FSc Notes Chemistry Part 1 Chapter 7 Thermochemistry Lecture 5

FSc Notes Chemistry Part 1 Chapter 7 Thermochemistry Lecture 5

Laws of Thermochemistry:

There are two laws of thermo chemistry.

According to this law:
The amount of heat evolved during the formation of one mole of a substance from its elements of heat  absorbed when this compound is broken down back into its reactants ( elements ).

For example:

H2(g) + 1/2 O2----------------H2O(l) ?H = - 285.9 KJ/ mol.

H2O(l) -----------------H2(g) O2(g) ?H = + 285.9KJ/ mol.

Hess's Law ( Law of Constant Heat Summation):

In 1840, Hess presented a law about the enthalpy ( heat content ) of a chemical reaction which is known as Hess's law or law of constant heat summation.

This law states that, " The enthalpy of a chemical reaction is carried out in single step or several steps, provided the initial and final conditions are the same"

So we can say that the heat change in a given reaction depends only on the initial and final states of the system and is independent of the path followed.

Lets consider an example

If a reactants. A is to be converted into the products " D". Then this reaction may be carried out in single step or several step.

i.e

Single steps:

A-------------- D ?H= a kj/ mole

Several steps:

A ---------B ?H1 = xKj/ mol

A --------C ?H2 = YkJ/ mol

C --------D ?H3 = zKj/ mol

The according to Hess's law:

?H = ?H1 + ?H2 + ?H3

A = x + y + z

Importance of Hess's Law:

The enthalpy of reactions which go to the completion without side reaction, can easily be measured by calorimetry. But those in majority of reaction, some sort of side reaction occurs. The enthalpy of such reactions cannot be measured directly. The enthalpy of such reactions is measured indirectly by using Hess's  law.

For example, consider the formation of CO2 from C and O2 .

This reaction can be carried out in two ways ie Direct Method & Indirect Method. in direct method, C is heated in excess of O2 ie

C(3) + O2(g)---------------------CO2(g) ?H = ---------I

In indirect method C is converted into Co2 in two steps which are

Co + 1/2 O2 (g) ----------------CO(g) ?H= --------II

CO(g) + 1/2 O2(g) -------------CO2(g) ?H2 = ------III

Not Result: C(g) + O2(g) --------CO2(g) ?H = ?H1 + ?H2

Hess's law.

In the above reaction: the enthalpy of reaction: II cannot be determined directly because during occurs this reaction, a side reaction occurs producing a small amount of Co2. Thus as a result of reaction II: the products ( CO ) certain a little Co2. Thus ? H1 is determined indirectly by using Hess's law.

The enthalpy changes for reaction I and III can be determined directly which are:

C(s) + O2(g)----------------CO2(g) ? H = -39 4 KJ/ mole -------I

&

CO(g) + 1/2 O2(g)-----------------CO2(g) ? H2 = - 284 KJ / mole

&

C(g) + 1/2 O2(g)----------------CO(g) ? H1 = ?

According to Hess's law:

? H = ? H1 + ? H2

Or

? H1 = ? H - ? H2 ====== - 394 – (-284 )

? H1 = - 394 + 284 ========= - 110 KJ / mole

Thus

C(s) + 1/2 O(g) ----------------CO(g) ? H1 = - 110 KJ / mole

Born-Haber Cycle:

The Born-Haber cycle: is technique which is used for applying Hess's law to the standard enthalpy changes, which occur when an ionic compound is formed. The lattice energy ( energy evolved when one mole of a crystal lattice is formed from its ions in gaseous state ) of an ionic compound, cannot be measured directly b/c we don’t have gaseous ions in the laboratory. Thus the lattice energy of a compound ( ionic ) is determined by Born-Haber cycle.

Ionization of Na(G):

mole gaseous sodium atoms absorbs a particular amount of energy to form one mole gaseous Na+ ion ( endothermic. The amount of energy required to evolve electrons from Na(g) atoms is known as ionization energy of Na ie

Na(g) -----------------Na+ (g) + ie ? Ho

I.E = 493.71KJ / mole

Bond Dissociation of Cl2:

One – half mole of Cl2 is dissociated into 1 mole gaseous chlorine atoms ( gaseous state ) by providing an amount of energy, known as bond dissociation energy of Cl2. ( endothermic process )

Ie 1/2 CL2(g) -------------------------------Cl(g) 1/2 ? Ho

B.D = 242.67/2 = 121 KJ / mole

Ionization  of Cl(g):

One mole of chlorine atoms ( in gaseous state ) gain one mole of electrons from sodium atoms, by evolving of a particular amount of energy which is known as electron affinity of Cl ( exothermic process )

Cl(g) + ie -------------Cl- (g) ? Ho

E.A = - 364.01KJ/ mole

Formation of Crystal Lattice of NaCL:

One mole gaseous sodium ions ( Na+ (g) ) and one mole of gaseous chloride ions ( Cl-(g) ) combine by means of ionic bond to form the crystal lattice one NaCl. The amount of energy evolved is known as lattice energy of NaCl and this cannot be measured directly. ( exothermic process ).

Written by: Asad Hussain