On the basis of numerous thermochemical observations, the Russian Chemist, Germain Henry Hess gave one of the most important generalisation of thermochemistry in 1840 which is known after his name as Hess’s Law. This generalisation is primarily based on the fact that enthalpy is a path independent thermodynamic function. In other words, the enthalpy change in a chemical reaction is same whether it occurs in one step or in more than one steps. This may be stated in the form of Hess’s law as follows. The standard enthalpy of reaction is the sum of algebraic sum of the standard enthalpies of reactions into which the overall reaction may be splitted or divided.

Consider a process involving the conversion of reactant A into B in one step (path I). The enthalpy change of a process is represented by MI. Now suppose the process is carried out in two steps involving a change from A to C and C to B according to path IT, (Fig. 17 .4). Let MI1 and ΔH_{2} be enthalpy changes from A to C and C to B respectively. Then, according to Hess’s law:

ΔH = ΔH_{1} + ΔH_{2}

Hess’s law is simply a corollary of the law of conservation energy. It implies that the enthalpy change of reaction depends only on the states of reactants and products and not on the manner by which the change is brought about. If Ml were not independent of the manner in which the corresponding change was brought about, it would have been 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 second path.

**ILLUSTRATION OF HESS’S LAW**

Let us consider the formation of carbon dioxide from carbon and oxygen. There are two ways by which the change can be brought about.

(a) Conversion of carbon to carbon dioxide

C(s) + O2 (g) à CO2 (g) ; ΔH = – 393.5 KJ

(b) Conversion of carbon to carbon monoxide and subsequent oxidation of carbon monoxide to carbon dioxide.

C(s) + 1/2 O2(g) à CO(g) ; ΔH_{1} = – 110.5 KJ

CO(g) + 1/2 O_{2} (g) à CO_{2}(g) ; ΔH_{2} = – 283.0 KJ

The two paths by which carbon can be converted into carbon dioxide have been shown in Fig. 17.5.

As is clear , ΔH = ΔH_{1} + ΔH_{2}

- 393.5 KJ = – 110.5 + ( -283.0) KJ

= – 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 the enthalpies of different reactions whose direct measurement is not possible. The thermochemical equations like algebraic equations can be added or substracted, multiplied or divided by any numerical factor. Some of the important applications of Hess’s law are discussed below:

**1. Determination of Enthalpy of formation**

There are large number of compounds such as methane, benzene, carbon monoxide, ethane, etc., which cannot be directly synthesized from their elements. Hence, their enthalpy of formation cannot be determined by the calorimetric method. In case of such substances the enthalpy of formation can be determined by an indirect method based upon Hess’s law. The unknown reaction is made one segment of the closed cycle of reactions. The enthalpy changes of the reactions which represent other segments of the cycle can be determined experimentally. Then by applying Hess’s law the desired value of Ml can be calculated.

For example, let us study the calculation of enthalpy of formation of so_{3}.

The enthalpy of formation of S0_{2} (ΔH_{1}) and enthalpy of combustion of so_{2} (ΔH_{2}) have been found to be -297.4 Kj moi^{-1} and – 97.9 kJ mol^{- 1} respectively. These changes have

been shown in Fig. 17.6.

Now according to Hess’s law,

ΔH = ΔH_{1} + ΔH_{2} = – 297.4 + (-97.9) = – 365.3 KJ.

The same result can be arrived at by means of thermochemical calculations

(i) Conversion of sulphur to sulphur dioxide

S + O_{2} à SO_{2} ; ΔH_{1} = – 297.4 KJ.

(ii) Combustion of sulphur dioxide to SO3

SO_{2 }+ 1/2 O_{2} à SO_{2} ; ΔH_{1} = – 297.4 KJ.

Add equations (i) and (ii),

S + 3/2 O_{2} à SO_{3} ;

ΔH = ΔH_{1} + ΔH_{2} = – 297.4 – 97.9 = – 395.3 KJ.

See solved example for further illustration.

**2. Determination of Enthalpy of Transition**

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

For example, let us calculate the enthalpy of the process SR àS_{M} from the enthalpies of combustion of monoclinic sulphur (S_{M}) and rhombic sulphur (S_{R}) which have been found to be- 296.4 and- 295.1 kJ mo11 respectively. Now Hess’s law cycle is constituted as S_{(R) }à SO2 (path I) and S_{(R) }às_{(M) }àSO2 (path II). These changes have been shown in Fig. 17.7

Now according to Hess’s law:

ΔH = ΔH_{1} + ΔH_{2}

ΔH = ΔH_{1} + ΔH_{2} = – 295.4 – (-296.9) = – 1.3 KJ.

The enthalpy of allotropic transformation of rhombic sulphur to monoclinic sulphur can also be calculated as follows by thermochemical calculations:

1. Combustion of rhombic sulphur

S_{(R) }+ O_{2 }(g) à SO_{2}(g) ; ΔH_{1} = – 295.1 KJ

2. Combustion of monoclinic sulphur

S_{(M) }+ O_{2}(g) à SO_{2}(g); ΔH_{2} = – 296.4 KJ

Subtracting equation (2) from equation (1)

S_{(R) }à S_{(M)} ; ΔH = ΔH_{1} – ΔH_{2}

= – 295.1 – ( -296.4) = + 1.3 KJ.

**3. Determination of Enthalpy of Hydration of Anhydrous Salt**

Hydration means the conversion of an anhydrous salt into its hydrate by combining with specific number of moles of water. For example, one mole of anhydrous copper sulphate combines with five moles of water to form one mole of hydrated copper sulphate.

Direct measurement of enthalpy of hydration is not possible. However, use of Hess’s law makes it possible as described below:

The enthalpies of dissolution of hydrated copper sulphate and that of anhydrous copper sulphate have been experimentally found to be + 11.70 kJ mol^{-1} and – 66.5 kJ mol^{-1} respectively. From these change the Hess’s law cycle is constituted according to Fig. 17 .8.

_{ }

According to Hess’s law,

ΔH = ΔH_{1 }+ ΔH_{2}

H_{1} = H – H_{2} = – 66.5 KJ – 11.70 KJ

= – 78.20 KJ.

**4. Determination of standard Heat of Reaction**

Hess’s law cycle can also be used for the determination of standard heat of the reaction from the knowledge of standard heats of formation of various reactants and the products. The Hess’s law cycle for elements in their standard states, reactants and products can be written as shown in Fig. 17.9.

= Sum of the standard enthalpy of formation of reactants

= Sum of the standard enthalpy of formation of the products

= Standard enthalpy change of the reaction Now according to Hess’s law,