UNIT-8 THERMOCHEMISTRY AND ENERGETICS OF CHEMICAL REACTIONS NEW_watermark.pdf

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1. •
• CHAPTER # 8
• THERMO CHEM ISTRY AND
•ENERGETICS OF CHEMICAL REACTIONS
6/27/2024 PROF:SIKANDER ALI QURESHI
1

2. •THERMOCHEMISTRY
•It is the study of thermal changes in chemical reactions , related to
heat released or absorbed as a result of chemical reactions. OR
•The study of heat changes accompanying a chemical reaction.
•THERMODYNAMICS:-
• The branch of science that deals with the relationship or conversion
of heat , work into other forms of energy and vice versa.
•It is study of all types of energy changes related to physical and
chemical changes . This branch consist of different laws which
describe how the different forms of energy can be interconverted. It
is the study based on conservation of energy.

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3. •It is macroscopic science which describes the initial and final state of
physical and chemical process, But does not study the rate of
reaction and constituents of substance .
•It also describes the conditions at which thermochemical process will
occur or not .

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4. •THERMODYNAMIC TERMS
• 1. SYSTEM:
•Any portion of the universe which is under observation in laboratory
or elsewhere is called system. OR
•The collection of material having well defined boundaries ,through
which matter and energy can be transferred is called system.
•System has some quantity of matter with particular macroscopic
properties.
•Water in beaker , oxygen gas in cylinder , tea in cup ,solution filled in
calorimeter are examples of system .
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5. •2. SURROUNDING :The area around the system is called
surrounding.
•The remaining part of universe other than system is called
surrounding . The energy or matter can be transferred from
surrounding to system.
•Ex : During the study of thermal decomposition of CaCO
3 in a
container. The area around the container where from heat is
supplied is called surrounding.


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6. •BOUNDARY:
•It is real or imaginary surface that divides system and
surrounding. Ex :Reaction between zinc dust and hydrochloric acid in
beaker . The beaker containing Zinc dust and dilute HCl is system and
it has specific characters like mass , volume , temperature etc. while
wall of beaker , the empty area of beaker and other part of universe
is surrounding.The surface which separates system from surrounding
is called boundary.
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7. •Types of thermodynamic system : Thermodynamic system is divided
into three types based on inter conversion of mass and energy .
•1.Open system : The system which can exchange both matter and
energy with surroundings is called an open system.It is also called
flow systems, because of the ability to exchange mass of a substance.
•Ex: A boiling tea kettle transfers heat and matter (steam) into
surrounding.
•Ex : Pot of boiling water on a stove.

•

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8. •Closed system
•The system which can exchange only energy not matter with
surroundings is called a closed system.
•Example :Pressure cooker , Cold bottle of water with cover.
•Isolated system
•The system in which neither energy nor matter can be exchanged
with surroundings is called an isolated system. An isolated system
may be a portion of larger systems, but they do not communicate
with outside in any way.
•Ex : Thermoflask , where any thing remains hot or cold , it does not
allow energy to transfer . Water cooler is also isolated system. .
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9. •MICRO SCOPIC PROPERTIES :
•The properties of single unit of particle (atom,ion, molecule) Which
are difficult to measure are called microscopic properties. EX: K.E,
E.A, E.N, atomic mass, bond lengths, bond energies, electron spin of
individual particles.
•These are the properties of building blocks( atoms, ions or
molecules) of matter. These constituents particles are invisible to
naked eye, therefore, they are measured by small units like
millimetres, micrometres, nanometers, picometers, etc.
•Macroscopic properties :
•The properties in bulk or general properties of system are called
macroscopic properties. They can be measured easily and can be
observed by naked eye . EX: Temperature, Mass , Volume , Density ,
Pressure ,amount of substance,color etc.
•Macroscopic properties are of two types


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10. •1.INTENSIVE PROPERTIES:
•The properties which are independent to the material of system are
called intensive properties. They do not change with the change in
the material of the system.
•Examples are density, pressure , temperature, viscosity, surface
tension, refractive index, melting point and boiling point.
•2.EXTENSIVE PROPERTIES:
•The properties which are directly proportional to the amount of the
system are called extensive properties.
•Ex; Mass, volume, mole number, enthalpy, entropy, the internal
energy and the Gibb's free energy.

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11. •ENTROPY (S):
•It is a thermodynamic function that describes the randomness and
disorder of molecules based on the number of different
arrangements available to them in a given reaction.
•Entropy is directly proportional to temperature .
•Ex: Gases have high entropy and solids have low entropy. A block of
ice will increase entropy as it melts. Entropy is a state function

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12. • STATES: The properties such as volume, pressure and temp: which
describe a system completely under given set of experimental
conditions are called states.
•A system is said to be in definite state when each of the properties
has definite value and is completely defined. The description of
system before it undergoes a chemical change is known as initial state
and after change is final state of system. The difference in initial and
final state is given by (∆).
•A chemical reaction is always described by the change of its states.
•Ex: A system composed of a gas in cylinder fitted with moveable
piston at initial pressure P
1 and has volume V
1, now if the pressure is
increased to P
2, then volume will be reduced to V
2 at constant
temperature.

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13. •The state of a system is the condition of a system. When any process
is performed on a system its state is altered.
•The change in pressure Δ P = P
2 – P
1,
•The change in volume Δ V = V
2 - V
1
•The change in energy Δ E = E
2 - E
l
•Thus
•Change in property =Value in the final state — value in initial state.
•

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14. •STATE FUNCTION:
•Any property that depends on initial and final states of system not on
the path of system which it follows for change. It depends on state
of system obtained but not on the way or path followed to attain that
state .
•Ex: If we heat sample of water from 0 ℃ to 25 ℃,the change in
temperature is equal to difference between initial and final
temperature not the way in which temperature has changed .
• T=T
2 - T
1
• = 25 ℃ − 0 ℃= 25 ℃
•EX:Enthalpy, free energy, Entropy , Pressure , Temperature , Internal
energy,density etc are state functions .
•It is macroscopic property of a system.
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15. •PATH FUNCTION:
•It depend on the path taken to reach that specific value. They are
properties whose values depend on the transition of a system from
the initial to the final state.
•Ex: heat , work and length. If different paths are chosen to reach
same point, the work done will be different ,but, you reach the same
point in each case.
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16. •Enthalpy of system( H)
•Enthalpy is property of the system also known as heat content of
system , which is measured at constant pressure because commonly
reactions are performed in open test tubes, conical flasks , beakers etc
It is thermodynamic quantity used to describe the total energy of
system containing both its internal energy and the energy associated
with its pressure and volume. Simply , enthalpy is sum of internal
energy(E) and the product of its pressure and volume (PV).
• H = E + PV
•Enthalpy is a state function. It is not possible to measure the
enthalpy of system in given state ,but, change in enthalpy (∆H ) is
measured for the state of system ∆ H = ∆E + ∆ (PV). It is measured
in calories or joules (1 Calorie = 4.184 joules )
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17. •THERMO CHEMICAL REACTIONS
•Such type of reactions in which change of energy takes place besides
change of material are called thermo chemical reactions and the
chemistry which deals with such reactions is called thermo chemistry.
•There are two types of thermos chemical reactions

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18. •1.EXOTHERMIC REACTIONS.
•heat is evolved from system to surrounding , so surrounding becomes
warmer .
•more bonds are formed than broken
• all combustion reaction are exothermic.
•product has less energy than reactant .So ∆H is negative (∆H< 0).
•The net potential energy of substances decreases.
•6. The temperature of surrounding Increases.
• C + O
2 CO
2 ΔH= -394Kj/mol
•CH
4 + 2O
2 CO
2 + 2H
20 ΔH =- 890 Kj/mol

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19. •2. ENDOTHER MIC REACTIONS
•Such type of chemical reactions in which heat is absorbed are called
endothermic reactions.
•Ex: C + 2S CS
2 ΔH = + 117Kj/mol
• H
2 + I
2 2HI ΔH = + 57 Kj/mol
• CH
4 C+ H
2 ΔH = +75.6 kJ/mol
• 6CO
2 + 6H
2O C
6H
12O
6 + 6O
2
•1. Where more bonds are broken than formed.
• 2. product has more energy than reactants. So ∆H is positive (∆H> 0)
•3. The net potential energy of substances increases.
•4. These reactions decrease the temperature of surrounding.
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20. •FIRST LAW OF THERMODYNAMICS: OR LAW OF CONSERVATION OF
ENERGY:
•This law was given by Helmholtz in 1847, it states that:
•"Energy can neither be created nor be destroyed but it can be
changed from one form to another form" However, it can exchange
energy with its surroundings in the form of heat and work.
• When system loses energy to the surroundings it will increase the
energy of the surroundings by that amount to keep the total energy
constant ,so it can also be defined that "The total energy of system
and its surroundings remains constant, but it may be changed from
one form to another form".
•It means that energy is transferred into or out of system in the form of
heat and work.

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21. •MATHEMATICAL DERIVATION OF FIRST LAW OF THERMODYNAMICS:
•Consider a system with internal energy E1, this system absorbs some
heat "q" from the surroundings and some work (w) is done on the
system, with the result energy changes to E
2, the change in energy is
given by E
2 - E
1 so according to first law of thermodynamics:
•ΔE=E
2 - E
1 =q+ w
•ΔE =q+ w
•Where q is + ve when system absorbs energy
•q is - ve when system evolves energy
•w is + ve when work is done on system
•w is - ve when work is done by system


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22. • EX 1: In a certain process 848j of heat is absorbed by a system, while
394j of work is done on the system. What is the change in energy ?
•Solution: ∆E =q+w
• = 848+394 =1242j
•EX 2: If 1800 calories, of heat is added to a system while system does
work equivalent to 2800 calories by expanding against the
surrounding. What is the value of ∆E for a system?
•Solution: ∆E = q+w
• = 1800+ (-2800)
• = 1800 – 2800=– 1000 calories

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23. •PRESSURE VOLUME WORK:
•The mechanical work due to expanding or squeezing of system at
constant pressure is called pressure- volume work
•Consider a cylinder of gas having cross section area “A" fitted with
frictionless , moveable piston having negligible weight . If the inside
pressure on piston is greater than external pressure then the piston
moves upward from height h
1 to h
2 , hence it does some work.
•W =F.d
•W = -F(h
2-h
1)---------(1)
• (- ve sign shows that external pressure opposes expansion of gas .
Now p=F/A so F=PA now put value of F in equation (1)
•W = - PA (Δh) ----------(2) 6/27/2024 PROF:SIKANDER ALI QURESHI
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24. •But A.Δh is equal to change in volume ΔV
•by putting the value of work ion eq: (2)
•W= -P ΔV
• In pressure volume work, force becomes pressure and distance
becomes volume change, where P is the external pressure and ∆V is
the change in volume. Work is path function.

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25. •Ex : A gaseous reaction is carried out in cylinder under constant
external pressure of 1 atm: If during the reaction volume increases
from 3dm
3
to 5 dm
3
by moving the piston upward .
•Calculate work done and express in K.J.
•P=1 atm:
•V= V
2-V
1= 5-3=2 dm
3
•W=?
•W= -PV= - 1X 2=-2 atm dm
3

•1 atm dm
3
= 101.325 J
•Now -2 x 101.325 = -202.65J or – 0.20265KJ
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26. •Internal energy E/U)
•It is sum of all the possible kinds of energies of all the particles of
system is called its internal energy ( E ). The internal energy of the
system is state function.
•A system containing some quantity of matter has definite amount of
energy present in it.
•This energy is the sum of kinetic as well as the potential energies of
the particles(atoms ,ions, molecules ) contained in the system.
• It is not possible, to measure the absolute value of internal energy of
a system,because it is state function, so change in energy (∆E ) is
measured for a change in the state of the system

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27. •Name ad define units of thermal energy
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28. •Ex : The burning of petrol in an automobile engine gives carbon
dioxide and water vapours. If the gases do 675J work in pushing the
piston outward and system loses 435J heat to the surrounding,
calculate the internal energy change in KJ
•Solution:
•q = - 435J
•W = - 675J
•E = ?
•E = q + w
•E = (-435) + (-675)
• = -435-675= - 1110 J= - 1.11 KJ


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29. •Applications of First law of thermodynamics
•(i)At constant volume (ii) At constant pressure
•1. The process at constant volume
• When reaction is carried out at constant volume , no work is done on
system or by system , therefore internal energy of reaction will be
equal to the heat change of the system .
•ΔE = q +w
•ΔE = q — PΔV (w=— PΔV )
•qv = ΔE + PΔV
•qv = ΔE + P (0)
•qv = ΔE It means that heat evolved or absorbed at constant volume
is equal to energy change.

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30. •2. PROCESS AT CONSTANT PRESSURE:
•In laboratory, chemical reactions are most commonly carried out at in
open vessel . It means they are carried out at constant pressure i.e
atmospheric pressure. So chemists are interested in heat change at
constant pressure. We know that:
•ΔE = q + w OR
• ΔE = q — PΔV OR
•qp = ΔE + PΔV = (1)
•If qp is the heat flow to or from system at constant pressure, ΔE is
change in energy and PΔV is work done with change in volume .
•
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31. •q
p = ΔE +PΔV therefore
•q
p = (E
2 — E
1) + P ( V
2 - V
1 )
•q
p =E
2 – E
1 + PV
2 – PV
1
•on re-arranging the equation
•q
p = (E
2 +PV
2) - (E
1- PV
1)
•q
p= E
2 +PV
2 - E
1+ PV
1
•(We known that H = E + PV )
•qp = H
2 — H
1
• qp = ΔH
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32. •HESS'S LAW OF CONSTANT HEAT OF SUMMATION:
•G.N Hess in 1840 gave an important generalization, known as Hess's
law, it states that:
•“If a chemical change takes place by several different routes, the
overall energy change is the same, regardless of the route by which
the chemical change occurs, provided the initial and final conditions
are the same”. Or
•“The amount of heat evolved or absorbed in a process during any
chemical reaction is always same, whether reaction is completed in
single or in many steps , directly or indirectly”
• It is because the change in enthalpy is state function, hence, it is
independent of the path taken ,but depends only on the initial and
final states of the reaction.

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33. •Explanation of Hess’s Law :
• Suppose in a process, the system changes from state A to stste D by
two ways . In path I, it directly changes from A to D and in 2
nd
Path, it
changes from A to D in three Steps proceedings from A to B, B to C
and finally from C to D but the change of heat will remain same.
Path1 : A ΔH D
•Path 2 : A ΔH
1 B ΔH
2 C ΔH
3 D
•Therefore total change in heat will bec
• ΔH = ΔH
1 +ΔH
2 + ΔH
3
•



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34. •Ex : CO
2 can be obtained from C and O
2(g) in two different ways, but
total change in heat remains same
•PATH – I : Carbon burns to form CO
2 directly
•C(graphite) + O
2(g) CO
2(g) ΔH° = -393.52KJ
•PATH - II : Secondly carbon first changes to carbon monoxide and
then changes to carbon dioxide
•C(graphite) + ½O
2 CO
(g) ΔH° = -110.5KJ
•CO
(g) + ½O
2(g) CO
2 ΔH° =-283.02KJ
•Total heat change in PATH-II = ΔH
1 + ΔH
2 = -110 +(-283) = –393.52KJ.
•Enthalpy change in both cases is same i-e in path I and path II , it
proves the Hess’s law of constant heat summation .
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35. •Standard enthalpy of formation (ΔH°
f
)
• It is the amount of heat absorbed or evolved when one mole of the
pure substance (compound) is formed from its elements.
•All the substances involved are in their standard physical states and
the reaction is carried out under standard conditions at 25°C (298 K)
and one atmospheric pressure.
•The enthalpy of formation for MgO
(s) is ( ∆H°f) =- 692 kJ /mol
•Mg
(s) + 1/2O
2(g) MgO
(s) ∆H°f = - 692 kJ /mol
•The enthalpy of formation for CO
2(g) is, ( ∆H°f) =– 393.7 kJ /mol
•C
(g) + O
2 (g) CO
2(g) ∆H°f = - 393.7 kJ /mol



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36. •Standard enthalpy of formation of some organic and in organic
substances in Kj/mol




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Organic
substances
∆H°f Inorganic
substances
∆H°f
CH
4 -74.85 CO -110.5
C
2H
6 -83 CO
2 -393.5
C
3H
8 -103.9 SO
2 -296.8
C
2H
4 +52.3 SO
3 -396
C
2H
2 +226.7 H
2S -20.17
C
6H
6 +49.03 NH
3 -46.19
CH
3OH -239 H
2O 285.8
C
2H
5OH -277.7 Fe
2O
3 -824.2
CH
3COOH -484.2 NaCl -411

37. •Standard Enthalpy of a Reaction (∆H°)
•It is defined as “the enthalpy change when the certain
•number of moles of reactants as indicated by the balanced chemical
equation, react together completely to give the products at STP”.
•The change in enthalpy in balanced chemical equation is given by
•∆H°(reaction)= np ∆H°f (product) nr ∆H°f (reactant)
•Use of Hess’s law in the determination of heat of Reaction
• The heat of reaction(∆H°) can be measured by two ways ,either by
using calorimeter or by Hess’s law in which sum of heat of reactants
is subtracted from sum of heat of product . .



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38. •Ex:Calculate enthalpy of combustion of propane at 25℃ by the given
information .
• C
3H
8(g) + 5O
2(g) 3CO
2(g) + 4H
2O
(g) (H° = ?)
•H
f of C
3H
8(g) = -103.9KJ/mol
•H
f of CO
2(
g) = -393.5KJ/mol
H
f of H
2O = -285.8KJ/mol
Solution:
The mathematical form of heat of reaction H is given as
H
reaction = [ ( np H°f
products ) – ( nr H°
reactants ]
= [ (-393.5 x3 ) + (-285.8x4)] – (-103.9x1)
= -2219.8 KJ/mol


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39. •Use of Hess’s law in the determination of heat of formation
•Ex: Oxidation of Sulphur to Sulphur trioxide
• S
(s) + 3/
2O
2(g) SO
3(g)
• Sulphur can not oxidize directly to SO
3 , so its heat of formation
can not be measured by calorimeter . It oxidizes in two steps
•1. S
(s)+ O
2
(g) SO
2
(g) ∆H
1= -296.8 Kj/mol
•2. SO
2(g) + 1/
2O
2(g) SO
3(g) ∆H
2= -99.2 Kj/mol
•So ΔH°f of SO
3 can be measured indirectly by adding eq: 1 and 2
• S
(s)+ O
2(g) SO
2(g) ∆H
1= -296.8 Kj/mol
• SO
2(g) + 1/
2O
2(g) SO
3(g) ∆H
2= -99.2 Kj/mol
• S
(s) + 3/
2O
2(g) SO
3(g) ΔH°f= -396 Kj/mol
•

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40. •APPLY HESS'S LAW TO CONSTRUCT ENERGY CYCLE
•EX: The Born-Haber Cycle
•Born Haber cycle was developed by German chemist Max Born and
Fritz Haber in 1919 to show the relationship between lattice energies
of ionic compounds with other energies in their formation . It states
that “It is cycle of enthalpy change in which an ionic solid is
theoretically formed from its elements in their standard states .”
•This cycle finds its special applications in Hess’s law. It states that
energy change in a cyclic process is always zero. It enables us, to
calculate the lattice energies of binary ionic compounds such as M
+
X
-

•The lattice energy of an ionic crystal is the enthalpy of formation of
one mole of the ionic compound from gaseous ions under standard
conditions.
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41. •Ex: Construct energy cycle to determine lattice energy of CsF
(s)
•The standard enthalpy of formation of CsF is = -553.5 Kj/mol
•Cs
(s)+1/
2F
2 CsF
(s) ΔH°&#3627408467; = -553.5 Kj/mol
•Step1: Sublimation of cesium: in this step cesium metal is converted
into gaseous atom . It is endothermic process
•Cs
(s) Cs
(g) ΔH=+76.5 Kj/mol
•Step2: Ionization of cesium atom: remove electron from cesium. It is
known as ionization energy . It is endo thermic.
•Cs
(g) Cs
+
(g) + e- ΔH=+375.5 Kj/mol
•
Step3:Dissociation of gaseous F
2:it is enthalpy of atomization
•1/
2F
2(g) F
(g) ΔH=+1/2x 157 Kj/mol
•
•
•
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42. •Step4:Formation of F
-
ion:F gains electron & releases energy called E.A
•F
(g) +1e
-
F
-
(g) ΔH=-328.2 Kj/mol
•Step5:Formation of of solid CSF:Opposite ions combine to form ionic
crystal and release energy called lattice energy
•Cs
+
(g) + F
-
(g) CSF
(s) ΔH=- ?
•to calculate lattice energy of CsF , Hess’s law will be applied .
•ΔH°&#3627408467; = ΔH
(sub) + ΔH
(I.E) +ΔH
(D) +ΔH
(E.A) +ΔH
(L.E)
•(-553.5) = (76.5) + (375.5) + (1/2x 157) + (-328.2) +(-ΔH
L.E)
• -553.5 =76.5 +375.5 +78.5-328.2= ΔH
L.E
• - 553.5 = 530.5-328.2 = ΔH
L.E
• -553.5 = 202.3 =ΔH
L.E
OR ΔH
L.E = -553.5-202.3=-755.8 Kj/mol
•

•
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43. •Determination of lattice energy of RbCl by using Born Haber cycle
•The standard enthalpy of formation of RbCl is = -430.5 Kj/mol
•Rb
(s)+1/2Cl
2 RbCl
(s) ΔH°&#3627408467; = -430.5 Kj/mol
•Step1: Sublimation of rubidium: in this step cesium metal is converted
into gaseous atom . It is endothermic process
•Rb
(s) Rb
(g) ΔH
=+86.5 Kj/mol
•Step2: Ionization of rubidium atom: remove electron from cesium. It
is known as ionization energy . It is endo thermic.
•Rb
(g) Rb+
(g) + e- ΔH=+403 Kj/mol
• Step3:Dissociation of gaseous Cl
2:it is enthalpy of atomization
•1/2Cl
2(g) Cl
(g) ΔH
=+1/2x242 Kj/mol
•
•
6/27/2024 PROF:SIKANDER ALI QURESHI
43

44. •Step4:Formation of F- ion:F gains electron & releases energy called E.A
•Cl
(g) +1e
-
Cl
-
(g) ΔH=-349 Kj/mol
•Step5:Formation of of solid RbCl:Opposite ions combine to form ionic
crystal and release energy called lattice energy
•Rb+
(g) + Cl-
(g) RbCl
(s) ΔH=- ?
•to calculate lattice energy of CsF , Hess’s law will be applied .
•ΔH°&#3627408467; = Δ
H(sub) + ΔH(I.E) +ΔH(D) +ΔH(E.A) +ΔH(L.E)
•(-430.5) = (86.5) + (403) + (1/2x242) + (-349 +(-ΔHL.E)
• -430.5 =86.5 +403 +121-349= ΔHL.E
• - 430.5 = 610.5-349 = ΔHL.E
• -430.5 = 261.5 =ΔHL.E OR ΔHL.E = -430.5-261.5=-692 Kj/mol
6/27/2024 PROF:SIKANDER ALI QURESHI
44

45. •Born haber cycle states that energy change in a cyclic process is
always zero define this statement
The statement "It states that energy change in a cyclic process is
always zero" refers to the principle of conservation of energy. In the
context of the Born-Haber cycle, this means that when you sum all the
energy changes associated with forming an ionic compound from its
constituent elements in a series of hypothetical steps, the total energy
change for the entire cycle will be zero.
•-553.5 = 202.3 =ΔHL.E OR
•ΔHL.E = -553.5-202.3=-755.8 Kj/mol
• = - 755.8=-755.8=0 kj/mol
•So the change in enthalpy is zero
•
6/27/2024 PROF:SIKANDER ALI QURESHI
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