TestBag

1

CHEMICAL ENERGETICS

Contents 1. Introduction

2. Exothermic and Endothermic Reactions

2.1 Exothermic Reactions 2.2 Endothermic Reactions.

3. Enthalpies of Reactions

3.1 Enthalpy Change in different Types of Chemical Reactions 3.1.1 Enthalpy of Formation 3.1.2 Enthalpy of Combustion 3.1.3 Enthalpy of Neutralization 3.1.4 Enthalpy of Solution 3.1.5 Enthalpy of Dilution 3.1.6 Enthalpy of Hydration

3.1.7 Lattice Enthalpy and Hydration Enthalpy 3.1.8 Enthalpy of Atomization 3.1.9 Enthalpy of Ionization 3.1.10 Enthalpy of formation of ions 3.1.11 Enthalpy of Hydrogenation

3.2 Enthalpy Change during Phase Transformation 3.2.1 Enthalpy of Fusion 3.2.2 Enthalpy of Vaporization 3.2.3 Enthalpy of Sublimation

3.2.4 Enthalpy of Allotropic Transformation

4. Hess’s Law of Constant heat summation

5. Bond Enthalpy

6. Calorimetry

TestBag

2

1. INTRODUCTION

Thermodynamics deals basically with energy changes accompanying all type of physical and chemical processes.

Energy is the capacity to do work or supply heat. It can be classified as either Kinetic or Potential.

Kinetic energy is the energy of motion.

K.E. = 1/2mv2

Potential energy is stored energy in an object because of its height or in a molecule because of its chemical bonds.

P.E. = mgh

For example, When the water is allowed to fall over a dam its potential energy is converted into kinetic energy. Propane and other substances used as fuels contain potential energy in their chemical bonds. When these substances undergo reaction with oxygen during burning, their potential energy is converted into mechanical energy. Some of the potential energy is also released as heat.

The SI unit for energy, 1 (kg m2)/s2 is given the name joule (J) after the English Physicist James Prescott Joule. Other unit commonly used is calorie, defined as the amount of energy required to raise the temperature of 1 g of water by 1° C.

1 cal = 4.814 J

Heat is the energy transferred between two objects as the result of a temperature difference. Chemical reactions are invariably associated with a transfer of energy and most frequently, energy transfer in chemical reactions take place in the form of heat. If the amount of energy transferred can be measured than it is possible to obtain vital information about interactions, which occur during a chemical interaction. Hence,

The branch of science, which deals with energy changes during chemical reactions, is called Chemical Energetics. The branch dealing with heat changes is called Thermochemistry.

Thermochemical Equation.

When a balanced chemical equation not only indicates the quantities of the different reactants and products but also indicates the amount of heat evolved or absorbed, it is called a thermochemical equation.

However, contrary to the usual practice about the balanced equations, fractional coefficients may be used in writing a thermochemical equation. For example, the formation of water is written as

TestBag

3

Thus, 285.8 kJ of heat is produced when 1 mole of hydrogen reacts with 0.5 mole of oxygen. If the quantities of reactants are doubled, the heat produced will also be doubled. For example, in the above case, we may write

2H2(g) + O2(g) → 2H2O (l) + 571.6 kJ mol-1

or 2H2(g) + O2(g) → 2H2O (l); ∆H = - 571.6 kJ mol-1

• If the coefficients of the substances are multiplied or divided by some number, the value of ∆H is multiplied or divided by the same number.

• If the reaction is reversed, the sign of ∆H changes but the magnitude remains the same.

Thermochemical Standard States.

The thermochemical standard state of a substance is its most stable state under standard pressure (one atmosphere) and at some specified temperature (25°C or 298 K unless otherwise specified). Under these conditions, any parameter is designated by ∆H°, ∆E°, ∆S° etc.

• For a gas, the standard state is the ideal gas at a pressure of one atmosphere and the given temperature; in a mixture of gases, its partial pressure must be one atmosphere.

• For a liquid, the standard state is the pure liquid at one atmosphere and the given temperature.

• For a substance in solid phase, the standard state is stable crystalline form at 1 atmosphere and a given temperature (eg, graphite form of carbon, rhombic form of sulphur).

• For a substance in solution, the standard state refers to one molar concentration.

2. EXOTHERMIC AND ENDOTHERMIC REACTIONS

2.1 Exothermic Reactions.

These are those reactions, which are accompanied by the evolution of heat.

The quantity of heat produced is shown along with the products with a ‘plus’ sign. A few examples of exothermic reactions are given below:

C(s) + O2(g) CO2(g) + 393.5 kJ

H2(g) + O2(g) H2O(l) + 285.8 kJ

N2(g) + 3H2(g) 2NH3(g) + 92.4 kJ

CH4(g) + 2O2(g) CO2(g) + 2H2O(g) + 890.4 kJ

C4H10(g) + O2(g) 4CO2(g) + 5H2O(g) + 2658 kJ

2.2 Endothermic Reactions.

Those reactions in which heat is absorbed are called endothermic reactions.

Although the heat absorbed can be written with a plus sign alongwith the reactants, however, it is usually shown alongwith the products with a minus sign. A few examples of endothermic reactions are given below: -

TestBag

4

N2(g) + O2(g) 2NO (g) – 180.7 kJ

C(s) + H2O(g) CO(g) + H2(g) – 131.4 kJ

C(s) + 2S(s) CS2(g) – 92.0 kJ

H2(g) + I2(g) 2HI(g) -52.5 kJ

2HgO(s) 2Hg(l) + O2(g) – 181.6 KJ

Exothermic and Endothermic reactions in terms of

The enthalpy change ( ) accompanying a reaction is given by

= heat content of products – heat content of reactants

= HProducts – HReactants

• A reaction is exothermic, i.e., heat is given out in a reaction if the total heat content of the reactants is more than that of the products, i.e., HR > HP.

But if it is so, then according to the above equation will be negative. Hence,

is negative for exothermic reactions.

Thus, the exothermic reactions given above may be written in terms of as

C(s) + O2(g) CO2(g); = - 393.5 KJ mol-1

H2(g) + O2(g) H2O(l); H = - 285.8 KJ mol-1

N2(g) + 3H2(g) 2NH3(g); = - 92.4 KJ mol-1

CH4(g) + 2O2(g) CO2(g) + 2H2O(l); = - 890.4 kJ mol-1

C4H10(g) + O2(g) 4CO2(g) + 5H2O(g); H = - 2658 kJ mol-1

• A reaction is endothermic, i.e., heat is absorbed in a reaction if the total heat content of reactants is less than that of products, i.e., HR < HP Then, according to the equation

H = HP - HR , H will be positive. Hence,

H is positive for endothermic reactions.

Thus, the endothermic reactions given earlier may be written in terms in H as,

N2(g) + O2(g) 2NO (g); H = +180.7 kJ mol-1

C(s) + H2O(g) CO(g) + H2(g); H= +131.4 kJ mol-1

C(s) + 2S(s) CS2(l); H = +92.0 kJ mol-1

H2(g) + I2(g) 2HI(g); H = +52.5 kJ mol-1

2HgO(s) 2Hg(l) + O2(g); H = +181.6 kJ mol-1

TestBag

5

Graphically, the exothermic and endothermic reactions may be represented as shown in fig.

• The physical state (s,g,l etc.) of the different substances must be mentioned because the heat evolved and absorbed depends upon the physical state.

• Unless otherwise mentioned, ∆H values are for the standard state of the substances (i.e., 298 K and 1 bar pressure).

3. ENTHALPIES OF REACTIONS

The amount of heat evolved or absorbed in a chemical reaction when the number of moles of the reactants as represented by the chemical equation have completely reacted, is called the Heat of Reaction or Enthalpy of Reaction or Enthalpy Change of Reaction. It is represented by rH.

Let us consider the following two examples:

(i) CH4(g) + 2O2(g) CO2(g) + 2H2O(l); H = -890.4 kJ mol-1

(ii) C(s) + H2O(g) CO(g) + H2(g); H = +131.4 kJ mol-1

The First equation indicates that when 1 mole of methane (i.e., 16g) combines completely with 2 moles of oxygen gas (i.e., 64 g), 890.4 kJ of heat is produced. Similarly, the second equation tells that when 1 mole of solid carbon (i.e., 12 g) reacts completely with 1 mole of steam (i.e., 18g) 131.4 kJ of heat is absorbed.

It is interesting to note that

Enthalpy of reaction, H = Sum of enthalpies of products – Sum of enthalpies of reactants.

= HProducts – HReactants

If the reaction is reversed, the sign of H changes, e.g,

For H2(g) + O2(g) H2O(l); H = -285.8 kJ mol-1

For H2O(l) H2(g) + O2(g); H = +285.8 kJ mol-1

• Standard Enthalpy of Reaction.

The enthalpy change of a reaction depends upon the conditions under which the reaction is carried out. Hence, it is essential to specify some standard conditions.

The standard enthalpy of a reaction is the enthalpy change accompanying the reaction when all the reactants and products are taken in their standard states.

TestBag

6

Standard conditions are denoted by adding the superscript to the symbol ∆H, ex. . The standard state of a substance at a specified temperature is its pure form at 1 bar.

• Factors on which the Enthalpy of Reaction depends.

The Enthalpy of Reaction (heat of a reaction) depends upon a number of factors as explained below:

(i) Conditions of constant pressure or constant volume, i.e., whether the reaction takes place at constant pressure or at constant volume.

∆H = ∆E + ∆ng RT Where, ∆H = heat of reaction at constant P ∆E = heat of reaction at constant V ∆ng = difference of coefficients of gaseous products and reactants i.e. ∆ng = np - nR

The difference in ∆H and ∆E is insignificant for solids and liquids; however in the case of gases, it is quite important.

(ii) Temperature. The heat of reaction depends upon the temperature at which the reactants and products are taken. The values are usually reported at 298 K.

Heat of reaction at constant P or constant V varies with temperature as given by Kirchhoff’s equation:

At constant P, ∆H2, = ∆H1 + ∆CP (T2 – T1)

At constant V, ∆E2 = ∆E1 + ∆CV (T2 – T1)

Where ∆H2 or ∆E2 is the heat of reaction at temperature T2 and ∆H1 or ∆E1 are at T1.

∆CP = [CP (products) – CP (reactants)] at constant P

∆CV = [CV (products) – CV (reactants)] at constant V

Thus, for the following gaseous phase reaction;

N2 (g) + 3H2 (g) → 2NH3(g)

∆CP = [2CP (NH3)] – [CP (N2) + 3CP (H2)]

∆CV = [2CV (NH3)] – [CV(N2) + 3CV (H2)]

∆ng = 2 – 4 = - 2

Generally CP = 2.5R, CV = 1.5R for monoatomic gases

CP = 3.5R, CV = 2.5R for diatomic gases

CP = 4.0R, CV = 3.0R for other gases

Hence, for the above case ∆CP = - 6R, ∆CV = -4R.

(iii) Physical state of the reactants and products. Since latent heat is involved in the change of state, therefore the physical state of the reactants (i.e. whether they are solids, liquids or gaseous) affects the heat of reaction. For example, when hydrogen and oxygen gases combine to form liquid water, the heat of reaction is different than when they combine to form water in the gaseous state.

H2(g) + O2(g) H2O(l); H = -285.8 kJ mol-1

H2(g)+ O2(g) H2O(g); H = -241.8 kJ mol-1

TestBag

7

Hence, it is essential to write the symbols s, l, g or aq. to indicate whether a particular substance is solid, liquid, and gaseous or an aqueous solution.

The difference between two ∆H° values arises due to difference in physical state of water.

Thus, H2O(l) → H2O(g), ∆H° = 44.1 kJ mol-1

It represents molar enthalpy of vaporization of water.

(iv) Allotropic modification. For elements like sulphur and carbon, which exist in different allotropic modifications, the heat of reaction is different if different allotropic form is involved in the reaction. For example,

C (diamond) + O2 (g) CO2(g); H = -380.4 kJ mol-1

C (graphite) + O2(g) CO2(g); H = - 393.5 kJ mol-1

The difference between two ∆H° values arises due to different allotropes of carbon.

Thus, C(diamond) → C(graphite), ∆H° = -13.1 kJ mol-1

It represents molar enthalpy of transition of carbon.

(v) Concentration of solutions. Heat change occurs when a solute is dissolved in a solvent to form a solution or when a solution is diluted. Therefore, if the solutions are involved in a reaction, their concentrations affect the heat of reaction.

(vi) Quantities of the reactants involved. If the quantities of the reactants are doubled, the

heat of reaction is also doubled.

3.1 Enthalpy Change in different Types of Chemical Reactions

‘Heat of reaction’ or ‘Enthalpy of reaction’ is a general term used for the enthalpy change accompanying any reaction. However, depending upon the nature (i.e., combustion, neutralization etc.), the enthalpy of reaction is named accordingly (i.e., enthalpy of combustion, enthalpy of sublimation, enthalpy of neutralization etc.). Similarly, depending upon the type of process involving a phase change such as fusion, vaporization, sublimation etc. the enthalpy change involved is named according (i.e., enthalpy of fusion, enthalpy of vaporization) etc.

A few important ‘Heats of reactions’ or ‘Enthalpies of reactions’ are as follows:

3.1.1 Enthalpy of Formation (∆fH)

The enthalpy change accompanying the formation of one mole of a compound from its elements in their most stable states of aggregation (also known as reference states) is called its enthalpy of formation (∆∆∆∆fH).

Example:

For formation of 2 moles of HCl, the enthalpy change is -92.3 kJ. Thus, Enthalpy of formation of HCl (∆Hf

0) is -92.3 /2 kJ mol-1.

TestBag

8

Standard Enthalpy of Formation

Standard enthalpy of formation of a substance is defined as the enthalpy change accompanying the formation of 1 mole of the substance in the standard state from it elements at 298 K and 1 bar pressure. It is usually represented by

For example, in the reaction

When 1 mole of CO2 (g) is formed from its elements, viz, C (s) and O2 (g) (all substances being taken in the standard state), 393.5 kJ of heat is produced. Hence, standard enthalpy of formation of gaseous CO2 is 393.5 kJ mol-1.

Caution points:

A. For elementary substances in the standard state, the standard enthalpy of formation ( ) is taken as zero.

B. The enthalpy change of the reaction given below does not represent the standard enthalpy of CaC3O (s).

This is because; the reactants in the given reaction are not its constituent elements. The reaction representing the standard enthalpy of formation of CaCO3 (s) would be

C. All the calculations are based on the enthalpy change, not an actual enthalpy values.

Enthalpies of formation of different compounds help to predict their relative stabilities:

(i) Only those compounds are more stable than their elements whose values are negative.

(ii) Greater the magnitude of the negative value, i.e., more is the energy released, more stable is the compounds.

3.1.2 Enthalpy of Combustion (∆cH)

The enthalpy of combustion of a compound is the decrease in enthalpy (∆H = -ve for exothermic process) accompanying complete combustion or oxidization in O2 of one mole of that compound.

Example:

CH4 (g) + 2O2 (g) → CO2 (g) + 2H2O (l) , ∆cH = 890.30 kJ mol-1

The significance of the word complete oxidation or combustion must be clearly understood.

The reaction does not represent the enthalpy of combustion of C since CO (g) can be further oxidized to CO2.

TestBag

9

Thus the enthalpy of combustion represented as,

Standard enthalpy of combustion

Standard enthalpy of combustion is the amount of heat evolved when one mole of the substance under standard conditions (298 K, 1 bar pressure) is completely burnt to form the products also under standard conditions. It is represented by

The standard enthalpy of combustion of butane, C4H10 (the main constituent of cooking gas), representing the combustion of 1 mole of butane, may be represented as follows:

C4H10 (g) + O2 (g) 4 CO2 (g) + 5 H2O (l),

= - 2658.0 kJ mol-1.

Example: How much energy is released when 6moles of octane is burnt in air? Given for CO2 (g), H2O (g) and C8H18 (g) respectively are -490, -240 and +160kJ/mol.

Solution:

=

∴ For 6 moles, enthalpy of combustion

Calorific values of foods and fuels

Different foods and fuels produce different amounts of heat on combustion. These are usually expressed in terms of their calorific values.

The calorific value of a fuel or food is the amount of heat in calories or joules produced from the complete combustion of one gram of the fuel or the food.

Thus, according to the above reaction, 1 mole of glucose produce heat = 2840 kJ.

Hence, calorific value of glucose = 2840/180 = 15.78 kJ g-1.

3.1.3 Enthalpy of Neutralization (∆neuH)

The amount of heat evolved when one gram equivalent of an acid is completely neutralized by one gram equivalent of a base in dilute solution is called enthalpy of neutralization of that acid.

The enthalpy of neutralization of a base by an acid is defined in the same manner.

Example:

HCl (aq) + NaOH (aq) NaCl (aq) + H2O (l) ; ∆neuH = 57.1 kJ

Hence, enthalpy of neutralization of HCl with NaOH or NaOH with HCl is 57.1 kJ.

TestBag

10

The enthalpy of neutralization of any strong acid (HCl, HNO3, H2SO4) with a strong base (LiOH, NaOH, KOH) or vice versa is always the same, i.e., 57.1 kJ (or 13.7 kcal). This is because the strong acids, strong bases and the salts that they from, are all completely ionized in dilute aqueous solution. Thus, the reaction between any strong acid and strong base, e.g., in the above case may be written as

Na+ + OH- + H+ + Cl- Na+ + Cl- + H2O; neut H = - 57.1 kJ mol-1

Or H+ (aq) + OH-(aq) H2O (l); neut H = - 57.1 kJ mol-1

Thus, neutralization is simply a reaction between the H+ ions given by the acid with the OH- ions given by the base to form one mole of H2O.

In case either the acid or the base or both are weak, the enthalpy of neutralization is usually less than 57.1 kJ. The reason for this may be understood by considering the neutralization of a weak acid like acetic acid with a strong base like NaOH. Acetic acid ionizes to a small extent whereas NaOH ionizes completely as

(i) CH3COOH CH3COO- + H+

(ii) NaOH Na+ + OH-

When H+ ions given by the acid combine with the OH- ions given by the base, the equilibrium (i) shifts to the right, i.e., more of acetic acid dissociates. A part of the heat produced during the combination of H+ ions and OH- ions is used up for the complete dissociation of acetic acid. The heat thus used up is called enthalpy of dissociation or enthalpy of ionization. It is 1.9 kJ for the acetic acid. Hence, the net heat evolved in the above reaction is 57.1 – 1.9 = 55.2 kJ.

Similarly,

We know that heat of ionization of acetic acid is 1.9 kJ and 6.9 kJ of heat has been absorbed in the ionization of CH3COOH and NH4OH together.

Hence, heat of ionization of NH4OH = 6.9 -1.9 =5.0 kJ mol-1

3.1.4 Enthalpy of Solution (∆solH)

It is a well known fact that dissolution of a solute in a solvent is exothermic or endothermic process depending on the nature of the solute.

(a) Integral enthalpy of solution

The change in enthalpy when one mole of a solute is dissolved in a specified quantity of a pure solvent at a given temperature to give a solution of desired concentration is called Enthalpy of Solution.

Example: KCl (s) + 200 H2O → KCl (200 H2O), ∆solH = +18.6 kJ mol-1

CuSO4 (s) + 200 H2O → CuSO4 (200 H2O) , ∆solH = -66.5 kJ mol-1

Thus, the first case is endothermic and second case is exothermic with enthalpies of solutions +18.6 and –66.5 kJ mol-1 respectively.

(b) Differential enthalpy of solution

However, if such a large volume of the solvent is used to dissolve solute that further addition of the solvent does not produce any more heat change, it is called enthalpy of solution at infinite dilution.

TestBag

11

BaCl2 (s) + (aq) → BaCl2 (aq)

Enthalpy of an ideal solution is taken as zero.

Hence, it is clear that the salts like copper sulphate, calcium chloride etc., when present in the hydrated state (i.e., CuSO4.5H2O, CaCl2.6H2O etc.) dissolve with the absorption of heat.

Thus, it can be generalized that the process of dissolution is usually endothermic for

(i) Salts, which do not form hydrates, like NaCl, KCl, KNO3 etc.

(ii) Hydrated salts like CuSO4.5H2O, CaCl2.6H2O.

3.1.5 Enthalpy of Dilution

It is defined as the enthalpy change when solution containing 1 mole of solute is diluted from one concentration to other concentration.

KCl (200 H2O) → KCl (400 H2O) ∆H = enthalpy of dilution

3.1.6 Enthalpy of Hydration.

The amount of enthalpy change (i.e., the heat evolved or absorbed) when one mole of the anhydrous salt combines with the required number of moles of water so as to change into the hydrated salt, is called the enthalpy of hydration or heat of hydration.

For example, the heat of hydration of copper sulphate is – 78.2 kJmol-1. This may be represented as

CuSO4 (s) + 5H2O CuSO4.5H2O (s), hyd H = - 78.2 kJ mol-1

3.1.7 Lattice Enthalpy and Hydration Enthalpy of ionic solutes

The dissolution of an ionic compound in water takes place in the following two steps:

Step 1. Dissociation of the ionic solid into free ions. The energy required for this process is called lattice energy of lattice enthalpy.

Step 2. Hydration of the ions. (or in general, called salvation of the ions, if solvent other than water is used). The energy released in this process is called hydration energy or enthalpy of hydration.

(i) Lattice enthalpy and its calculation (Born-Haber cycle).

Lattice enthalpy of an ionic compound is the enthalpy change that occurs when one mole of the ionic compound dissociates into ions in the gaseous state.

Max Born and Fritz Haber in 1919 put forward a method based on Hess’s law for the calculation of lattice energy (which could not be measured directly) and hence for prediction the stability of the ionic compound formed (because larger is the negative value for lattice energy, greater is the stability of the compound formed). The method is known as Born-Haber cycle. It is briefly described below taking the example of NaCl(s).

The heat of formation of NaCl (s) is found to be – 410 kJ mol-1, i.e.,

Na(s)+ Cl2 (g) → NaCl(s); = -4101 kJ mol-1

We can imagine that one mole of NaCl (s) can be prepared by the following series of steps:

(i) Crystalline sodium metal is sublimed to form gaseous atoms

Na (s) → Na (g), ∆H = ∆SubH = S

(ii) One-half mole of gaseous Cl2 molecules is dissociated to form 1 mole of gaseous atoms

TestBag

12

Cl2(g) → Cl (g), ∆H = ∆disso.H = D

(iii) The gaseous Na atoms are ionized to form gaseous sodium ions

Na (g) → Na+ (g) + e-,

∆iH = I.E. (Ionization enthalpy)

(iv) The gaseous chlorine atoms are converted into gaseous chloride ions by adding electrons

Cl (g) +e- → Cl-(g),

∆eg H = Electron gain enthalpy/ Electron affinity = E.A.

(v) The gaseous Na+ and Cl- ions combine to form 1 mole of crystalline sodium chloride

Na+ (g) +Cl—(g) → NaCl (s),

∆lattice H = U (Lattice energy/enthalpy)

The complete process involving the formation of one mole of NaCl (s) directly and through a series of steps indirectly may be represented by a cyclic process as under:

Applying Hess’s law (according to which the heat change is same whether one mole of the compound is formed directly or indirectly through a number of steps), and using the symbols without proper signs, we have

∆f H° = S+ D+I.E.+E.A.+U

Example: Calculate the lattice enthalpy of NaCl using Born-Haber cycle.

Solution: Various values for NaCl are as follows:

Heat of formation of NaCl (∆fH) = -410.0 kJ mol-1

Heat of sublimation of sodium (S) = 109.0 kJmol-1

Dissociation Energy of chlorine (D) = 244.0 kJmol-1

Ionization energy of sodium (I.E.) = 496 kJ mol-1

Electron affinity of chlorine (E.A.) = -349 kJ mol-1

As heat of formation

∆fH = S + D/2 +I.E. +E.A. + U

∴Lattice Energy of NaCl,

U = ∆fH – S – D/2 – I.E. – E.A.

TestBag

13

= -410.0 – 109.0 – 244/2 – 496 – (-349)

= -788 kJmol-1

(ii) Hydration Enthalpy of ions

The energy released in the process of hydration or salvation of dissociated ions is known as enthalpy of hydration or Hydration Enthalpy of ions. This is generally an exothermic change.

Thus, sol H = lattice H + hyd H

For example, for one mole of NaCl (s),

Lattice enthalpy = + 788 kJ mol-1

Hydration energy = - 784 kJ mol-1

sol H = + 788 kJ mol-1 – 784 kJ mol-1

= + 4 kJ mol-1

As mentioned above, for most of the ionic compounds, sol H is +ve. Hence, dissolution of most

of the salts increases with increase in temperature. If the lattice enthalpy of a salt is very high, the dissolution of the compound may not take place at all. It is for this reason that fluorides are less soluble than chlorides.

Note:

• ∆Hsol= (∆Hhydration-lattice energy)

• If ∆Hhydration > lattice energy, solute dissolves in water

• If ∆Hhydration < lattice energy, solute doesn’t dissolve in water

• If ∆Hhydration = lattice energy, solute remains in equilibrium with given solvent, i.e., H2O.

If integral heat of solution of the hydrated and anhydrous salt is known, then heat of hydration can be calculated.

For example,

Thus on subtracting both equations,

3.1.8 Enthalpy of Atomization (∆aH)

The change in enthalpy when one mole of atoms in gaseous state is formed from its element is called enthalpy of atomization.

TestBag

14

Example:

H2 (g) → 2H (g), ∆aH0 = 435.0 kJ mol-1

CH4 (g) → C (g) + 4H (g), ∆aH0 = 1665 kJ mol-1

Na(s) → Na(g), ∆aH0 = 108.4 kJ mol-1

In the first example, enthalpy of atomization is same as bond dissociation enthalpy. In the third example, enthalpy of atomization is same as enthalpy of sublimation. The second reaction represents only enthalpy of atomization neither bond energy nor sublimation energy.

3.1.9 Enthalpy of Ionization

When a covalent compound on dissolution in water splits to produce ion in the solution, the enthalpy change accompanying the process is called enthalpy of ionization.

For example,

HCl(g) + aq H+(aq) + Cl- (aq), ion H = - 75.2 kJ mol-1

Thus, enthalpy of ionization of HCl (g) is -75.2 kJ mol-1. In fact, the enthalpy of ionization of a covalent compound is the same as its enthalpy of solution or enthalpy of dissolution ( ion H ).

3.1.10 Enthalpy of formation of ions

When an ionic solid is dissolved in water, free ions are produced in the aqueous solution. For the calculation of enthalpy of formation of an ion in the aqueous solution, enthalpy of formation of H+ ion in the aqueous solution is taken as zero. For example, f H for chloride ion in aqueous solution can be calculated from the following data:

(i) H2 (g) + Cl2 (g) HCl (g),

f H = - 92.8 kJ mol-1

(ii) HCl (g) + aq H+ (aq) + Cl- (aq),

diss H = - 75.2 kJ mol-1

From eqn. (ii), r H = { f H [H+ (aq) ]

+ f H [Cl- (aq)]} - f H (HCl)

= - 75.2 = 0 + H f [Cl- (aq)] – (- 92.8)

Or f H [Cl- (aq)] = - 75.2 – 92.8

= - 168.0 kJ mol-1

3.1.11 Enthalpy of Hydrogenation

The amount of enthalpy change that takes place when one mole of an unsaturated organic compound is completely hydrogenated is called enthalpy of hydrogenation.

For example, enthalpy of hydrogenation of ethylene is the enthalpy change for the reaction,

CH2 = CH2 + H2 → CH3-CH3

∆H = ∆Hhydrogenation

TestBag

15

3.2 Enthalpy Change during Phase Transformation

3.2.1 Enthalpy of Fusion (∆fusH)

The change in enthalpy when one mole of a solid substance is melted at its melting point is defined as enthalpy of fusion of a substance. It is also called Molar Enthalpy of Fusion.

Example:

H2O (s) → H2O (l) , ∆fusH = +6.0 kJmol-1

Ice Water

A freezing is reverse of fusion; the enthalpy of freezing (or enthalpy of solidification) has same value as the enthalpy of fusion but has the opposite sign. Thus,

H2O (l) → H2O (s) , ∆freezingH = -6.0 kJmol-1

Water Ice

3.2.2 Enthalpy of Vaporization (∆vapH)

The change in enthalpy when one mole of a liquid is converted into vapours at its boiling point is called enthalpy of vaporization. It is also called Molar Enthalpy of Vapourisation.

Example:

H2O (l) → H2O (g) , ∆vapH = +40.7 kJmol-1

Water Steam

As condensation is reverse of vaporization, the enthalpy of condensation has the same value as the enthalpy of vaporization but has opposite sign. Thus,

H2O (g) → H2O (l) , ∆conH = -40.7 kJmol-1 Steam Water

3.2.3 Enthalpy of Sublimation (∆subH)

The change in enthalpy when one mole of a solid substance is converted directly into its vapours at a temperature below its melting point is called the enthalpy of sublimation. It is also called Molar Enthalpy of Sublimation.

Example:

CO2 (s) → CO2 (g) , ∆subH = +25.2 kJmol-1

Dry ice carbon dioxide

I2 (s) → I2 (g) , ∆subH = +62.39 kJmol-1

Most solids that sublime readily are molecular solids, e.g., iodine, naphthalene etc.

The heat of sublimation is related to heat of fusion and heat of vaporization as:

∆subH = ∆fusH + ∆vapH

Thus, it is clear that sublimation is nothing but fusion and vaporization is carried out in one step.The magnitude of enthalpy change for a phase transition depends upon the strength of intermolecular forces, e.g., ∆vapH for H2O is much larger than that of acetone because the former has intermolecular hydrogen bonding.

TestBag

16

3.2.4 Enthalpy of Allotropic Transformation

The enthalpy change that takes place when one mole of one form of an allotropic modification changes to another is called Enthalpy of Allotropic Transformation.

For example,

C (graphite) C (diamond), H = transform

S (Monoclinic) S (Rhombic), H = transform

4. HESS’S LAW OF CONSTANT HEAT SUMMATION

Hess’s Law states as follows:

The total amount of heat (enthalpy) evolved or absorbed in a reaction is the same whether the reaction takes place in one step or in a number of steps.

Enthalpy is a state function; therefore the change in enthalpy is independent of the path between initial states and final states. The heat change in a particular reaction is the same and it is independent of the path or the manner by which this change is brought about.

If a reaction takes place in several steps then its standard reaction enthalpy is the sum of the standard enthalpies of the intermediate reactions into which the overall reaction may be divided at the same temperature.

Illustration:

Consider the Chemical reaction between the carbon and oxygen which is carried out in two ways,

1) C (graphite) + O2 (g) CO2 (g); H

2) C (graphite) CO (g) CO2 (g)

It is observed that, H= H1+ H2

Thus, the enthalpy change in a system is independent of the path followed.

Applications of Hess’s Law:

1. Using this law, the enthalpy changes for the reactions, which are experimentally not possible, can be calculated.

2. Bond enthalpies can also be calculated.

3. The calculations are based upon the following consequences of Hess’s Law :

The thermochemical equations can be treated as algebraic equations which can be added, subtracted, multiplied or divided.

4. Calculation of heat of reaction:

∆Hreaction= ∑ (Heat of formation - ∑ (Heat of formation

TestBag

17

of products) of reactants)

∆Hreaction= ∑ (Heat of Combustion - ∑ (Heat of combustion

of reactants) of products)

Example: The enthalpy changes for the following processes are listed below

Given that the standard states for iodine and chlorine are (s) and , Find the standard enthalpy of formation of ICl (g).

Solution:

5. BOND ENTHALPY (OR BOND ENERGY)

Bond energy is defined as follows:

Bond energy is the amount of energy released when one mole of bonds are formed from the isolated atoms in the gaseous state or the amount of energy required to dissociate one mole of bonds present between the atoms in the gaseous molecules. It is represented by ∆bH or ∆bondH.

1. In chemical reactions the formation of a chemical bond is accompanied by the release of energy. Conversely energy has to be supplied for the breaking of a bond.

2. Bond strengths are commonly described by their bond dissociation energy which is the energy required to break one mole of a bond of particular type. This is a definite quantity and is expressed in kJ mol-1.

3. For diatomic molecules the bond dissociation energy is same as bond energy, whereas in polyatomic molecules the bond energy is taken as the mean average of the various bond dissociation energies of the bonds of a given type.

4. The thermochemical data is useful in determining the bond energies of different bonds.

For example, the bond energy of C-H bond in methane can be calculated form its heat of formation. The heat for formation of methane from carbon and hydrogen has been found to be .

Methane has four C-H bonds and the energy required to break all the four C-H bonds is 1663 kJ. Therefore the average C-H bond energy is

TestBag

18

Difference between Bond Energy and Bond Enthalpy

Strictly speaking, bond energies should be ∆U° values. However, the normally reported values are bond enthalpies, i.e., ∆H° values. Bond dissociation energies are obtained at 0 K by spectroscopic method whereas bond enthalpies (∆H° values) are calculated by considering contributions from heat capacities and P∆V terms. For Example, Bond enthalpy of H2 at 298 K is 435 kJmol-1, i.e.,

H2 (g) → 2H (g) , ∆H° = 435 kJmol-1

The corresponding bond energy (∆U°) for this reaction is 430.8 kJmol-1. Thus, there is very small difference and the two are used interchangeably.

An important use of the bond enthalpy is that it helps to calculate the enthalpy of formation of atoms. For Example,

Enthalpy of formation of H-atoms = 435/2 = 217.5 kJmol-1

Bond enthalpy is also helpful to know about resonance energy. Resonance energy is the difference of heat of formation based on experimental and theoretical values(using BE values).

Calculation of Bond Enthalpy

For polyatomic molecules, the bond enthalpy of a particular bond is found from the values of the enthalpies of formation .Similarly the bond enthalpies of heteronuclear diatomic molecules like HCl, HF etc can be obtained directly from experiments or may be calculated from the bond enthalpies of homonuclear diatomic molecules. Formulas Used:

(i) ∆rH = ∑ ∆fH (Products) - ∑ ∆fH (Reactants)

(ii) ∆rH = ∑ Bond Energies or Enthalpies of Reactants - ∑ Bond Energies or Enthalpies of Products

= ∑ B.E. (Reactants) - ∑ B.E. (Products)

Example: If bond energy of Cl-Cl bond, H-H bond and H-Cl bond are 243, 435 and 431 kJmol-1, what is of HCl? Solution:

First, calculate ∆H of this reaction using BE values.

Two moles of HCl(s) are formed from its elements hence

TestBag

19

6. CALORIMETRY

In the laboratory, heat changes in physical and chemical processes are measured with a calorimeter, which is an insulated (closed) container and is used specifically for this purpose. Calorimetry-the measurement of heat changes, requires the concept of

• Specific heat (s), which is the amount of heat, required to raise the temperature of one gram of the substance by one degree.

• Heat capacity (C), which is the amount of heat, required to raise the temperature of given quantity of the substance by one degree.

C=ms

Where, m is the mass of the substance in grams.

Specific heat has the unit J/ C and heat capacity has the unit J/˚C. q = ms ∆ t

Q = C ∆ t

∆t (change in temperature) = ( )

For endothermic change, q is positive and for exothermic change, q is negative.

(i) Constant-Volume Calorimetry

Heat of combustion is usually measured by placing a known mass of a compound in a steel container called a constant volume bomb calorimeter, which is filled with oxygen about 30 atm of pressure. This is immersed in known amount of water. On ignition of the sample electrically, there is evolution of heat, which can be calculated accurately by recording the rise in temperature of the water.

Heat lost by the sample=Heat gained by water

We assume that the system is perfectly insulate (isolated) so that no heat is lost to the surroundings during measurements. Thus,

= + + = 0

∴ = - ( + )

Also, = m (water) s (4.184 J/˚C) ∆ t

= ∆t

Thus, qreaction (heat of reaction) is calculated.

(ii) Constant- Pressure Calorimetry

A simple device than the constant-volume calorimeter is the constant-pressure calorimeter used to determine the heat change for non combustion reactions (e.g., neutralization, solution, hydration etc.). Because, the pressure is constant, the heat change for the process is equal to ∆H.

Example: A quantity of 100.0 mL of 0.5 M HCl is mixed with 100.0 mL of 0.5 M NaOH in a constant pressure calorimeter that has a heat capacity of 335 J/°C. Initially NaOH and HCl are at 22.50°C and final temperature of the mixed solution is 24.90°C. Calculate the heat change for the neutralisation reaction.

TestBag

20

NaOH (aq)+HCl (aq) → NaCl (aq)+H2O(l)

Assume densities of solution is 1.00 g mL-1 and specific heats of the solutions 4.184 J/g °C. Also calculate heat of neutralisation.

Solution: ∆Hsystem =∆Hsolution+∆Hcalorimeter+∆Hreaction=0

∴ ∆Hreaction= - (∆Hsolution+∆Hcalorimeter)

To calculate ∆∆∆∆Hsolution

∆Hsolution = ms ∆t

Mass of mixture = 200mL × 1 g mL-1 = 200 g

s = 4.184 J/s°C

∆t = 24.90 – 22.50 = 2.40°C

∴ ∆Hsolution = 200 × 4.184 × 2.40 J

= 2008.32 J = 2.008 kJ

To calculate ∆∆∆∆Hcalorimeter ’

∆Hcalorimeter = c × ∆t = 335 J°/C × 2.40°C

= 804 J = 0.804 kJ

∆Hreaction = -(2.008+0.804)

=-2.812 kJ for the given mixture

To calculate heat of neutralisation when 1 g equivalent of an acid is neutralised by 1 g equivalent of a base (= - mol of each in this case) we consider

100 mL of 0.5M HCl = 100 × 0.5 millimol HCl

= 0.05 mol HCl

100 mL of 0.5M NaOH = 100 × 0.5 millimol NaOH

= 0.05 mol NaOH

∴ Heat of neutralisation =

= 56.24 kJ mol-1