Need of thermodynamics and the Laws of Thermodynamics. Important principles and definitions of thermochemistry. Concept of standard state and standard enthalpies of formations, Integral and differential enthalpies of solution and dilution. Calculation of bond energy, bond dissociation energy and resonance energy from thermochemical data. Variation of enthalpy of a reaction with temperature – Kirchhoff’s equation. Statement of Third Law of thermodynamics and calculation of absolute entropies of substances. JFI -just for information

1. F. Y. B. Sc.
SEMESTER- II
CORE COURSE: DSC-2B
(Physical Chemistry & Organic Chemistry)
CHEMICAL ENERGETICS
Dr. Mithil S. Fal Desai
Assistant Professor in Chemistry
SHREE MALLIKARJUN
&
Shri Chetan Manju Desai College
Canacona Goa
1

2. Chemical Energetics
• Need of thermodynamics and the Laws of Thermodynamics.
• Important principles and definitions of thermochemistry.
• Concept of standard state and standard enthalpies of
formations,
• Integral and differential enthalpies of solution and dilution.
• Calculation of bond energy, bond dissociation energy and
resonance energy from thermochemical data.
• Variation of enthalpy of a reaction with temperature –
Kirchhoff’s equation.
• Statement of Third Law of thermodynamics and calculation
of absolute entropies of substances.
2

3. What is thermodynamics?
The study that deals with the relations
between heat and other forms of energy
as it undergoes physical or chemical
transformation.
3

4. Need of Thermodynamics
This study is of everyday importance as it improves our understanding
about chemical and physical process in which there is change in the
energy in the form of heat into other forms.
Scope: 1) Most physical chemistry laws are derived from
thermodynamics. eg. Van’t Hoff law, phase rule and distribution law.
2) Can predict whether a chemical change is possible in given set of
conditions. (T, P and C, mainly)
3) Can predict extend of chemical change. (chemical equilibrium)
Limitations:
1) Applicable to macroscopic system. (energy of bulk is considered)
2) It concern with only initial and final state. (no rate at which energy
change is considered)
4

5. Laws of Thermodynamics.
• First Law of Thermodynamics
• Second Law of Thermodynamics
• Third Law of Thermodynamics
5

6. Important terms in thermodynamics.
• System + Surrounding
=Universe
• Open/close/isolated system
• Internal energy (E)
• Heat (q)
• Heat capacity (c)
• Work (w)
• Pressure (p)
• kJ/mol
• Enthalpy(H=E+PV)
• Entropy(S)
• Exothermic/Endothermic
• Chemical potential (μ)
• Helmholtz free energy(A)
• Gibbs free energy (G)
• Adiabatic process(dq=0)
• Isothermal process (dT=0)
• Isobaric process(dP=))
• Isochoric process (dV=0)
• Standard state
• Reference state
• STP
6

7. Heat capacity (c)
• Capacity to absorb heat!
In a system if calories of heat (q) absorbed by a
mass (m) and temperature changes from T1 to
T2 the heat capacity is given by
𝑐 =
𝑞
𝑚 (𝑇2 − 𝑇1)
Heat capacity is defined as heat absorbed by
unit mass raising the temperature by 1 ◦C or 1
K.
7

8. Joule-Thomson effect (JFI)
• The phenomenon of lowering of temperature
when a gas is made to expand adiabatically
from a region of high pressure to region of low
pressure.
8
One mole
N2 at 10
atm in V1
at T1
Half mole
N2 at 1 at
in V2 at T2
T2 < T1 accordingly

9. First Law of Thermodynamics
• Total energy of an isolated system remains constant.(may change
from one form to other)
Or
• Total energy in the universe is constant.
Consider a system is changed from A to B consider the energy change
from Ea to Eb
ΔE = Eb - Ea
This energy change can occur by evolution or absorption of heat (q)
and with and without work (w) being done on the system.
ΔE = q – w (as, w = P X ΔV)
ΔE = q - PΔV
9

10. Some easy considerations and
calculation
• Cyclic isothermal process (ΔE=0) and w=q
• Isochoric process (ΔV=0) and ΔE=qv
• Adiabatic process (q=0) and ΔE= -w
Simple problem: A gas cylinder fitted with a piston expands against a
constant pressure of 1 atm from volume of 5 L to 10 L. This system
absorbs 400 J thermal energy from its surrounding determine change
in internal energy of system.
w= -P(V2-V1)=1 atm (10L – 5L) = 5 L atm (as, 1 L atm =101.2 J)
w= -506 J
ΔE= q - w
ΔE= 400 J - (-506 J)
ΔE= 906 J
10

11. Second Law of Thermodynamics
• A spontaneous change is accompanied by an increase in
the total the total entropy of the system ad its surrounding
ΔSsystem + ΔSsurrounding > 0
𝑤
𝑞2
=
𝑇2 − 𝑇1
𝑇2
Some other well known explanations
Lord Kelvin ‘ It is impossible to take heat from a hotter
reservoir and convert it completely into work by a cyclic
process without transferring a part of heat to cooler reservoir.’
Clausius ‘It is impossible for a cyclic process to transfer heat
from a body at a lower temperature to one at higher
temperature without at the same time converting some work
to heat.’
11

12. Calculation for understanding
An engine operating between 423 K and 298 K takes
500 J heat from a high temperature reservoir.
Assuming that there is negligible friction loss
calculate the work that can be done by this engine,
Efficiency =
𝑤
𝑞2
=
𝑇2 − 𝑇1
𝑇2
=
𝑤
500
=
423 − 298
423
𝒘 = 147.75 J
12

13. Third Law of Thermodynamics
At absolute zero, the entropy of a pure crystal is
zero (at T= 0 K, S=0)
Clausius introduced numerical definition of entropy
‘entropy of a system not undergoing any changes is
a constant quantity.’
Thus, for a change in entropy (ΔS) is equal to heat
(q) energy absorbed or evolved divided by the
temperature (T).
ΔS =
𝑞
𝑇
or =
Δ𝐻
𝑇
13

14. Calculation for understanding
Calculate the entropy change when one mole of
ethanol is evaporated at 351 K. The molar heat
of vaporisation of ethanol is 39840 J mol-1.
ΔS =
Δ𝐻
𝑇
=
39840
351
= 113.5 J K-1 mol-1
14

15. Concept of Standard State and reference
state
• The standard state of a substance at specified
temperature and 1 bar pressure is its pure
form.
• The reference state is most stable state of a
substance at specified temperature and 1 bar
pressure.
15

16. Standard Enthalpy of formation
Standard Enthalpy of formation is change in enthalpy that
takes place when one mole of compound is formed from all
substances being there in standard states(298 K and 1 atm).
H2(g) + ½ O2(g) H2O(g) (ΔH◦f= -57.84)
Calculate ΔH◦ for the reaction
CO2(g) + H2(g)  CO(g) + H2O(g)
Given that ΔH◦f for CO2(g), CO(g) and H2O(g) are -393.5, -111.31
and 241.80 kJ K-1 mol-1, respectively.
ΔH◦ = ΔH◦f (product) -ΔH◦f (reactants)
ΔH◦ =[(ΔH◦f CO(g)) + (ΔH◦f H2O(g))] - [(ΔH◦f (CO2(g)) + (ΔH◦f (H2(g))]
ΔH◦ =[(-111.3)+ (-241.8)] - [(-393.5) + (0)]
ΔH◦ = 40.4 kJ K-1 mol-1
16

17. Integral enthalpies of solution and
dilution
Solution
A + (A+B) = AB
(Results in concentrated solution)
Dilution
B + (A+B) = AB
(Results in concentrated solution
dilute solution)
A- Solute
B- Solvent
(A+B)- Solution 17
A
A + B B
A + B

18. Integral enthalpy of solution (Δ𝐻 𝑠𝑜𝑙)
• Heat absorbed or released when a particular amount solute
is dissolved in a definite amount of solvent
OR
• Enthalpy change accompanying with pure solvent of
particular volume with addition of definite amount of
solute.
18
HCl (g)+ 10 H2O(l)  [HCl+10 H2O](aq) ;Δ𝐻 𝑠𝑜𝑙= – 69.0 kJ mol-1
HCl (g)+ 25 H2O(l)  [HCl+25H2O](aq) ;Δ𝐻 𝑠𝑜𝑙= – 72.0 kJ mol-1
HCl (g)+ 200 H2O(l)  [HCl+25H2O](aq) ;Δ𝐻 𝑠𝑜𝑙= – 74.0 kJ mol-1

19. Integral enthalpy of dilution
If a infinite amount of solvent is added to a solution with a
known concentration of solute, the corresponding change of
enthalpy is called as integral heat of dilution on infinite dilution.
19
Mathematically,
Integral enthalpy of dilution(Δ𝐻) = Enthalpy change with addition of solvent to solution (Δ𝐻𝑖) −
Enthalpy change associated with preparation of solution Δ𝐻°𝑖
Δ𝐻 = Δ𝐻𝑖 − Δ𝐻°𝑖
Eg. Calculate the integral heat of dilution at 298 K for the addition of 195 moles of water to 1 mole
of HCl in 5 moles of water. The enthalpy of formation of 1 mole of HCl in 5 moles of water is – 37.37
kcal and enthalpy of formation in 200 moles of water is -39.79 kcal.
HCl (g)+ 5H2O(l)  [HCl+5H2O](aq) ;Δ𝐻°𝑖 = – 37.37 kcal
[HCl+5H2O](aq) + 195H2O (l)  [HCl+195H2O](aq) ;Δ𝐻°𝑖 = – 39.79 kcal
Δ𝐻 𝑎𝑡 298𝐾
= Δ𝐻𝑖 1 𝑚𝑜𝑙 𝑜𝑓 𝐻𝐶𝑙 𝑖𝑛 200 𝑚𝑜𝑙 𝑜𝑓 𝑤𝑎𝑡𝑒𝑟 − Δ𝐻°𝑖 1 𝑚𝑜𝑙 𝑜𝑓 𝐻𝐶𝑙 𝑖𝑛 5 𝑚𝑜𝑙 𝑜𝑓 𝑤𝑎𝑡𝑒𝑟
Δ𝐻 𝑎𝑡 298𝐾 = −39.79 − −37.37 = −𝟐. 𝟒𝟐 𝒌𝒄𝒂𝒍

20. To summarize Integral enthalpies of solution
and dilution
20
0 20 40 60
-1000
-500
0
Integral enthalpy of dilution
enthalpychange(Jmol
-1
)
moles of H2
O
Integral enthalpy
Integral enthalpy of solution
Consider one
mole acid is
diluted with
different moles
of water and
enthalpy change
is plotted
against the
moles of water
added.

21. Differential enthalpies of solution and
dilution
21

22. Differential enthalpy of solution (ΔHsol)
It is the partial derivative of the total heat of solution with
respect to the molar concentration of one component of the
solution, when the concentration of the other component or
components, temperature, and pressure are constant.
𝑑𝑞
𝑑𝑛
= 𝐻𝑖 − 𝐻𝑖° → ΔHsol
𝑑𝑞 -change in heat when solute is added
𝑑𝑛 -change in number of moles of solute
22

23. Differential enthalpy of dilution(ΔHdil)
• Heat evolved or absorbed when a unit weight (one
mole) of solvent is added to an infinite weight of
solution at a constant temperature and concentration.
ΔHdil=[ 𝑑ΔHdil
𝑑ΔnA
]T,p,nB,nC….=HA- HA*
HA-Partial molar enthalpy of solvent in the solution
HA* - molar enthalpy of solvent
𝑑Δn= infinitesimal change of mole number of dilution
23Y. C. Wu and T. F. Young, JOURNAL OF RESEARCH of the National Bureau of Standards, 85, 1980
http://www.personal.psu.edu/mrh318/Wu-Young-JRNBS-1980.pdf

24. Just for information
y or f(x) = 2𝑥
y= 𝑎
𝑏
𝑓 𝑥 𝑑𝑥 ; integrating f(x)
y= 0
2
2 𝑥 𝑑𝑥
y= 2(2)2- 2(0)2
y= 8
24
0.0 0.5 1.0 1.5 2.0 2.5
0
1
2
3
4
5
y
x
Area = 8
x y= 2x
0 0
0.5 1
1 2
1.5 3
2 4
Integration: finding area under the curve
Differentiation: finding slope of the curve
0.0 0.5 1.0 1.5 2.0 2.5
0
1
2
3
4
5
y
x
Slope =
𝑑𝑦
𝑑𝑥
=
𝑦2−𝑦1
𝑋2−𝑋1
Slope =
4−2
2−1
Slope = 2

25. Calculation
of bond energy, bond
dissociation energy and
resonance energy from
thermochemical data
25

26. Bond energy and bond dissociation
energy
• Bond energy is defined as average amount of
energy required to break all bonds of a
particular type in one mole of substance.
• Bond dissociation energy is the amount of
energy needed to break down a particular
bond.
26

27. Problems on bond energy
• Given that energies for H-H, O=O and O-H are 104, 118 and 111 kcal
mol-1, respectively. Calculate heat of the following reaction.
H2(g) + ½ O2(g)  H2O(g)
Solution: In above reaction, one H-H and half O=O bond are broken
and two O-H bonds are formed.
ΔH = ΔHproducts - ΔHreactant
Δ𝐻 = 2Δ𝐻(O−H) - [1 Δ𝐻(H−H) + ½ Δ𝐻(O−H)]
Δ𝐻 = 2 × 111 - [ 1 × 104 + ½ × 118]
Δ𝐻 = 222 - [ 104 + 59]
Δ𝐻 = 59 kcal mol−1
The heat of reaction is 59 kcal mol-1
27

28. Resonance energy
28
cyclohexane
cyclohexene
cyclohexa-1,4-diene
cyclohexa-1,3,5-triene
Energyinkcal
+ H2
+ 2H2
+ 3H2
+ 3H2
benzene
28.6 obs. 33.4 obs.
57.2 calc.
85.8 calc.
36 kcal is
resonance
energy
The difference in energy
for dehydrogenation of
cyclohexene to cyclohexa -
1,3,5- triene in calculated
and observed value is
Resonance Energy.
*Not to scale

29. Variation of enthalpy of a reaction with
temperature - Kirchhoff’s equation
Kirchoff’s equation is can be stated as
d(∆𝐻) = ∆𝐶𝑝 𝑑𝑇
or
∆𝐻2 − ∆𝐻1 = ∆𝐶𝑝 𝑇2 − 𝑇1
Where ∆𝐻1 𝑎𝑛𝑑 ∆𝐻2 are differential heat content (enthalpy) of
reactant and product, respectively.
∆𝐶𝑝 is difference in heat capacity of product and reactant at constant
pressure.
T1 and T2 are two different temperatures.
Change in heat of reaction at constant pressure per degree change of
temperature is equal to difference in heat capacities of product and
reactants at constant pressure.
29

30. Problem on Kirchhoff’s equation
The heat of reaction below reaction at 27 ◦C is -22 kcal. Calculate the heat of
reaction at 77 ◦C. The molar heat capacities at constant pressure at 27 ◦ C for
hydrogen, chlorine and hydrochloric acid are 6.82, 7.70 and 6.80 cal mol-1,
respectively.
½ H2 + ½ Cl2  HCl
∆𝐶𝑝= Heat capacity of product – heat capacity of reactant
∆𝐶𝑝 = 6.80 −[
1
2
× 6.62 +
1
2
× (7.70)]
∆𝐶𝑝 = −0.46 𝑐𝑎𝑙
∆𝐶𝑝 = −0.46 𝑋10−3 𝑘𝑐𝑎𝑙
∆𝐻2 − ∆𝐻1 = ∆𝐶𝑝 𝑇2 − 𝑇1
∆𝐻2 − −22.1 = −0.46 𝑋10−3 350 − 300
∆𝐻2 = -22.123
Heat of reaction at 77 ◦C is -22.123 kcal
30

31. • Statement of Third Law of
Thermodynamics
• Refer slide 13
31

32. Calculation of absolute entropies of
substances
32

33. 33
Eg 1. Calculate the standard entropy of formation, of CO2
gas. Given the standard entropies of CO2(g), C(g)and O2(g)
are 213.6, 5.740 and 205.0 JK-1, respectively.
Solution:
C(g) + O2(g)  CO2(g)
ΔS◦f = S◦product – S◦ reactants
ΔS◦f = S◦(CO2) – [S◦(C) + S◦(O2)]
ΔS◦f = 213.6– [5.740 + 205.0]
ΔS◦f= 2.86 JK-1

34. Eg 2 . Calculate the entropy change in the
surroundings when 1.0 mol H2O(l) is formed from its
element under standard conditions at 298 K. (Given
ΔH◦f H2O(l)=-286 kJ mol-1)
Solution: As enthalpy change is negative the energy
is released from the system in the form of heat is
286 kJ.
∆𝑆 =
∆𝐻
𝑇
∆𝑆 =
286
298
∆𝑆 = 𝟎. 𝟗𝟔𝟎 𝐤𝐉 𝐊−1
34