CH110 General Chemistry for B.Sc.

Chemistry Ist year Semester I

THERMOCHEMISTRY
INTRODUCTION
THERMO implies HEAT
DYNAMICS implies MOVEMENT OR CHANGE
There are two factors that determine whether or not we will observe a particular reaction.
Thermodynamics tells us whether a change is possible. It also tells us what the composition
of equilibrium will be when the composition of the reaction mixture ceases to change.
Thermodynamics is the study of energy and its changes or transformations.
Thermochemistry is the study of the relationships between chemical reactions and energy
changes involving heat.
COMMONLY USED TERMS

System:
A system is a portion of the universe on which we wish to focus our attention. We call
everything else the surroundings
surroundings. A system and its surroundings are separated from each
other by a boundary or interface. Energy exchange occurs across this boundary. Usually the
boundary is real and apparent. For example, the surface of a solution or the walls of a beaker
is the boundary.
y. Sometimes the boundary is imaginary and chosen simply for convenience.

Figure 5.11: Thermodynamic System Diagram

Adiabatic Process:
An adiabatic process is a process in which heat cannot be transferred across the interface
between the system and its surroundings. For example,
1. A reaction carried out in an insulated container like a thermos bottle.
2. An explosive reaction that occurs so rapidly that heat energy produced cannot be
readily dissipated. The heat build
build-up raises the products to very high

Dr. Tanweer Ahmad, Senior Lecturer, Department of Chemistry, CBU

 CH110 General Chemistry for B.Sc.Chemistry Ist year Semester I

temperatures, and these products fly apart rapidly, pushing walls, ceilings, etc
(the surroundings) before them.
Isothermal Process:
An isothermal process is when thermal contact between the system and its surroundings is
kept at constant temperature while the change takes place. For example, biochemical
reactions in the human body are isothermal processes.
State of system:
Some particular set of conditions like pressure, temperature, number of moles of each
component and their physical forms (for example, gas, liquid, solid or crystalline form) that
precisely define a system’s properties before and after the change occurs. Once these
variables are specified, all the properties of the system are fixed. Knowledge of these
quantities permits us to define unambiguously the properties of a system. For example, two
samples of pure liquid water, each of 1 mole and each at the same temperature and pressure.
We know that all properties of each sample will be identical.
State function or state variable:
These are quantities like P, V, and T that determine the physical state of a given system.
When in a particular state, values of a state function do not depend on the prior history of the
sample.
For example, the volume of 1 mole of water at 25oC and 1 atm (101.325 kPa) does not
depend on what its temperature or pressure might have been at sometime in the past.
Furthermore, in going from one state to another, the changes in P, V and T do not depend on
how the sample is treated.
Equation of state:
There are instances in which the inter relations between the state functions can be expressed
in equation form to give an equation of state. The equation of state for an ideal gas, PV=nRT,
is an example.
Heat capacity:
This is the quantity of heat energy required to raise the temperature of a given quantity of a
substance by one degree Celsius. The units of heat capacity are JoC-1.
Specific heat:
This is the heat capacity per gram, that is, the quantity of heat necessary to raise the
temperature of 1 g of a substance by 1oC. The specific heat of water is 4.184 J g-1oC-1.
Molar specific heat:

Dr. Tanweer Ahmad, Senior Lecturer, Department of Chemistry, CBU Page 2

 CH110 General Chemistry for B.Sc.Chemistry Ist year Semester I

This is the heat needed to raise the temperature of 1 mole of a substance by 1oC. Since the
molar mass of water is 18.0 mol-1, the molar heat capacity is 18.0×4.184=75.3 J mol-1 oC-1.
Example 5.1: Calculating the Heat Capacity of a Water Bath

What would be the heat capacity expressed in kJ oC-1 of a water bath containing 4.00 dm3 of
water? The specific heat of water is 4.814 J g-1 oC-1.

We are to calculate the number of joules needed to raise the temperature of the water in the
bath by 1.00oC. Assuming the density of water to be 1.00 g cm-3, the mass of water is 4000 g.
Therefore, heat capacity of bath is

4.814 J 1 kJ
4000 g × × = 19.3 kJ o C −1
g o C 1000 J

THE FIRST LAW OF THERMODYNAMICS (OR LAW OF CONSERVATION OF

ENERGY)
In thermodynamics, we study the energy changes that occur when systems pass from one
state to another. Observations of many scientists have led to the conclusion that in any
process, energy is neither created nor destroyed. This conclusion is the basis of the law of
conservation of energy which is stated as the first law of thermodynamics that says: if a
system undergoes some series of changes that ultimately brings it back to its original state,
the net energy change is zero.

INTERNAL ENERGY

Thermodynamics defines a quantity called internal energy, E, that is used in describing

changes when they take place in chemical or physical systems. Internal energy is the total
energy of the system that it possesses as a consequence of kinetic energy of its atoms,
ions, or molecules, plus all the potential energy that arises from binding forces between
the particles making up the system.

A change in internal energy, ∆E , is defined as

∆E = E final − E initial where Efinal and Einitial are not measurable.

Though Efinal and Einitial are not measurable, there is no inconvenience in this lack of
knowledge since we are only interested in how the energy changes and this we can measure.
When a system changes from one state to another, there are two ways for it to exchange
energy with its surroundings, namely,

Dr. Tanweer Ahmad, Senior Lecturer, Department of Chemistry, CBU Page 3

 CH110 General Chemistry for B.Sc.Chemistry Ist year Semester I

1. gain or lose heat energy or

2. do work or have work done on it.

The energy book keeping for both heat and work is taken care of by the Zumdahl and
Zumdahl 9th Edition equation on pp 250 that specifies that

ΔE = q + w

where q is defined as heat absorbed from the surroundings by a system when it undergoes a
change and w is defined as the work done by the system on its surroundings.

The above equation is also a form of the first law of thermodynamics and simply states
that: the change in the internal energy is equal to the difference between the quantity of
energy gained by the system in the form of heat and the quantity of energy removed in the
form of work that is performed on the surroundings.

In summary,

q is positive (q>0) if heat is added to the system

q is negative (q<0) if heat is evolved (removed from) the system
w is positive (w>0) if the surrounding performs work, that is, energy is added
to the system and the system contracts
w is negative (w<0) if the system performs work, that is, energy is removed
from the system and the system expands.
∆E is a state function while q and w are not, that is, q and w are path-
dependent functions.

WORK

This examines how the system does work. A form of work important in chemical change is
accomplished when a system expands its volume against an opposing pressure where by
definition

= × and
( ) = ( / )

When considering a gas in a cylinder of cross-sectional area A with gas column of height h,
the volume of gas, V, is given by

V=A×h

Dr. Tanweer Ahmad, Senior Lecturer, Department of Chemistry, CBU Page 4

 CH110 General Chemistry for B.Sc.Chemistry Ist year Semester I

During a change in volume,

∆V = Vf − Vi or

∆V = Ah f − Ah i
= A(h f − h i )
= A∆h

The product of pressure and volume change is given as

F
P∆V = ⋅ A(∆h) = F∆h
A
Which shows that P∆V is equivalent to force (F) times distance ( ∆h ) and, therefore equals
work

w = P∆V

This expansion is performed by any system that expands against an external pressure
imposed by the surrounding. Conversely, when a system contracts under the influence of an
external pressure, work is performed on the system.

SI units of Pressure in Pascals (Newtons per square metre)

1 Pa = 1 N/m2 and volume is expressed in cubic metres, m3

Therefore units of P∆V are

N
Pa ⋅ m3 = 2
⋅ m3 = N ⋅ m but since 1 N ⋅ m = 1 J there by
m

1 Pa ⋅m3 = 1 J

What the 1st Law of Thermodynamics tells us is that, any particular change whether it be
physical or chemical, has a certain net energy change associated with it that is determined
solely by the energies of the initial and final states. Once we have picked what these states
are, the value of ∆E is fixed. But there are many different sets of q and w that produce the
same ∆E .

Example:

Expansion of an ideal gas against a constant opposing pressure of 100 kPa

(a) initial state Pgas = 100 kPa, Vgas = 1.00 dm3

(b) Final state Pgas=100 kPa, Vgas = 10 dm3

Dr. Tanweer Ahmad, Senior Lecturer, Department of Chemistry, CBU Page 5

 CH110 General Chemistry for B.Sc.Chemistry Ist year Semester I

w = P∆V , since ∆V = 9 dm3 ,

(
w = (100 kPa ) 9.0 dm3 )
( )(
= 100 ×103 N m − 2 9.0 ×10 −3 m3 )
= 900 N ⋅ m = 900 J

HEAT AND WORK IN CHEMICAL SYSTEMS

The work we get in an automobile battery depends on how we discharge the battery. If we
simply take a steel bar and place it across the terminals, sparks fly and the battery gets very
hot – all the energy produced is heat and we have done no work. On the other hand, if we
connect the battery to an electric motor, like the starter of a car, we can use some of the
energy of the reaction in the battery to crank the engine. This rotates parts and moves pistons.
Work is accomplished because the motor offers some resistance to the flow of electricity.
The quantity of work that can be done by a battery depends on how fast it is discharged. If
discharged quickly by offering no resistance, heat is produced and no work is done. As we
increase the resistance, we slow down the energy removal rate but more work appears.
A very important concept is that the maximum work available from any change is
obtained if the change occurs by a reversible process.

HEATS OF REACTION - THERMOCHEMISTRY

Chemical reactions occur with either absorption or evolution of energy. This reflects
differences between potential energies associated with the bonds in the reactants and the
products. For example,

H +H 
→ H2 + energy evolution
Total potential energy > Potential energy of molecule
of isolated atoms

Energy evolved during Energy that molecule should

=
molecule formation absorb to produce isolated atoms
If we do a reaction in a closed container of fixed volume, ∆V = 0 and P∆V = 0. Then the heat
absorbed at constant volume is given by the equation
∆E = q v

That is, ∆E is equal to heat absorbed or evolved by the system under conditions of constant
volume (that is, heat of reaction at constant volume).

Under these reaction conditions,

• an endothermic reaction will be such that q and ∆E are positive while
• an exothermic reaction will be such that q and ∆E are negative.

Dr. Tanweer Ahmad, Senior Lecturer, Department of Chemistry, CBU Page 6

 CH110 General Chemistry for B.Sc.Chemistry Ist year Semester I

Experimentally, ∆E can be measured using a device called a bomb calorimeter. This is a

steel bomb in which reactants are placed in. The bomb is placed in an isolated water bath
whose quantity is precisely known. A reaction set off in the bomb evolves heat whose
temperature raises that of the water bath as measured by a thermometer at the new
equilibrium temperature. Knowledge of the temperature change and heat capacity of the
calorimeter (bomb + water bath) allows calculation of heat evolved in the chemical
reaction.

Example 5.2: Calculating the heat of reaction determined in a bomb calorimeter.

H2 (0.100g) and O2 (0.800g) is compressed into 1.00 dm3 bomb calorimeter whose initial
temperature, final temperature and heat capacity are, Ti=25.000oC, Tf=25.155oC and
Cp=90.8 kJ oC-1. What is ∆ in this reaction?

Reasoning and Solution:

∆T = 0.155 o C . From the given heat capacity, evovled energy is
kJ
Heat evolved (q)=Heat capacity (Cp) × ΔT = 90.8 × o
× (0.155o C )
C
∴ q = −14.1 kJ
Hence ∆E = q = −14.1 kJ
In this example ∆ depends on the amount of H2 and O2 reacted, that is, ∆ is an
extensive quantity. We can convert this to an intensive quantity, one that is characteristic of
the reaction between any amounts of H2 and O2 by calculating the heat evolved per mole of
product formed. In the above example 0.0500 mol of H2O was produced. Therefore, we say
∆E = -14.1 kJ/0.0500 mol of H2O or ∆E = -282 kJ mol-1

REACTIONS AT CONSTANT PRESSURE, THE ENTHALPY CHANGE

Heat evolved at constant volume permits us to compute ∆E . However, most changes of

interest occur in open containers at constant atmospheric pressure. To avoid considering
PV work in such reactions, we define a new thermodynamic function called heat content, or
enthalpy, H. This is give as
H = E + PV whose change at constant pressure is
∆H = ∆E + P∆V
If only PV work is involved in the change,
∆E = q − P∆V so that

Dr. Tanweer Ahmad, Senior Lecturer, Department of Chemistry, CBU Page 7

 CH110 General Chemistry for B.Sc.Chemistry Ist year Semester I

∆H = (q − P∆V ) + P∆V
∆H = q p

Thus we see that ∆ equals to heat, qp, absorbed or evolved at constant

pressure.

Enthalpy of a system is a state function just like internal energy so that the
magnitude of ∆H depends only on the enthalpies of the initial and final states
according to the equation
∆H = ∆H final − ∆Hinitial

DIFFERENCES BETWEEN ∆ AND ∆

Often differences in ∆E and ∆H are small

(i) Reactions in which reactants and products are liquids or solids of necessity
involve only very small P∆V and thus ∆H has nearly the same value as ∆E
(that is, ∆H ≈ ∆E )
(ii) Reactions in which gases are either consumed or produced have much larger
volume changes and the P∆V term is also much greater. Even in these cases
however, ∆E is usually much larger compared to the P∆V term so that ∆E
and ∆H are still nearly the same.

Example10.3: Converting between ∆H and ∆E

When 2.00 mole of H2 and 1.00 mole of O2 at 100oC and 1 atm react to produce 2.00 mol of
gaseous water at 100oC and 1 atm, a total of 484.5 kJ are evolved. What are
(a) ∆H and
(b) ∆E for the production of a single mole of H2O(g)
Solution:

(a) Since the reaction 2H2 ( g ) + O2 ( g ) 

→ 2H2O( g ) is occuring at constant
pressure,
− 484.5 kJ
q p = ∆H =
2 mol H2O
The minus sign, remember, signifies that the reaction is exothermic. For production of 1 mole
of water,
∆H = −242 .2 kJ mol −1 .

Dr. Tanweer Ahmad, Senior Lecturer, Department of Chemistry, CBU Page 8

 CH110 General Chemistry for B.Sc.Chemistry Ist year Semester I

(b) To calculate ∆E from ∆H , we have to use the equation for the definition of change in
enthalpy
∆H = ∆E + P∆V
Solving for ∆E we have
∆E = ∆H − P∆V .
For simplicity we assume that the gases in the reaction behave as ideal gases. Thus, taking Vi
and ni to be the initial volume and initial number of moles of reactants at a given P and T, we
obtain,
PVi=niRT
Similarly, for the final state (for the products),
PVf=nfRT
The pressure-volume work in the process is given by
(
PVf − PVi = P Vf − Vi = P∆V)
This is equal to
P∆V = n f RT − ni RT
= (n f − ni )RT
= (∆n)RT
The quantity ∆n =(number of moles of gaseous products)−(number of moles of gaseous
reactants). Using the coefficients in the chemical equation in part (a), we see that there are
two moles of gaseous products and three moles of gasesous reactants. Therefore
∆n = 2.00 mol - 3.00 mol = - 1.00 mol.
Next, taking R=8.314 J mol K-1 to obtain the desired energy units, we get
-1

( )
P∆V = (− 1.00 mol ) 8.314 J mol −1 K −1 (373 K )
= −3100 J = −3.10 kJ (when rounded )
What we have calculated, based on the chemical equation, is the P∆V work for the formation
of 2 mol of H2O(g). For 1 mol, P∆V = -1.55 kJ. Now we can calculate ∆E in kJ mol-1 as
∆E = ∆H − P∆V
(
∆E = −242.2 kJ mol −1 − − 1.55 kJ mol −1 )
= − 240.6 kJ mol −1
Notice that ∆H and ∆E are nearly not different, even in this reaction involving a change
in the number of moles of gas. They differ by only 0.6%.

In this last example we see that PV work involved in a chemical reaction in which gases are
either consumed or produced can be calculated by the simple expression

Dr. Tanweer Ahmad, Senior Lecturer, Department of Chemistry, CBU Page 9

 CH110 General Chemistry for B.Sc.Chemistry Ist year Semester I

− = (∆ )!"
Remember that ∆n is the change in the number of moles of gas on going from reactants to
products. The above equation does not apply to reactions in which only liquids and solids are
involved. For such reactions the volume changes are extremely small and the P∆V work is
usually negligible compared to other energy changes that take place. For such reactions ∆E
and ∆H are therefore essentially identical.

HESS’S LAW OF HEAT SUMMATION

Since enthalpy is a state function, the magnitude of ∆ for a chemical reaction does not
depend on the path taken by the reactants as they proceed to form the products. As an
example, lets consider the conversion of 1 mol of liquid water at 100oC and 1 atm to 1 mol of
vapor at 100oC and 1 atm. This process absorbs 41 kJ of heat for each mole of H2O vaporized
and hence ∆H = +41 kJ. This reaction can be represented as

H 2 O(l) 
→ H 2 O(g) ∆H = +41 kJ
An equation written in this manner, in which the energy change is also shown, is called
a thermochemical equation and is nearly always interpreted on a mole basis. Here, for
instance, it can be seen that 1 mol of H2O(l) is converted to 1 mol of H2O(g) by the
absorption of 41 kJ.

The value of ∆H for this change will always be +41 kJ, provided that we refer to the same
pair of initial and final states.

We could even go so far as to first decompose the 1 mol of liquid into gaseous hydrogen and
oxygen and then recombine the elements to produce H2O(g) at 100oC and 1 atm. The net
change in enthalpy would still be the same, +41 kJ. Consequently, it is possible to look at
some overall change as the net result of the sequence of chemical reactions. The net value of
∆# for the overall process is merely the sum of all the enthalpy changes that take place
along the way. These last statements constitute Hess’s Law of Heat Summation. An
alternative statement of Hess’s Law is – If a reaction is carried out in a series of steps, the
∆# for the reaction will be equal to the sum of the enthalpy changes for the individual
steps.

The thermochemical equations serve as a useful tool for applying Hess’s Law. For
example, the thermochemical equations corresponding to the indirect path described above
for the vaporisation of water are

H 2O(l) 
→ H 2 (g) + 12 O 2 (g) ∆H = +283 kJ

Dr. Tanweer Ahmad, Senior Lecturer, Department of Chemistry, CBU Page 10

 CH110 General Chemistry for B.Sc.Chemistry Ist year Semester I

H 2 (g) + 12 O 2 (g) 
→ H 2O(g) ∆H = −242 kJ

Fractional coefficients in thermochemical equations are allowed because they refer to moles
not atoms or chemical species.

The net change (the vapourisation of one mole of water) is obtained by adding the two
chemical equations together (the way we add simultaneous equations) and then cancelling
any quantities that appear on both sides of the arrow (or chemical equation). Thus we have

H 2O(l) + H 2 (g) + 12 O 2 (g) 

→ H 2O(g) + H 2 (g) + 12 O 2 (g) or
H 2O(l)  → H 2O(g)
The heat of the overall reaction is equal to the algebraic sum of the heats of reaction for the
two steps.
∆H = +283 kJ + (− 242 kJ )
= + 41 kJ

Thus, when we add thermochemical equations to obtain some net change, we also add
their corresponding heats of reaction.
To describe the nature of these thermochemical changes, we can also illustrate them
graphically using an enthalpy diagram as shown in Fgiure 5.2 below

Figure 5.2: Enthalpy Diagram for the Evaporation of Water

Example 5.4: Applying Hess’s Law by Combining Thermochemical Equations

The thermochemical equation for the combustion of acetylene, a fuel in torches, is given by
equation (1).

(1) 2C2H 2 (g) + 5O2 (g) 

→ 4CO2 (g) + 2H 2O(l) ∆H1 = −2602 kJ

Dr. Tanweer Ahmad, Senior Lecturer, Department of Chemistry, CBU

 CH110 General Chemistry for B.Sc.Chemistry Ist year Semester I

Ethane, another hydrocarbon fuel, reacts as follows:

(2) 2C 2 H 6 (g) + 7O 2 (g) 

→ 4CO 2 (g) + 6H 2O(l) ∆H 2 = −3123 kJ

while hydrogen and oxygen combine following this equation

(3) H 2 (g) + 12 O 2 (g) 

→ H 2O(l) ∆H 3 = −286 kJ

All these data correspond to the same temperature and pressure, that is, RTP where 25oC and
1 atm. Use these thermochemical equations to calculate the heat of hydrogenation of
acetylene,

(4) C 2 H 2 (g) + 2H 2 (g) 

→ C 2 H 6 (g) ∆H 4 = ?

Reasoning and Solution:

To solve this problem we must combine equations (1), (2) and (3) in such a way that when
they are added, everything cancels except the formulars in the desired equation (4). This
involves keeping an eye on the final equation as we rearrange the given equations so that they
add up correctly. For example, we have 1C2H2(g) on the left hand side (LHS) of equation (4),
so we want to use equation (1) with its coefficients divided by two (to give equation 5). We
must also have 2H2(g) on the LHS, so we have to take equation (3) multiplied by 2, and we
have to multiply its ∆H by 2 (this gives equation 6). Finally, we have 1 C2H6(g) on the RHS
of equation (4). To get it there, we must reverse equation (2) and divide it coefficients by 2
(to give equation (7)). When we reverse an equation we change the sign of its ∆H because it
changes an exothermic reaction into an endothermic one. In this case, we must divide by 2 .
This gives us, therefore,
− 2602 kJ
(5) C2 H 2 (g) + 52 O2 (g) 
→ 2CO2 (g) + H 2O(l) ∆H5 = = − 1301 kJ
2
(6) 2H 2 (g) + O 2 (g) 
→ 2H 2O(l) ∆H 6 = 2(− 286 kJ ) = − 572 kJ
+ 3123
(7) 2CO2 (g) + 3H2O(l) 
→ C2H6 (g) + 72 O2 (g) ∆H7 = kJ = + 1561kJ
2
Adding equations 5, 6 and 7 gives
C2H2(g) + 2H2(g) + 7 O (g)
2 2
+ 3H2O(l) + 2CO2(g) 
→

2CO2(g) + 3H2O(l) + C2H6(g) + 7 O (g)

2 2
Cancelling elements that are on both sides, we get
C 2 H 2 (g) + 2H 2 (g) 
→ C 2 H 6 (g)

Dr. Tanweer Ahmad, Senior Lecturer, Department of Chemistry, CBU Page 12

 CH110 General Chemistry for B.Sc.Chemistry Ist year Semester I

Which is the equation we want. Since it is obtained by adding equations (5), (6) and (7), its
∆H ( ∆H4 in the statement of the problem) is obtained by adding the ∆H s of (5), (6) and (7).
∆H 4 = ∆H 5 + ∆H 6 + ∆H 7
= (− 1301 kJ ) + (− 572 kJ ) + (1561 kJ )
= − 312 kJ

HEATS OF FORMATION

A particularly useful type of thermochemical equation corresponds to the formation of one

mole of a substance from its elements. The enthalpy changes associated with these reactions
are called heats of formation or enthalpies of formation and are denoted as ∆H f . For

example, thermochemical equations for the formation of liquid and gaseous water at 100oC
and 1 atm are, respectively

H 2 (g) + 12 O 2 (g) 
→ H 2O(l) ∆H f = −283 kJ mol −1

H 2 (g) + 12 O 2 (g) 
→ H 2O(g) ∆H f = −242 kJ mol −1

We use these equations to obtain the heat of vapourisation of water by first reversing the first
equation and then adding it to the second. When reversing the first equation we should also
reverse the sign of ∆H as shown below.

Exothermic H 2 (g) + 12 O 2 (g) 

→ H 2O(l) ∆H f = −283 kJ mol −1

Endothermic H 2O(l) 
→ H 2 (g) + 12 O 2 (g) ∆H = − ∆H f = +283 kJ mol −1

When this last equation is added to that of formation of H2O(g), we obtain

H2O(l) + H2(g) + 1 O (g) 

→ H2O(g) + H2(g) + 1 O (g)
2 2 2 2

H2O(l) 
→ H2O(g)
And heat of reaction is
∆H = ∆H%,'()(*) − ∆H%,'()(+)
∆H = −242kJ − (−283kJ) = +41kJ

∆H reaction = (sum of ∆H f of products ) − (sum of ∆H f of reactants )

STANDARD STATES

The magnitude of ∆H f depends on the conditions of temperature, pressure and physical state

(gas, liquid, solid, crystalline form) of the reactants and products. For instance, at 100oC and

1 atm, ∆H f H O (l) = −283 kJ while at 25oC and 1 atm, ∆H f H O (l) = −286 kJ.
2 2

Dr. Tanweer Ahmad, Senior Lecturer, Department of Chemistry, CBU Page 13

 CH110 General Chemistry for B.Sc.Chemistry Ist year Semester I

To avoid the necessity of always having to specify the conditions for which ∆H f is recorded

and to permit comparisons between ∆H f for various compounds, a standard set of conditions

is chosen, usually 25oC and pressure of 1 atm (101.325 kPa) as explained in Zumdahl and
Zumdahl 9th edition on page 265. Under these conditions a substance is said to be in its
standard state. Heats of formation of substances in their standard states are indicated as ∆H of .

For example, the standard heat of liquid water ∆H fo H 2 O (l) = −286 kJ/mol and represents the

heat liberated when H2 and O2, each in their natural form at 25oC and 1 atm, react to produce
H2O(l) at 25oC and 1 atm.

A table that contains standard heats of formation permits us to calculate standard heats of

reaction ∆H o for different chemical changes. In performing these calculations, we arbitrarily

take ∆H of for an element in its natural, most stable form at 25oC and 1 atm to be equal to zero.

Example 5.5: Calculating ∆H o for a reaction from standard heats of formation

Many careful cooks keep sodium bicarbonate (baking soda) handy because it is a good
extinguisher of oil or grease fires. Its decomposition products help smother the flames. The

→ Na 2CO 3(s) + H 2O (g) + CO 2(g) . Calculate ∆H o for the reaction.

reaction is 2NaHCO 3(s) 

Reasoning and Solution:

Our important equation is

∆H o =(sum of ∆H of products ) - (sum of ∆Hof reactants )

This means that we must add up all the heat evolved during the formation of the products
from their elements and then subtract the heat evolved by the formation of the reactants from
their elements. Using data from Table 12.1 in Brady and Humison, 4th Edition or Appendix 3
of the Raymond Chang, Chemistry 10th Edition, we see that the total enthalpy of formation
of the products is
 − 1131 kJ 
1 mol Na 2CO3(s) ×   = −1131 kJ
 1 mol Na 2CO3(s) 
 

 − 394 kJ 
1 mol CO 2(g) ×   = −394 kJ
 1 mol CO 2(g) 
 
 − 242 kJ 
1 mol H 2O(g) ×   = −242 kJ
 1 mol H 2O(g) 
 

Dr. Tanweer Ahmad, Senior Lecturer, Department of Chemistry, CBU Page 14

 CH110 General Chemistry for B.Sc.Chemistry Ist year Semester I

Total ∆Hof of products = −1767 kJ

 − 947.7 kJ 
For the reactant we have 2 mol NaHCO3(s) ×   = −1895.4 kJ
 1 mol NaHCO3(s) 
 

Recalling that ∆H o = (sum of ∆H of products ) - (sum of ∆Hof reactants ) gives

∆H o = −1767 kJ − (− 1895 kJ )
= + 128 kJ
Comment:
Notice that in calculating ∆H o for the overall reaction we have multiplied each ∆Hof by the
appropriate coefficient from the equation. This gives the total heat of reaction for the number
of moles specified by the chemical reaction.
Example 5.6: Calculating ∆H o for a reaction from standard heats of formation
Calculate ∆H o for the reaction

2Na 2O 2(s) + 2H 2O(l) 

→ 4NaOH (s) + O 2(g) .

How many kilojoules are liberated when 25.0 g of Na2O2 react according to the equation?

Reasoning and Solution:

We use the equation is

∆H o = (sum of ∆H of products ) – (sum of ∆Hof reactants )

Which gives

∆H o = [ 4∆H fo NaOH (s) + ∆H fo O 2 (g) ] – [ 2∆H of Na 2 O 2 (s) + 2∆H fo H 2 O (g) ]

All the data are from Table 12.1 in Brady and Humison, 4th Edition except ∆H fo O 2 (g) which we

know (or should know) that heat of formation of a pure element is zero. Alternatively using
Appendix 3 of the Raymond Chang, Chemistry 10th Edition, gives all data for the above
equation. Therefore
 − 426.8 kJ 
∆H o = [4 mol ×   + 0.00] −
 1 mol NaOH(s) 
 
 − 504.6 kJ   
[2 mol ×   + 2 mol ×  − 286 kJ 
 1 mol Na 2O 2(s)   1 mol H 2O(l) 
   
= [−1707 kJ] − [−1009.2 kJ − 572 kJ]
= −1707 kJ + 1581 kJ
= − 126 kJ

Dr. Tanweer Ahmad, Senior Lecturer, Department of Chemistry, CBU Page 15

 CH110 General Chemistry for B.Sc.Chemistry Ist year Semester I

To calculate the number of kilojoules liberated by 25.0 g Na2O2, we have to realise that the
∆H o that we calculated is the energy given off when 2 mol of Na2O2 react. Therefore, 2 mol
Na2O2 gives -126 kJ. The molar mass of Na2O2 is 78.0 g mol-1 so that

   
25.0 g Na2O2 ×  1 mol Na 2O 2  ×  − 126 kJ  = − 20.2 kJ
 78.0 g Na O   2 mol Na O 
 2 2  2 2

Thus, the reaction of 25.0 g of Na2O2 liberates 20.2 kJ.

Comment:
25
25 g = mol of Na 2O2 = 0.32 mol
Notice that 78 .
-126 kJ is liberated by 2 mol
-63 kJ is liberated by 1 mol
Therefore -63 kJ/mol x 0.32 mol is liberated by 0.32 mol
and -63 x 0.32 kJ = -20.2 kJ

INDIRECT DETERMINATION OF ∆H of

It is often impossible to measure directly the heat of formation of a compound; for example,
formation of methane, CH4, from gaseous hydrogen and graphite. Therefore, in order to
determine ∆H of for compounds such as methane, indirect methods must be used. One

technique that works well with many carbon-containing compounds is to burn the compounds
in a bomb calorimeter and measure the heat of combustion. In these reactions, the heats of
formation of the products are generally known, so the unknown value of ∆H of for the

reactant can be calculated as shown in the example below.

Example 5.7: Using the heat of combustion to calculate

The combustion of 1 mol of benzene, C6H6(l), to produce CO2(g) and H2O(l) liberates 3271
kJ when the products are returned to 25oC and 1 atm. What is the standard heat of formation
of C6H6(l) expressed in kilojoules per mole?

Reasoning and Solution:

The equation for the combustion of 1 mol of C6H6 is

C 6 H 6(l) + 7 12 O 2(g) 
→ 6CO 2(g) + 3H 2O (l)

The standard heat of reaction, ∆H o =3271 kJ. From the equation

∆H o = (sum of ∆H of products ) – (sum of ∆Hof reactants )

Dr. Tanweer Ahmad, Senior Lecturer, Department of Chemistry, CBU Page 16

 CH110 General Chemistry for B.Sc.Chemistry Ist year Semester I

we have
∆H o = [ 6∆H fo CO 2 (g) + 3∆H fo H 2 O (l) ] – [ ∆H fo C 6 H 6 (l) + 7 12 ∆H fo O 2 (g) ]

Since ∆H fo O 2 (g) = 0.0 by definition,

∆H o = [ 6∆H fo CO 2 (g) + 3∆H fo H 2 O (l) ] – [ ∆H fo C 6 H 6 (l) ]

Solving for the heat of formation of benzene, we have

[ ∆H fo C 6 H 6 (l) ] = [ 6∆H of CO 2 (g) + 3∆H of H 2 O (l) ] – ∆H o

From Appendix 3 Thermodynamics data in Raymond Chang’s Chemistry, 10th Edition, we

can obtain the heats of formation of CO2(g) and H2O(l). Therefore,
[ ∆H fo C 6 H 6 (l) ] = 6(− 394 ) kJ + 3(− 286 ) kJ – (− 3271) kJ

[ ∆H fo C 6 H 6 (l) ] = + 49 kJ

Since 1 mole of C6H6 is involved, ∆H of = +49 kJ mol −1

BOND ENERGIES
When we started talking about the 1st Law of Thermodynamics, we introduced the term
internal energy, ∆E , that is used to describe energy changes of chemical or physical
systems. We said internal energy is the total energy of the system that it possesses as a
consequence of its kinetic energy of its atoms, ions or molecules, potential energy that arises
from binding forces between the particles making up the system.

Heats of reactions can be associated to changes in potential energy resulting from breaking
and forming chemical bonds. Strictly speaking, we should use ∆E for this purpose, however,
since P∆V is small in ∆H , we can use ∆H in place of ∆E and obtain reasonable results.
Thus, the terms bond energy and bond enthalpy can be used interchangeably.
Bond energy is defined as the energy required to break a bond to produce neutral fragments.

For a complex molecule the energy needed to break all the bonds and reduce gaseous
molecules to neutral gaseous atoms in the atomisation energy. Its value is the sum of all the
bond energies in the molecule. Simple diatomic molecules like H2, Cl2, HCl or O2 possess
only one bond, so the atomisation energy is the same as the bond energy.

As an example, we will consider the molecule of CH4 whose standard heat of formation has
been experimentally determined to be -74.9 kJ mol-1. This corresponds to the enthalpy

change, ∆Hof , for the reaction

Dr. Tanweer Ahmad, Senior Lecturer, Department of Chemistry, CBU Page 17

 CH110 General Chemistry for B.Sc.Chemistry Ist year Semester I

C(s,graphite) + 2H 2(g) 
→ CH 4(g)

An alternative path to take us from free elements to the methane follows the series of
successive reactions below:

1. C (s, graphite) 
→ C (g) ∆H1

2. 2H 2(g) 
→ 4H (g) ∆H 2

3. C (g) + 4H (g) 
→ CH 4(g) ∆H 3

The sum of these reactions will give the overall reaction so that the sum of their ∆H s must

equal to the ∆H of the overall reaction (that is, ∆Hof for CH4)

∆Hof = ∆H1 + ∆H 2 + ∆H3

Step 1 and Step 2 involve formation of gaseous atoms from their elements as found in tables

like Appendix 3 in Raymond Chang’s Chemistry, 10th Edition, ∆Hof for CH4 has been
experimentally measured so that only ∆H 3 is unknown and its understood to be the negative
of the atomisation energy (negative because Step 3 involves the formation of chemical
bonds).
∆H3 = −∆Hatomisation for CH 4

Substituting and solving for the atomisation energy gives

∆H atom = ∆H1 + ∆H 2 − ∆H fo of CH 4

Using values from tables, we have

∆H1 = +715 kJ - heat of formation of gaseous carbon atoms
∆H 2 = 4(+ 218 kJ ) - four times the heat of formation of 1 mole of
gaseous H atoms
∆H of of CH 4 = −74 .9 kJ - heat of formation of CH4

∆Hatom = (+ 715 kJ ) + (872 kJ ) − (- 74.9 kJ )

= + 1662 kJ
This is the total energy needed to break all 4 C-H bonds in 1 mole of CH4.

Therefore average bond energy per mol of C-H bonds in CH4 is

∆Hatom 1662 kJ
= = 415 kJ mol−1
4 4
Example 5.8: Calculating bond energies from thermodynamic data
The heat of formation of ethylene C2H4, is +51.9 kJ/mol. The structure of the molecule is

Dr. Tanweer Ahmad, Senior Lecturer, Department of Chemistry, CBU Page 18

 CH110 General Chemistry for B.Sc.Chemistry Ist year Semester I

Assuming the C-H bond energy is 415 kJ/mol, calculate the C=C bond energy

Reasoning and Solution:

Let us examine the two alternative paths shown below that take us from the reactants [C(s)
and H2(g)] to the product [C2H4(g)]. The sum of the energy changes to steps 1, 2 and 3 must be

the same as that for the direct path, ∆Hof .

2C(g) + 4H (g)
↑ ∆H1 ↑ ∆H 2 ∆H 3
2C(s) + 2H 2(g) 
o→ C 2 H 4(g)
∆H f

→ ∆Hof
∆H1 + ∆H 2 + ∆H 3 

Step 1 ∆H1 is twice the heat of formation of a mole of carbon atoms.

∆H1 = 2(715 kJ ) = 1430 kJ
Step 2 ∆H 2 is four times the heat of formation of a mole of hydrogen atoms.
∆H 2 = 4(218 kJ ) = 872 kJ
Step 3 ∆H 3 is the negative of the atomisation
∆H 3 = − ∆H atom

Therefore
1430 kJ + 872 kJ + ( − ∆H atom ) = +51.9 kJ

Solving for ∆H atom yields

∆Hatom = 1430 kJ + 872 kJ − 51.9 kJ

= 2250 kJ
This is the energy required to break 4 moles of C-H bonds plus one mole of C=C bonds, that
is,
∆H atom = 4 ∆H C − H + ∆H C = C

Substituting and solving for ∆H C = C , we get

∆H C = C = ∆H atom − 4 ∆H C − H
= 2250 kJ − 4(415 kJ)
= 590 kJ per mole of C = C bonds

It is worth noting that this value is within 3% of the accepted average C=C bond energy of
607 kJ mol-1

Dr. Tanweer Ahmad, Senior Lecturer, Department of Chemistry, CBU

 CH110 General Chemistry for B.Sc.Chemistry Ist year Semester I

Average bond energies in tables can be used to calculate heats of formation with a fair degree
of accuracy. This is so because nearly all C-H bonds are pretty much alike, whether in a small
molecule such as CH4 or a large complex one such as C46H86.

Dr. Tanweer Ahmad, Senior Lecturer, Department of Chemistry, CBU Page 20