In this article, we shall study change in enthalpy for different chemical processes.

Enthalpy of Formation (Δ_fH° or Δ_formationH°):

The change in enthalpy of a chemical reaction at a given temperature and pressure, when one mole of the substance is formed from its constituent elements in their standard states is called the heat of formation.

Explanation: Consider the following reaction

C_(s) + O_2(g)  → CO_2(g)   ,  Δ_formationH°  =  -395.39 kJ mol^-1

Since one mole of carbon dioxide gas is formed we can say that the heat of formation of carbon dioxide gas is -395.39 kJ.

Notes:

• The standard state of the element is that stable state of the element in which it exists at 1 atm. Pressure and 298 K
• Enthalpies of elements in their standard states are arbitrarily taken as zero.
• Enthalpy of a compound is equal to its heat of formation.
• When solving problems on the heat of formation make sure that the product side of the thermochemical equation has one mole of the substance whose heat of formation is to be calculated.

Enthalpy of Dissociation (Δ_dissociationH°):

Change in enthalpy of a chemical reaction at a given temperature and pressure, when one mole of a substance is dissociated into its constituent elements is called the heat of dissociation.

Explanation: Consider the following reaction

H_2(g) →  2H_(g),    Δ_dissociationH° = + 435.136 kJ mol^-1

Since one mole of hydrogen gas is dissociated the heat of dissociation of hydrogen gas is + 435.136 kJ

Enthalpy of Combustion (Δ_cH° or Δ_combustionH°):

Change in the enthalpy of a chemical reaction at a given temperature and pressure when one mole of a substance is combusted (burn) completely in excess of oxygen is called the heat of combustion.

Explanation: Consider the following reaction

C_(s) + O_2(g)  → CO_2(g)   , ΔH  =  -395.39 kJ mol^-1

Since one mole of carbon is combusted completely the heat of combustion of carbon is – 395.39 kJ.

Enthalpy of Neutralization (Δ_{neutralization}H°):

Change in the enthalpy of a chemical reaction at a given temperature and pressure when one gram equivalent weight of acid is completely neutralized by one gram equivalent weight of the base is called the heat of neutralization.

Explanation: Consider following reaction

HCl_(aq) + NaOH_(aq)  →   NaCl  +    H₂O     Δ_{neutralization}H°  =  -56.9 kJ

The heat of neutralization of HCI by NaOH is -56.9 KJ.

For all strong acids and bases, the heat of neutralization is the same because, in their neutralization reaction, there is a combination of H+ ions of an acid with OH- ions of the base to produce un-dissociated water.

Change of Phase:

Change of phase involves the change in the physical state of matter. During the phase change, the chemical properties of the substance do not change but physical properties change. The following are the types of phase changes.

Fusion: This is the process in which the matter changes from solid-state to liquid state. It is endothermic process.

e.g. melting of ice H₂O_(s) → H₂O_(l)

Vapourization: This is the process in which the matter changes from liquid state to gaseous state. It is an endothermic process.

e.g. boiling of water H₂O_(l) → H₂O_(g)

Sublimation: This is the process in which the matter changes from the solid-state into a gaseous state directly. It is an endothermic process.

e.g. heating of camphor Camphor_(s) → Camphor_(g)

Characteristics of Change of Phase:

• The phase change always takes place at constant pressure and temperature.
• During the phase transition, there is an equilibrium between the two phases. Thus both the phases exist simultaneously.
• The Change in temperature takes place only when completion of phase transition.

Enthalpy of Fusion (Δ_fusH°):

The enthalpy-change that accompanies the fusion of one mole of a solid without the change in temperature at constant pressure is called its enthalpy of fusion.

Explanation: Consider the following reaction

H₂O_(s) → H₂O_(l),               Δ_fusionH  = +6.01  kJ mol^-1

Thus, the equation indicates that when one mole of ice melts at 0° C at 1 atm pressure, the enthalpy-change is 6.01 kJ. i.e. 6.01 kJ of energy is absorbed.

Enthalpy of Freezing (Δ_freezeH°):

The enthalpy change that accompanies the freezing of one mole of a liquid without a change in temperature at constant pressure is called its enthalpy of freezing.

Explanation: Consider the following reaction

H₂O_(l) → H₂O_(s),               Δ_freezeH  = +6.01  kJ mol^-1

Thus, the equation indicates that when one mole of water freezes at 0° C at 1 atm pressure, the change in enthalpy is -6.01 kJ. i.e. 6.01 kJ of energy is released.

Enthalpy of Vaporization (Δ_vaporizationH°):

The enthalpy change that accompanies the vaporization of one mole of a liquid without a change in temperature at constant pressure is called its enthalpy of vaporization.

Explanation: Consider the following reaction

H₂O_(l) → H₂O_(g),   Δ_{vapourization}H  = + 40.7  kJ mol^-1

Thus, the equation indicates that when one mole of water vaporizes at 100° C at 1 atm pressure, the change in enthalpy is + 40.7 kJ. i.e. 40.7 kJ of energy is absorbed.

Enthalpy of Condensation (Δ_condensationH°):

The enthalpy change that accompanies the condensation of one mole of a liquid without a change in temperature at constant pressure is called its enthalpy of condensation.

Explanation: Consider the following reaction

H₂O_(l) → H₂O_(g),   Δ_condensationH  = – 40.7  kJ mol^-1

Thus, the equation indicates that when one mole of water vapours condense at 100° C at 1 atm pressure, the enthalpy-change is – 40.7 kJ. i.e. 40.7 kJ of energy is released.

Enthalpy of Sublimation (Δ_sublimationH°):

The direct conversion of solid to vapour without going through the liquid state is called sublimation. The enthalpy-change that accompanies the condensation of one mole of a solid directly into vapours at a constant temperature at constant pressure is called its enthalpy of sublimation.

Explanation: Consider the following reaction

H₂O_(s) → H₂O_(g),       Δ_sublimationH  = +51.08  kJ mol^-1

Thus, the equation indicates that when one mole of ice sublimes at O° C at 1 atm pressure, the enthalpy change is + 51.08 kJ. i.e. 51.8 kJ of energy is absorbed.

It is to be noted that Δ_sublimationH = Δ_fusionH + Δ_{vapourization}H

Atomic or Molecular Changes:

Enthalpy of Ionization (Δ_ionizationH°):

The enthalpy change that accompanies the removal of an electron from each atom or ion in one mole of gaseous atoms or ions is called its enthalpy of ionization.

Explanation: Consider the following reaction

Na_(g)   →   Na⁺_(g)+ e^– ,    Δ_ionizationH  = 494  kJ  mol^-1

Thus, the equation indicates that when one mole of gaseous sodium atom ionizes to Na⁺_(g) ion, the change in enthalpy is + 494 kJ. i.e. 494 kJ of energy is absorbed.

Enthalpy of Atomization (Δ_atomizationH°):

The enthalpy change that accompanies the dissociation of all the molecules in one mole of gas-phase substance into gaseous atoms is called its enthalpy of atomization.

Explanation: Consider the following reaction

Cl_2(g) → Cl_(g)+ Cl_(g),          Δ_atomizationH  = 242  kJ mol^-1

Thus, the equation indicates that when one mole of gaseous chlorine molecule dissociates completely into its atomic form in the gaseous state then the change in enthalpy is + 242 kJ. i.e. 242 kJ of energy is absorbed.

Enthalpy of Solution (Δ_solutionH°):

Change in enthalpy of chemical reaction at a given temperature and pressure, when one mole of a solution is dissolved in a specified quantity of solvent so as to form a solution of particular concentration is called as enthalpy of a solution.

Explanation: Consider the following reaction

KOH_(s)   +   H₂O_(l) → KOH_(aq)        Δ_solutionH   = -58.57 KJ mol^-1

Thus, the equation indicates that when one mole of potassium hydroxide (solute) dissolves in one mole of water (solvent) to form one mole of potassium hydroxide solution in water then the change in enthalpy is -58.57 kJ. i.e. 58.57 kJ of energy is released

Bond Enthalpy (Bond Energy):

Chemical reactions involve the breaking and making of chemical bonds. Energy is required to break a bond and energy is released when a bond is formed. It is possible to relate the heat of reaction to changes in energy associated with the breaking and making of chemical bonds. With reference to the enthalpy changes associated with chemical bonds, two different terms are used in thermodynamics. (i) Bond dissociation enthalpy (ii) Mean bond enthalpy. Let us discuss these terms with reference to diatomic and polyatomic molecules.

Diatomic Molecules:

Consider the following process in which the bonds in one mole of dihydrogen gas (H2) are broken:

H_2(g) →  2H_(g) ;   ΔH–HHO = 435.0 kJ mol^-1

The enthalpy change involved in this process is the bond dissociation enthalpy of H–H bond.

The bond dissociation enthalpy is the change in enthalpy when one mole of covalent bonds of a gaseous covalent compound is broken to form products in the gas phase. Note that it is the same as the enthalpy of atomization of dihydrogen. This is true for all diatomic molecules.

Polyatomic Molecule:

In the case of polyatomic molecules, bond dissociation enthalpy is different for different bonds within the same molecule. Let us consider a polyatomic molecule like methane, CH4.  The overall thermochemical equation for its atomization reaction is given below:

CH_4(g) →  C_(g) + 4H_(g) , ΔH = +1665 KJ.

In methane, all the four C – H bonds are identical in bond length and energy. However, the energies required to break the individual C – H bonds in each successive step differ. In such cases, we use mean bond enthalpy

CH_4(g) →  CH_3(g) + H_(g) , ΔH = +427 KJ.

CH_3(g) → CH_2(g) + H_(g), ΔH = +439 KJ

CH_2(g) → CH_(g) + H_(g), ΔH = +452 KJ

CH_(g) → C (g) + H_(g),   ΔH = +347 KJ

Adding above reactions we get

CH_4(g) →  C_(g) + 4H_(g) , ΔH = +1665 KJ

We find that mean C–H bond enthalpy in methane as 1664/4 = 416 kJ/mol. Using Hess’s law, bond enthalpies can be calculated.

Enthalpy of Reaction from Bond Enthalpy:

The reaction enthalpies are very important quantities as these arise from the changes that accompany the breaking of old bonds and the formation of the new bonds. We can predict the enthalpy of a reaction in the gas phase if we know different bond enthalpies. The standard enthalpy of reaction is related to bond enthalpies of the reactants and products in gas phase reactions as:

ΔH = ∑ Bond enthalpies _reactants  –     ∑ Bond enthalpies _products

Remember that this relationship is approximate and is valid when all substances (reactants and products) in the reaction are in a gaseous state.

The values of given bond enthalpy can be used to calculate bond enthalpies of specific bonds in the molecule.

Crystal Lattice Energy (Δ_LatticeH°):

Crystal lattice energy is defined as the enthalpy change or energy released accompanying formation of 1 mole of crystalline solid from its constituent ions in the gaseous state at a constant temperature.

Explanation:

M⁺_(g)  + X^–_(g) → M⁺X^–_{(s) ,}  ΔH = – x KJ  mol^-1

Thus, the equation indicates that when one mole of ionic compound M+X-(s) is formed from its constituent ions the change in enthalpy is – x kJ i.e. x kJ of energy is evolved. Crystal lattice energy is always negative.

The sequence of actions involved in the formation of 1 mole of an ionic compound in its standard state from its constituent elements in their states at constant temperature and pressure is called Born-Haber cycle.

Factors Affecting Crystal Lattice Energy:

Crystal lattice energy depends on the interionic distance in the crystalline solid. As the distance decreases, the crystal lattice energy increases. Crystal lattice energy depends on the charge of constituent cations and anions.

Enthalpy of Hydration (Δ_HydrationH°):

It is defined as the heat evolved when one mole of gaseous ions dissolve in water by hydration to give infinitely dilute solution at constant temperature and pressure.

Explanation: Consider the following reaction

Na⁺_(g) + aq → Na⁺_(ag) ,     ΔH = – 390 KJ mol^-1

Thus, the equation indicates that when one mole of sodium ion in the gaseous state is dissolved in water the change in enthalpy is – 390 kJ. i.e. 390 kJ of energy is released.

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The branch of chemistry which deals with the quantitative study of thermal or heat changes in various chemical reactions is known as thermochemistry. In this article, We shall discuss very important concept of chemistry i.e. heat of reaction.

Thermochemical Equation:

An equation which indicates the heat changes in a chemical reaction at a certain temperature and pressure with an indication of states of reactants and products is called a thermochemical equation

Example:

C_(s)  + O_2(g)  → CO_2(g)   , ΔH  =  -395.39 kJ

This thermochemical equation indicates that when one mole of solid carbon reacts with one mole of gaseous oxygen at constant pressure one mole of gaseous carbon dioxide is obtained. During this reaction, 395.39 kJ of heat is evolved.

Guidelines for Writing Thermochemical Equation:

• The equation must be automatically balanced like a chemical equation.
• The value of ΔH must be indicated. ΔH is negative for exothermic reaction and positive for an endothermic reaction.
• The physical state of each reactant and each product must be indicated. Symbols such as (s) for solid-state, (l) for the liquid state, (g) for gaseous state and (aq.) for an aqueous solution should be indicated.
• The thermochemical equation can be reversed. During the reversal of the sign of ΔH must be changed.
• The temperature of the reaction can be written as a suffix to ΔH

Example:

C_(s)  + O_2(g)  → CO_2(g)   , ΔH  =  -395.39 kJ

This thermochemical equation indicates that when one mole of solid carbon reacts with one mole of gaseous oxygen at constant pressure one mole of gaseous carbon dioxide is obtained. During this reaction, 395.39 kJ of heat is evolved.

The necessity of Mentioning of State of a Substance:

It is important to mention the physical state of substances in the thermochemical equation because the change of physical state is also accompanied by the enthalpy change.

Example: Consider the following thermochemical equations

H_2(g) +    1/2O_2(g)  →    H₂O_(l) , ΔH =  – 286 kJ

Thus, when 1 mole of hydrogen gas reacts with half a mole of oxygen gas to form one mole of liquid water, 286 kJ of heat is produced.

H_2(g) +    1/2O_2(g)  →    H₂O_(g) , ΔH =  – 249 kJ

Thus, when 1 mole of hydrogen gas reacts with half a mole of oxygen gas to form one mole of water vapours, 249 kJ of heat is produced. Hence, whenever there is a state change, there is a change in enthalpy of reaction. Hence in thermochemical reaction, the state of each substance involved should be mentioned.

Heat of Reaction OR Enthalpy of Chemical Reaction:

The difference between the sum of enthalpies of products and the sum of enthalpies of reactants at a given temperature is called as the heat of reaction. It is denoted by ΔH or ΔU

Explanation:

Consider a general chemical reaction 

A    +    B →  C     +   D.

Let H_A , H_B, H_C, and H_D be the enthalpies of A, B, C and D respectively then the heat of reaction is given by

ΔH  =   (H_C +  H_D)  –  (H_A +  H_B)

The heat of reaction can be determined either at constant pressure or at constant volume.

Heat of Reaction at Constant Pressure:

The difference between the sum of enthalpies of products and the sum of enthalpies of reactants at a given temperature and constant pressure is called the heat of reaction at the constant pressure at a given temperature.

It is denoted by ΔH. Normally heat of reaction at constant pressure is specified at 298 K and 1 atm. Pressure. This heat of reaction is called the standard heat of reaction.

Thus at constant pressure heat of reaction is given by

ΔH = ∑ ΔH_Products  –     ∑ ΔH_Reactants

ΔH is negative for exothermic reaction and positive for an endothermic reaction.

Heat of Reaction at Constant Volume:

The difference between the sum of internal energies of products and the sum of internal energies of reactants at a given temperature and constant volume is called the heat of reaction at the constant volume at a given temperature. It is denoted by ΔE.

Thus at constant pressure heat of reaction is given by

ΔU = ∑ ΔU_Products  –     ∑ ΔU_Reactants

Factors Affecting Heat of Reaction:

• Physical states of substances involved.
• The amount of substance involved.
• Way of carrying out the reaction i.e. in a case of gaseous reactions heat of reaction depends on whether the reaction is carried out at constant pressure or at constant volume.
• The pressure of reactants and products.
• The temperature (as explained by Kirchhoff’s equations)

Different Types of Chemical Reactions on the Basis of Change in the Enthalpy:

On the basis of change in the enthalpy chemical reactions are classified into exothermic reactions and endothermic reactions

Exothermic Reactions:

The chemical reactions which involve the evolution of heat are called as exothermic reactions.

Example:

C_(s)+ O_2(g)    →  CO_2(g) , ΔH = -395.39 kJ

In this case, the enthalpy of products is less than the enthalpy of reactants. For such reactions, the change in enthalpy is always negative.

Characteristics of Exothermic Reactions:

• The chemical reactions which involve the evolution of heat are called exothermic reactions.
• For such reactions, the change in enthalpy is always negative.
• In this case, the enthalpy of products is less than the enthalpy of reactants
• Products are more stable than the reactants.

Endothermic Reactions:

The chemical reactions which involve absorption of heat are called endothermic reactions.

Example :

2C_(s)+ H_2(g)    →  C₂H_2(g) , ΔH = + 225.94kJ

In this case, the enthalpy of products is more than the enthalpy of reactants. For such reactions, the change in enthalpy is always positive.

Characteristics of Endothermic Reactions:

• The chemical reactions which involve absorption of heat are called endothermic reactions.
• For such reactions, the change in enthalpy is always positive.
• In this case, the enthalpy of products is more than the enthalpy of reactants.
• Products are less stable than the reactants.

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