Thermochemistry

It’s the branch of physical chemistry which deals with the thermal or heat changes caused by chemical
reactions. When the internal of reactants (Er) is greater than the internal energy of the products (Ep),
the difference of internal energy,∆𝐸, is released as heat energy.
∆𝐸 = 𝐸𝑝𝑟𝑜𝑑𝑢𝑐𝑡 − 𝐸𝑟𝑒𝑎𝑐𝑡𝑎𝑛𝑡𝑠
∆𝐸 = 𝐸𝑝 − 𝐸𝑟.
Such a reaction is called exothermic reaction. If the internal energy of the products (Ep) is greater than
that of products (Er), heat is absorbed from the surroundings. Such a reaction is called endothermic
reaction. The amount of heat released or absorbed in a chemical reaction is termed the heat of reaction.
The energy changes in chemical reactions are largely due to the breaking of existing bonds between the
atoms and the formation of new bonds. Thus thermochemistry provides useful information regarding
the bond energies.
UNITS OF ENERGY
The energy is expressed as the calorie or joule.
1cal=4.184J
Enthalpy of a reaction.
Thermochemical measurements are made either at a) constant volume or b) constant pressure. The
magnitudes of changes observed under the two conditions are different. The change in internal energy
(∆E) is the heat change accompanying a chemical reaction at constant volume because no external work
is performed. However, at constant pressure not only does the change in internal energy take place but
work is also involved because of expansion or contraction. The enthalpy of a system is defined as the
sum of the internal energy and the product of its pressure and volume.
H=E+PV
Where E= internal energy, P=pressure, V=volume. It is also called heat content.
Just like internal energy, enthalpy is also a function of the state and it is not possible to measure its
absolute value. However a change in enthalpy (∆𝐻) accompanying a process can be measured
accurately.
∆𝐻 = 𝐻𝑝𝑟𝑜𝑑𝑢𝑐𝑡𝑠 − 𝐻𝑟𝑒𝑎𝑐𝑡𝑎𝑛𝑡𝑠.
∆𝐻 = 𝐻𝑝 − 𝐻𝑟
∆𝐻 = ∆𝐸 + 𝑃 × ∆𝑉 = ∆𝐸 + ∆𝑁𝑅𝑇
Therefore, while the heat change in a process is equal to its change in internal energy ∆E at constant
volume, it gives at constant pressure the enthalpy change ∆H.
Heat of reaction or enthalpy of reaction can be defined as the enthalpy change in the transformation of
the reactants at a given temperature and pressure into the products at the same temperature and
pressure. The heat change accompanying a reaction taking place at 298k and one atmospheric pressure
is called the standard heat change or standard enthalpy change. It is denoted by∆𝐻°.
EXOTHERMIC AND ENDOTHERMIC REACTIONS.
An evolution of heat to surroundings means that the system as a whole loses heat and therefore,
finishes with lower energy content. That is why ∆H is written as negative for an exothermic reaction and
positive for an endothermic one.
 Reactants

∆𝑯 > 𝟎 Products
Increasing Heat released ∆𝑯 > 𝟎
enthalpy Heat
absorbed

Products Reactants

Exothermic Endothermic

Enthalpy diagram for an exothermic and endothermic reaction

QUESTION
The heat of combustion of ethylene at 𝟏𝟕℃ and at constant volume is – 𝟑𝟑𝟐. 𝟏𝟗𝑲 Cal. Calculate the
heat of combustion at constant pressure considering water to be in liquid state (R=2Cal
𝒅𝒆𝒈𝒓𝒆𝒆−𝟏 𝒎𝒐𝒍𝒆−𝟏 )
Soln
𝑦𝑖𝑒𝑙𝑑𝑠
𝐶2 𝐻4 (𝑔) + 3𝑂2 (𝑔) → 2𝐶𝑂2 (𝑔) + 2𝐻2 𝑂(𝑙)
1mole 3moles 2moles negligible volume
No. of moles of the products is=2
No. of moles of reactants=4
∆𝑛 = 2.4 =−2
∆𝐻 = ∆𝐸 + ∆𝑛𝑅𝑇, ∆𝐸 = −332.19 𝐾𝐶𝑎𝑙

T= 273 +17 = 290K, R= -2Cal= 2 × 10−3Kcal

∆𝐻 = −332.19 + 2 × 10−3 × −2 × 290

= −333.3 𝑘𝑐𝑎𝑙
VARIATION OF HEAT (ON ENTHALPY WITH TEMPERATURE)
Change In heat of reaction at constant volume per degree change in temperature is equal to the
difference in heat capacities at constant volume of products and reactants
∆𝐸2 − ∆𝐸1 = ∆𝐶𝑉 ( 𝑇2 − 𝑇1 ) … … . (𝑖)
Change in heat of reaction at constant pressure per degree change of temperature is equal to difference
in heat capacities of products and reactants at constant pressure
∆𝐻2 − ∆𝐻1 = ∆𝐶𝑃 (𝑇2 − 𝑇1 ) … … … … … … (𝑖𝑖)
Equation (i) and (ii) are called Kirchoff’s equations
Question
1 1
The heat of reaction 2 𝐻2 + 2 𝐶𝑙2 → 𝐻𝐶𝑙 at 27°𝑐 𝑖𝑠 − 22.1 𝑘 𝑐𝑎𝑙. Calculate the heat of reaction at 77°𝑐.
The molar heat capacities at constant pressure at 22°𝑐 for hydrogen, chlorine and HCl are 6.82, 7.70 and
6.80 𝑐𝑎𝑙 𝑚𝑜𝑙𝑒 −1 𝑟𝑒𝑠𝑝𝑒𝑐𝑡𝑖𝑣𝑒𝑙𝑦
Soln
1 1
𝐻2 + 𝐶𝑙2 → 𝐻𝐶𝑙 ∆𝐻 = −22.1 𝑘 𝑐𝑎𝑙
2 2
∆𝐶𝑃 = 𝐻𝑒𝑎𝑡 𝑐𝑎𝑝𝑎𝑐𝑖𝑡𝑖𝑒𝑠 𝑜𝑓 𝑝𝑟𝑜𝑑𝑢𝑐𝑡𝑠 − 𝐻𝑒𝑎𝑡 𝑐𝑎𝑝𝑎𝑐𝑖𝑡𝑖𝑒𝑠 𝑜𝑓 𝑟𝑒𝑎𝑐𝑡𝑎𝑛𝑡𝑠
1 1
= 6.80 − { (6.82) + (7.70)}
2 2
−0.46 × 10−3 𝑘 𝑐𝑎𝑙
𝑇2 = 𝑁273 + 77 = 350, 𝑇1 = 273 + 27 = 350 𝐾
𝑇2 − 𝑇1 = 350 − 300 = 50 𝐾
∆𝐻2 − ∆𝐻1 = ∆𝐶𝑃 (𝑇2 − 𝑇1 )
∆𝐻2 = ∆𝐻1 + ∆𝐶𝑃 (𝑇2 − 𝑇1
=−22.1 + −0.46 × 10−3 𝐾 𝑐𝑎𝑙 × 50
= −22.123 𝑘 𝑐𝑎𝑙
Given the following heats of reactions at 25℃.
C2H4 (g) + 3O2 (g) →2CO2 (g) +2H2O (l) ∆𝐻 = −337.3𝑘𝑐𝑎𝑙
2H2 (g) +O2 (g)→ 2H2O∆𝐻 = 136.6𝑘𝑐𝑎𝑙
2C2H6 (g) +7O2 (g)→ 4CO2 (g) +6H2O∆𝐻 = −745.6𝑘𝑐𝑎𝑙
Calculate for the reaction
C2H4 (g) +H2 (g)→ C2H6 (g) at 25
Soln:
2C2H4 (g) + 6O2 (g) →4CO2 (g) +4H2O (l) ∆𝐻 = −674.6𝑘𝑐𝑎𝑙
2H2 (g) +O2 (g)→ 2H2O∆𝐻 = 136.6𝑘𝑐𝑎𝑙
4CO2 (g) +6H2O→ 2C2H6 (g) + 7O2 (g) ∆𝐻 = −745.6𝑘𝑐𝑎𝑙
Adding
2C2H4 (g) +2H2 (g) →2C2H6 (g) ∆𝐻 = 65.6𝑘𝑐𝑎𝑙
C2H4 (g) +H2 (g) →C2H6 (g) ∆𝐻 = 32.8𝑘𝑐𝑎𝑙
Hence the heat of reaction of hydrogenation of C2H4 is +32.8𝑘𝑐𝑎𝑙.
DIFFERENT TYPES OF HEAT (ENTHALPY) OF REACTION.
a) Enthalpy of formation.
Enthalpy of formation is defined as the change involving the formation of a mole of compound from its
elements. If the elements are in their standard state the enthalpy of formation is called the standard
enthalpy of formation and is denoted by 𝐻𝑓°.
The standard enthalpy of formation of a compound is equal to the enthalpy content of the compound. A
positive value of enthalpy of formation would mean that the compound is less stable than its elements.
The enthalpy of formation of a compound may be obtained either by direct measurement of ∆𝐻 for the
reaction or from enthalpies of reaction involving the compound. The change in enthalpy that takes place
when one mole of a compound is formed from its elements, all substances being in their standard states
(298k and 1 atm pressure) is called standard heat of formation.
Standard heat of reaction (∆𝑯°) from standard heat of formation (∆𝑯𝒇°).
The standard heat of reaction is equal to the standard heat of formation of products minus the standard
heat of formation of reactants.
∆H° = [total standard heat of formation of products − total standard heat of formation of reactants]
QUESTION.
The standard heat of formation of C2H5OH, CO2 (g), H2O (l) are −277.0, −393.5 𝑎𝑛𝑑 − 285.5𝐾𝐽𝑚𝑜𝑙-1
respectively. Calculate the standard heat change for the reaction.
C2H5OH+3 CO2 (l) →2 CO2 (g)+ 3 H2O (l)
Soln:
∆H° = ∆𝐻𝑓°(𝑝𝑟𝑜𝑑𝑢𝑐𝑡𝑠) − ∆𝐻𝑓°(𝑟𝑒𝑎𝑐𝑡𝑎𝑛𝑡𝑠)
∆H° = [2 × ∆H°f(CO2 (g)) + 3 × ∆H°f (H2O (l)) − ∆H°f(C2H5OH(l)) + 3∆H°f(O2(g))]
= [2 × (−395.5) + 3 × (−285.5)] − [−277.0 − 0]
= −1643.5 − (−277)
= −1366.5KJ
NOTE: the standard heat of formation of all elements i.e. O2, H2, etc. is assumed to be zero.
HEAT OF COMBUSTION
It’s defined as the change in enthalpy of system when one mole of the substance is completely burnt in
excess of air or oxygen. It is denoted by ∆𝐻𝐶 . The heat of combustion of a substance is always negative.
Heat energy is evolved during the process of combustion i.e. ∆𝐻𝐶 = −𝑣𝑒.
Application of the heat of combustion.
 Calculation of heat of formation.
Since the heats of combustion of organic compounds can be determined with considerable ease, they
are employed to calculate their heat of formation. The direct determination of these is often impossible.
 Calorific value of foods and fuels.
The calorific value is defined as the amount of heat produced in calories (or joules) when one gram of
substance is completely burnt. It is expressed in calg-1 or Jg-1.
 Deciding constitution.
Heat of combustion of organic compounds is to a large extend additive property, as shown by the fact
that in a homologous series the difference between the heats of combustion of successive members is
nearly constant and is equal 158cals. Constants corresponding to the heats of combustion of various
atoms and linkages have been worked out. The heat of combustion of an organic substance can be
calculated from its probable structural formula by adding up the values of the constants corresponding
to the atoms and linkages involved therein. If the value so obtained comes out to be the same as the
experimental value of the heat of combustion of the compound, the assumed formula must be correct.
Question.
Calculate the standard heat of formation of propane if its heat of combustion is -2220.2kJmol-1. The heat
s of formation of CO2 (g), and H2O (l) are -395.5 and -285.8KJmol-1 respectively.
Soln:
We have been given:
 (i) C3H8(g) + 5O2(g) →3CO2(g)+ 4H2O(l) ∆𝐻𝐶 = −2220.2𝐾𝐽
1
(ii) C(s)+ O2→CO2(g) ∆𝐻 = −393.5𝐾𝐽
2
1
(iii) H2(g) + O2(g) →H2O(l) ∆𝐻 = −285.8𝐾𝐽
2
We should manipulate this question in a way so as to get the required equation.
3C(s) + 4H2(s) → C3H8 (g) ∆𝐻𝐶 =?
Multiplying equation (ii) by 3 and equation (iii) by 4 and adding up we get
3C(s) +3 O2→ 3CO2 (g) ∆𝐻 = −1180.5𝐾𝐽
4H2 (g) +2O2 (g) →4H2O(l) ∆𝐻 = −1143.2𝐾𝐽
3C(s) + 4H2 (g) +5O2 (g) → 3CO2 (g)+ 4H2O(l) ∆𝐻 = −2323.7𝐾𝐽 (iv)
Subtracting equation (i) from equation (IV) we have
3C(s) + 4H2 (g) + 5O2 (g) → 3CO2 (g) + 4H2O(l)∆𝐻 = −2323.7𝐾𝐽 −
𝐶3 𝐻8 (𝑔) + 5𝑂2 (𝑔) → 3𝐶𝑂2 (𝑔) + 4𝐻2 𝑂(𝑙) ∆𝐻𝐶 = −2220.2𝐾𝐽
3𝐶(𝑠) + 4𝐻2 (𝑔) − 𝐶3 𝐻8 (𝑔) → 0 ∆𝐻 = −103.5
3𝐶(𝑠) + 4𝐻2 (𝑔) → 𝐶3 𝐻8 (𝑔) ∆𝐻 = −103.5
The heat of formation of propane is −103.5𝑘𝐽𝑚𝑜𝑙-1.
Heat of solution
The heat of solution is defined as the change in enthalpy when one mole of substance is dissolved in a
specified quantity of solvent at a given temperature. The change in enthalpy when a solution from one
specified concentration is diluted to some other specified concentration is called the integral enthalpy of
dilution. The differential enthalpy of a solution is defined as the enthalpy change when one mole of a
solute is added to such a large quantity of the solution that its concentration does not change
appreciably. It’s also known as the partial molar enthalpy of the solution. likewise, the enthalpy change
accompany the addition of a mole of solvent to a large volume of the solution without affecting the
concentration of the solution is referred to as the differential enthalpy of dilution or partial molar
enthalpy of dilution.
Question
Calculate the enthalpy of solution when a mole of HCl (g) dissolves in 100 moles of H2O (l) at 298k. Given
∆𝐻°𝑓(𝐻𝐶𝑙) = −92.3𝐾𝐽𝑚𝑜𝑙 −1 and∆𝐻°𝑓(𝐻𝐶𝑙. 100𝐻2 𝑂) = −168.1𝐾𝐽𝑚𝑜𝑙 −1 .
SOLN:
HCl (g) + 100H2O (l) →HCl.100H2O
∆H° = ∆𝐻𝑓°(𝑝𝑟𝑜𝑑𝑢𝑐𝑡𝑠) − ∆𝐻𝑓°(𝑟𝑒𝑎𝑐𝑡𝑎𝑛𝑡𝑠)
= [(−168.1) − (−92.3)]
= −75.8KJ
ENTHALPY OF NEUTRALIZATION
Enthalpy of neutralization is the enthalpy change when a mole of H3O+ ions in dilute solution are
completely neutralized by OH- ions forming undissociated water.
H3O+ (aq) + OH- (aq) →2H2O (l) ∆H° = −57.32KJmol-1
Dilute solutions should be used to avoid enthalpy changes due to mixing of acid and base if one mole of
a strong monoprotic acid (HCl, HNO3, etc.) is mixed with 1 mole of a strong base (NaOH, KOH, etc.),
neutralization takes place as shown above. Since these acids and bases are completely ionized in dilute
solutions, the enthalpy change is always -57.32KJmol-1.
For a weak acid or a weak base the enthalpy of neutralization is not the same, but less than -57.32KJ, in
these cases neutralization involves ionization of the acid or the base in addition to the neutralization.
The heat required to ionize the weak acid or the weak base is known as the Enthalpy of ionization.
Question.
The enthalpy of neutralization of ammonium hydroxide by the hydrochloric acid is−51.46KJmol-1.
Calculate the enthalpy of ionization of ammonium hydroxide.
Soln:
1. 𝑁𝐻3 (𝑎𝑞) + 𝐻 + (𝑎𝑞) = 𝑁𝐻4+ (𝑎𝑞) ∆𝐻 ° = 51.46 𝐾𝐽 𝑚𝑜𝑙 −1
2. 𝐻 + (𝑎𝑞) + 𝑂𝐻 − (𝑎𝑞) → 𝐻2 𝑂(𝑙) ∆𝐻 ° = −57.32 𝐾𝐽𝑚𝑜𝑙 −1
3. 𝑁𝐻3(𝑎𝑞) + 𝐻2 𝑂(𝑙) → 𝑁𝐻4+ (𝑎𝑞) + 𝑂𝐻 − (𝑎𝑞) ∆𝐻 ° = ∆𝐻 ° 𝐼𝑂𝑁𝐼𝑍𝐴𝑇𝐼𝑂𝑁
Adding equations 2 &3

𝑁𝐻3 (𝑎𝑞) + 𝐻 + = 𝑁𝐻4+ (𝑎𝑞)

𝐻 + (𝑎𝑞) + 𝑂𝐻 − (𝑎𝑞) → 𝐻2 𝑂(𝑙)

𝑁𝐻3 (𝑎𝑞) + 𝐻 + (𝑎𝑞) + 𝐻 + (𝑎𝑞) + 𝑂𝐻 − (𝑎𝑞) → 𝑁𝐻4+ (𝑎𝑞) + 𝐻2 𝑂(𝑙)

𝑁𝐻3 (𝑎𝑞) + 𝐻 + (𝑎𝑞) + 𝐻2 𝑂(𝑙) → 𝑁𝐻4+ (𝑎𝑞) + 𝐻2 𝑂(𝑙)
𝑁𝐻3 (𝑎𝑞) + 𝐻 + (𝑎𝑞) → 𝑁𝐻4+ (𝑎𝑞) ∆𝐻 ° = −57.32 + ∆𝐻 ° 𝑖𝑜𝑛𝑖𝑧𝑎𝑡𝑖𝑜𝑛
From equation 1 ∆𝐻 ° = −51.46 𝐾𝐽 𝑚𝑜𝑙 −1
(−57.32 + ∆𝐻 ° 𝑖𝑜𝑛𝑖𝑧𝑎𝑡𝑖𝑜𝑛 ) KJ𝑚𝑜𝑙 −1 = −51.46𝐾𝐽𝑚𝑜𝑙 −1
∆𝐻 ° 𝑖𝑜𝑛𝑖𝑧𝑎𝑡𝑖𝑜𝑛 = (−51.46 + 57.32)𝐾𝐽𝑚𝑜𝑙 −1
= 5.86𝐾𝐽𝑚𝑜𝑙 −1

ENTHALPIES OF FORMATION OF IONS IN SOLUTION

The enthalpy change in the formation of an ion at unit activity from its elements in aqueous solution is
known as the enthalpy of formation of the ion in solution.
The absolute value of enthalpy of formation of individual ions in aqueous solution is not possible but the
difference of enthalpy of formation of two ions can be obtained.
Enthalpy of Hydration
The enthalpy change when a mole of a hydrous salt combines with the required number of moles of
water to form a specified hydrate is known as the enthalpy of hydration.
Enthalpy of precipitation
Enthalpy change involved in the formation of a mole of precipitate upon mixing dilute solutions of
relevant electrolytes is called the enthalpy of precipitation.
Enthalpy of phase transition
Enthalpy change in the transformation of a substance from one sate to another is known as the enthalpy
of transition.
a) Heat of fusion
It is defined as the heat change (or enthalpy change) when one mole of a solid substance is converted
into the liquid state at its melting point.
𝐻2 𝑂(𝑠) → 𝐻2 𝑂(𝑙) ∆𝐻 = +1.43 𝑘 𝑐𝑎𝑙
Ice water
Greater the heat of fusion of a substance higher the magnitude of inter molecular forces.
b) Heat of vapourization
The heat of vapourization is defined as the heat change (or enthalpy change)When one mole of liquid is
considered into vapour or gaseous state at its boiling point.
𝐻2 𝑂(𝑙) → 𝐻2 𝑂(𝑔) ∆𝐻 = +9.71𝑘 𝑐𝑎𝑙
Water steam
c) Heat of sublimation
It’s the heat change (or enthalpy change)
When one mole of a solid is directly converted into gaseous state at a temperature below its melting
point,
𝐼2 (𝑠) → 𝐼2 (𝑔) ∆𝐻 = +14.92 𝑘 𝑐𝑎𝑙
Heat of transition
It’s the change in enthalpy which occurs when one mole of an element changes from one allotropic form
to another.
Diamond→ 𝑐𝑎𝑚𝑜𝑟𝑝ℎ𝑜𝑢𝑠 ∆= +3.3 𝑘 𝑐𝑎𝑙
Heat of atomization
The heat required to convert 1 gramme – a tom or 1 mol of an element from its normal state at 25°𝐶
and 1 atms into free atoms.
Heat of hydrogenation
The heat change when 1 mole of an unsaturated compound is completely converted into the
corresponding saturated compound by reaction with gaseous hydrogen
𝐶2 𝐻4 (𝑔) + 𝐻2 (𝑔) → 𝐶2 𝐻6 (𝑔) ∆𝐻 = −137.5𝐾𝐽.

The Heat of solution at infinite dilution

This is defined as the heat change when 1 mol of a substance dissolves in such a large volume of solvent
that addition of more solvent produces no further heat of dilution.
Hydration energy
The energy change accompanying the hydration of a mole of gaseous ions.
Lattice energy
Its energy liberated when one mole of an ionic crystal lattice is formed from its constituents ions in the
gaseous state.
𝑁𝑎+ (𝑔) + 𝐶𝑙 − (𝑔) → 𝑁𝑎+ 𝐶𝑙 − (𝑠) ∆𝐻 = 𝐿𝐴𝑇𝑇𝐼𝐶𝐸 𝐸𝑁𝐸𝑅𝐺𝑌
ENERGY RELATIONSHIPS DURING SOLUTIONS OF AN IONIC SOLID

+
𝑀(𝑔) + 𝐴+
(𝑔)
−∆𝐻𝑙𝑎𝑡𝑡 +
∆𝐻ℎ𝑦𝑑 = ∆𝐻𝑠𝑜𝑙

Lattice energy of
+
MA 𝑀(𝑎𝑞) + 𝐴−
(𝑎𝑞)

∆𝐻𝐿𝑎𝑡𝑡
∆𝐻 𝐻𝑒𝑎𝑡 𝑜𝑓 𝑆𝑜𝑙𝑢𝑡𝑖𝑜𝑛

𝑀+ 𝐴+
(𝑠)

Conditions governing values of heat of reactions

Most important amongst these conditions is to explicit of state the stoichiometric equations related to
the reaction. Other important conditions are:
a) Temperatures prevailing during the reaction
b) Physical allotropic and crystalline states of reactants and products of the reaction
c) Concentration and solvents of solution involved
d) Pressures of involved gases
e) Condition of constancy of pressure or volume.
Measurement of the heat of Neutralization of molar HCL with molar NaOH
One mole of HCl of known volume is allowed to react completely with one mole of NaOH of known
volume in calorimeter.
The rise in temperature is the determined using Beckmann thermometer as the reaction proceeds.
The values of temperature are plotted against the time.

𝑇2

Temperature

𝑇1

Time

Extrapolate to obtain final temperature i.e. 𝑇2

Rise in temperature ∆𝑇 = 𝑇2 − 𝑇1
Knowing the heat capacity of calorimeter together with the bulb of the thermometer immersed the
masses of the acid and alkali solutions and their specific heats, the amount of heat liberated can be
measured.
With the known volume and concentration of HCL solution, the heat of neutralization can be calculated.
If Q is the amount of heat evolved when V (ml) of HCL solution of concetration C g – equivalents/litre is
neutralized completely, then,

𝑄
Heat of neutralization= 1000× 𝑉 × 𝐶
Example
The measured heat of neutralization on per mole of acetic acid, formic acid, hydrocyanic acid and
hydrogen sulphide are 55.23, 56.07, 12.13 and 15.90 KJ/ mole respectively. Arrange the acids in
decreasing order of strength. Assume that for the reaction
𝐻 + + 𝑂𝐻 − → 𝐻2 𝑂, ∆𝐻 = 57.32 𝐾𝐽.

Soln
The acid with the lowest positive value of heat of ionization will be the strongest acid.
The neutralization of a weak acid can be regarded as a two-step reaction

∆𝐻(𝑛𝑒𝑎𝑢𝑡𝑟𝑎𝑙𝑖𝑠𝑎𝑡𝑖𝑜𝑛) = ∆𝐻(𝑖𝑜𝑛𝑖𝑧𝑎𝑡𝑖𝑜𝑛) + ∆𝐻(𝐻 + + 𝑂𝐻 − )

(i) Ionization
(ii) 𝐻 + + 𝑂𝐻 − → 𝐻2 𝑂
∆𝐻(𝑛𝑒𝑎𝑢𝑡𝑟𝑎𝑙𝑖𝑧𝑎𝑡𝑖𝑜𝑛)𝑓𝑜𝑟 𝑡ℎ𝑒 𝑎𝑐𝑖𝑑 ℎ𝑎𝑠 − 𝑣𝑒 𝑠𝑖𝑔𝑛
∆𝐻 𝑓𝑜𝑟 𝑖𝑜𝑛𝑖𝑧𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑎𝑐𝑒𝑡𝑖𝑐 𝑎𝑐𝑖𝑑 =
−55.23 − −57.32 = +2.09 𝐾𝐽/𝑚𝑜𝑙
∆𝐻 𝑓𝑜𝑟 𝑖𝑜𝑛𝑖𝑧𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑓𝑜𝑟𝑚𝑖𝑐 𝑎𝑐𝑖𝑑 =
−56.07 − −57.32 = +1.25 𝐾𝐽/𝑚𝑜𝑙.
∆𝐻 𝑓𝑜𝑟 𝑖𝑜𝑛𝑖𝑧𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 ℎ𝑦𝑑𝑟𝑜𝑔𝑒𝑛𝑖𝑐 𝑎𝑐𝑖𝑑 = −12.73 − −57.32 = +45.19𝐾𝐽/𝑚𝑜𝑙.
Formic acid, acetic acid, hydrogen sulphide.

CHARACTERISTICS OF ∆𝑯 𝑭𝑶𝑹 𝑨 𝑹𝑬𝑨𝑪𝑻𝑰𝑶𝑵

1. If a reaction is reversed, the sign of ∆𝐻 𝑖𝑠 𝑎𝑙𝑠𝑜 𝑟𝑒𝑣𝑒𝑟𝑒𝑠𝑒𝑑.
2. The magnitude of ∆H is directly proportional to the quantities of reactants and productions in a
reaction. If the coefficients in a balanced reaction are multiplied by an integer, the value of ∆H is
multiplied by the same integer.

THE LAWS OF THERMOCHEMISTRY

1. LAVOISIER AND LAPLACE
It states that, the quantity of heat which must be supplied to decompose a compound into its elements
is equal to the heat evolved when the compound is formed from its elements.
Diamond → amorphous ∆𝐻 = +3.3 𝑘 𝑐𝑎𝑙

2) HESS’S LAW OF CONSTANT HEAT SUMMATION

The law states that the total enthalpy change for a reaction is the same whether the reaction takes place
in a single step or in several steps.

∆𝐻1
A+B C+D

∆H1 = ∆H2 + ∆H3

∆𝐻2 ∆𝐻3

From the thermochemical reactions calculate the enthalpy change for the reaction
3 3
𝐶𝐻3(𝑔) + 𝐶𝑙2 (𝑔) → 𝐶𝐻𝐶𝑙3 (𝑙) + 𝐻2 (g)
2 2

5 1 3
(i) 𝐶𝐻𝐶𝑙3 (𝑔) + 𝑂2 (𝑔) → 𝐶𝑂2 (𝑔) + 𝐻2 𝑂(𝑙) + 𝐶𝑙2 (𝑔) ∆𝐻 ° = −373.3 𝐾𝐽
4 2 2
(ii) 𝐶(𝑔𝑟𝑎𝑝ℎ𝑖𝑡𝑒) + 2𝐻2(𝑔) → 𝐶𝐻4)(𝑔) ∆𝐻 ° = −74.8 𝐾𝐽
(iii) 𝐶(𝑔𝑟𝑎𝑝ℎ𝑖𝑡𝑒) + 𝑂2 (𝑔) → 𝐶𝑂2 (𝑔) ∆𝐻 ° = −393.5 𝐾𝐽
1
(iv) 𝐻2 (𝑔) + 2 𝑂2 (𝑔) → 𝐻2 𝑂(𝑙) ∆𝐻 ° = −285.8𝐾𝐽
SOLN
1 1
On substracting eqn (iii) from eqn (i) we get (v) 𝐶𝐻𝐶𝑙3 (𝑙) + 4 𝑂2 (𝑔) − 𝐶(𝑔𝑟𝑎𝑝ℎ𝑖𝑡𝑒) → 2 𝐻2 𝑂(𝑙) +
3
𝐶𝑙 (𝑔) ∆𝐻 ° = +20.2 𝐾𝐽
2 2

Now adding equation (ii) to equation (v) gives

1 1 3
(VI) 𝐶𝐻𝐶𝑙3 (𝑙) + 𝑂2 (𝑔) + 2𝐻2 (𝑔) → 𝐶𝐻4(𝑔) + 𝐻2 𝑂(𝑙) + 𝐶𝑙2 (𝑔) ∆𝐻 ° = −54.6𝐾𝐽
4 2 2
Dividing equation (iv) by 2 and substracting it from (vi), we get
 3 3
𝐶𝐻𝐶𝑙3 (𝑙) + 𝐻2 (𝑔) → 𝐶𝐻4 (𝑔) + 𝐶𝑙2 (𝑔) ∆𝐻 ° = 88.3 𝐾𝐽
2 2

OR

3 3
𝐶𝐻4 (𝑔) + 𝐶𝑙2 (𝑔) → 𝐶𝐻𝐶𝑙3 (𝑙) + 𝐻2 (𝑔) ∆𝐻 ° = −88.3 𝐾𝐽
2 2

APPLICATIONS OF HESS’S LAW

(1) Determination of heat of formation of substances which otherwise cannot be measured

experimentally
The substances like methane, carbon monoxide, benzene, etc. cannot be prepared by uniting their
elements. Therefore, it is not possible to measure the heats of formation of such compounds directly.
These can be determined indirectly by using HESS’S law.
(2) Determination of heat of transition
The heat of transition of one allotropic form to another can also be calculated with the help of HESS’S
law.
(3) Determination of heats of various reactions
By using HESS’S law we can calculate the heats or enthalpies of many reactions wehich otherwise cannot
be measured directly.

BOND ENERGY
The bond energy is defined as the average amount of energy required into break all bonds of a
particular type in one mole of substance. Bond energy is a measure of strength of the bond. In other
words, bond energy is the force with which the atoms are bonded together. It depends upon the size of
the atom, Electronegativity, Bond length.
Knowledge of bond enthalpy is useful for calculating heats of reaction for gaseous reactions for which
no thermal data is available and which involve substances having covalent bonds.

Enthalpy
(heat
content) Energy supplied to Energy released as bonds form
break bonds

Reactants

Energy difference=∆𝐻
heat of reaction

Products

Reaction path
The bond breaking is an endothermic process whereas bond making is exothermic, Assume 𝐸(𝐶 − 𝐻) =
415.5𝐾𝐽

2𝐶(𝑔) + 6𝐻(𝑔)

∅
∆𝐻𝑎𝑡 (𝐶2 𝐻6 (𝑔)) =
∅
6× ∆𝐻𝑎𝑡 (𝐻2(𝑔) ) = 6 × 218𝐾𝐽 𝐸(𝐶 − 𝐶) + 6𝐸(𝐶 − 𝐻)

2𝐶(𝑔) + 3𝐻2(𝑔)

∅
2× ∆𝐻𝑎𝑡 (𝐶(𝑔𝑟𝑎𝑝ℎ𝑖𝑡𝑒) )= 2× 715𝐾𝐽

2𝐶(𝑔𝑟𝑎𝑝ℎ𝑖𝑡𝑒) + 3𝐻2(𝑔)

∆𝐻𝑓∅ (𝐶(𝑔𝑟𝑎𝑝ℎ𝑖𝑡𝑒) ) = −85𝐾𝐽

𝐶2 𝐻6 (𝑔)

Equating the two routes 𝐶2 𝐻6 (𝑔) 𝑡𝑜 2𝐶(𝑔) + 6𝐻(𝑔)

𝐸(𝐶 − 𝐶) + 6𝐸(𝐶 − 𝐻) = −(−85) + 2 × 715 + 6 × 218𝐾𝐽
𝐸(𝐶 − 𝐶) + 6 × 415.5 = 2823𝐾𝐽
𝐸(𝐶 − 𝐶) = 2823 − 2493 = 330𝐾𝐽
Therefore, The 𝑐 − 𝑐 bond energy in ethane, 𝐸(𝐶 − 𝐶) = 330𝐾𝐽𝑚𝑜𝑙 −1

Uses of bond energies

1. Comparing the strengths of bonds
2. Understanding structure and bonding
3. Estimating the enthalpy changes in reactions
4. Understanding the chemical reaction mechanisms
Energy changes in forming ionic substances
The lattice energy of an ionic crystal is the heat of formation for one mole of the ionic compound from
gaseous ions under standard conditions.
𝑁𝑎+ (𝑔) + 𝐶𝑙 − (𝑔) → 𝑁𝑎+ 𝐶𝑙− (𝑠) ∆𝐻𝑙𝑎𝑡𝑡 (𝑁𝑎+ 𝐶𝑙 − (𝑠)
BORN- HABER CYCLE
It’s a thermodynamic cycle derived by application of HESS’S law. Commonly used to calculate lattice
energies of ionic solids and average bond energies of covalent compound e.g. for NaCl.

𝑁𝑎+ (𝑔) + 𝑒 − + 𝐶𝑙 − (𝑔)

∆Hat (Cl2(g) ) = +121KJ ∆𝐻𝑒 (𝐶𝑙) = −364𝐾𝐽

+ 1
𝑁𝑎(𝑔) + 𝑒 − + 2 𝐶𝑙2(𝑔)
+ −
𝑁𝑎(𝑔) + 𝐶𝑙(𝑔)
∆𝐻𝑖 (𝑁𝑎(𝑔)) = +500𝐾𝐽

−
∆𝐻𝑙𝑎𝑡𝑡 (𝑁𝑎+ 𝐶𝑙(𝑠) ) =?

1
𝑁𝑎(𝑔) + 𝐶𝑙2 (𝑔)
2

∆𝐻𝑎𝑡 (𝑁𝑎(𝑠)) = +108𝐾𝐽

1
𝑁𝑎(𝑠) + 2 𝐶𝑙2(𝑔)

−
∆𝐻𝑓 (𝑁𝑎+ )𝐶𝑙(𝑠) ) = −411𝐾𝐽

The heat of atomization of sodium

∆𝐻𝑎𝑡 (𝑁𝑎(𝑠))
𝑁𝑎(𝑠) → 𝑁𝑎(𝑔)
The first ionization energy of sodium
∆𝐻1 (𝑁𝑎(𝑔) )
+
𝑁𝑎(𝑔) → 𝑁𝑎(𝑔) + 𝑒−
The heat of atomization of chlorine
1 ∆𝐻𝑎𝑡 (𝐶𝑙2(𝑔) )
𝐶𝑙2(𝑔) → 𝐶𝑙(𝑔)
2
The electron affinity of chlorine
∆𝐻𝑒 (𝐶𝑙)
−
𝐶𝑙(𝑔) + 𝑒 − → 𝐶𝑙(𝑔)
The lattice energy for sodium chloride can be obtained.
−
∆𝐻𝑙𝑎𝑡𝑡 (𝑁𝑎+ 𝐶𝑙(𝑠) ) = (364 − 121 − 500 − 108 − 411)𝐾𝐽𝑚𝑜𝑙 −1 = −776𝐾𝐽𝑚𝑜𝑙 −1

SODIUM HYDRATION AND LATTICE ENERGY

When sodium chloride is dissolved in water, the overall change can be represented as:
+ −
𝑁𝑎𝐶𝑙(𝑔) + (𝑎𝑞) → 𝑁𝑎(𝑎𝑞) + 𝐶𝑙(𝑎𝑞)
For one mole of the solute and the formation of an infinite dilute solution, this process is described as
the heat of solution of sodium chloride.
∆𝐻𝑠𝑜𝑙𝑛 (𝑁𝑎𝐶𝑙(𝑠) )
∆𝐻(𝑠𝑜𝑙𝑛) (𝑁𝑎𝐶𝑙(𝑠) ) = +5𝐾𝐽
The separation of the solid ionic crystal into monatomic gaseous ions is the reverse of the lattice energy
process:
+
𝑁𝑎𝐶𝑙(𝑠) → 𝑁𝑎(𝑔) − ∆𝐻𝑙𝑎𝑡𝑡 = +776𝐾𝐽
The solvation (hydration) of these gaseous ions by water molecules is known as the hydration energy
∆𝐻ℎ𝑦𝑑
+ − + −
𝑁𝑎(𝑔) + 𝐶𝑙(𝑔) + 𝑎𝑞 → 𝑁𝑎(𝑎𝑞) + 𝐶𝑙(𝑎𝑞)

The relationship between lattice energy, hydration energy and heat of solution of sodium chloride

+ −
𝑁𝑎(𝑔) + 𝐶𝑙(𝑔) + (𝑎𝑞)

−∆𝐻(𝑙𝑎𝑡𝑡) (𝑁𝑎𝐶𝑙(𝑠) ) = 776𝐾𝐽 ∆𝐻ℎ𝑦𝑑 = −771𝐾𝐽

+ −
𝑁𝑎(𝑎𝑞) + 𝐶𝑙(𝑎𝑞)

∆𝐻𝑠𝑜𝑙𝑛 (𝑁𝑎𝐶𝑙(𝑠) ) = +5𝐾𝐽

𝑁𝑎𝐶𝑙(𝑠) + 𝑎𝑞

∆𝐻𝑠𝑜𝑙𝑛 = −∆𝐻𝑙𝑎𝑡𝑡 + ∆𝐻ℎ𝑦𝑑

Q
Calculate the bond enthalpy for a 𝐶 − 𝑂 bond in methanol from the following data:
1
I. 𝐶(𝑠) + 2𝐻2(𝑔) + 2 𝑂2(𝑔) → 𝐶𝐻3 𝑂𝐻(𝑔) ∆𝐻 ° = −200.0𝐾𝐽𝑚𝑜𝑙 −1
II. 𝐶(𝑔) → 𝐶(𝑔) ∆𝐻 ° = 716.8𝐾𝐽𝑚𝑜𝑙 −1
III. 2𝐻2(𝑔) → 4𝐻(𝑔) ∆𝐻 ° = 872.0𝐾𝐽𝑚𝑜𝑙 −1
 1
IV. 𝑂
2 2(𝑔)
→ 𝑂(𝑔) ∆𝐻 ° = 249.0𝐾𝐽𝑚𝑜𝑙 −1

Solution
The enthalpy change for the dissociation of 𝐶𝐻3 𝑂𝐻(𝑔) 𝑖𝑠 𝑔𝑖𝑣𝑒𝑛 𝑎𝑠:
𝐶𝐻3 𝑂𝐻(𝑔) → 𝐶(𝑔) + 4𝐻(𝑔) + 𝑂(𝑔) ∆𝐻 ° =?
Adding (ii) (iii) and (IV) gives:
1
𝐶(𝑠) + 2𝐻2(𝑔) + 𝑂2(𝑔) → 𝐶(𝑔) + 4𝐻(𝑔) + 𝑂(𝑔) ∆𝐻 ° = 1837.9𝐾𝐽
2
Now substracting equation (i) from the above equation gives:
𝐶𝐻3 𝑂𝐻(𝑔) → 𝐶(𝑔) + 4𝐻(𝑔) + 𝑂(𝑔) ∆𝐻 ° = 2037.8𝐾𝐽
Since in methanol, there are three 𝐶 − 𝐻 𝑎𝑛𝑑 𝑜𝑛𝑒 𝐶 − 𝑂 and one 𝑂 − 𝐻 bonds
Therefore,
° ° °
2037.8KJ=3(∆𝐻𝐶−𝐻 + 1(∆𝐻𝑂−𝐻 ) + 1(𝐻𝐶−𝑂 ))
°
2037.8𝐾𝐽 = 3(415) + 463.6 + ∆𝐻𝐶−𝑂
°
(∆𝐻𝐶−𝑂 ) = 329.0𝐾𝐽𝑚𝑜𝑙 −1

MEAN BOND ENTHALPY

This is the average value of energy which results from the breaking and making of chemical bonds during
a reaction.
Explanation on how Bond Enthalpy Data can be used to determine the enthalpy change in a reaction.
The change in enthalpy ∆𝐻 represents the net change per mole.
The energy must be supplied to break the covalent bonds in the reaction and the energy must be given
out when the bonds in the product formed.
The standard enthalpy of a reaction is the difference between the sum of the average standard bond
enthalpies of the product and the sum of the average standard bond enthalpies of the reactants.
Illustration:
Given that the average standard of enthalpies 𝐶 − 𝐶 and 𝐶 − 𝐻 bonds are 348 and 413 𝐾𝐽𝑚𝑜𝑙 −1
1
respectively and the standard enthalpies of carbon and 2 ℎ𝑦𝑑𝑟𝑜𝑔𝑒𝑛 𝑔𝑎𝑠 are 718 and
218 𝐾𝐽𝑚𝑜𝑙 −1 𝑟𝑒𝑠𝑝𝑒𝑐𝑡𝑖𝑣𝑒𝑙𝑦. 𝐶𝑎𝑙𝑐𝑢𝑙𝑎𝑡𝑒 𝑡ℎ𝑒 𝑠𝑡𝑎𝑛𝑑𝑎𝑟𝑑 𝑒𝑛𝑡ℎ𝑎𝑙𝑝𝑦 𝑜𝑓 𝑓𝑜𝑟𝑚𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑒𝑛𝑡ℎ𝑎𝑛𝑒.
Solution
2C(graphite) + 3H2(g) → C2 H6(g) ∆H =?
C(graphite) → 𝐶(𝑔) ∆𝐻 = +718𝐾𝐽𝑚𝑜𝑙 −1
1
𝐻 →𝐻 ∆𝐻 = +218 𝐾𝐽𝑚𝑜𝑙 −1
2 2(𝑔)
𝐻2(𝑔) → 2𝐻(𝑔) ∆𝐻 = +2 × 218 = 436 𝐾𝐽𝑚𝑜𝑙 −1
C(graphite) → 2𝐶(𝑔) ∆𝐻 = +718 × 2 = +1436 𝐾𝐽𝑚𝑜𝑙 −1
3H2(g) → 6𝐻(𝑔) ∆𝐻 = +436 × 3 = +1308 𝐾𝐽𝑚𝑜𝑙 −1
Thus the total energy required to break bonds in forming a molecule of ethane from graphite and
hydrogen is sum of (i) and (ii) = 1436 + 1308 = +2742𝐾𝐽𝑚𝑜𝑙 −1
Now the bond energy of 𝐶 − 𝐶 𝑖𝑠 − 348𝐾𝐽𝑚𝑜𝑙 −1
(iii)
H H

H C C H 6 𝐶 − 𝐻 𝑎𝑛𝑑 𝐶 − 𝐶

H H
So the energy of 6(𝐶 − 𝐻) bond is 6 × (−413) = −2478𝐾𝑗𝑚𝑜𝑙 −1 (iv)
Thus the total energy released by making bonds in a molecule of ethane is the sum of (iii) and (iv)
= −348 ± 2478 = −2826𝐾𝐽𝑚𝑜𝑙 −1
Therefore, The net energy change = +2742 + −2826 = −84𝐾𝐽𝑚𝑜𝑙 −1
Hence the heat of formation of ethane is −84𝐾𝐽
ENERGY LEVEL DIAGRAMS OF REACTIONS
Energy changes in reactions can be shown in diagrams where the X- axis is used to indicate the enrgy.
Thus the reaction 𝐶𝐻4 + 𝐶𝑙2 → 𝐶𝐻3 𝐶𝑙 + 𝐻𝐶𝑙 ∆𝐻298 =
−98.3𝐾𝐽 𝑡𝑎𝑘𝑖𝑛𝑔 𝑝𝑙𝑎𝑐𝑒 𝑢𝑛𝑑𝑒𝑟 1 𝑎𝑡𝑚 𝑎𝑡 25℃ 𝑐𝑎𝑛 𝑏𝑒 𝑟𝑒𝑝𝑟𝑒𝑠𝑒𝑛𝑡𝑒𝑑 𝑎𝑠 𝑏𝑒𝑙𝑜𝑤

𝐶𝐻4 + 𝐶𝑙2
Reactants

°
∆𝐻295 = −98.3𝐾𝐽
Energy
𝐶𝐻3 𝐶𝑙 + 𝐻𝐶𝑙

Progress of reaction
Example
Calculate ∆𝐻 for the synthesis of diborane from its elements, according to the equation
2𝐵(𝑠) + 3𝐻2(𝑔) → 𝐵2 𝐻6(𝑔)
Using the following data
3
2𝐵(𝑠) + 𝑂2(𝑔) → 𝐵2 𝑂2(𝑠) ∆𝐻 = −1273𝐾𝐽
2
𝐵2 𝐻6(𝑔) + 3𝑂2(𝑔) → 𝐵2 𝑂2(𝑠) + 3𝐻2 𝑂(𝑔) ∆𝐻 = −2035𝐾𝐽
1
𝐻2(𝑔) + 𝑂2(𝑔) → 𝐻2 𝑂(𝑙) ∆𝐻 = −286𝐾𝐽
2
𝐻2 𝑂(𝑙) ∝→ 𝐻2 𝑂(𝑔) ∆𝐻 = −44𝐾𝐽

Soln
2𝐵(𝑠) + 3𝐻2(𝑔) → 𝐵2 𝐻6(𝑔) ∆𝐻 = +132𝐾𝐽
The ∆𝐻 for the synthesis of 1 mole diborane from the elements is +36𝐾𝐽.

Q.
Calculate the heat of the reaction
𝐶𝑂2(𝑔) + 𝐻2(𝑔) → 𝐶𝑂(𝑔) + 𝐻2 𝑂(𝑔)
Under standard conditions from the following data at 25℃
 1
𝐶 + 𝑂2 → 𝐶𝑂 ∆𝐻𝐹° = −110.5𝐾𝐽𝑚𝑜𝑙 −1
2
𝐶 + 𝑂2 → 𝐶𝑂2 ∆𝐻𝐹° = −393.7𝐾𝐽𝑚𝑜𝑙 −1

Soln.

𝐶 + 𝐻2

1
𝑂
2 2
∆𝐻𝑓° = −110.5
𝑂2 ∆𝐻𝑓 = −393.7
𝐶𝑂 + 𝐻2
𝐶𝑂2 + 𝐻2

∆𝐻𝑓° = −241.8

1
𝑂
2 2
? ∆𝐻𝑓 =?

𝐶𝑂 + 𝐻2 𝑂

−393.7 + ∆𝐻𝑓 = −110.5 + 241.8

∆𝐻𝑓 = 393.7 − (110.5 + 241.8)
393.7 − 352.3
= 41.4 𝐾𝐽𝑚𝑜𝑙 −1

BOND ENERGY
The bond energy is defined as the average amount of energy required to break all bonds of a particular
type in one mole of the substance.
How bond energy can be used to calculate the heat of reaction for gaseous
Knowledge of bond enthalpy is useful for calculating heat of reaction for gaseous reactions for which no
thermal data is available and which involve substance having covalent bonds.
Suppose we desire to determine the bond energy of 𝐶 − 𝐻 bond in methane. For this purpose we need
to know the enthalpy change for the reaction.
𝐶(𝑔) + 4𝐻(𝑔) → 𝐶𝐻4(𝑔)
This is obtained by combining the heat of formation of methane from 𝐶(𝑔) + 𝐻2(𝑔) with the heat of
sublimation of carbon i.e.𝐶(𝑠) → 𝐶(𝑔) and the heat of dissociation of hydrogen into atoms I.e. 𝐻2(𝑔) →
2𝐻(𝑔) , which can be determined by spectroscopic methods. The value so obtained represents the bond
energy of four 𝐶 − 𝐻 bonds. Since all the bonds in methane are identical the bond energy of 𝐶 − 𝐻
bond is obtained by dividing this value by four.

FACTORS WHICH BOND ENERGY DEPEND ON:

1. Size of the atom
2. Electronegativity
3. Bond length
Q.
Calculate the bond enthalpy for a 𝐶 − 𝑂 bond in methanol from the following data:
 1
I. 𝐶(𝑠) + 2𝐻2(𝑔) + 2 𝑂2(𝑔) → 𝐶𝐻2 𝑂𝐻(𝑔) ∆𝐻 ° = 200.0𝐾𝐽𝑚𝑜𝑙 −1
II. 𝐶(𝑠) → 𝐶(𝑔) ∆𝐻 ° = 716.8 𝐾𝐽𝑚𝑜𝑙 −1
III. 2𝐻2(𝑔) → 4𝐻(𝑔) ∆𝐻 ° = 872.0 𝐾𝐽𝑚𝑜𝑙 −1
1
IV. 𝑂
2 2(𝑔)
→ 𝑂(𝑔) ∆𝐻 ° = 249.0 𝐾𝐽𝑚𝑜𝑙 −1
The bond enthalpy for 𝐶 − 𝐻 bond is 413 𝐾𝐽𝑚𝑜𝑙 −1 and for 𝑂 − 𝐻 bond is 463.6 𝐾𝐽𝑚𝑜𝑙 −1
SOLN
The enthalpy change for the dissociation of 𝐶𝐻3 𝑂𝐻(𝑔)𝑖𝑠 𝑔𝑖𝑣𝑒𝑛 𝑏𝑦
𝐶𝐻3 𝑂𝐻(𝑔) → 𝐶(𝑔) + 4𝐻(𝑔) + 𝑂(𝑔) ∆𝐻 ° =?
1
Adding (ii), (iii) and (iv) gives 𝐶(𝑠) + 2𝐻2(𝑔) + 2 𝑂2(𝑔) → 𝐶(𝑔) + 4𝐻(𝑔) + 𝑂(𝑔) ∆𝐻 ° =?
Now substracting eqn (i) from 1837.9KJ from the above eqn gives
𝐶𝐻3 𝑂𝐻(𝑔) → 𝐶(𝑔) + 4𝐻(𝑔) + 𝑂(𝑔) ∆𝐻 ° = 2037.8 𝐾𝐽𝑚𝑜𝑙 −1
Since in methanol, there are three 𝐶 − 𝐻, and one 𝑂 − 𝐻 bonds, therefore,
° ° ° °
2037.8𝐾𝐽𝑚𝑜𝑙 −1 = 3[∆𝐻𝐶−𝐻 ] + 1[∆𝐻𝑂− 𝐻
] + 1[∆𝐻𝐶−𝑂 ] = 3(415) + (463.6) + ∆𝐻𝐶−𝑂
°
∆𝐻𝐶−𝑂 = 329.0 𝐾𝐽𝑚𝑜𝑙 −1

Explain the following statements

‘’The neutralization of all bases by acids do not necessarily occur at 𝑃𝐻 7’’
Neutralization is the reaction between an acid and a base. The product is a salt. At the
equivalence point, no excess acid or base is present and it is thus expected that the 𝑃𝐻 may be
7. However, due to hydrolysis, acid or base may be different from 7. For example, in case of
titration of weak acid by a strong base the salt contains the conjugate strong base which
accepts proton from water producing basic solution.
Similarly, in case of titration of weak acid and weak base, both the cation and anion are strong
conjugate base and acid respectively and thus they will interact with water. The 𝑃𝐻 depends on
degree of interaction.
However, if we consider the neutralization of a strong base and strong acid, then the salt
contains cations and anions, which are weak conjugates and there will be no interaction and 𝑃𝐻
remains 7.
PROPERTY OF ENTHALPY WHICH MAKES HESS’S LAW POSSIBLE
Because ∆𝐻 ° is a state function dependent only on the initial and final states and is independent of the
path, the values of ∆𝐻 ° for both routes should be identical.
The results obtained using different routes are chemically and thermochemically identical which shows
that ∆𝐻 ° is a state function.

Q.
Given the following heats of reaction at 25℃
𝐶2 𝐻4(𝑔) + 3𝑂2(𝑔) → 2𝐶𝑂2(𝑔) + 2𝐻2 𝑂(𝑙) ∆𝐻 = −337.3 𝑘𝑐𝑎𝑙
2𝐻2(𝑔) + 𝑂2(𝑔) → 2𝐻2 𝑂(𝑙) ∆𝐻 = −136.6 𝑘𝑐𝑎𝑙
2𝐶2 𝐻6(𝑔) + 7𝑂2(𝑔) → 4𝐶𝑂2(𝑔) + 6𝐻2 𝑂(𝑙) ∆𝐻 = −745.6
Calculate the ∆𝐻 for the reaction
𝐶2 𝐻4(𝑔) + 𝐻2(𝑔) → 𝐶2 𝐻6(𝑔) at 25℃
SOLN
 2𝐶2 𝐻4 + 2𝐻2

2∆𝐻2 = 2 × −337.3
6𝑂2 = −674.6

2∆𝐻1 4𝐶𝑂2 + 4𝐻2 𝑂 + 2𝐻2 𝑂

𝑂2

4𝐶𝑂2 + 4𝐻2 𝑂 + 2𝐻2 𝑂

7𝑂2

∆𝐻4 = −745.6
2𝐶2 𝐻6

2∆𝐻1 = 2∆𝐻2 + ∆𝐻3 − ∆𝐻4

= −674.6 − 136.6 + 745.6
= −811.2 + 745.6
= 65.6
65.6
∆𝐻1 = − = −32.8
2
Q.
Explain using the energy diagram of KI; what you understand by the following lattice energy, heat of
solution and hydration energy.
+ −
𝐾(𝑔) + 𝐶𝑙(𝑔) + 𝑎𝑞

𝐻𝑦𝑑𝑟𝑎𝑡𝑖𝑜𝑛 𝑒𝑛𝑒𝑟𝑔𝑦 = ∆𝐻𝐻𝑦𝑑

𝐿𝑎𝑡𝑡𝑖𝑐𝑒 𝑒𝑛𝑒𝑟𝑔𝑦 = ∆𝐻𝑙𝑎𝑡

+ −
𝐾(𝑎𝑞) + 𝐶𝑙(𝑎𝑞)

Heat of solution

𝐾𝐶𝑙(𝑠) + 𝑎𝑞
The lattice energy of an ionic crystal is the heat of formation for one mole of the ionic
compound from gaseous ions under standard conditions (25℃, 1 𝑎𝑡𝑚)
Heat of hydration of a substance is the thermal energy change accompanying formation of one
mole of a specified hydrate from one mole of the anhydrous substance on combination with
the stoichiometric quantity of water.
Heat of solution of a substance is the heat evolved in making a solution of specified
concentration in a specified solvent at a specified temperature by completely dissolving one
mole of the substance.
Q.
Calculate the hydration energy of potassium iodide assuming its heat of solution is
+31𝐾𝐽𝑚𝑜𝑙 −1 and its lattice energy is −746𝐾𝐽.
Soln
∆𝐻𝐻𝑦𝑑 = ∆𝐻𝑠𝑜𝑙𝑛 + ∆𝐻𝑙𝑎𝑡
= +31 + 746
= −715𝐾𝐽𝑚𝑜𝑙 −1

Q.
Two gaseous compounds in form of pollutants in auto exhaust are CO and NO. An
environmental chemist is studying ways to convert them to less harmful gases through the
following equation.
1
𝐶𝑂(𝑔) + 𝑁𝑂(𝑔) → 𝐶𝑂2(𝑔) + 𝑁2(𝑔) ∆𝐻 =?
2
Given the following information, calculate the unknown ∆𝐻 in the above equation.
1
Equation A: 𝐶𝑂(𝑔) + 2 𝑂2(𝑔) → 𝐶𝑂2(𝑔) ∆𝐻𝐴 = −283.0𝐾𝐽
Equation B:𝑁2(𝑔) + 𝑂2(𝑔) → 2𝑁𝑂(𝑔) ∆𝐻𝐵 = +180.6𝐾𝐽
Solution
2𝐶𝑂2(𝑔) + 𝑁2

𝑂2 ∆𝐻𝐵 = +180.6𝐾𝐽

−∆𝐻1 2 𝐶𝑂2 + 2𝑁𝑂

𝑂2 2∆𝐻𝐴 = 2 × −283
= 566
2CD +2NO
−2∆𝐻1 = ∆𝐻𝐵 − 2∆𝐻𝐴
= +180.6 − (−566)
= 746.6 𝐾𝐽
∆𝐻1 = −373.3 𝐾𝐽

Q.
When 1 mole of liquid benzene is completely burnt in oxygen to form liquid water and carbon
dioxide gas ∆𝐻 is−78.10 𝑘𝑐𝑎𝑙 𝑎𝑡 25℃. Calculate the heat of this reaction at constant volume
at the same temperature (R= 2 cal)
Soln
1
𝐶6 𝐻6(𝑙) + 7 𝑂2(𝑔) → 3𝐻2 𝑂(𝑙) + 6𝐶𝑂2(𝑔) ∆𝐻 = −781 𝑘𝑐𝑎𝑙
2
∆𝐻𝑃 = ∆𝐸 + ∆𝑛𝑅𝑇 ∆𝑛 = 𝑛𝑃𝑟𝑜𝑑𝑢𝑐𝑡 − 𝑛𝑟𝑒𝑎𝑐𝑡𝑎𝑛𝑡 = 6 − 7.5 = −1.5
−781 × 103 𝑐𝑎𝑙 = ∆𝐸 + 1.5 × 2 × 298
∆𝐸 = 1.5 × 2 × 298 − 781 × 103
= 894 − 781000
= 780106 𝑐𝑎𝑙
= −780.106 𝑘𝑐𝑎𝑙

Q.
Dibrone (𝐵2 𝐻6) is highly reactive Boron hydride. Calculate ∆𝐻 (enthalpy change) for the
synthesis of diborane from its elements according to the equation:
2𝐵2(𝑠) + 3𝐻2(𝑔) → 𝐵2 𝐻6(𝑔)
Using the following data:
3
2𝐵(𝑠) + 𝑂2(𝑔) → 𝐵2 𝑂3(𝑠) ∆𝐻 = −1273𝐾𝐽
2
𝐵2 𝐻6(𝑔) + 3𝑂2(𝑔) → 𝐵2 𝑂3(𝑠) + 3𝐻2 𝑂(𝑔) ∆𝐻 = −2035𝐾𝐽
1
𝐻2(𝑔) + 𝑂2(𝑔) → 𝐻2 𝑂(𝑙) ∆𝐻 = −286𝐾𝐽
2
𝐻2 𝑂(𝑙) → 𝐻2 𝑂(𝑔) ∆𝐻 = +44𝐾𝐽
Soln
2𝐵(𝑠) + 3𝐻2(𝑔)

∆𝐻2 = −1273𝐾𝐽

3
𝑂2 𝐵2 𝑂3(𝑠) + 3𝐻2(𝑔)
2

∆𝐻3 = 3 × −286 = −858𝐾𝐽

∆𝐻1 𝐵2 𝑂3(𝑠) + 3𝐻2 𝑂(𝑔)

∆𝐻4 = 3 × 44 = +132𝐾𝐽

𝐵2 𝑂3(𝑠) + 3𝐻2 𝑂(𝑔)

3𝑂2 ∆𝐻5 = −2035𝐾𝐽

𝐵2 𝐻2(𝑔)

∆𝐻1 = ∆𝐻2 + ∆𝐻3 + ∆𝐻4 − ∆𝐻5

= −1273 + 858 + +132 − −2035
= +136𝐾𝐽.

Q.
Calculate the enthalpy of the reaction as given by:
∆H° = [Sum of bond enthalpies of all bonds in the reactants]
− [Sum of bond enthalpies af all bonds in the products]

H H H H

𝐶 = 𝐶(𝑔) + 𝐵𝑟 − 𝐵𝑟(𝑔) = 𝐵𝑟 − 𝐶 − 𝐶 − 𝐵𝑟(𝑔)

H H H H
° ° ° °
= (∆𝐻𝐶=𝐶 + ∆𝐻𝐵𝑟−𝐵𝑟 ) − (∆𝐻𝐶−𝐶 + 2∆𝐻𝐶−𝐵𝑟 )
= (6100 + 193.0)𝐾𝐽 − (348.0 + 2 × 276.0)𝐾𝐽
= (803 − 900)𝐾𝐽
= −97.0𝐾𝐽.
Q.
The enthalpy change for the reaction
𝐶𝐻4(𝑔) + 𝐶𝑙2(𝑔) = 𝐶𝐻3 𝐶𝑙(𝑔) + 𝐻𝐶𝑙(𝑔) 𝑖𝑠 − 104.6𝐾𝑗
The bond enthalpy of 𝐶 − 𝐻 𝑖𝑠 − 83.7𝐾𝐽𝑚𝑜𝑙 −1 greater than the bond enthalpy of 𝐶 − 𝐶𝑙. If
the bond enthalpy of 𝐻 − 𝐻 and 𝐻 − 𝐶𝑙 are almost the same in magnitude, calculate the
enthalpy change for the reaction:
𝐻2(𝑔) + 𝐶𝑙(𝑔) → 2𝐻𝐶𝑙

Soln
∆H° = [Sum of bond enthalpies of all bonds in the reactants]
− [Sum of bond enthalpies af all bonds in the products]

° ° °
∆H° = ∆HH−H + ∆HCl−Cl ) − (2∆HH−Cl ) =? … … . (i)
Also
° °
∆𝐻𝐶−𝐻 − ∆𝐻𝐶−𝐶𝑙 = 83.7 … … … . (𝑖𝑖)

H H
𝐻 − 𝐶 − 𝐻 + 𝐶𝑙 − 𝐶𝑙 = 𝐻 − 𝐶 − 𝐶𝑙 + 𝐻 − 𝐶𝑙
H H

° ° ° ° °
∆H° = [4∆HC−H + ∆HCl−Cl ] − [3∆HC−H + ∆HC−Cl + ∆HH−Cl ] = −104.6

= [∆H°C−H + ∆H°Cl−Cl ] − [∆H°C−Cl + ∆H°H−Cl ] = −104.6

= [∆H°C−H − ∆H°C−Cl ] + [∆H°Cl−Cl − ∆H°H−Cl ] = −104.6

° °
83.7 + [∆HCl−Cl − ∆HH−Cl ] = −104.6

° °
∆HCl−Cl − ∆HH−Cl = −104.6 − 83.7

° °
∆HCl−Cl − ∆HH−Cl = −188.3 … … … (iii)

But it’s given that the bond enthalpies of 𝐻 − 𝐻 and 𝐻 − 𝐶𝑙 is almost the same in magnitude.
° °
Therefore, 𝐻𝐻−𝐶𝑙 ≅ ∆𝐻𝐻−𝐻

° ° °
∆H° = (∆HH−Cl + ∆HCl−Cl ) − (2∆HH−Cl )

° ° °
∆H° = ∆HH−Cl + ∆HCl−Cl − 2∆HH−Cl

° °
∆H° = ∆HCl−Cl − ∆HH−Cl … … … … … … … … . . (iv)

Comparing equations (ii) and (IV)

The enthalpy change of H2(g) + Cl2(g) = 2Hcl(g) is − 188.3KJ.
Q.
Use Hess’s law to calculate the following enthalpy of reaction of the major process of steam reforming.
𝐶𝐻4(𝑔) + 𝐻2 𝑂(𝑔) ⇌ 𝐶𝑂(𝑔) + 3𝐻2(𝑔)
Given the separate reactions of carbon dioxide and hydrogen gas and methane decomposition
𝐶𝑂(𝑔) + 𝐻2(𝑔) → (𝐶𝑔𝑟𝑎𝑝ℎ𝑖𝑡𝑒 + 𝐻2 𝑂(𝑔) ∆𝐻 = −131.3𝐾𝐽)
𝐶𝑔𝑟𝑎𝑝ℎ𝑖𝑡𝑒 + 2𝐻2(𝑔) → 𝐶𝐻4(𝑔) ∆𝐻 ° = −74.8𝐾𝐽

Soln

∆𝐻2 = −131.3
∆𝐻1 = −74.8

𝐶𝑂(𝑔) + 𝐻2(𝑔) + 2𝐻2(𝑔)

∆𝐻3
𝐶𝐻4(𝑔) + 𝐻2 𝑂(𝑔)

∆𝐻3 + ∆𝐻2 = −∆𝐻1

∆𝐻3 + 131.3 = +74.8
∆𝐻3 = 131.3 + 74.8 = 206.1𝐾𝐽

Q.
From the following information on bond enthalpy data, calculate the enthalpy of formation of
gaseous isoprene

𝐶𝐻3 = 𝐶 − 𝐶𝐻 = 𝐶𝐻2

𝐶𝐻2
Bond enthalpy of 𝐻 − 𝐻, 𝐶 − 𝐻, 𝐶 − 𝐶 𝑎𝑛𝑑 𝐶 = 𝐶
𝑖𝑠 435.94 𝐾𝐽𝑚𝑜𝑙 −1 , 415.8 𝐾𝐽𝑚𝑜𝑙 −1 , 347.7 𝐾𝐽𝑚𝑜𝑙 −1 𝑎𝑛𝑑 600.7 𝐾𝐽𝑚𝑜𝑙 −1 𝑟𝑒𝑠𝑝𝑒𝑐𝑡𝑖𝑣𝑒𝑙𝑦.
The enthalpy of sublimation of carbon is 716.68𝐾𝐽𝑚𝑜𝑙 −1 .
If the enthalpy of formation of gaseous isoprene obtained from the combustion data is
8.79 𝐾𝐽𝑚𝑜𝑙 −1 , how would you account in the two values?

Soln
4𝐻2(𝑔) + 5𝐶(𝑠) → 𝐶𝐻2 = 𝐶 − 𝐶𝐻 = 𝐶𝐻2

𝐶𝐻3

∆𝐻𝑓° = (4∆𝐻𝐻−𝐻 + 5∆𝐻𝑠° ) − (8∆𝐻𝐶−𝐻 + 2∆𝐻𝐶=𝐶 + 2∆𝐻𝐶−𝐶 )

∆𝐻𝑓° = (4 × 435.94 + 5 × 716.68) − (8 × 415.8 + 2 × 600.7 + 2 × 347.7)
= (1743.76 + 3583.4) + (3326.4 + 1201.4 + 695.4)
= 103.96𝐾𝐽
The difference in the two values (103.96 − 8.79 = 95.17) is attributed due to resonance.

Q.
From the measured enthalpy of the reaction
1 3
𝐶𝑔𝑟𝑎𝑝ℎ𝑖𝑡𝑒 + 𝐶𝑙2(𝑔) + 𝐻2(𝑔) → 𝐶𝐻3 𝐶𝑙(𝑔) ∆𝐻 ° = −82.01𝐾𝐽𝑚𝑜𝑙 −1
2 2
Calculate the enthalpy of the reaction
𝐶(𝑔) + 𝐶𝑙(𝑔) + 3𝐻(𝑔) → 𝐶𝐻3 𝐶𝑙(𝑔)
𝐶𝑔𝑟𝑎𝑝ℎ𝑖𝑡𝑒 → 𝐶(𝑔) ∆𝐻 ° = 716.68 𝐾𝐽𝑚𝑜𝑙 −1
𝐻2(𝑔) → 2𝐻(𝑔) ∆𝐻 ° = 435.94𝐾𝐽𝑚𝑜𝑙 −1
𝐶𝑙2(𝑔) → 2𝐶𝑙(𝑔) ∆𝐻 ° = 242.15𝐾𝐽𝑚𝑜𝑙 −1

Soln
𝐶(𝑔) + 𝐶𝑙(𝑔) + 3𝐻(𝑔)
3 3
∆𝐻 ° = 2 × 435.94 𝐾𝐽𝑚𝑜𝑙 −1
2

3
𝐶(𝑔) + 𝐶𝑙(𝑔) + 2 𝐻2(𝑔)

1 1
∆𝐻 ° = 2 × 242.15 𝐾𝐽𝑚𝑜𝑙 −1
2

1 3
𝐶(𝑔) + 2 𝐶𝑙2(𝑔) + 2 𝐻2(𝑔)

∆𝐻 ° = 716.68 𝐾𝐽𝑚𝑜𝑙 −1

1 3
𝐶𝑔𝑟𝑎𝑝ℎ𝑖𝑡𝑒 + 𝐶𝑙2(𝑔) + 𝐻2(𝑔)
2 2

∆𝐻 ° = −82.01 𝐾𝐽𝑚𝑜𝑙 −1

𝐶𝐻3 𝐶𝑙(𝑔)

3 1
∆𝐻 ° = − × 435.94 + × 242.15 − 716.68 + −82.01
2 2
= −653.91 + −121.075 + −716.68 − 82.01
= −1573.675 𝐾𝐽𝑚𝑜𝑙 −1
Q.
Calculate the hydration energy of potassium iodide, assuming that its heat of solution is
+21 𝐾𝐽𝑚𝑜𝑙 −1 and its lattice energy is −642 𝐾𝐽

Soln
∆𝐻ℎ𝑦𝑑 = ∆𝐻𝑠𝑜𝑙𝑛 + ∆𝐻𝑙𝑎𝑡
= +21 + −642
= −621𝐾𝐽𝑚𝑜𝑙 −1

CONDITIONS THAT HAVE A SIGNIFICANT INFLUENCE ON THE HEAT OF A REACTION

1. Temperature prevailing during the reaction
2. Physical, allotropic and crystalline states of reactants and products of the reaction
3. Concentration and solvents of solutions involved
4. Pressures of involved gases
5. Condition of constancy of pressure.
Q.
a. Draw a complete fully labeled Born-Haber Cycle for the formation of potassium
bromide.
b. Using the information in the table below, calculate the lattice energy of potassium
bromide.

Reaction ∆𝑯/𝑲𝑱𝒎𝒐𝒍−𝟏
1 −
𝐾(𝑠) + 𝐵𝑟2(𝑙) → 𝐾 + 𝐵𝑟(𝑠) − 392
2

𝐾(𝑔) → 𝐾(𝑔) + 90

+
𝐾(𝑔) → 𝐾(𝑔) + 𝑒− + 420

1
𝐵𝑟 → 𝐵𝑟(𝑔) + 112
2 2(𝑙)
−
𝐵𝑟(𝑔) + 𝑒 − → 𝐵𝑟(𝑔) − 342
c. The values of the lattice energies of the other potassium halides are:

KF KCl KI
−1
Compound lattice energy/ 𝐾𝐽𝑚𝑜𝑙 −813 −710 −643

What explanation can you give for the trend in these values?
a.
Solution
+
𝐾(𝑔) + 𝑒 − + 𝐵𝑟(𝑔)
Ionization of Bromine
Atomization of (Electron affinity for Bromine)
Brimine

+ 1
𝐾(𝑔) + 𝑒 − + 𝐵𝑟2(𝑙)
2

Ionization of potassium

+ −
𝐾(𝑔) + 𝐵𝑟(𝑔)

1
𝐾(𝑔) + 2 𝐵𝑟2(𝑙)

Atomization of potassium

Lattice energy

1
𝐾(𝑠) + 𝐵𝑟2(𝑙)
2

Heat of reaction

−
𝐾 + 𝐵𝑟(𝑠)

b. The sum of all these enthalpies is equal to the heat of the reaction
𝐻𝑒𝑎𝑡 𝑜𝑓 𝐴𝑡𝑜𝑚𝑖𝑧𝑎𝑡𝑖𝑜𝑛 𝐼𝑜𝑛𝑖𝑧𝑎𝑡𝑖𝑜𝑛 𝐴𝑡𝑜𝑚𝑖𝑧𝑎𝑡𝑖𝑜𝑛 𝐼𝑜𝑛𝑖𝑧𝑎𝑡𝑖𝑜𝑛
( )=( )+( )+( )+( )
𝑟𝑒𝑎𝑐𝑡𝑖𝑜𝑛 𝑜𝑓 𝐾 𝑜𝑓 𝐾 𝑜𝑓𝐶𝑙 𝑜𝑓 𝐵𝑟

𝐿𝑎𝑡𝑡𝑖𝑐𝑒
+( )
𝑒𝑛𝑒𝑟𝑔𝑦
392 = +90 + +420 + +112 + −342 + 𝐿𝑎𝑡𝑡𝑖𝑐𝑒 𝑒𝑛𝑒𝑟𝑔𝑦
−392 = +280 + 𝐿𝑎𝑡𝑡𝑖𝑐𝑒 𝑒𝑛𝑒𝑟𝑔𝑦
𝑇ℎ𝑒𝑟𝑒𝑓𝑜𝑟𝑒, 𝐿𝑎𝑡𝑡𝑖𝑐𝑒 𝑒𝑛𝑒𝑟𝑔𝑦 = −672𝐾𝐽𝑚𝑜𝑙 −1
 c. As the difference in electronegativity between metal ion and halide increases, lattice
energy also increases.

Q.
Use the following data and answer the questions that follow:

𝑙𝑖(𝑠) → 𝑙𝑖(𝑔) ∆𝐻 = +159.6𝐾𝐽

1
𝐶𝑙 → 𝐶𝑙(𝑔) ∆𝐻 = +121.8𝐾𝐽
2 2
+
𝐿𝑖(𝑠) → 𝐿𝑖(𝑔) ∆𝐻 = +520.8𝐾𝐽
− −
𝐶𝑙(𝑔) + 𝑒 → 𝐶𝑙(𝑔) ∆𝐻 = −369𝐾𝐽
1
𝐿𝑖(𝑠) + 𝐶𝑙2(𝑔) → 𝐿𝑖𝐶𝑙(𝑠) ∆𝐻 = −411.6𝐾𝐽
2

a) Draw a fully labelled diagram of an energy cycle for formation of Lithium Chloride.
b) Calculate the lattice energy of lithium chloride.
Solution
a.
+
𝐿𝑖(𝑔) + 𝐶𝑙(𝑔) + 𝑒 −

Ionization energy of lithium −369𝐾𝐽 Electron affinity of chlorine

+520.8
𝐾𝐽
+ −
𝑙𝑖(𝑔) + 𝐶𝑙(𝑔) 𝐿𝑖(𝑔) + 𝐶𝑙(𝑔)
Heat of atomization of chlorine or
+121.8 Dissociation energy of
𝐾𝐽 Chlorine

1
𝐿𝑖(𝑔) + 2 𝐶𝑙2(𝑔)
Lattice energy of
Heat of Lithium Chloride
+159.6 Atomization of Lithium
𝐾𝐽
1
𝐿𝑖(𝑠) + 2 𝐶𝑙2(𝑔)
−411.6
𝐾𝐽 Heat of formation of lithium
Chloride

𝐿𝑖 + 𝐶𝑙− Ionic Lattice

b. +411.6 + 159.6 + 121.8 + 520.8 + − 369 + 𝐿𝑎𝑡𝑡 𝑒𝑛𝑒𝑟𝑔𝑦

𝑡ℎ𝑒𝑟𝑒𝑓𝑜𝑟𝑒, 𝐿𝑎𝑡𝑡𝑖𝑐𝑒 𝑒𝑛𝑒𝑟𝑔𝑦 = 844.8𝐾𝐽
Q.
Given that, the bond enthalpies for, 𝐶 − 𝐻, 𝐶𝑙 − 𝐶𝑙, 𝑎𝑛𝑑 𝐻 − 𝐶𝑙 is 99, 58, 78 and 103 kcal
respectively. Determine ∆𝐻 for the reaction below using energy diagram:

𝐶𝐻4(𝑔) + 𝐶𝑙2(𝑔) → 𝐶𝐻3 𝐶𝑙(𝑔) + 𝐻𝐶𝑙(𝑔)

𝐶(𝑔) + 4𝐻(𝑔) + 2𝐶𝑙(𝑔)

𝐶𝑙 − 𝐶𝑙 = 58 𝑘𝑐𝑎𝑙
3(𝐶 − 𝐻) = 3 × −99
𝐶(𝑔) + 4𝐻(𝑔) + 𝐶𝑙2(𝑔) = 375 𝑘𝑐𝑎𝑙

𝐶 − 𝐶𝑙
4(𝐶 − 𝐻) = 99 × 4 = 396 𝑘𝑐𝑎𝑙
𝐶𝐻3 𝐶𝑙(𝑔) + 𝐻(𝑔) + 𝐶𝑙(𝑔)

𝐶𝐻4(𝑔) + 𝐶𝑙2(𝑔) 𝐻 − 𝐶𝑙 = −103 𝑘𝑐𝑎𝑙

∆𝐻𝑟𝑥𝑛

𝐶𝐻3 𝐶𝑙(𝑔) + 𝐻𝐶𝑙(𝑔)

−∆𝐻𝑟𝑥𝑛 + 396 + 58 + −375 + −103 = 0

∆𝐻𝑟𝑥𝑛 = −24 𝑘𝑐𝑎𝑙.