F.Y. B. Sc., Semester -I Physical Chemistry Savitribai Phule Pune University, Pune

1. Arts, Commerce & Science College, Tryambakeshwar (Nashik)
Email ID : Manohar@mvptryambakcollege.ac.in

2. Heat or Enthalpy of Neutralization
• Definition :
The heat of neutralization of an acid by base is defined as the heat change
when one gram equivalent of the acid is neutralized by one gram equivalent
of base, the reaction carried out in dilute aqueous solution.
Example:
NaOH + HCl NaCl + H2O ∆H = 57.1 kJ
1 gram equivalent of HCl is neutralized by 1 gram equivalent of NaOH , it is
produced 57.1 kJ heat.
Heat of neutralization of any strong acid ( HCl. HNO3, H2SO4 etc. ) with
strong base ( KOH, NaOH ) or vice versa, is always same i.e. 57.1 kJ

3. • If acid or base is weak than heat of neutralization is usually less than 57.1
kJ
• Example: reaction between Acetic acid (CH3COOH) and base sodium
hydroxide (NaOH) heat of neutralization is less.
• Problem 1.4 :
Determine the amount of heat released when :
a) 0.75 mole of nitric acid is neutralized by 0.75 mole of sodium hydroxide.
b) 0.75 mole of hydrochloric acid is neutralized by 0.5 mole of potassium hydroxide.
c) 900 ml of 0.5 M hydrochloric acid solution is mixed with 500 ml of 0.4 M sodium
hydroxide solution. Assuming the specific heat of water as 4.18 JK-’g-1 also calculate
the rise in temperature.
d) 200 ml of 0.2 M sulphuric acid is mixed with 300 ml of 0.1 M sodium hydroxide
Calculate the rise in temperature.

4. Solution :
a) The reaction is :
(0.75 moles) HNO3 + (0.75 moles) NaOH NaNO3 + H2O
When one mole of HNO3 is neutralized by 1 mole of NaOH,
heat released is 57.1 kJ. Therefore, when 0.75 mole of HNO3 is
neutralized by 0.75 mole of NaOH,
heat released will be 57.1 x 0.75 = 42.8 kJ.
b) Out of 0.75 mole of HCI, 0.5 mole will react with 0.5 mole of KOH and
0.25 mole of HCI will be left unreacted.
Heat released = 57.1 x 0.5 = 28.5 kJ.
c) 1000 ml of 1 M HCl = 1 mole
therefore 900 ml of 0.5 M HCI = 900 x 0.5/1000
= 0.45 mole

5. • Similarly,
500 ml of 0.4 M NaOH = 500x0.4 /1000 = 0.2 mole
Out of 0.45 mole of HCI, 0.2 mole of HCl reacts with 0.2 mole of NaOH
.: Heat evolved = 57.1 x 0.2 = 11.4 kJ
To calculate the rise in temperature,
use the relation Mass x Sp. heat x Rise in temp. = Heat evolved
Heat evolved is 11.4 kJ or 11400 J, mass of the mixture is 900 g + 500 g (assuming the densities of
the solution as unity) and sp. heat is 4.18 J.
Rise in temp. = Heat evolved/ (Mass x Sp.heat )
= (11.4 x1000) / (1400x4.18) = 1.950
d) 1 molecule of H2SO4 releases two H+ ions
200 ml. of 0.2 M H2SO4 = = ( 2 x 200 x 0.2 ) / 1000 moles of H+ = 0.08 moles of H+
300 ml of 0.1 M NaOH = 0.03 mole OH-
Out of 0.08 mole of H+, 0.03 mole H+ will react with 0.03 mole OH-
Heat evolved = 57.1 x 0.03 = 1.7 kJ
:. Rise in temperature =1.7x1000 /500 x 4.18= 0.80

6. Different kinds of Heat (or Enthalpy ) of
Reaction
• There are Five major kinds of heat of reaction
1. Heat of Solution (∆H)
a) Integrated heat of solution
b) Differential heat of solution
2. Heat of Hydration (∆H)
3. Heat of Fusion (∆H)
4. Heat of Vaporization (∆H)
5. Heat of Sublimation (∆H)

8. 1. Heat of Solution (∆H)
• Heat of solution of substance is defined as when 1 mole of substance
is dissolved in such a large volume solvent that further addition of
the solvent does not produce any more heat change
1. Integral heat of solution:
It is the heat change that occurs when a specific quantity of solute is dissolved in a
specified quantity of a specified solvent under condition of constant temperature and
pressure
2. Differential heat of solution :
Differential heat of solution of a solute at a specified concentration of a specified
temperature and pressure condition is the heat change observed per further mole of solute
dissolved in such a large quantity of solution that no significant change in concentration
occurs.

9. • The thermochemical equations for the dissolution of KCl and CuSO4 in
water may represented as,
KCl (s) + aq. KCl (aq), ∆H = +18.6 kJ
CuSO4 (s) + aq. CuSO4 (aq), ∆H =- 66.5 kJ
• Thus,
the heat of solution of KCl = + 18.6 kJ mol-1
and heat of solution of CuSO4 = – 66.5 kJ mol-1
• It is interesting to note that the salts like copper sulphate, calcium
chloride, etc., when present in the hydrated state (i.e., CuSO4.5H20,
CaCl2. 6H2O, etc) dissolve with the absorption of heat. For example,
CuSO4.5H2O + aq CuSO4 (aq), AH =+ 11.7 kJ
Thus, it can be generalized that th- process of dissolution is usually endothermic for:
i) salts which do not form hydrates like NaCI, KCI, KNO, etc.
ii) hydrated salts like CuSO4.5H20, CaCl : 6H2O, etc

10. 2. Heat of Hydration (∆H)
• The amount of heat change (i.e., the heat evolved or absorbed)
when one mole of the anhydrous salt combines with the required
number of moles of water so as to change into the hydrated salt, is
called the heat of hydration.
For example, the heat of hydration of copper sulphate is - 78.2 kJ mol-1.
This is represented as follows:
CuSO4(s) + 5H20 CuSO4.5H20 (s) ∆H =- 78.2 kJ

11. 3. Heat of Fusion (∆Hfus)
• Heat of fusion is the heat change that takes place when one
mole of solid substance changes into its liquid state at its
melting point.
• Heat of fusion (∆H fus) of ice (m. p. = 273 K) is 6.0 kJ mol-1
It may be represented as follows :
H2O (s) H2O (l) ∆H fus= + 6.0 kJ mol-1
Ice Water

12. 4. Heat of Vaporization (∆Hvap)
It is the heat change that takes place when one mole
of a liquid changes into its gaseous state at its boiling
point.
For example,
The heat of vaporization (∆Hvap) of water into its gaseous state
at the boiling point (373 K) is 40.7 kJ. It may be represented as
follows:
H2O (s) H2O (l) ∆H vap= + 40.7 kJ mol-1
Water Steam

13. 5. Heat of Sublimation (∆H)
• Heat of sublimation of a substance is the heat change that
takes place when 1 mole of a solid changes directly into
vapour phase at a given temperature below its melting point.
• Sublimation is a process in which a solid on heating changes directly into
gaseous state below its melting point.
• For example, the heat of sublimation of iodine is 62.39 kJ mol-1.
• It is represented as follows:
I2 (s) I2 (g), ∆H = + 62.39 kJ mol-1
Most solids, that sublime readily are molecular solids, e.g. iodine and
naphthalene, etc.

14. 1.4 Bond Enthalpy or Bond Energy
Definition: The bond energy of a particular bond is defined as the
average amount of energy released when one mole of bonds are
formed from isolated gaseous atoms or the amount of Energy required
when I mole of bonds are broken so as to get the separated gaseous
atoms.
• For diatomic molecules (like H2, HCl, etc.), the bond energy is equal to the
dissociation energy of the molecules but for a polyatomic molecule like CH4 the
bond dissociation energies of the four C-H bonds are different. Hence, an
average value is taken.
• Heat of reaction can be calculated from bond energy data (Table 1.1) .
• The following formula can be used directly in these calculations.
• ∆H reaction = ∑ Bond energies of reactants – ∑ Bond energies of products

15. Bond energies of some common bonds are given in Table 1.1
Bond
Bond Energy
(kJ mol-1)
Bond
Bond Energy
(kJ mol-1)
H-H 436 H-I 297
C-H 414 F-F 155
H-F 565 Cl-Cl 242
O-H 463 Br-Br 190
H-Br 364 I-I 149
N-H 389 O=O 494
C-C 347 NΞN 941
H-CI 431 C=C 619
C-O 335 CΞ C 812
C=O 707 C-Cl 326
C-N 293 CΞN 879
C=N 616

16. Problem's on Calculation of Bond
Energy
•
Problem 1.6 : Calculate the enthalpy change for the reaction
H2(g) + Br2 (g) 2HBr (g)
Given that the bond energies of H – H, Br – Br and H - Br are 435,
192 and 364 kJ mol' respectively.
Solution :
First method. By calculating the total energy absorbed and released
• Energy absorbed in the dissociation of 1 mole of H-H bonds = 435 kJ
• Energy absorbed in the dissociation of 1 mole of Br - Br bonds = 192 kJ
• Total energy absorbed = 435 + 192 = 627 kJ
• Energy released in the formation of 1 mole of H-Br bonds = 364 kJ
• Energy released in the formation of 2 moles of H - Br bonds= 2 x 364 = 728kJ
• Energy released is greater than energy absorbed

17. Hence, net result is the release of energy.
Energy released = 728 kJ - 627 kJ = 101 kJ Thus, for the given
reaction,
∆H reaction = - 101 kJ
• Second method.
• Using the formula directly
• ∆H reaction = ∑ Bond energies of reactants - ∑ Bond energies of products
= [Bond energy (H-H) + Bond energy (Br - Br)] - [2 x Bond energy (H - Br)]
= 435 + 192 - (2 x 364)
= 627-728
∆H reaction = - 101 kJ.