Standard Enthalpy Of Formation, Combustion And Bond Dissociation

As we know all reactions result in the formation of products from the reactants. Of all the reactions that take place, some absorb energy while other results in the evolution of energy. Hence, we always experience a change in enthalpy whenever a reaction takes place. This enthalpy change is described as the enthalpy of reaction. Here, we are going to deal with few other enthalpy changes like enthalpy of formation, enthalpy of bond dissociation and enthalpy of combustion.

Standard enthalpy of formation:

Standard enthalpy of formation is defined as the enthalpy change when one mole of a compound is formed from its elements in their most stable state of aggregation (stable state of aggregation at temperature: 298.15k, pressure: 1 atm).  For example formation of methane from carbon and hydrogen:

\(C (graphite, s) +2H_2 (g) \rightarrow CH_4 (g)\); \(Δ_fH°\) = \(-74.81kJmol^{−1}\)

Enthalpy of formation is basically a special case of standard enthalpy of reaction where two or more reactants combine to form one mole of the product. Let us take an example of formation of hydrogen bromide from hydrogen and bromine.

\(H_2 (g) + Br_2 (l) \rightarrow 2HBr (g)\); \(Δ_rH°\) = \( -72.81kJmol^{−1}\)

As we can see in this case two moles of hydrogen bromide is produced. Hence, enthalpy of reaction cannot be taken as enthalpy of formation of hydrogen bromide rather we can say:

\(Δ_r H°\) = \( 2Δ_fH° \)

\(Δ_r H°\) = \(enthalpy~ of ~reaction\)

\(Δ_fH°\) = \(enthalpy ~of~ formation\)

bond dissociation enthalpy:

Enthalpy of bond dissociation is defined as the enthalpy change when one mole of covalent bonds of a gaseous covalent compound is broken to form products in the gaseous phase. Generally, enthalpy of bond dissociation values differ from bond enthalpy values which is the average of sum of all the bond dissociation energy in a molecule except, in case of diatomic molecules. For example:

\( Cl_2(g) \rightarrow 2Cl(g)\);                 \(Δ_{Cl–Cl}H^0\) = \(242 kJmol^{-1}\)

Standard enthalpy of combustion:

Standard enthalpy of combustion is defined as the enthalpy change when one mole of a compound is completely burnt in oxygen with all the reactants and products in their standard state under standard conditions (298K and 1 bar pressure). For example:

\(H_2 (g) + \frac{1}{2} O_2 (g) \rightarrow H_2O (l); Δ_cH°\) = \(-286 kJmol^{-1}\)

\(C_4 H_{10} (g) + \frac{13}{2} O_2 (g) \rightarrow 4CO_2 (g) + 5H_2O (l)\); \(Δ_cH°\) = \(-2658 kJmol{-1}\)

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Practise This Question

For the reactions (i) H2(g) + Cl2(g)2HCl(g) + xKJ
(ii) H2(g) + Cl2(g)2HCl(l) + yKJ

Which of the following statements is correct?