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JEE Main 2026 Preparation: Question Papers, Solutions, Mock Tests & Strategy Unacademy » JEE Study Material » Chemistry » Third Law of Thermodynamics

Third Law of Thermodynamics

The third law of thermodynamics focuses on entropy changes with the change in the temperature of the system.

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Standard energy change of formation is known as a change in enthalpy when one mole of a substance is produced in a standard state from its pure elements in the same conditions. It is also referred to as standard enthalpy of formation and is equal to standard enthalpy change of formation of products and reaction. Standard energy change of formation is equal to the sum of standard enthalpies of products to the sum of standard enthalpies of reactants. 

What is the standard energy change of formation?

Standard energy change of formation or standard enthalpy of formation is a change in energy when a mole of a compound is formed in its most standard condition from its pure state. The standard energy change of formation is represented as ∆Hf°. 

Where, 

∆ is a change in enthalpy

° is the degree of change

f represents the formation of elements. 

Standard energy change of formation formula is 

∆Hf°= ∆H° (products) – ∆H° (reactants)

The equation states that the standard energy change of formation is equal to the sum of enthalpies of formation of products subtracted by standard enthalpies of formation of reactants. 

The standard enthalpy of formation is zero for an element in its stable form under 1 bar of pressure. One exception to this is phosphorus. Rest all elements in their stable state have an enthalpy of formation as zero, as there is no change involved in their formation. 

Bond enthalpy of formation

Bond enthalpy of formation describes the amount of energy in a bond between atoms in a molecule. It is that energy that is added for homolytic or symmetrical cleavage of a bond that occurs during the gas phase.

It is important to see that breaking a bond is an endothermic process (because energy is needed to break the bonds). And the energy change related to the breaking of bonds is always positive.

On the other hand, the formation of chemical bonds is always an exothermic process. In such cases, the enthalpy change is always negative. 

Standard enthalpy of combustion

Standard enthalpy of combustion is defined as energy change that takes place when one mole of a compound is completely burnt in oxygen with all the reactants and products in standard states at standard conditions. 

Many readily available substances with high enthalpies of combustion, such as hydrogen and carbon (as coal or charcoal), and hydrocarbons (compounds containing only hydrogen and carbon), such as methane, propane, and the significant components of gasoline, are used as fuels.

Solved examples of standard energy of formation

Question 1

Calculate ΔHf for the following reaction:

8 Al(s) + 3 Fe3O4(s) → 4 Al2O3(s) + 9 Fe(s)

Solution:

Standard energy change of formation ∆Hf for a reaction is equal to the sum of the heats of formation of the product compounds subtracted from the total of the heats of formation of the reactant compounds:

ΔHf = Σ ΔHf products – Σ ΔHf reactants

Substituting the value in the given equations. 

ΔH = 4 ΔHf Al2O3(s) – 3 ΔHf Fe3O4(s)

The values of standard energy change of formation can be found in the table. After substituting the value in the equation, we get:

ΔH = 4(-1669.8 kJ) – 3(-1120.9 kJ)

ΔH = -3316.5 kJ

Question 2

Determine ΔHf for the given equations

HBr(g) → H+(aq) + Br-(aq)

Solution:

Standard heat of formation can be calculated by substituting the values in the equation of enthalpy of formation. 

ΔHf = Σ ΔHf products – Σ ΔHf reactants

We know that the standard heat of formation is 0 for H+ so the equation becomes

ΔHf = ΔHf Br–(aq) – ΔHf HBr(g)

You can substitute the values by going through the standard energy change of formation table for ions. After substituting the values, the equation becomes:

ΔH = -120.9 kJ – (-36.2 kJ)

ΔH = -120.9 kJ + 36.2 kJ

ΔH = -84.7 kJ

Question 3

Define the standard enthalpy of formation of H2 and why?

Solution: 

The standard enthalpy of formation for any pure metal in its natural form is 0 J. Technically speaking; however, this is arbitrary and relative. Because bond formation is an exothermic process, and some heat is undoubtedly lost when H2 is formed from individual H atoms, the enthalpy of formation of any substance can never be zero.

Conclusion

Standard energy change of formation is the enthalpy change when one mole of a compound is formed from its elements in their standard stages. Enthalpy of formation is a special case of standard enthalpy of reaction where two or more reactants mix to form one mole product. 

faq

Frequently Asked Questions

Get answers to the most common queries related to the JEE Examination Preparation.

What do you understand by the term entropy in thermodynamics? Explain the entropy of solid, liquid, and gas.

Ans:  Entropy of a substance or system is defined as the ran...Read full

Briefly explain a spontaneous process.

Ans : The process which does not require any external help to...Read full

What is the absolute zero temperature in thermodynamics? How can it be achieved?

Ans :  Absolute zero temperature is the coldest temperature which caus...Read full

What is a pure crystalline system as explained by the third law of thermodynamics? Name the molecules which do not follow the law.

Ans : In a pure crystalline system, atoms are arranged in the same order havin...Read full

Ans:  Entropy of a substance or system is defined as the randomness of the movement of molecules in that system. In solids, the molecules are arranged properly, experiencing less movement, and thus have a low value of entropy. In gases, the movement of molecules is fast, and thus they experience more randomness, exhibiting a high value of entropy. In liquids, the entropy lies in between the solids and gases.

Ans : The process which does not require any external help to occur is known as a spontaneous process. A spontaneous process can be predicted by simply calculating the entropy of the universe (∆S(universe).

Examples of the spontaneous process:
Melting of ice, cooling of tea, the flow of water from higher to lower level, falling of an object on the ground. All these processes happen automatically.

Ans :  Absolute zero temperature is the coldest temperature which causes atoms and molecules of the system to freeze. They are not able to undergo any changes. Absolute zero temperature is zero kelvin. Absolute zero temperature is not possible to achieve.

Ans : In a pure crystalline system, atoms are arranged in the same order having similar shapes and sizes. The third law of thermodynamics says that if this type of system has zero Kelvin temperature, the absolute entropy of the system will be zero or nearly zero.

Carbon monoxide and hydrochloric acid do not obey the third law of thermodynamics. In CO, the kinetic energy will be zero at zero kelvin. But the crystal formation causes disorder in the structure. They will have some residual entropy at the absolute zero temperature. So, according to the third law of thermodynamics, the entropy of the system will not be zero at zero kelvin.

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