## Hess's Law of Constant Heat Summation

G.H.Hess proposed a law regarding the heats or enthalpies of reaction in 1840 called the Hess's law. This law states that 'the heat change in a particular reaction is the same whether it takes place in one step or several steps'.

For example, a reactant 'A' changes to a product 'B' in one step and the heat change during this process is DH. If the reaction is carried out in two steps where 'A' first changes to 'C' an intermediate stage and then 'C' changes to 'B' in the following step then let the heat change during the formation of 'A' to 'C' beDH_{1}and that from 'C' to 'B' be DH

_{2}

_{.}From Hess's law the heat change for the reaction is given as:

DH = DH_{1} + DH_{2}

Fig: 5.4 - Illustration of Hess's law

This means that the amount of heat evolved or absorbed in a chemical reaction depends only upon the energy of initial reactants and the final products. The heat change is independent of the path or the manner in which the change has taken place.The formation of carbon dioxide from carbon and oxygen can be illustrated as follows. Carbon can be converted into carbon dioxide in two ways. Firstly solid carbon combines with sufficient amount of oxygen to form CO_{2}. The same reaction when carried in presence of lesser amount of oxygen gives carbon monoxide which then gets converted to CO_{2} in step two in the presence of oxygen.

DH = DH

_{1}+ DH

_{2}

Thus, one can conclude that thermochemical equations can be added, subtracted or multiplied like algebraic equations to obtain the desired equation.

## Application of Hess's Law

Hess's law has been useful in determining the heat changes of reactions, which cannot be measured directly with calorimeter. Some of its applications are:

### Applications and problems

### Determination of heat of formation

Compounds whose heats of formation cannot be measured directly using calorimetric methods because they cannot be synthesised from their elements easily e.g. methane, carbon monoxide, benzene etc are determined using Hess's Law.

For example, the heat of formation of carbon monoxide can be calculated from the heat of combustion data for carbon and carbon monoxide as shown above.### Determination of heat of transition

The heats of transition of allotropic modification of compounds such as diamond to graphite, rhombic sulphur to monoclinic sulphur, yellow phosphorous to red phosphorous etc. can be determined using Hess's Law.

For example, the heat of transition of diamond to graphite can be calculated from the heat of combustion data for diamond and graphite, which is -395.4 kJ and -393.5 kJ respectively.The thermochemical equations showing the combustion reaction of diamond and graphite are:

##### The conversion that is required is:

This can be obtained by subtracting the second equation from the first one.

### Determination of heat of hydration

The heats of hydration of substances is calculated using Hess's law.

For example the heats of hydration of copper sulphate can be calculated from the heats of solution of anhydrous and hydrated salts of copper. The heats of solution of CuSO_{4}and CuSO

_{4}.5H

_{2}O are -66.5 and -11.7 kJ mol

^{-}

^{1}. The corresponding thermochemical equations are:

The process of hydration can be expressed as:

According to Hesss law, DH_{1} = DH + DH_{2}

_{1}- DH

_{2}

= -66.5 11.7 = -78.2 kJ/mol

### Determination of heats of various reactions

Hess's law is useful in calculating the enthalpies of many reactions whose direct measurement is difficult or impossible.

### Problems

19. Calculate the standard heat of formation of carbon disulphide (l). Given that the standard heats of combustion of carbon (s) sulphur (s) and carbon disulphide (l) are 393.3, -293.72 and -1108.76kJ mol^{-1}respectively.

### Solution

The given data can be written in thermochemical equation form as:

The required equation is:

Multiplying equation (ii) by 2 and adding to equation (i) we get,

Subtracting equation (iii) from the above equation we have,

20. Calculate lattice energy for the change,

Given that DH

_{s}

_{u}

_{b}

_{l}. of Li = 160.67 kJ mol

^{-}

^{1}, DH

_{D}

_{i}

_{s}

_{s}

_{o}

_{c}

_{i}

_{a}

_{t}

_{i}

_{o}

_{n}of

Cl_{2} = 244.34 kJ mol^{-}^{1}, DH_{i}_{o}_{n}_{i}_{s}_{a}_{t}_{i}_{o}_{n} of Li(g) = 520.07 kJ mol^{-}^{1},

_{E}

_{.}

_{A}of Cl(g) = - 365.26 kJ mol

^{-}

^{1}, DH

^{o}

_{f}of LiCl(s) = - 401.66 kJ mol

^{-}

^{1}.

### Solution

Considering the different changes that occur in the formation of solid lithium chloride based on the data given the lattice energy of the above can be constituted as:

or

= - 839.31 kJ mol

^{-}

^{1}

## 1 comment:

I think it is good to simplify rather than making it hard.instead of saying that we subtract equation iii from equation above, point out why subtracting.however,we subtract there respective anthalpy rather than entire equation.

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