Made by:
Chemistry Lecturer: Gokarna Nepal
1
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General introduction:
The branch of science which deals with the study of different form of
energy and the quantitative relationship between them is known as
thermodynamics.
When we confine our study to chemical changes and chemical
substances only, then that branch of thermodynamics is known as
chemical thermodynamics.
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Some basic terms and concepts:
1. System: A system means that part of Universe which is under study.
For example: a certain amount of solid, liquid or gas contain in a
system.
2. Surrounding: The rest of the universe or environment around the
system is called surrounding.
So, the surrounding means all other things which can interact
with the system.
3. Boundary: Anything that separates the system from the
surrounding is called boundary.
Thus, universe = system + surrounding
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Isolated system – neither
mass nor energy can cross
the selected boundary
Example (approximate): coffee in
a closed, well-insulated thermos
bottle
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Closed system – only energy
can cross the selected
boundary
Examples: a tightly capped cup of
coffee
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Open system – both mass and
energy can cross the selected
boundary
Example: an open cup of coffee
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4. State function: A physical quantity is said to be state function if its value
depends only upon the state of the system and doesn’t depend upon the
path by which this state has been attained.
For example: potential energy, internal energy, enthalpy, free energy etc.
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For concept only (not necessary to write)
• let us consider two points at the peak and a base of a hill. The climbers can
move from base to the peak of the hill through different path. Work done
depend upon the distance travelled by the climber (may be longer and a
shorter route). So, it is not a state function. However a climbers standing on
the peak of hill has a fixed value of PE, irrespective of the fact that whether
he reached by stairs or lift. Thus PE of the person is a state function
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Thermodynamic properties:
i. Intensive properties: The properties of the system which do not
depend upon the amount of the substance present is called
intensive properties. For example: temperature, pressure, viscosity,
density, surface tension, etc.
The boiling point of water is 100 oc at one atmosphere whatever
its mass.
ii. Extensive properties: The properties of the system which depend
upon the amount of substance present is called extensive
properties. For example, mass, volume, energy, work, entropy, etc.
let us consider a glass of water, if we double the mass of water,
the volume, no. of moles, internal energy etc. also get doubled so
they are extensive properties.
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Types of thermodynamic process:
Isothermal and Adiabatic process:
A process is said to be isothermal, if the temperature of the system
remains constant during each stage of the process. Such type of
process may be achieved by placing the system in thermostat.
Here, dT = 0, keeping temperature constant.
A process is said to be adiabatic, if no heat enters or leaves the system
during any step of the process i.e. the system is completely insulated
from the surrounding.
Here, dQ= 0, keeping heat constant.
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For concept not necessary to write………..
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Isochoric and isobaric process:
A process is said to be isochoric, if the volume of the system remains
constant during each step of the process.
Here, dv = 0.
A process is said to be isobaric if the pressure of the system remains
constant during each step of the process.
Here, dp = 0
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Internal energy (E) :
The energy of the thermodynamic system is called internal energy. It
includes all the possible forms of energy of the system. The sum of different
forms of energy associated with the molecule is called internal energy.
Internal energy = Translational energy + rotational energy + vibrational
energy + electrical energy + nuclear energy + bonding energy……..
The internal energy of the system depends upon the state of the system but not upon
the path on which the system attains another state. So, internal energy is a state
function. Therefore, the absolute value internal energy cannot be determined.
However, the change in internal energy can be calculated as:
ΔE = E2 – E1
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In a chemical reaction,
ΔE = Eproduct – Ereactant
If Eproduct > Ereactant, heat is absorbed by the
system and hence reaction is endothermic.
If Eproduct < Ereactant, heat is released by the
system and hence reaction is exothermic.
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Sign convention:
1. ∆H = -ve, then energy is evolved.
2. ∆H = +ve, then energy is absorbed
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First law of thermodynamics:
It states that, “ Energy can be neither be created nor destroyed
although it may be converted from one form to another ”
OR
The total amount of energy in the Universe remains constant although
it may undergo transformation from one form to another.
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Mathematical formulation of first law of
thermodynamics:( relation between internal
energy, work and heat)
The internal energy of the system can be increased in 2 ways.
1. By supplying heat to the system
2. By doing work on the system
in board…………….
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Sign convention:
Heat absorbed by the system: q is +ve
Heat evolved by the system: q is –ve
Work done on the system: W = +ve
Work done by the system: W = -ve
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Enthalpy:
• H = E +PV
• Heat content of system at
constant pressure
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GN Chemistry
Enthalpy or heat content:
• Enthalpy is also a state function and its absolute value cant be
determined. However, change in enthalpy ∆𝑯 𝑯𝒑 − 𝑯𝑹 of a
chemical process can be measured accurately in a calorimeter which
is kept open to the atmosphere.
• The enthalpy or heat content of the elements at their standard state
is taken as zero.
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GN Chemistry
Enthalpy of reaction( Heat of reaction)
• It is defined as the amount of heat evolved or absorbed (i.e change in
enthalpy) when all moles of the reactant converted into product.
• For example:
• N2(g) + 3H2(g) → 2NH3(g) ∆𝐻 = −22𝑘𝑐𝑎𝑙
• Since, the value of heat of reaction depends on pressure and
temperature, the value are generally given under standard conditions
and the heat of reaction is known as standard enthalpy of reaction
and represented by ∆𝐻 0
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GN Chemistry
Type of Enthalpy of reaction( Heat of reaction)
1. Enthalpy of solution: It is quantity of heat change involved when 1 mole
of a substance is dissolved in an excess of solvent at a given temperature
and pressure so that further dilution of solution produce no heat change.
Eg,
−1
HCl(g) + aq HCL(aq) ∆𝐻 = −72.7𝐾𝐽𝑚𝑜𝑙
The process for enthalpy of solution is a physical process.
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GN Chemistry
Type of Enthalpy of reaction( Heat of reaction)
2. Standard enthalpy of formation (∆𝐻𝑓 0)
• It is defined as the amount of heat evolved or absorbed when one mole of the
substance is formed from its constituents elements under given condition of
temperature and pressure.
• Normally the temperature and pressure are chosen as 250C and 1 atm is
called standard state.
• For example:
• C(s) + O2(g) → CO2(g); ∆𝐻𝑓 0= - 94kcal
• Note:Balance the eqn in such a way that it represent one mole of substance
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GN Chemistry
Continue on enthalpy of formation
• Higher the negative value of enthalpy of formation, greater is the
stability of the compound.
• In case of allotropes, the enthalpy of formation of the most stable
allotrope is taken as Zero. For example heat of formation of graphite
is zero, while that of diamond is not zero.
1
• H2+ O2 → H2O (right)
2
• 2H2+ O2 → 2H2O (wrong)
• H++ OH- → H2O (wrong)
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GN Chemistry
3. Standard enthalpy of Combustion(∆𝑯𝑪 0):
• It is the amount of heat evolved (i.e. decrease in enthalpy) when one mole of the
substance is completely burnt in air or oxygen.
• For example:
• CH4 + O2 → CO2 + H2O ∆𝐻 = −34𝑘𝑐𝑎𝑙
• Heat of combustion is always negative.
• C, H and S are respectively oxidized to CO2 , H2O and SO2 respectively.
• Take one mole of substance whose heat of combustion is to be determined and
take required mole of oxygen as required for balancing the equation.
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GN Chemistry
4. Standard enthalpy of neutralization(∆𝐻𝑛 0):
• It is the amount of heat evolved(change in enthalpy) when one gram equivalent
of an acid(base) is neutralized by one gram equivalent of a base(acid) in dilute
solution.
• For example:
• HCl(aq) + NaOH(aq) → NaCl (aq)+ H2O ∆𝑯 = −𝟏𝟑. 𝟕𝒌𝒄𝒂𝒍
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GN Chemistry
Hess’s law of constant heat summation
• The law states that the total heat change accompanying a chemical
reaction is the same whether the reaction takes place in single step or
in multiple step
• It means that the heat of a reaction depends only on the initial
reactant and final product not on intermediate products that may be
formed.
• It includes only the initial reactant and final product but independent
of intermediate steps.
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GN Chemistry
∆𝐻 = 𝑝 + 𝑞 + 𝑟 = 𝑥
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GN Chemistry
Illustration:
• Carbon can be converted into carbon dioxide either directly or via the
formation of carbon monoxide. The heat change involved in both the
processes is found to be same as indicated below:
• Formation of CO2 directly:
C + O2 → CO2(g) ∆𝐻 = −94.0𝑘𝑐𝑎𝑙
Foramtion of CO2 via formation of carbonmonoxide:
1
• C (g) + O2 (g) → CO(g) ∆𝐻 = −26.0𝑘𝑐𝑎𝑙
2
1
• CO (g) + O2 (g) → CO2(g) ∆𝐻 = −68.0𝑘𝑐𝑎𝑙
2
• ∆𝐻= ∆𝐻1+ ∆𝐻2 = -26.0 – 68.0 =-94.0kcal.
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GN Chemistry
Application:
1. Determination of enthalpy of formation:
Enthalpy of formation of some compounds can be calculated by using
Hess’s law of constant heat summation. Which is difficult to calculate
directly. Enthalpy of formation of CO can be calculated as,
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2. Determination of formation of bond: One atom of C combines with
four atoms of hydrogen to give methane gas.
C(s) + 4H→ CH4 ∆𝐻=-398kcal
So the average energy per mole of C-H bond can be calculated as
C-H bond = -398/4 =-99.5kcal
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Formula to calculate Enthalpy of reaction:
(heat of reaction)
• ∆𝐻=Enthalpy of product –Enthalpy of reactant
• ∆𝐻= enthalpy of combustion of reactant –enthalpy of combustion of
product
• ∆𝐻 = Bond energy of reactants – bond energy of product
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GN Chemistry
Bond energy:
• When a bond is formed between the two free atoms in a gaseous stae to form a
molecular product in a gaseous state, some heat is always evolved which is
known as the bond formation energy or the bond energy. The bond energy may
be referred to as heat of formation of the bond
• Also, Bond energy may be defined as the average amount of energy required to
dissociate(break) bonds of that type in one mole of the compound. This bond
energy of C-H in CH4 is the average value of the dissociation energies of the four
bonds.
• ∆𝐻 = Bond energy of reactants – bond energy of product
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GN Chemistry
Here are some numerical problems we will
discuss it in class
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The standard heat of formation at 298K for CCl4 (g), H2O
(g), CO2(g) and HCl(g) are -25.5, 57.8,-94.1 and 22.1
kcal/mole respectively. Calculate the ∆𝐻for the reaction.
CCl4(g) + 2H2O(g) → CO2(g) +4HCl(g)
• Solution:
• ∆𝐻=Enthalpy of product –Enthalpy of reactant (Hp-HR)
• ∆𝐻= (HCO2+4 x HHCl) – (HCCl4 +2xHH2O)
• ∆𝐻 = −94.1 + 4x22.1 − −25.5 + 2x − 57.8
• ∆𝐻 = −41.4kcal
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GN Chemistry
Calculate the heat of formation of naphthalene from the following data
C(s) + O2(g) → CO2(g) ∆𝐻 = −94.405𝑘𝑐𝑎𝑙
1
H2(g)+ O2 → H2O (l) ∆𝐻 = −68.3𝑘𝑐𝑎𝑙
2
C10H8(s) + 12O2 (g) → 10 CO2 (g) +4H2O (l) ∆𝐻 = −1231.6𝑘𝑐𝑎𝑙 ans=14.35kcal
• Soln: here formation of naphthalene is to find so put it in product side and
change the sign of enthalpy change
• C(s) + O2(g) → CO2(g) ∆𝐻 = −94.405𝑘𝑐𝑎𝑙
1
• H2(g)+ O2 → H2O (l) ∆𝐻 = −68.3𝑘𝑐𝑎𝑙
2
• 10 CO2 (g) +4H2O (l) → C10H8(s) + 12O2 (g) ∆𝐻 = +1231.6𝑘𝑐𝑎𝑙
• 10 C (s) + 4 H2(g) → C10H8 (s) ∆𝐻=14.35kcal (formation of naphthalene)
• Note: Multiply the equation by that stoichiometric coefficient to remove
unwanted compound
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GN Chemistry
Q. The enthalpy of reaction for
N2(g) + 3H2 (g) → 2NH3 (g) is - 92.4 kJ Calculate the enthalpy of
formation of ammonia. [2] 2072
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Q. Distinguish between extensive and intensive properties giving one example of each
Q. Calculate the standard enthalpy of formation of water in the following
reaction:
2H2 (g+ 02(g) → 2H20 (l), ∆H = - 136 Kcal
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Q. Define state function and give any two correct examples of it
Q. State the first law of thermodynamics and write its mathematical
relation.
Q. Define the terms:
i. Extensive properties ii. Internal energy
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Q. Calculate the enthalpy of formation of NH3 from the following
equation. N2(g)+ 3H2(g) → 2NH3 (g),∆H = —186kJ.
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2076 set B
Define bond dissociation energy. The bond dissociation energy of H2 (g)
and Cl2 (g) are 435KJ/mol and 243 KJ/mol respectively. The enthalpy of
formation of HCl (g) = —92 KJ/mol. Calculate the bond dissociation
energy of HCI (g). +431KJ/mole
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2076 set C
State enthalpy of combustion. If heat of formation of CO2 , H2O and
C6H12O6 are -395 KJ/mole,—269.4 KJ/mol and —1169 KJ/mo
respectively. Calculate the heat of combustion of glucose.
-2815KJ/mole
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2076 set C
State Hess’s law of constant heat summation. Heat of combustion of
benzene (C6H6) is -3280 kJ. Heat of formation of C02 and water are
-395 kJ and -286 kJ respectively. Calculate the heat of formation of
benzene. Ans: +52 kJ
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The standard enthalpy of formation Of H2O(I), C02 (g) and C6H6
(I) are —286, —393.5 and +49.02 KJ respectively at 298 K.
Calculate the standard enthalpy of Combusti0n of C6H6 (l) at the
given temperature. Ans: - 3268.02 KJ
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What is meant by enthalpy of formation? Calculate the enthalpy of formation of
ethane at 298k, if the enthalpies of combustion of C, H and C2H6 are - 94.14. - 68.47
and —373.3 K cal. respectively. ans: 20.39 Kcal
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Define enthalpy of combustion. Enthalpy of formation of benzene is
55KJ, enthalpy of formation of water and carbondioxide are -395KJ
and -285KJ respectively. Calculate the enthalpy of combustion of
benzene. Ans; -2950KJ
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Define bond enthalpy. The bond enthalpy of gases H2, Cl2 and HCl
are 104KCal mol-1 , 58kcal mol-1 and 103 Kcalmol-1 respectively.
Calculate enthalpy of formation of HCl
Ans= -22Kcal
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The enthalpies of formation of CO2(g) , H2O(l) and CH4(g) are -
393.5, -286.2 and 74.8 k j mol-1 respectively. Calculate the
enthalpy of combustion of methane? Ans=
891.1KJ/mol
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Define heat of formation. Heat of combustion methane, carbon
and hydrogen are -210, -94 and -68 Kcal respectively. Calculate
the heat of formation of methane. Ans; -20Kcal
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Heat of formation of ethyl alcohol, water and carbondioxide are -
64.1 Kcal, -68.5 Kcal and -95 kcal. Calculate the heat of
combustion of ethyl alcohol? Ans:-331.4Kcal
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Heat of formation of ethyl alcohol, water and carbondioxide are -
64.1 Kcal, -68.5 Kcal and -95 kcal. Calculate the heat of
combustion of ethyl alcohol? Ans:-331.4Kcal
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Calculate the enthaloy of formation of benzene if enthaloy of
combustion of benzene and carbon are -3280 KJ/mol and -395
KJ/mol respectively. The enthalpy of formation of water is -
285KJ/mol 2070 Set C
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Estimate the enthalpy change for the reaction
H2(g) + Cl2(g) → 2HCl Given
bond energy of H-H =435Kj/mol
Bond energy of Cl-Cl = 243 Kj/mol
Bond energy of H-Cl = 430 Kj/mol ans= -182 kj/mol
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