THERMOCHEMISTRY

TEACHING NOTES
LECTURE-1

Thermochemistry is the branch of physical chemistry which deals with the transfer of heat between a
chemical system and its surrounding when a change of phase or chemical reaction takes place within the
system. Depending upon the conditions under which the reaction is carried out, the quantity of heat
transferred is related to energy or enthalpy change due to changes of states which occur in the system.

In this chapter we will introduced enthalpies of some specific reaction. Like, Enthalpy of formation
(Hf), Enthalpy of combustion (Hcomb), Bond dissociation of enthalpy (HBDE) & Enthalpy of
Neutralisation (Hneutralization)

Enthalpy change, rH of a reaction-Reaction enthalpy

The enthalpy change accompanying a reaction is called the reaction enthalpy. It may also be defined at
the amount of heat lost or gained in the chemical reaction when all the reactants and products are
maintained at the same temperature and pressure.
The enthalpy change of a chemical reaction may be given as
(H)reaction = qp

rH = (sum of enthalpies of products)-(sum of enthalpy of reactants)

= (· H)products – S( · H)reactants
where  is the stoichiometric coefficients of reactants and products.
Types of Reactions :
(i) Exothermic Reactions : Heat is evolved during the reaction. For such reaction H is negative, which
implies that
  P H (products) <   R H (reactants)
(ii) Endothermic Reactions : Heat is absorved during the reaction. For such reaction H is positive,
which implies that
  P H (products) >   R H (reactants)
Unit of heat of reaction is expressed as kJ or kJ/mol (per mol extant of reaction as per stoichiometry)

FACTORS AFFECTING H OF THE REACTIONS ARE :

(1) Reaction condition
(a) Most of the chemical reaction are carried out at constant temperature with either pressure or volume
constant.
At constant pressure q p  H reaction
At constant volume q v  U reaction
or H = U + (PV)
If reaction involve only ideal gases an occur set constant temperature
H = U + ng· RT

In most of the reaction sign of H & U will be same sign as PV is small compare to H & U

Thermochemistry [1]
(b) In case adiabatic rigid container used for a reaction where U = –ve (exothermic reaction)
A + B  C E = – 30 kJ/mol
then q = 0, w = 0 thus U (overall) = 0
the heat release in the reaction is used to increased the temperature of final reaction constituent.
Tfinal > Tinitial

(2) Temperature at which reaction occurs

H 2  H1
T2  T1 = CP)r (Kirchoffs Equation)

Where C P   C P ( products)   C P ( reac tan ts )

U 2  U1
T2  T1 = CV)r

(3) Pressure change in reaction

For a reaction involving only ideal gas (H)reaction will be independent of pressure since H of any/ideal
gas is function of temperature only.
However if the reaction involve non-deal substance than 'H' of substance will be pressure dependent.
 H 
In the absence of appropriate data   , the variation in enthalpy can be neglected
 P T
A+BC H P1 & H P2
(a) For ideal gas H P1  H P2
(b) For non ideal gas H P1  H P2

(4) Physical states of reactants & products

1
For products : H 2( g )  O 2(g )  H 2 O ( I) ; H  68.3kcal
2
1
H 2(g )  O 2(g )  H 2 O (g ) ; H  57.0kcal
2
(5) Allotropic forms of elements
For reactants : C Diamond  O 2 ( g )  CO 2 ( g ) H  94.3kcal
C Amorphous  O 2 ( g )  CO 2 ( g ) ; H  97.6kcal

(6) Depends on type of solvent and concentration of solution

NaOH (aq) + HCl (aq)  NaCl(aq) + H2O(l)
4M 2M H 1
.004M .002M H 2

H1  H2

In order to take care of all the factors involved in heat of reaction.

We express heat of reaction as standard enthalpy reaction.

Thermochemistry [2]
 Standard enthalpy of reaction, H° :
As enthalpy of a reaction depends on the conditions under which a reaction is carried out, it is necessary
to specify some standard conditions. The standard enthalpy of reaction is the enthalpy change for a
reaction when all the participating substance (reactants and products) are in their standard condition.

The standard condition are :

* Solid / liquid / gas should be at 1 bar
* for substance dissolved in solution concentration should be 1M.

Hess's law of constant heat summation:

According to Hess’s law (a consequence of first law), if a set of reactants is converted into a set of
product by more than one sequence of reactions, the total enthalpy change will be the same for every
sequence.
As such, the chemical equations can be treated ordinary algebraic expressions and can be added or
subtracted to yield the required equation. The corresponding enthalpy changes are also manipulated in
the same way to obtain the enthalpy change for the desired equation.

Some standard enthalpy changes of specific reactions

(i) Enthalpy of formation, fH :

It is the enthalpy change when one mole of a substance is formed from its elements in their most abundant
naturally occurring form (also called reference states). For example, the reference state of oxygen,
carbon and sulphur are O2 gas, Cgraphite and Srhombic , respectively some reactions with standard molar
enthalpies of formation are :
1
H2(g) + O (g)  H2O(l) ; fH° = – 285.8 kJ mol–1
2 2
Note :
1. By convention, enthalpy of formation fH, of an element in reference state is taken as zero.
2. The enthalpy of formation can be used to determine the enthalpy change of any reaction as

rH =  a i  f H(Products) –  bi  f H(reactants)

i i

where ai and bi represent the coefficients of the products and reactants in the balanced chemical equation.
3. Hf data can be used to compare stability of isomer and allotropes
4. The reference state of commonly used elements are :

Elements Reference state

C C(graphite)
S S8(Rhombic) (Rhombic sulphur is energy
wise more stable as compared to monoclinic sulphur)

P P4(white)
O O2(g)
H H2(g)
Br Br2(l)
Metal M(s)[except Hg(l)]
P P4 (white)

Thermochemistry [3]
Examples for enthalpy of formation
1
(i) H2(g) + S (r) + 2O2 (g)  H2SO4 (l), heat of reaction =  f H 0H 2SO 4 ,l 
8 8
(ii) Cgraphite + O2(g)  CO2 (g),  f H 0CO 2 g 
3 1
(iii) H 2 g   P4 s   2O 2 g   H3PO4 (l),  f H 0H PO ,l 
2 4 3 4

1
(iv) H (g)  H+ (g) + e¯  f H 0H  g   0
2 2
1
(v) H (g)  H+ (aq) + e¯ , H 0f of H  aq   0 , S0f   G 0f  0
2 2
(vi) I2(S)  I2(g) H > 0
(vii) Br2(l)  Br2(l) Hf = 0

Application of Hf
(i) Calculation of H of any general reaction.
Let us considered a general reaction

aA + bB  cC + dD
Hreaction = H f products   H f reac tan t 
= cH f C   dH f D   – aH  f A  bH f B  

Illus1 How much heat will be required at constant pressure to form 1.28 kg of CaC2 from CaO(s) & C(s) ?
Given : fH° (CaO, s) = – 152 kcal / mol.
fH° (CaC2, s) = – 14 kcal / mol
fH° (CO, g) = – 26 kcal / mol
(A) + 112 kcal (B) 224 kcal (C) 3840 kcal (D*) 2240 kcal
[Sol. CaO(s) + 3C(s)  CaC2(s) + CO(g)
fH° = (– 14 – 26) – (– 152) = + 112 kcal / mol
 1280 
Total heat required =  × 112  2240 kcal ]
 64 
Illus2 The f H° (N2O5, g) in kJ/mol on the basic of the following data is :
2NO(g) + O2 (g)  2NO2 (g) rH° = –114 kJ/mol
4NO2 (g) + O2 (g)  2N2O5 (g) rH° = –102.6 kJ/mol
f H° (NO,g) = 90.2 kJ/mol
(A*) 15.1 (B) 30.2 (C) – 36.2 (D) none of these
1 1
[Sol. N2(g) + O2(g)  NO (g) fH° = 90.2
2 2
N2(g) + O2(g)  2NO(g) rH° = 90.2 × 2 ... (1)
2NO(g) + O2(g)  2NO2(g) rH° = – 114 ... (2)
1  102.6
2NO2(g) + O (g)  N2O5(g) rH° = = – 51.3 ... (3)
2 2 2
(1) + (2) + (3)

Thermochemistry [4]
 5
N2(g) + O (g)  N2O5(g) fH° (N2O5, g) = 15.1 kJ/mol ]
2 2

Illus3. When 1 mole of a strong monoprotic acid (HA)reacts with one mole of mole of nonacidic base (BOH),
55kJ of heat is liberated at constant pressure.
Calculate Hf of OH¯ ion if Hf of H2O(l) is – 290 kJ.
HA (aq) + BOH (aq)  H2O + AB (aq)
Strong acid and strong bases dissociate in their aq. solution without requirement of any extra energy.

Illus4. Calculate amount of iron which can obtained by giving 20 kJ of heat to Fe3O4. Given |Hf| Fe3O4 =
200 kJ if
(i) Heat is given at constant pressure
(ii) Heat is given at constant volume

Illus5. When 2 moles of ethane undergo complete combustion at 300K in a closed rigid vessel, 1400 kJ of heat
is liberated if HfCO2 = – 390 kJ
HfH2O(l) = – 285 kJ
HfH2O(g) = – 245 kJ

Calculate Hf of ethane ?

(ii) Comparision of stability of isomers or allotropes

Hf data are often referred as heat content of the system, therefore among isomer & allotropes, the
more stable species will have lower. Value of heat content of the system (Hf lower value with sign)
Example

< CH2 = CH – CH3

Propene
Cyclopropane Hf – 60 kJ/mole
Hf – 40 kJ/mole

This concept should not be used for comparing stability of two different molecules or substance

eg. Hf CO2 (g) = – 390 kJ

Hf H2O (l) = – 285 kJ
Stability cannot be compared

Thermochemistry [5]
 LECTURE-2

* Enthalpy of combustion, CH :

It is the enthalpy change (always negative) when one mole of the substance undergo complete combustion.
Elements Combustion products
C CO2 (g)
N (From hydrocarbon) N2
H H2O(l)/H2O(g)
S SO2(g)
P P 4O10
Metal Metal oxide

Example
Cgraphite + O2 (g)  CO2(g) ; Hcomb. (graphite) = Hf CO2(g)
Cdiamond + O2 (g)  CO2(g) ; Hcomb. (diamond), the heat exchange is not
equal to enthalpy formation of CO2(g) as
reactant is not present in standard state
1
Cgraphite + O (g)  CO(g) ; Hf CO(g), the heat exchange is not equal to
2 2
enthalpy of combustion of Cgrahite.
1
Cdiamond + O (g)  CO(g) ; The heat exchange in the reaction is neither
2 2
equal to enthalpy of combustion nor equal to
enthalpy of formation.
1
H2(g) + O (g)  H2O (l); Hcomb H2 (g) = Hf H2O (l)
2 2
S8 + 8O2 (g)  8SO2 8 × Hcomb S
5 1
P+ O2  P4 O10 Hcomb P
4 4
CH4(g) + 2O2(g)  CO2(g) + 2H2O(l) ; H = – 890.8 kJ mol–1 at 298 K.

All combustion reaction are always exothermic.

Application of H combustion data

(i) Calculation of Carlorific value : Calorific value of a fuel is definition as amount of energy released by
combustion of 1 gm or 1 lit of the fuel.
H comb .
Calorific value = kJ/gm
mol. mass
In case of hydrocarbons, alkanes which have maximum percentage of Hydrogen have maximum calorific
value.
Calorific value of C2H6 > C2H4 > C2H2
(ii) Calculation enthalpy of reaction using enthalpy of combustion reaction constituents.

For any general reaction

aA + bB  cC + dD
for the reaction Hcomb for A, B, C & D is given
Hreaction =  Hcomb. (reactant) –  Hcomb. (product)

Thermochemistry [6]
 In case all the substance are not showing combustion. If substances other than H 2O are present in
the reaction. Which doe not undergo combustion then their Hcomb. can be taken as zero.
A + CO2  C + D
Hcomb = 0
In case of H2O is present.
Always check the state of H2O as per combustion data & reaction data.

In case they are same Hcomb = zero

In case the physical state are diff. then use phase transformation.

Illust. Calculate Hreaction

C3H8(g) + H2 (g)  C2H6(g) + CH4(g)
Given
Hcomb. H2(g) = – 285 kJ
Hcomb. CH4(g) = – 890 kJ
Hcomb. C2H6(g) = – 1560 kJ

Hcomb. C(graphte) = – 393 kJ

DHformation C3H8 = – 104 kJ

(iii) To find final temperature which the system attains if the changes in the system are carried out
under adiabatic conditions.

Illust. For a reaction A(g) + B(g)  2C(g) ; H = – 40 kJ

If the reaction is carried out in an adiabatic rigid container initially at 300 K by taking 2 moles of A & 3
moles of B, then calculate final temp. of the reaction. Make appropriate assumption A, B & C are all
monoatomic.
CV A = 8 cal/mol.
CV B = 9 cal/mol.
CV C = 7 cal/mol.

(iv) Comparison of stability of isomers or allotropes

For any combustion reaction Hbomb < 0

Cdiamond + O2 (g)
Enthalpy

Cgraphite + O2 (g)

CO2 (g)

State of reaction

Unstable isomer / allotrope will have more enthalpy of combustion.

Thermochemistry [7]
(v) Measurement of U and H : calorimetry
The process of measurement of energy changes in the physical or chemical processes is carried out in a
vessel called Calorimeter, which is immersed in a known volume of a liquid, knowing the heat capacity of
the Calorimeter and the liquid in which Calorimeter is immersed, it is possible to determine the heat
absorbed or evolved in the process by measuring temperature changes. Measurements are made under
two different conditions :
(i) At constant volume, qv or U or E
(ii) At constant pressure, qp or H

(i) U measurement : For chemical reactions, heat absorbed at constant volume, is measured in a bomb
Calorimeter. In this Calorimeter, a steel vessel (the bomb) is immersed in a water bath. A combustible
substance is burnt in pure oxygen supplied in the bomb. Heat evolved during the reaction is transferred
to the water around the bomb and its temperature is monitored. Since the bomb Calorimeter is sealed,
its volume does not change, i.e., the energy changes associated with reactions are measured at constant
volume.

(ii) H measurement : Measurement of heat change at constant pressure (generally at atmospheric pressure)
can be done in a Calorimeter shown in the figure. In this case, the Calorimeter is left open to atmosphere.
As the reaction occurs in the Calorimeter, the temperature change is noticed and then heat of reaction is
measured with the knowledge of heat capacity of Calorimeter system.

Thermochemistry [8]
Illust. Calculate Hf (CH4) using following information
Hcomb(CH4) = – 890 kJ/mol
Hf(CO2) = – 400 kJ/mol
Hf(H2O(l)) = – 240 kJ/mol

Enthalpy changes during phase transformations :

(i) Enthalpy of Fusion, fus.H : It is the enthalpy change that accompanies melting of one mole of a
solid substance at constant temperature (melting point of solid) and pressure.
for example,
H2O(s)  H2O(l) ; fusH = + 6.01 kJ mol–1

(ii) Enthalpy of Vaporisation, vapH :

It is the amount of heat required to vapourise one mole of a liquid substance at constant temperature
(boiling point of liquid) and pressure for example :
H2O(l)  H2O(g) ; vapH = + 40.79 kJ mol–1

(iii) Enthalpy of Sublimation, subH :

It is the amount of heat required to sublime one mole of a solid substance at constant temperature
(sublimation temperature of solid) and pressure. For example
CO2(s)  CO2(g) ; subH = +25.2 kJ mol–1

Illust. For a normal substance Hfusion at 200 K normal melting pt. is 2.7 kcal if normal boiling point is 300
k & Cp,m liquid = 20 cal/mol k, Cp,m gas = 10 cal/mol k
Calculate Hsublimation
(Trouton's rule) : The molar entropy of vaporization of most liquids which do not involve hydrogen
bonding and also do not possess boiling point less than 150 K is about 10.5 R.

Thermochemistry [9]
 LECTURE -3

Bond Enthalpies (Bond energies), bondH :

The bond enthalpy of diatomic molecules like H2, Cl2, O2 etc. may be defined as the enthalpy change
(always positive) when one mole of covalent bonds of a gaseous covalent substance undergo homolytic
fission to form products in the gas phase, under conditions of constant pressure and temperature. For
example
Cl2(g)  2Cl(g) ; Cl–ClH = +242 kJ mol–1
O2(g)  2O(g) ; O=OH = +428 kJ mol–1

Hatomisation is enthalpy change when 1 mole of a substance undergo complete dissociation into gaseous
atoms.
Example : H2(g)  2H (g); Hreaction = HBDE H–H = 2 Hf H(g)
= Hatomisation of H2 (g)

I2 (s)  2I(g) Hreaction  HBDE = 2 Hf H(g)

= Hatomisation of I2 (s) (g)

In case of polyatomic molecules, bond dissociation enthalpy is different for different bonds within the
same molecule. In such cases, mean bond enthalpy is used. Mean bond enthalpy may be defined as the
average enthalpy change to dissociate a particular type of bond in the compounds.
CH4 (g)  CH3(g) + H(g) H 1
CH3 (g)  CH2(g) + H(g) H 2
CH2 (g)  CH1(g) + H(g) H 3
CH (g)  C (g) + H(g) H 4

CH4 (g)  C(g) + 4H(g) Hatomisation

Hatomisation = H1 + H2 + H3 + H4 = 4 × Average bond dissociation enthalpy of C–H

H atomisatio n
HBDE =
4
Application of bond dissociation enthalpy

Calculate of H reaction
(i) In gas phase reactions, the standard enthalpy of reaction, rH°, is related with the bond enthalpies of
reactants and products as

rH° =  bond enthalpies (reactants) –  bond enthalpies (products)

Illus For the reaction

N2H4 (g)  N2H2 (g) + H2(g) rH° = 109 kJ / mol
Calculate the bond enthalpy of N = N.
Given : B.E. (N–N) = 163 kJ/mol, B.E. (N–H) = 391 kJ/mol, B.E. (H–H) = 436 kJ/mol
[Sol. The equation :
H = 109 = N  N 4 N H  H  H 2 N H  N  N
N N = 163 + 2 × 391 – 436 – 109 = 400 kJ/mole ]

Thermochemistry [10]
(ii) Let us consider a general reaction
aA + bB  cC + dD
H = Erequired – Ereleased

Erequired involve bond breaking , phase transformation (Non gaseous  Gaseous substance), resonance
energy of the molecule of the reactant side.
Ereleased involve bond formation, phase transformation (gaseous  non gaseous substance), resonance
energy of the molecule on product side.
Illus Calculate heat released in combustion of 1 mole of CH3OH (l) in a rigid vessel at 300 K if
Bond energy (C – H) = 410 kJ/mole
(C – O) = 350 kJ/mole
(O – H) = 465 kJ/mole
(O = O) = 495 kJ/mole
(C = O) = 710 kJ/mole
Hvap.CH3OH(l) = 35.5 kJ/mole
Hvap. H2O (l) = 40.6 kJ/mole

Illus What is the Bond energy (In kJ/mole) of C–H in Methane from the following data?
Hf [CO2 (g)] = – 394 kJ / mole
Hf [H2O (l)] = – 285 kJ / mole
Hsublimation {Carbon (graphite)} = + 716 kJ/mole
Hcombustion [CH4(g)] = – 890 kJ / mole
Bond energy (H–H) = 435 kJ/mole [Ans. 415]

(iii) In case bond energy is to be calculated

Illus Hf PCl5(g) = – 375 kJ/mole
Hf P(g) = 315 kJ/mole
Hf Cl(g) = 122 kJ/mole
Calculate average bond dissociation of P–Cl bond in PCl5.

Resonance energy
It is the enthalpy change when 1 mole of a resonance hybrid is converted with its most stable resonating
structure.

Resonance Most stable HRE > 0

hybrid Resonating Structure

It case reaction involve molecules which show resonance, then

(a) If molecules showing resonance is on reacting side then |HRE| is energy required.
(b) If it is on product side, then |HRE| is energy released.

Illus Calculate resonance energy of C6H6 (g).

Given : Hf [C6H6(g)] = – 360 kJ mol–1
HSub [C(graphite)] = 716 kJ mol–1
B.E.H–H = 437 kJ mol–1
B.E.C=C = 620 kJ mol–1
B.E.C–C = 340 kJ mol–1
B.E.C–H = 490 kJ mol–1

Thermochemistry [11]
[Sol. For C6H6
6 C(s) + 3H2 (g)  C6H6 ; Hexp = – 360 kJ mol–1
Hcal = – [3 (C–C) + 3(C = C) + 6 (C–H)] + [6CS g + 3 (H–H)]
= – [3 × 340 + 3×620 + 6 × 490] + [6 × 716 + 3 × 437]
= – 5820 + 5607 = – 213 kJ mol–1
 Resonance energy = Exp. Hf – Calculated Hf
= – 360 – (–213) = –360 + 213 = – 147 kJ mol–1 ]

Illus The standard molar enthalpies of formation of cyclohexane (l) and benzene (l) at 25°C are – 156 and
+ 49 kJ mol–1 respectively. The standard enthalpy of hydrogenation of cyclohexene (l) at 25° C is
– 119 kJ mol–1. Use these data to estimate the magnitude of the resonance energy of benzene.
[Sol. Given
C6H10 (l) + H2(g)  C6H12 (l) ; H = – 119 kJ
and C6H6 (l) + 3H2 (g) C6H12 (l) ; H = 3 × (–119) = – 357 kJ
Also given
6 C(s) + 6 H2 (g)  C6H12 (l) ; H = – 156 kJ
C6H6 (l) + 3H2(g)  C6H12 (l) ; H = – 357 kJ
-------------------------------------------------------------------
Subtracting 6 C(s) + 3H2 (g)  C6H6 (l) ; H = + 201 kJ
 Resonance energy = 49 – 201 = – 152 kJ
 Magnitude of resonance energy = 152 kJ Ans.]

Ionisation Enthalpy (HIE)

It is the enthalpy change (always positive) when an electron is remove from an isolated gasesous atom in
its ground state under conditions of constant temperature and pressure.
X(g)  X+ (g) + e¯ ; HIE
Ionisation energy is also energy required for the above process calculated at 0 K.
Thus HIE , 0K = IE
X(g)  X+ (g) + e¯ ; HIE

C pm 5R 5R 5R
2 2 2

5R
Since all the species involved are monoatomic (3 degrees of freedom), so (nCpm)reaction =
2
HIE, TK – HIE, 0K = (nCpm)reaction (T – 0)
5R
HIE, TK – HIE, 0K = (T – 0)
2
5R
HIE, = IE +
2

Electron Gain Enthalpy (HEG)

It is the enthalpy change when an electron is added to a neutral gaseous atom to convert it into a negative
ion under conditions of constant temperature and pressure.
X(g) + e¯  X¯(g); HEG < 0 (In most of the cases)

Thermochemistry [12]
 Electron affinity (EA) is also defined for same process, however it is energy released the at 0 K
HEG, 0K = – EA
5RT
H EG ,TK  H EG , 0 K  
2
5RT
H EG ,TK   EA   
2
5RT
H EG ,TK   EA 
2
Enthalpy of Hydration, hydH :
For ion : It is the enthalpy change when 1 mole of a isolated gaseous ion is mixed with large amount of
water, to convert into completely hydrated aqueous state.
A+(g) + H2O  A+ (aq) (completely solvated state)
B¯(g) + H2O  B¯ (aq) HHE (< 0)

For substances - It is the enthalpy change (always negative) when one mole of an anhydrous (or partly
hydrated) compound combines with the required number of moles of water to form a specific hydrate at
the specified temperature and pressure. For example :
CuSO4(s) + 5H2O(l)  CuSO4 · 5H2O(s) ; (hyd)H = – 78.20 kJ mol–1

Lattice Enthalpy (latticeH)

The lattice enthalpy of an ionic compound is the enthalpy change which occurs when one mole of an ionic
compound dissociates into its ions in gaseous state under conditions of constant temperature and pressure.
Na+Cl¯(s)  Na+(g) + Cl¯(g) ;
lattice H = + 788 kJ mol–1
Enthalpy of Atomisation, aH :
It is the enthalpy change (always positive) when one mole of a substance is completely dissociated into
atoms in the gaseous state, under constant pressure and temperature condition.
For example
H2(g)  2H(g) ; aH = 435.0 kJ mol–1
CH4(g)  C(g) + 4H(g) ; aH = 1665 kJ mol–1

Enthalpy of Transition
It is the enthalpy change when one mole of one allotropic form changes to another under conditions of
constant temperature and pressure. For example
C(graphite)  C(diamond) trsH = 1.90 kJ mol–1

Illus Cesium chloride is formed according to the following equation

Cs(s) + 0.5Cl2(g)  CsCl(s) .
The enthalpy of sublimation of Cs, enthalpy of dissociation of chlorine, ionization enthalpy of Cs &
electron gain enthalpy of chlorine are 81.2, 243.0, 375.7 and –348.3 kJ mol 1. The enthalpy change
involved in the formation of CsCl is  388.6 kJ mol1 . Calculate the lattice enthalpy of CsCl.
Ans. 618.7 kJ mol1

Thermochemistry [13]
 LECTURE -4

Enthalpy of Solution, (Hsol):

When a solute is dissolved in solvent there is an evolution or absorption of heat. The enthalpy change
per mole of solute dissolved is not constant; it usually varies with concentration of the of the solution. IF
HCl gas is dissolved in water then depending upon amount of solvent added, different enthalpy change
is observed.
HCl (g) + H2O  HCl(aq)
1 mole 10 mole H1 = – 69 kJ/mole
1 mole 25 mole H2 = – 72 kJ/mole
1 mole 40 mole H3 = – 73 kJ/mole
1 mole 200 mole H4 = – 74 kJ/mole
1 mole Excess H5 = – 75 kJ/mole

Enthalpy of solution or integral enthalpy of solution

It is defined as the enthalpy change when one mole of a substance is dissolved in a specified amount of
solvent, to form a solution of specific concentration under conditions of constant temperature and pressure
therefore, Hsolution HCl (g) (1 mole in 20 mole of H2O) = – 72 kJ/mole.

Ethalpy of infinite diluted solution (infinite diluted solution )

Enthalpy change when 1 mole of a solute is mixed with large amount of solvent (so that further addition
of solvent does not cause any further enthalpy change) to form infinite diluted solution.
infinite diluted solution HCl = – 75 kJ/mole
In general Hsolution is used for enthalpy of infinite diluted solution.

Enthalpy of dilution (Hdilution)

Enthalpy change when solution of specified concentration undergo dilution by addition appropriated
amount of solvent to form another solution of specific concentration.
HCl [1 mole in 10 mole of water]  HCl (1 mole in 25 mole of water)
Hdilution = – 3 kJ/mole

If Hdilution = zero, then solution should be in infinite diluted state.

Illus 100 gm of anhydrous CuSO4, when dissolved in excess of water produces 42 kJ of heat. The same
amount of CuSO4. 5H2O on dissolving in large excess of water absorbed 4.60 kJ. What is the heat of
hydration CuSO4?
100
[Sol. 100 gm of CuSO4 = mole
159.5

42  159.5
Heat of solution of CuSO4 per mole = – = – 66.99 kJ  – 67 kJ
100

100
100 gm of CuSO4. 5H2O = mole
249.5

Thermochemistry [14]
 249.5
Heat of solution per mole (CuSO4. 5H2O) = 4.6 × = 11.477 kJ
100
Now, CuSO4(s) + aq.  CuSO4 (aq) ; rH = – 67 kJ mol–1
CuSO4(aq)  CuSO4 . 5H2O + aq ; rH = – 11.477 kJ mol–1
---------------------------------------------------------------------------
CuSO4(s) + 5 H2O  CuSO4 .5H2O ; rH = – 78.477 kJ mol–1
 Enthalpy of hydration of CuSO4(s) ; H = – 78.477 kJ mol–1 Ans. ]

Variation of enthalpy with amount of solute, solvent can be plotted as

Enthapy change
H 1 H1
H sol. 
ns

ns

Moles of solute/1000 gm of solvent

Variation of Hsolution with amount of solvent H (Fixed amount of solute)

Enthapy change

Enthlpy of infinite dilution

Weight of solvent

Enthalpy of Neutralisation neutH :

It is the enthalpy change (always negative) when one g-equivalent of an acid / base is neutralised with
one g-equivalent of a base / acid in solution phase to form salt & water.
Alternatively - It is enthalpy change when one mole of H+ (from acid) react with one mole of OH¯ (from
base) to form water.
It is observed that Hneutralisation is dependent on two factors
(i) Type of acids / base (Strong / weak)
(ii) Concentration (infinite dilution or not)

(i) Type of acid & base : -

(a) Strong acid / Strong base (infinite diluted)

Thermochemistry [15]
 HA(aq) + BOH(aq)  BA(aq) + H2O
SA SA
For acid / base Hionisation is defined as enthalpy change involved in complete dissociation of 1 mole
acid / base into constituent ion in the infinite diluted state.

For strong acid & strong base Hionisation = 0

HA(aq) + BOH(aq)  BA(aq) + H2O
H+ + A¯ (aq) + B+ (aq) + OH¯ (aq)  B+ (aq) + A¯(aq) + H2O
H+ + OH¯  H2O; Hneutralisation = –13.7 kcal/eq–1
For all infinite diluted solutions of strong acid and strong base enthalpy of neutralisation is
–13.7 kcal/eq–1
HCl(aq) + NaOH(aq)  NaCl(aq) + H2O(l) ; neutH = – 57.7 kJ eq–1

(b) Weak acid / Strong base (infinite diluted)

HA(aq) + NaOH(aq)  NaA(aq) + H2O

Hion Hion = 0 Hion = 0

H+ + A¯ (aq) + Na+ (aq) + OH¯ (aq)  Na+ (aq) + A¯(aq) + H2O

As Hion of weak acid / base is positive, therefore energy released would be less in case of neutralisation
of weak acid/base
Hneutralisation (WA/SB) = Hion (WA) –13.7
Exception : Hionisation of HF is negative

Illus Heat of neutralization (H) of NH4OH and HF are – 51.5 and –68.6 kJ respectively. Calculate their
heat of dissociation?
(i) HCl (aq) + NaOH (aq)  NaCl (aq) + H2O ; H = – 57.3 kJ
(ii) HCl (aq) + NH4OH (aq)  NH4Cl (aq) + H2O ; H = – 51.5 kJ
(weak base)
[Sol.  The heat of dissociation of NH4OH,
H = – 51.5– (–57.3) = 5.8 kJ
Similarly we have
HF (aq) + NaOH (aq)  NaF (aq) + H2O ; H = – 68.6 kJ
 The heat of dissociation of HF,
H = – 68.6 – (–57.3) = – 11.3 kJ ]

Illus It is known that enthalpy of neutralisation (SA/SB) = – 13.7 kcal eq–1

Hneutralisation (WA/SB) = – 12.7 kcal eq–1
Hneutralisation (WB/SA) = – 10 kcal eq–1
Find the value of Hneutralisation (WA/WB).

Thermochemistry [16]
Effect of concentration Hneutralisation
Depending upon whether the solutions of acid/base are infinite diluted or having some specific
concentration. Hneutralisation can be calculated as

NaOH + HCl  NaCl + H2O (l)

(4M, 250 ml) (2M, 500 ml) (M, 750 ml)

Hdil. of NaOH (4M, ) Hdil. of HCl (2M, ) –H dil. of NaCl (xM, )

NaOH HCl NaCl + H2O (l)

(infinite diluted) (infinite diluted) (infinite diluted)

Hion = 0 Hion = 0 Hion = 0

Na+(aq) + OH¯ (aq) + H+ (aq) + Cl¯ (aq)  Na+ + (aq) Cl¯ (aq) + H2O (l)

Illus Calculate heat released when 250 ml of 8M CH3COOH solution is mixed with 1 litre of 2 M NaOH
Given
Hion CH3COOH = 1.2 Kcal/mol
Hdil. CH3COOH (8M, ) = – 0.5 Kcal/mol
Hdil. NaOH (2M, ) = + 1.5 kcal/mol
Hdil. CH3COONa (M, ) = 0.75 kcal/mol
Hneutralisation SA/SB infinite dilution = – 13.7 kcal/equivalent

Illus 2M H2SO4 solution is mixed with 1 molar NaOH solution. If total volume of the solution is one litre,
then calculate maximum possible heat released ?

Thermochemistry [17]