Hess Law

What Is Hess Law?

Hess’ law, also known as Hess’s law of constant heat summation, states, “at constant temperature, heat energy changes (enthalpy – ΔHrec) accompanying a chemical reaction will remain constant, irrespective of the way the reactants react to form product”.

Hess’ law is based on the state function character of enthalpy and the first law of thermodynamics. Energy (enthalpy) of a system (molecule) is a state function. So, enthalpy of reactant and product molecules is a constant and does not change with origin and path of formation.

The first law of thermodynamics states that the total energy of the substances before and after any (physical or chemical) change should be equal. According to the law, the total energy of the reactant should be equal to the total energy of the product. Any difference in the energy between the reactants and products is also fixed at a particular temperature and will not change with the path followed by the reactants to form products. Hence, heat energy also can be considered as a reactant or product of the reaction and be included in the reaction.

Hence, exothermic reactions can be written as: A + B → C + D + ΔH

Similarly, endothermic reactions become: A + B + ΔH → C + D

This allows reactions containing reactants and products to be treated as algebraic equations and carrying out mathematical operations on them. It should be remembered, that an exothermic reaction in one direction will be endothermic in the reverse direction and vice-versa.

Importance of Hess Law

Every substance (atom/ molecules) possesses energy within. The internal energy depends on the nature of force existing in the substance and the temperature. When the substance undergoes chemical reactions, some bonds connecting some atoms are broken and some bonds are made new. The breaking and making of bonds involve energy.

So, in reactions, product substances may have either less or the same or more energy than the reacting substances. Reactions accordingly may release heat to become exothermic or absorb heat and endothermic. Reactants may further react to give the product;

1. In single-step or
2. In multi-steps or
3. Along with other products

Knowledge of the energy changes in any reaction is essential for the manipulation of the reactants and products in a chemical process to our requirement.

Heat energy changes of reactions measured at constant volume are called internal energy change ΔE and energy measured at constant pressure is called enthalpy change ΔH.

The experimental measurements give only the net value of all reactions or products formed. It is not possible to measure experimentally the enthalpy change of an intermediary reaction step or any intermediary product.

For example, carbon reacts with oxygen to form carbon dioxide in excess oxygen. Carbon and oxygen combine to form carbon dioxide directly or in two steps -initially form carbon monoxide and then go to carbon dioxide. Measurement will give the energy changes for the formation of carbon dioxide only and not for carbon monoxide.

Similarly, measuring the enthalpy of formation of benzene, from carbon and hydrogen is not possible, because carbon and hydrogen may combine to form not only benzene, but also other types of hydrocarbons in the given conditions.

Hess’s law is useful and is the only way of calculating such non-measurable enthalpy changes in physical and chemical changes.

Forms of Hess Law

Hess’ law is stated in many ways.

For multi-step reactions:

If reactants react to form products not in a single step but in a number of consecutive steps involving many intermediary products, the sum of all the reactants, products and the corresponding energy changes will give the reactant, products and heat energy changes of the overall reaction. So, like molecules, heat energy changes also can be subjected to mathematical operations.

For multi-different reactions:

If the reactants and products of a required chemical reaction can be obtained by the summation of many other chemical reactions, the enthalpy of the required reaction of reactants to the products also can be obtained by the sum of the enthalpy changes of all those chemical reactions.

a) Hess law and multi-step reaction:

Reactant can form product B by following three different steps. C, D and E are intermediates in the other stepwise reactions. Hess’ law states that the enthalpy of the reaction (ΔH1) is the same irrespective of the path.

So, the enthalpy of direct single-step reaction and other paths giving intermediates C, D and E should be the same. ΔH1 = ΔH2+ ΔH3 = ΔH4 + ΔH5 + ΔH6.

Example: Carbon reacts with oxygen to form carbon dioxide releasing  94.3 kcals of heat in a single step. Carbon can also react in a two-step process of forming an intermediate carbon mono-oxide, which again is converted to carbon dioxide. ( ΔH = – Heat released)

C + O₂ → CO + 26.0 kcals

CO + O₂ → CO₂ + 68.3kcals

On adding the two reactions, C + O₂ → CO₂ + 94.3 kcals

As per Hess law, ΔH = ΔH1 + ΔH2 = -26.0 + 68.3 = 94.3 kcals

Net reaction enthalpy of both reactions is the same as that of single-step formation. So, enthalpy of reaction does not change on the path followed by the reactants.

b) Hess law and multi-different reactions:

Combustion of carbon, sulphur and carbon disulphide are exothermic with an enthalpy of- 393.5kJ, -296.8kJ and -1075kJ.

Reactions are-

C(s) + O₂ (g) → CO₂ (g) + 393.5 kJ ……..(1)

S(s) + O₂ (g) → SO₂ (g) + 296.8 kJ .…….(2)

CS₂ (l) + 3O₂(g) → CO₂(g) + 2SO₂(g) + 1075.0 kJ ..…….(3)

These reactions and enthalpy changes can be treated as algebraic equations to get the heat of formation of carbon disulphide even without doing experiments.

Equation 1: C(s) + O₂ (g) → CO₂ (g) + 393.5 kJ

2x equation 2:2S(s) + 2O₂ (g) → 2SO₂ (g) + 593.6 kJ

Reverse of equation3: CO₂(g) + 2SO₂(g) → CS₂ (l) + 3O₂(g) -1075.0 kJ

Adding the three reactions: C (s) + 2S (s) → CS₂ (l) -87.9 kJ

Formation of carbon disulphide is an endothermic reaction.

Application of Hess Law of Heat Summation

Hess’ law of heat summation is an efficient way to estimate heat changes that cannot be measured experimentally.

1. Enthalpy change in a physical change

Carbon and diamond are allotropes of carbon. But measuring the energy change in the conversion of graphite to diamond cannot be determined, as the process cannot be carried out. Still, the heat changes for this hypothetical physical change can be calculated using Hess law.

Graphite and diamond combine with oxygen with the heat of reaction as -393.4kJ and – 395.4kJ, respectively.

C (graphite) + O₂ → CO₂ ΔHgr = -393.4kJ

C (diamond) + O₂ → CO₂ ΔHdi = -395.4kJ

Reversing the combustion reaction of diamond as-

CO₂ → C (diamond) + O₂ ΔHdi = + 395.4kJ

Adding,

C (graphite) + O₂ → CO ₂ ΔHgr = – 393.4kJ

C (graphite) → C (diamond) ΔHtr = +2.kJ

Enthalpy change in the allotrope transition of graphite to diamond is endothermic of 2KJ.

2. Enthalpy change of chemical reaction

The bond energy of hydrogen, iodine and hydrogen iodide are 218, 107kJ and 299kJ, respectively.

Estimate the enthalpy of hydrogen iodide formation. Is the reaction endothermic or exothermic?

Formation of hydrogen iodide from hydrogen and iodine follows the reaction-

Enthalpy of formation of hydrogen iodide is the heat changes occurring when one atom of hydrogen and one atom of iodine react to form one mole of hydrogen iodide in standard conditions (as gas). To get one atom of hydrogen or iodine, the molecular bond has to be broken.

Heat of formation = Bond energy of HI – Bond dissociation of H₂ – Bond dissociation energy of I₂.

= 299 – (218 + 107) = 299-325 =-26kJ

As the heat of formation is negative, the reaction is exothermic.

3. Enthalpy of formation

When carbon combines with hydrogen, many hydrocarbons can be formed. Hence, the heat of formation of benzene cannot be determined experimentally. The heat change can be calculated by Hess law.

6C + 3H₂ → C₆H₆ ΔH C₆H₆ = ?

Heat of formation of carbon dioxide and water are -393.5kJ and -285.8KJ, respectively. Heat of combustion of benzene is -3301kJ.

C + O₂ → CO₂ ΔH1 = -393.5kJ…..1

H₂ + O₂ → H₂ O ΔH2 = -285.8kJ……2

C₆H₆ + 9O₂ → 6CO₂ + 3H₂ O ΔH3 = -3301kJ …….3

6 x Reaction 1: 6C + 6O₂ → 6CO₂ 6ΔH1 = -2361kJ…..1

3 x Reaction2: 3H₂ + 3O₂ → 3H₂ O 3ΔH2 = -857.4kJ……2

Reverse of reaction 3: 6CO₂ + 3H₂O → C₆H₆ + 9O₂ -ΔH3 = +3301kJ …….3

Adding the three reactions- 6C + 3H₂ → C₆H₆ ΔH= +82.6kJ

Heat of formation of benzene is 82.6kJ.

4. Bond energy

5. Lattice energy