In this article, we shall study Hess’s law and its applications in thermochemistry.

Statement :

It states that the change in enthalpy accompanying a chemical reaction is independent of the pathway between initial and final states of the chemical reaction.

Explanation:

Let us suppose substance ‘A’ is converted into D directly.

A   →   D,                ΔH =   Q kJ

Suppose the same change is brought about in different steps,

i) A   →  B,               ΔH₁     =   Q₁ kJ

ii)  B → C,                ΔH₂    =   Q₂ kJ

iii)  C →D ,                ΔH₃   =   Q₃ kJ

Representation:

Then, according to Hess’s law,

ΔH = ΔH₁ + ΔH₂ + ΔH₃

Q   =   Q₁   +   Q₂   + Q₃

Thus, Hess’s law implies that change in enthalpy of a chemical reaction depends upon the initial and final state of a chemical reaction irrespective of the number of steps involved in a chemical reaction.

Illustration:

The change in enthalpy of formation of carbon dioxide can be determined in two ways.

One-step preparation:

Solid carbon is burnt in excess of oxygen.

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

Preparation in two steps:

Step – 1: Solid carbon is burnt in limited supply of oxygen to form carbon monoxide,

C_(s) + 1/2O_2(g)  →  CO_(g)   , Δ H₁ = -111.7 kJ

Step – 2: Carbon monoxide is then burnt in excess of oxygen to from carbon dioxide,

CO_(g)+ 1/2O_2(g)  → CO_2(g , ΔH₂=  -283.67 kJ

According to Hess’s law,

Δ H₁ + ΔH₂ =   -111.7 kJ  –  283.67 kJ  = -395.39 kJ   = ΔH

Thus Hess’s law is illustrated.

Applications of Hess’s Law:

• Thermochemical equations can be added subtracted or multiplied like ordinary algebraic equations.
• Hess’s law is useful to calculate heats of many reactions which do not take place directly.
• It is useful to find out heats of extremely slow reaction.
• It is useful to find out the heat of formation, neutralization, etc.

Numerical Problems Based on Hess’s Law:

Example – 01:

Calculate the enthalpy of formation of CO from given data

i) C_(s) + O_2(g)  →   CO_2(g)              ΔH° = -393.5 KJ 

ii) CO_(g)+ ½O_2(g)  →   CO_2(g)      ΔH° = – 283.0 KJ 

Solution:

The formation of carbon monoxide is represented by following thermochemical equation

C_(s) + ½O_2(g)  →   CO_(g)      ΔH° = ?

Keeping given equation (i) as it is and reversing equation (ii) we get

i) C_(s) + O_2(g)  →   CO_2(g)              ΔH₁° = -393.5 KJ 

iii) CO_2(g)  →    CO_(g)+ ½O_2(g)    ΔH₂° = + 283.0 KJ 

Adding equations (i) and (iii) and by Hess’s Law we get

C_(s) + ½O_2(g)  →   CO_(g)      ΔH° = ΔH₁° +  ΔH₂° =  -393.5 KJ + 283.0 KJ

C_(s) + ½O_2(g)  →   CO_(g)      ΔH° =  -110.5 KJ

Ans: Hence enthalpy of formation of CO is – 110.5 kJ mol^-1

Example – 02:

Calculate the enthalpy of formation of CH₄ from given data

i) C_(s) + O_2(g)  →   CO_2(g)                                   ΔH° = -393.5 KJ 

ii) H_2(g) + ½O_2(g)  →   H₂O_(l)                           ΔH° = – 285.8 KJ 

iii) CH_4(g) + 2O_2(g)  →   CO_2(g)  +  2 H₂O_(l)    ΔH° = – 890.3 KJ 

Solution:

The formation of CH₄ is represented by following thermochemical equation

C_(s) + 2H_2(g)  →   CH_4(g)      ΔH° = ?

Keeping given equation (i) as it is, multiplying equation (ii) by 2 and reversing equation (ii) we get

i) C_(s) + O_2(g)  →   CO_2(g)                                      ΔH₁° = -393.5 KJ 

iv) 2 H_2(g) +  O_2(g)  →   2 H₂O_(l)                           ΔH₂° = – 571.6 KJ 

v)   CO_2(g)  +  2 H₂O_(l)   →  CH_4(g)+ 2O_2(g)     ΔH₃° = + 890.3 KJ 

Adding equations (i) (iv) and (v) and by Hess’s Law we get

C_(s) + 2H_2(g)  →   CH_4(g)    ΔH° = ΔH₁° +  ΔH₂° +  ΔH₃° = -393.5 KJ – 571.6 KJ  + 890.3 kJ

C_(s) + 2H_2(g)  →   CH_4(g)     ΔH° =  – 74.8 KJ

Ans: Hence enthalpy of formation of CH₄ is – 74.8 kJ mol^-1

Example – 03:

Calculate the ΔH° for the reaction

2 ClF_(g) + O_2(g)  →   Cl₂O_(g)    + OF_2(g)

From following equations

i) F_2(g)  +   ClF_(g)  →   ClF_3(l)                                       ΔH° = – 139.2 KJ 

ii) 2 ClF_3(l)  + 2 O_2(g)  →   Cl₂O_(g)  + 3 OF_2(g)         ΔH° = + 533.4 KJ 

iii) F_2(g) +½O_2(g)  →   OF_2(g)                                      ΔH° = + 24. 7 KJ 

Solution:

Required reaction is

2 ClF_(g)+ O_2(g)  →   Cl₂O_(g)    + OF_2(g)

Multiplying equation (i) by 2, keeping equation (ii) as it is, and multiplying equation (iii) by 2 and reversing it.

iv) 2 F_2(g)  +   2 ClF_(g)  →   2 ClF_3(l)                            ΔH° = – 278.4 KJ 

ii) 2 ClF_3(l)  + 2 O_2(g)  →   Cl₂O_(g)  + 3 OF_2(g)         ΔH° = + 533.4 KJ 

v) 2 OF_2(g)     →   2 F_2(g)+    O_2(g)                              ΔH° = – 49.4  KJ 

Adding equations (iv) (ii) and (v) and by Hess’s Law we get

2 ClF_(g)+ O_2(g)  →   Cl₂O_(g)    + OF_2(g)   ΔH° = ΔH₁° +  ΔH₂° +  ΔH₃° = – 278.4 KJ  + 533.4 KJ  – 49.4 kJ

2 ClF_(g) + O_2(g)  →   Cl₂O_(g)    + OF_2(g)   ΔH° = + 205.6 KJ

Hence ΔH° for the reaction is 205.6 kJ

Example – 04:

Calculate the ΔH° for the reaction between ethene with water to form ethanol from the following data.

From following equations

i) C₂H₅OH_(l)  +   3 O_2(g)  →   2CO_2(g)  + 3 H₂O_(l)     ΔH° = – 1368 KJ 

ii) C₂H_4(g)  +   3 O_2(g)  →   2CO_2(g)  + 2 H₂O_(l)          ΔH° = – 1410 KJ 

Is  ΔH° calculated the enthalpy of formation of liquid ethanol?

Solution:

The reaction between ethene and water is represented by

C₂H_4(g)  +   H₂O_(l)   →   C₂H₅OH_(l)         ΔH° = ?

Reversing equation (i), keeping equation (ii) as it is

iii) 2CO_2(g)  + 3 H₂O_(l)  → C₂H₅OH_(l)  +   3 O_2(g)    ΔH° = + 1368 KJ 

ii) C₂H_4(g)  +   3 O_2(g)  →   2CO_2(g)  + 2 H₂O_(l)          ΔH° = – 1410 KJ

Adding equations (iii) and (ii) and by Hess’s Law we get

C₂H_4(g)  +   H₂O_(l)   →   C₂H₅OH_(l)         ΔH° = ΔH₁° +  ΔH₂° = 1368 KJ  -1410 KJ 

C₂H_4(g)  +   H₂O_(l)   →   C₂H₅OH_(l)         ΔH° = – 42 kJ

Hence ΔH° for the reaction is – 42 kJ

This cannot be the heat of formation of ethanol, because it is not obtained from its constituent elements in their standard state.

Example – 05:

Calculate the ΔH° for the reaction       2C_(graphite)  +   3 H_2(g)  →   C₂H_6(g)  

From following equations

i) C₂H_6(g)  +   7/2 O_2(g)  →   2CO_2(g)  +3 H₂O_(l)          ΔH° = – 1560 KJ 

ii)  H_2(g)  +  1/2 O_2(g)  →   H₂O_(l)                                   ΔH° = – 285.8 KJ 

ii)  C_(graphite)  +  O_2(g)  →   CO_2(g)                                  ΔH° = – 393.5 KJ 

Solution:

The required reaction is

2C_(graphite)  +   3 H_2(g)  →   C₂H_6(g)          ΔH° =?

Reversing equation (i), multiplying equation (ii) by 3 and multiplying equation (iii) by 2 we get

iv) 2CO_2(g)  +3 H₂O_(l)   →   C₂H_6(g)  +   7/2 O_2(g)          ΔH° = + 1560 KJ 

v)  3H_2(g)  + 3/2 O_2(g)  →   3 H₂O_(l)                                   ΔH° = – 857.4 KJ 

vi)  2C_(graphite)  +  2O_2(g)  →   2CO_2(g)                               ΔH° = – 787.0 KJ 

Adding equations (iv), (v) and (vi) and by Hess’s Law we get

2C_(graphite)  +   3 H_2(g)  →   C₂H_6(g) , ΔH° = ΔH₁° +  ΔH₂° +  ΔH₃° = 1560 KJ  -857.4 KJ  – 787 kJ

2C_(graphite)  +   3 H_2(g)  →   C₂H_6(g)  ΔH° = – 84.4 kJ

Ans: Hence ΔH° for the reaction is – 84.4 kJ

Example – 06:

Calculate the standard enthalpy for the reaction      

2Fe_(s) +   3/2 O_2(g)  →  Fe₂O_3(s)  

From following equations

i) 2 Al_(s)  +  Fe₂O_3(s)    →   2Fe_(s)  + Al₂O_3(s)         ΔH° = – 847.6 KJ 

ii) 2 Al_(s)    +  3 /2 O_2(g)  →   Al₂O_3(s)                       ΔH° = – 1670 KJ 

Solution:

The required reaction is

2Fe_(s)  +   3/2 O_2(g)  →  Fe₂O_3(s)           ΔH° =?

Reversing equation (i), keeping equation (ii) as it is we get

iii) 2Fe_(s)  + Al₂O_3(s)    →  2 Al_(s)  +  Fe₂O_3(s)          ΔH° =  847.6 KJ 

ii) 2 Al_(s)    +  3 /2 O_2(g)  →   Al₂O_3(s)                       ΔH° = – 1670 KJ 

Adding equations (iii) and (ii) we get

2Fe_(s)  +   3/2 O_2(g)  →  Fe₂O_3(s)      ΔH° = ΔH₁° +  ΔH₂° = 847.6 KJ  – 1670 KJ

2Fe_(s)  +   3/2 O_2(g)  →  Fe₂O_3(s) ,   ΔH° = – 822.4 kJ

Hence standard enthalpy of the reaction is – 822.4 kJ

Example – 07:

Given the following equations and ΔH° at 25 °C     

i)  Si_(s)  + O_2(g)     →    SiO_2(s)                  ΔH° = – 911 KJ 

ii) 2C_(graphite)  +  O_2(g)  →   2CO_(g)         ΔH° = – 221 KJ 

iii)   Si_(s)    +  C_(graphite)   →   SiC_(s)           ΔH° = – 65.3 KJ 

Calculate ΔH° for the reaction SiO_2(s)    + 3 C_(graphite)   →   SiC_(s)    +    2CO_(g) 

Solution:

The required reaction is

SiO_2(s)    + 3 C_(graphite)   →   SiC_(s)    +    2CO_(g)         ΔH° =?

Reversing equation (i), keeping equations (ii) and (iii)  as it is we get

iv)  SiO_2(s)       →    Si_(s)  + O_2(g)           ΔH° =  911 KJ 

ii) 2C_(graphite)  +  O_2(g)  →   2CO_(g)         ΔH° = – 221 KJ 

iii)   Si_(s)    +  C_(graphite)   →   SiC_(s)           ΔH° = – 65.3 KJ 

Adding equations (iv), (ii) and (iii) and by Hess’s Law we get

SiO_2(s)    + 3 C_(graphite)   →   SiC_(s)    +    2CO_(g)         ΔH°  = ΔH₁° +  ΔH₂° +  ΔH₃° = 911 kJ  -221 kJ -65.3 kJ

SiO_2(s)    + 3 C_(graphite)   →   SiC_(s)    +    2CO_(g)         ΔH°  = + 624.7 kJ

Hence ΔH°   for the reaction is 624.7 kJ

Example – 08:

Given the following equations and ΔH° at 25 °C     

i)  2H₃BO_3(aq)     →    B₂O_3(s)    +   3 H₂O_(l) ΔH° = + 14.4 KJ 

ii) H₃BO_3(aq)    →   HBO_2(aq)    +  H₂O_(l)            ΔH° = – 0.02 KJ 

iii)   H₂B₄O_7(s)     →   2 B₂O_3(s)    +  H₂O_(l) ΔH° = 17.3  KJ 

Calculate ΔH° for the reaction H₂B₄O_7(s)   + H₂O_(l)    →   4 HBO_2(aq)

Solution:

The required reaction is

H₂B₄O_7(s)   + H₂O_(l)    →   4 HBO_2(aq)       ΔH° =?

Reversing equation (i) and multiplying by 2, Multiplying equations (i) by4 keeping equation (iii)  as it is we get

iv)  2B₂O_3(s)    +   6 H₂O_(l)    →    4 H₃BO_3(aq)   ΔH° = – 28.8 KJ 

v) 4  H₃BO_3(aq)    →   4 HBO_2(aq)    +  4 H₂O_(l)        ΔH° = – 0.08 KJ 

iii)   H₂B₄O_7(s)     →   2 B₂O_3(s)    +  H₂O_(l) ΔH° = 17.3  KJ

Adding equations (iv), (v) and (iii) and by Hess’s Law we get

H₂B₄O_7(s)   + H₂O_(l)    →   4 HBO_2(aq)       ΔH° = ΔH₁° +  ΔH₂° +  ΔH₃° = – 28.8 kJ  – 0.08 kJ + 17.3kJ

H₂B₄O_7(s)   + H₂O_(l)    →   4 HBO_2(aq)       ΔH° = – 11.58 kJ

Hence ΔH°   for the reaction is -11.58 kJ

Previous Topic: Concept of Bond Enthalpy

Next Topic: Concept of Entropy of a System

Science > Chemistry > Chemical Thermodynamics and Energetics > Hess’s Law and its Applications

The post Hess’s Law and its Applications 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.

Previous Topic: Enthalpy of Reaction

Next Topic: Bond Enthalpy

Science > Chemistry > Chemical Thermodynamics and Energetics > Heat of Reaction Of Different Processes

The post Enthalpy Of Different Processes appeared first on The Fact Factor.