Enthalpy of Physical changes and Hess's Law

Enthalpy of Physical Changes

The three states of matter solid, liquid and gas differ from one another in the arrangement of their constituents particles. The magnitude of intermolecular forces acting between the particles in three states are so different.

It is well known that when a solid is converted into liquid state, energy is to be supplied. This energy is used to break the intermolecular forces in the solid which are of high magnitude.

Whenever, there is a change in the state of matter (solid \(\rightarrow\) liquid or liquid\(\rightarrow\) gas), the process is called phase change or transition. It is also accompanied by the change in enthalpy or heat content of the system.

1. Enthlapy (heat) of Fusion :

It is defined as the heat (enthalpy change when one mole of solid substance is converted into the liquid state at its melting point.

For example, when one mole of ice melts at its melting points 0⁰C or 273K, the process can be represented as

H₂O(s) \(\longrightarrow\) H₂O(l); \(\Delta\)H = +1.43 kcals.

Ice Water

and is accompanied by the absorption of 1.43kcals of heat.

The enthalpy of fusion depends upon the intermolecular forces of attraction between the molecules of different solids. Ionic solids such as sodium chloride, barium chloride, etc. have high enthalpies of fusion because of strong electrostatic forces of attraction between their constituents. On the other hand, molecular solids such as oxygen, bromine and helium have low enthalpies of fusion because the attractive forces between their molecules are the weak Van der Waals’ forces of attraction.

Thus, from the values of heats of fusion of various substances we can compare their magnitudes of intermolecular forces. Greater is the heat of fusion of as substance, higher is the magnitude of intermolecular forces.

2. Heat (enthalpy) of vaporization :

It is defined as the heat change ( or enthalpy change ) when one mole of liquid is converted into vapours or gaseous state at its boiling point.

For example, when one mole of water is converted into steam at 100⁰C or 373K, the heat absorbed is 9.71 kcal, which is the heat of vaporization of water. The change can be represented as

H₂O(l) \(\longrightarrow\) H₂O(g) ; \(\Delta\)H₃₇₃ = +9.71 kcals.

The heat ofa vaporization of ethyl alcohol (C₂H₅OH) and C₆H₆ are +7.29 kcal/mole and +7.36 kcal/mole respectively. The value of heats of vaporization can also be used for the comparision of the magnitude of intermolecular forces of attraction in liquids.

3. Heat (enthalpy) of Sublimation :

Sublimation is a process in which a solid on heating changes directly into gaseous state without changing into liquid state. It occurs at a temperature below the melting point of the solid. Heat of sublimation is defined as 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. For example, heat of sublimation of iodine is +14.92 kcal/mole. It can be represented as

I₂(s) \(\longrightarrow\) I₂(g); \(\Delta\)H = +14.92 kcals.

Hess’s Law of Constant Heat Summation

In 1840, G.H Hess stated one of the most important generalization of thermochemistry known after his name as Hess’s law. The law states that ‘ the enthalpy change in a chemical or physical process is same whether the process is carried out in one step or in several steps.’ If the reaction takes place in more than one step, the enthalpy of the reaction is equal to the algebraic sum of the enthalpies of various step reactions.

Let us consider a process involving the conversion of reactant A into B in one step (Path I). Let the enthalpy change of the process is \(\Delta\)H. Now suppose, the same process is carried out in two steps involving a change from A to C and C to B respectively. Then, according to Hess’s law

\(\Delta\)H = \(\Delta\)H₁ + \(\Delta\)H₂

Hess’ law is simply a corollary of the law of conservation of energy. It implies that the enthalpy change of reaction depends only on the states of reactants and products and not on the manner in which the change is brought about. If \(\Delta\)H were not independent of the manner in which the corresponding change was brought about, it would be possible to create or destroy energy by taking a system from A to B by one path and then returning it to state A by the second path.

Illustration of Hess’ law

Let us consider the formation of sulphur trioxide from sulphur and oxygen. There are two wasy by which the change can be brought about.

1. By the direct combination of sulphur and oxygen.

S(s) + 3/2 O₂(g) \(\longrightarrow\) SO₃(g); \(\Delta\)H = -395.3 kJ.

1. By converting sulphur to sulphur dioxide and subsequent oxidation of sulphur dioxide to sulphur trioxide.

S(s) + O₂(g) \(\longrightarrow\) SO₂(g); \(\Delta\)H₁ = -297.4 kJ.

SO₂(g) +1/2 O₂(g) \(\longrightarrow SO₃(g); \(\Delta\)H₂ = -97.9kJ.

Therefore, \(\Delta\)H = \(\Delta\)H₁ + \(\Delta\)H₂

or, -395.3 = -297.4 + (- 97.9) = -395.3 kJ

As another illustration let us consider the formation of carbon dioxide from carbon and oxygen through two paths.

1. Conversion of carbon to carbon dioxide.

C(s) + ½ O₂(g) \(\longrightarrow\) CO(g); \(\Delta\)H₁ = -110.5 Kj.

CO(g) + ½ O₂(g) \(\longrightarrow\) CO₂(g); \(\Delta\)H₂ = -283.0 Kj.

The two paths by which carbon can be converted to carbon dioxide have been shown above.

Therefore, \(\Delta\)H = \(\Delta\)H₁ + \(\Delta\)H₂

or, -393.5 = -110.5 + (-283.0) = -393.5 Kj.

Applications of Hess’s law

The practical utility of Hess’s law lies in the fact that it allows us to carry out thermochemical calculations to predict enthalpies of different reactions whose direct measurement is not possible. The thermochemical equations like algebraic equations can be added, substracted, multiplied and divided by any numerical factors. Some of the important applications are listed below :

1. Determination of heat of formation

There is a large number of compounds such as methane, benzene, carbon monoxide, ethane, etc, which cannot be determined by the calorimetric method. In such cases, the heat of formation cannot of substances can be determined by an indirect method based upon Hess’s law. Let us study the determination of the heat of the formation of carbon monoxide.

1. Formation of CO

C(s) + ½ O₂(g) \(\longrightarrow\) CO(g) ; \(\Delta\)H = ?

1. Combustion of carbon

C(s) + O₂(g) \(\longrightarrow\) CO₂(g); \(\Delta\)H₁ = -393.5 Kj.

• Combustion of CO

CO(g) + ½ O₂(g) \(\longrightarrow\) CO₂(g); \(\Delta\)H₂ = -283.0 Kj.

Substracting equation iii) from ii), we have

C(s) + ½ O₂(g) \(\longrightarrow\) CO(g); \(\Delta\)H = \(\Delta\)H – (\(Delta\)H₂ = -393.5 – (- 283.0)

= -393.5 + 283.0 = -110.5 Kj

Thus heat of formation of CO is -110.5Kj.

2. Determination of enthalpy of transition : Transition implies the conversion of one allotrophic form of asubstanceto another. For example, the change of graphite to diamond, red phosphorus to yellow phosphorus, rhombic sulphur to monoclinic sulphur, etc. Such reactions are very slow and enthalpy changes accompanying them cannot be measured directly. However, Hess’s law is quite helpful in determining enthalpy of transition.

The enthalpy of allotropic transformation of rhombic sulphur to monoclinic sulphur can be calculated as follows :

1. Combustion of rhombic sulphur

S_(rhombic) + O₂(g) \(\longrightarrow\) SO₂(g); \(\Delta\)H₁ = -295.1 kJ.

1. Combustion of monoclinic sulphur

S_(monoclinic) + O₂(g) \(\longrightarrow\) SO₂(g); \(\Delta\)H₂ = -296.4Kj.

Substracting equation ii) from equation i), we have

S_(rhombic) – S_(monoclinic) = 0; \(\Delta\)H = \(\Delta\)H₁ - \(\Delta\)H₂

= 295.1- (-296.4)

= -295.1 + 296.4 Kj

= +1.3 Kj

Thus, enthalpy of transition of rhombic sulphur to monoclinic sulphur is +1.3 Kj.