Standard Enthalpy of Formation Combustion and Bond Dissociation

Reactions are the result of chemical compositions. To make the reaction possible, reactants are necessary. They collage together to form new products. Every reactant absorbs energy during its chemical collages. 

Some of the reactions absorb energy, whereas others take part in the evolution of energy. We know that the change in enthalpy is obvious in many chemical reactions. Without it, the process is incomplete. 

You can describe the change of enthalpy as the enthalpy of reaction. This article is all about how you should describe standard enthalpy of formation, standard enthalpy of combustion, and enthalpy of bond dissociation.

Define Enthalpy of Formation

We can define standard enthalpy of formation just by mentioning the enthalpy change. It is possible when a compound’s one mole is created from its associated elements within their stable state of aggregation state. 

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The stable state of aggregation is considered when the temperature is at 298.15 K, and atmospheric pressure is at 1 atm.

What Is the Enthalpy of Formation?

The enthalpy of formation definition can be understandable with the examples. Let’s take an example to elaborate it briefly. We can consider the formation of methane from hydrogen and carbon:

C(graphite, s) + 2H₂(g)→ CH₄(g); Δ_fH^o = - 74.81kJmol^-1

Can you answer What Is Standard Enthalpy Of Formation? Enthalpy of formation comes under the category of a special case of standard enthalpy of reaction. In this process, two or more reactants are involved. They combine together to create one mole of the product. 

The example of the formation of hydrogen bromide from bromine and hydrogen can be the best example. Here is the expression:

H₂(g)  + Br₂(l) ⟶ 2HBr(g) ; Δ_rH^o = - 72.81kJmol^-1

As per the above expressions, it is clear that two moles of hydrogen bromide are available. Therefore, standard enthalpy of formation can be taken as the enthalpy of reaction, and not as the enthalpy of formation of hydrogen bromide.

We can say that 

Δ_fH^o = 2 Δ_rH^o

Δ_fH^o =  Enthalpy of formation

Δ_rH^o = Enthalpy of reaction

Enthalpy of Combustion

We can say that the enthalpy of combustion is only possible when one mole of a compound is burnt completely to give rise to oxygen at the end. All of the processes are taken into consideration when all the reactants and products are in the standard state and under standard conditions (1 bar pressure and 298K).

For example:

H₂(g) + 1/2 O₂(g) ➝ H₂O(l) ; Δ_cH^o = - 286 kJmol^-1

C₄H₁₀(g) + 13/2 O₂(g) ➝4CO₂(g) + 5H₂O(l) ;  Δ_cH^o = -2658 kJmol^-1

Standard enthalpy of combustion is a positive value as combustion is always exothermic. When a chemical substance comes under the process of combustion, it generates energy to outside. So, the change in enthalpy for the exothermic reactions is negative.

However, in the convention process, the molar heat of combustion (also molar enthalpy of combustion) is considered as a positive value.

We can calculate the enthalpy of combustion with ease. The process is very simple. We do it by calculating the difference between the mass of the fuel before the boiled water and the mass of the fuel. 

The workout energy of a substance can be given as 1 mole. The unit that stands to show the measurement of enthalpy of combustion is known as Joule per mole (or Kilojoule per mole).

If you need your answer in KJ (kilojoule) format, you just need to devise the result by 1000. 

Bond Dissociation Enthalphy

This is a type of change in enthalpy where one mole of covalent bonds of a gaseous compound is taken apart to manufacture different gaseous phase products.

In general, the enthalpy of bond dissociation is always different from the bond enthalpy values. In a molecule, it is the average of some of all the bond dissociation energy.

A few examples of diatomic molecules that come under the bond dissociation enthalpy process:

• Cl₂(g) ➝2Cl(g)

• Δ_Cl-ClH^o = 242kJmol^-1