In this article, we shall study the concept of bond enthalpy. During the formation of the bond, energy is evolved or released. Thus the bond formation reaction is exothermic. Due to the release of energy molecule formed is at a low energy level and more stable than isolated atoms. The energy released when a covalent bond is formed is called the energy of bond formation.

Bond Enthalpy:

Chemical reactions involve the breaking of bonds in the reactants and the formation of new bonds in the products with the liberation of energy. Whenever a chemical bond is formed, in a molecule, the energy is liberated and therefore to break the same bond an equivalent amount of energy must be given (supplied) to the molecule. Hence bond enthalpy can be defined as “the average amount of energy per mole required to break a particular bond in a gaseous molecule producing free gaseous atoms or radicals”.

Example:

CH_4(g) →   C_(g) + 4H_(g), ΔH° = 1664 KJ mol^-1

The meaning of the above equation is that to break 1 mole of methane in the gaseous state into carbon (in a gaseous state) and hydrogen in a gaseous state, 1664 KJ  of energy is required. This is the energy required to break 4 C-H bonds in Methane.

Note: It should be noted that in the above case the products are C_(g) and 4H_(g) and not 2H_2(g) (g), C_(s).

Now, a molecule of methane contains 4 C — H bonds. Hence the average C —H bond energy in methane = 1664 /4 = 416 KJ mol^-1.

Bond Dissociation Enthalpy:

It is defined as the energy required per mole to break a specific bond in a specific molecule, in a gaseous state.

For diatomic molecules, bond energy and bond dissociation energy will be the same. But for polyatomic molecules bond energy and bond dissociation energy will be different.

Example of Poly Atomic Molecule:

The different bond dissociation energies involved in methane are as follows :

CH_4(g) →   CH_3(g) + H_(g), DΔH°_CH3-C = 435 KJ mol^-1

CH_3(g) →   CH_2(g) + H_(g), DΔH°_CH2-C = 443 KJ mol^-1

CH_2(g) →   CH_(g) + H_(g), DΔH°_CH-C = 443 KJ mol^-1

CH_(g) →   C_(g) + H_(g), D  ΔH°_CH-C = 343 KJ mol^-1

Thus to break a particular C — H bond, the energy required is different than that required for another C — H bond in the same molecule. But the average C — H bond energy in methane = (435 + 443 + 443 + 343) /4 = 1664 /4 = 416 kJ

Heat of Reaction from Bond Enthalpy:

A chemical reaction involves breaking and forming of chemical bonds. During the formation of the bond, energy is released and during the breaking of a bond, energy is absorbed.

The enthalpy changes involving gaseous reactants and gaseous products having covalent bonds can be calculated with the help of bond enthalpies of reactants and products using following formula

Heat of Reaction = ∑ ΔH° (reactant bonds) – ∑ ΔH° (products bonds)

Numerical Problems on Bond Enthalpy:

Example – 01:

Calculate the enthalpy change of the reaction

CH_4(g) +  Cl_2(g)    →   CH₃Cl_(g) +   HCl_(g), 

ΔH°_C-H = 414 KJ mol^-1, ΔH°_Cl-Cl = 243 KJ mol^-1, ΔH°_C-Cl = 330 KJ mol^-1, ΔH°_H-Cl = 431 KJ mol^-1,

Solution:

Heat of Reaction = ΔH° = ∑ ΔH° (reactant bonds) – ∑ ΔH° (products bonds)

∴  ΔH° = (4 × ΔH°_C-H + 1 × ΔH°_Cl-Cl ) – (3× ΔH°_C-H + 1 × ΔH°_C-Cl + 1 × ΔH°_H-Cl )

∴  ΔH° = (4 mol × 414 KJ mol^-1 + 1 mol × 243 KJ mol^-1 ) – (3 mol × 414 KJ mol^-1 + 1 mol  × 330 KJ mol^-1 + 1 mol × 431 KJ mol^-1)

∴  ΔH° = (1656 KJ + 243 KJ) – (1242 kJ + 330 KJ +  431 KJ)

∴  ΔH° = 1899 KJ – 2003 kJ = – 104 kJ

Ans: Hence Reaction enthalpy is – 104 kJ

Example – 02:

Calculate the N-N bond enthalpy in the reaction

N₂H_4(g) +  H_2(g)    →   2NH_3(g)             ΔH° = – 184 KJ

ΔH°_N-H = 389 KJ mol^-1, ΔH°_H-H = 435 KJ mol^-1, 

Solution:

Heat of Reaction = ΔH° = ∑ ΔH° (reactant bonds) – ∑ ΔH° (products bonds)

∴  ΔH° = (4 × ΔH°_N-H + 1 × ΔH°_N-N + 1 × ΔH°_H-H ) – 2× (3× ΔH°_N-H )

∴  – 184 kJ = (4 mol × 389 KJ mol^-1 + 1 mol × ΔH°_N-N + 1 mol × 435 KJ mol^-1) – 2× (3mol × 389 KJ mol^-1 )

∴  – 184 kJ = 1556 KJ + 1 mol × ΔH°_N-N + 435 KJ – 2334 KJ

∴  – 184 kJ = – 343 KJ + 1 mol × ΔH°_N-N

∴   1 mol × ΔH°_N-N  =  -184 kJ + 343 kJ

∴   1 mol × ΔH°_N-N  =  159 kJ

∴   ΔH°_N-N  =   + 159 kJ mol^-1

Ans: The N-N bond enthalpy in the reaction is 159 kJ mol^-1

Example – 03:

Calculate the N-H bond enthalpy in the reaction

N_2(g) +  3H_2(g)    →   2NH_3(g)             ΔH° = – 83 KJ

ΔH°_N≡N = 946 KJ mol^-1, ΔH°_H-H = 435 KJ mol^-1, 

Solution:

Heat of Reaction = ΔH° = ∑ ΔH° (reactant bonds) – ∑ ΔH° (products bonds)

∴  ΔH° = (1 × ΔH°_N≡N + 3 ×  ΔH°_H-H ) – 2× (3× ΔH°_N-H )

∴  – 83 kJ = (1 mol × 946 KJ mol^-1 +  3 mol × 435 KJ mol^-1) –  6 mol × ΔH°_N-H

∴  – 83 kJ =  946 KJ  +  1305 KJ  –  6 mol × ΔH°_N-H

∴  – 83 kJ =  2251 KJ   –  6 mol × ΔH°_N-H

∴   6 mol × ΔH°_N-H  =  2251 KJ  +  83 kJ 

∴   6 mol × ΔH°_N-H  =  2234 KJ

∴   ΔH°_N-H  =  2234 KJ  / 6 mol

∴   ΔH°_N-H  = + 389 KJ mol^-11

Ans: The N-H bond enthalpy in the reaction is 389 kJ mol^-1

Example – 04:

Calculate the standard enthalpy ΔH° of the reaction

CH_4(g) +  O_2(g)    →   CH₂O_(g) +   H₂O_(g), D

ΔH°_C-H = 414 KJ mol^-1, ΔH°_O=O = 499 KJ mol^-1,  ΔH°_C=O = 745 KJ mol^-1, ΔH°_O-H = 464 KJ mol^-1,

Solution:

Heat of Reaction = ΔH° = ∑ ΔH° (reactant bonds) – ∑ ΔH° (products bonds)

∴  ΔH° = (4 × ΔH°_C-H + 1 ×  ΔH°_O=O ) –  (2× ΔH°_C-H + 1× ΔH°_C=O +  2× ΔH°_O-H )

∴  ΔH° = (4 mol × 414 KJ mol^-1+ 1 mol ×  499 KJ mol^-1 ) –  (2 mol × 414 KJ mol^-1+ 1 mol 745 _{KJ mol}^-1+  2 mol × 464 KJ mol^-1 )

∴  ΔH° = (1656 kJ + 499 kJ) –  (828 kJ + 745 _kJ+  928 kJ)

∴  ΔH° = 2155 kJ –  2501 kJ = – 346 kJ

Ans: The standard enthalpy ΔH° of the reaction is – 346 kJ

Example – 05:

Calculate C-Cl bond enthalpy from the following data

CH₃Cl_(g) +  Cl_2(g)    →   CH_2 Cl2(g) +   HCl_(g), DΔH°  = – 104 kJ

ΔH°_C-H = 414 KJ mol^-1, ΔH°_Cl-Cl = 243 KJ mol^-1,  ΔH°_H-Cl = 431 KJ mol^-1,

Solution:

Heat of Reaction = ΔH° = ∑ ΔH° (reactant bonds) – ∑ ΔH° (products bonds)

∴  ΔH° = (3 × ΔH°_C-H + 1 ×  ΔH°_C-Cl + 1 ×  ΔH°_Cl-Cl ) –  (2× ΔH°_C-H + 2× ΔH°_C-Cl +  1× ΔH°_H-Cl )

∴  – 104 kJ= (3 mol × 414 KJ mol^-1 + 1 mol ×  ΔH°_C-Cl + 1 mol ×  243 KJ mol^-1 ) –  (2 mol × 414 KJ mol^-1+ 2 mol × ΔH°_C-Cl¹+  1 mol × 431 KJ mol^-1 )

∴  – 104 kJ= (1242 KJ  + 1 mol ×  ΔH°_C-Cl + 243 KJ) – (828 KJ  + 2 mol × ΔH°_C-Cl+  431 KJ )

∴  – 104 kJ= (1485 KJ  + 1 mol ×  ΔH°_C-Cl ) – (1259 KJ  + 2 mol × ΔH°_C-Cl )

∴  – 104 kJ= 1485 KJ  + 1 mol ×  ΔH°_C-Cl  – 1259 KJ  – 2 mol × ΔH°_C-Cl 

∴  – 104 kJ= 226 KJ  – 1 mol ×  ΔH°_C-Cl 

∴ 1 mol ×  ΔH°_C-Cl = 226 KJ  + 104 kJ = 330 kJ

∴ ΔH°_C-Cl =  330 kJ mol^-1 

Ans: The C-Cl bond enthalpy is 330 kJ mol^-1 

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Science > Chemistry > Chemical Thermodynamics and Energetics > Bond Enthalpy

The post Bond Enthalpy appeared first on The Fact Factor.

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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