Enthalpy Change in reaction or Enthalpy of reaction( ∆RH)

Internal energy Change in reaction or Internal energy of reaction( ∆RU)

The amount of heat is absorbed or released during chemical reaction

carried out at constant V
Relation between Enthalpy of reaction( ∆RH) and
Internal energy of reaction( ∆RU)
Enthalpy Change in reaction or Enthalpy of reaction( ∆RH)
∆rH = Sum of enthalpies of products – Sum of enthalpies of reactants
 Thermochemical Equation :
Step 1: Equation must be balanced
C2H5OH + 3 O2 2 CO2 + 3 H2O
Thermochemical Equation :
C2H5OH + 3 O2 2 CO2(g)+ 3 H2O (l)
(l) (g)

Step 2
Physical state of reactants & products should be mentioned
 Thermochemical Equation :
C2H5OH + 3 O2 2 CO2(g)+ 3 H2O (l) H = – 1430 kJ mol–1
(l) (g)
 Thermochemical Equation :

C2H5OH + 3 O2 2 CO2(g)+ 3 H2O (l) H = – 1430 kJ mol–1

(l) (g)

Step 4

If the reaction conditions are not mentioned, then the

conditions are 298 K & 1 atm.
Thermochemical Equation :
C2H5OH + 3 O2 2 CO2(g)+ 3 H2O (l) H = – 1430 kJ mol–1
(l) (g)

2 CO2 (g) + 3 H2O (l) C2H5OH (l) + 3 O2(g) H = + 1430 kJ mol–1

Thermochemical Equation : H = – 1430 kJ mol–1
C2H5OH + 3 O2 2 CO2(g)+ 3 H2O (l)
(l) (g)

2 CO2 (g) + 3 H2O (l) C2H5OH (l) + 3 O2(g) H = + 1430 kJ mol–1

Step 6

Enthalpy of an element in its standard state is considered to be zero.

Hess’s Law of constant Heat Summation:
According to this law the total enthalpy change of a reaction is same
whether the reaction takes place in a single step or in several steps.
Applications of Hess’s Law :
Applications of Hess’s Law :
➢ Calculation of heat of extremely slow reaction.
➢ Calculation of heat of extremely fast reaction.
➢ Helps to calculate heat of ionization of weak acids and weak bases .

➢ It is used to calculate the calorific value of fuels .

 Question:
Calculate the heat of the reaction CO2(g) + H2(g) CO(g) + H2O(g)
from the following data :
i) C(s) + ½ O2(g) CO (g) H = – 110.5 kJ
ii) C(s) + O2(g) CO2 (g) H = – 394.5 kJ
iii) H2(g) + ½ O2(g) H2O(g) H = – 242.6 kJ
Ans :
The following two reactions are known :
Fe2O3(s) + 3CO(g) → 2Fe(s) + 3CO2(g); ΔH = –26.8 kJ
FeO(s) + CO(g) → Fe(s)+ CO2(g); ΔH = –16.5 kJ
The value of ΔH for the following reaction
Fe2O3(s) + CO(g) → 2FeO(s) + CO2(g) is;
(a) +6.2 kJ (b) +10.3kJ (c) –43.3 kJ (d) –10.3 kJ

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Standard enthalpy of reactions(∆H0):
Standard enthalpy of formation:
NOTE:
Which of the following chemical reaction defining (∆fH0)

(a)C(Diamond) + O2(g) → CO2(g)

(b) H2(g) + ½O2(g) → H2O

(c)N2(g) + 3H2(g) → 2NH3(g)

(d)CO(g) + 1/2O2(g) → CO2(g)

(e) Ca(s) + 3/2 O2 + Cgraphite → CaCO3(s)

Calculation of ∆r𝐇𝚯 from ∆𝐟𝐇𝚯
Calculation of ∆r𝐇𝚯 from ∆𝐟𝐇𝚯
The values of heat of formation of SO2 and SO3 are –298.2kJ and –98.2 kJ. The heat
of formation of the reaction SO2 + (1/ 2)O2 → SO3 will be
(a) –200 kJ (b) –356.2 kJ (c) + 200 kJ (d) – 396.2 kJ

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For the reaction :
H2(g) + Cl2(g) → 2HCl(g), ΔH at 298K = –44.12 kcal. At 298 K, the enthalpy
of formation of HCl will be
(a)–44.12 (b) –22.06 (c) +44.12 (d) –10.06

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Diborane is a potential rocket fuel which undergoes combustion according to the
equation B2H6(g) + 3O2(s) → B2O3(s) + 3H2O(g)
Calculate the enthalpy change for the combustion of diborane. Given
(i) 2B(s) + 3/2 O2(g) → B2O3(s); ΔH= –1273 kJ per mol
(ii) H2(g) + ½ O2(g)→ H2O(l); ΔH = –286 kJ per mol
(iii) H2O(l)→ H2O(g); ΔH = 44 kJ per mol
(iv) 2B(s) + 3H2(g) → B2H6(g); ΔH = 36 kJ per mol
(a) +2035 kJ per mol (b) –2035 kJ per mol
(c) +2167 kJ per mol (d) –2167 kJ per mol

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Enthalpy of combustion:
Calculation of ∆r𝐇𝚯 from ∆𝐜𝐇𝚯
The heat of combustion of C2H5OH(l) is -300 Kcal. If the heat of formation of CO2(g) and H2O(l)
are -94.3 Kcal and -68.5 K cal respectively. Calculate the heat of formation of C2H5OH(l)
C2H5OH(l) + 3O2(g) → 2CO2(g) + 3H2O(l)

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If the standard molar enthalpy change for combustion of graphite powder is –2.48 × 102
kJ mol–1, the amount of heat generated on combustion of 1 g of graphite powder is
(a) 26 kJ (b) 21 kJ (c) 23 kJ (d) 24 kJ

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The standard enthalpy of formation of CO2(g), CO(g), and H2O(g) are -393.5 KJ/mol, -
110.5 KJ/mol, and -241.8 KJ/mol respectively. The heat exchanged by the reaction at
constant volume at standard condition (….KJ) for the reaction
CO2(g) + H2(g) → CO(g)+ H2O(g)

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Calorific value (C.V):
In an experiment, 4.5 kg of a fuel was completely burnt. The heat produced was
measured to be 180,000 kJ. Calculate the calorific value of the fuel.

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 Enthalpy of neutralization:
The heat evolved when one equivalent of acid is completely neutralized by one
equivalent of a base in dilute solution is called as heat of neutralization (∆N𝑯Θ ).
34
35
The enthalpy of neutralization of NaOH and NH4OH by HCl is -13680 cal and -12270
cal respectively. Find enthalpy of ionization of NH4OH.

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The heats of neutralization
Cl2CHCOOH by NaOH is 12830 cal,
HCl by NaOH is 13680 cal,
NH2OH by HCl is 12270 cal,
The heat of neutralization of dichloro acetic acid by NH4OH will be
(a) -13680 cal (b) -11420 cal (c) -14120 cal (d) -12830 cal

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Calculate the amount of heat that will be released when 300 mL of 0.2 M HCl mixed
with 200 mL of 0.2 M NaOH. (Given: Enthalpy of neutralization of HCl with NaOH is
57 KJmol-1.

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Calculate the amount of heat that will be released when 300 mL of 0.2 M H2SO4 is
mixed with 200 mL of 0.2 M NaOH.

39
200 mL of 0.2 M HCl is mixed with 300 mL of 0.1 M NaOH. The molar heat of
neutralization of this reaction is –57.1 kJ. The increase in temperature in °C of the system
on mixing is x × 10–2. The value of x is
(a) 82 (b) 72 (c) 83 (d) 80

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Enthalpy of Solution: (∆𝒔𝒐𝒍 𝑯𝟎 )
➢ The enthalpy change that takes place when 1 mole of a solute
dissolves in a solvent to form an ‘infinitely’ dilute solution.
NaCl(s) + H2O(l) → NaCl(aq) ; ∆𝒔𝒐𝒍 𝑯Θ = +5 kJ mol-1

Where ∆𝒔𝒐𝒍 𝑯Θ - Enthalpy of Solution

∆𝒔𝒐𝒍 𝑯Θ can be exothermic or endothermic
➢ Enthalpy of Solution is the sum of two imaginary steps: reverse
of the Lattice Enthalpy and the Hydration Enthalpy.

∆𝐬𝐨𝐥 𝐇Θ = - ∆𝐋𝐄 𝐇Θ + ∆𝐡𝐲𝐝 𝐇Θ

Enthalpy diagram
Na+(g) + Cl-(g)

-HLE =+776 kJ mol-1 Hhyd = -771 kJ mol-1

Na+(aq) + Cl-(aq)

Hsol = +5 kJ mol-1
NaCl(s) + H2O(l)

Hsol = -HLE + (Hhyd(cation) + Hhyd(anion)))

Hsol = - (-)776 + (-)771 = + 5 kJ mol-1

Enthalpy of dilution :
➢ It is defined as the enthalpy change that occurs when a solution of
one concentration is diluted further to form solution of another
concentration.

HCl(l) + H2O HCl (aq)

∆H(dilution) = – 0.76kJ/mole
Lattice Enthalpy:
➢ The enthalpy change that takes place when 1 mole of a solid ionic
lattice forms from its gaseous ions.
Na+(g) + Cl– (g) → Na +Cl – (s) LEHΘ = - 788 kJ/mol

➢ HLE is defined exothermically.

➢ The more closely ions pack together in the solid lattice the more
exothermic is HLE.
Born – Haber Cycle
➢ Since it is impossible to determine lattice enthalpies by experiment,
we use an indirect method, we construct an enthalpy change
diagram called a Born-Haber Cycle.

Na+(g) + Cl– (g) → Na+ Cl– (s); HLE = – 788 kJ mol-1

Born – Haber Cycle

The importance of the cycle is that, the sum of

the enthalpy changes round a cycle is zero.
 Na+ (g) + Cl (g)
kJ mol–1
½ bondH
121

kJ mol–1
egH

-348.6
Na+ (g) + ½ Cl2 (g)
kJ mol–1

iH
+495.6

Na (g) + ½ Cl2 (g)

Na+ (g) + Cl– (g)
kJ mol–1

subH
108.4

Na (s) + ½ Cl2 (g) latticeH

kJ mol–1

fH
+411.2

NaCl (s)
1) ∆𝐬𝐨𝐥 𝐇 Θ is equal to…

a) - ∆𝐋𝐄 𝐇Θ + ∆𝐡𝐲𝐝 𝐇Θ

b) - ∆𝐋𝐄 𝐇Θ − ∆𝐡𝐲𝐝 𝐇Θ

c) ∆𝐋𝐄 𝐇Θ − ∆𝐡𝐲𝐝 𝐇Θ

d) None of these
2) ∆𝐬𝐨𝐥 𝐇 Θ can be…

a) endothermic
b) exothermic
c) both
d) None of these
Enthalpy Changes During Phase Transformations:

Melting /
Fusion Vapourization

Freezing Condensation

Sublimation
Enthalpy of fusion ( ∆fusH ) :
Enthalpy of freezing (∆freezH) :
Enthalpy of Vapourization ( ∆vapH ) :
Enthalpy of Condensation ( ∆ConH ) :
Enthalpy of sublimation ( ∆SubH ) :
The sublimation energy of I2(s) is 57.3 kJ/mol and the enthalpy of fusion is 15.5
kJ/mol. The enthalpy of vaporisation of I2 is
(a) 41.8 kJ/mol (b) –41.8 kJ/mol (c) 72.8 kJ/mol (d) –72.8 kJ/mol

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Consider the following data
H2O(s) → H2O(l) ΔH = 6.05 kJ
H2O(l) → H2O(g) ΔH = 43.7 kJ
At 273 K, the sublimation energy of H2O(s) → H2O(g) will be
(a) 56.75 (b) 75.20 (c) 49.75 (d) 100.20

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What is the internal energy (kJ) change occurs when 36 g of H2O(l) at 100°C
converted to H2O(g)?
ΔH°(vapourisation) = 40.79 kJ/mol
(a) 75.38 (b) 80.98 (c) 70.98 (d) 45.89

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Standard enthalpy of vapourisation ΔvapH° for water at 100°C is 40.66 kJ mol–1. The
change in internal energy of vaporisation of water at 100°C (in kJ mol–1) is :
(a) + 37.56 (b) – 43.76 (c) + 43.76 (d) + 40.66
(Assume water vapour to behave like an ideal gas)

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Enthalpies for different types of reactions
Standard Enthalpy of Combustion:(∆c𝑯Θ )
The enthalpy change when one mole of a substance is completely
burnt in its standard state is known as standard enthalpy of
combustion (∆c𝑯Θ ).

13
C4H10(g) + O2 (g) → 4CO2 (g) + 5H2O (l);
2
∆c 𝑯Θ = -2658.0 kJ mol-1
Enthalpies for different types of reactions

Enthalpy of Atomization: (∆a𝑯Θ )

The enthalpy change on breaking one mole of bond completely to obtain
atoms in the gas form is known as Enthalpy of atomization (∆a 𝑯Θ ).

H2 (g) → 2H (g) ; ∆a𝑯Θ = 1665 kJ mol-1

Enthalpies for different types of reactions
Enthalpy of ionization (∆ionH) :
The enthalpy change that accompanies the removal of an electron from
each atom or ion in one mole of gaseous atoms or ions is called as
enthalpy of ionization.
–
Ca(g) Ca+ (g) + e ∆ionH = + 590 kJ/mole

Ca+(g) Ca+2(g) + e– ∆ionH = + 1150 kJ/mole

1st ionization enthaply < 2nd ionization enthalpy

 Bond Enthalpy:
➢ Chemical reactions involve the breaking and making chemical bonds.
Energy is required to break a bond and energy is released when a bond is
formed. It is always positive.

∆rHΘ = ∑ bond enthalpiesreactants –∑ bond enthalpiesproducts

Two types of bond enthalpies are:

➢ Bond dissociation enthalpy

➢ Mean bond enthalpy

Bond dissociation enthalpy:
Change in enthalpy when 1 mole of covalent bond of a gaseous
covalent compound is broken to form products in the gas phase.
Example
For diatomic molecules
Cl2 (g) → 2Cl (g); C-ClH = 242 kJ mol–1
O2 (g) → 2O (g); O-OH = 428 kJ mol–1
Mean bond enthalpy
CH4 (g) → C (g) + 4H (g);
It is the mean value of bond enthalpies. aH = 1665 kJ mol–1
Example For polyatomic molecules
CH4 (g) → CH3 (g) + H (g); bondH = +427 kJ mol–1

CH3 (g) → CH2 (g) + H (g); bondH = +439 kJ mol–1

CH2 (g) → CH (g) + H (g); bondH = +452 kJ mol–1
CH (g) → C (g) + H (g); bondH = +347 kJ mol–1
CH4 (g) → C (g) + 4H (g); bondH = 1665 kJ mol–1
Mean bond enthalpy
CH4 (g) → C (g) + 4H (g);
It is the mean value of bond enthalpies. aH = 1665 kJ mol–1
Example For polyatomic molecules

To calculate C-H H ,we use mean bond enthalpy of C-H bond

𝟏
C-H H =
 H
𝟒 a

= 𝟏
×1665 kJ mol–1
𝟒
=
416 kJ mol–1
Calculation of enthalpy of reaction from B.E data

68
The B.E of gaseous H2, Cl2, and HCl are 104, 58, and 103 kcal/mol respectively. Calculate
the enthalpy of formation of HCl(g)?

69
The B.E of C=C, C-C, C-H and H-H are 147, 83, 99, and 104 kcal/mol. Calculate the heat
of hydrogenation of ethene?

70
The dissociation energy of CH4 is 360 kcal/mol and dissociation energy of ethane is 620
kcal/mol. Calculate the B.E of C-C.

71
Calculate the enthalpy of formation of ammonia from the following bond energy data.
N-H bond = 389 kJ/mol; H-H bond = 435 kJ/mol; N=N bond = 945.36 kJ/mol.

72
Calculate the standard enthalpy change (in kJ mol–1) for the reaction
H2(g) + O2(g) → H2O2(g),
given that bond enthalpy of H–H, O=O, O–H and O–O (in kJ mol–1) are respectively
438, 498, 464 and 138.
(a) –130 (b) –65 (c) +130 (d) –334

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Given that bond energies of H–H and Cl–Cl are 430 kJ mol–1 and 240 kJ mol–1
respectively and ∆Hf HCl is – 90 kJ mol–1, bond enthalpy of HCl is
(a) 380 kJ mol–1 (b) 425 kJ mol–1 (c) 245 kJ mol–1 (d) 290 kJ mol–1

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