Enthalpy of Reaction: The enthalpy change when reactants in their standard states (p = 1 bar; usually T = 298 K) convert to products in their standard states is known as the standard enthalpy of reaction in chemistry. This is the standard heat of reaction at constant pressure and temperature, however, it may be measured using calorimetric methods even if the temperature changes throughout the measurement, as long as the initial and final pressure and temperature are equal.

Introduction to Enthalpies of Different Types of Reactions

The sum of internal energy of the system and the product of its pressure and volume is its enthalpy. It is also known as heat content. That is,

• \({\text{H}} = {\text{E}} + {\text{PV}}\)
• Where \({\text{E=}}\) Internal energy of the system
• \({\text{P=}}\) Pressure of the system
• \({\text{V=}}\) Volume of the system

Enthalpy is a state function, and its absolute value cannot be determined. A change in enthalpy \(\left({\Delta {\text{H}}} \right)\) associated with a process, on the other hand, can be measured precisely and is provided by the expression.

\(\Delta {\text{H}} = {{\text{H}}_{{\text{products }}}} – {{\text{H}}_{{\text{rectants }}}}\)

\( = {{\text{H}}_{\text{p}}} – {{\text{H}}_{\text{r}}}\)

The change in enthalpy noticed will be the sum of the internal energy change \(\left({\Delta {\text{E}}} \right)\) and the work done in expansion or contraction if \(\Delta {\text{V}}\) is the change in volume in a reaction at constant temperature and pressure. That is

\(\Delta {\text{H}} = \Delta {\text{E}} + {\text{P}} \times \Delta {\text{V}}\,\,\,\,\,\,\left( 1 \right)\)

Let us consider a reaction involving gases. If \({{\text{V}}_{\text{A}}}\) is the total volume of the gaseous reactants, \({{\text{V}}_{\text{B}}}\) is the total volume of the gaseous products, \({{\text{n}}_{\text{A}}}\) is the number of moles of gaseous reactants and \({{\text{n}}_{\text{B}}}\) is the number of moles of gaseous products, all at constant pressure and temperature, then using the ideal gas law, we write,
\({\text{p}}{{\text{V}}_{\text{A}}} = {{\text{n}}_{\text{A}}}{\text{RT}}\)
and \({\text{p}}{{\text{V}}_{\text{B}}}{\text{ = }}{{\text{n}}_{\text{B}}}{\text{RT}}\)
Thus, \({\text{p}}{{\text{V}}_{\text{B}}} – {\text{p}}{{\text{V}}_{\text{A}}} = {{\text{n}}_{\text{B}}}{\text{RT}} – {{\text{n}}_{\text{A}}}{\text{RT}} = \left( {{{\text{n}}_{\text{B}}} – {{\text{n}}_{\text{A}}}} \right){\text{RT}}\)
or \({\text{p}}\left( {{{\text{V}}_{\text{B}}} – {{\text{V}}_{\text{A}}}} \right) = \left( {{{\text{n}}_{\text{B}}} – {{\text{n}}_{\text{A}}}} \right){\text{RT}}\)
or \({\text{p}}\Delta {\text{V}} = \Delta {{\text{n}}_{\text{g}}}{\text{RT}}\)
Here, \(\Delta {{\text{n}}_{\text{g}}}\) refers to the number of moles of gaseous products minus the number of moles of gaseous reactants.
Substituting the value of \({\text{p}}\Delta {\text{V}}\) from equation \(1,\) to get \(\Delta {\text{H}} = \Delta {\text{U}} + \Delta {{\text{n}}_{\text{g}}}{\text{RT}}\)
The above equation is useful for calculating \(\Delta {\text{H}}\) from \(\Delta {\text{U}}\) and vice versa.

Depending on the nature of the reaction, enthalpy or enthalpy changes accompanying chemical reactions are expressed in various ways. These are discussed below.

Enthalpy of Formation

The change in enthalpy that arises when one mole of a compound is produced from its elementsis defined as the enthalpy of formation. It is denoted by \(\Delta {{\text{H}}_{\text{f}}}.\) The enthalpy of formation of ferrous sulphide and acetylene, for example, can be stated as:

\({\text{Fe}}\left({\text{s}} \right) + {\text{S}}\left({\text{s}} \right) \to {\text{FeS}}\left({\text{s}} \right)\,\,\,\,\Delta {{\text{H}}_{\text{f}}} = – 100.41\,{\text{kJ}}\,{\text{mo}}{{\text{l}}^{ – 1}}\)

\(2{\text{C}}\left({\text{s}} \right) + {{\text{H}}_2}\left({\text{g}} \right) \to {{\text{C}}_2}{{\text{H}}_2}\left({\text{g}} \right)\,\,\,\,\Delta {{\text{H}}_{\text{f}}} = 222.3\,{\text{kJ}}\,{\text{mo}}{{\text{l}}^{ – 1}}\)

Standard Enthalpy of Formation

The enthalpy change that occurs when one mole of a compound is produced from its elements when all substances are in their standard states (\(298\,{\text{k}}\) and \(1\) atm pressure) is defined as the standard enthalpy of formation of a compound. It is denoted by \(\Delta {\text{H}}_{\text{f}}^ \circ .\) The standard enthalpy of formation of all elements in their standard state is assumed to be zero by convention.

Standard Enthalpy of Reaction from Standard Enthalpy of Formation

From the values of standard enthalpies of the formation of various reactants and products, we can determine the reaction enthalpy under standard conditions. The standard enthalpy of reaction equals the standard enthalpy of product formation minus the standard enthalpy of reactant formation.

\(\Delta {{\text{H}}^ \circ } = \left[ \begin{gathered} {\text{Total}}\,{\text{standard}}\,{\text{heat}} \hfill \\ {\text{of}}\,{\text{formation}}\,{\text{of}}\,{\text{products}} \hfill \\ \end{gathered} \right] – \left[ \begin{gathered} {\text{Total}}\,{\text{standard}}\,{\text{heat}} \hfill \\ {\text{of}}\,{\text{formation}}\,{\text{of}}\,{\text{reactants}} \hfill \\ \end{gathered} \right]\)

\(\Delta {{\text{H}}^ \circ } = \Delta {\text{H}}_{\text{f}}^ \circ \left({{\text{products}}} \right) – \Delta {\text{H}}_{\text{f}}^ \circ \left({{\text{reactants}}} \right)\)

Let us consider a general reaction:

\({\text{aA}} + {\text{bB}} \to {\text{cC}} + {\text{dD}}\)

The standard enthalpy of reaction is given by

\(\Delta {{\text{H}}^ \circ } = \Delta {\text{H}}_{\text{f}}^ \circ \left({{\text{products}}} \right) – \Delta {\text{H}}_{\text{f}}^ \circ \left({{\text{reactants}}} \right)\)

\( = \left[{{\text{c}} \times \Delta {\text{H}}_{\text{f}}^ \circ \left({\text{C}} \right) + {\text{d}} \times {\text{H}}_{\text{f}}^ \circ \left({\text{D}} \right)}\right] – \left[{{\text{a}} \times \Delta{\text{H}}_{\text{f}}^ \circ \left({\text{A}} \right) +{\text{b}} \times \Delta {\text{H}}_{\text{f}}^ \circ \left({\text{B}}\right)} \right]\)

Enthalpy of Combustion

The change in enthalpy of a system when one mole of a substance is entirely burned in excess of air or oxygen is known as the enthalpy of combustion.It is denoted by \(\Delta {{\text{H}}_{\text{c}}}.\)The following reaction shows enthalpy of combustion of methane:

\({\text{C}}{{\text{H}}_4}\left({\text{g}}\right) + 2{{\text{O}}_2}\left({\text{g}} \right) \to {\text{C}}{{\text{O}}_2}\left({\text{g}} \right) + 2{{\text{H}}_2}{\text{O}}\left({\text{l}} \right)\,\,\,\,\Delta{{\text{H}}_{\text{c}}} = – 890.3\,{\text{kJ}}\,{\text{mo}}{{\text{l}}^{ – 1}}\)

Enthalpy of Solution

When a substance is dissolved in a solvent, enthalpy changes are commonly observed. The enthalpy of the solution of reactants and products must be considered when a reaction takes place in the solution. The change in enthalpy when one mole of a substance is dissolved in a particular quantity of solvent at a given temperature is known as the enthalpy of solution.

The enthalpy absorbed is \(78.5\,{\text{kJ}}/{\text{mol}}\) when one mole of copper sulphate is dissolved in water to produce one molar solution. There will be another change in enthalpy if the solution is further diluted. If we continue to dilute the solution, we will reach a point where additional dilution has no thermal effect. The state of infinite dilution is the name given to this condition. To avoid the quantity of the solvent, we must include the concept of infinite dilution in our definition, which can be stated as follows: the enthalpy of solution is the change in enthalpy that occurs when one mole of a substance is dissolved in a solvent, such that further dilution causes no change in enthalpy. The enthalpy of a solution can also be expressed as :

\({\text{KCl}}\left({\text{s}} \right) + {{\text{H}}_2}{\text{O}}\left({\text{l}} \right) \to {\text{KCl}}\left({{\text{aq}}} \right)\,\,\,\Delta {\text{H}} = – 18.4\,{\text{kJ}}\,{\text{mo}}{{\text{l}}^{ – 1}}\)

\({\text{MgS}}{{\text{O}}_4}\left({\text{s}} \right) + {{\text{H}}_2}{\text{O}}\left({\text{l}} \right) \to {\text{MgS}}{{\text{O}}_4}\left({{\text{aq}}} \right)\,\,\,\Delta {\text{H}} = – 84.8\,{\text{kJ}}\,{\text{mo}}{{\text{l}}^{ – 1}}\)

Enthalpy of Neutralisation

The change in enthalpy of a system when one gram equivalent of an acid is neutralised by one gram equivalent of a base or vice versa in dilute solution is known as the enthalpy of neutralisation. The enthalpy of neutralisation can be illustrated using the following example.

\({\text{HCl}}\left({{\text{aq}}} \right) + {\text{NaOH}}\left({{\text{aq}}} \right) \to {\text{NaCl}}\left({{\text{aq}}} \right) + {{\text{H}}_2}{\text{O}}\left({\text{l}} \right)\,\,\,\Delta {\text{H}} = – 57.1\,{\text{kJ}}\,{\text{mo}}{{\text{l}}^{ – 1}}\)

The enthalpy of neutralisation of a strong acid and strong base is \( – 57.1\,{\text{kJ}}\,{\text{mo}}{{\text{l}}^{ – 1}},\) regardless of which acid or base is used, according to the given statistics. With the help of ionisation theory, this regularity has been adequately explained. If HA and BOH represent any strong acid and base, respectively, and equivalent amounts of these are mixed in a dilute solution, we get

\({{\text{H}}^ + }\left({{\text{aq}}} \right) + \overset{ – }{\mathop {\text{A}}} \left({{\text{aq}}} \right) + \overset{ + }{\mathop {\text{B}}} \left({{\text{aq}}} \right) + {\text{O}}{{\text{H}}^ – } \to \overset{ – }{\mathop {\text{A}}} \left({{\text{aq}}} \right) \to \left({{\text{aq}}}\right) + \overset{ + }{\mathop {\text{B}}} \left({{\text{aq}}} \right) + {{\text{H}}_2}{\text{O}}\left({\text{l}} \right)\,\,\Delta {\text{H}} = – 57.1\,{\text{kJ}}\,{\text{mo}}{{\text{l}}^{ – 1}}\)

Ignoring the ions which are present on both sides of the equation, we get

\({{\text{H}}^ + }\left({{\text{aq}}} \right) + {\text{O}}{{\text{H}}^ – }\left({{\text{aq}}} \right) \to {{\text{H}}_2}{\text{O}}\left({\text{l}} \right)\,\,\Delta {\text{H}} = – 57.1\,{\text{kJ}}\,{\text{mo}}{{\text{l}}^{ – 1}}\)

The enthalpy of neutralisation of an acid and a base is thus equal to the enthalpy of water production from hydrogen and hydroxyl ions. The enthalpy of neutralisation varies substantially from \( – 57.1\,{\text{kJ}}\,{\text{mo}}{{\text{l}}^{ – 1}}\) when weak acids or weak bases are neutralised by strong bases or strong acids, respectively.

Enthalpy of Hydration

The enthalpy of hydration is defined as the enthalpy change when one mole of anhydrous salt form hydrated salt by combining with the required number of moles of water.

Enthalpy of Fusion

The enthalpy change when one mole of a solid substance is transformed into the liquid state at its melting point is known as enthalpy of fusion. Take the melting of one mole of ice at its melting point of \({0^ \circ }{\text{C}}\) or \(273\,{\text{k}}\) as an example. The procedure can be described as follows:

\(\underset{{{\text{ice}}}}{\mathop {{{\text{H}}_2}{\text{O}}}} \left({\text{s}} \right) \to \underset{{{\text{Water}}}}{\mathop {{{\text{H}}_2}{\text{O}}}} \left({\text{l}} \right)\Delta{\text{H}} = 6.0\,{\text{kJ}}\,{\text{mo}}{{\text{l}}^{ – 1}}\)

A total of \(6.0\,{\text{kJ}}\,{\text{mo}}{{\text{l}}^{ – 1}}\) of enthalpy is absorbed during this process. We can compare the magnitudes of intermolecular forces between distinct substances based on their fusion values. The amount of intermolecular forces increases as the enthalpy of fusion of a substance increases.

Enthalpy of Vaporisation

When one mole of liquid is transformed into a vapour or gaseous state at its boiling point, the enthalpy change is defined as the enthalpy of vaporisation. When one mole of water is turned into steam at \({100^ \circ }{\text{C}},\) the enthalpy absorbed is \(40.6\,{\text{kJ}}\,{\text{mo}}{{\text{l}}^{ – 1}},\) which is the enthalpy of vaporisation of water. The process takes place as shown below:

\(\underset{{{\text{Water}}}}{\mathop {{{\text{H}}_2}{\text{O}}}} \left({\text{l}} \right) \to \underset{{{\text{steam}}}}{\mathop {{{\text{H}}_2}{\text{O}}}} \left({\text{g}} \right)\Delta {\text{H}} = 40.6\,{\text{kJ}}\,{\text{mo}}{{\text{l}}^{ – 1}}\)

Enthalpy of Sublimation

Sublimation is the process by which a solid transform into a gaseous state without first becoming liquid. It occurs at a temperature below the solid’s melting point. The enthalpy changes when one mole of a solid is directly transformed into the gaseous state at a temperature below its melting point is known as enthalpy of sublimation. Iodine, for example, has a sublimation enthalpy of \(62.4\,{\text{kJ}}\,{\text{mo}}{{\text{l}}^{ – 1}}.\) It can be written as:

\({{\text{I}}_2}\left({\text{s}} \right) \to {{\text{I}}_2}\left({\text{g}} \right)\,\,\,\,\,\Delta {\text{H}} = 62.4\,{\text{kJ}}\,{\text{mo}}{{\text{l}}^{ – 1}}\)

Enthalpy of Transition

The change in enthalpy that happens when one mole of an element moves from one allotropic form to another is known as the enthalpy of transition. The transformation of diamond into amorphous carbon, for example, can be depicted as

\({{\text{C}}_{{\text{diamond}}}} \to {{\text{C}}_{{\text{amorphous}}}}\,\,\,\,\,\Delta {\text{H}} = 13.8\,{\text{kJ}}\,{\text{mo}}{{\text{l}}^{ – 1}}\)

Bond Enthalpy

The formation of a bond between two atoms results in the release of energy. When the bond is broken, an equal amount of energy is absorbed. The bond energy is the average amount of energy necessary to break all bonds of a specific type in one mole of a substance. As a result, the bond energy of the \({\text{H-H}}\) bond is the amount of energy necessary to break all of the bonds in a single mole of gas. It is measured in \({\text{kcal}}/{\text{mol}}\) or \({\text{kJ}}/{\text{mol}}{\text{.}}\) The bond energy of the \({\text{H-H}}\) bond, for example, is \({\text{433}}\,{\text{kJ}}/{\text{mo}}{{\text{l}}^{ – 1}}\) or \(103.58\,{\text{kcal}}/{\text{mo}}{{\text{l}}^{ – 1}}.\) The bond energies of some common bonds are listed below:

Bond	Bond Energy
\({\text{Cl}} – {\text{Cl}}\)	\(243\,{\text{kJ}}\,{\text{mo}}{{\text{l}}^{ – 1}}\)
\({\text{O}} – {\text{O}}\)	\(499.0\,\,{\text{kJ}}\,{\text{mo}}{{\text{l}}^{ – 1}}\)
\({\text{C}} – {\text{H}}\)	\(414.0\,\,{\text{kJ}}\,{\text{mo}}{{\text{l}}^{ – 1}}\)
\({\text{O}} – {\text{H}}\)	\(460.0\,\,{\text{kJ}}\,{\text{mo}}{{\text{l}}^{ – 1}}\)

Summary

The sum of a system’s internal energy and the product of its pressure and volume is its enthalpy. Enthalpy is a function of the state, and its absolute value cannot be determined. A change in enthalpy \(\left({\Delta {\text{H}}} \right)\) associated with a process, on the other hand, can be measured precisely. Depending on the nature of the reaction, enthalpy or enthalpy changes accompanying chemical reactions are expressed in various ways.

FAQs on Enthalpies of Different Types of Reactions

Q.1. What is enthalpy, and what are its types?
Ans: Enthalpy of a system is defined as the sum of a system’s internal energy and the product of its pressure and volume. Enthalpy is a function of the state, and its absolute value cannot be determined. A change in enthalpy \(\left({\Delta {\text{H}}} \right)\) associated with a process, on the other hand, can be measured precisely. Depending on the nature of the reaction, enthalpy or enthalpy changes accompanying chemical reactions are expressed in a variety of ways, such as Enthalpy of Formation, Enthalpy of Combustion and Enthalpy of Neutralisation etc.

Q.2. What are various types of enthalpies?
Ans: Some of the important types of enthalpies are given as:
1. Enthalpy of Formation
2. Enthalpy of Combustion
3. Enthalpy of Solution
4. Enthalpy of Neutralisation
5. Enthalpy of Hydration
6. Enthalpy of Fusion
7. Enthalpy of Vaporisation
8. Enthalpy of Sublimation
9. Enthalpy of Transition
10. Bond Enthalpy

Q.3. How can we calculate the standard enthalpy of reaction from the standard enthalpy of formation?
Ans: We can calculate the standard enthalpy of reaction from the standard enthalpy of formation of various reactants and products. The standard enthalpy of reaction equals the standard enthalpy of product formation minus the standard enthalpy of reactant formation.

\(\Delta {{\text{H}}^ \circ } = \left[ \begin{gathered} {\text{Total}}\,{\text{standard}}\,{\text{heat}} \hfill \\ {\text{of}}\,{\text{formation}}\,{\text{of}}\,{\text{products}} \hfill \\ \end{gathered} \right] – \left[ \begin{gathered} {\text{Total}}\,{\text{standard}}\,{\text{heat}} \hfill \\ {\text{of}}\,{\text{formation}}\,{\text{of}}\,{\text{reactants}} \hfill \\ \end{gathered} \right]\)
\(\Delta {{\text{H}}^ \circ } = \Delta {\text{H}}_{\text{f}}^ \circ \left({{\text{products}}} \right) – \Delta {\text{H}}_{\text{f}}^ \circ \left({{\text{reactants}}} \right)\)
Let us consider a general reaction:
\({\text{aA}} + {\text{bB}} \to {\text{cC}} + {\text{dD}}\)
The standard enthalpy of reaction is given by
\(\Delta {{\text{H}}^ \circ } = \Delta {\text{H}}_{\text{f}}^ \circ \left({{\text{products}}} \right) – \Delta {\text{H}}_{\text{f}}^ \circ \left({{\text{reactants}}} \right)\)
\( = \left[{{\text{c}} \times \Delta {\text{H}}_{\text{f}}^ \circ \left({\text{C}} \right) + {\text{d}} \times {\text{H}}_{\text{f}}^ \circ \left({\text{D}} \right)}\right] – \left[{{\text{a}} \times \Delta{\text{H}}_{\text{f}}^ \circ \left({\text{A}} \right) +{\text{b}} \times \Delta {\text{H}}_{\text{f}}^ \circ \left({\text{B}}\right)} \right]\)

Q.4. What is the enthalpy of neutralisation?
Ans: The change in enthalpy of a system when one gram equivalent of an acid is neutralised by one gram equivalent of a base or vice versa in dilute solution is known as the enthalpy of neutralisation. The enthalpy of neutralisation of an acid and a base equals the enthalpy of water production from hydrogen and hydroxyl ions. The enthalpy of neutralisation varies substantially from \(- 57.1\,{\text{kJ}}\,{\text{mo}}{{\text{l}}^{ – 1}}\) when weak acids or weak bases are neutralised by strong bases or strong acids, respectively.

Q.5. What is bond enthalpy?
Ans: The formation of a bond between two atoms results in the release of energy. When the bond is broken, an equal amount of energy is absorbed. The bond energy is the average amount of energy necessary to break all bonds of a specific type in one mole of a substance. As a result, the bond energy of the \({\text{H-H}}\) bond is the amount of energy necessary to break all of the bonds in a single mole of gas. It is measured in \({\text{kcal}}/{\text{mol}}\) or \({\text{kJ}}/{\text{mol}}{\text{.}}\) The bond energy of the \({\text{H-H}}\) bond, for example, is \({\text{433}}\,{\text{kJ}}/{\text{mo}}{{\text{l}}^{ – 1}}\) or \(103.58\,{\text{kcal}}/{\text{mo}}{{\text{l}}^{ – 1}}.\)

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