of Definition
are some examples of intensive properties.
6. S tate of system. The condition of existence of a system
Function
Internal energy
Function
U*orE dU or dE = dLJ - dw
---
when its macroscop_ic p roperties have a definite value, is called
the s tate of th e system. With the change in one or more Entropy s ds = dq,ev
macroscopic properties, system undergoes a change in state . T
When a process occurs, the state of the system changes. The Enthalpy H H=E+PVorU +PV
firs t and the last state of the system are called initial state and Gibb's Free energy
fmal state respectively.
7. State variables. Those macroscopic properties, which
Helmholtz free
energy (work function)
G
A =
G=H-TS
A E - TS or U - TS

,~~SllilQg~~ c~use a change in the state of the syste_m are call_ed

slate variables.The four most common macroscopic properties Chemical Potential µ - (-OG)
which are sufficient to define the state of a system are composition, µ i - anj T,P,nl i
pressure, volume and temperature. If these properties of the
system are fixed, all other physical properties of the system get
automatically fixed. Further, if the system is homogeneous and is
Heat capacity at

constant pressure
Cp Cp =(!~1
composed of a single substance, the composition is fixed (100%). u'
Hence, the state of the system can be defined by the use of pressure,
temperature and volume. In actual practice, it is not necessary to
Heat capacity at
C. C.= ("E)
aT
or(~h
V
constant volume
~
spec ify even th e se three variables becaus~ these are
interdependent. The relationship between them 1s called the *Accordin g to Late st IUPAC
recommendations Internal
equation of state. For example, the equation of state for an ideal gas represented by u. '
 -,JCAL THERMODYN AMIC S 6/5
,~fl"'
O vario us types of therm od ynan11c .
J. . b . b pro 1_g ~o! c~- A
tion winch nngs a out the changes. th cesse s. The force !s on]y infinitesimally greate!J!!_an_!! t~,1?,POSi! ibnum
ssion of equil
t1itr;; -<I the process. The different proce s~e m estat e of the system reversible process proceeds through a succe
mj s comm only encou ntered steps, each of which is an equilibrium state. A rever sible proce ss
~ell cthe study of chemical therm odyna
cs are as follows · enoug h time is allow ed
ol)llllg • · muSl be carried out in such a way that
ttziisotherntal process. A process • . erma l if between each change for the properties to come
to a c~nSla?t
m rema in is said to be isoth
c/rnpcrature of the syste and canno t be reahsed 10
we1c h s cons tant dur·
various value. Reversible processes are ideal
eration s. In_ sue a process, heat tan flow fro th mg practice. This is because a reversible process requires
infinite time
ermi c r . m e system to the
op ounding m case of exoth tor its completion and the proce ss is to be adjus ted by_an
s an~ fr?m the
:~undings to the system in case of endo ~:=n mfinitesimal amount at each stage. For this reaso n a reversible
an . th c reacti ons m order
1
rnaintain a constant temperature. For iso ermaI process df'=0 process is also called as quas istatic proce ss. Neve rthele ss, the
• • s to
iO
' · and serve s as a mean
{P) Adiabatic process. A process is s .d tO b . concept of reversibility is highly usefu l
to chem ists.
een ~h e adiab atic if no are of imme nse impo rtance
bea( excbange take s plac e betw e syste m and the study many processes which
t increasing
d' d .
surroun mgs urm~ any s ep of the process. In case of an a . . A reversible process can be reversed at any stage by
comp letely insul ated from th dia~atic the op~ g force by an infini tesim al small amou nt.
rocess, the system IS e surro undings
al t If • irreversibJe
P ·
andd_q is ~thiu. othzero. a process is exothermic, the heat
evolved ,Ab) Irreversib]e process. A process is said to be
l' (inste ad it is carri ed
rernruns w1 n e syste m. Cons equen tly the t if it is not carried out iqfmitesimaUy slow1
system rises. If on the other hand, the pro~ess ise:npdeortahture _of ththe out rapidly) so that the system does not e~.,_! £4. ~ !2,!_~ n
d t th . . ermic, e e proce ss canno t be rever sed without
·
heat requrre o carry out e reaction Is supplied bY e system
th equilibrium. An irreversfo
y and witho ut chang ing the prope rties
itself. The temperature of the system, therefore, falls. ·ffiehelpo i an external agenc
: of the surroundings.
From the above discussion, it may be concluded that the difference
co t t A few examples which may be used to illustrate
In an isothermal process, the temperature remains between reversib]e and irreversible proce sses are as follows :
the syste m can freely exchns an
during each step of the process. and ange an airtight,
'th th adiab atic proce ss, on the (i) Consider a gas enclosed in a cylinder fitted with
lhe energy w1 e surroundmgs. In an press ure P on
exchange weightless and frictionless piston (Fig. 6.2). Let the
other hand, the temperature may change because the free the piston be exactly equaJ to the int(?rnal press ure of the gas. In
undin gs is not possible.
of energy between the _syst~m and the surro such a situation, the piston will neither move upwa rd nor down ward .
impra cticab le becau se it is
However, a perfect adiabatic change is Consequently, there wil1 be no chang e in volum e of the gas
system from the surrou nding s.
not possible to completely insulate a (Fig. 6.2a). Now suppose the press ure on the piston is lower ed by
baric process. A process is said to be isobaric if the ntdP. The pressu re, P-dP on the piston
~ an infinitesimally small amou
d of the the piston
pressure of the system remains const ant during the perio b~ing infinitesimally smaller than the pressure of the gas,
298 Kand I
change. For example, conversion of 1 mole of water at , the gas will expan d (Fig. 6.2b) by an infini tesimal
will move up. Hence
is an isoba ric at P-dP , the
atrn pressure into vapour at 373 K and 1 atm pressure smal] amount. If the pressure on the gas is maintained
accom panie d y, i.e. , in a
process. For an isobaric process, dP=O but it may be gas wil1 continue to expand infinitesimal1y slowl
by change of volume. thermodynamically reversible manner.
oric if is equal
cJft)-fsochoric process. A process is said to be isoch , , ~en ~e pressure on the piston (external pressure)
const ant durin g each step of the s~a11 er ~an the intern al press ure of the gas,
the volume of the system remains to p w~ch 1s much
of the
sion
process, Combustion of a subst ance in a bomb calori meter is the the gas will expand rapidly (Fig. 6.2c) . Hence, the expan
For an isochoric process, dV=O gas wiU be irreversible .
example of an isochoric process.
but it may be accompanied by change of pressure.
number
In case of reactions in which there is no change in the
comb inatio n of
of moles of gaseous molecules, as for example, the (P-dP)
chloride :
hydrogen and chlorine to give gaseous hydrogen
H2 + Cl 2 - 2HC1,
th
tbere will be no chang e in volum e as well as pressure. Hence, e p«P

process W ilj, be isochoric as well as isobaric.

proces~ if
a {v.,Cyclic process. A proces_s is said to be a cyclic
ges retur ns to its
~Y~tes.?:.after-ompleting a series of chan
0
ngmal stat .
h · f m the~-~ initiaJ state to
;;.:;.;,......... (a) (b) (c)
th I at . Se ue c ste s startrn ro
Any partic uJar path
efinal th' t .
. e s s em 1s - · .. _ . Of Fig. 6.2 Reversible and irrev ersib le
out s_tate a expansion of a gas.
of many different possible ways from initia] to final
;stern is called a process. The path of a cyclic proce
ss ts called '------ ----~:.--
U!) To take_ano~her example, consider the working of
a
cycle. F. If the
directly or Galv ~ic cell w~1ch 1s coupled t.o an opposing E.M.
in The fmal state of a system can be attained either oppo~mg_E .M.~. ts ~qua_l to the ~-~ -F. of the cell, no curre nt will
are two types of paths of the sys:m: •
Costages..According]y, there e· flow m either direction m the c!fcu1 t (Fig 6 3a) If th 0
tly, there are two differ ent types of processes. These small er (very. ~lightl · e ) pthposmg
E.M.F . is made infinjtesim alJy 1
. , "d t0 be a rever•sible cell E•M•p ., a very smaJI current 1s .
· given Y ess an the
Pr ~ a) Reversible process. A process 1s sa• _so that the dnvm g
• out by the cell (F.1g . 6.3b) .
Oce.-ss ifit is carried out infinitesimalll:'. s1o~
......-,.r., - ·- -.___...
 6/6 ISC CHEMISTRY
r.:=:::================~~~====::::::::::::::::::::P~l\l,
The cell under these conditions is said to work reversibly. Now if (iii) Volume (i~) Density . t
the opposing E:M.F. is slig~tly greater than the cell E.M.F., a v~ry [Ans. (z) Surface tension (ii) De .
small current will start fl?wm~ in the opposing direction (Fig. 6.3c). . . ns1tyJ I
PRoBLEM 3. Which of the foUowmg JS an extensive llron.. _
The flow of current 10 this case is also taking place in a ~ cti ·
(u..) Re,ra
thennodynamically reversible manner. ;\ v· •ty ve mdex ..-cl't),?•
(t, JSCOSI
(iii) Heat capacity (iv) Specific heat
~ the above example, if the opposing E.M.F. is made
appreciably smaller or appreciably larger than the E.M.F. of the [Ans. Heat capa .
c~ll, ~en an appreciable amount of current will be flowing in one l.b===================================~;;c1~]
~ect:J.o~ or the other and the process becomes thennodynarnically 6.5. ENERGY
urevers1ble.
The energy is related to the capacity of the system to do
~ - Systems c~ h~veenergy m two ~ndamental ways};neuc
and Potential. Kinetic energy or K.E. 1s the energy of rn .
. k.I" 11 Olton
associated with mecharucal wor t is ca cu ated from a mo .
object's mass (m) and velocity (u) by the equation, VUtg
1 2
. . =-mv
KE 2
Potential energy or ~.E. is energy "in ~torage", existing
because there are in the umverse natural attractions or repulsio
that objects or their parts exp~rienc~ for each other. One impo~
property of potential energy 1s that it can be converted into kinetic
energy. Likewise, kinetic energy can be converted into potential energy,
Potential energy increases whenever objects that attract
each other are pulled apart, as in lifting a book to a higher level
above earth. Potential energy also increases whenever objects that
,L It occurs at a very slow It takes place rapidly and does naturally experience a repelling force are forced together, _as in
speed and involves a series not involve a series of
of equilibrium states. · equilibrium states. compressing a spring.
v "'2. The opposing force and the The opposing force and the Potential energy decreases when objects which attract each
driving force are nearly driving force differ widely. other come closer, as when a book falls towards earth. Potenti~
energy also decreases when things that repel each other are
equal.
allowed to move farther apart, as when a compressed spring pushes
3. Work obtainable in · a Work obtainable is always less
reversible process is than work obtainable in a objects apart.
maximum reversible process. Energy can be converted into work or can be obtained from
· 4. It can not be realised in Most of the processes occuring work. If work is done on the system its energy will be increased.On
actual practice and is only in nature and laboratory are the other hand, if the work is done by the system, its energy willlM:
theoretic~. irreversible. lowered.
5. The · reversible process Irreversible process does not The SI unit of energy is called the joule (J) and corresponds
occurs in an infinite number involve many steps and gets to the amount of kinetic energy possessecfoy a 2 kg object moving
of steps and infinite time is completed in a short time. at a speed of 1 meter per second.
required for its completion. 2
6. All the changes occurring in In an irreversible process,, the 1J = .!_ (2 kg)
2
(lm)
ls
ms-
or 1J = 1 kg 2 2 ; lkJ = 103 J
- - -- 2
the direct process can be changes taking place in direct 2
reversed in the reverse process cannot be reversed. In C.G.S. units, the unit of energy is erg (1 erg= lgcm r l•
process. lJ = 107 erg
Another unit of energy is calorie (cal).
, 1 cal= 4.184 J; 1 kcal= 103 cal= 4.184 kJ

PROBLEM I . Select the open, closed and isolated system from 6 .6 . MODES OF TRANSFERENCE O~
the following : .. ENERGY BETWEEN SYSTEM A~
(i) A thermos flask containing ice c~ld water (t~) A cup SURROUNDINGS (WORK AND HEA )
· hmaYbl
containing boiling water (iii) Hot tea put mto a closed insulated Every system has definite amount of energy wtuc cAll
vessel (iv) A china dish containing e~her placed on a table. defined as the capacity of a system to perform work. A sySleill 1bt
(v) A chemical reaction taking place ma corked flask. lo . y from f
se ene_rgy . to the surroundings or gain energ JTlodeSo
[Ans. (i) Isolated system (ii) Open system surroundmgs m a number of ways. The two important toll' :
(iii) Isolated system (iv) Open system ( v) Closed system] transference of energy are heat and work as described t,e iJie
wan
PROBLEM z. Which of the following are intensive properties ? (a) . Heat {q). lf a system is at higher temperature 10 tlll
5
surroundings, then energy is transferred from the sy ieJ11 ysie~
(i) Surface tension (ii) Entbalpy
--===============..l surroundings. This leads to the fall in temperature of t11e s
 CJ\I. THERMODYNAMICS
tfi1l 6/7
~ . ternperature of the surroundings. Th'
; ~se Ill 1ecular motion till thermal equilibrium i_s takes place . the number of
(ii) the volume of the system changes, i.e. , ber of moles
,41gb moture of the system and surroundings bec1s established, moles of gaseous reactants is different from the num
,. _,,,nera . omes equal If
l· tw¢e of the system 1s 1ower than that of surroundin s · of gaseous products.
~pe~tS transferred from the surroundings to the system. g · then (iii) the pressure remains constant.
Y The
~ changed between a system and the surro d" .
. .tfV eJ. cliff • -- U!!.., IQffli l!bfn Calculation of Pressure-Volume work
61".leai"e!~ es are erent IS 1f:Po>D? as heat. In ot~r Words (Mechanical work)
~ oftieat from a body at higher temperature to th bod ·
~~ ture ~ y Consider a gas enclosed
ioier teDlpera • Force = Pressure x Piston area
~ 11 is irnpcrtant to note tha~ heat is not a substance and a in a cylinder fitted with a
frictionless piston having area
+ =p Xa
e does not have a defirute amount of heat· However, 1t .
bstaD C f of cross-section = A sq. cm.
~a definite amount o energy. Flow _of heat means that the energy (Fig. 6.4).
~i,eing exchanged ~ecaus~ of the difference in temperature.
Volume of gas enclosed
Unit. In SI units, unit of heat energy !_; Joule and erg in in the cylinder= V
c.G.S. units. Pressure acting on the
Sign conven~ion: q is ~ositive if heat is absorbed by the piston = P (constant) <
system and is negatJ. ve 1f heat 1s evolved. dV=a dl
Internal pressure of the gas.
Piston
For example, if we supply 10 kJ of heat energy to the system, Distance through l'Q'.iri:it.'iitJ:tP',-...area
we writeq= 10 kJ. If heat given out to the surrounding is 10 kJ, then which the piston moves = dl = 'a' sq.cm
wewriteq = - 10 kJ When the internal
(b) Work .(W ). Another mode of transfer of energy is work. pressure of gas is slightly
For example, if a gas, enclosed in a cylinder fitted with a piston, is more than the external
al higher pressure than the surroundings, the piston moves pressure, the gas expands and Fig. 6.4 Pressure-
upwards. This process continues till the pressure of the gas piston moves. Let the distance volume work.
liecomes equal to the outside pressure. The energy transfer that through which the
gas expands or distance through which the piston moves be ' di' .
lakes place is U~rk.
Change in volume, dV = a.ell.
is said to be done whenever the point of Force
pplication of a force is displaced in the direction We know that Pressure = - -
of force. If F is the magnitude of force and d is the Area
displacemen t of the point of application in the or Force = Pressure x Area
direction In which the force acts, then the work done
Therefore, force on the piston F = P.A.
is given by
W=Fxd
If the small work done by the movement of the piston is dw,
then
In general, work is expressed as the product of two facto~s- Work = Force x Distance
an intensity factor and a capacity factor. If there is no opposmg
dw=-P.a. di
force, then the motion itself produces no work. On the other hand,
·Ill the absence of motion, even the strongest oppo s mg · t:orce cannot (Negative sign is used because the work is done by the system)
But a x di= dV, a small increase in the volume of the gas.
generate any work.
,., . all d intensity factor] Therefore, the small amount of the work (dw) done by the gas is
nork done= [A generalized force, c e given by
x [A generaliud displacement, called capacity factor] dw=-PxdV
th e which are used
. There are many kinds of work. Some o f es If the gas expands from initial volume V 1 to the final volume
10
lhennodynamics are : V 2, then the total work done (w) can be obtained by integrating the
. k · said to be done equation dw = - PdV (when pressure is constant) over the volume
( 1) Electrical Work This type of wor is u·a1
When • · uit If the paten change from V I to V 2 • Thus,
d' an electric current flows through a circ · . . factor)
ifference causing the flow of current is E volts (i~tenSi~ •s Q V2
illld th fl
e quantity of electricity that ows 1
·n a given ume 1 w = - J PdV =- P (V2 - V 1) = - P ..'.\V
couJo b . •
V1
Ill s (capacity factor) , then
Electrical work (W) = E.M.F. x Quantity of electnc1ty . [·.· ~V=V 2 -Vi]
Consequently, no work 1s done if there is no change in
= Ex Q volt coulomb or joule volume, i.e., when/:::.Vis zero.
= EI.t joule . .
(I) x ume m se
conds (t)] . If the system loses energy, we say that the work is done b
[· . Q _ .
· - current strength m amperes k) In most the system. On the other hand, if system gains energy, the work i~
(") • J wor • done on the system.
Pr0cesI! Pr~ssure-Volume work (MechaDlf me or mechanical
\I/ark ~-of mterestto a chemist is pres s ure-vo u are involved If the external pressure (P) is slightly more than th e pressure
.
illld · • uis type of work is said to be done, if gases of the ga~, the gas will contract and the work will be done b the
on exp . surroundmgs on the system. In that case, V will b l th y
(.\ ans1on :
11 th
. th rmal equilibrium;
an d u"V 1s .
· negat:J.ve. H ence 2 e ess an v I
e system and the surroundings are 10 e
 6/8 ISC CHEMISTR y
. P~lll.
. kinetic energy and move to and fro 10 all ct·
w=P6V molecules gam
considered as amode which stimulates
·
th
trecrion.l
It may be mentioned here that Pis the external pressure and Thus, heat can be etaniJo 1·
hence is sometimes written as Pext. so that motion of molecules. . IJl
e work is done on the piston to push it d
w =Pex1.XilV When so m . th d' .
s start moving m e 1rect1on of the p·oWn, t~
Thus. we conclude that : gaseous molecu le d h . iston S
b
work may e co nsidered as a rno et at stimulates the Organl~ ·: o,
(i) Jl;the gas expands, V 2 > V and work is done by
~)J'e system then w is negative 1 motion.
\)H1" ~~ the gas contracts V 2 < V 1 and work is done on Th US, beat is a random form of energy whereas Work .l, ~
the system then w is positive. organised form of energy. . .
Wor k done in free expansion of a gas agamst vacuutn
Sign Convention. According to the latest IUPAC .
recommendations, positive sign is assigned to the work done on If the gas is allowed to expand agamsl vacuum, then lhe
. pressure, · p is zero, the work done will be
opposing
the system and a negative sign is assigned to the work done by the i.e., ext zero.
system. This.convention puts energy and work on the same footing. W ---Pext x~V=0x~ V=0
The work done on the system, like heat added to the system, Thus, for free expansion of a gas against vacuum, the "ort
increases the internal energy of the system and thus, is assigned a done is zero.
positive sign. However, according to the old convention, positive Work done in a reversible process is maximum work. 1'hii
sign is assigned to the work done by the system and negative sign is discussed in section 6.12.
is assigned to the work done on the system.
To summarise, we have 6. 7 . EQUIVALENCE BETWEEN HEAT ANO
Work done on the system= w positive ; Work done by the MECHANICAL WORK
system= w negative It is a common observation that wheneve~ some mechanical
Heat absorbed by the system= q positive ;Heat evolved by work is done, heat is produced. Joule earned out the exact
the system= q negative measurements of the work done and heat produced. On the basis
Units of heat and work. (a) Heat is measured in calories (cal), of his experiments, he concluded that in a cyclic process the net
kilocalories (kcal.), Joules (J) or kilojoules (kJ). mechanical work done (w) by the system must be proportional to
heat produced (Q) \
These are related to each other as
l..e., \l w oc Q. or w = JQ ,
lcal=4.184 J; 1 kcal=4.184kJ ,
where J is a constant of proportionality and is called Joule
S.I. units of heat are .Lgg!.t-_91 JiilQjmile mechanical equivalent of heat. Its numerical value is taken as 4.1 84
where Joule is equal to Newton-meter (Nm) i.e., 1 J = Nm. x f(Viirg; 4. (84 joule. Thus, with the expenditure of 4.184 x10;
(b) Work is measured in ergs or joules where 1 Joule= 1 kg erg or4.184 joule of mechanical energy, 1 calorie of heat is produced.
m2 r 2 = 1 Nm= 107 ergs
6.8. INTERNAL ENERGY, E OR U
1 erg = 1 g cm 2 s-2
Every system is made up of atoms, ions or molecules. Energy
,~~~ §.I. unit of work isloul~ (J) . . is stored in atoms, ions and molecules in different fonns suchas
""'~ Work is not a state function. Like heat, work 1s not a state potential energy, kinetic energy, nuclear energy, electronic energy.
function. It is because the work depends upon the path followed. chemical energy, etc.
For example,-the work done by a person for reaching the top of a I'
: The total energy stored in a system (atom, ion or molecukl
building by using a lift is not the same as going on foot by using in different forms such as ROtential energy. kinetic energy,
steps of the stair. nuclear energy, .chel)lical energy, etc. is called its internal ener?J
or intrinsic energy. '
~~~1 NEY ff/CTS ~~
1
~ Accordingt olatest IUPAC recommendations, internal energy
is denoted by the letter, U. However, in some books it is represented
1. Heat and Work are both transient phenofmhenatw. Systbe~~ byE.
never possess heat or work, but either o t e o or o
of these cross the boundary of the system when the system The value of internal energy of a substance depends u~
undergoes a change of state. factors such as nature of the substance, pressure, tempera(Ult
..
etc. For e_xample, the internal energy of one mole of CO2is di~~
2. Both heat and work are algebraic quant1t1es. f from the mternal energy of one mole of H 0 under similar condi~
3. Both heat and work originate only on the boundary o 2
oftempe~a~re and pressure. Similarly, the internal energyo',t~
the system. mole of hqu1d water at 373 K is differnt from the internal en~
4. Both heat and work arise only due to an effect on the one m~ .of steam at the same tempearature. .
surroundings. . ( AhsolntJ: value-of internal energy cannot be dete~
5. In addition to heat and work, radiant energy and electrical This 1s because . . · " , · ,e
energy are also modes of transference of energy. th, d"f"' 11
"~~Y..1 2a2~ ent;!_gies ored in a syste . FortunatelY· reacu~
6. Work has the same units as those of energy.
II ,, are mamJy mte~ested m change in energy unng a chcllllcal rgieS
It may be considered as the diffcrc,11.: e between the internal ene
Difference between heat and work. Consider a gaseo~s or products and reactants. i.e.,
.
system fitted with a piston. Wh en hca, ti·s added to the gas, its
..._-A,tr"= UI' - UR

.I
 6/ 9
AL r ttE RM OD YN AM ICS st
e,ttC sta nds for 'd' 1·n G ~k olv es two eps . yst em ·
di . pro nou nce d as 'de lta' and Procedure. The pro ced ure inv alo rim ete r s hea t of
~~ aci ty Of th e c U) .
f . difference) . (z) To kno w the hea t cap
ene rgy cha nge (~ i.e. ,
· Rea ctio ns wit h negati
;~~ is negative wh en Up <ic UR ns and pro cee d wi :
(ii) To kno w the inte rna l
. t yst em . A
~ 60 are cal led exo the rm rea ctio e.
com bus tion at con stan t vol um · 0 f the cal on me er stin um cup
ty _ d . the pla
,Jue . JJ of heat. (z) To kno w th e hea t cap aci and its
h U p > U R· React10 · ns wit h pos itiv e pou nd is pla ce mal •m ete r
01000 • •
!' ~E J is pos1t1ve w en kno wn wei ght of a.known com • tak en in the c oning an ele ctn. c
tak e pla ce with ass
J{of~V are cal led end oth erm ic rea ctio ns
and (C) . A kno wn wei ght of wate~ 1s . . . . ~ bso rbe d due
The rea ctio n IS rmn ate d by
temperature is noted. 0
:;e app ara tus .
1
rioll of heat. F. Th e hea t evo ~e
of ene rgy is jou fo (J). In C.G.S. cur ren t through the fila me nt, ·ry (i· e ,
j?IDf!l(Jnits of U. In SI u~ ts, the uni t to the reaction ' cha nge s the
tem per atu re of e en
the hea t cap ac1ori me.ter .
. the unit of ene rgy 1s erg . per atu re . cha nge , of the cal
7 4J. Fro m the obs erv ed tem fall m tem per atu re)
~~.
= lJ= 10 erg ; 1 cal ori e=4 .18 heat change per deg ree rise or
o~ a sys tem is the inherent Th e abo ve
Cbafacleristics. (i) Inte rna l ~ne rgy can be calculated. 6 U). d
of Its pos itio n. It i.s-a-state (ii) To kno w the int ern al
ene rgy cha nge (
resent in the sys tem by vJrtue .th th kn·ow n qua ntit y of the com po un
1oergY P fall ) 1·s not ed
wnction. exp erim ent is rep eat ed w1
e
per atu re (ns
.
e or d
rgy of a sys tem can not be under examination and cha nge
in tem t
~ bsolute val ue of inte rna l ene t cap aci ty of cal ori me ter syS em
an
sib le to det erm ine the exa ct values Sub stit utin g the val ue of hea
1,
nni,ned because it is not pos
h I . 0na 1 ene rgy , nuc lea r rela tion
,.ie .
of constituent ene rgi es sue
as tran s at1 cha nge in tem per atu re in the
M
tllergy, vibrational ene rgy . etc . tlU. =Z X6 TXm-
~ is an ext ens ive pro per ty. t vol um e, i.e. , ~V can be cal
cul ate d.
the heat of c~mbustion at con stan
RM IN ATIO N OF tion at con sta nt vol um e.
6,9, EXPERIMENT AL DE TE Here ,:\U = He at of com bus
EN ER GY ( ~ U)
CHANGE IN IN TE RN AL z = Hea t cap aci ty of the cal ori me ter
sys tem .
nt tem per atu re and con stan t
ff a reaction tak es pla ce at con sta to ~ T = Ch ang e in tem
per atu re.
~U , of the rea ctio n is equal
1olume, the internal ene rgy cha nge refo re, the inte rna l m = Ma ss of the sub sta
nce tak en
ndi ngs . The
rbeheat exchanged with the sur rou det erm in~ d ~y M = Mo lec ula r ma ss of the
sub sta nce .
pro ces s can be
energy change acc om pan yin g a is
I
D_YN AM IC S .
orb ed pro vid ed the rea cuo n
I

measuring the heat evo lve d or abs

and con stan t vol um e. The val ue .&:!O. FI RS T LAW OF TH ER MOthe sys tem and the
carried out at con stan t tem per atu re rgy bet we en
ofaU can be me asu red by usi ng a bom b calorimeter (Fig. 6.5). ----F The tran sfe ren cebyoftheenelaw of con ser vat ion of ene rgy . Th
is
ter is ma de up of thr ee sur rou ndi ngs is gui ded 0). It is
Construction. Bo mb cal ori me Ma yer and He lmh olt z ( 184
~d law was first of all stat ed by
ich is ma de of stee l, wat er 1 al obs erv atio ns and hum an exp
eri enc e.
roocemric shells i.e. bom b wh e~ : pur ely based upon exp erim ent
ost shell in_rea ctio ? chamb one of the foll ow ing wa ys :
insulated outer shell. The inn erm t big This law ma y be stated in any
stee l so tha t 1t can withS and cre ate d nor des tro yed by
an y
called bomb. It is ma de of hea vy ~e Jt( Energy can nei the r bewev
ctio n at con stan t vol um e. At er, one for m of ene rgy can be
pressures resulting fro m the rea physical or che mic al change.
Ho
10P,tbere is an inlet for oxy gen . The re is a
thin fila me nt (F) wf1!~h t of the oth er for m.
iate the rea ctio n. ~e 'bo mb
is cha nge d into equ iva len t am oun is
ran be heated ele ctri cal ly to init of ene rgi es in iso lat ed sys tem
ing a kno wn wei ght of wat er. t,,(ii) The sum of all the for ms
surrounded by wat er bat h con tain re. IA
to me asu re the tem per atuwh constant.
There is a sensitive the rmo me.ter . ter bath · The oe str uct a per pet ua l mo tio
n
lliech3IUc· al stin er (S) is pro vid ed m the wa . . . th (iii ) It is imp oss ibl e to con ene rgy .
bat h) is ins ula ted to mm mu se
e wo rks wit hou t con sum ing
!apparatus, (i.e ., Bo mb and wa ter machine, ,i.e. , a ma chi ne wh ich
ene rgy is con sta nt in the
oss of heat by rad iati ons . _ffi) The tota l sum of ma ss and
erc onv eni ble (V = mc2).
I universe though the se are int
SCR EW CAP be con ver ted int o ene rgy and
It is no ': ~o wn tha t ma ss can sio n :
to eac h oth er by the exp res
THE RM OM ETE R the two qua ntit ies are rela ted
V= mc 2
ss m and
ed by the des tru ctio n of ma
wh ere U is the epe rgy pro duc ·
/
c is the vel oci ty of ligh t. al
m, the law sta tes tha t ''th e tot
INSULATED The refo re, in the mo dif ied for nge d" .
d sys tem * rem ain s un cha
OUT ER mass and energy of an isolate w of
Ac cor din g to fir st .la
-- VES SEL
Ma tbe ma ti~ al for m. des tro y~d but
nei the r be cre ate d nor
PLATINUM the rmo dyn am ics, ene rgy can am oun t of the
nge d into equ iva len t
CUP one for m of ene~gy can be cha cha nge d b
rgy of a sys tem can be
WATER oth er form . The mte ma l ene : Y any
ing me tho ds
one of the fol low
(i) By hea ting or coo ling
the sys tem .

Fig. 6 .5 Bo mb Ca lorim
·
ete:r;.. ,;,;....:;11~=-=~
 6/10 ISC CHEMIST,ty

(ii) When work is done on the system or work is done by the 6 _10 _3 _ some special Fo~ms of the First
(i) For an lsochoric Process, i.e. , for a process taltin \lj
1./
pl\'tl

system. Consider a system having internal energy, U 1• Suppose

heat q is supplied to the system. Thus, its energy increases from at constant volume, 6 V = zero gPloee
U1 to U1 +q. Further, suppose that work (w) is done on the system. P6V=zero
As a result, its internal energy further increases and becomes equal
Hence, from equation of the first law of thennoctyncllltic
to U2. Here U2 is the energy of the final state. Thus:
6U=~ '
Al2 =U 1+q+w; U 2 -U 1 =q+w
is the heat absorbed at constant volume. 'T'L
'1fiu=q+w] ... (1) w here q v . . ' uus, h ,
lied to a system under constant vo1ume 1s used up m inc _e;i
i.e., Change in internal energy= Heat added to the system+ Work ww ~~
done on the system internal energy. g
(ii) For an Isotherm Process,
_ .
i.e., for a process !akin
gplace
It is the mathematical form of first law of thermodynamics.
at constant temperature - -
The values of q and w depend upon the path by which a 6U=0
change is carried out. So, both of these are not state functions. On
Hence, from equation (2) q = - w
the other hand, internal energy is a state function and the value of
dU depends upon the initial and final state. It may be noted that (ii i) For an Adiabatic Process, i.e., for .a process in Which no
whatever may be the process, dU is always equal to q + w, thus, heat enters or leaves the system,
q + w is also a state function. · q=0and6V=w or-~U=-w
When dU = 0, i.e., when internal energy does not change, i.e. , work is done at the cost of its internal energy.
Then from relation (l) we have: (iv) For a Cyclic Process, 6U = 0
·o=q+w or q=-w ...(2) Hence q=-w
It means that when dU = 0, heat absorbed by a system is (v) For an Isobaric Process, ~p ;= 0 consider a system
equal to negative of work done on the system or equal to work showing increase in volume from V I to V 2 at a constant pressureP
done by ~ system . during absorption of heat q. The expansion work or work done b;
. ._ li_/n<f". Change In Internal Energy and First the system is w = - P6 V
-~ Law of Thermodynamics Fromeq. (1), qp =6U-(- P6V)=6U +P.~V
According to first law of thermodynamics, ~U = q + w. , =Ui-U 1 +P(V 2 -V 1)
If the process is such that work done by the system is only ~ = (U2 +PV 2 )-(U 1 +PVi)=H2 -H 1
pressure-volume work, then, w = -P~ V. Thus: where H 2 and H I are the enthalpies of the system in final and initial
dU=q-PdV state respectively. --
If the process takes place at constant volume, i.e., ~ V = 0. 6.11. WORK DONE DURING EXPANSION
Then Pd V = 0 and thus dU = qv· Hence,
OF , AN IDEAL GAS
"Change in internal ene__rgy, ~U is~ _to h~ t (qv1al;,sQrb._ed
(Q_,Work done during Isothermal Expansion of an Ideal Ga-
or evolved at constant v.QlUDl.t:.and.constant-temper:a~e."
Let us consider ' n' moles of an ideal gas enclosedinacylinoo-
6.10.2. Physical Significance of Internal fitted with a frictionless piston.
Energy Let V 1 = Volume of the gas at constant temperature T.
(I_~ _Q!J ~ cal _significance of internal energy can be P = External pressure acting over the gas.
understoodj n a be!!_<e_r way by taking an example from everyday The work done by the gas for a small volume change iN
Tifu."Th;-food taken by us ~ts COJ!\;'.erted into !!_eat _e ~~rgy (q) . A should be given by
part of this--energy is spent in doing work (- ve) and the rest is
dw = - f P.dV
stored in the body in the form of internal energy. During growth
Integrating between limits of V = V and V = V2 , wegel ll1f
period, a child takes more energy in the fon:11 of food as co~pared 1
total work of expansion as :
to work done and the balance (q- w = dU) 1s added to the mtemal
V2
energy and the child grows in a healthy_manner. How~ver,_after w = - f P.dV
forty, any accumulation of extra energy 1s dangerous, smce 1t can V1
lead to diabetes. Therefore, to keep dU equal to zero: q must be
equal to the work done. This work may be external physical work or For an ideal gas, PV = nRT ,· p. = !!_
V RT
it may be inte;nal work such as internal m~vement of the heart ~d From equation (i) and (ii), we get
stomach, etdJ3 asal metabolism (the chemical processes occumng
V2 nRT V2
in an organism at rest) r egmres 30QJ~l.._<2f_~ gx.,per hour. If no w = f - - dV = - nRT . In -
food is taken then 300 kJ per hour consumption of energy will be Y1 V V1
supplied from the reserve internal energy, which would cause a
or w = - 2.303 n RT Jog V 2 X,./.
loss of weight.
V1 v
*Isolated s)!.St means combination of system and surroundings so that A · 1· · V, _!i
·
any energy lost by the system 1s · d by the surroundings and vi ce gam or an 1deaJ gas p V - p V so ~ ::: p
game '' I I- 2 2 ' V1 2
versa.

I
 1 ~L. 1HERM00YN AM1cs

"'"j
~r
w = - 2.303 nRT log
P2 ~
P
_1
6 / ll.
If V2 < VI then w- is + ve ,· e compression work or work
'" ts of ' w' will depend upon th . done on th . irr ' • •'
. eunitsofR e system 1s positive.
rk done d urmg Adiabatic E . · At constant volume , w•l IT = O•
. . "Pansaon Of
for an adiabatic process, dlJ = 6w an Ideal Gas. _R_E --
_ O_F WO R-K- AND HEAT
6 - 1 2. -N-A-T U
dV
rJoW, df = Cv or dlJ = Cv clI' (a) Work-a Path Dependent Function
Work is a path dependent function and has greater value for
flJUS, for 1 mole of a gas, 6w = Cv clI'
~ reversible process compared to the same process carried out
for n moles, 6w = nCv clI' ~eversibly. This can be illustrated with the help of an indicator
f hUS, total work done, diagram. For an isothermal reversible process the variation of
T2 volume with pressure may be represented by an indicator diagram
W = f nCv af = nCv(T2 -T)
... (1)
as shown in Fig. 6.6. Let the initial pressure and volume of the gas
T1 I
be P1 and V I respectively. For a slight release of pressure (fromc to
e cv is the molar heat capacity* at constant volume d) if the increase in volume is dV (from a to b) then the work done
wber by the system (-dw) = (P-dP)dV approxPdV> Since for reversible
cP change the pressures at points c and d differ only slightly, PdV
Also cp- CV= R or -c
V
- 1 = RIC V may be taken equal to the area of strip abdc. The total work w done
by the gas for the reversible expansion from an initial volume V 1 to
R R a final volume V2 will be equal in magnitude to the sum of the areas
or y- 1 = - or C =- - •. Cp
[ . ~=y
] of all such strips as abdc. Since the change is slow and continuous
Cv v Y- l
the work done is given by the integral :
Substituting the value of C v in (i), V2
w = nR (T2 - T 1) \._){./ - w = J PdV = Area ABCD
... (ii) V1
y - 1
Case 1. If T2 > T 1, w is positive and we say that work is done
on the system.
Case 2. If T 2 < T 1, w is negative and in this case, work is
REVERSIBLE WORK
done by the system.

liE'IFIICTS w
a:
=>
(/)
1. Durin~ expansion ?f gas w will be negative and therefore, (/)
w
LiU will have negative value i.e. there will be a decrease in a:
a..
the internal energy of the system and hence its temperature
will fall. B
I
2. During compression , w will be positive and therefore, ~U 1 V2
will have a positive value, i.e., there will be increase in
internal energy and hence temperature of the system will D a b C
rise. In this case, the work is done by surroundings on the VOLUME
system which is stored as the internal energy. (a)
3. For an isothermal process, PV = constant.
4. For an adiabatic process, PVr = constant,

Where y = specific heat ratio, i.e., Y =

5· Work done in the isothermal and reversible expansion of
c:
C

w
~ gas is always greater than the work done in an a:
=> IRREVERSIBLE WORK
irreversible isothermal expansion ~f a g~s. Thus, work th
(/)
en
done by a system is not a state function. It 1s related to e w = P {Vz--V 1 )
th to a:
ma~n~_r in which the process is carried out rather an a..
I
the tn1t1al and final states of the system. I
P2 - - -,- - - - - - B
I
(iii) Work done in irreversible Isothermal Process V1
I
I I V2
td . Suppose an ideal gas expands against external pressure p D C
veils volume changes by an amount dV then work done w can be VOLUME
nby w =-PdV. (b)
For a finite change V 1 to V 2, ~
Tota} work done, w =-P (V2 - V1) orJwirr::-P (V z- V1) Fig. 6 .6 Wo~k a path function (a) Reversible work
of expans1~n from V 1 and V 2 . (b) Irreversible
1 thelfV2> V1 then wirris- ve, i. e., expansi11 work OI workdcre work done 1f the external pressure is sudd I 1

~ mis negative. - _ pped to the final value p 2 _

l.:kz==:c:=d-ro en Y
lfdeta'J
1
s refer to section 6.14.
f

6 / 12 ISC CHEMISTrty
. t t f . P41ll,l
In an isothermal irreversible expansion the pressure suddenly Since internal energy 1s_ as a e uncllon: l'.\U has a ,
drops from P, to P2 and the gas expands against the new pressure value. However, the work be10g a path function, the vaJ~fl!li~
from volume VI to :V2·_Since P2 is constant during expansion, depends upon the path followed . Therefore, for a given e of 1,
therefore, total work 1s given by the area BCDE and is equal to dU, the value of q will.also depend up.on tl:te P.!.th follo Value01
Y2 · DI?_t a s t_a_le f unc!I0.I_L
process. In other _w_grds,_9 is · - · w~v,
- -.._,_!he
P2 f dV = P2 (V2 - y ) ·- Important conclusion which c~ be drawn from th
Th · ·
Y1
·
1
discussion is that t'U_is a ~tate function but n~~q n; ~e
followed
.
us, It is qwte clear that the work depends upon the path
for the change • F or iso
· th ermal reversible
. . the
expansion
magrutude_ of ~ork is greater than that for an irreversible isothermal
~ e functiQn. Io mail:iematical terms, ai1'terential of~
differential whereas the differentials of q and w are
differentials. Therefore, for a small change in U, q and III th e~aq
n: e~ a

change as is evident from Fig. 6.6 (a) and 6 _6(b). law of thermodynamics takes the form ' efirQ
s·mce p.is a fu nction of Y2 dU =Oq +ow
volume, the integral f PdV as such
Conditions under w~ch 'q' and 'w' become state fun~
V1
c~not be evaluated. The dependence of P on V is different for (i)' For a process taking place at constant volume PdV
diffe~nt processes. Hence, the numerical value of w will be different
. . '
Hence, from the first law of thermodynamics, l'.\U = qv (for an
:::()

for different processes shown as follows : isochoric process).

. (i) Expansion into vacuum. In this case, the gas is expanding Here subscript vindicate~ constant volume. Since dU is a
agamst a zero external pressure and hence it cannot do any work. state function, hence q also must be a state function under th
.. ese
. w=-PdV =0 ·.· pext =0 cood1t100s.
(ii) Expansion against a constant external pr essure. A (ii) For a process carried out at constant pressure, (Isobaric
constant pressure has no volume dependence and hence work process), the work of expansion done by the system is given by
clone by the gas is given by w=-P.~V
w =-PdV =-P(Vi- V 1) =-P~V Therefore, under these conditions, w is also a state functio~
(iii) Reversible work of expansion (Maximum work). For a Since dU = qP + w therefore
rever~ible exp~sion of the gas, the internal pressure of the gas qP =dU -w = ~U +P~V = ~H (a state function)
(Pin 1) is only slightly greater than the Pext and the_wqrk_done by ,a Hence, qP also must be a state function under the conditioru
sy.!!:m is the maximum work if the process is carried under of constant pressure.
reversible conditions. -
(iii) For an adiabatic process, q = 0, .·. ~ U = w. Hence, 'w' ~
We know that the work done by a system= - Pext x ~ V. a state function under this condition.
Therefore, for a given increase in volume (~ V), the work
done depends upon the value of Pext· Since the gas is in equilibrium NUMERICAL PROBLEMS BASED ON :
with internal pressure, it will expand only when the external prssure
is less than the internal pressure of the gas. The gas will not expand FIRST LAW OF THERMODYNAMI~
if the external pressure is grealer than the internal pressure of the For mulae and Units
gas itself. At most, the external pressure can be infinitesimally
smaller than the Pint· Thus the work obtained will be maximum Ac~ord ing to First Law of Thermodynamics, dU =q +»'
when the Pext and Pint differ only by an infinitesimally small amount Umts of q, ~U and ware Joules (IL atm. = 101 .3 J).
from each other. But this is the condition for a reversible process. For a sy st em involving pressure volume work only,
Hence, the work done is maximum only if the process is carried w=-P~V;~U =q-P.~V
out revers~. Therefore, 1.n a reversible process, w is sometimes Heat evolved by the system, q = - ve . Heat absorbed by tit
replaced by Wmax during expansion by the system. system, q = + ve. '
Similarly, work of contraction (or compression) done on the Work done by the system w = - ve . work done on dt
system is minimum if carried out reversibly. system, w = + ve. ' ,
From the above discussion, it is obvious that the work q- EXAMPLE 1
obtained in an irreversible process is less than wrnax. It is also
important to note that the work required to reverse the process in
case of a reversible change is equal to Wrnax but opposite in sign.
t
equal i;~~n~bsorbs 200 joules of heat and performs wo~
of the system~o es. Calculate the change in tlu~interual cner1
From the examples discussed above, we see that work done Solution Am -vstelll
by the system is different in different cases and hence we conclude = 200 joules. · ount of heat (q) absorbed by rhe s,
that w depends on how the process is carried out. In other words,
work done by th ,
w is a path dependent function. e system = - w = - 100 joules.
th
We know at ~U = q + w (First law of thennodynarnics)
(b) Heat- a path Dependent Functiou
Heat is not a state function and thi s can be derived from the :. l\U = 200 - 100 = 100 Joules
mathematical form of first law of thermodynamics, viz., 11'.1" E XAMPLC 2
~V=q+w Calcula te lhc a . · Jeof~
'·• I
lu l.'ll g ,lS ()l'c11cnt in
• mount of work done when one mo P~"r ...Al
- a vessel of 2 litrc capacity at one atlJIO~
 L r ttERMODYNAMics
~icA
~~ . ])owed to enter into an evacuated 6/13
.tJl'eiS a vessel of 10 litre
·/ itY· tiOO• Since expansion takes place i
· solU rnal pressure, P = 0
t)le exte
n an evacuated
q- EXAMPLE 6 cc=:::...::::=-===----------
One mole of an ideal gas is expanded isothermally aga?15t a
constant pressure of 3 atomspberes from 10 litres to 35 htres.
,.sel, 1r W = P X 6..V
,;, 6111 wor~ Calculate the work done, change in internal energy a nd ~;~~
=
W 0 x 6.. V zero = absorbed during the process. (I.S.C. 2
Solution. Work done, w = - P.~ V .
fµ\r,1PLE 3
(·.· Workdone by C the system is negauve)
I . J)loles of an ideal gas at 300 K and l0 t
fi"~sotberroal reversible expansion to doub~ ~pressure ~ V = 35 - 10 = 25 litre ; Pressure p = 3 atm
~dergo:ntlYit is isothermally compressed to its ori~ Ivo~ume. w=-3 x 25=-75 Latm
~l)Se~~be net beat absorbed b y the gas and what is th a vot ume. =-75x 101.3J=-7597.5J
\\tat IS ? e ne work
~oe bythe~as. s· th . For an isothermal process,~U =0 [·: 1 L atm=l0l.3 J]
solution. mce e syst~m returns to its original state, the According to First law of thermodynamics, U = q + w
5 is a cyclic process. In this process, no net heat is absorbed :. 0=q-7597.5Jorq=7597.5J
~ere is no net work done by the gas.

tfXA"1PLE 4 =====.::::::..=:~.......,-----
Calculate the work done in Joules when 3 moles of an ideal
Positive sign of q indicates an endothermic reaction.
q, EXAMPLE 7
Calculate the maximum work done when 64 g of oxygen gas
atz7°C expands isothermally and reversibly from 10 atm to occupying a volume of 7 litres is expanded isothermally ao d
ftm (1atro = 1.013 x 1()5 Nm- 2). What wilJ be the work done if reversibly to 14 litres at 27°C [R = 1.987 cal] (I.S.C. 20_0 8 )
:eexpansion is against a constant pressure of 1 atm? Solution. Maximum work done in isothermal reverSible
Solution. According to the given data : expansion process is given by
n = 3 ; T = 27 + 273 = 300 K V V2
P 1 = 10 atm ; P 2 = 1 atm
w = - nRTln-1._ = - 2.303 nRTlog
V1
V
l
(i) Work done during isothermal reversible expansion is Molar mass of 0 2 = 32 g mo1- 1
~ven by : 64
p 10 = 64 g o 2 = = 2 moles or n = 2
w=-2.303nRTlog - 1 =-2.303x3x8.314x3001og J 32
P2 1 T = 27 + 273 = 300 K; V 1 = 7 L ; V 2 = 14 L
=-2.303 X 3 X 8.314 X 300 X 1J 14
w =-2.303 X 2 X 1.987 X 300log?
=-17232.4 J =-17.232 kJ
= - 2.303 X 2 X 1.987 X 300 log 2
(ii) Work done when the gas expands against constant
=-2.303 X 2 X 1.987 X 300 X 0.3010cal
pressure=-P~V =-P (V 2- V 1)
=-826.4 cal
nRT nRT
V1=-p ;V2=-p u:.- EXAMPLE 8
l . 2
A sample of gas is compressed at a pressure of 0.5 atm.
V = 3 X 0.0821 X 300 = 7 .389 L from a volume of 400 cm3 to 200 cm3 . During the process 8.0 J of
I 10 beat flows to the surroundings. Calculate the change in the internal
_ 3 X 0.0821 X 300 = 73.89 L energy of the system. (I.S.C . 2002)
V2- Solution. (i) Work done when the gas expands against a
~ 66
w =- 1 (73.89 7 .389) litre atm =- :~~~~~:
constant pressure is given by :
w=-P (V 2 -V 1)
=-66501 X 101.32J=- 6737 ·88 J- .
. [·.· 1 litreatrn= 101.321] According to the given data:
P = 0.5 atrn.; V 1 = 400 cm3 = 0.4 L
It EXAMPLE 5 :;.:.-.:.::.:-.::.::.-::.-::.-::..·--=== . V 2 = 200 cm3 = 0.2 L
5 e contained ma 100 w =-0.5 (0.2-0.4) = 0.1 Latin.
litt n10Ies of nitrogen at 5 atm pres s ur ·t expanded to 200
61tecylinder absorbed 30.26 kJ of beat whe!3 •. ternal energy of =0.1 X 101.32J [·: 1 Latm.= 101.32 J]
th es at 2 atm pressure. What is the change lil lil = 10.132J
egas?
(ii) According to first law of thermodynamics,
. ds from 100 litre to
200rSolution. Work done when gas expan. given by ~u =q +w= (-8 + 10.132) J =2.132J
tires against a constant pressure of 2 atm is litr s Its' EXAMPLE 9
w::: - p (V 2 - VI) = - 2 (200 - 100) = 200 atmkJ .e A gas cylinder of 5 lit.rc capacit)r cont:,ining 4 kg of helium
==-200 X 101.32J=-20264J=- 20•
26 gas at 27°C develo1>ed a leak leading lo the escape of gas into the
lieat absorbed, q = 30.26 kJ · atmosphere. lf the atmospheric pressure was 1.0 atmosphere,
Fro fi · calculate the work done by IJtc gas, assuming ideal behaviour.
q::: :n1rst law of thermodynaID.1 ~s kJ _ _ = 10 kJ (l.S.C. 2004)
c. -w or 6..E =q + w = 30 •2
20 26
 ISC CHEMISTR.y I>
6/14 "d .
. f ~l\l
To avoid the necessity o cons1 enng PV work Wh ·1
Solution. Mass of helium gas = 4 kg = 4 x 103g of reaction are measured at constant pressure, we definen !tea~
4 3 thermodynamic property calJed heat content or enth e anc,
Enthalpy is defined.~ the total heatmatmt_of~ t~ . ~
Numberofmolesofhelium(n) = IO -lOOOmoles
X
4
Gas in cylinder expands against a constant pressure of 1 atm. ~ e . Mathemaucally : ~
H=U+PV
V 1 =SL; T = 273 + 27 = 300 K
Thus, enthalpy may also be defmed as the sum or1.n . .(,j
- nRT 1000 X 0.0821 x 300
V2 - p= l = 24630 L energy (U) and pressure volume energy (PV) of a systern. ~
Characteristics. (i) Absolute value of enthalpy of
Work done (w)=-P.6.V =-P(V 2-V ,) cannot be determined just like the internal energy. a system
= - l (24630- 5) = - 24625 lit atm. (ii) The enthalpy of a system is a state function th
AH) d , eref0r
=-24625 X 101.3]=-2494512.5] the magnitude of entha1PY change ( l..l epends only c,
= - 2495.5125 kJ enthalpies of the initial and final states. Thus, we can Writeon the
AH = Hfinal - ~nitial ,,,(i,)
- i•l•!•$911l:.iQ;1•l=!•:J~b~ (iii) Enthalpy is an extensive property.

6.13.1. Change of Enthalpy in a chemical

~ OBLIJI 4. What happens to the total energy (decreases or reaction
mcreases) content of a system when :
Let a chemical system has internal energy U I and volum y
(i) heat is added to the system 1
Let it react at constant temperature and constant pressure 10 e.
(ii) heat is transferred to the surroundings another chermc · al system w1"th mtern
· al energy U2 and volumeSIie y
(iii) work is done on the system Let HI be the enthalpy of the first system and H2 the enthalpy ~f
(iv) work is done by the system? second system.
[Ans. (i) Increases (ii) Decreases Then, H, =U1 +PV1 ...(iiiJ
(iii) Increases (iv) Decreases]
H2=U2+PV2 ...(r.)
PROBLEM 5. In a process. 701 J of heat is absorbed by a system
~H=H2-H1 =(U2+PV:z.)-(U 1 +PV 1)
~d 394 J of work is done by the system. What is the change in
mternal energy ? [Ans. 307 J] = (U2- U 1) +P (Vi-V 1)
PRDBLEM &. A gas present in a cylinder fitted with a frictionless Thus, ~=~U+P!!:,.V ...(1'!
piston expands against a constant pressure of 1 atm from a But ~U = q + w [First law of thennodynamicsj
volume of 2 litre to a volume of 6 litre. In doing so, it absorbs Suppose only expansion work is involved in the change.We
800 J heat from surroundings. Determine increase in internal know that
energy of process. [Ans. 394.8 J]
w=-P~V
PlloaLEM 7. Calculate the maximum work for the expansion of
one mole of an ideal gas from 1 L to 10 Lat a temperature of ~U=q-PxV
0°C. (I.S.C. 2005) [Ans. - 5.227 kJ] [First law of thermodynamics] ...(111
PRoBLEM 8. One mole of a gas is allowed to expand isothennally Substituting the value of ~U from equation (vi) in (v), wege1
and reversibly from a volume of 1 dm3 to 50 dmJ at 273 K. . Al!~ q - f.~ V_+ J>~y ;::,-u-- -
Calculate w, ~ (or ~U) and q assuming ideal behaviour of the
: . \ ~H = qP (at constant pressure). Thus,
gas. (I.S.C. 2000) [Ans. w = - 8.880 kJ, ~U = o, q = + 8.880 kJ]
PROBLEM 9. What will be the volume change if 500 J of work is
in
Change enthalpy ,~11 IS equal toJ:iea!.(11 ) absorbed or
evolved at constant pressure. - P -
done by a system containing an ideal gas ? The surrounding
exerts constant pressure of 5 atmosphere (1 atm = 101.3 J) To summarise, we say if q is the amount of heal absorbed ((
[Ans.0.987L] evolved by the system, then
PRoBLEM I 0. A system absorbs 5 kJ of heat at constant volume
and its temperature rises from 15°C to 30°C. Calculate the q (a\constantvolume) = AU; i.e. , 9- =AU
. V
valueofq, w and~U. [Ans.q= 5 kJ ; ~U = SkJ and w=0] q (at constant pressure) = un
Au , 1
•.e., qp-Llll
_ Au
PRoaLEM 11. 750 cm3 of a sample of an ideal gas is compressed
that fin gedneral, if ~Hr is the enthalpy of the reactants and~
by an average pressure of 1 atmospheric pressure to 250 cm3• 0 pro ucts, then
Doring this process, 20 J of heat flows out to the surroundings.
Calculate the change in internal energy of the system. Af1=~H -~H =q
E . P r P
[Ans. 30.65 J] be meas:pen~ental ~etermination of 6H. Like 6U. 6Hcan ~"
red ma calonmeter show n .In p·1£.' 6 .7. I

with Th the calorimete r consists ,· . of. a loamed

. polys1yrcn~· "·up.fit!~11~,;
6.13. ENTHALPY, H a ennometer
. and · - d Out 1n ,.J""
s11rrer. Th1.· ren,:1ion is i.:arne
Heat change at constant volume permits us to calculate ~U. open vessel I e calo · · .rcCO((J"
with th h ~,· · ·: nmcter. 171t· change in 1cmpcrn1ure •~ tbC
However, most of the reacti ons arc carried out in open vessels al . e_ e _P of ll very_sensi1i ve thcnnomett.'r After tindingout re
hca l (capuc ,t y (Z) 0 f , . .· · . bstall'
essentially constant atmospheric pressure. Under these co nditions ( , ) the . lite l:lllo, 11nc1cr syslcm. mass of the_su aJ1 ~
rather sizeable volume changes can occ ur and work done is cquol .111 ~ · cnlllulpy chongc per mole (L~ H) of the reacuon'
1:11 1cul oted with th. I , 1 . .
toP~V. · c le P of the tnllowing relation :
 6/1 5
THE RMO MET ER
and ~ · eactants and prod ucts are in the solid an
not m gaseous states. As volu mes of solid
· d ti 'd t tes
qut
· ' d ds0 anot
ch~nge appreciably, therefore, AV = s and hqu t s . .
STIR RER 0 . For exam ple whe n mtn c
acid reacts with sodium hydr oxid e solu
FOA MED POLYSTYRENE tion , AV = O
CUP . HN 3 (aq) + NaO H (aq )- NaN 03 (aq) + H20 ([)
tant s are equ ~
Whe n num ber of mol es of gase ous reac
to number of moles of gase ous prod ucts
, i.e. , Mg = O.
For example,
H2( g) + Cl (g) - 2HC1 (g) (~( g)= 2-( 1 + l) =0)
REACTION MIXTURE 2
I mole 1 mole 2 mole
Con ditio ns whe n AH >AU .AH is
1 calo rim eter for mea sur in h . grea ter than L\U whe n
fi9t 6c~nstant pre ssu re (atm osp ~ <s) is positive, i.e., whe n num ber
h:ri c eat changes of mol es of gase ous prod ucts
a iq ~reater tnan the num ber of mol es of
M pre ssu re) gase ous reac tant s.
For example,
,a::Z x bT X - whe re Mis the mol ar mas s ofth
~• m b PCl 5(g) -
e su stance. I mole
PC1 3 (g) + Cl 2 (g)
Reta tion sbip betw eenA flan dAU . 1 mole I mole

WeknoW that for a chem ical reac tion , [~8 =(1 + 1)- 1 =+ 1 mol e]
~=A U+ PA V N 2 0 5 (g) - N 2 0 4 (g) + _!_ 0 2 (g)
· 2
· 1· • I mole !mol e 1
for reactions mvo vmg soli ds and liqu ...(z)
ids, the val f -2 mole
.. : u . 'fi
!llrl6Udonotw11ersigm 1can tly.H owe ver, Au
they diffe rsigues
nificoantl
un
y [ 1
Ang = ( 1 + - ) - 1 = + 1 mol e]
incase of gaseous reac tant s and prod ucts
!)at takes place at cons tant tem pera
. Con side r a reaction 2 2
ture and pressure. Let nr and
are the number of mol es of reac tant s and Con ditio ns whe n AH <AU .AH is less
prod ucts respectively. than AU whe n&i (g) is
1{and VP are
the volu me of reac tant s and prod ucts negative, i.e., whe n num ber of mol es
respectively . of gase ous prod ucts is less
Then for an ideal gas, than the num ber of mol es of gase ous
reac tant s. For exam ple,
2 NO(g) + 02(g> -
PV = nRT 2N0 2(g) [An g= 2- (2 + 1) = - 1 mol
e]
In case of reac tant s : / 2 mole I mole
~ AH is mor e sign ifica nt than AU. Mos t of
2 mole s

I>_l_ace in open vess els,i .e,::. at _co~ t~nt the. reac tion s take
PVr = nr RT (at cons tant T and P) m e~ ure. The refo re, the
... (ii) ene ~y chan~~ a~':?mPU!lY.:!f!.g_a_£roc~
In case of prod ucts : ~~ -~-U :~:n1~tant pres_sure , i.e.,
AH 1s more sigm fican t than AU.
PVP = np RT (at cons tant T and P) ·
...(iii) To ~ummarise, we hav e ·
Subtracting equa tion (ii) from (iii), we
get Alt8 =0
AH =AU whe n
PVP -PV r=n pRT -nrR T whe n Ang = + ive
AH
AH>AU
P(VP- V r) = (np -nr) RT <AU whe n Ang =-i ve
PD .V= ~ RT
8
[&1 is the diffe renc e betw een num ber NU ME RIC Ai. PA OB I.E MS BA
8 of mol es of gaseous SED ON :
products and reac tants ]
RELATION BETWEEN L\u and L\H
Thus , equation (i) chan ges to the form
D.H = D.U + D.n(g) RT For mu lae and Un its
or [Energy chan ge at] [En ergy chan qP = qv + An(g ) RT or AH = AU + 6.n(
constant pres sure = cons tant volu ge at] g) RT
me An(g) = (np - nr) gase ous ; R = 8.31
4 JK-1 mo\-1

T
Cha nge in the num ber of moles of] x
+ [ gase ous prod ucts and reac tants
osummarise, we hav e .
RT 1W EXA MPL E 10 c::==============:::::::::-
Calculate the difference betw een heat s
ot· rn•a c ti"on at con sum t
. pres sure an d con stan t volu me for the '-<

reac tion at 25°c in kJ

MI =D.U + PAV 2C6H 6(l) + 150 2 (g)- -+ 12CO z(g) +
MI= D.U + An RT 6H2O (l) .
Th ... (iv) Solution. Hea t of reac tion at cons tant
. ese relations help tog con vert ~ into . cons tant volu me (AE ) are rela ted as : pres sure (6.H ) d
D.U and vice versa.
~q Since Ml= q and D.U = q rela tion an nt
,+i\n RT (iv) can be written as qP AH =AU +A,1 RT
p V' 8
I II 2C6 H6( l) + lS0 2(g ) -
andc tis a relation betw een heat of reac 12C 0 2(.i:) + M-- -l,0 (/ )
tion at cons tant pres sure 15 n-ole
onstant , I I 2 1110k -
Co .• o ume . AH -AU =i'.\1111RT
'llhenl'l 1ltons whe n AH = AU. Foll owi .. lwh crcL\J,~ = I.,
ng are the cond1uons T=2 5+2 73= 298 K : R = 8.314JK - ~ml - I 5 - 1
:::~u : -- . - -. mol e)
vo rr} iWhen reaction take s plac e in a clos = - '.\m olcX K .3 14J K X29 8K
ed vessel of ftxe d ~H - AU = - 74:\'2.7 J = -7.4 33 kJ
, .e., When 6. V = 0.
 6 / 16 ISC CHEMISllty
q- E XAMPLE 11 _____ _,_ - - - - - - - - · - - r===== ==:===:==~~=~=::===:=:::::C-:~~~l'l
PlloaLEM 14. The beat of combusti on or ethyl aJCObot
The reaction of cynamidc, NH2CN(s) with dioxygen was carried and one atmosphe ric press~e is-1363 kJ. C~ J{.
out in a bomb calorimet er and 8U was found to be- 742.7 kJ moJ-1 internal energy change for therr reaction. ~ !;
at 298 K. CaJcuJate MJ at 298 K for the reaction. (Ans. 8 U ==- 136()
3 2
2
NHzCN( s) + Oz(g)- + Nz(g) + COz(g) + HzO(l) PlloaL£M 15. The beat of combusti on of one gram or ~ 1:1] !.
· · in a bomb calorime ter was found to be -14.57 kJ at 2S~
Solution For the given
•
3
equation, Numbero f moles of gaseous Calculate the enthalpy of combusti on of acetic acid at ri
25
.~c.
reactants (n 1) = ,
2 (Aos.--874.2kJJ_i
Number of moles of gaseous products (n 2) = I + 1 = 2
3 6.14. HEAT CAPAC ITY OF A SYSTEM
~=n2-n 1 =2- =0.5
2 Heat cap~~ity _or_~ accurat ely m~h,
T=298K ; R=8.314J K-I mo1-1 ca.f!W1y_.j,1_~ t e m, between any two teni •t
Hence, 8H = 8V + ~RT ratures, is defined as the amount of heat requip"
to raise the t e m p e r a t u _ ~ ' l d
=- 742.7 .kJ + 0.5 X 8.314 X 10-3 X 298 kJ loWer to the ~igher tempera ture divided bv t~
tempera tare differen ce. ._,_ e
=- 742.7.kJ + 1.2388 kJ =-741.46 kJ.
Es" E XAMPLE _
12 r - - - - - - -------:.,-:.,-:.,-_-_-:_-_-:_ Thus, ~ is- the amo-u~t of heat supplied to a system and
if
result, the temperature of the s_yst~m rises from T1 to T2, tbent
The heat liberated on complete combustion of7.8g benzene is
327 kJ measured at constant volume at 27°C. CalcuJate the heat
of combusti on of benzene at constant pressur e. R = 8.3J
heat capacity C of i:':le-"'-;
give: b! • \

1 - (T2 - T1) - d T .../i)

Solution. Combusti on of benzene is represented as
15 In simple words, heafcapaciiy may be defined as t ~
(g) + 3H 2O(l)
C6H5(/) +
2 02 (g)--+ 6CO 2 6mole
of heat e!lergy require4__to raise the tempe_r_at.Jge_o f 1 ~1
7.5 mole by 1 K. If the mass of the substance is lg, the heat capacity li
Molar mass ofC6H 6 = 6 x 12 + 6 x I= 78 called specific heat capacity.
7 .8 g of benzene on combustio n liberates heat = 327 k
Specific heat is the amount of heat energy required to rait
the temperature of lg of the substance through 1K and the311Qm
78 g of benzene on combustio n liberates heat= 3270 kJ of heat energy required to raise the temperature of 1moleoft!'i
8U =-3270.k J; T=27 +273=30 0K substance through lK is called molar heat capacity.
But 8H =8U +8ngRT =-3270- 1.5 x 0.0083 x 300 Molar heat capacity = Specific heat x Molar mass of !Ir
substance .
==-3270 -3.735 =-3273.7 35 kJ moJ-1
Heat capacity varies with temperatu re. Therefore, the valueof
11'¥ EXAMPLE 13 C has to be considere d for a very narrow range of temperature.
The enthalpy change (8H ) for th e reaction : Hence, the true heat capacity is defined by the differential equatioo
N2 (g) + 3H2 (g) - 2NH3 (g) is- 92.38 kJ a t 298 K. W hat C= oq .
...(n1
is 8 U at 298 K ? a['
Solution. 8H and 8U are related as LIB= 8V + 8n (g) RT w~ere Oq is the small amount of heat absorbed by a system which
For the reaction, N2 (g) + 3H2 (g) --+ 2NH 3 (g) raises the temperatu re of the system by a small amount al' (sai
from T to T + elf).
8n(g) = 2-(1 + 3) =-2 mol
Since q is not a state function and depends upon the path follo~ed
T=298K ;R=8.31 4JK-1 mol- 1 by the system, therefore, heat capacity is also not a state funcllOIL
Llli=-92 .38 kJ =-92380 } Hence, to evaluate C, the condition s such as constant volwne (l
:. -923801 =dU + (-2mol) x (8.314JK - mol- ) x 298K
1 1 constant pressure have to be specified in order to define the pam.
-92380J =dU-49 55J Thus,_ there are two different types of heat capacities. These are:
or
(z) Heat capacity at constant volume C
8U = [-92380 +4955] J =-87425 J =-87.42 5kJ
(ii) Heat capacity at constant press~e ~
N according ' Pof thermodynaJIII·cs·
· to the equation of first law
FOLL OW UP PROBLEMS · ow
Oq =dU + PdV; ... C = dU + PdV ...(iiil
a['
PROBLEM 12. One mole of naphthal ene (C10Hs) was burnt in When th e volume is kept constant, dV = Oand
oxygen gas a t constant volume to give carbon dioxide gas and
liquid water at 25°C. The heat evolved was found to be
.-. c (au)
= . tii•I

5139.0 k,J. Calculate the heat of reaction at constant pressure.

(R =8.314J K-1 moJ-1) _ [Ans.-51 44.0kJ]
PROBLEM 13. The enthalpy change (811) for the reaction is
f,
or or an
1·de
~ .':.rp,at
Cv=-
aT v
dU
capacity can be given by I.he
.
'""
relaO"'
)i i

- 92.38 kJ at 298 K. What is the value of dU at 298 K ? . ~ ~

"'For an ideal as-IIl temaI energy change is independent Of voturne
de010
f
N (g) +3H (g) -2NH3 (g) [Ans. dU =-87.425 .kJJ
2 2
depends only on temperature. Enthalpy of an ideal gas is indepen
pressure and volume and is a function of temperature only.
 c,\L. THERMODYN AMICS 6/17
~•d
~ eat capacity of a system at con
~11f, 11 defined a-s the rate of cha·n stant_volume From equations (i) and (iii), we get
~at be with temperature at constanle o,f internal 1 1
, .•,gY vo ume. Cp-Cv = R=8.314JK- mot"
· change in v
e~~
nstant pressure, there 1s 01 The above relationshi is callectl).feyer~s g!atiqnsbip.
A!c%one. from equation (iii) , we have, ume and some Alternative Derivation. No work is done at constant volume
15
(P.~ V = 0, because~Vis zero) whereas at constant pressure, some
1~r~ _ (au)
Cp- + p (av)
-
aT P aT ... (vz) work is done as a result of increase in volume.
- p
H==U+PV Thus, Cp-Cv = work done by one mole of gas due to expension
NoW of gas when heated through 1°C.
.fferentiating the above equation w.r.t. Tat constant pressure, ...(z)
D1 For one mole of an ideal gas, PV = RT.
it get On raising the temperature by 1°C, temperature changes from
( aH) = (au) +
P (av)
... (vii)
T (T + 1) and the volume changes from V to (V + ~V)
aT P aT P aT P ...(ii)
Thus, at(T+l)°C, P(V +~V)=R(T+ 1)
proroequations (vi ) and (vii) , we get

Cp == ( ~~ t or for an ideal gas, .. .( viii)

or
Subtracting (i) from (ii),
P(V +,1.V)-PV=R (T+ 1)-RT
P~V=R
aH
Cp = aT .. .(ix) But PL\V is the work of expansion done by one mole of the gas
when heated through 1°C.
The above relation can also be obtained as follows :
Thus,§-Cv= R- J
Cp = (<5q)p
af ..
. ,ofh eat capac1tJes,
R atlo y = (Cp)
,C v
But (Dq)p == dH (according to the def. of enthalpy change) .
\' _ (i) Formonoatomic gases(He, Ne, Ar, etc.),y = 1.66
. . lrCp=~ J · ; ,.,,- (ii) For diatomic gases (H 2, 0 2, N2, etc.), y = 1.40
Thus, heat capacity of a system at constant (iii) For triatomic gases(O 3, CO 2, H 2S, etc.), y = 1.30
pressure may be defined as the rate of change of \..,
Specific heat is an intensive property. It is an inherent thermal
enthalpy with temperature at constant pressure.
property and is unique for each substance. For example ,
Unitsofheatcapacity.Cal deg- 1 mol- 1 (C.G.S.) specific heat of iron at 25°C is 0.45 J g- 1 C- 1 and that of silver is
0

1
0.235 J g- C- .
0 1
or JK- 1 mol- 1 (S.I.)
Heat capacity on the other hand, is an extensive property because
, 6.1 5, RELATION SHIP BETWEE N Cp and C v i~ _epen!;!Lupon..the. s~ _of..the..s.aµiple. We could not for example,
IN GASEOUS SYSTEM S speak of the heat capacity of water but only of the heat capacity of
if the volume of the system is kept constant during the addition a specific quantity of water (say a glassful or a bucketful).
of heat to a system, then no work is done. Thus, the heat added to Heat exchanged = Heat capacity x L\ T
lliesystem is used up completely to increase the internal energy of = Heat capacity x (T1 Ti)
lliesystem. Again if the pressure is kept constant during the addition
= specific heat x mass x (Ti-Ti)
of heat to a system, then heat absorbed also does some work of
expansion in addition to the increase in internal energy• Thus, if the where Ti is the initial temperature and T1 is the final temperature.
~lllperature of the system in case of a process at constant pressure
ij to be raised by the same value as constant volume, then some NUMERICAL PAOBI.EMS BASED ON :
extra heat is required for doing work of expansion. Hence Cp > Cv·
HEAT CAPACITY
We know that Cv = (~)and Cp = (~) ff' EXAM P LE 14
Calculate the member of kJ ne
... (!) temperatureof 60 0gofalumin" f cessary to raise th e
Cp-Cv = ( ~ ) - ( ~ ) . . ·. mm rom35toSS°C M I . 1
capacity of alununmm is 24 J mol- l K- l · o a1 1cat
Also (by definition) Solution. Atomic mass of Al =27
H=U+PV.
anct (for one mole of an ideal gas)
PV=RT 60 mol
: • number of moles of Al, n =
I)· H=U +RT 27
· with respect to temperature, T
lfferenti· ating
q=nXCX~T
dH dlJ ... (it)
-=-+R =(~~mo)) X (24 J mol-1 K-1) X (55 - 35) K
dT af
dH dU ... (iii)
= 1066.7 J = 1.07 kJ
df-dT=R
' £

6/18 ISC CHEMISTR.y p

t\1\1,1
~ E XAMPL E 15 ~H=H p-Hr = + ve
Calcu late the enthaJpy chang e on freezing 1.0 mol of water Reactions which are accompanied by abs~t ion.--Qf
at e positiv value of MI are known as endo•t:--~1-
l0.0°C to ice at-10.0°C, Clru H = 6.03 kJ mo1-1 at 0°C Cp
5 [H2O( l)] e~~ and hav ~~-~~::,.e.;.:;;; :;.;;..;--- - - - -'!ltl'ltlic
= 75.3 J mo1- K - 1, CP [H 20 (s)] = 36.8 J mo1-1 K- 1
1 reactions, e.g.,
_ .--. N (g) + o2 (g) - 2NO (g) - 180.5 kJ
Solution. Tota1Af1 = (1 mol water at 10° - 1 mol of water 2
ooq at or N2 (g) + Oz (g) - 2 NO (g) ;~= + 180.SkJ
+ 1 mol water at 0°C-- + 1 mol ice at 0°C) Some other examp les of endoth ermic reactions are :
+ ( 1 mol ice at 0°C-- + 1 mol ice at- l 0°C) C(s) + H2O (g)- C0 (g) + H2 (g); ~ =+ 131.4 kJ
H (g)+I2 (g)-2 ID(g ) ;~= +52.2 kJ
=Cp[H 2O(l)] X ~T+Af1treezing+Cp[H O(s)] x ClT 2
2 SnO (s)+2C (s) -sn(s )+2C O(g) ;Af1=+360kJ
=(75 .3JK- l mol- 1) (10-0 ) K+ (-6.03 x 103 JmoI-1) 2 • All\
,.,,,~-. Exothermic reactions (Negativ~ LU 1.,. In many reactio~
+ (36.8 JK- 1 mol-) x (0- 10) K th-"{heat energy is evolve d when the reacuo n occurs. Heat
evolved
= (753- 6030- 368) JK- 1 moJ- 1 =-564 5 J =-5.6 45 kJ lowers the enthalpy. Conse quentl y, the enthalpy of products
.
less than that of reactants [Fig. 6.9] 15

FOL LOW UP PRO BLE MS ~=H p-Hr =-ve

PROBL EM 16. Calcu late the energ y neede d to PRODUCTS (Hp)

raise the
tempe rature of 10 g of iron from 25°C to 500°C if the specifi
--.- ---- -- ----
c 1

heat capac ity of iron is 0.45 JK_-1 1 r [Ans. 2.137kJ] I

PROBLEM 17. In one experi ment, a studen t mixed 50 mL of 1.0 I

M HCl at 25°C with 50 mL of 1.0 MNaO H also at 25°C HA> Hp

in a t.H=-Ve
caJori meter . After the mixtu re was stirred , the tempe rature
quickl y rose to 3 l.9°C. What is the energy evolved ? Assum
e
that the specif ic heats of soluti on are close to that of
pure
water , i.e., 4.18J g-10 C-1• The density ofl.0M NaOH is
1.04
g mL- and that of 1 MHC l is 1.02g mL-1• [Ans. 2.97 kJ]
PROBLEM 18. CaJcu late the atomi city of an
eleme ntary gas REACTANTS (HA)
havin g molar mass 40 and specific heat 0.313 Jg-1 K- 1
.
PROGRESS OF REACTION
[Ans. Monoatomic]

6.16 . ENT HAL PY CHA NGE S DUR ING CHE - Fig . 6 .9 . Entha lpy chan ge during an
exoth ermic react ion.
MIC AL REA CTIO NS-E NDO THE RMI C
AND EXO THE RMI C REA CTI ONS Reactions which are accompanied by evolution of.beatene~
(SIG N OF ~ H) and have negative vaJue of~ are known as exothermic reaction1
e.g.,
,
1. E ndoth ermic reacti ons (Posi tive ClH). Many reactio C (s) + 02 ( s ) - CO (g) + 393.5 kJ
ns 2
are accom panied by absorption of heat energy. Heat
absorbed or C(s) + 02 (g) - CO2 (g) : 6.H=- -393.5 kJ
raises the enthal py. Thus, enthalpy of products is greate
r than the Some other examp les of exoth ermic reactions are :
enthal py of the reactants [Fig. 6.8].
Ni(g) + 3H2 (g) -2N H (g); ~H = -92.3 kJ
3
• CH4 (g) + 202 (g)- CO (g) + 2H O(l) ;MI= - 890.4~
PRODU CTS (Hp)
2 2
--.- ---- ---- ---- -- 13
C 4H10(g) +
1 .; 2 O2- 4CO (g)+S H O(l) ;MI= - 2876~
2 2
H+ (aq_) + OH- (aq) - H20 (l); ~H =- 57.1 kJ
Thus, It may be conclu ded that :
Hp>HR
6H=+V e
For endothermic reactions : Ml= +ve
For exothermic reactions . Ara-_
•L.ll1
.--Ve
Simila rly for · . perali¢
' reacti ons taking place
and
f volum_ e ' .~u is. +ve "tor endot hermi at constant ternAU1·s , 1.,'
or exothe rmic reactio ns. c reactions and u

REACT ANTS (HR)

PROGR ESS OF REACT ION
- #•HMW1 11il~;t❖ l!#:e!lf,
PR:BLEM 19. Cl~ssify the follow ing reacti ons
into exothert!Uc 1\
Fig. 6.8. Enthalpy chan ge durin g an
an endothenruc reactions: '
endothermic reaction. OH
1 1
2 (g) + 2 02 ( g ) - H2 O(g); ~H =- 286 k.J
 AL -rtt ERM OD YN AM ICS 6/1 9
,~~JC~~ '.g(g))-~lli<~~I ~~ ~~===:::::===:::::::-,
#•1•!•) '.♦l •l :IQ ;3•1=1 • 4►~ S-
(g) + Clz - 2HC (g) ; AfI = - 185 kJ

-
I
I 2
1Jil th (Of{)z.8H20 (s) + 2NH4Cl (s): __ BaC R20 (s)
+ 2NH 3 (aq ) + 88 0 (I) _2•
l~J JJa 2 ,M l-6 3S kJ
O ou {I) • - • PROBLEM 20.C alcu late the enth
alpy cha nge for the reac tion :

~·
)
11C(S) + 2H2 (g + 2 ( g ) - CH 3CO
'
LlH = - 491 8 kJ
H2 (g) + Ch (g )- 2HCI (g) ; ~rH =?
d'Oz (g) + 6H20 (I)- + C6" 120 6 (s) + 602 (g ,
). · if bon d ener gy of H - H bon d = 430 kJ
, (f)I)\.; bon d ener gy of Cl - Cl bon d = 242 kJ 2
2840 kJ [An s. ~rH = - l 8 kl]
(Ans· (I) Exo ther mic (ii)
E
Exo ther mic (. -~) = bon d ener gy ofH - Cl bon d= 427 kJ.
m ndothemu
( • Exo ther mic (v) Endothennic~
iv) EQ UA TIO NS
6.1 8. TH ER MO CH EM ICA L
fi7. , OR IG IN OF EN TH AL PY CHAN
A RE AC TIO N
GE IN A che mic al e uat ion wh
l'.I e~ taki ng~plac e durJ
ic ,..i.n cJ11 des_ t.b.e ._.h.~~t
ag-. .a C em lc~L fea.ci:l.o.a:LS
~ ~ !:.~ ~ c~!. equ a ·
tion ed, a che mic al reac t·Ion invo . the the rmo che mic al
As already men lves the Important conventions for wri ting
. h tant s and form atio f
_.,; 0g of bon ds m t e reac n o new bonds
Th . k th e b onds eqwttions are :
orei1JU
e ene rgy Is requ ired to brea e and for
10 f
orm products. • .
~ For exo ther mic reactions, LlrH is neg ativ
1 the form atio n of bonds.
while the energy IS re ease d d~n ng tive .
tion s are related to bond endotflermic reactions, ..1,H is posi
fherefore, enthalp~ cha nge s dun ng reac in a give n reac tion , ~H
itse lf inte ract s with the 2. Unless otherwise mention ed,
energies. In solu tion s, the solv ent · of a subs tanc e, i.e., whe n
reactants and the .pro ducfts and in soli ds, ther e are interact Ions values are given for standard state
. The refo osp heri c pre ssur e.
between the part ic 1es o neig hbo urin
g mol ecul es. reactions occur at 298 Kan d one atm
ds and solu tion s. Hence, we sh:~i ces of the che mic al
situation is comp~icate~ in soli 3. The coe ffic ient s of the sub stan
alpy cha nge of a reaction mol es of eac h sub stan ce
consider the relat10nsh1p betw een enth equ atio ns 'ind icat e the num ber of
the reac tion s in gase ous valu es corr esp ond to thes e
and bond ene rgie s by con side ring involved in the reaction. The ~rH
phase. coefficients.
will be equal to the ical equ atio n for a
In gas phase, the enth alpy of a reaction 4. Whi le writ ing the ther moc hem
minus the energy released tant s and pro duc ts mus t
energy required to brea k all the bon ds reaction, the phy sica l state of the reac
ple, in the case of formation a sub stan ce is indi cate d with
during the formation of bonds. For exam be mentioned. The phy sica l state of
kJ mo1- 1 • Thus, heat of aq for soli d liqu id, gas and
ofHCl(g) from H2 and Cl 2, LlH = - 185 the help of designations s, l, g and
the energy required to
reaction should be the diff eren ce betw een aqueous states respectively.
the energy released due to
break the bonds betw en H 2 and Cl 2 and l stat e of reac tant s and
. It is essential to indicate the phy sica
!he formation of two mol ecul es of HCI end upo n the phy sica l stat e
products because dH values also dep
H- H+ 43 7k J-H +H e.
for reactants and products. For exa mpl
Cl- Cl+ 244 kJ- Cl+ CI 1 - 286 kJ
H2( g)+ 2 O2( g) -H 2O (l) ;~n -1=
H+ Cl- HC l+4 33k J ' .
1
or 2H + 2C l- 2HCl + 866 kJ H2 (g) + 02 (g) - H2O (g) ; LlrH = - 242 kJ
2
Ll,tt =43 7 + 244 -86 6=- 185 kJ equ atio n are mut ipli ed
5. When the coefficients in a chem ical
1 6 c::.=.-:.-:.-:..-:..---::::~----:_::..=.:::.:::--z:~:::i.::
:::"!2:m:z::i.'il mus t also be mul tipl ied or
rt EXAMPLE
0
or divi ded by a factor, LlrH valu e
ple,
reac tion : divided by the sam e factor. For exam
Calculate the enth alpy cha nge for the l
H2 + F 2 - 2HF ; MI = ? 2
H2 (g) + 02 (g) - H2O (l) ; ~rH = - 286 kJ
d= 43 4 kJ moJ -l
Given that ; Bon d ene rgy of H - H bon if coefficients are mul tipli ed by 2, we
writ e the equ atio n
Bond ener gy ofF - F bon d= 158
kJ mo1 - I = - 57'2 kJ
2H2 (g) + 02 (g) - + 2H2O (l) ; LlrH
Bond ener gy of HF bon d= 565 kJ
mo1 - I. rsed , the sign of' L1
6. When a chem ical equ atio n is reve
. , one H- H bond and one d mag n,·r, ,de <J·.f· LJ,-H . · _rH
urion. In the give n reaction value is also changed. How ever' the
A
So1 · 1ema 111s
F,i:: ds are to be form e ·
whi le two H-F bon unaffected. For example,
' 1b bond is to be brok en
erefore, SO 3 (g); drH = - 97.9 kJ
SO2 (g) + 1/20 2 (g) - (not hcrm ic)
d]
ii,1-1::: [- 2 x bon d ene rgy of H - F bon
+ bon d ener gy of H - H bond SO3 ( g ) - SO2 (g) + 1/2 0 2 (g); LlrH
= + 97.9 kJ
(t:11duthcrmic)
+ bon d ener gy of F - F bond an end othe · ·
Thus, an exothermic reaction will be nnic reac tion
565 ) + 434 + 158] kJ in the backward direction .
::: [-(2 X

==- 113 0 + 592 =-5 38 kJ.

 T
6/20
Table 6.2 Standard enthalpies of some substa
6.19. ENTHALPY (HEAT) OF REACTION 298 K (kJ/mol). llce, at
The t otal a m ou nt ._qt._ heat~eULUY c,voJv__§.d or Substance
~ sorbed when t he number of males of reactants
- fo-alv'ett, · as give~ Tn
L.:===+--:::-:-::--i--;;-;~
H20 (g)
~--1-~~~
- 24l.82
AICI3 (s)
t- - - - e m ca equation Is known as ~t - 1675.7 H20 (l) -285.83
AI20 3 (s)
Al2(S04)3 -3441.0 HF(g) - 271.1
Since, most of the reactions take place at constant pressure,
therefore, enthalpy change, 6H, is the heat of reaction, e.g., BaC03(s) -1216.3 HCI(g) -92.3
C(s) +02 (g)-C0 2 (g) ;6rH =-393.5 kJ BaCl2(s) -860.2 HBr(g) - 36.40
Thus, - 393.5 kJ is the heat of reaction. BaO(s) -553.5 KCI (s) -436.74
At constant volume, 8rH = 6U. Therefore, internal energy Ba(NOJh m .o KBr(s) - 393,s
change (6U) is the heat of reaction at constant volume. Br2 (l) 0 MgO(s) -«>1.7
As al.ready mentioned, heat of reactions or 8rtt values also Br2(g) 30.91 Mg(OH)2 -924.54
depend upon the physical state of reactants and products. It also CaO(s) -635.o<J MgCl2 (s) -641.8
depends upon the temperature of the reaction. Therefore, 8rtt MgClz.2H20 (s) -1280.o
CaC03(s) -1206.92
values are reported in a standard or a reference state. According to
thermodynamic convention, a substance is said to be in standard C(graphite>
0 NaF(s) -573.65
state when it is present in its pure form at J bar at a specified C(diarnond) 1.89 NaCl(s) -411.75
temperature. 715.0 NaBr(s) -361.~
c(gas>
For example, the standard state of liquid ethanol at 298 K is C~(g) -74.81 Nal(s) -287.78
pure liquid ethanol at 298 K and 1 bar.
C2~(g) 52.26 N2(g) 0.0
The standard enthalpy of a reaction is the enthalpy change
C2H2 (g) 54.19 NH3(g) -46.11
for a reaction whrp 311 the ceacmnts (elements and compounds)
are in their standard state. - Cz!¼(g) -84.68 NI¼Cl(s) -314.4
It is represented by 8rH 8 where the superscript '8' indicates C3Hs (g) -103.2 NO(g) ~15
the standard state. n-C4H10(g) -888.0 N02 (g) 33.18
CJ-16([) 49.03 l(g) 25.48
6.20. DIFFERENT TYPES OF ENTHALPY
CH30H(l) -238.66 N20(g) 815
(HEAT) OF REACTIONS
C2HsOH(l) -277.69 HN03 (L) -174.1
The heat changes taking place during chemical reactions
can be expressed in different ways depending upon the nature of CH3COOH (l) -487.0 PCl3 (L) -319.70
the reaction. Some common types of the enthalpy of reactions are CC4(l) -134 PC15 (s) -443.5
described below : CO(g) -110.525 Si02 (s) (quartz) -910.94
6.20.1. Standard Enthalpies of Formation C02(g) -393.5 Sil4 (g) 34.0
A formation reaction is one in which 1 mol of a substance is C02(aq) -413.8 SiC14 (g) -657.0
formed from the elements in their most stable forms at I bar and Cl2(g)} 0.0 SnC12 (s) -325.l
usually at a temperature of 298K. Cl(g) 121.3 SnC4 ([) -511.3
Enthalpy of formation is defined as : HgS(s)red -58.2 S(s) rhombic 0
The enthalpy (heat) change wh~n one mole of su,b~nce is H2 (g) 0 S(s) monoclinic - 71
formed from its elements. H202(l) -187.8 S02 (g) -296.83
It IS denoted b ~ . For example, 2 moles of hydrogen gas lll(g) 26.48 S0 3 (g) -395.72
react with one mole of oxygen gas to produce 2 moles of water
H(g) 218.0 H2S (g) -20.6
(liquid). The reaction is exothermic and proceeds with the evolution
of 572 kJ of heat. If may be expressed as : 12 (g) 62.4 H 2so4 (l) - 813.8
2H 2 (g) +0 2 (g)- 2H20 (L) ; 6H =-572kJ ' ~
I.t may be noted that standard molar enthalpy of formao~
Hence, heat of formation ofH 20 (/), i.e., a special case of 8r H 9 where one mole of a compound is fornieJ
from its elements. e.g.
61 H = - 5~2 kJ = - 286 kJ
C(s)+02(g)-C02 (g) ;6 He=- 393 .5 kJ
1
Similarly, heats of formation of carbon dioxide, ammonia and . H · the enthalpy of formation of COi, H0we\'11·
ere - 393 .5 kJ 1s
sulphur dioxide are, - 393.5 kJ, -46.19 kJ and-297 kJ. m the reaction,
Standard enthalpy (heat) of formation . It is defined as: CaO (s) + 02 (g) - CaC03 (s) 6 H9 = - I78.3 kJ
fC~
The enthalpy change when 1_!llOl~_of a substance is formed r .
Here - 178.3 kJ is not the enthalpy of fonnatton r,J °
from its elements in their standard states. beca~se CaC03 has been formed from other compouods and
The standard enthalpy of formation is denoted by 8r ~e ~d from its elements. .
't ll
standard enthalpies of some common compounds are given m In calculating the 61 H0 values given in Table 8.6, ,
Table6.2. assumed that the standard enthalpies of elements are -u,ro.
~
e~JC
AL THERMODYNAMICS
•
f formation and Stability
- 6/21
~t o A .ue . d'
~e os1·o·ve value. of ur& m 1cates that the th
en alpy of th
q- EXAMPLE 18
P dformed is greater th an the sum of the en . e 0 (g) N O (g) and
th Enthalpies of formation of CO (g ), C 2 ' ~ 1 Find
,rt1p0llllfro!Il which the compound is fonned C alpies of the N2O4 (g) are-110, - 393, 81 and 9.7 kJ moJ-l respective Y·
ry- nts bl ha . . onsequentl th
I~ dis less sta e t n Its elements. Nitro en y, e the value of Ll,H for the reaction :
,it00IJTI t very high temperature of an electrt· g and oxygen
/V'"'.bille a .
\~~ NO,The heat of fo~atlon of nitric Oxide is 89.9 kJ
o,11d0 of heat is a?sorbed m the fonnation of one mole of ~l ~r
~re
c arc to fonn • . N2O4(g) +3CO (g) --- N2O (g) + 3C0 2 (g)
Solution. A, H = "J..6. H (Products) -IA1H (Reactants)
1
= [Ll1H <N2O) +3A1H(CO2)]-[h1H (N2O4) + 36.tH (CO)]
~.9 that NO 1s less stable than N2 and o and .t . ~ · This
ests .
¢~S roe other compounds with positive h1He are th ·ct e
2 1 1s 1ound to b = [81 + 3 (-393)]-[9.7 + 3 (-110)] kJ =- 777 ·7 kJ
° cs C H Th e oxi es of
. ,.,.,,i~•••l:14;t•)=1 • ➔ ~~s•
~e· 0 2 and 2 2· e compounds having posi't•1ve va1ue of
·lfOge , d h •
~ "7 re called en ot emuc compounds and a 1 stable
,1rW a re ess
"1 ntheir elements. On the oth_er hand, negative value of L1 He
1M. ates that the compound 1s associated with less en ergy 'f
as PaoBLEM 21.Calculate the standard internal energy chan~~
.
for the reaction OF2(g) + H20(g)- .02~)kJ+ ~~~e ·
~1c
d to the constituent atoms. Consequently the c nd 2
(Ompa re . , ompou
. estable than its elements. For example, Li'!Hefor H O( ) . 298 K. The standard enthalpies of formation m mo ·
~,nor J Th. . d' h 2 g is
,iB6kl mo/- . . ts m tcates t at bonds in H2o are stronger OF2(g) = 20, H2O (g) = -250 and HF (g) = - 270
u,.Dft H_H bonds m H 2 and O - 0 bonds in 0 2. Compounds like (R=8.314JK-1 mot-1) [Ans.hu8=-3 12 -5 k1]
PROBLEM 22. Calculate 6. u0 for chloride ion from the
8
HOhaving negative values of L11H are called exothermic 1
2
01pounds and are
more stable than their elements. In general, following data.
8 , the more stable the
:ehigher the negative va!ue of L11H
compound is. Conversely, higher the positive value or lower the ½Hz(g)+ ½ctz(g)-HCl(g); A1He=-92.4kJ
negative J1H8 value, more unstable the compound is. HCl (g) +nH2o- ff+(aq) + CJ-(aq); h,H8 =-74.8 kJ,
~rHe of a reaction At Heu+ (aq) = 0.0 kJ [Ans.-167.2:kJ]
Standard enthalpies] [Standard enthalpies] PROBLEM 23. The enthalpy of formation of Fe2O3 (s) is- 824.2
= of formation of all - of formation of all kJ moJ-. Calculate the enthalpy change for the reaction.
[ products all reactants
4Fe (s) + 302(g)-2Fe2O3 (s) [Ans.-1648.4:kJ]
or ~rH8 = Ih1H8 (Products) -I6.JI8 (Reactants) PROBLEM 24. Calculate the standard enthalpy change (6.118)
Where I is known as summation and means sum of. For a and standard internal energy change (AU8 ) for the following
general reaction : reaction.
2H2O2 ([)- 2H20 (l) + 02 (g)
aA+bB-cC +dD
8
At H at 25°C for H2O2 (l) = -188 kJ mol-
~,H8 = IA1 He (Products)-I A1 He (reactants)
H2O(l) =-286 kJ moJ- and R = 8.314 J K-mol-
=[ch1 He (C)+dh1 H8 (D)]
[Ans.-198.5 kJ mol- 1]
- [a h1He (A)+ b A1He (B)]

ff EXAMPLE 17 t:====~==--=.::-:::;-:;-:;-.:_•:.,.-_-_-_-_--~~~=-=-=-=-:::i 6.20.2. Enthalpy (Heat) of solution

Calculate the enthalpy change for the reaction : It is defined as the enthalpy or heat change whe~le-of ,
a solute is (fissolve~ mJar.g~~x<;ess of solv_ent so that O!Lfurther
C (graphite) + 2H2 (g)--- CH4 (g) dilution no J!eatcJiang~occurs,e.g., .
Given that (i ) C {graphite) + 0 2 (g)--- CO2 (g) NaCl (s) + aq- NaCl (aq); A501H6 = + 5.0 kJ
6.1 n0= _ 393.5 kJ moJ-l
Nl4NO3 (s) +aq-~O3 (aq); L\0 1H8 =26.0kJ
(ii) H2 (g) + 0.5 0 , (g)--- H2O (l) Integral heat of solution is the new tenn used now-a-days. It
1
• 6.t 8 0 = _ 285.8 kJ moJ- is defined as the enthalpy change when 1 mole of a solute is dissolved
in a specific amount of solvent, e.g.
(iii) CH4 (g) + 202(g)- CO2 (g) + 2820 (l) ;
moJ-1 (I.S.C. 2010) (1) H2SO4 (l) + H2O ([)- H2SO4 (aq) ; Llsot He=- 30.5 kJ
t::.r H0 = - 890.3 kJ (ii) H2SO4 ([) + lOH2O([)- H2SO4 (aq); hsot He =-68.0 kJ
Solution. Required equation is 6.20.3. Integeral Heat of Dilution
C{graphite) + 2H2 (g)-CH4 ; AtH~tt,i) ==? It is defined as the net change of enthalpy when a solution
According to the given data: con · · lmolofsoluteataknownconcentration(C 1) ~
c14 (g) + 202 (g)- CO2 (g) + 2H2O ([) b adding more solvent to get a solution of another known
concentration (C2J It is in fact, equal to the differencel>etween the
~,lie=A He(C02)+26.1He(H2O)] e
integral heats u~on at the two concentrations. For example,
f - [At He (CH4) +2 t::._,H (02)]
when 20 mol of water 1s added to a solution containing 1 mo) of HCI
Or e (Cu .) + 2X0] in 5 mol of water, the accompanying enthalpy change of-8.182 kJ
~ 890.3==-393.5-2x285.8 -[t:.1H 114 moJ-1
is the integral heat of dilution .
74 8
. . i\H0 (CH4) = 890.3-393.5-571. 6 ==- · kJ
 6/22 ISC CHEMISTR.y P
. .
(i) HCI (g) + 5H2O({)-+ HCl.5H O; Llli=-63.995 kJ moJ-l Similarly. heat of neutral 1~a_t10n of NH4OH (weak bill~~l-t
2 HCI (strong acid) is -5 l.5_kJ. !his ,_s ~cause 5.6 k.J (57. l - s~lby
(ii) HCJ (g) + 25H2O-+ HCL25H O; Lili=- 72.177 kJ mo1- 1
2 heat energy is consumed m d1ss~c1at1_ng the weak base (NJi ,5)of
(iii) HCl.5H2O + 20H2O-+ HCl.25H O ; Lili dilution
2 The heat of neutralization of aceuc acid an~ ammoniuin hydr~O~).
=- 72.177-(-63.99 5)=-8.182 kJ mo1- 1 l·s 49 6 kJ because 7 .5 kJ ( 1.9 + 5.6) energy 1s used to dissoc· ~dc
1
·
6 .20 .4. Enthalpy (Heat) of Hydration weak acid (acetic acid) and we ak 6ase (ammonium . •ate,~
hydroxide) "'
Ut is defined as the enthalpy change when one mole of an
anhydrous oi;.natlmlly_hydrated salt combin~ with water tolo.mi
Experimental Determination of Heat of Neutralizau ·
Strong Acids and Strong Bases (S~y HCl and NaOII). A~:~
@liydraWe.g., -- - - ~ volume of HCI of known concentration (say I_ oo mL of IN Hci;t
taken in the calorimeter (a thennos flask) and lts temperature( /
CuS04 (s) + 5H2O ([)-+ CuSO4.5H2O (s)
noted. Then aknown_volume ofNa<?H (say lOOmLo!INNao~::
L\Hyd H8 = - 78.2 kJ taken. It is added mto the calonmeter after notmg down h
BaCl2 (s) + 2H2O ([)- BaCI2.2H2O (s) temperature of alkali solution (say tz). !he mixture is stirred and~:
L\HydH 6 =- 29.41 kJ highest temperature reached (say t3) 1s noted.
6.20.5. Enthalpy (Heat) of Neutralizati on The reaction between HCI and Na~)H is_exothermic and the
heat evolved on mixing the two. solutions _is absorbed by the
U!_
is defined as the change in enthalpy when one gr~m calorimeter and the solution present m the calonmeter, thermometer
e~uivalent of -~n a~.ru:ut.I:aliz ed b~~e oj'.J'..i.r.e.xersa,Jn. stirrer, etc. Taking the specific heat and density of the solution ~
dilute solutionEf."g., 1, total heat evolved due to the neutralization of 100 mL of IN HCi.
HCI (aq) + NaOH (aq)- NaCl (aq) + H2O (I) = heat absorbed by ( 100 + 100) mL of solution present in the
L\,,H 6 =- 57.1 kJ calorimeter+ Heat absorbed by the calorimeter
HNO3 (aq) + KOH(aq)- KNO3 (aq) + H2O (l)
L\nH6 =-57.l kJ
= (100+100) (, 3 -
11
;
12
)+MS (r 3 -
11
;'
2
)cal.=QC~.
HNO3 (aq) + NaOH (aq)- Na,1\/O 3(aq) + H2O ({) Hence, the heat of neutralization, i.e., heat evolved by the
L'inH 6 =-57.1 kJ neutralisation of 1000 mLof lN HCI (lg. eq). = IOQCal
Fron1 the above examples, it is clear that the enthalpy of 19
neutralization of all strong acids by strong bases and vice versa
1W EXAMPLE TI
are nearly- 57. l kJ. This is because all strong acids and bases are Given that: ff+ (aq) + OH- (aq)-H2O(1)+ 57.lkj
almost completely ionized. Neturalisation is a reaction between H+ Calculate the heat evolved when
ions (from an acid) and OH- ions (from a base) to form unionized (i) 0.5 mol of HCI solution is neutralised by 0.5 mol or
water molecules. For example, the reaction between NaOH and HCI NaOH solution.
can be represented as :
(ii) 0.5 mol ofllNO 3 is mixed with 0.25 mol of KOH.
HCl ~ W + CJ-
(iii) 300 cm 3 of0.2 M HCI solution is mixed with200cm3of
N aOH ~Na++ OH- 0.1 M KOH solution.
H ++OH- - H 2O ; L'inH 6 =-57 .lkJ (iv) 200 cm 3 of0.2 MH 2SO-' is mixed with400cm3 of0.1M
l gm eq . I gm. eq . NaOH solution.
Thus, neutralization is the reaction between one mole ofH+ Solution. (i) HCJ is a strong acid and NaOH is a strong
ions and one mole of OH- ions in the aqueous solutions to form base. Therefore, these ionise completely in solution.
one mole.of water. Hence, heat of neturalization of any strong acid
and strong base is always the same. 0.5 mol ofHCJ = 0.5 mo! ofH+ ions and0.5 molofNa0H=
0.5 mol of OH- ions
Neutralization of weak acids and weak bases. Weak acids
and weak bases do not undergo complete ionizatio~ i~ediat~ly On mixing the two solutions. the net reaction is
on dissolution in water. On the other hand, these get 10mzed du~ng W (0.5 mol) + OH- (0.5 mo!)- H O (0.5 mol)
2
the reaction of neutralization. A part of heat _ev~lv~d dunng According to the given data, 57. l kJ of heat is released w~
·zation is consumed in causing complete 10mzatlon of the 1 molofH+ionscombinewith 1 molofOH-ionstofonn I molofHP·
neu tra ll
weak acid or weak base. As a result, the heats ?f neutral'1zat1_·
on of :. Heat released in the formation of0.5 mo! of water is
acids with a strong base or weak bases with a strong acid ~e
wea k . . . f ak d = 57. l X 0.5 =28.55 kJ
less than _ 57. I kJ. The heat of net~abzat1on m case o a we ac1
(ii) 0.5 mol of HNO 3 x 0.5 mo! of H+ ions
or wea k b ase 1·s equal to the difference between the heat of
. r·10n for strong acids and .bases and the heat of
ne utra I1za 0.25 mo! of KOH= 0.25 mol of OH- ions
. . · · ('on,·zation) of the weak acid or weak base, e.g. , the 0-25 -mole of KOH neutralise only 0.25 mol of HN01111 f;'(1!!
d1ssoc1at10n 1 ·d · h ct·
heat of neutralization of acetic aci~ (weak ac1 ) wit so mm 0.25 mo! of water and 0.25 mo! of H+ ions remain unrcacieJ. i)
h droxide (strong base) is-55.2 kJ. It_1s bec_au~e 1.9 kJ (57 .1-55.~)
:. Heat released in the formation of 0.25 ,nol of wnier
o~ heat energy is consumed in dissociating the weak acid = 57. l X 0.25 = 14.275 k.J
(CH3COOH). +
. CH COOH + Na++OH---+ H2O+CH3COO- +Na
(iii) 1000cm3 of0.2 M HCI = 0.2 mol W ions
3
~,,H0=-55 .2kJ 300 cm3 of0.2 M HCI = _Ql_ x 300 =0.06rno1WioOi
CH 3COOH~ CH3COO- +W ; ~,,H 0 =+ 1. 9 kJ 1000
1000 cm 1 ol' 0. I M KOH '= 0. I molOWions
 c.AL THE RMO DYN AMI CS 6/2 3
.,tr,tl 0.1
,,. o~en , sulfu r and
OO cm3 of 0.1 M KOH = WOO x 200 = 0 _02 l _. Similarly, the heat of comb ustio n of hydr
mo OH Ions 890 kJ respe ct1 vely •
Z methane are_ 286 kJ _ 297 kJ and_
..,, J of oH- ions neut raliz e o 02 1 of H+ · The term com~ lete oxida tion or comb usti~
n in d~fin ition is
0 02 •··0 H O and 0.04 (0.0 6-0.02)· molmof H+ . Ions to ation 1_s poss ible, e.g.,
rnol of 2 used to avoid confusion when parti al oxid
.02d
fo/111 o
o Ions remain
heat of comb ustio n of carbo n relat es to the
react ion,
8cte ·
.
~ . Heat relea sed rn the form ation of o·02 mol of wate r
1
02- -+ CO
·~ 0.02 = 1.142 kJ. C + Oz - CO2 and not to react ion, C + 2
,jJ.1 . tr d'b · efore 1 1 Since, combustion is alwa ys acco mpa nied
by evol ution of
, (iv) HzSO4 ts as ong 1 as1c acid. Ther
' mo of H2S04 tive.
•UJllisheS cwo mol of H+ ions. heat, the enthalpy of combustion is alwa ys nega t
n. The heat of comb uS ion
I !OOOcm3of 0.2M H2S O4 =
0.2 x 2=0. 4mo lofH +ion s Significance of heat of comb ustio
havi ng a high valu e of
values are useful inlating_the..fuels. A fuel
= 1000 l d supe rior as com pare d
3 0.2 M H2SO4 0.4 X 200 -- 0 08 enthalpy (heat) of comb ustio n is cons idere
200 cm of · mo ofH+
to others.
jons ls
JOOOcm 3 of0. 1 MN aOH =0.l molo fOH -ion
s Calo rific Val ues of Foo ds and Fue
prod uce heat are
Most comm on chem ical react ions used to
3MN O -- 1000 0.1
X 400 =0.0 4mol ofOH -ions y relea sed when a fuel or food is
400cm of0.1 a H combustion reactions. The energ
burnt is know n as its fuel value or calor ific
valu e. It is defin ed as :
H+ ions to
0.04 mol of OH- ions neutr alize 0.04 mol of y_J_h~ _£_omp_lete
) ofH+ ions are left The amo unt of heat ener gy relea sed .b_
{or1110.04mol of wate r and 0.04 mol (0.08 -0.04 the fuel.
co~t i_Q n of one gram of
unreacted. their asso ciate d
s of wate r It is comm on to repor t calor ific value s with out
:. Heat liber ated in the form ation of 0.04 mole all comb ustio n react ions are exot herm ic.
negative sign beca use
=57. 1 X 0.04 =2.2 48kJ on a gram rathe r than
Furthermore, calorific value s are repo rted . Thes e do
usua lly mixt ures
mole basis beca use fuels and food s are
FO LL OW~(lP ',;:P·R <l'B LE MS not have any fixed mole cular weig ht.
of octan e, C8H J 8,
Ther moch emic al equa tion for comb ustio n
sed when : a comp onen t of gaso line is :
PROBLEM 25. Wha t wouJ d be the heat relea 0 (/)
is neutr alize d by 2CsH1s (/) +25 0 2 (g) - 16CO2 (g) + 18H2
(I) 0.4 mole of hydr ochl oric acid in solut ion ~=- 10,9 2Qk J
2 X 114
0.4mole of KOH solu tion
mole of NaO H 10 920 47.9 kJ
(ii) 0.5 mole of HN0 3 is mixe d with 0.2 •
Therefore, calorific value ofC 8H 18 ([) is = 2 X 114
solution?
600 cm3 of 0.1 M [·. · Mole cular weig ht of C 8H 18 is 114]
(iii) 400 cm3 of 0.2 MH2S0 4 is mixe d with
fuels are give n
KOH solut ion ? Calorific value s of some impo rtant food s and
3
400 cm of 0.4 M in Table 6.3.
(iv) 500 cm3 of 0.5 M HCI is mixe d with
com mon .
NaOH solut ion ? . Table 6.3 Calorif!c valu es of som e
[Ans. (i) 22.84 kJ (ii) 11.42 kJ (iii) 3.426 kl
(iv) 9.l36 kl] . . . food s and fuel s .
~

Food Calorific Fuel Calorific

6,20,6. Ent halp y (He at) of com bus tion ~ cH value (kJ/g) value (kJ/g)
one mole of
It is defined as the enth alpy (heat) change when Curd 2.5 Dun g cake 6-8
.,
asubstance undergoes com plete comb ustio n,e.g Mille 3.2 Woo d 17
C(s) +O 2 (g) -CO 2 ; ~JI= -393 .5kJ 7.3 Char coal 33
0 th Egg
defined as e
Slandard enth alpy of com bust ion (~cH ) is nd Mea t 12.0 Fuel oil 45
the reac tan~ ~
;!lhalpy change per mole of the subs tanc e, all at the spec ified Hone y 13.3 Kero sene 48
le oducts bein g in thei r stan dard state
s
Rice 14.5 Meth ane 55
lllJ>erature.
Dun· . fb . (mai or comp onen t Butte r 30.4 Buta ne (LPG ) 55
of coo . ng comp lete comb ust10 n o utane 'J

Ghee 37.6 Hyd roge n 150

king gas), 2658 kJ of beat is relea sed.
C4H1 0 (g) + 13/2 0 2 (g)- -+ 4CO 2
(g) + 5 Hz0 (I) . .
1._mong the food s, carb ohyd rates are the pnnc 1haJ sour ~es
~Jfe = - 2658.0 kl of energ y due to their high calor ific value H
· o~ev er, l e calo nfic
Comb . 1 ·
ter gives 2802 kl/ value s of prote ins is quite low · The othe ·r cons rnue nts of our food
ust1on of gluc ose in a bom b ca onm e h •tam· al .
lllol ofh sue as v1 ms, s ts, etc. do not prov ide mu h
eat. build ing mate rials cfo:: ~~~~ p~?t erns
_1 are used by body main ly as 100
?f
+ 6 HzO (/)
CJf 12O6 (s) + 60 2 (g)- -+ 6CO2 ~(g)cHe play a . c
= _ 2802 kJ mol organ walls , skin, hair etc . Vita mins also n •?1p ortan t role rn
life proce sses. Thus , a balan ced diet must
. body during the ydrar es fats prot . cdont am all the majo r
lne our ct'> are fonned cons tillle nts-c arboh ems an a smal l amo unt of
&enetation same overall proce ss takes place _m
after ase . of energy from food. However, the fin~
react ions mvo vmg
P'f'~ ·
enzymes.
,
vitamins. · ·'

nes of comp lex bio-c bemi cal

6 / 24 ISC CHEMISTR\' p
4
''
change jn internal ener gy per d ay . If ·t b e energy lost Was rtl,t
.
The amount of energy, the body requires, varies considerably
depending upon such factors as body weight, age and muscular as sucrose (1632 kJ per 100 g), how many days should it t~Oted
activity. The average adult requires about 2500-3000 kcal (10,000- lose 1 kg? (Ignore water loss). ~ct0
13,000 kl) a day. This is about the same amount of energy as that Solution. Energy taken by a man= 10000 kJ
consumed by a 100-watt bulb operating for a 24 hour period. Energy consumed in doing work = 12500 kJ
Among the fuels, hydrogen has the highest calorific value. Change in internal energy per day= 12500- lOOoo::: lXl kJ
25
However, due to some technical problems such as storage and The energy is lost by the man as he expends more energy
~portatiog_,.it is not co~only used as domestic or industrial than he takes,
fu~ ' - . Now 100 g of sucrose =
1632 kJ loss in energy ; IO(X) g
Calorific value of a fuel can be determined experimentally by sucrose= 16320 kJ
burning a known weight of the fuel or food in a bomb calorimeter : . Number of days required to lose 1000 g of weight
as described in section 8.9. 16320
Q> EXAMPLE 20
- -,~-- ------ ------
Gobar gas obtained by bacterial fermentaion of animal
or 16320 kJ of energy =
2500
= 6.5 days.

refuse contains mainly methane. The heat of combustion of

methane to CO 2 and water as gas is given b y
CI-Li (g) + 202 (g ) - CO2 (g) + 2H2O (g) + 809 kJ PROBLEM 26. (i) A cylinder of gas contains 14 kg ofbutane.tt
How much gobar gas would have to be produced per day for a normal family needs 20000 kJ of energy per day for cooking,
a small village community of 100 families, if we assume that each bow long will the cylinder last ? Heat of combustion of butane
family has to be supplied 20000 kJ of energy per day to meet all is 2658 kJ mo1-1•
its needs and that the methane gas content in gobar gas is 80 %
(ii) If the air supply to the burner is insufficient (i.e., you have
b y weight ?
a yellow instead of a blue Dame), 30 per cent of the gas escapes
Solution. Daily energy consumption of one faimly without combustion. How long would the cylinder last, if the
=20000kJ gas is burnt this way? [Ans. (i) 32 days (ii) 22.5 days]
Total number of families = I 00 PROBLEM 27. The heat evolved in the combusiton of glucose ls
The energy consumption of 100 families shown in the following equation :
= 20000 X 100 = 2000000 kJ C~12O6 (s) + 602 (g)- 6C02 (g) + 6H2O (g);
The combustion equation for methane is :
~H = - 2840kJ
CH4 (g) + 202 ( g ) - CO2 (g) + 2H2O (g) + 809 kJ What is the energy requirement for the production of0.36g
(16g)
of glucose for the reverse reaction? [Ans. 5.68kJ]
809 kJ of energy is evolved from methane= 16 g
PROBLEM 28. The thermochemica l equation for solid and liquid
2000000 kJ of energy is evolved from methane rocket fuels are given below:
= _!&_ X 2000000=3955 5g ~ 39.6kg
809 (i) 2Al (g) +¾02 (g)- Al20 3 (s) ; ~H = - 1667.8 k.J
Since, methane content of gobar gas is only 80%, therefore,
the weight of gobar gas is (ii) H2 (g) ½02 (g)- H20 (I) ; ~H = - 285.9 kJ
39.6 x 100 = _ kg (a) If equal~ of aluminium and hydrogen are used, wbidJ
= 49 5
80 is better rocket fuel ?
H" EXAMPLE 21 _ (b) Determine L\11 for the reaction,
A healthy person needs about 10000 kJ energy per day.
How much carbohydrates (in mass) will he have to consume, 1
Al2O3 (s ) - 2Al (s) + 1 - 0 2 (g) I
· g that al) his energy needs are met only by carbohydrates, 2 ,.n I
assumm
the form of glucose ? The enthalpy of com b ustion · ofl
in g ucose 1·s [Ans. (a) Hydrogen (b) 1667.8~
2900 kJ mo1- 1.
Solution. Molecular mass of glucose, C~1206 6 · 21 · ENTHALP Y CHANGES DURING
= 6 X 12 + 12 X 1 + 6 X 16 = 180 PHASE TRANSFOR MATIONS
1
Combustion of glucose can be represented as : . . Phas. e tr ··
. ~s1t1ons ·
involve the conversion ol- a so 1·d 1 in1°rJ
CJ-112O6 (s) + 602 (g)- 6CO2 (g) + 6H2O (g) + 2900 kJ h~md, a hqu1d mto a gas or direct change of a solid into 3 ~ 35dlt
2900 kJ energy is produced by 180g of glucose vic~ versa. All these transitions requjre energy to brin~ 1~
desired change. The heat changes involved ttre reported 111 villi
10,000 kJ energy is produced by glucose of ent~alpy change of_ that process . In :1 proce~s j~ vo~ 1~
180
= - - X )0 000=620.7g absorption of heat, ~H is positive and 6H is negauve if Ii 8
2900 ' evolved during the process.
I • J"~uthalpy
n- EXAMPLE 22 (I kn1) of fu-, ion /\ 11 (->. It is defiJJt' d os~
wt
change h, cnthal
Jf a man takes diclcqui valcntLo 100001<.f pcr d:1y nnd c:xp1·1u.ls PY wI' Ich takt's placetl.L,·w hen one 010 Je of, llv· ,g.,
ener gy .m all forms
. . to a t ot•,1l o f12500'•"·•fvHda y, w hnt11, U1(' 1.uh1,IJ1 nrc is converted h1to Its liquid state at ils mclting pOJfl
-....
NAM ics
,~xcAL fH ERMODY
H2O ([) ; ffE>::: + 6 L\fus
,~ HzO (s)_-
.02 kJ 6 / 25
~~ He 1s the enthalpy off us1on
.
or
f{ere us d 98.0 kJ mo1-1 .
. the standar state. tnolar entb 1
a PY of solid ct· respecuvel y. Therefore, enthalpy of sublim ation of
.
•JJill so mm at 298 K is written as :
f '[!!US, enthalpy of fus10 n of Water is 6.02 He
oJved when one mole of liquid WaterkJ. Same am ,:\sub He = Afus He + Llvap
A

K chan °unt of 1
1is ev
,(J (ice) at 273 . ges into solid = 2.6 kJ moI- 1 + 98 kJ rnol- 1 = 100.6 kJ rnoI-
on and
,j1er HzO (l)-H zO(s ) ; .::\er ffE> h Like enthalpy of fusion , enthalpy of vapourizati
~ . eez :::_ 6.02 kJ matio n also depen d upon the stren gth of
T.B
. ~nt alpy of subli
~fusHe =.t1fiq~d - solid r:,,t_elting of a solid is have strong er
Stand d endothermic , so mtermolecular forces . For example ' water molecules .
alpies of fusio .n are posit ive. attra · forces due to hydrogen bonding as compared to orgaruc
JeDth . . ar entha1 li u·cttv~
PY changes of lpy of
. and vapounzatio
/Jl on .
n are given in Tabl e 6 .4.
. q ids hke acetone. Therefore, water has higher entha
. · uon
vap ounza · as compared to acetone. (Table 6.4).
fhe enthalpies of fusion depend upon the Int
. b th ofd'f ti errnolecular n and
, sof attraction etween e mole cules Table 6.4 Standard Enthalpy Changes of Fusio
~ di hl ·ct iideerent sol'ids. Ionic
uds such as so . um c on e, barium chlor Vapourization.
PlhalPies of fusion beca use of stron g el ' etc. have high Substance B.P., Avap
0
Hw
M.P., Atus He
~ction between their molecules . On the :tro~ tatic forces of T, (K) (KJ mo,1) Tb (K) (kJ mol-1)
0
.uds such as oxygen, brom ine and He hav erl and, molecular 5.59
• e ow entha1p·ies of N2 63.15 0.72 77.35
,,,jon beeause the attractive forces between th err
'P . 188.0 16.15
molecules are HO 159.0 1.992
I"' W 1s ' forces of attraction. 30.0
we w eak van der aa CC4 250.16 2.5 349.69
d 23.35
2. Enthalpy (Hea t) of vapo uriza tion · It is defime as the NH3 195.40 5.65 239.73
h of a liquid is conve rted • to 177.8 5.72 329.4 29.l
lfl!balPY change.w en one mole
.. m vapours CTl3COCH3
6.01 373.15 40.79
alils boibng pomt, e.g., H2O 273.15
8 m 68.0 6.836 82.00 6.04
H2O (l)- H2O (g) ; L\vap H =40.79kJ 30.8
of water CJ!6 278.65 9.83 353.25
Same amount of heat is liberated when one mole 1081.0 28.8 1(565.00 170.0
NaO
iapours condense into liquid state .
8
H2O (g) - H2O (l); ~ condH =-40 .79kJ B' EXAMPLE 23
a process Stand ard vapou rization entha lpy of benzene at
its boilin g
3. Enthalpy (Heat) of sublimation. Sublimation is a 100 W electr ic beate r
urs below its melting point. A point is 30.8 kJ mo1-i , for how long would
mwhich a solid changes into vapo that temp eratu re?
chloride and have to opera te to vapou rize a 100g samp le at
number of solids such as iodin e , amm oniu m
(Power= energytrim e and 1 W = J s- )
1
heati ng. There fore, enthalpy
naphthalene undergo sublimation on
Solution. Molar mass of benzene,
ofsublimation is defined as :
changes CJ!6 =6 X 12+6 X 1 =78
The enthalpy (heat) change when one mole of solid 1
eratu re below its melti ng point, e.g., Power of electric heater= 100 W = 100 Js-
. ilirectJy into its vapours at a temp mol- 1
Standard vapourization enthalpy of benzene= 30.8 kJ
l2(s )- 12 (g) ; ~ sub He= 62.34 kJ Heat required to vapourize 78 g benzene= 30.8 x 1000
J
Thus, enthalpy of sublimation of 12 is 62.34 kJ.

l
Heat required to vapourize 100 g benzene
CO2 (s)- CO 2 (g) ; ~ sub He= 25.2kJ (at 195 K) 30.8 X 1000
73 = - - - - X 100 = 39487 J
C10Hg(S) -c1 0H s (g) ; ~sub He= kJ 78
Naphth alene
ation 100 J heat is given by 100 W heater in one sec
. enthalpy is a state property, th e en thalpy of sublim
Smce 39487 J heat is given by 100 Wheater in
ran be expressed as 39487 39487 .
= ~ sec = 100 x 60 nun = 6.6 min
~sub He = ~fus He + ~ vap He
ur EXAMPLE 24 I • , - ... ; • w
Vapour joule s
When lg ofliquid naphthalene (C10H8) solidifies, 149
hthalene.
of heat is evolved. Calculate the enthalpy offusion ofnap
>, e Solution. Molecular mass of naphthalene
.Q.. . dvapH
.c
(0 = 10 x 12+8 x 1 = 128a.m.u .
'E
w Liquid .·. One mole of naphthalene = 128 g
Amount of heat evolved when lg of CIOH s solidifies
= 149 joules
r dfusH
e :. Amount of heat evolved when I mole (or 128g) of
solidifies= 149 x 128= 19072 joule s =l9.0 72k.f .
C,oHs

on of ·,
- Since 19.072 kJ of heat is evolved in the solidi ficati
amount of heat will be
J mole of liquid naphthalene, the same
Solid 0 naphthalen e.
He + ~ vaPH absorbed in the fusion of I mole of solid
F"
19, 6.10 A5 ubH = 0 ~ fu 5. metal ArusH 8 = 19072 J or 19.072 k.J
. of sodiu m
F
¾d or example , at 298 K, entha lpy of fusion 2 6 kl mot- I and
enth . . diurn are .
alpy of vapourization of liquid so