THERMOCHEMISTRY

Introduction: This is the study of the energy changes accompany

chemical reactions. One of the main features of chemical reactions is

the associated energy changes. The main form of this energy is heat.

This is often referred to as enthalpy change, denoted as ΔH. The heat

change may be positive or negative. A chemical reaction associated

with positive ΔH is termed an endothermic reaction, while a

chemical reaction associated with negative ΔH is referred to as

exothermic reaction.
 Endothermic Reaction

Energy
Products

ΔH = +ve
Reactants

Reaction pathway
 Exothermic reaction

Energy
Reactants

ΔH = -ve
Products

Reaction pathway
 Definitions of some heat changes

(1) Standard Heat Change of Formation: It is the heat change associated

with the formation of one mole of a substance from its constituent

elements under standard condition. It is designated as ΔH°f, and is

always exothermic. Some examples are

1/2N2(g) + 3/2H2(g) NH3(g); ΔH°f (NH3)

2Al(s) + 3/2O2(g) Al2O3(s);ΔH°f (Al2O3)

1/2N2(g) + 2H2(g) + 1/2Cl2(g) NH4Cl(s); ΔH°f (NH4Cl)

(2) Standard Heat Change of Combustion: It is the heat change identified

with the complete combustion of one mole of a substance in oxygen, under

standard condition. It is denoted as ΔH°c, and is always exotherthermic.

Some example as shown below

CH4(g) + 2O2(g) CO2 (g) + 2H2O(l); ΔH°c (CH4)

C2H5OH(l) + 3O2(g) 2CO2(g) + 3H2O(l); ΔHºc (C2H5OH)

N2H4(l) + 3O2(g) 2NO2(g) + 2H2O(l); ΔH°c (N2H4)

(3) Standard Heat Change of Hydration: It is the heat change that occurs when

one mole of a gaseous dissolves in water to give an infinitely dilute solution,

under standard condition. It is denoted by ΔH°h, and it is always exothermic.

The higher the charge density of the ion, the more exothermic the enthalpy of

hydration. Some examples are given below

Li+(g) + H2O(l) Li+(aq); ΔHºh (Li+)

K+(g) + H2O(l) K+(aq); ΔH°h (K+)

F-(g) + H2O(l) F-(aq); ΔHºh (F-)

Mg2+(g) + H2O(l) Mg2+(aq); ΔHºh (Mg2+)

(4) Standard Heat Change of Solution: It is the heat change that occurs when one

mole of a substance dissolves in water to give an infinitely dilute solution. It is

denoted as ΔH°sol, and may be negative (exothermic) or positive (endothermic). It is

positive, if the final temperature of the solution is less than the initial temperature,

while is exothermic, if the final temperature is greater than the initial temperature.

Some examples are given below

NaCl(s) + H2O(l) Na+(aq) + Cl-(aq); ΔH°sol (NaCl)

MgSO4(s) + H2O(l) Mg2+(aq) + SO42-(aq); ΔHºsol (MgSO4)

(5) Standard Heat Change of Neutralization: It the enthalpy change that

accompanies the complete neutralization of one mole of hydrogen ion, from

an acid, by hydroxide ion from a base. It is designated by ΔHºn, and it is

always negative (exothermic). The major reaction is the formation of water

Molecule: H+(aq) + OH-(aq) H2O(l); ΔHºn

Its value is constant for a given pair of strong acid and base.

(6) Atomization Energy: It the enthalpy associated with the formation of one mole of

a gaseous atom of a given element, under standard condition. It is denoted as ΔH°at.

Its value is always positive (endothermic). Some examples are given below

C (s) C(g); ΔHºat (C)

1/2Cl2(g) Cl(g); ΔHºat (Cl)

(7) Bond Energy: It is the energy required to break one mole of a particular

bond. It is often referred to as bond dissociation energy. The term average

bond energy is used if two or more of the same kind of bonds are broken,

successively, in a given molecule. Bond energy is always positive. This is

illustrated below

(i) Cl2(g) 2Cl(g); ΔHºreaction = 2 × B.E. (Cl-Cl)

(ii) CH4 (g) C(g) + 4H(g); ΔHºreaction = 4 × B.E. (C-H)

The bond energy of C-H bond in CH4 is correctly designated as average bond

energy, since more than one of the same type of bond are broken, in succession.
(8) Electron Affinity: It is the heat change that accompanies the gain of one

mole of electron by one mole of a gaseous atom or ion of a given element.

The first electron affinity of an element is exothermic, while subsequent values are

endothermic. This is as a result of the force of repulsion, associated with gain of

electron by a negatively charged species.

Cl(g) + e- Cl-(g) ; 1st E.A. (Cl)

O-(g) + e- O2-(g); 2nd E.A. (O)

(9) Lattice Energy: Is the energy released when one mole an ionic crystal is formed

from its constituent gaseous ion, under standard condition. Lattice energy is always

exothermic. The degree of exothermicity is dependent on the charge densities of the

constituent ions.
The higher the charge densities of the ions, the more exothermic the lattice

energy.

Na+(g) + Cl-(g) NaCl(s); ΔHºlatt (NaCl)

2Al3+(g) + 3O2-(g) Al2O3(s); ΔHºlatt (Al2O3)

Mg2+(g) + O2-(g) MgO(s); ΔH°latt (MgO)

Li+(g) + Cl-(g) LiCl(s); ΔH°latt (LiCl)

 BORN HABER CYCLE

It is an energy cycle that gives details of all the enthalpy changes associated with a

given chemical reaction. The information obtainable from the cycle can be used to

calculate the value of an unknown enthalpy change. This is achieved by forming a

simple equation, using the knowledge of Hess’ law. The states that the heat

change in a given chemical reaction is constant, and is independent of the routes

taken by the reaction. The law treats the enthalpy change of a reaction as a state

function. Some energy diagrams or cycles are discussed.

 (A)Energy Cycles for the determination of Enthalpy Changes

that cannot be found by experiment

(I) Energy cycle incorporating enthalpy changes of combustion

ΔH°f (CH4)
C(s) + 2H2(g) CH4(g)

ΔH°c(C) 2×ΔHºc(H2) ΔHºc(CH4)

CO2(g) + 2H2O(l)

Application of Hess’ law gives the equation below

ΔHºf(CH4) + ΔH°c(CH4) = ΔH°c(C) + 2ΔH°c(H2)

Example 1: Ketene, C2H2O, is a member of class of unsaturated organic compound

that is widely used in pharmaceutical research for the synthesis of organic

compounds

(i) Draw a Born-Haber cycle incorporating the enthalpy change of formation of

keten and the data shown below

ΔH°c(CO2) = -395 kJ mol-1

ΔH°c(H2) = -286 kJ mol-1

ΔH°c(C2H2O) = -1028 kJ mol-1

(ii) Use the data and the equation derivable from the Born-Haber cycle to

calculate the standard enthalpy change of formation of keten

 Solution:

ΔH°f (C2H2O)
(i) 2C(s) + H2(g) + 1/2O2(g) C2H2O(l)

2ΔH°c(C) ΔHºc(H2)
ΔH°c (C2H2O)

2CO2(g) + H2O(l)

(ii) Application of Hess’ law gives the equation below

ΔH°f (C2H2O) + ΔH°c (C2H2O) = 2ΔH°c(C) + ΔHºc(H2)

ΔH°f (C2H2O) = 2ΔH°c(C) + ΔHºc(H2) - ΔH°c (C2H2O)

ΔH°f(C2H2O) = 2(-395) + (-286) – (-1028) = -48 kJ mol-1

 (II) Energy cycle incorporating bond energy values and atomization

energies
ΔHºf(C2H5OH)
2C(s) + 3H2(g) + 1/2O2(g) C2H5OH(l)

2ΔH°at(C) 6ΔH°at(H) ΔH°at(O) B.E.(C-C) + 5B.E.(C-H) +

B.E.(C-O) + B.E.(O-H)

2C(g) + 6H(g) + O(g)

Application of Hess’ law gives the equation

ΔH°f(C2H5OH) + B.E.(C-C) + 5B.E.(C-H) + B.E.(C-O) + B.E.(O-H) =

2ΔH°at(C) + 6ΔHºat(H) + ΔHºat(O)

Example 2: The standard enthalpy change of formation of SiCl4(g) is -610 kJ mol-1.

The standard enthalpy changes of atomization of the elements silicon and chlorine

are +334 and +122 kJ mol-1, respectively.

(i) Draw an energy cycle incorporating these data

(ii) Use the equation derivable from the cycle to calculate the average bond energy

of the Si-Cl bond

Solution:
(i)
ΔHºf(SiCl4)
Si(s) + 2Cl2(g) SiCl4(l)

ΔH°at(Si) 4×ΔH°at(Cl)
4×B.E.(Si-Cl)
Si(g) + 4Cl(g)
(ii) Applying Hess’ law, the equation below can be derived from the energy

cycle drawn

ΔHºf(SiCl4) + 4× B.E.(Si-Cl) = ΔH°at(Si) + 4×ΔH°at(Cl)

4×B.E.(Si-Cl) = ΔH°at(Si) + 4×ΔH°at(Cl) – ΔH°f(SiCl4)

4×B.E.(Si-Cl) = [334 + 4(122)] – (-610)

B.E.(Si-Cl) = 358 kJ mol-1

 (B) Energy cycle for the formation of ionic crystal

An example of this is shown for the formation of a metal halide, MX(s)

M+(g) + X(g)

1st I.E.(M) 1st E.A.(X)

M+(g) + X-(g)
M(g) + X(g)

ΔHºat(X)

M(g) + 1/2X2(g)

ΔHºat(M)
L.E.(MX)
M(s) + 1/2X2(g)
0
ΔHºf(MX)
MX(s)
By applying Hess’ law, the below equation can be obtained

L.E. (MX) = ∆Hºf (MX) – [∆Hºat (M) + ∆Hºat (X) + 1st I.E. (M) + 1st E.A. (X)]

where L.E. (MX) = Lattice energy of (MX)

∆Hºf (MX) = Heat of formation of (MX)

∆Hºat (M) = Heat of atomization of M

∆Hºat (X) = Heat of atomization

1st I.E.(M) = First ionization energy of M

1st E.A.(X) = First electron affinity of X

Example 3:

Use the following data to construct a Born-Haber cycle and calculate the lattice

energy of barium sulphide, BaS.

Standard enthalpy change of formation of BaS(s) = -460 kJ mol-1

Standard enthalpy change of atomization of Ba(s) = +180 kJ mol-1

Standard enthalpy change of atomization of S(s) = +279 kJ mol-1

First and second ionization energies of Ba(g) = +1468 kJ mol-1

First electron affinity of sulphur atom = -200 kJ mol-1

Second electron affinity of sulphur atom = +640 kJ mol-1

Solution:
Ba2+(g) + S2-(g)
Ba2+(g) + S(g)
2nd E.A. (S)
1st E.A. (S)
2nd I.E. (Ba)
Ba2+(g) + S-(g)
Ba+(g) + S(g)

1st I.E. (Ba)

Ba(g) + S(g)
ΔH°latt(BaS)

ΔH°at(S)
Ba(g) + S(s)

ΔH°at(Ba)

Ba(s) + S(s)
0
ΔH°f (BaS)
BaS(s)
The equation below can be obtained from the energy cycle drawn above

L.E. (BaS) = ∆Hºf (BaS) – [∆Hºat (Ba) + ∆Hºat (S) + 1st I.E. (Ba) + 2nd I.E. (Ba)

+ 1st E.A. (S) + 2nd E.A. (S)]

L.E. (BaS) = -460 – [180 + 279 + 1468 + (-200) + 640] kJ mol-1

L.E. (BaS) = -2827 kJ mol-1

Example 4:

(a)Write the equation which represents the lattice energy of calcium fluoride

(b) (i) Construct a Born-Haber cycle which can be used to calculate the lattice

energy of calcium fluoride .

(ii) Calculate the lattice energy of calcium fluoride using the following data

Enthalpy change of formation of calcium fluoride = -1228 kJ mol-1

Enthalpy change of atomization of calcium = +178 kJ mol-1

Enthalpy change of atomization of fluorine = +79 kJ mol-1

First electron affinity of fluorine = -328 kJ mol-1

First ionization energy of calcium = +590 kJ mol-1

Second ionization energy of calcium = +1150 kJ mol-1

Solution
(a) Ca2+(g) + 2F-(g) CaF2(s)
(b) (i)

Ca2+(g) + 2F(g)

2nd I.E. (Ca) 2×1st E.A. (F)

Ca2+(g) + 2F-(g)
Ca+(g) + 2F(g)

1st I.E. (Ca)

Ca(g) + 2F(g)
ΔH°latt(CaF2)
2×ΔH°at(F)
Ca(g) + F2(g)

ΔH°at(Ca)
Ca(s) + F2(g)
0
ΔH°f(CaF2)
CaF2(s)
(b) (ii) L.E. (CaF2) = ∆Hºf (CaF2) – [∆Hºat (Ca) + 2×∆Hºat (F) + 1st I.E. (Ca) + 2nd I.E. (Ca)

+ 2×1st E.A. (F)]

L.E. (CaF2) = -228 – [178 + 2(79) + 590 + 1145 + 2(-328)]

L.E. (CaF2) = -2643 kJ mol-1

 INTRODUCTORY CHEMICAL THERMODYNAMICS
Chemical thermodynamics is the study of the energy transformations

associated with chemical reactions. There are various types of energy, but

the energy of interest is heat energy.

Definitions of Some Terms:

(a) Universe: This is made of the surrounding and the system

(b) System: This refers to the substance undergoing a change in a

thermodynamic process

(c) Surrounding: This refers to all object acting on a system in a

thermodynamic process.

(d) Open System: A system in which mass and energy are transferred

between the system and the surrounding

(e) Closed System: A system in which only energy is transferred between the system

and the surrounding

(f) Entropy (S): This is a measure of the degree of disorderliness of a system.

(g) Internal Energy (U): It is the energy associated with the internal processes

occurring in a substance. This energy becomes available when a substance

undergo a chemical or physical change.

(h) Heat energy (q): This is the internal energy transferred from one substance to

another due to temperature difference.

GIBB’S FREE ENERGY

If a reaction is carried out at constant pressure, the quantity of heat, q, given to

the system is the same as the enthalpy, H, of the system. The portion of H that is
unavailable for work is the product of temperature, T, and the entropy, S, of the

system, TS. The remaining portion, that is available for work is called the Gibb’s Free

Energy, G.

Hence, Gibb’s free energy is expressed as

G = H – TS 1.1

ΔG = ΔH –TΔS 1.2

ΔG = Change in Gibb’s free energy, its unit is J mol-1

ΔH = Enthalpy change, its unit is J mol-1

T = Temperature of the system in Kelvin, K

ΔS = Entropy change, its unit is J mol-1 K-1

Under standard condition, you have

ΔG° = ΔH° - TΔS° 1.3

Example : Calculate the standard free energy change at 300 K for the reaction

CO(g) + H2O(g) CO2(g) + H2(g)

given the following data:

Substance CO(g) H2O(g) CO2(g) H2(g)

ΔH°f/kJ mol-1 -110.5 -241.8 -393.5 0
S°/J K-1 197.9 188.7 213.8 130.5

Solution:

First of all calculate ΔH° and ΔS° for the reaction

ΔH°reaction = ΔH°f(Products) - ΔH°f(Reactants)

ΔH°reaction = [ΔH°f(CO2(g) + ΔH°f(H2(g)] – [ΔH°f(CO(g)) + ΔH°f(H2O(g)]

ΔH°reaction = (-393.5 + 0)- (-110.5 + (-241.8) = -393.5 +352.3 = -41.2 kJ mol-1

ΔS°reaction = Sº (Products) - S°(Reactants)

ΔSºreaction = [S°(CO2(g)) + Sº(H2(g))] – [S°(CO(g)) + Sº(H2O(g)]

ΔSºreaction = (213.8 + 130.5) – (197.9 + 188.7) = 344.3 – 386.6 = -42.3 J K-1 mol-1

ΔSºreaction = -0.0423 kJ K-1 mol-1

ΔGº = ΔHº - TΔSº = [-41.2 – (300 ×(-0.0423)] = -41.2 + 12.69 = -28.51 kJ mol-1

The value of ΔGº is negative, which implies the reaction is spontaneous. A positive ΔGº

indicates the reaction in question is non-spontaneous. It is a common phenomenon that

a system tends to move towards a state of low energy. Equation 1.2 or 1.3 shows that

such state is encouraged by a more negative ΔH and/or a more positive ΔS. This implies

that a process, in which the system move from high energy to low energy state, tends to

proceeds spontaneously. In the same manner, a process in which the system moves from

a state of low disorder to that of high disorder tends to proceeds spontaneously.