Hess’s Law of Constant Heat Summation

Russian chemist Germain Henri Hess gave an important law about the enthalpy change. He said that enthalpy change of one compound to the other compound is always same, even if it occurs in one step or more steps.

Enthalpy is a state function that means it doesn’t depend on the path taken by the system. Let’s take an example to understand it better:

A + B ⟶ D      Δ_rH

This reaction may take place into more steps like:

A + B ⟶ C         Δ_rH₁       .....equation 1                 

C ⟶ D             Δ_rH₂       .....equation 2

Then you can get Δ_rH by simply applying mathematical operations by adding equation 1 and 2.

[A + B ⟶ C Δ_rH₁] + [ C ⟶ D Δ_rH₂] = A + B ⟶ D  

  

 So, Δ_rH = [Δ_r H₁ + Δ_rH₂]

Let’s take another example: if you have given two different reactions:

C(graphite) + O₂(g) ⟶ CO₂(g) Δ_r H= -393.5 kJ mol^-1    ........equation 1                 

CO(g) + ½ O₂(g) ⟶ CO₂(g)      Δ_r H= -283 kJ mol^-1           .........equation 2                 

And asked you to calculate the Δ_r H of the following reaction

C(graphite) + ½ O₂(g) ⟶ CO(g)           Δ_r H =?

It’s puzzle time; try to rearrange the above two equation so that you can get the desired equation. We want the carbon and oxygen on the reactant side and CO on the product side. We can take equation 1 as it is given, but we have to reverse the equation 2 to get the CO on the product side. When you reverse the equation 2, sign of Δ_rH also gets reversed. And you get a new equation 3.

CO₂(g) ⟶ CO(g) + ½ O₂(g)     Δr H= +283 kJ mol^-1          .........equation 3                 

Now if we add the equation 1 and 3 we can get the desired equation.

Δ_r H = [Δ_rH₁ -393.5 kJ mol^-1] + [Δ_r H₃ + 283 kJ mol^-1]

Δ_r H = -110.5 kJ mol^-1

So you can state the law of constant heat summation as ‘If a reaction takes place in several steps then its Δ_r H^ө is the sum of standard enthalpies of the intermediate reactions into which the overall reaction may be divided at the same temperature.’

 Δ_r H^ө= Δ_r H^ө₁ + Δ_r H^ө₂+ Δ_r H^ө₃ + ....Δ_r H^ө_n

Hess’s Law of Constant Heat Summation

Bond Dissociation Enthalpy Δbond H^ө

Now you have learnt to calculate enthalpy of reaction by using standard enthalpy of formation and by using standard enthalpies of intermediate reactions. There is one more way to calculate the enthalpy of reaction. When any reaction occurs, some bonds are broken and some new bonds are formed. Energy is needed to break the bonds and energy is released when new bonds are formed. It means if we calculate the bond enthalpies of reactant and products we can get the enthalpy of reaction.

Δ_r H = Ʃ Δbond H^ө(reactants) - Ʃ Δbond H^ө(products)

Bond enthalpy or bond dissociation enthalpy is the enthalpy change when one mole of covalent bond of a gaseous covalent compound is broken into gaseous products.

H₂(g) ⟶ 2H(g) ....... Δ_H-H H^ө = 435.0 kJ mol^-1

Cl₂(g) ⟶ 2H(g) ....... Δ_H-H H^ө = 435.0 kJ mol^-1

The above examples are of simple diatomic molecules. What happens in polyatomic molecule? Let’s take an example of methane CH₄. It has four C-H bond. C is sp³ hybridised, all C-H bonds are identical and have same energy. But each successive step requires different energy to break C-H bond. Because for each C-H bond there is different environment. In first step there are three more C-H bonds while in second step there are two more C-H bonds and so on.

CH₄(g) ⟶ CH₃(g) + H(g)          ........ Δ_C-HH^ө = 427.0 kJ mol^-1

CH₃(g) ⟶ CH₂(g) + H(g)          ........ Δ_C-HH^ө = 439.0 kJ mol^-1

CH₂(g) ⟶ CH(g) + H(g)           ........ Δ_C-HH^ө = 452.0 kJ mol^-1

CH(g) ⟶ C(g) + H(g)               ........ Δ_C-H H^ө = 347.0 kJ mol^-1

In such cases we take the mean value for the bond dissociation enthalpy. So Δ_C-H H^ө will be:

Δ_C-H H^ө = ¼ (427 + 439 + 452 + 347) kJ mol^-1

Δ_C-H H^ө = 416 kJ mol^-1

We have calculated Δ_C-H H^ө for methane. It can be different for other molecules but it will be nearer to the mean value so we can take it as a reference for any C-H bond.

Lattice enthalpy

Bond dissociation enthalpy is useful for covalent bonds but not for ionic compounds. In ionic compounds, ions are trapped in lattice and we need to overcome the lattice energy to set these ions free. You know that ionic compounds dissociate into its constituent ions only when they get dissolved in a solvent. Let’s try to understand the process of dissolution of any ionic compound in a solvent. This process occurs in two steps, in the first step lattice gets broken and ions are freed, and in the second step solvationoccurs. If the solvent is water then solvation is known as hydration.

MX(s) ⟶ M⁺(g) + X^-(g) ⟶ M⁺(aq) + X^-(aq)

So the enthalpy change for the overall reaction is called the enthalpy of solution Δ_solution H^ө which is the sum of lattice enthalpy Δ_lattice H^ө and solvation enthalpy Δ_hydrationH^ө.

Δ_solution H^ө = Δ_lattice H^ө+ Δ_hydration H^ө

It means if you want to know the lattice enthalpy of NaCl you must know the values of Δ_solution H^ө and Δ_hydrationH^ө.  Two German scientists Max Born and Fritz Haber developed an indirect way to calculate the lattice enthalpy of NaCl. They constructed an enthalpy diagram by applying Hess’s law of constant heat summation. We will learn about it in the next post of Thermodynamics. 

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