Enthalpies for Different Types of Reactions

Thermodynamics is a field of physics that studies the relationship between heat, work, and temperature, as well as their relationships with energy, entropy, and the physical properties of matter and radiation. The four principles of thermodynamics regulate the behaviour of these quantities, which provide a quantitative description using quantifiable macroscopic physical quantities but may be described in terms of microscopic elements using statistical mechanics. Thermodynamics is a branch of science and engineering that covers a vast range of topics, including physical chemistry, biochemistry, chemical engineering, and mechanical engineering, as well as other complex fields like meteorology.

Enthalpies of Different Types of Reactions

The enthalpy of a system is the product of its internal energy and the product of its pressure and volume. It’s also referred to as heat content. That is to say,

H = E + PV

Where,

E = Internal energy of the system

P = Pressure of the system

V = Volume of the system

The absolute value of enthalpy cannot be calculated because it is a state function. On the other hand, a change in enthalpy (ΔH) associated with a process may be measured precisely and is supplied by the formula.

ΔH = H_products – H_reactants

      = H_p – H_r

If ΔV is the change in volume in a reaction at constant temperature and pressure, the change in enthalpy will equal the sum of the internal energy change (ΔE) and the work done in expansion or contraction. That is correct.

ΔH = ΔE + P × ΔV

Enthalpy or enthalpy changes accompanying chemical processes are expressed in a variety of ways, depending on the nature of the reaction. These will be explored further down.

Enthalpy of Formation

The enthalpy of formation is the change in enthalpy that occurs when one mole of a compound is made from its constituent parts. ΔHf is the symbol for it. For example, the enthalpy of production of ferrous sulphide and acetylene can be expressed as:

Fe(s) + S(s) → FeS(s)   ΔH_f = –100.41kJmol^–1

2C(s) + H₂(g) → C₂H₂(g)   ΔH_f = 222.3kJmol^–1

Standard Enthalpy of Formation

The standard enthalpy of production of a compound is defined as the enthalpy change that occurs when one mole of a compound is created from its elements when all components are in their standard states (298k and 1 atm pressure). ΔH^∘_f is the symbol for it. By convention, the standard enthalpy of production of all elements in their standard state is taken to be zero.

Standard Enthalpy of Reaction from Standard Enthalpy of Formation

We can calculate the reaction enthalpy under standard conditions using the values of standard enthalpies of the formation of various reactants and products. The standard enthalpy of reaction is equal to the product formation enthalpy minus the reactant formation enthalpy.

ΔH^∘ = [Total standard heat of formation of products] – [Total standard heat of formation of reactants]

ΔH^∘ = ΔH^∘_f (products) – ΔH^∘_f (reactants)

Let us consider a general reaction:

aA + bB → cC + dD

The standard enthalpy of reaction is given by

ΔH^∘ = ΔH^∘_f (products) – ΔH^∘_f (reactants)

       = – [a × ΔH^∘_f(A) + b × ΔH^∘_f(B)]

Enthalpy of Combustion

The enthalpy of combustion is the change in a system’s enthalpy when one mole of a substance is completely burnt in excess of air or oxygen. ΔH_c is the symbol for it. The enthalpy of methane combustion is seen in the following reaction:

CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l)       ΔH_c = –890.3kJmol^–1

Enthalpy of Solution

Enthalpy changes are prevalent when a material is dissolved in a solvent. When a reaction occurs in a solution, the enthalpy of the solution of reactants and products must be addressed. The enthalpy of solution is the change in enthalpy that occurs when one mole of a substance is dissolved in a specific amount of solvent at a specific temperature.

When one mole of copper sulphate is dissolved in water to generate one molar solution, the enthalpy absorbed is 78.5kJ/mol. If the solution is diluted more, the enthalpy will change again. We will reach a point where further dilution has no thermal effect if we continue to dilute the fluid. This circumstance is known as the state of endless dilution. To avoid including the quantity of the solvent in our definition, we must incorporate the concept of infinite dilution, which can be stated as follows: The change in enthalpy that occurs when one mole of a chemical is dissolved in a solvent so that further dilution has no effect on the enthalpy. The enthalpy of a solution can also be expressed as :

KCl(s) + H₂O(l) → KCl(aq)   ΔH = –18.4kJmol^–1

MgSO₄(s) + H₂O(l) → MgSO₄(aq)   ΔH = –84.8kJmol^–1

Enthalpy of Neutralisation

The enthalpy of neutralisation is the change in a system’s enthalpy when one gram equivalent of an acid is neutralised by one gram equivalent of a base or vice versa in dilute solution. The enthalpy of neutralisation can be demonstrated using the example below.

HCl(aq) + NaOH(aq) → NaCl(aq) + H₂O(l)      ΔH = –57.1kJmol^–1

According to the data, the enthalpy of neutralisation of a strong acid and strong base is –57.1kJmol^–1, regardless of which acid or base is utilised. This pattern has been properly described using ionisation theory. We get HA and BOH when we mix equivalent amounts of any strong acid and base in a dilute solution.

H⁺(aq)+A^–(aq)+B⁺(aq)+OH^– → A^–(aq)→(aq)+B⁺(aq)+H₂O(l)     ΔH = –57.1kJmol^–1 

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

H⁺(aq)+OH^–(aq)→H₂O(l)  ΔH = –57.1kJmol^–1

The enthalpy of neutralising acid and a base is thus the same as the enthalpy of producing water from hydrogen and hydroxyl ions. When weak acids or weak bases are neutralised by strong bases or strong acids, the enthalpy of neutralisation varies substantially from –57.1kJmol^–1.

Enthalpy of Hydration

The enthalpy of hydration is defined as the change in enthalpy that occurs when one mole of anhydrous salt is combined with the requisite number of moles of water to generate hydrated salt.

Enthalpy of Fusion

Enthalpy of fusion is the change in enthalpy that occurs when one mole of a solid substance is converted into a liquid state at its melting point. Consider the melting of one mole of ice at its melting point of 0^∘C or 273^∘F. The following is a description of the procedure:

H₂O_ice(s)→H₂O_Water(l)  ΔH=6.0kJmol^–1

During this procedure, a total of 6.0kJmol^–1 of enthalpy is absorbed. The magnitudes of intermolecular forces between different substances can be compared using their fusion values. As the enthalpy of fusion of a substance increases, so does the amount of intermolecular forces.

Enthalpy of Vaporisation

This process consumes a total of 6.0kJmol^–1 of enthalpy. Based on the fusion values of different substances, we can compare the magnitudes of intermolecular forces between them. As the enthalpy of fusion of a substance rises, so does the amount of intermolecular forces.

H₂O_Water(l)→H₂O_steam(g)   ΔH=40.6kJmol^–1

Enthalpy of Sublimation

Sublimation is the transformation of a solid into a gaseous state without becoming liquid first. It occurs at a temperature below the melting point of the solid. Enthalpy of sublimation is the change in enthalpy that occurs when one mole of a solid is directly converted into a gaseous state at a temperature below its melting point. The sublimation enthalpy of iodine, for example, is 62.4kJmol^–1. It’s possible to write it as:

I₂₍s) → I₂(g)   ΔH = 62.4kJmol^–1

Enthalpy of Transition

The enthalpy of transition is the change in enthalpy that occurs when one mole of an element changes from one allotropic state to another. For example, the transformation of diamond into amorphous carbon might be illustrated as

C_diamond→C_amorphous    ΔH = 13.8kJmol^–1

Bond Enthalpy

The establishment of a bond between two atoms causes energy to be released. The same quantity of energy is absorbed when the link is severed. The average amount of energy required to break all bonds of a certain type in one mole of a substance is known as bond energy. As a result, the H-H bond’s bond energy is equal to the energy required to break all of the bonds in a single mole of gas. It is expressed in terms of kcal/mol or kJ/mol. The H-H bond, for example, has a bond energy of 433 kJ/mol^–1 or 103.58 kcal/mol^–1. The following are the bond energies of several common bonds:                                                              

Bond	Bond Energy
Cl–Cl	243 kJmol–1
O–O	499.0 kJmol–1
C–H	414.0 kJmol–1
O–H	460.0 kJmol–1

Sample Questions

Question 1: What is enthalpy, and what are its types?

Answer    

The sum of a system’s internal energy and the product of its pressure and volume is called enthalpy. The absolute value of enthalpy cannot be established because it is a function of the state. On the other hand, a change in enthalpy (H) associated with a process can be precisely measured. Enthalpy or enthalpy changes accompanying chemical reactions are stated in a variety of ways depending on the nature of the reaction, such as Enthalpy of Formation, Enthalpy of Combustion, and Enthalpy of Neutralization.

Question 2: What are various types of enthalpies?

Answer

The following are some of the most common types of enthalpies:

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

Question 3: From the standard enthalpy of formation, how can we compute the standard enthalpy of reaction?

Answer

The standard enthalpy of reaction can be calculated using the standard enthalpy of formation of various reactants and products. The standard enthalpy of reaction is equal to the product formation enthalpy minus the reactant formation enthalpy.

ΔH^∘ = [Total standard heat of formation of products] – [Total standard heat of formation of reactants]

ΔH^∘ = ΔH^∘_f(products) – ΔH^∘_f(reactants)

Let us consider a general reaction:

aA+bB→cC+dD

The standard enthalpy of reaction is given by

ΔH° = ΔH°_f(products) – ΔH°_f(reactants)

       = – [a×ΔH^∘_f (A)+b×ΔH^∘_f (B)]

Question 4: What is the enthalpy of neutralisation?

Answer

The enthalpy of neutralisation is the change in a system’s enthalpy when one gram equivalent of an acid is neutralised by one gram equivalent of a base or vice versa in dilute solution. The enthalpy of neutralising an acid and a base is the same as the enthalpy of producing water from hydrogen and hydroxyl ions. When weak acids or weak bases are neutralised by strong bases or strong acids, the enthalpy of neutralisation differs significantly from 57.1kJmol^–1.

Question 5: What is bond enthalpy?

Answer

The release of energy is caused by the establishment of a bond between two atoms. A same quantity of energy is absorbed when a link is severed. The bond energy is the average amount of energy required to break all of a substance’s bonds of a particular type in one mole. As a result, the H-H bond’s bond energy is the energy required to break all of the bonds in a single mole of gas. It is expressed in either kcal/mol or kJ/mol units. The H-H bond, for example, has a bond energy of 433 kJ/mol^–1 (103.58 kcal/mol^–1).

Question 6: What is enthalpy of vaporisation ?

Answer

The enthalpy of vaporisation is the enthalpy shift that occurs when one mole of liquid is turned into a vapour or gaseous state at its boiling point. The enthalpy absorbed when one mole of water is converted to steam at 100°C is 40.6 kJmol^–1, which is the enthalpy of water vaporisation. The following is how the process works:

H₂O_Water(l) → H₂O_steam(g)  ΔH = 40.6 kJmol^–1