CBSE Class 11 Chemistry Notes For Chapter 6 Chemical Thermodynamics

CBSE Class 11 Chemistry Notes For Chapter 6 Chemical Thermodynamics Introduction

Energy is the ability of a physical system to do work. It always remains conserved in any event, i.e., it cannot be created or destroyed. However, it can take different forms such as mechanical energy, chemical energy, electrical energy, heat energy, light energy, sound energy, etc. Energy can be converted from one form to another.

Thermodynamics is the study to find out the conditions of interconversion of different forms of energy as well as to predict the extent of their conversions.

The branch of science which deals with the interconvertibility of different forms of energy (mainly heat and work) is called thermodynamics. The Greek words ‘thermo’ means heat and ‘dynamics’ means power.

Thermodynamics is based on four fundamental rules or laws namely zeroth law, first law, second law, and third law of thermodynamics. These laws are based on natural experience gathered over centuries. As the laws of thermodynamics are derived from the direct human experience, it is sometimes called axiomatic science.

Class 11 Chemistry Notes For Chapter 6 Chemical Thermodynamics

Energy which we generally use in our daily lives is mostly obtained from chemical reactions. Thus the study of energy changes in chemical reactions is considered to be an important topic in chemistry.

The branch of thermodynamics that deals with energy evolved or absorbed during physical or chemical transformation is called chemical thermodynamics or chemical energetics.

Importance Of Thermodynamics

Some important applications of thermodynamics are:

• In thermodynamics, we study the inter-relationships among various macroscopic variables like pressure, temperature, volume, etc., and the changes in these variables as a result of various processes.
• Energy changes associated with physical or chemical transformations can be explained using thermodynamics.
• Under a given set of conditions, the feasibility of a physical or chemical transformation can be predicted from thermodynamics.
• Thermodynamics can predict the extent to which a physical or chemical transformation occurs before it reaches an equilibrium state.
• The relative amount of the reactants and products at the equilibrium of a reaction as well as the value of the equilibrium constant of the reaction can be determined by thermodynamics.
• Under a given set of working conditions, the maximum efficiency of a heat engine can be determined by thermodynamics.
• The ideal condition(s) for the transformation of different types ofenergy can be determined by thermodynamics.

Limitations of thermodynamics

1. Classical thermodynamics applies only to the macroscopic system (For example in the case ofa few grams of ice) and not to the microscopic system (For example in the case of a few molecules or atoms). So it cannot give us any idea regarding the structure of matter.
2. Thermodynamics does not give us any information about the rate or velocity ofa process (such as chemical reaction, osmosis, etc.). Also, it cannot give us any idea about the mechanism or path ofa chemical reaction.

Class 11 Chemistry Notes For Chapter 6 Terms And Concepts Related To Thermodynamics

Some terms are frequently used in thermodynamics. One should have a good knowledge of these terms before going through the subject of thermodynamics. Here are these terms and their brief explanations.

System, Surroundings, and Boundary

System:

In thermodynamics, a system is defined as the part of the universe under study, which is separated from the rest of the universe by real or imaginary boundaries- System

System, Surroundings, and Boundary Explanation:

If we study something 2 about the human body, then the M human body is considered to be the system. Similarly, if we perform an experiment living cell or one mole of water, then the living cell or one mole of water will be the system.

Surroundings:

Everything outside the system In the universe Is called the surroundings. Once the system Is defined, the surroundings will be defined automatically.

System + Surroundings=Universe

Boundary:

Heal or Imaginary surface that separates a system from its surroundings is called the boundary of the system.

Explanation:

If a certain amount of water taken In a beaker is considered to be the material of our experiment, then water will be our system.

• The rest of the universe including the beaker will be the surroundings of the lire system.
• The interface between water and glass and that between water and air are the boundaries between the system and its surroundings.
• If a gas enclosed in a cylinder fitted with a piston is considered to be the object of our analysis, then the gas will be our system, and the rest of the universe including the cylinder and the piston will constitute the surroundings of the system.

1 lore, the inner surfaces of the walls of the die cylinder, and the piston are the boundaries between the system and its surroundings.

Role of boundary:

The boundary of a system plays an important role when there occurs an interaction between the system and its surroundings. During the interaction, a system exchanges energy or matter or both matter and energy with Its surroundings through its boundary.

Characteristics of the boundary of a system:

• The boundary of the n system may be real or imaginary.
• It may be rigid or non-rigid, in the case of a rigid boundary, the volume of the system does not change, while In the case of a non-rigid boundary, the volume of the system can change.
• Tim boundary of a system may be permeable or Impermeable. In the case of a permeable boundary, the exchange of matter takes place between the system and its surroundings through the boundary, whereas in the case of an impermeable Imundury, no exchange of matter can take place through the boundary.
• Exchange of beat may take place between a system and US surroundings through the boundary. A boundary that permits the cycling of heat between a system and Its surroundings Is called an antennal or diathermic boundary. On the other hand, a boundary that does not permit the exchange of heat between the system and its surroundings is called an adiabatic boundary.
• In reality, a perfect adiabatic boundary is not possible. The wall of a Dewar flask nearly behaves as an adiabatic boundary. According to the definition, the surroundings mean everything outside the system.
• But in thermodynamics, the surroundings are considered as the portions of the universe around the system, upto which a change occurring in the system has its influence, For Example, the burning of a candle produces light and heat.
• When we consider the effect of light, the tire space enclosing the candle, which is illuminated by Uie light, will be the surroundings. On the other hand, if we consider the effect of heat (produced), a narrow space affected by the produced heat, will be the surroundings

Types of system

Depending on the nature of its boundary, a system may or may not exchange matter energy, or both with its surroundings. Based on the exchange of matter and energy with the surroundings, systems may be classified as open systems, closed systems, and isolated systems.

1. Open system:

A system that can exchange both energy and matter with its surroundings is called an open system.

Examples: Every living being in nature:

Every living being (system) takes food (matter) and excretes waste materials (matter) to the surroundings. They (systems) also exchange heat (energy) with the surroundings.

Some water (or any other liquid) in an open container:

• In an open atmosphere, water (system) continuously evaporates and water vapor (matter) escapes from the container to the air (surroundings). Also, O₂ or CO₂ (matter) from the air may dissolve into water (system).
• So, the exchange of matter takes place between the system and its surroundings.
• If the temperature of water (system) is different from that of its surroundings, then there occurs an exchange of heat (energy) between water and its surroundings.

The ocean:

The Ocean Is a perfect example of an open system. Water (matter) evaporates from the ocean to the atmosphere and then again is added to it in time of rain. The ocean also absorbs heat (solar energy) and releases its energy in the form of latent heat.

2. Closed system:

A system that can exchange energy with its surroundings but not matter is called a closed system.

Examples: Boiling of water in a closed glass/metallic container:

• During boiling, water vapor (matter) cannot escape to the surroundings from the container, also any matter from the surroundings cannot enter into the system.
• So, the exchange of matter is not possible between the system and its surroundings. But if the temperature of the water (system) is different from the surroundings, then heat (energy) will be exchanged between the system and its surroundings through the boundary wall of the container.
• So, the exchange of energy is possible between the system and its surroundings.

A gas enclosed in an impermeable metallic cylinder fitted with a piston:

Since the cylinder and piston are impermeable, matter will not be exchanged between the system and its surroundings.

• But heat (energy) will be exchanged between the system and its surroundings through the wall and piston of the cylinder.
• If a pressure greater than the pressure of the system is applied on the piston, work is done on the system. As a result, energy in the form of work is transferred to the system from its surroundings.
• On the other hand, if the pressure on the piston is kept lower than that of the system, the system does the work on its surroundings, and energy in the form of work is transferred from the system to its surroundings.

Isolated system:

A system that can exchange neither energy nor matter with its surroundings is called an isolated system.

An isolated system does not interact with its surroundings as its boundary is impervious to all matter and does not permit energy to pass through it As a result, any change occurring in an isolated system does not influence the surroundings, and vice-versa

Examples: Hot tea kept in a sealed & thermally insulated flask (Dewar flask):

Since the flask is closed, matter cannot be transferred from the system (hot tea) to its surroundings, and vice-versa. As the walls ofthe flask are thermally insulated, the system cannot exchange energy with its surroundings.

Water taken in a closed container having rigid, impermeable, and thermally insulated walls:

Since the container is closed and its walls are impermeable, no matter can exchange between the system (water) and its surroundings.

Furthermore, the container is insulated. So, no heat can flow into or out of the system. As the walls are rigid, there will be no change in the volume of the system. So, no energy in the form of work can be exchanged between the system and its surroundings. A perfectly isolated system is a hypothetical concept because no wall is perfectly adiabatic.

Based on physical properties and chemical composition, a system may further be classified as—

• Homogeneous system: If the physical properties and chemical compositions are uniform throughout a system, then it is called a homogeneous system. For example, pure solid, liquid or gaseous substance, gas mixture, 2 completely miscible liquids like water, alcohol, etc.
• Heterogeneous system: If the physical properties and the chemical compositions are different in different parts of a system, then it is called a heterogeneous system. For example, the mixture of two immiscible liquids (water and benzene), a mixture of two solids (sugar and NaCl) etc.
• Extensive and intensive properties of a system: A system consisting of a large number of atoms, ions, or molecules is called a macroscopic system. For example, a certain amount of water, a certain volume of solution, a certain mass of sodium chloride, etc., are macroscopic systems. The properties associated with a macroscopic system are called macroscopic properties. For example, temperature, pressure, concentration, mass, density, composition, etc., are the properties ofa macroscopic system.

The macroscopic properties of a system can be classified into two categories:

1. Extensive properties
2. Intensive properties.

Extensive property:

Tile property which depends upon the mass (or size) of the system i.e., the quantity of matter present in the system is called an extensive property. Extensive properties are additive, i.e., the total value of an extensive property of a system is equal to the sum of the extensive properties of different parts ofthe system.

Examples:

Mass, volume, internal energy, enthalpy, entropy, heat capacity, Gibbs free energy, etc., are the extensive properties of a system.

The volume of a system:

If V is the volume of lg substance (system) at a particular temperature and pressure, then the volume of 5 g of the same substance will be 5 x V. So the volume of a system is an extensive property.

The number of moles ofa system:

If the number of moles of1 g of a substance (system) is ‘ri then the number of moles for 2 g of that substance will be ‘ 2n! So, several moles of a system is an extensive property.

The internal energy of a system:

When water transforms to ice, its internal energy in the form of heat is liberated. Under an identical set of conditions, the heat liberated in the transformation of 5 g of water to 5 g of ice is found to be five times as much as that liberated when 1 g of water transforms into 1 g of ice. This means that a 5 g sample of water contains five times as much internal energy as a 1 g sample of water does. Thus, the internal energy ofa system is an extensive property.

Intensive property:

The property that does not depend upon the mass (or size) of the system i.e., the amount of matter present in the system, is called an intensive property. Intensive property has the same magnitude at every point in a homogeneous system under equilibrium.

Examples:

Temperature, pressure, density, viscosity, molar heat capacity, refractive index, boiling point, freezing point, surface tension, Viscosity coefficient, molar volume, molar internal energy, molar enthalpy, molar entropy, molar free energy, mole fraction, color, concentration, thermal conductivity, specific rotation, standard reduction potential, etc.

The boiling point of a liquid:

If we take different amounts of pure water in two different containers and determine their boiling points at a given pressure, we get the same boiling point for both although the amount of water is different. Thus, the boiling point ofa liquid does not depend upon the amount of the liquid, implying that it is an intensive property.

The density of a substance:

At a given temperature, the density of 1 kg of a pure sample of copper is the same as the density of lg of the same sample. Thus, the density of a substance does not depend upon the amount of substance, indicating that it is an Intensive property.

The concentration of a homogeneous solution:

If the concentration of a homogeneous solution is 1 g.L-1 at a particular temperature, then the concentration ofa drop of that solution will also be 1 g.L-1 at that temperature. Thus the concentration of a homogeneous solution does not depend upon the amount of solution. So, it is an intensive property.

Important points regarding intensive & extensive properties:

1. The ratio of two extensive properties is always an intensive property.

Explanation:

• The mass (m) and volume (V) of a system are extensive properties. But the ratio of these two, i.e., density \(\left(\frac{m}{V}=d\right)\) is an intensive property.
• Extensive property becomes intensive when It Is expressed in terms of per unit mass or unit mole.

Explanation:

Ifx is an extensive property for moles of a system, then the value of X per mole, \(X_m=\frac{X}{n}\), will be an intensive property because it denotes the value of X for 1 mol of the system and is independent of the amount of substance.

• The internal energy of a system is an extensive property, but internal energy per mole (molar internal energy) or internal energy per gram (specific internal energy) is an intensive property.
• The volume of a system is an extensive property, but volume per mole (molar volume) or volume per gram (specific volume) is an intensive property.
• Similarly, the heat capacity of a system is an extensive property, but heat capacity per mole (molar heat capacity) or heat capacity per gram (specific heat capacity) is an intensive property.

Class 11 Chemistry Notes For Chapter 6 Chemical Thermodynamics – Internal Energy And Its Change

Heat is released during the solidification of a liquid (for example when 1 g of water is transformed into ice at 0°C and late pressure, it releases 80 cal heat.) When coal burns, it produces both light and heat.

Mechanical work is done during the expansion of water vapor from high pressure to low pressure. We have many other examples like these, where we see that the system does work or produces heat or other form of energy out of its own intrinsic or inherent energy, without taking aid from an external energy source.

These observations indicate that every system contains some amount of energy intrinsically associated with it This intrinsic energy associated with every system or substance, is called its internal energy. Internal energy is denoted by the symbol either U or E.

Internal Energy And Its Change Definition

Every system, for its existence, is associated with an amount of energy in exchange for which it can do work or produce heat or another form of energy without the help of external energy. This energy is termed internal energy.

Origin of internal energy:

The constituent particles in a system possess kinetic and potential energies arising from different internal modes of motion such as translational motion, vibrational motion, rotational motion, and electronic motion. In addition, they also possess nuclear energy, bond energy, and energy due to intermolecular attractions or repulsions.

• All these energies contribute to the internal energy ofthe system.
• Therefore, the sum of all forms of energies ofthe constituent particles in a system gives rise to the internal energy ofthe system.
• The absolute value of the internal energy of a system cannot be measured experimentally because it is not possible to determine all the types ofenergy associated with the internal energy of a system.
• However, in a process, the change in the internal energy (AU) of a system can be determined experimentally.

Some important points about internal energy

1. Internal energy is an extensive property:

Explanation:

Internal energy ofa system is an extensive property because it increases as the amount of substance present in the system increases. For example, the internal energy of 5g of water is five times that of lg of water.

2. Internal energy depends upon the nature of the system:

Explanation:

The constituent particles (atoms, ions, and molecules) are different for different systems. This makes magnitudes of translational energy, rotational energy, vibrational energy, electronic energy, binding energy, etc. different for different systems.

Therefore, the values of internal energy will be different for different systems, even under identical conditions. For example, under identical conditions of temperature and pressure, the internal energy of one mole of O₂ gas Is different from that of one mole of N₂ gas.

3. Internal energy Increases with an increase in temperature:

Explanation:

With increasing temperature, the magnitudes of the translational motion, rotational motion, vibrational motion, etc., of constituents of a system increase. This results in an increase in energies associated with these motions. Consequently, the internal energy of the system increases.

4. Internal energy ofa system is a state function:

Explanation:

In a process, the change in internal energy (All) ofa system only depends upon the initial and final states of the system. It does not depend upon the path followed for carrying out the process. A process with a given initial and final states can be carried out in different paths but the change in internal energy will be the same in all paths.

For example, the combustion of glucose,

⇒  \(\mathrm{C}_6 \mathrm{H}_{12} \mathrm{O}_6(s)+6 \mathrm{O}_2(\mathrm{~g}) \rightarrow 6 \mathrm{CO}_2(\mathrm{~g})+6 \mathrm{H}_2 \mathrm{O}(l)\),

Can be carried out either by its direct combustion in a reaction vessel, or by oxidizing it in presence of enzyme. However, the change in internal energy in each case will be the same as the initial and final states for both processes are identical.

Change in the internal energy of a system in a process Definition

Process, the change in internal energy of a system is defined as the difference in internal energies between the final and initial states of the system in the process.

Let us consider a process in which the internal energy of the system at the initial and final states are U₁ and U₂, respectively. Therefore, the change in internal energy of the system, ΔU = U₂-U₁

Change in internal energy of a system in a chemical reaction:

In a chemical reaction, the system is considered to be made up of reactants and products involved in the reaction.

At the beginning of the reaction, the system contains only reactants, and after the completion of the reaction, the system contains only products. Therefore, the change in internal energy in a chemical reaction equals the difference between the internal energies of the products and the reactants.

Let us consider a reaction:

A →B The change in internal energies of the reaction is given by,

⇒ \(\Delta \boldsymbol{U}=\overline{\boldsymbol{U}}_{\boldsymbol{B}}-\overline{\boldsymbol{U}}_A ; \text { where } \bar{U}_A \text { and } \bar{U}_B\)

Are the molar internal energies (internal energies per mole) of the product (B) and reactant (A), respectively

1. If then ΔU = negative, indicating an exothermic reaction.
2. If then ΔU = positive, indicating an endothermic reaction.

Both calorimeters are used to determine the change in internal energy in a combustion reaction.

• The change In internal energy in an isothermal expansion or compression of an ideal gas (system) is zero.
• When an ideal gas undergoes isothermal expansion or compression, the average distance between the molecules in the gas changes, but this change does not affect the internal energy of the gas because the molecules of an ideal gas do not experience intermolecular forces of attraction.
• Therefore, an isothermal expansion or compression of an ideal gas (system) does not cause any change in the internal energy of the gas.
• The change in internal energy of an ideal gas depends only upon temperature. At a constant temperature, the internal energy of an ideal gas is independent of its volume or pressure

Class 11 Chemistry Notes For Chapter 6 Chemical Thermodynamics – Heat Change In A Chemical Reaction

Like other processes, heat exchange with the surroundings also occurs in case of chemical reactions. The amount of heat absorbed or given off during a reaction depends on the conditions under which the reaction is carried out.

Generally, reactions are carried out either under the condition of constant volume or under the condition of constant pressure. However, the fact is that carrying out a reaction at constant volume is not as convenient as that at constant pressure.

In the laboratory, reactions are carried out in containers open to the atmosphere, so they occur under the condition of constant atmospheric pressure. According to the first law of thermodynamics, if a process involving only pressure-volume work occurs at constant volume, on heat change (qv) in the process equals the change in internal energy (AU) of the system.

Therefore, ΔU = qv or, = -qy. On the other hand, if a process occurs at constant pressure according to the first law of thermodynamics, the heat change in the process (qp) becomes equal to the change in enthalpy (ΔH) of the system. Therefore, ΔH = qp or, -ΔH= -qP

In a process, the change in internal energy is ΔU = U₂-U₁, and the change in enthalpy is ΔH – H₂– H₁ where subscripts 1 and 2 denote the initial state and the final state of the system, respectively, in the process. In a reaction system, only reactants are present at the beginning of the reaction and only products are present at the end of the chemical reaction.

Therefore, the internal energy or enthalpy at the beginning of a reaction means the total internal energy or enthalpy ofthe reactants undergoing the reaction. Similarly, the internal energy or enthalpy at the end of a reaction means the total internal energy or enthalpy of the products formed in the reaction. Therefore,

1. For a chemical reaction occurring at constant volume, the heat change (qv) = AU = (total internal energy of the products – total Internal energy of the reactants) \(=U_{\text {products }}-U_{\text {reactants }}=U_P-U_{R^*}\)

2. For a reaction occurring at constant pressure, the heat Change \(\left(q_P\right)=\Delta H=H_{\text {products }}-H_{\text {reactants }}=H_P-H_R \text {. }\)

As most of the reactions are carried out at constant pressure and heat change associated with constant pressure equals the change in enthalpy, the heat change of the reaction in a reaction usually signifies the change in enthalpy (ΔH) is the reaction unless the constant-volume condition is stated.

Exothermic and endothermic reactions

Exothermic reaction:

Reactions associated with the evolution of heat are called exothermic reactions. In an exothermic reaction, heat is released from the reacting system to the surroundings.

Thus, when we touch the reaction container (which is the part of the surroundings), in which the exothermic reaction is taking place, we feel warm. Since heat is released in an exothermic reaction, the total energy of the reactants is greater than that of the products.

Let us consider an exothermic reaction:

A + B→C + D. If the heat released in the reaction is’ q ‘ then, the total energy of the reactants = total energy of the products + q t.e., the total energy of A and H = the total energy of C and D + q.

If we imagine ‘q’ as the part of the die product, then the above equation can be written as

A + B →C + D + q The amount of heat evolved is usually written with a positive sign on the right-hand side of the balanced equation of the reaction.

Endothermic reaction) Reactions Associated with the absorption of hoot are called endothermic reactions, In an endothermic reaction. In the absorbed by the reading system to the surroundings. Thus, when we touch the reaction container (which Is the pan of the surroundings), in which an endothermic reaction Is taking place, we feel cold.

Since the absorption of bent occurs In mi endothermic reaction, the total energy of products will be greater than the total energy of the reactants. Let us consider an endothermic reaction; A + B→C+D.

Heat absorbed In this reaction he then the total energy of the reactants s the total energy if the products total energy of A and If = total energy of C and D- q.

If imagine ‘q’ as the part ofthe reactant, then the above equation can be written as:

A+B+QC+D Or, A+B+C+D-q

The amount of heat absorbed is usually written with a negative sign on the right-hand side of the balanced equation ofthe reaction.

Enthalpy change in exothermic reactions:

In an exothermic reaction, heat is released by the reacting system. Therefore, for any exothermic reaction occurring at constant pressure qp<0.

Since qp=ΔH, ΔH<0 Or, \(\Sigma H_{\text {products }}-\Sigma H_{\text {reactants }}<0 \quad \text { or, } \quad \Sigma H_{\text {products }}\) Thus the total enthalpy ofthe products is less than that ofthe reactants.

In an exothermic reaction, the enthalpy of the reaction system decreases (ΔH < 0). Enthalpy change in endothermic reactions:

In an endothermic reaction, heat is absorbed by the reaction system. Therefore, for any endothermic reaction occurring at constant pressure, qo>0.

Since \(q_p=\Delta H, \Delta H>0 \text { or, } \Sigma H_{\text {products }}-\Sigma H_{\text {reactants }}>0\text { or, } \Sigma H_{\text {products }}>\Sigma H_{\text {reactants }}\)

In an endothermic reaction, the enthalpy of the reaction system Increases (ΔH > 0).

Enthalpy diagram of exothermic and endothermic reactions In an exothermic reaction

⇒ \(\Sigma H_{\text {products }}<\Sigma H_{\text {reactants }}\) So, in its enthalpy diagram ,\(\Sigma H_{\text {products }}\) lies below the \(\Sigma H_{\text {reactants }}\) hand,

In an endothermic reaction,

⇒ \(\Sigma n_{\text {products }}>\Sigma H_{\text {reactants }}\) So, in Its enthalpy diagram, \(\Sigma H_{\text {products }}\) lies above the \(\Sigma H_{\text {reactions }}\)

A few examples of exothermic and endothermic reactions

1. Exothermic reactions and the values of All (at 25°C and 1 atm)

CH₄(s) + 2O₂ (g)→ CO₂ (g) + 2H₂O(l) ; ΔH = -890 kJ

C(graphite, s) + O₂(g) → CO₂ (g) ; ΔH = – 393.5 kJ

2H₂(g) + O₂ (g)→ H₂O(l) ;ΔH = – 571.6 kJ

H₂ (g) + Cl₂(g) → 2HCl(g) ; ΔH = -185 kJ

2. Endothermic reactions and the values of All (at 25°C and 1 atm)

N₂ (g) + O₂ (g)→ 2NO(g) ; ΔH = +180.5 kJ

CaCO₃(s) → CaO(s) + CO₂ (g) ; ΔH = +178.3V 0

C(coal, s) + H₂O(g)→ CO(g) + H₂(g) ; ΔH = +130 kJ

Thermochemical equations

Thermochemical equations Definitions

The thermochemical equation is a balanced chemical equation, in which the physical states of the reactant(s) and product(s) as well as the amount of heat evolved or absorbed in the reaction are mentioned.

Conventions for writing a thermochemical equation:

To indicate the physical states ofthe reactant (s) and the product(s), symbols s, l, and g are used for solid, liquid, and gaseous states, respectively. These symbols are to be placed within parentheses just after the chemical formulae ofthe substances concerned.

For any reactant or product dissolved in aqueous solution, the term ‘ aq ’ (short form of the word aqueous) is to be placed within a parenthesis just after its formula.

The amount of heat evolved or absorbed (or the enthalpy change) in a reaction is to be written with a proper sign (+ or -) on the right-hand side immediately after the balanced equation.

In an exothermic reaction, if x kj of heat is evolved, then +x kj or ΔH = -x kj is to be written on the right-hand side immediately after the balanced chemical equation.

Example:

⇒  \(\mathrm{CH}_4(\mathrm{~g})+2 \mathrm{O}_2(\mathrm{~g}) \rightarrow \mathrm{CO}_2(\mathrm{~g})+2 \mathrm{H}_2 \mathrm{O}(l)+890.3 \mathrm{~kJ}\)

Or, \(\text { or, } \mathrm{CH}_4(\mathrm{~g})+2 \mathrm{O}_2(\mathrm{~g}) \rightarrow \mathrm{CO}_2(\mathrm{~g})+2 \mathrm{H}_2 \mathrm{O}(\mathrm{l})\)

Δ H =-890.3 kJ

In an endothermic reaction, if +x kj of heat is absorbed, then -x kj or AH = +xkj is to be written on; the right hand immediately after the balanced equation.

Example:

⇒\(\mathrm{N}_2(g)+\mathrm{O}_2(g) \rightarrow 2 \mathrm{NO}(g)-180.5 \mathrm{~kJ}\) or, \(\mathrm{N}_2(g)+\mathrm{O}_2(g) \rightarrow 2 \mathrm{NO}(g)\)

Δ H=+180.5 kJ

The coefficients of the reactants and the products in a thermochemical equation indicate their respective number of moles This allows us to use fractional coefficients for reactants and products. Because ΔH is an extensive property, when the balanced equation is multiplied by a factor, the value of ΔH will also be multiplied by that factor.

⇒ \(2 \mathrm{H}_2(g)+\mathrm{O}_2(g) \rightarrow 2 \mathrm{H}_2 \mathrm{O}(l)\)

ΔH=-571.6 kJ,

Or, \(\mathrm{H}_2(\mathrm{~g})+\frac{1}{2} \mathrm{O}_2(\mathrm{~g}) \rightarrow \mathrm{H}_2 \mathrm{O}(l), \Delta H=-285.8 \mathrm{~kJ}\)

Unless otherwise stated the value of ΔH mentioned in a thermochemical equation is considered to be the value at standard states.

A thermochemical equation can be written in this manner. When this is done, the magnitude of AH remains the same but its sign becomes the opposite

Example:

⇒ \(\mathrm{N}_2(g)+\mathrm{O}_2(\mathrm{~g}) \rightarrow 2 \mathrm{NO}(\mathrm{g})\)

ΔH=+180.5 kJ

Or,  \(2 \mathrm{NO}(g) \rightarrow \mathrm{N}_2(\mathrm{~g})+\mathrm{O}_2(\mathrm{~g}) ; \Delta H=-180.5 \mathrm{~kJ}\)

Class 11 Chemistry Notes For Chapter 6 Chemical Thermodynamics Numerical Examples

Question 1. At a fixed temperature & pressure, the heat released in the formation of 3 mol SO₃(g) from SO₂(g) and 2(S) is 291 kj. What will be the change in enthalpies in the formation of1 mol & 4 mol SO₃(g)?
Answer:

The thermochemical equation for the formation of 3 mol of SO₃(g) from the reaction between SO₂(g) and O₂(g) is given

⇒ \(3 \mathrm{SO}_2(\mathrm{~g})+\frac{3}{2} \mathrm{O}_2(\mathrm{~g}) \rightarrow 3 \mathrm{SO}_3(\mathrm{~g}), \Delta H=-291 \mathrm{~kJ}\)

The thermochemical equation for the formation of 1 mol of

⇒  \(\mathrm{SO}_3(g) \text { is: } \mathrm{SO}_2(\mathrm{~g})+\frac{1}{2} \mathrm{O}_2(\mathrm{~g}) \rightarrow \mathrm{SO}_3(\mathrm{~g}) \text {, }\)

⇒ \(\Delta H=\frac{1}{3}(-291) \mathrm{kJ}=-97 \mathrm{~kJ}\)

Therefore, the change in enthalpy for the formation of lmol and 4mol SO₃(g) will be -97 kJ and 4 ×(-97)

=- 388 kj, respectively.

Question 2. At a particular temperature and pressure, the heat produced in the formation of 2 mol of C₂H₆(g) from the reaction between C₂H₂(g) and H₂(g) is 626 kj. What amount of H₂(g) will react with the required amount of C₂H₂(g) to produce 939 kj of heat at the same temperature and pressure?
Answer:

The Thermochemical equation for the formation of 2 mol of 2H₆(g) from the reaction between C₂H₂(g) and H₂(g) is

⇒ \(2 \mathrm{C}_2 \mathrm{H}_2(\mathrm{~g})+4 \mathrm{H}_2(\mathrm{~g}) \rightarrow 2 \mathrm{C}_2 \mathrm{H}_6(\mathrm{~g}) ; \Delta H=-626 \mathrm{~kJ}\)

∴ 626 kJ of heat = 4 mol H₂(g)

∴ \(939 \mathrm{~kJ} \text { of heat } \equiv \frac{4}{626} \times 939 \equiv 6 \mathrm{~mol} \mathrm{H}_2(\mathrm{~g})\)

= 6 × 2

= 12g H₂(g)

Therefore, if 12 g H₂(g) reacts with the required amount of C₂H₂(g) to produce C₂H₆(g), then 939 kJ of the heat is evolved.

Question 3. At a particular temperature and pressure, N₂(g) and O₂(g) react to form 4 mol of N₂O. The heat absorbed in this reaction is 328 kj. What would the change in enthalpy be due to the formation of 2 mol of N₂(g) and 1 mol of O₂(g) from N₂O(g) at the same temperature and pressure?
Answer:

The thermochemical equation for the formation of

⇒ \(\mathrm{N}_2 \mathrm{O}(g): 4 \mathrm{~N}_2(g)+2 \mathrm{O}_2(\mathrm{~g}) \rightarrow 4 \mathrm{~N}_2 \mathrm{O}(g) ; \Delta H=+328 \mathrm{~kJ}\)

Writing this equation in the reverse manner, we obtain,

⇒ \(4 \mathrm{~N}_2 \mathrm{O}(\mathrm{g}) \rightarrow 4 \mathrm{~N}_2(\mathrm{~g})+2 \mathrm{O}_2(\mathrm{~g})\)

ΔH=-328 kJ

Therefore, the thermochemical equation for the formation of 2 mol of N₂(g) and 1 mol of O₂(g) from N₂O(g) will be:

⇒  \(2 \mathrm{~N}_2 \mathrm{O}(\mathrm{g}) \rightarrow 2 \mathrm{~N}_2(\mathrm{~g})+\mathrm{O}_2(\mathrm{~g});\)

⇒\(\Delta H=-164 \mathrm{~kJ}\)

Thus, the change in enthalpy for the formation of 2 mol of N₂(g) and 1 mol of O₂(g) is -164 kJ.

Class 11 Chemistry Notes For Chapter 6 Chemical Thermodynamics – Various Types Of Heat Of Reaction Or Enthalpy Of Reaction

Heat or enthalpy of the formation of a compound

Heat or enthalpy of the formation of a compound Definition

At a given temperature and pressure, the enthalpy of formation of a compound is defined as the enthalpy change for the reaction in which one mole of the compound is formed from its constituent elements

The enthalpy of formation of a compound is denoted by the symbol ΔHf, where subscript ‘f’ stands for formation. To compare the enthalpy of formation for different compounds, the enthalpy of formation at a standard state is calculated for the compounds. The enthalpy of formation for any compound in the standard state is known as the standard enthalpy of formation of that compound.

Standard enthalpy of formation:

The standard enthalpy of formation of a compound is defined as the change in enthalpy for the reaction in which one mole of the compound in its standard state (i.e., at a particular temperature and 1 atm pressure) is produced from its constituent elements in their standard states

The standard enthalpy of formation for any compound is denoted by \(\Delta H_f^0\), where subscript ‘f’ stands for formation and superscript ‘0’ indicates the standard state. The value of \(\Delta H_f^0\) may be positive or negative

In the thermochemical equation representing the formation reaction of a compound, one mole of the compound is formed. For this reason, the unit of ΔH0f is expressed in kJ.mol-1 (or J.moI-1) or kcal.mol^-1 (or cal. mol^-1).

Example:

⇒ \(\mathrm{H}_2(g)+\frac{1}{2} \mathrm{O}_2(g) \rightarrow \mathrm{H}_2 \mathrm{O}(l) ; \Delta H_f^0=-285.8 \mathrm{~kJ} \cdot \mathrm{mol}^{-1}\)

Generally, the standard enthalpy of formation \(\Delta H_f^0\) of a compound is quoted at 25°C (or 298 K). If temperature is not mentioned, then 25°C temperature is to be considered. At a given temperature, the standard state of an element or a compound is the most stable and purest state of that element or compound at 1 atm pressure and at that temperature. The most stable forms of some elements at 25°C and 1 atm are given below.

1. Dihydrogen: H₂(g)
2. Dioxygen: O₂(g)
3. Dinitrogen: N₂(g)
4. Sodium: Na(s)
5. Chlorine molecule: Cl₂(g)
6. Bromine molecule: Br₂(Z)
7. Iodine molecule: I₂(s)
8. Carbon: C (graphite, s)
9. Sulfur: S (rhombic, s), etc.

The constituent elements of a compound must be present in their standard states in the equation representing the formation reaction of the compound.

At 25° nC, the standard enthalpy of formation of water =-205.8kI.mol^-1:

At 25 °C and 1 atm, the heat evolved due to the formation of 1 mol of H₂O(/) from the reaction between 1 mol of H₂(g) and 1/2 mol of O₂(g) is 285.8 kJ. Alternatively, it can be said that the change in enthalpy of the given reaction at 25 °C and 1 atm is -285.8 kj

⇒ \(\underbrace{\mathrm{H}_2(g)+5^{\circ} \mathrm{C} \& 1 \mathrm{~atm}}_{25^{\circ} \mathrm{C} \& 1 \mathrm{~atm}} \underbrace{\frac{1}{2} \mathrm{O}_2(\mathrm{~g})} \rightarrow \Delta H_f^0=-285.8 \mathrm{~kJ} \cdot \mathrm{mol}^{-1}\)

At 25°C, The standard enthalpy of formation of nitric acids

⇒ \([\mathrm{NO}(g)]=+90.3 \mathrm{~kJ} \cdot \mathrm{mol}^{-1}\)

At 25°C and 1 atm, the heat evolved due to the formation of 1 mol of NO(g) from the reaction between 1/2 mol of N₂(g) and 1/2 mol of 02(g) is 90.3kJ. mol^-1 . Alternatively, it is said that the change in enthalpy of the following reaction at 25°C and I atm is +90.3 kj.

⇒ \(\underbrace{\frac{1}{2} \mathrm{~N}_2(g)+\frac{1}{2} \mathrm{O}_2(g)}_{25^{\circ} \mathrm{C} \& 1 \mathrm{~atm}} \rightarrow \underbrace{\mathrm{NO}(g)}_{25^{\circ} \mathrm{C} \& 1 \mathrm{~atm}}\)

⇒ \( \Delta H_f^0=+90.3 \mathrm{~kJ} \cdot \mathrm{mol}^{-1}\)

Important points about standard enthalpy of formation:

The standard enthalpy of formation of a compound is not always equal to the value of AH° indicated in the thermochemical equation representing the reaction in which the compound in its standard state is formed from its stable constituent elements in their standard states.

Explanation:

At 25°C and 1 atm, two chemical equations for the formation of H₂O(Z) from their stable constituent elements H₂(g) and O₂(g) are given below.

⇒ \(\mathrm{H}_2(g)+\frac{1}{2} \mathrm{O}_2(g) \rightarrow \mathrm{H}_2 \mathrm{O}(l) ; \Delta H^0=-285.8 \mathrm{~kJ}\)

⇒ \(2 \mathrm{H}_2(g)+\mathrm{O}_2(g) \rightarrow 2 \mathrm{H}_2 \mathrm{O}(l) ; \Delta H^0=-571.6 \mathrm{~kJ}\)

The ΔH° in equation [1] indicates the standard enthalpy of the formation of H₂O(Z) because, in this reaction, one mole of H₂O(l) is formed from its stable constituent elements H₂(g) and O₂(g). On the other hand, in equation (2), two moles of H₂O(Z) are formed from the stable constituent elements H₂(g) and O₂(g).

Thus, according to the definition of the enthalpy of formation, the value of AH° from equation [2] does not indicate the standard enthalpy of formation of H₂O(Z).

The standard enthalpies of formation of all elements in their standard states are conventionally taken as zero. In the case of an element having different allotropes, the standard enthalpy of formation of the most stable allotropes form in the standard state is considered zero.

Explanation:

The most stable forms of hydrogen, oxygen, nitrogen, sodium, etc. at 25 °C and 1 atm are H₂(g), O₂(g), N₂(g), Na(s), etc. Thus, the standard enthalpies of formation of H₂(g), O₂(g), N₂(g), Na(s), etc. are zero.

According to the definition, at 25°C and 1at, the standard enthalpy of formation of hydrogen is the same as the standard enthalpy change ofthe following reaction \(\mathrm{H}_2\left(\mathrm{~g}, 25^{\circ} \mathrm{C}, 1 \mathrm{~atm}\right) \rightarrow \mathrm{H}_2\left(\mathrm{~g}, 25^{\circ} \mathrm{C}, 1 \mathrm{~atm}\right)\)

No change has occurred in this process. Hence ΔH°= 0. Therefore, at 25°C, for hydrogen \(\Delta H_f^0=0\).

For the same reason \(\Delta H_f^0\) is zero for N₂(g), O₂(g), etc. Diamond and graphite are the two allotropic forms of solid carbon.

Between these two forms, graphite is the most stable form at 25°C and 1 atm. Thus the standard enthalpy of formation

⇒  \(\Delta H_f^0\) of solid graphite [C (graphite, s)] at 25°C is zero. But, for diamond [C (diamond, s)], at 25°C \(\Delta H_f^0 \neq 0\) A

At 25°C, the standard enthalpy of formation of diamond

⇒  \(\Delta H_f^0 \neq 0\) is equal to the standard enthalpy of reaction for the following change \(\left.\mathrm{C} \text { (graphite, } s, 25^{\circ} \mathrm{C}, 1 \mathrm{~atm}\right) \longrightarrow \mathrm{C} \text { (diamond, } s, 25^{\circ} \mathrm{C}, 1 \mathrm{~atm} \text { ) }\)

In this transformaton, ΔH° = +1.9kJ.

Hence, the standard enthalpy of formation of diamond, \(\Delta H_f^0\) =+1.9kJ.

The most stable form of sulfur at 25°C and 1 atm is solid rhombic sulfur [S (rhombic,s)]. Thus, at 25°C, \(\Delta H_f^0\) for rhombic sulphur is zero.

Determination of the standard enthalpy of reaction (AH0) from the value of the standard enthalpy of formation \(\left(\Delta H_f^0\right)\):

At a particular temperature, the standard reaction enthalpy ofa reaction (AH0) = total enthalpy of formation of the products – total enthalpy of formation of the reactants at the same temperature.

∴ \(\Delta H^0=\sum n_i \Delta H_{f, i}^0-\sum n_j \Delta H_{f, j}^0\)

Where \(\Delta H_{f, i}^0\) and \(\Delta H_{f, i}^0\) are the standard enthalpies of formation of i -th product and j -th reactant, respectively, and nt and nj are the number of moles of i -th product and j -th reactant respectively in a balanced chemical equation.

 The standard heat of formation of some compounds (H°_f) at 25° C:

Heat of combustion or enthalpy of combustion

The heat of combustion or enthalpy of combustion Definition

At a particular temperature and pressure, the change in enthalpy associated with the complete combustion of 1 mol of a substance above oxygen is termed the enthalpy of combustion of the substance at that temperature and pressure.

Generally, the enthalpy of combustion is denoted by ΔH_c.

Standard enthalpy Of combustion:

The standard enthalpy of combustion of a substance is defined as the enthalpy change for the reaction in which one mole of the substance is completely burnt in oxygen when all the reactants and products are in their standard states.

ΔH°c denotes the standard enthalpy of combustion. the combustion reactions are exothermic, the values ofthe enthalpy of combustion are always negative.

Examples:

The thermochemical equations of the combustion of some substances (elements or compounds) at 25°C and 1 atm pressure are given below.

Combustion of graphite:

⇒ \(\mathrm{C}(\mathrm{s}, \text { graphite })+\mathrm{O}_2(\mathrm{~g}) \rightarrow \mathrm{CO}_2(\mathrm{~g}) ; \Delta H_c^0=-393.5 \mathrm{~kJ} \cdot \mathrm{mol}^{-1}\)

Equation:

Also indicates the formation reaction of CO₂(g). So, the standard enthalpy of combustion of graphite is the same as the standard enthalpy of formation of CO₂(g).

Combustion of methane:

⇒ \(\mathrm{CH}_4(\mathrm{~g})+2 \mathrm{O}_2(\mathrm{~g}) \rightarrow \mathrm{CO}_2(\mathrm{~g})+2 \mathrm{H}_2 \mathrm{O}(l)\)

⇒ \(\Delta H_c^0=-890 \mathrm{~kJ} \cdot \mathrm{mol}^{-1}\)

Combustion of sucrose:

⇒ \(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}(\mathrm{~s})+12 \mathrm{O}_2(\mathrm{~g}) \rightarrow 12 \mathrm{CO}_2(\mathrm{~g})+11 \mathrm{H}_2 \mathrm{O}(l)\)

⇒ \(\Delta H_c^0=-5644 \mathrm{~kJ} \cdot \mathrm{mol}^{-1}\)

At 25°c, the standard enthalpy cf of C₄H₁₀(g) is -2878 kj. mol^-1:

This means that at 25°C and 1 atm pressure when 1 mol of C₄H₁₀(g) is completely burnt in the presence of oxygen according to the following reaction, 2878 kj of heat is released.

⇒ \(\mathrm{C}_4 \mathrm{H}_{10}\left(\mathrm{~g}, 25^{\circ} \mathrm{C}, 1\mathrm{~atm}\right)+\frac{13}{2} \mathrm{O}_2\left(\mathrm{~g}, 25^{\circ} \mathrm{C}, 1 \mathrm{~atm}\right)\)[

⇒ \( 4 \mathrm{CO}_2\left(\mathrm{~g}, 25^{\circ} \mathrm{C}, 1 \mathrm{~atm}\right)+5 \mathrm{H}_2 \mathrm{O}\left(\mathrm{l}, 25^{\circ} \mathrm{C}, 1 \mathrm{~atm}\right)\)

The standard heat of combustion of some compounds (25°C):

Applications of the enthalpy of combustion:

Calculation of the enthalpy of formation:

The enthalpy of formation of many substances cannot be directly measured; for example, CH₄ cannot be directly prepared from its constituent elements by the reaction:

⇒ \(\mathrm{C} \text { (graphite, } s)+2 \mathrm{H}_2(\mathrm{~g}) \rightarrow \mathrm{CH}_4(\mathrm{~g}) \text {. }\)

Similarly, glucose is not formed by the reaction:

⇒  \(\mathrm{C} \text { (graphite, } s)+6 \mathrm{H}_2(\mathrm{~g})+3 \mathrm{O}_2(\mathrm{~g}) \rightarrow \mathrm{C}_6 \mathrm{H}_{12} \mathrm{O}_6(\mathrm{~s})\)

The determination of heat (or enthalpy) of the formation of such types of compounds is possible from the known values ofthe heat of combustion of different substances.

Example 1.  Calculate the standard enthalpy of formation of CH₄(g). Given: The standard heat of combustion of CH₄ =- 890 kj.mol^-1 and the standard heat of formation of H₂O(J) and CO₂(g) -285.8 kj.mol^-1 and -393.5 kj. mol^-1, respectively.
Solution:

The thermochemical equation for the combustion of

⇒  \(\mathrm{CH}_4 \text { is } \mathrm{CH}_4(\mathrm{~g})+2 \mathrm{O}_2(\mathrm{~g}) \rightarrow \mathrm{CO}_2(\mathrm{~g})+2 \mathrm{H}_2 \mathrm{O}(\mathrm{l})\)

⇒ \(\Delta H_c^0=-890 \mathrm{~kJ} \cdot \mathrm{mol}^{-1}\)  …………………..(1)

The formation reactions for H₂O(1) and CO₂(g) are

⇒ \(\mathrm{H}_2(\mathrm{~g})+\frac{1}{2} \mathrm{O}_2(\mathrm{~g}) \rightarrow \mathrm{H}_2 \mathrm{O}(l),\)

⇒ \(\Delta H_f^0\left[\mathrm{H}_2 \mathrm{O}(l)\right]=-285.8 \mathrm{~kJ} \cdot \mathrm{mol}^{-1}\)…………………(2)

⇒ \(\mathrm{C} \text { (graphite, } s)+\mathrm{O}_2(\mathrm{~g}) \rightarrow \mathrm{CO}_2(\mathrm{~g})\)

⇒ \(\Delta H_f^0\left[\mathrm{CO}_2(\mathrm{~g})\right]=-393.5 \mathrm{~kJ} \cdot \mathrm{mol}^{-1}\) …………………(3)

The formation reactions for CH₄(g) is

⇒ \(\mathrm{C}(\text { graphite, } s)+2 \mathrm{H}_2(\mathrm{~g}) \rightarrow \mathrm{CH}_4(\mathrm{~g})\)

To obtain equation (4), we can write equation (3) + 2 x equation (2)- equation (1). This given

⇒ \(\Delta H^0=\Delta H_f^0\left[\mathrm{CO}_2(g)\right]+2 \times\left(\Delta H_f^0\left[\mathrm{H}_2 \mathrm{O}(l)\right]\right)-\Delta H_c^0\)

=[- 393.5 + 2(-285.8)- (-890)] kJ = -75.1 kJ …………………(4)

Thus, the standard heat of formation of CH₄(g) =-75.1 kJ. mol^-1

Fuel efficiency:

Based on the data of heat of combustion of various fuels (For example coal, kerosene, petrol, etc.), the efficiency of each fuel with the same amount, the one that liberates a larger amount of heat on considered to be a better fuel.

So, the heat of combustion is very important regarding the selection ofa fuel. The amount of heat produced due to the combustion of l g of a fuel is known as the calorific value of that fuel.

Determination of calorific value of foods:

Energy is obtained by the oxidation of carbohydrates and fats present in the food that we consume. These carbohydrates and fats are oxidized into CO₄ and H₂O along with the liberation of heat This heat of combustion maintains our body temperature and the strength ofthe muscle.

⇒ \(\mathrm{C}_6 \mathrm{H}_{12} \mathrm{O}_6(\mathrm{~s})+6 \mathrm{O}_2(\mathrm{~g}) \rightarrow 6 \mathrm{CO}_2(\mathrm{~g})+6 \mathrm{H}_2 \mathrm{O}(l)\)

⇒ \(\Delta H^0=-2808 \mathrm{~kJ}\)

The amount of heat liberated in the combustion (or oxidation) of 1 g of an edible substance (food) is known as the calorific value of that food. Based on this calorific value, we can easily prepare a balanced food chart. As the calorific values of fats are high, thus in cold countries, foods containing fats are consumed in larger quantities.

Calorific values of some common foods and fuels:

Numerical Examples

Question 1. Calculate the standard enthalpy of reaction at 25 temperature for the following reaction:

⇒ \(\mathrm{C}_6 \mathrm{H}_6(l)+\frac{15}{2} \mathrm{O}_2(g) \rightarrow 6 \mathrm{CO}_2(g)+3 \mathrm{H}_2 \mathrm{O}(l)\)

 Given: The standard enthalpy of formation of C₆H₆(f), CO₂(l), and H₂O(l) -393.5 kj.moH and -285.8 kj. mol^–1respectively.

Answer:

⇒ \(\mathrm{C}_6 \mathrm{H}_6(l)+\frac{15}{2} \mathrm{O}_2(g) \rightarrow 6 \mathrm{CO}_2(g)+3 \mathrm{H}_2 \mathrm{O}(l)\)

The standard heat of reaction for this reaction is

⇒ \(\Delta H^0=6 \times \Delta H_f^0\left[\mathrm{CO}_2(\mathrm{~g})\right]+3 \times \Delta H_f^0\left[\mathrm{H}_2 \mathrm{O}(l)\right]\)

⇒ \(-1 \times \Delta H_f^0\left[\mathrm{C}_6 \mathrm{H}_6(l)\right]-\frac{15}{2} \Delta H_f^0\left[\mathrm{O}_2(g)\right]\)

⇒ \(6 \times(-393.5)+3 \times(-285.8)-1 \times(49.0)-\frac{15}{2} \times 0\)

⇒ \(-3267.4 \mathrm{~kJ}\left[\Delta H_f^0\left[\mathrm{O}_2(g)\right]=0\right]\)

Question 2. Calculate standard enthalpy of reaction at 25°C for the reaction: \(\mathrm{CCl}_4(g)+2 \mathrm{H}_2 \mathrm{O}(g) \rightarrow \mathrm{CO}_2(g)+4 \mathrm{HCl}(g) .\) Given: The standard heat of formation of CCl₄(g) , H₂O(g), CO₂(g) and HCl(g) are -25.5, -57.8, -94.1 and -22.1 kcal-mol^-1.
Answer:

The standard heat of reaction for the given reaction,

⇒ \(\Delta H^0=\Delta H_f^0[\left.\mathrm{CO}_2(g)\right]+4 \times \Delta H_f^0[\mathrm{HCl}(g)]\)

⇒ \(\Delta H_f^0\left[\mathrm{CCl}_4(\mathrm{~g})\right]-2 \times \Delta H_f^0\left[\mathrm{H}_2 \mathrm{O}(\mathrm{g})\right]\)

= \(-94.1+4 \times(-22.1)-(-25.5)-2 \times(-57.8)\)

=  -41.4 kcal

3. ΔH values for the given reactions at 25°C are:

⇒ \(\mathrm{C}_3 \mathrm{H}_8(\mathrm{~g}) \rightarrow 3 \mathrm{C}(s)+4 \mathrm{H}_2(\mathrm{~g}) ; \Delta H^0=103.8 \mathrm{~kJ} \cdot \mathrm{mol}^{-1}\)

 ⇒ \(2 \mathrm{H}_2(\mathrm{~g})+\mathrm{O}_2(\mathrm{~g}) \rightarrow 2 \mathrm{H}_2 \mathrm{O}(\mathrm{l}) ; \Delta H^0=-571.6 \mathrm{~kJ} \cdot \mathrm{mol}^{-1}\)

⇒ \(\mathrm{C}_2 \mathrm{H}_6(\mathrm{~g})+\frac{7}{2} \mathrm{O}_2(\mathrm{~g}) \rightarrow 2\)

⇒ \( \mathrm{CO}_2(\mathrm{~g})+3 \mathrm{H}_2 \mathrm{O}(l) \)

⇒ \(\Delta H^0=-1560 \mathrm{~kJ} \cdot \mathrm{mol}^{-1}\)
Answer:

⇒ \(\mathrm{C}_3 \mathrm{H}_8(\mathrm{~g}) \rightarrow 3 \mathrm{C}(\mathrm{s})+4 \mathrm{H}_2(\mathrm{~g})\)

⇒  \(\Delta H^0=103.8 \mathrm{~kJ} \cdot \mathrm{mol}^{-1}\) ………………………..(1)

⇒ \(2 \mathrm{H}_2(\mathrm{~g})+\mathrm{O}_2(\mathrm{~g}) \rightarrow 2 \mathrm{H}_2 \mathrm{O}(\mathrm{l})\)

⇒ \( \Delta H^0=-571.6 \mathrm{~kJ} \cdot \mathrm{mol}^{-1}\) ………………………..(2)

⇒ \(\mathrm{C}_2 \mathrm{H}_6(\mathrm{~g})+\frac{7}{2} \mathrm{O}_2(\mathrm{~g}) \rightarrow 2 \mathrm{CO}_2(\mathrm{~g})+3 \mathrm{H}_2 \mathrm{O}(\mathrm{l})\)

⇒  \(\Delta H^0=-1560 \mathrm{~kJ} \cdot \mathrm{mol}^{-1}\)………………………..(3)

⇒ \(\mathrm{CH}_4(g)+2 \mathrm{O}_2(g) \rightarrow \mathrm{CO}_2(g)+2 \mathrm{H}_2 \mathrm{O}(l)\)

⇒ \(\Delta H^0=-890 \mathrm{~kJ} \cdot \mathrm{mol}^{-1}\)………………………..(4)

⇒ \(\mathrm{C}(s)+\mathrm{O}_2(g) \rightarrow \mathrm{CO}_2(g)\)

⇒ \(\Delta H^0=-393.5 \mathrm{~kJ} \cdot \mathrm{mol}^{-1}\)………………………..(5)

Multiplying equation (2) by \(\frac{5}{2}\)

⇒ \( \Delta H^0=-393.5 \mathrm{~kJ} \cdot \mathrm{mol}^{-1}\) ………………………..(6)

⇒ \(\mathrm{C}(\mathrm{s})+\mathrm{O}_2(\mathrm{~g}) \rightarrow \mathrm{CO}_2(\mathrm{~g}) ; \Delta H^0=-393.5 \mathrm{~kJ} \cdot \mathrm{mol}^{-1}\)

Reversing equations (3)and (4), we obtain,

⇒ \(2 \mathrm{CO}_2(g)+3 \mathrm{H}_2 \mathrm{O}(l) \rightarrow \mathrm{C}_2 \mathrm{H}_6(g)+\frac{7}{2} \mathrm{O}_2(g)\)

ΔH° = 1560 KJ………………….(7)

⇒ \(\mathrm{CO}_2(\mathrm{~g})+2 \mathrm{H}_2 \mathrm{O}(\mathrm{l}) \rightarrow \mathrm{CH}_4(\mathrm{~g})+2 \mathrm{O}_2(\mathrm{~g})\)

∴ ΔH° =+ 890 KJ………………….(8)

Multiplying equation (5) by 3, we obtain

⇒ \(3 \mathrm{C}(\mathrm{s})+3 \mathrm{O}_2(\mathrm{~g}) \rightarrow 3 \mathrm{CO}_2(\mathrm{~g})\)

⇒ \(\Delta H^0=3 \times(-393.5)=-1180.5 \mathrm{~kJ}\)…………………….(9)

Adding equations (1), (6), (7), (8), and (9), we obtain,

⇒ \(\mathrm{C}_3 \mathrm{H}_8(\mathrm{~g})+\mathrm{H}_2(\mathrm{~g}) \rightarrow \mathrm{C}_2 \mathrm{H}_6(\mathrm{~g})+\mathrm{CH}_4(\mathrm{~g})\)

⇒ \(\Delta H^0=[103.8+(-1429)+1560+890+(-1180.5)]\)

∴ ΔH° = -55.7 KJ

For the given reaction, ΔH° = -55.7 KJ

Question 4. Calculate ΔH° for the following reaction at 298K: \(\) Given: At 298 K temperature,

⇒ \(\mathrm{C}_2 \mathrm{H}_3 \mathrm{CO}_2 \mathrm{H}(\mathrm{l})+3 \mathrm{O}_2(\mathrm{~g}) \rightarrow 3 \mathrm{CO}_2(\mathrm{~g})+2 \mathrm{H}_2 \mathrm{O}(\mathrm{l})\),ΔH° = – 1368 kJ 

⇒ \(2 \mathrm{C}_2 \mathrm{H}_2(g)+5 \mathrm{O}_2(g) \rightarrow 4 \mathrm{CO}_2(g)+2 \mathrm{H}_2 \mathrm{O}(l)\), ΔH° = – 2600 kJ

⇒ \(2 \mathrm{CO}(g)+\mathrm{O}_2(g) \rightarrow 2 \mathrm{CO}_2(g) ; \Delta H^0=-566 \mathrm{~kJ}\), ΔH° = – 566 kJ

Answer:

⇒ \(\mathrm{C}_2 \mathrm{H}_3 \mathrm{CO}_2 \mathrm{H}(\mathrm{l})+3 \mathrm{O}_2(\mathrm{~g}) \rightarrow 3 \mathrm{CO}_2(\mathrm{~g})+2 \mathrm{H}_2 \mathrm{O}(\mathrm{l})\)

ΔH° = – 1368 kJ ………..(1)

⇒ \(2 \mathrm{C}_2 \mathrm{H}_2(g)+5 \mathrm{O}_2(g) \rightarrow 4 \mathrm{CO}_2(g)+2 \mathrm{H}_2 \mathrm{O}(l)\)

ΔH° = – 2600 kJ ………..(2)

⇒ \(2 \mathrm{CO}(g)+\mathrm{O}_2(g) \rightarrow 2 \mathrm{CO}_2(g) ; \Delta H^0=-566 \mathrm{~kJ}\)

ΔH° = – 566 kJ ………..(3)

Reversing equation (1), we get,

⇒ \(3 \mathrm{CO}_2(g)+2 \mathrm{H}_2 \mathrm{O}(l) \rightarrow \mathrm{C}_2 \mathrm{H}_3 \mathrm{CO}_2 \mathrm{H}(l)+3 \mathrm{O}_2(g)\)

⇒ \(\Delta H^0=+1368 \mathrm{~kJ}\) …………………………(4)

Dividing each of equations (2) and (3) by 2, we obtain

⇒ \(\mathrm{C}_2 \mathrm{H}_2(g)+\frac{5}{2} \mathrm{O}_2(\mathrm{~g}) \rightarrow 2 \mathrm{CO}_2(\mathrm{~g})+\mathrm{H}_2 \mathrm{O}(l)\)

⇒ \(\Delta H^0=-1300 \mathrm{~kJ}\)…………………………..(5)

⇒ \(\mathrm{CO}(\mathrm{g})+\frac{1}{2} \mathrm{O}_2(\mathrm{~g}) \rightarrow\mathrm{CO}_2(\mathrm{~g})\).

⇒\(\Delta H^0=-283 \mathrm{~kJ}\)………………………..(6)

Adding equations (4),(5), and (6), we obtain,

⇒ \(\mathrm{C}_2 \mathrm{H}_2(g)+\mathrm{CO}(g)+\mathrm{H}_2 \mathrm{O}(l) \rightarrow \mathrm{C}_2 \mathrm{H}_3 \mathrm{CO}_2 \mathrm{H}(l)\)

⇒ \(\Delta H^0=[1368+(-1300)+(-283)] \mathrm{kJ}=-215 \mathrm{~kJ}\)

∴ ΔH° for the given reaction =-215 kJ.

Question 5. Calculate the standard enthalpy of formation of C₆H₆(l) at 25°C temperature using the given data
Answer:

Reaction to the formation of C₆H₆(l):

⇒ \(6 \mathrm{C}(s \text {, graphite })+3 \mathrm{H}_2(g) \rightarrow \mathrm{C}_6 \mathrm{H}_6(l)\)

Given:

⇒ \(\mathrm{C}_6 \mathrm{H}_6(l)+\frac{15}{2} \mathrm{O}_2(g) \rightarrow 6 \mathrm{CO}_2(g)+3 \mathrm{H}_2 \mathrm{O}(l)\)

⇒ \(\Delta H^0 =-781 \mathrm{kcal}\)……………(1)

⇒ \(\mathrm{H}_2(g)+\frac{1}{2} \mathrm{O}_2(g) \rightarrow \mathrm{H}_2 \mathrm{O}(l)\)

⇒ \( \Delta H^0 =-68.32 \mathrm{kcal}\) ……………(2)

C(s, graphite) + O₂(g)→ CO₂(g); ΔH° = -94.04 kcal

⇒ \( \Delta H^0 =-94.04 \mathrm{kcal}\) ……………(3)

Reversing equation (1), we obtain

⇒ \(6 \mathrm{CO}_2(g)+3 \mathrm{H}_2 \mathrm{O}(l) \rightarrow \mathrm{C}_6 \mathrm{H}_6(l)+\frac{15}{2} \mathrm{O}_2(\mathrm{~g})\)

⇒ \(\Delta H^0=+781 \mathrm{kcal}\)……………(4)

Multiplying equation (2) by 3 & equation (3) by 6, we get

⇒ \(3 \mathrm{H}_2(g)+\frac{3}{2} \mathrm{O}_2(g) \rightarrow 3 \mathrm{H}_2 \mathrm{O}(l)\)

⇒ \(\Delta H^0=-204.96 \mathrm{kcal}\) ……………(5)

And \(6 \mathrm{C}(s, \text { graphite })+6 \mathrm{O}_2(g) \rightarrow 6 \mathrm{CO}_2(\mathrm{~g})\)

⇒ \(\Delta H^0=-564.25 \mathrm{kcal}\) ……………(6)

Now adding equations (4), (5) and (6), we obtain

⇒ \(\mathrm{C}(s, \text { graphite })+3 \mathrm{H}_2(g) \rightarrow \mathrm{C}_6 \mathrm{H}_6(l)\)

⇒ \(\Delta H^0=+781+(-204.96)+(-564.24)=+11.8 \mathrm{kcal}\) ……………(7)

Equation (7) indicates a thermochemical equation for the formation of C₆H₆ (l) at the standard state.

Therefore, the standard heat of formation of C6H6(Z) =+ 11.8 kcal.mol-1

Question 6. At 25°C temperature, the heat of combustion of sucrose, carbon, and hydrogen is -5644 kj. mol^-1 , -393.5 kj.mol^-1 & -285.8 kJ – mol^-1 respectively. Determine the heat of the formation of sucrose at 25°C.
Answer:

.C₁₂H₂₂O₁₁(s)+12O₂(g)→  12CO₂(g)+ 11H₂O(Z);

ΔH°= -5644 kj mol^-1  ……………………………….(1)

C(graphite, s) +O₂ (g)→ CO₂(g)

ΔH° = -393.5 kJ.mol^-1  ……………………………….(2)

H₂(S) + ½O₂(g)→ H₂O

ΔH° =- 285.8 kJ.mol^-1   ……………………………….(3)

The formation reaction of sucrose:

⇒ \(12 \mathrm{C} \text { (graphite, } s)+11 \mathrm{H}_2(g)+\frac{11}{2} \mathrm{O}_2(g) \rightarrow \mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}(s)\)

Reversing Equation (1), we get

12CO₂(g) + 11H₂O(g) → C₁₂H₂₂O₁₁(s) + 12O₂(g) ;

ΔH° = +5644 kJ   ……………………………….(4)

Multiplying equation (2) by 12 and equation (3) by 11.

12C(graphite, s) + 12O₂(g)→12CO₂(g);

ΔH° = 12 ×(-393.5) = -4722 kJ ……………………………….(5)

11H₂(g) +\(\frac{11}{2}\)O₂(g)  →12H₂O(l)

ΔH° = 11 ×(-285.8)

= -3143.8kJ……………………………….(6)

By adding equations (4), (5) and(6) and their corresponding ΔH° values, we obtain

⇒ \(12 \mathrm{C}(s, \text { graphite })+11 \mathrm{H}_2(g)+\frac{11}{2} \mathrm{O}_2(\mathrm{~g}) \rightarrow \mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}(s)\)

∴ \(\Delta H^0=+5644+(-4722)+(-3143.8)=-2221.8\)  ……………………………….(7)

Equation (7) indicates, the thermochemical equation for the formation of C₁₂H₂₂Ou(s) at standard conditions. Therefore, the standard enthalpy of the formation of

⇒  \(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}(\mathrm{~s})=-2221.8 \mathrm{~kJ} \cdot \mathrm{mol}^{-1}\)

Question 7. At 25°C temperature, the standard heat of formation of CH₄(g), CO₂(g) and H₂O(g) are -74.8kJ.mol^-1 , -393.5kJ.mol^-1 & -241.6kJ.mol^-1 respectively. How much heat will be evolved during combustion of lm3 CH₄(g) at 25°C temperature and 1 atm pressure? Consider CH₄(g) behaves like an ideal gas
Answer:

Combustion reaction of CH₄(g) is given by:

⇒ \(\mathrm{CH}_4(g)+2 \mathrm{O}_2(g) \rightarrow \mathrm{CO}_2(g)+2 \mathrm{H}_2 \mathrm{O}(l)\)

∴ Standard heat of combustion of CH₄(g)

⇒ \(\Delta H_c^0=\left[1 \times \Delta H_f^0\left[\mathrm{CO}_2(g)\right]+2 \times \Delta H_f^0\left[\mathrm{H}_2 \mathrm{O}(l)\right]\right]\)

⇒ \(-\left[1 \times \Delta H_f^0\left[\mathrm{CH}_4(\mathrm{~g})\right]+2 \times \Delta H_f^0\left[\mathrm{O}_2(\mathrm{~g})\right]\right]\)

=-\(393.5+2 \times(-241.6)-1 \times(-74.8)+2 \times 0=-801.9 \mathrm{~kJ}\)

Since In standard state for stable and pure element \(\left.\Delta H_f^0=0 \text {; thus, } \Delta H_f^0\left[\mathrm{O}_2(g)\right]=0\right)\)

lm³ CH₄(g) = 10³L CH₄(g)

Number of moles of 103L CH4(g) at 25°C and 1 atm pressure

n =\(\frac{P V}{R T}=\frac{1 \times 10^3}{0.0821 \times 298}=40.87\)

⇒  \(\text { pressure, } n=\frac{P V}{R T}=\frac{1 \times 10^3}{0.0821 \times 298}=40.87\)

∴ Heat evolved in combustion = 40.87 ¹ 801.9 = 32773.6 kj

Question 8. Calculate the value of enthalpy of combustion of cyclopropane at 25°C and 1 atm pressure. Given: Standard enthalpy of formation of CO₂(g), H₂O(I) & propene (g) at 25°C are -393.5 kJ.mol^-1, -285.8 kJ.mol^-1 & 20.4 kj.mol^-1 respectively. Also, the standard enthalpy change for isomerization reaction: Cyclopropane(g)y=yPropene(g) is -33.0kJ.mol^-1.
Answer:

Given:

\(\mathrm{C}(\text { graphite, } s)+\mathrm{O}_2(\mathrm{~g})\rightarrow \mathrm{CO}_2(\mathrm{~g}) \)

⇒ \( \Delta H_f^0=-393.5 \mathrm{~kJ} \cdot \mathrm{mol}^{-1}\) ………………….. (1)

⇒ \(\mathrm{H}_2(\mathrm{~g})+\frac{1}{2} \mathrm{O}_2(\mathrm{~g}) \rightarrow \mathrm{H}_2 \mathrm{O}(l)\)

⇒ \(\Delta H_f^0=-285.8 \mathrm{~kJ} \cdot \mathrm{mol}^{-1}\) ………………….. (2)

3C(graphite, s) + 3H₂(g)→ C₃H₆(g) (propene);

⇒ \(\Delta H_f^0=+20.4 \mathrm{~kJ} \cdot \mathrm{mol}^{-1}\)………………….. (3)

⇒ \(\text { Cyclopropane }\left[\mathrm{C}_3 \mathrm{H}_6(\mathrm{~g})\right] \rightarrow \text { Propene }\left[\mathrm{C}_3 \mathrm{H}_6(\mathrm{~g})\right]\)

⇒\(\Delta H^0=-33.0 \mathrm{~kJ} \cdot \mathrm{mol}^{-1}\)………………….. (4)

Reversing equation(3), we get,

⇒ \(\left.\mathrm{C}_3 \mathrm{H}_6 \text { (propene, } g\right) \rightarrow 3 \mathrm{C}(\text { graphite, } s)+3 \mathrm{H}_2(g)\)

⇒ \( \Delta H_f^0=-20.4 \mathrm{~kJ} \cdot \mathrm{mol}^{-1}\)  …………………. (5)

Multiplying equations (1) and (2)by (3), we get,

⇒ \(3 \mathrm{C} \text { (graphite, } s)+3 \mathrm{O}_2(\mathrm{~g}) \rightarrow 3 \mathrm{CO}_2(\mathrm{~g}) \text {; }\)

⇒\(\Delta H^0=-1180.5 \mathrm{~kJ}\) …………………. (6)

⇒ \(3 \mathrm{H}_2(g)+\frac{3}{2} \mathrm{O}_2(g) \rightarrow 3 \mathrm{H}_2 \mathrm{O}(l)\)

⇒ \(\Delta H^0=-857.4 \mathrm{~kJ}\) …………………. (7)

Adding equations (4), (5),(6) and (7) we obtain,

⇒ \(\left.\mathrm{C}_3 \mathrm{H}_6 \text { (cyclopropane, } g\right)+\frac{9}{2} \mathrm{O}_2(g) \rightarrow 3 \mathrm{CO}_2(g)+3 \mathrm{H}_2 \mathrm{O}(l)\)

⇒  \(\Delta H^0=-33+(-20.4)+(-1180.5)+(-857.4)=-2091.3 \mathrm{~kJ}\)

So, the standard enthalpy of combustion of lmol cyclopropane (g) at 25°C and 1 atm pressure = -2091.3 kJ.mol^-1

Question 9. At 25°C, the heat evolved due to the complete combustion of 7.8g of C₆H₆(Z) is 326.4 kJ. Calculate the heat evolved due to complete combustion of the same amount of C₆H₆(l) at the same temperature and constant pressure of 1 atm.
Answer:

7.8 \(\mathrm{~g} \mathrm{C}_6 \mathrm{H}_6=\frac{7.8}{78}=0.1\)mol

∴ M_C6H6 = 78

The combustion reaction of C₆H₆(l):

⇒ \(\mathrm{C}_6 \mathrm{H}_6(l)+\frac{15}{2} \mathrm{O}_2(g) \rightarrow 6 \mathrm{CO}_2(g)+3 \mathrm{H}_2 \mathrm{O}(l)\)………………………………………(1)

∴ \(\Delta n=6-\frac{15}{2}=-\frac{3}{2}\)

From the given information, the heat evolved in complete combustion of 7.8 g or 0.1 mol of C₆H₆(Z) at 25°C temperature and at constant volume

=-326.4 kj.

Therefore, at 25°C and constant volume, the heat evolved due to the complete combustion of lmol C₆H₆(Z) =-30264 kj.

Now, the heat of reaction at constant volume = ΔU.

∴ ΔU = -3264 kj.mol^-1

Again we know, ΔH = ΔU + ΔnRT

For reaction [1]

⇒ \(\Delta H=\left[-3264+\left(-\frac{3}{2}\right) \times\left(8.314 \times 10^{-3}\right)(298)\right]=3267.71 \mathrm{~kJ}\)

Therefore, at 25°C and 1 atm pressure, the heat evolved due to the complete combustion of1 mol C₆H₆(l) =-3267.71kj.

∴ At 25°C and 1 atm pressure, the heat evolved due to complete combustion of 7.8 g or 0.1 mol of C₆H₆(l) =- 326.771 kj.

Enthalpy change due to phase transition

A physical process in which a substance undergoes a change from one physical state to another, but its chemical identity remains the same is called phase transition.

Fusion, sublimation, and vaporization are endothermic processes. So, these processes involve the absorption of heat. On the other hand, solidification, condensation, and deposition are exothermic processes, involving the evolution of heat.

Enthalpy of fusion:

Enthalpy of fusion Definition:

At constant pressure, the amount of heat required by one mole of a solid substance for its complete liquefaction at its melting point is called the enthalpy of fusion of that solid.

The enthalpy of fusion of a substance is the same as its molar latent heat of fusion. As fusion is an endothermic process, the value of enthalpy of fusion of a substance is positive.

Example: Heat absorbed during the transformation of 1 mole of ice into 1 mole of water at 0°C is 6.02 kj. Therefore, the enthalpy of fusion of ice at 0°C and 1 atm pressure,

Significance of the enthalpy of fusion:

The enthalpy of fusion (ΔH_fus) of a solid substance is a measure of the interparticle forces of attraction in the solid. The stronger the interparticle forces of attraction in a solid, the larger the value of its

⇒ \(\Delta H_{\text {fus }}\). The interparticle forces of attraction in ionic solids [like— NaCl, MgCl₂, etc.] are stronger than those in molecular solids [like ice, I2(s), etc.]. This is why the values of ionic solids are found to be greater than those of molecular solids.

For example, the values of \(\Delta H_{f u s}\) for Nacl, which is an ionic solid, is +28.8kJ. mol^-1 where for .ice, which is a molecular solid it is +6.02kj. mol^-1

Solidification or freezing is the reverse process of fusion because during solidification liquid phase is transformed into the solid phase, while the solid phase is transformed into the liquid phase during the fusion process. Hence, at a particular temperature and pressure, the enthalpy of solidification ofa substance =(-) its enthalpy of fusion.

For example, at 0°C temperature and 1 atm pressure, the enthalpy of fusion of ice = + 6.02 kj.mol^-1 and the enthalpy of solidification of water = -6.02kj. mol^-1

Enthalpy of vaporisation:

Enthalpy of vaporization Definition:

At constant pressure, the amount of heat required for the complete vaporization of 1 mole of a liquid at its boiling point is termed the enthalpy of vaporization of that liquid.

The enthalpy of vaporization of a liquid is the same as its molar latent heat of vaporization. The enthalpy of vaporization is a positive quantity because vaporization is an endothermic process.

Example:

At 100°C and 1 atm pressure, 40.4 kj of heat is required to completely convert 1 mol of water into 1 mol of water vapor. Thus, at 100°C and 1 atm, the enthalpy of vaporization of water \(\left(\Delta H_{\text {vap }}\right)=+40.4 \mathrm{~kJ} \cdot \mathrm{mol}^{-1}\)

⇒ \(\mathrm{H}_2 \mathrm{O}(l) \rightarrow \mathrm{H}_2 \mathrm{O}(\text { vap }), \Delta H_{v a p}=+40.4 \mathrm{~kJ} \cdot \mathrm{mol}^{-1}\)

Significance of the enthalpy of vaporization:

The value of the enthalpy of vaporization of a liquid is a measure of the intermolecular forces of attraction in the liquid. The stronger the intermolecular forces of attraction the larger the values of ΔH_vap

For example, ΔH_vap for water is +40.4 kj.mol^-1, while that for benzene is +30.5 kj.mol^-1, indicating that the intermolecular forces of attraction are stronger in water than in benzene. Thus, the amount of heat required to vaporize 1 mol of benzene is less than that required to vaporize 1 mol of water.

Condensation is die reverse process of vaporization because during condensation vapour phase transforms into a liquid phase whereas during vaporization liquid phase transforms into a vapor phase. Thus, the magnitude of the enthalpy of vaporization

⇒ \(\left(\Delta H_{\text {vap }}\right)\) and enthalpy of condensation \(\left(\Delta H_{\text {condensation }}\right)\) are the same but the opposite in sign ., i., e ΔH

For example:

At 100°C temperature and 1 atm pressure, for water

⇒ \(\Delta H_{\text {vap }}=+40.4 \mathrm{~kJ} \cdot \mathrm{mol}^{-1} \text { and }\)

⇒ \( \Delta H_{\text {condensation }}=-40.4 \mathrm{~kJ} \cdot \mathrm{mol}^{-1}\)

Enthalpy of sublimation:

Enthalpy of sublimation Definition:

Enthalpy of sublimation is defined as the amount of heat required by lmol of a solid substance for its complete vaporization at a given condition of temperature and pressure.

The value of the enthalpy of sublimation is always positive because sublimation is an endothermic process.

Example:

At ordinary temperature and pressure, 62.3 kj heat is required to convert 1 mol of solid I₂ to I₂ vapor. Thus, at ordinary temperature and pressure, the enthalpy of sublimation of iodine

⇒ \(\left(\Delta H_{s u b}\right)=+62.3 \mathrm{~kJ} \cdot \mathrm{mol}^{-1}\)

As the enthalpy is a state function, the change in enthalpy in the following two processes (1 and 2) will be the same.

The total change in enthalpy \(=\Delta H_{\text {fus }}+\Delta H_{\text {vap }}\)

⇒ \((\Delta H)_{\mathrm{I}}=(\Delta H)_{\mathrm{II}} \text { i.e., } \Delta H_{\text {sub }}=\Delta H_{\text {fus }}+\Delta H_{\text {vap }}\)

Numerical Examples

Question 1. Calculate The enthalpy change in the fusion of 100 g ice at 0°C temperature and 1 atm pressure The enthalpy change in the vaporization of 10 g water at 100°C temperature and 1 atm pressure. Given: Latent heat of ice at 0°C temperature and 1 atm pressure = 6.02 kJ.mol^-1 and latent heat of vaporisation of water at 100°C temperature = 40.4 kj.mol^-1
Answer:

Latent heat of fusion of ice at 0°C temperature and 1 atm pressure = 6.02 kj. mol^-1 Therefore, the enthalpy of fusion of 18g ice = 6.02 kj.

∴ Enthalpy of fusion 100 ice \(=\frac{6.02}{18} \times 100=33.44 \mathrm{~kJ}\)

∴ Change in enthalpy of fusion of 100 g ice = + 33.44 kj

Latent heat of vaporization of water at 100°C = 40.4 kj. mol^-1 .

Therefore, the enthalpy of vaporization of 18g water = 40.4 kJ

∴ Enthalpy of vapourisation of 10g water \(=\frac{40.4}{18} \times 10=22.44 \mathrm{~kJ}\)

Question 2. The heat required to completely vaporize 7.8 g of benzene at 1 atm pressure and 80°C temperature (boiling point of benzene) is 3.08 kj. What is the value of the enthalpy of vaporization of benzene? What will be the change in enthalpy if 54.6 g of benzene vapor is condensed at 1 atm pressure and 80°C temperature?
Answer:

⇒ \(7.8 \mathrm{~g} \text { benzene }=\frac{7.8}{78}=0.1 \mathrm{~mol} \text { benzene }\) since \(\mathrm{M}_{\mathrm{C}_6 \mathrm{H}_6}=78\)

∴ Change in enthalpy of fusion of100 g ice = + 33.44 kJ

∴ Latent heat of vaporization of water at 100°C = 40.4 kj.mol-1. Therefore, the enthalpy of vaporization of18g water = 40.4 kJ

∴ Enthalpy of vapourisation of 10g water \(=\frac{40.4}{18} \times 10=22.44 \mathrm{~kJ}\)

Question 3. The heat required to completely vaporize 7.8 g of benzene at 1 atm pressure and 80°C temperature (boiling point of benzene) is 3.08 kJ. What is the value of the enthalpy of vaporization of benzene? What will be the change in enthalpy if 54.6 g of benzene vapor is condensed at 1 atm pressure and 80°C temperature?
Answer:

⇒ \(7.8 \mathrm{~g} \text { benzene }=\frac{7.8}{78}=0.1 \mathrm{~mol} \text { benzene }\)

1. As per the given data, the heat required for complete vaporization of 0.1 mol of benzene at 1 atm pressure and 80°C temperature is 3.08 kj. So, at the same temperature and pressure, the heat required for complete vaporization of1 mol of benzene is 30.8 kJ

Hence, according to the definition, enthalpy of vaporization of benzene at atm pressure and 80°C temperature = + 30.8 kJ

2.  54.6g of benzene \(=\frac{54.6}{78}=0.7 \mathrm{~mol}\)

Therefore, enthalpy of vaporization of 0.7 mol benzene at 1 atm pressure and

⇒  \(80^{\circ} \mathrm{C}=\frac{0.7 \times(+3.08)}{0.1}=+21.56 \mathrm{~kJ}\)

1 atm pressure and 80°C \(=\frac{0.7 \times(+3.08)}{0.1}=+21.56 \mathrm{~kJ} \text {. }\)

Again, enthalpy of condensation =(-) enthalpy of vaporization. So, the change in enthalpy of condensation of 54.6g benzene at 1 atm pressure and 80°C temperature =-21.56 kJ

Heat or enthalpy of neutralization

Heat or enthalpy of neutralization Definition:

The change in enthalpy that occurs when 1 gram equivalent of an acid is completely neutralized by 1 gram equivalent of a base or vice-versa in a dilute solution at a particular temperature is called the enthalpy (or heat) of neutralization.

The change in enthalpy that occurs when 1 mol of H⁺. ions react completely with lmol of OH^– ions in a dilute solution to form 1 mol water at a particular temperature is known as the enthalpy (or heat) of neutralization.

The enthalpy of neutralization is denoted as ΔHN, where subscript TV ‘indicates ‘neutralization’.

Neutralization of strong acid and strong base:

If both the acid and base are strong, then the value of heat of neutralization constant is found to be almost and this value is -57.3 kJ

⇒ \(\text { Examples: } \mathrm{HCl}(a q)+\mathrm{NaOH}(a q) \rightarrow \mathrm{NaCl}(a q)+\mathrm{H}_2 \mathrm{O}(l)\)

⇒ \(\Delta H_N=-57.3 \mathrm{~kJ}\)

⇒ \(\mathrm{HNO}_3(a q)+\mathrm{KOH}(a q) \rightarrow \mathrm{KNO}_3(a q)+\mathrm{H}_2 \mathrm{O}(l)\)

⇒ \(\Delta H_N=-57.3 \mathrm{~kJ}\)

Explanation:

Consider the neutralization reaction involving HCl and NaOH in a dilute aqueous solution.

⇒ \(\mathrm{HCl}(a q)+\mathrm{NaOH}(a q) \rightarrow \mathrm{NaCl}(a q)+\mathrm{H}_2 \mathrm{O}(l)\) As HCl NaOH and NaCl all are strong electrolytes, they completely dissociate in aqueous solution.

Hence, the above neutralization reaction can be written as: ^-1.

⇒ \(\begin{array}{r}
\mathrm{H}^{+}(a q)+\mathrm{Cl}^{-}(a q)+\mathrm{Na}^{+}(a q)+\mathrm{OH}^{-}(a q) \longrightarrow \\
\mathrm{Na}^{+}(a q)+\mathrm{Cl}^{-}(a q)+\mathrm{H}_2 \mathrm{O}(l)
\end{array}\)

Cancelling the species that appear on both sides, we have

⇒ \(\mathrm{H}^{+}(a q)+\mathrm{OH}^{-}(a q) \rightarrow \mathrm{H}_2 \mathrm{O}(l)\)

Therefore, the neutralization of strong acid and strong base is essentially, the combination of H+ ions and OH- ions to form water. This is the only reaction that occurs during the neutralization of strong acid and strong base.

This is why the heat of neutralization of all strong acids and strong bases is virtually constant and equal to -57.3 kj.mol-1

Neutralization of a strong acid by the weak base, weak acid by the strong base, and weak acid by weak base:

If either acid or base is weak or both are weak, then the heat of neutralization value will be different in each case

Examples:

\(\mathrm{CH}_3 \mathrm{COOH}(a q)+\mathrm{NaOH}(a q)\)→\(\mathrm{CH}_3 \mathrm{COONa}(a q)+\mathrm{H}_2 \mathrm{O}(l) ; \Delta H_N=-55.9 \mathrm{~kJ}\)

⇒\(\mathrm{HCN}(a q)+\mathrm{NaOH}(a q) \rightarrow \mathrm{NaCN}(a q)+\mathrm{H}_2 \mathrm{O}(l)\)

⇒  \(\Delta H_N=-12.1 \mathrm{~kJ}\)

⇒\(\mathrm{HCl}(a q)+\mathrm{NH}_4 \mathrm{OH}(a q) \rightarrow \mathrm{NH}_4 \mathrm{Cl}(a q)+\mathrm{H}_2 \mathrm{O}(l)\)

⇒ \(\Delta H_N=-51.4 \mathrm{~kJ}\)

Explanation:

A weak acid or weak base partially ionizes in aqueous solution. The ionization of a weak acid or base keeps on going during its neutralization process. Heat is absorbed in the ionization process. A part of the heat evolved during the reaction between H+ and OH- ions is utilized for the ionization of weak acid or weak base.

Hence, the value of heat of neutralization of strong acid and strong base is numerically greater than that associated with a neutralization process in which either acid or base or both are weak.

As the heats of ionization of different weak acids or weak bases are different, the value of the heat of neutralization of a weak acid by a strong base is different for a different weak acid. Similarly, the heat to be liberated in the neutralization of a weak base by a strong acid depends on the nature ofthe weak base.

In an acid-base neutralization, if the acid or base is weak, then the heat of neutralization = the heat of ionization of the weak acid (or weak base) + the heat of reaction for the reaction;\(\left[\mathrm{H}^{+}(a q)+\mathrm{OH}^{-}(a q) \rightarrow \mathrm{H}_2 \mathrm{O}(l)\right]\) Using this equation the heat of ionization of the weak acid or weak base can be determined.

Numerical Examples

Question 1. Determine the heat of neutralization for the following neutralization reactions: 100 mL of 0.2 (M) HCl solution is mixed with 200 mL of 0.15 (M) NaOH solution. 200 mL of 0.4 (M) H₂SO₄ solution is mixed with 300 mL of 0.2 (M) KOH solution.
Answer:

The amount of HCI in 100 mL of 0.2 (M) HCI solution

= \(\frac{0.2}{1000} \times 100=0.02\) mol and the amount of NaOH in 200 of 0.15 (M) NaOH solution

= \(\frac{0.15}{1000} \times 200=0.03 \mathrm{~mol}.\) 0.02 mol of H+ ions are produced in the ionization of 0.02 (M) aqueous HCl solution & 0.03 mol OH ions are produced in the ionization of 0.03(M) aqueous NaOH solution.

Therefore, if 100 mL of 0.2 (M) HCl is mixed with 200 mL of 0.15 (M) NaOH, then 0.02 mol of H₂O(/) will be formed from a die reaction between 0.02 mol H⁺ ions and 0.02 mol OH^–ions. Thus heat evolved (i.e., heat of neutralisation) for tills reaction = 0.02 × 57.3 = 1.146 kJ.

⇒ \(\text { (2) } 200 \mathrm{~mL} \quad 0.4 \text { (M) } \mathrm{H}_2 \mathrm{SO}_4=\frac{0.4}{1000} \times 200=0.08 \mathrm{~mol}\)

⇒ \(\mathrm{H}_2 \mathrm{SO}_4 \& 300 \mathrm{~mL} 0.2(\mathrm{M}) \mathrm{KOH}=\frac{0.2}{1000} \times 300=0.06 \mathrm{~mol} \mathrm{KOH.}\)

The amount of H+ ions formed from the complete ionization of O.OOmol of aqueous H2SO4 = 2 × 0.00

=0.16mol

since 2 mol H⁺ is formed from 1 mol of H₂SO₄] and that of OH^– ions formed due to dissociation of 0.06 (m) KOH =0.06 mol since 2 mol H⁺ is formed from 1 mol of H₂SO₄] and that of OH^– ions formed due to dissociation of 0.06(M) KOH=0.06 mol

Therefore, if 200 mL of 0.4 (M) H₂SO₂ solution is mixed with 300 mL of 0.2 (M) KOH solution then effectively 0.06 mol H₂O(Z) will be formed from the reaction between 0.06 mol OH- ions and the same amount of H+ ions.

Thus the amount of heat produced (i.e., the heat of neutralization) of this reaction = 0.06 × 57.3

= 3.438 kJ

Question 2. The heat of neutralization of acetic acid and NaOH is 55.9 kj. If the heat of neutralization of all strong acids and strong bases is 57.3 kj, then calculate the heat of ionization of acetic acid.
Answer:

The neutralization reaction of the strong acid-strong base in an aqueous solution is:

⇒ \(\mathrm{H}^{+}(a q)+\mathrm{OH}^{-}(a q) \rightarrow \mathrm{H}_2 \mathrm{O}(l)\) The change in enthalpy in this reaction =-57.3 kj.

The heat of neutralization in the reaction of a weak acid (acetic acid) and strong base = heat of ionization of acetic acid + heat of reaction for

⇒ \(\left[\mathrm{H}^{+}(a q)+\mathrm{OH}^{-}(a q) \rightarrow \mathrm{H}_2 \mathrm{O}(l)\right].\).

Given, -55.9kJ=heat of ionization of acetic acid -57.3 kj. Therefore, heat of ionisation of acetic acid = 57.3- 55.9 = 1.4 kJ.mol^-1

Heat or enthalpy of solution

Generally, when a solute is dissolved in a solvent, heat is evolved or absorbed. At a given temperature the amount of heat absorbed or evolved depends upon the amount and nature of both solvent and solute.

Heat or enthalpy of solution Definition:

At a given temperature, the heat (or enthalpy) change associated with the dissolution of 1 mol of a solute in a specified amount of solvent so that (further addition of solvent will not produce any significant thermal effect) is termed as heat (or enthalpy) of solution of that solute at that temperature.

Therefore, if 100 mL of 0.2 (M) HCl is mixed with 200 mL of 0.15 (M) NaOH, then 0.02 mol of H₂O(/) will be formed from a die reaction between 0.02 mol H⁺ ions and 0.02 mol OH^– ions.

Thus heat evolved (i.e., heat of neutralisation) for tills reaction = 0.02 × 57.3 = 1.146 kJ.

Explanation:

The Addition Of More Solvent to the solution (i.e., alter dissolution of solute) causes dilution of the solution and enthalpy changes In a continuously decreasing manner. At 25°G, the heat absorbed for the dissolution of 1 mol of KCl in 50 mL of water Is 171191 J. Addition of another 50 mL of water to the tills solution caused absorption of 401. 1J of heat.

If we again add 50 mL of water to the solution, then 142.3 of heat is absorbed. Further addition of 50 mL of water leads to an absorption of 100.4 J of heat. If water is again added to the solution, no change in enthalpy is found to occur. Therefore, the heal of solution of KCl at 25°C = (17091 +401.1 + 142.2+ 100.4) f =10614.7 J.

The dissolution of I mol of KCl in a sufficient amount of water can be expressed by the thermochemical equation as \(\mathrm{KCl}(s)+a q \rightarrow \mathrm{KCl}(a q); \Delta H=+18.61 \mathrm{~kJ}\) In this equation, ‘ aq ‘ indicates a large amount of water and KCl(ag) indicates infinitely diluted aqueous solution of KCl.

⇒ \(\text { Similarly, } \mathrm{CuSO}_4(s)+a q \rightarrow \mathrm{CuSO}_4(a q) ; \Delta H=-66.5 \mathrm{~kJ}\)

⇒ \(\mathrm{H}_2 \mathrm{SO}_4(l)+a q \rightarrow \mathrm{H}_2 \mathrm{SO}_4(a q) ; \Delta H=-96.2 \mathrm{~kJ}\)

⇒ \(\mathrm{HCl}(g)+a q \rightarrow \mathrm{HCl}(a q) ; \Delta H=-74.8 \mathrm{~kJ}\)

Generally, the heats of the solution are found to be positive for the hydrated salts (like CuSO₄-5H₂O, MgSO₄-7H₂O, FeSO₄-7H₂O, etc.) and the salts which cannot form stable hydrates (like NaC₁, KC₁, NH₄C_1, etc.). For anhydrous salts such as CuSO₄, CaCl₂, MgCl₂, etc., the heat of the solution is normally negative.

Integral heat Of solution:

At a specified temperature, the change in enthalpy associated with the dissolution of one mole ofa solute in a specified amount of a solvent is known as the integral heat solution. For example,

The values of ΔH in equations (1) and (2) indicate the integral heats of the solution when the amounts of water are 12 mol and 100 mol respectively. As the integral heat of the solution depends upon the amount of solvent, it is necessary to specify the amount of solvent while reporting the integral heat of the solution.

Heat or enthalpy of dilution

The dilution of a solution increases with the addition of more and more solvent to the solution, and consequently, heat is either absorbed or evolved in the process.

Integral heat of dilution

Integral heat of dilution Definition:

It Is defined as the change In heat (or enthalpy) when the concentration of a solution containing one one mole of the solute is changed (diluted) by adding more solvent. The heat (or enthalpy) of dilution is equal to the difference between the integral heats of the solution at the two concentrations.

Integral heat of dilution Explanation:

At 25°C when 1 mol of HCl(g) is dissolved In 25 mol of water, 72.3 kj of heat is evolved,

⇒ \(\mathrm{HCl}(\mathrm{g})+25 \mathrm{H}_2 \mathrm{O}(l) \rightarrow \mathrm{HCl}\left(25 \mathrm{H}_2 \mathrm{O}\right)\)

\(\Delta H=-72.3 \mathrm{~kJ}\) ………………………………..(1)

Again, if 1 mol of HCl(g) is dissolved in 40 mol of water, then 73.0 kj of heat is evolved

⇒ \(\mathrm{HCl}(g)+40 \mathrm{H}_2 \mathrm{O}(l) \rightarrow \mathrm{HCl}\left(40 \mathrm{H}_2 \mathrm{O}\right); \Delta H=-73.0 \mathrm{~kJ}\)………………………………..(2)

⇒ \(\mathrm{HCl}\left(25 \mathrm{H}_2 \mathrm{O}\right)+15 \mathrm{H}_2 \mathrm{O}(l) \rightarrow \mathrm{HCl}\left(40 \mathrm{H}_2 \mathrm{O}\right)\)

⇒ \(\Delta H=(-73.0+72.3)=-0.7 \mathrm{~kJ}=-700 \mathrm{~J}\)

Thus, the addition of 15 mol of water to the solution containing 1 mol of dissolved HC1 in 25 mol of water evolves 700J of heat.

Therefore, for the process \(\mathrm{HCl}\left(25 \mathrm{H}_2 \mathrm{O}\right) \rightarrow \mathrm{HCl}\left(40 \mathrm{H}_2 \mathrm{O}\right) \text {, }\) the heat of dilution = -700 J.

Hess’s Law Of Constant Heat Summation

Hess’s Law Of Constant Heat Summation Law:

If the reaction is carried out in one step or a series of steps, then the change in enthalpy in both cases will be the same provided that the initial and final states are the same in both cases

Explanation: Let us consider, for example, the reaction A → D, which is carried out by two different processes. In process 1, A Is directly converted into D, and the enthalpy change in the process is A. In process 2, which consists of two steps, A is first converted into C, and then C is converted into D . If the enthalpy changes in two steps (A→C) and (C→D) of process 2 are ALL, and AH₃, respectively, then according to the Hess’s law, AH₂ = A H₂ + AH₃.

Examples: CO₂ can be prepared from carbon in two ways: Direct oxidation of graphite into CO₂ (one step): C(graphite, s) + O₂(g)→CO₂(g); A = -393.5 kj

C is first oxidized to CO, and then CO is oxidized to CO₂ (two steps):

⇒ \(\mathrm{C}(\text { graphite, } s)+\frac{1}{2} \mathrm{O}_2(g) \rightarrow \mathrm{CO}(g)\)

⇒ \(\Delta H_2^0=-110.5 \mathrm{~kJ}\)

⇒ \(\mathrm{CO}(g)+\frac{1}{2} \mathrm{O}_2(g) \rightarrow \mathrm{CO}_2(g) ; \Delta H_3^0=-283.0 \mathrm{~kJ}\)

According to the Hess’s law

⇒ \(\Delta H_2^0+\Delta H_3^0=[-110.5+(-283.0)] \mathrm{kJ}=-393.5 \mathrm{~kJ}\)

= \(\Delta H_1^0\)

Thermodynamic explanation of Hess’s law:

As enthalpy is a state function, for any chemical reaction, the change in enthalpy (AH) depends only on the states of the reactants (initial state) and the products (final state), and not on how the change is brought about. This means that the change in enthalpy does not depend upon the number of intermediate states in a process.

Therefore, if a reaction is carried out by a process involving one step or by a process involving multiple steps, the change in enthalpy will be the same in either case. Hess’s law is simply a corollary of the first law of thermodynamics: Let us consider the reaction A + B→+D, which is carried out by two alternative processes I and II as given below.

Process-1: A + B → D; ΔH₁ = -xkJ

Process 2: A + B→ C; ΔH₂ = -ykJ

C → D; ΔH₃ = -zkJ

According to Hess’s law, if the initial state and final state are the same, then x = y + z. Let (y + z) > x. This means that preparing D from A and B by process 2 and converting D back into A and B by following the process will lead to an evolution of [(y + z)- x] amount of heat without any input of energy from outside. This contradicts the first law of thermodynamics. So, (y + z) & x must be equal, which corroborates the Hess’s law.

So, Hess’s law regarding heat change in a chemical reaction is a corollary of the first law of thermodynamics.

Applications of Hess’s law

The importance of Hess’s law lies in the fact that thermochemical equations can be treated as algebraic equations. They can be added or subtracted as we do with algebraic equations.

There are many reactions whose heats of reaction are not possible to be measured directly by experiment. However, we can determine the heat of reaction for their reactions in an indirect way by making use of Hess’s law. A few applications of Hess’s law are given below.

Calculation of the heats of formation of the compounds whose heats of formation cannot be determined directly:

The formation of many compounds such as CH₄(g), C₂H₂(g), and C₂H₅OH(Z) from their constituent elements is practically impossible.

So, we cannot directly determine the heat of formation of these compounds. However, we can do so indirectly by applying Hess’s law.

Example: Determination of the standard heat of formation of C₂H₂(g) from the following information:

⇒ \(\mathrm{C}_2 \mathrm{H}_2(\mathrm{~g})+\frac{5}{2} \mathrm{O}_2(\mathrm{~g}) \rightarrow 2 \mathrm{CO}_2(\mathrm{~g})+\mathrm{H}_2 \mathrm{O}(l) ;\)

⇒ \( \Delta H^0=-1300 \mathrm{~kJ}\)……………………(1)

⇒ \(\mathrm{C}(\text { graphite, } s)+\mathrm{O}_2(\mathrm{~g}) \rightarrow \mathrm{CO}_2(\mathrm{~g})\)

⇒ \(\Delta H^0=-393.5 \mathrm{~kJ}\)……………………(2)

⇒ \(\mathrm{H}_2(\mathrm{~g})+\frac{1}{2} \mathrm{O}_2(\mathrm{~g}) \rightarrow \mathrm{H}_2 \mathrm{O}(l)\)

⇒ \(\Delta H^0=-285.8 \mathrm{~kJ}\) ……………………(3)

Reversing equation (1)_ and multiplying equation [2] by 2,

⇒ \(2 \mathrm{CO}_2(g)+\mathrm{H}_2 \mathrm{O}(l) \rightarrow \mathrm{C}_2 \mathrm{H}_2(\mathrm{~g})+\frac{5}{2} \mathrm{O}_2(\mathrm{~g})\)

⇒ \(\Delta H^0=+1300 \mathrm{~kJ}\) ……………………(4)

⇒ \(\text { and } 2 \mathrm{C}(\text { graphite, } s)+2 \mathrm{O}_2(\mathrm{~g}) \rightarrow 2 \mathrm{CO}_2(\mathrm{~g})\)

⇒ \(\Delta H^0=2 \times(-393.5)=-787 \mathrm{~kJ}\) ……………………(5)

Adding equations (3), (4) and (5), we obtain,

⇒ \(2 \mathrm{C} \text { (graphite, } s)+\mathrm{H}_2(\mathrm{~g}) \rightarrow \mathrm{C}_2 \mathrm{H}_2(\mathrm{~g})\)

⇒ \(\Delta H^0=[-285.8+1300-787] \mathrm{kJ}=+227.2 \mathrm{~kJ}\) ……………………(6)

Since equation [6] is the formation reaction of C₂H₂(g), the ΔH° of this reaction represents the standard heat of formation of C₂H₂(g). Hence, the standard enthalpy of
formation

⇒ \(\mathrm{C}_2 \mathrm{H}_2(\mathrm{~g})=+227.2 \mathrm{~kJ} \cdot \mathrm{mol}^{-1} \text {. }\)

Calculation of heat of reaction for the reactions whose heats of reaction cannot be determined directly:

The heats of reaction for many chemical reactions cannot be determined directly. \(\mathrm{C}(\text { graphite, } s)+\frac{1}{2} \mathrm{O}_2(\mathrm{~g}) \rightarrow \mathrm{CO}(\mathrm{g}),\) cannot be determined directly because when 1 mol of solid graphite reacts with 0.5 mol of O₂(g), CO₂(g) along with CO(g) is produced.

However, it is possible to calculate the heat of the reaction indirectly by applying Hess’s law

Example: Let us consider (the determination of heat of a reaction in which 1 mol of CO(g) is formed from the reaction of solid graphite with oxygen;

⇒ \(C(g r a p h i t e, s)+\frac{1}{2} \mathrm{O}_2(g) \rightarrow CO(g);

\Delta H^0\) = ? …………………………….(1)

CO₂(g) he prepared by the complete combustion of both solid graphite and CO(g). From the heats of reactions of the following two reactions, the heat of reaction for the reaction can easily be calculated.

⇒ \(\text { C(graphite, s) }+\mathrm{O}_2(g) \rightarrow \mathrm{CO}_2(g) ; \Delta H^0=-393.5 \mathrm{~kJ}\) …………………………….(2)

\(\mathrm{CO}(g)+\frac{1}{2} \mathrm{O}_2(g) \rightarrow \mathrm{CO}_2(g) ; \Delta H^0=-283.0 \mathrm{~kJ}\) …………………………….(3)

Adding equation (2) to the reverse of equation(3)we get

⇒ \(\mathrm{C}(\text {graphite,}s)+\mathrm{O}_2(\mathrm{~g})+\mathrm{CO}_2(\mathrm{~g})\longrightarrow\)

⇒ \(\mathrm{CO}_2(\mathrm{~g})+\mathrm{CO}(\mathrm{g})+\frac{1}{2}\mathrm{O}_2(\mathrm{~g}) \)

⇒ \(\Delta H^0=[+283.0-393.5] \mathrm{kJ}=-110.5 \mathrm{~kJ}\)

∴ \(\mathrm{C}(\text { graphite, } s)+\frac{1}{2} \mathrm{O}_2(\mathrm{~g}) \rightarrow \mathrm{CO}(\mathrm{g})\)

⇒ \(\Delta H^0=-110.5 \mathrm{~kJ}\)

Therefore, the heat of reaction for reaction = -110.5kJ

Determination of heat of transition:

Some elements (like C, S, P, etc,) exist in two or more allotropic forms. During the allotropic transformation of such elements, heat is generally absorbed or evolved. The enthalpy change that occurs in an allotropic transformation is known as heat of transition.

At a given temperature and pressure, the heat or enthalpy change that occurs when 1 mol of an allotropic form of an element transforms into another form is called the heat of transition (AHtrn).

Example:

The two important allotropes of sulfur are rhombic sulfur and monoclinic sulfur. By applying Hess’s law, the heat of transition of these two allotropes can easily be calculated from their heats of combustion data.

⇒ \(\mathrm{S}(\text { monoclinic }, s) \rightarrow \mathrm{S}(\text { rhombic }, s) ; \Delta H^0=?\)

The heats of combustion of rhombic and monoclinic sulphur at 25°C temperature and 1 atm pressure are, -296.9 and -297.2 kj – mol^-1, respectively.

⇒ \(\mathrm{S}(\text { monoclinic, } s)+\mathrm{O}_2(\mathrm{~g})\rightarrow \mathrm{SO}_2(\mathrm{~g})\)

⇒ \(\Delta H^0=-297.2 \mathrm{~kJ} \cdot \mathrm{mol}^{-1}\) ………………(1)

⇒ \(\mathrm{S}(\text { rhombic, } s)+\mathrm{O}_2(\mathrm{~g})\rightarrow \mathrm{SO}_2(\mathrm{~g})\)

\(\Delta H^0=-296.9 \mathrm{~kJ} \cdot \mathrm{mol}^{-1}\) ………………(2)

Subtracting equation (2) from equation (1), we obtain, S(monoclinic, s) -+ Sfrhombic, s)

⇒\(\Delta H_{t r n}^0=[-297.2-(-296.9)] \mathrm{kJ} \cdot \mathrm{mol}^{-1}=-0.3 \mathrm{~kJ} \cdot \mathrm{mol}^{-1}\)

Therefore, at 25°C temperature and 1 atm pressure 0.3 kJ heat is evolved due to the transformation of lmol monoclinic sulfur to 1 mol rhombic sulfur.

Determination of heat of hydration:

Hess’s law can be applied to calculate the enthalpy changes associated with the hydration of several salts to form their corresponding hydrates.

Heat Of Hydration:

At a given temperature and pressure, the change in enthalpy accompanying the formation of l mol of a hydrate from the anhydrous form of the compound is called the heat of hydration of that anhydrous compound.

For example, at 25°C temperature and 1 atm pressure, the combination of1 mol of anhydrous MgSO₄(s), and 7 mol H₂O(Z) produces lmol of MgSO₄-7H₂O(s). It is accompanied by the liberation of 105 kj of heat. Therefore, the heat of hydration of MgSO₄(s) = -105 kj-mol^-1

⇒ \(\mathrm{MgSO}_4(s)+7 \mathrm{H}_2 \mathrm{O}(l) \rightarrow \mathrm{MgSO}_4 \cdot 7 \mathrm{H}_2 \mathrm{O}(s)\)

⇒ \(\Delta H_{\text {hyd }}^0=-105 \mathrm{~kJ} \cdot \mathrm{mol}^{-1}\)

The heat of hydration of a salt cannot be determined directly. However, it can be determined by applying Hess’s law from the known values of the heat of the solution of the hydrate and the anhydrous form of the salt.

Example:

⇒ \(\mathrm{CuSO}_4(s)+5 \mathrm{H}_2 \mathrm{O}(l) \rightarrow \mathrm{CuSO}_4 \cdot 5 \mathrm{H}_2 \mathrm{O}(s) ; \Delta H_{\text {hyd }}^0=?\)

At 25°C temperature and 1 atm pressure, the heat of the solution of CuS04(s) and CuS04-5H20(s) are -66.1 kj.mol^-1 and +11.5 kJ.mol^-1, respectively.

⇒ \(\mathrm{CuSO}_4(s)+n \mathrm{H}_2 \mathrm{O}(l) \rightarrow \mathrm{CuSO}_4(a q) ;\) [n is a very large number]

⇒ \(\Delta H_{\text {sol }}^0=-66.1 \mathrm{~kJ}\)

⇒ \(\mathrm{CuSO}_4+5 \mathrm{H}_2 \mathrm{O}(s)+(n-5) \mathrm{H}_2 \mathrm{O}(l) \rightarrow \mathrm{CuSO}_4(a q)\)

⇒ \(\Delta H_{\text {sol }}^0=+11.5 \mathrm{~kJ}\)

Subtracting equation (2) from equation (1), we obtain,

⇒ \(\mathrm{CuSO}_4(s)+5 \mathrm{H}_2 \mathrm{O}(l)\rightarrow \mathrm{CuSO}_4 \cdot 5 \mathrm{H}_2 \mathrm{O}(s)\)

⇒ \(\Delta H_{h y d}^0=[-66.1-(+11.5)] \mathrm{kJ}=-77.6 \mathrm{~kJ}\)

Therefore, at 25°C temperature and 1 atm pressure, the heat of hydration of anhydrous CuSO₂(s) = -77.6kJ.mol^-1

Standard Heat Of Formation Of An Ion

Heat or enthalpy change occurs when an ionic compound dissociates into ions in solution. So, like a compound, an ion also has the heat of formation.

Standard Heat Of Formation Of An Ion Definition:

Enthalpy or heat change associated with the formation of one mole of an ion in an infinitely dilute solution at standard state (i.e., at 1 atm pressure and a specified temperature) is termed as the standard heat of formation of that ion The determination of standard heat of formation of an ion is usually done at 25°C temperature. By convention, the standard heat of formation (or enthalpy of formation) of H+ ion in aqueous solution is taken to be zero.

Therefore \(\frac{1}{2} \mathrm{H}_2(g) \rightarrow \mathrm{H}^{+}(a q) ; \Delta H_f^0\left[\mathrm{H}^{+}(a q)\right]=0\left[25^{\circ} \mathrm{C}\right]\)

Example: Determination ofthe heat of formation of OH^–(aq):

1.  \(\mathrm{H}^{+}(a q)+\mathrm{OH}^{-}(a q) \rightarrow \mathrm{H}_2 \mathrm{O}(l) ; \Delta H^0=-57.3 \mathrm{~kJ} \cdot \mathrm{mol}^{-1}\)

2. \(\mathrm{H}_2(g)+\frac{1}{2} \mathrm{O}_2(g) \rightarrow\mathrm{H}_2 \mathrm{O}(l)\)

⇒ \( \Delta H_f^0\left[\mathrm{H}_2 \mathrm{O}(l)\right]=-285.8 \mathrm{~kJ} \cdot \mathrm{mol}^{-1}\)

By writing an equation in a reverse manner and then adding it to an equation, we obtain,

⇒ \(\mathrm{H}_2(g)+\frac{1}{2} \mathrm{O}_2(g) \rightarrow \mathrm{H}^{+}(a q)+\mathrm{OH}^{-}(a q)\)

⇒ \(\Delta H^0=-285.8+57.3=-228.5 \mathrm{~kJ} \cdot \mathrm{mol}^{-1}\)

Therefore \(\Delta H^0=\Delta H_f^0\left[\mathrm{H}^{+}(a q)\right]+\Delta H_f^0\left[\mathrm{OH}^{-}(a q)\right]\)

⇒ \( -\Delta H_f^0\left[\mathrm{H}_2(g)\right]-\frac{1}{2} \Delta H_f^0\left[\mathrm{O}_2(g)\right]\)

⇒ \(\text { or, }-228.5=0+\Delta H_f^0\left[\mathrm{OH}^{-}(a q)\right]-0-0\)

[In standard state, the heat of formation of any stable pure dement is taken as 0. So

⇒ \(\left.\Delta H_f^0\left[\mathrm{H}_2(g)\right]=\Delta H_f^0\left[\mathrm{O}_2(g)\right]=0\right]\)

∴ \(\Delta H_f^0\left[\mathrm{OH}^{-}(a q)\right]=-228.5 \mathrm{~kJ} \cdot \mathrm{mol}^{-1}\)

Alternative method: From the equation we get,

⇒ \(\Delta H^0=\Delta H_f^0\left[\mathrm{H}_2 \mathrm{O}(l)\right]-\Delta H_f^0\left[\mathrm{H}^{+}(a q)\right]-\Delta H_f^0\left[\mathrm{OH}^{-}(a q)\right]\)

[/latex]\text { or, }-57.3 \mathrm{~kJ}=-285.8 \mathrm{~kJ}-0-\Delta H_f^0\left[\mathrm{OH}^{-}(a q)\right][/latex]

∴ \(\Delta H_f^0\left[\mathrm{OH}^{-}(a q)\right]=-228.5 \mathrm{~kJ}\)

Heat Of Solution Of Ionic Compounds

The dissolution of an ionic compound (like NaCl, KC1, etc.) in a polar solvent (like water) can be considered as the combination ofthe following two processes:

Breaking of the crystal lattice of the compound into gaseous constitutes ions Interaction of resulting ions with the solvent molecules i.e., solvation of the ions (or hydration if water is used solvent). Process (1) is endothermic, while the process (2) is exothermic.

In an ionic compound [MX(s)], the energy required to transform one mole of the ionic crystal into its gaseous constituent ions (M+ and X-), separated by infinite distance, is equal to the reaction enthalpy of the following reaction and is called lattice enthalpy (ΔHl).

⇒ \(\mathrm{MX}(\mathrm{s}) \rightarrow \mathrm{M}^{+}(\mathrm{g})+\mathrm{X}^{-}(\mathrm{g}) \text {; reaction enthalpy }=\Delta H_L \quad \cdots[1]\)

The change in enthalpy accompanying the hydration (AHhy(J) ofthe cations [M+(g)] and anions [X'(g)], that formed on the dissociation of 1 mol of MX(s), is equal to the change in enthalpy of the following hydration process.

⇒ \(\mathrm{M}^{+}(g)+\mathrm{X}^{-}(g)+\text { water } \rightarrow \mathrm{M}^{+}(a q)+\mathrm{X}^{-}(a q)\)

\(\text { Change in enthalpy }(\Delta H)=\Delta H_{h y d}\)…………………………..(2)

The change in enthalpy accompanying the dissolution of 1 mol of an ionic compound such as MX(s) is termed the enthalpy of solution (AI1sol) ofthe compound.

⇒ \(\mathrm{MX}(s)+\text { water } \rightarrow \mathrm{M}^{+}(a q)+\mathrm{X}^{-}(a q)\) \(\text { Enthalpy change }(\Delta H)=\Delta H_{\text {sol }}\)…………………………..(3)

Adding equations(1) and (2) results in equation(3) Thus, Δ H_sol equals the sum of ΔH₂ and ΔH_hyd. This is per Hess’s law.

The value of ΔHl is always positive and that of ΔHhyd is always negative. Depending upon the magnitude of these two, the sign of the ΔH_sol will be either positive or negative. In the case of most ionic salts, the ΔH_sol is positive This is why the solubilities of such salts, increase with temperature rise.

Lattice Energy (Or Lattice Enthalpy) And Born-Haber Cycle

Lattice Energy (Or Lattice Enthalpy) And Born-Haber Cycle Definition:

The lattice energy of a solid ionic compound it The. the energy required to break 1 mol of the compound (at a particular temperature and pressure) into its gaseous ions, separated by an infinite distance.

Example:

The lattice enthalpy of NaCl at 25°C and ] atm pressure = +788 kj .mol^-1 means that at 256C temperature and 1 atm pressure, 788 kJ of heat is required to break mol of NaCl(s) into lmol of Na⁺g) & I mol of Cl^–(g) ions, separated by infinite distance.

Therefore, the value of the lattice enthalpy for MX(s) type of the compound is equal to the value of the enthalpy change for the process

⇒  \(\mathrm{MX}(\mathrm{s}) \rightarrow \mathrm{M}^{+}(\mathrm{g})+\mathrm{X}^{-}(\mathrm{g})\)

Determination of lattice enthalpy of an ionic compound a standard state by applying Hess’s law:

To illustrate how to determine the lattice energy of a solid ionic compound, let us consider, for example, the determination of the lattice energy of NaCl. NaCl can be prepared in a single step by reacting to its constituent elements or by an indirect process involving multiple steps.

Preparation of 1 mol of NaCl in a single step:

1 mol of metallic Na is reacted with 0.5 mol of Cl₂ gas to form lmol of solid NaCl. The enthalpy change in the reaction is equal to the standard enthalpy of the formation of solid sodium chloride (NaCl).

⇒ \(\mathrm{Na}(s)+\frac{1}{2} \mathrm{Cl}_2(g) \rightarrow \mathrm{NaCl}(s) ; \Delta H_f^0(\mathrm{NaCl})=-411.2 \mathrm{~kJ} \cdot \mathrm{mol}^{-1}\)

According to Hess’s law, if the initial and final states remain fixed, then the change in enthalpy for the transformation from the initial state to the final state is the same, regardless of whether the reaction is completed in one step or several steps.

Hence \(\Delta H_f^0(\mathrm{NaCl})=376.5-\Delta H_L^0 \text { or, }-411.2=376.8-\Delta H_L^0\)

∴ \(\Delta H_L^0=+788.0 \mathrm{~kJ} \cdot \mathrm{mol}^{-1}\)

Thus, the lattice enthalpy of solid NaCl at standard state = + 788 kj. mol^-1

Born-Haber cycle

Bom-Haber cycle is a thermochemical cycle consisting of a series of steps that describe the formation of an ionic solid from its constituent elements. It is based on Hess’s law and establishes a relation between the lattice energy of an ionic compound and the enthalpy changes that occur in various steps associated with the formation ofthe compound. This cycle is very useful for calculating lattice enthalpy and electron gain enthalpy (electron affinity), which are difficult to measure experimentally.

As enthalpy is a state function, the total change in enthalpy for a cyclic process is zero. Thus,

⇒ \(+411.2+108.4+121+496-348.6-\Delta H_L^0=0\)

∴ \(\Delta H_L^0=+788 \mathrm{~kJ} \cdot \mathrm{mol}^{-1}\)

Atomisation Enthalpy

Standard enthalpy of automation of an element:

The enthalpy change associated with the formation of 1 mol of gaseous atoms from a stable element when all substances are in their standard states is called the enthalpy of atomization \(\left(\Delta H_{\text {atom }}^0\right)\) of the element.

The atomization of any substance is an endothermic process. So, for any substance, \(\Delta \boldsymbol{H}_{\text {atom }}^0\) is always positive. The unit of \(\Delta H_{\text {atom }}^0\) kj mol-1 or kcal- mol-1.

Explanation:

The standard enthalpy of atomization of hydrogen at 25°C

⇒ \(\frac{1}{2} \mathrm{H}_2(\mathrm{~g}) \rightarrow \mathrm{H}(\mathrm{g}); \Delta H^0{ }_{\text {atom }}=+218 \mathrm{~kJ}\)

The standard enthalpy of atomization of graphite at 25°C, ΔH°fom = +717 kj-mol^-1: T

his means that at 25°C and 1 atm pressure, +717 kj heat (or enthalpy) is required to produce 1 mol of carbon vapor [C(g)] from graphite. Thus the change in enthalpy for the process, C(graphite, s) -C(g) at 25°C and 1 atm pressure, ΔH° = +717 kj-mol^-1.

At a particular temperature, for homonuclear diatomic gases, the heat (or enthalpy) of atomization is equal to half the heat of dissociation, i.e., the standard enthalpy of atomization \(=\frac{1}{2} \times \text { enthalpy of dissociation. }\)

At a particular temperature, for a monoatomic solid, the heat (or enthalpy) of atomization is equal to its heat of sublimation at that temperature.

Bond Dissociation Energy (Or energy enthalpy)(Or Enthalpy)

Chemical reactions are associated with the formation and dissociation of bond(s). During dissociation of a bond heat or energy is required, so it is an endothermic process. On the other hand, bond formation is associated with the evolution of heat. Hence it is an exothermic process. At a particular temperature and pressure, the amount of heat required to break a bond is equal to the amount of heat released to form that bond at the same conditions.

Bond Dissociation Enthalpy

Bond Dissociation Enthalpy Definition:

The Amount of energy required to dissociate one mole of a specific type of bond in a gaseous covalent compound to form neutral atoms or radicals is called bond dissociation energy (ΔHbond) of that bond. As bond dissociation is an endothermic process, the value of the bond dissociation energy is always positive. The bond dissociation energy is generally expressed in kJ. mol-1 or kcal^-1 mol^-1. Here the term ‘mol’ indicates per mole of bond.

The bond dissociation energy of a diatomic gaseous molecule:

The change in standard enthalpy (ΔH°) for the process

⇒ \(\mathrm{Cl}_2(\mathrm{~g}) \rightarrow 2 \mathrm{Cl}(\mathrm{g}) \text { at } 25^{\circ} \mathrm{C}\) and 1 atm pressure = 248 kj.mol^-1. In this process, the gaseous Cl is formed by the dissociation of 1 mol of Cl—Cl bonds atoms in a gaseous state.

Therefore, this standard enthalpy change is equal to the bond dissociation energy (or enthalpy) of the Cl—Cl bond \(\left(\Delta H_{\mathrm{Cl}}^0-\mathrm{Cl}\right)\) The change in standard enthalpy (ΔH°) for the process O₂(g) → 2O(g) at 25°C temperature and 1 atm pressure = 498 kj. mol^-1

In this process, the gaseous O atoms are formed by the dissociation of lmol of O = Obonds in the gaseous state at 25°C and 1 atm pressure. Therefore, this standard enthalpy change is equal to the bond dissociation energy of the O=O bond

⇒ \(\left(\Delta H_{\mathrm{O}}^0=0\right)\).

The bond dissociation energy of the bond in a multi-atomic molecule: If a molecule has more than one bond of a particular type, then the stepwise dissociation of these bonds requires different amounts of energy.

For example, in the NH₃ molecule, although three N—H bonds are equivalent stepwise dissociation of these bonds requires different amounts of energy.

⇒ \(\mathrm{NH}_3(\mathrm{~g}) \rightarrow \mathrm{H}(\mathrm{g})+\mathrm{NH}_2(\mathrm{~g}) ; \Delta H_{\text {bond }}^0=+431 \mathrm{~kJ} \cdot \mathrm{mol}^{-1}\)

⇒ \(\mathrm{NH}_2(\mathrm{~g}) \rightarrow \mathrm{H}(\mathrm{g})+\mathrm{NH}(\mathrm{g}) ; \Delta H_{\text {bond }}^0=+381 \mathrm{~kJ} \cdot \mathrm{mol}^{-1}\)

⇒ \(\mathrm{NH}_2(\mathrm{~g}) \rightarrow \mathrm{H}(\mathrm{g})+\mathrm{NH}(\mathrm{g}) ; \Delta H_{\text {bond }}^0=+381 \mathrm{~kJ} \cdot \mathrm{mol}^{-1}\)

If more than one similar type of bond is present in a molecule, then the average bond dissociation energy of those bonds is expressed as the bond energy of that bond.

Therefore, the bond energy of the N—H bond in NH₃ molecule \(=\frac{431+381+360}{3}=391 \mathrm{~kJ} \cdot \mathrm{mol}^{-1} \text {. }\) = 391 KJ. mol^-1

⇒ \(\mathrm{CH}_4(\mathrm{~g}) \rightarrow \mathrm{H}(\mathrm{g})+\mathrm{CH}_3(\mathrm{~g}) ; \Delta H_{\text {bond }}^0=+427 \mathrm{~kJ} \cdot \mathrm{mol}^{-1}\)

⇒ \(\mathrm{CH}_3(\mathrm{~g}) \rightarrow \mathrm{H}(\mathrm{g})+\mathrm{CH}_2(\mathrm{~g}) ; \Delta H_{\text {bond }}^0=+439 \mathrm{~kJ} \cdot \mathrm{mol}^{-1}\)

⇒ \(\mathrm{CH}_2(\mathrm{~g}) \rightarrow \mathrm{H}(\mathrm{g})+\mathrm{CH}(\mathrm{g}) ; \Delta H_{\text {bond }}^0=+452 \mathrm{~kJ} \cdot \mathrm{mol}^{-1}\)

⇒ \(\mathrm{CH}(\mathrm{g}) \rightarrow \mathrm{H}(\mathrm{g})+\mathrm{C}(\mathrm{g}) ; \Delta H_{\text {bond }}^0=+347 \mathrm{~kJ} \cdot \mathrm{mol}^{-1}\)

⇒ \(\mathrm{CH}(\mathrm{g}) \rightarrow \mathrm{H}(\mathrm{g})+\mathrm{C}(\mathrm{g}) ; \Delta H_{\text {bond }}^0=+347 \mathrm{~kJ} \cdot \mathrm{mol}^{-1}\)

Therefore, the bond energy ofthe C —H bond in CH4(g) , molecule

⇒  \(\text { molecule }=\frac{427+439+452+347}{4}=416.25 \mathrm{~kJ} \cdot \mathrm{mol}^{-1} \text {. }\)

Definition Of Bond Energy

The average value of the dissociation energies of all the similar types of bonds present in a gaseous compound is called the bond energy of that type of bond

Standard enthalpy of dissociation [Standard bond dissociation energy) of diatomic molecules at 25 °C:

Standard bond enthalpy or bond energy of some chemical bonds at 25°C:

Determination of bond enthalpy or bond energy

The standard bond enthalpy of a bond in a compound can be calculated from the known value of the standard enthalpy of formation of that compound and the value of the standard enthalpy of atomization of the elements constituting the compound. For example, consider the determination of bond enthalpy ofthe C—H bond in methane molecule.

In 1 mol of CH₄(g) molecule, the energy required to break 4 mol of C— H bonds is equal to the standard enthalpy of reaction for the following change (ΔH°).

CH₄(g) C(graphite, s) + 4H(g); ΔH°………………..(1)

The value of ΔH° can be calculated from the value of the standard enthalpy of formation of CH₄(g) and the standard enthalpy of atomization of C(graphite.s) and Ha(g).

The formation reaction of CH₄(g) and the atomisation reaction of C(graphite.s) and Ha(g) are as follows

⇒ \(\mathrm{C}(\text { graphite, } s)+2 \mathrm{H}_2(\mathrm{~g}) \rightarrow \mathrm{CH}_4(\mathrm{~g})\)

⇒ \( \left.\Delta \mathrm{H}_f^0\left[\mathrm{CH}_4(\mathrm{~g})\right]=-7.8 \mathrm{k}\right] \cdot \mathrm{mol}^{-1}\) …………………………………….(2)

⇒ \(\mathrm{C}(\text { graphite, } s) \rightarrow \mathrm{C}(\text { graphite, } g)\)

⇒ \( \left.\Delta H_{\text {atom }}^0=717 \mathrm{k}\right] \cdot \mathrm{mol}^{-1}\)……………(3)

⇒ \(\frac{1}{2} \mathrm{H}_2(\mathrm{~g}) \rightarrow \mathrm{H}(\mathrm{g})\)

⇒ \(\Delta H_{\text {atom }}^0=218 \mathrm{~kJ} \cdot \mathrm{mol}^{-1}\)

Equation (3) + 4 × equation (4)- equation (2) gives,

CH₄(g)→ C(graphite, g) + 4H(g); ΔH° = [717 + 4 ×  218- (-74.8)]kJ = 1663.8k]

Therefore, in 1 mol of CH₄(g) molecule, the energy required to break 4 mol of C —H bonds = 1663.8k).

Energy required to break1 mol of C — H bonds

⇒\(\frac{1}{4} \times 1663.8 \mathrm{~kJ}=415.95 \mathrm{~kJ} \cdot \mathrm{mol}^{-1}\)

So, the bond energy of the C— H bond = 415.95 kJ. mol^-1

Determination of the standard enthalpy of a gaseous reaction from the values of bond enthalpies of the reactant(s) and product(s)

In general, chemical reactions are associated with the breaking of bonds (on the reactant side) and the formation of new bonds (on the product side). When a chemical bond is formed energy is released.

On the other hand, energy is required to break a bond. The standard enthalpy change in a reaction in the gaseous state is related to the bond energies of the reactants and products.

If the total bond energy of all the bonds in reactant molecules is reactant tie clane in enthalpy for breaking all the bonds in reactant molecules

⇒ \(\Delta H_1^0=\sum(\mathrm{BE})_{\text {reactant }} \text {. }\) Similarly if the total bond energy of the product(s) be

⇒  \(\text { (BE) product, }\), then the change in enthalpy for the formation of bonds in product molecules

⇒ p\(\Delta H_2^0=-\sum(\mathrm{BE})_{\text {product }} \) [-ve sign indicates that heat is released during the formation.

⇒ \(\Delta H^0=\Delta H_1^0+\Delta H_2^0=\sum(\mathrm{BE})_{\text {reactant }}-\sum(\mathrm{BE})_{\text {product }}\)

∴ \(\Delta H^0=\sum(\mathrm{BE})_{\text {reactant }}-\sum(\mathrm{BE})_{\text {product }}\)

Using equation [1],

It is possible to determine the standard reaction enthalpy of a gaseous reaction from the bond energy data ofthe reactants and products. \(\text { If } \sum(\mathrm{BE})_{\text {reactant }}>\sum(\mathrm{BE})_{\text {product }} \text {, then } \Delta H^0>0 \text {. }\)

This means that if the total bond energy of all the bonds in reactant molecules is greater than that of all the bonds in product molecules, the reaction will be endothermic.

On the other hand, if

⇒ \(\sum(\mathrm{BE})_{\text {reactant }}<\sum(\mathrm{BE})_{\text {product, }} \text { then } \Delta \boldsymbol{H}^{\mathbf{0}}<\mathbf{0}\)

⇒ \(\Delta H^0<0\) and the willl be exothedrmic.

Numerical Examples

Question 1. Calculate The bond energy of O-H Bond In H₂O(g) at the standard state from the following data;

⇒ \(\mathrm{H}_2(g) \rightarrow 2 \mathrm{H}(\mathrm{g}) ; \quad \Delta H^0=436 \mathrm{~kJ} \cdot \mathrm{mol}^{-1}\)

⇒ \(\frac{1}{2} \mathrm{O}_2(g) \rightarrow O(g) ; \quad \Delta H^0=249 \mathrm{~kJ} \cdot \mathrm{mol}^{-1}\)

⇒ \(\mathrm{H}_2(g)+\frac{1}{2} \mathrm{O}_2(g) \rightarrow \mathrm{H}_2 \mathrm{O}(g)\)

⇒ \(\Delta H_f^0\left[\mathrm{H}_2 \mathrm{O}(g)\right]=-241.8 \mathrm{~kJ} \cdot \mathrm{mol}^{-1}\)

Answer: Equation + equation (2)- equation gives,

⇒ \(\mathrm{H}_2(g)+\frac{1}{2} \mathrm{O}_2(g)-\mathrm{H}_2(g)-\frac{1}{2} \mathrm{O}_2(g) \rightarrow 2 \mathrm{H}(g)+\mathrm{O}(g)-\mathrm{H}_2 \mathrm{O}(g)\)

⇒\(\Delta H^0=(436+249+241.8) \mathrm{kJ}\)

Or, \(\mathrm{H}_2 \mathrm{O}(\mathrm{g}) \rightarrow 2 \mathrm{H}(\mathrm{g})+\mathrm{O}(\mathrm{g})\)

⇒ \(\Delta H^0=+926.8 \mathrm{~kJ}\)

The equation indicates the dissociation of O—H bonds Present in 1 mol of H₂O(g). The standard enthalpy change for this process (ΔH°) = +926.8kJ

No. of O— H bonds present in lmol H₂O =2mol.

Thus, the energy required to break 2mol O—H bonds =+926.8 kJ.

∴ The energy required to break 1 mol O — H = + 463.4 kj So, bond energy of O—H bond = + 463.4 kj.mol^-1

Question 2. Calculate the S—F bond energy in the SF₆ molecule. Given: Enthalpy of formation for SF₆(g) , S(g),F(g) are -1100, 275, 80 kj.mol-1 respectively.
Answer:

⇒ \(\frac{1}{8} \mathrm{~S}_8(s) \rightarrow \mathrm{S}(g) ; \Delta H_f^0[\mathrm{~S}(g)]=275 \mathrm{~kJ} \cdot \mathrm{mol}^{-1}\)

⇒  \(\frac{1}{2} \mathrm{~F}_2(g) \rightarrow \mathrm{F}(g) ; \Delta H_f^0[\mathrm{~F}(g)]=80 \mathrm{~kJ} \cdot \mathrm{mol}^{-1}\)

⇒ \(\frac{1}{8} \mathrm{~S}_8(s)+3 \mathrm{~F}_2(g) \rightarrow \mathrm{SF}_6(g)\)

⇒ \(\Delta H_f^0\left[\mathrm{SF}_6(g)\right]=-1100 \mathrm{~kJ} \cdot \mathrm{mol}^{-1}
\)

No. of moles of S—F bonds in lmol SF₆ molecule = 6. In 1 mol of SF₆, the breaking of S- the reaction gives F bonds:

SF₆(g) -S(g) + 6F(g) Standard enthalpy for this reaction is

⇒ \(\Delta H^0=\Delta H_f^0[\mathrm{~S}(g)]+6 \Delta H_f^0[\mathrm{~F}(g)]-\Delta H_f^0\left[\mathrm{SF}_6(g)\right]\)

= \([275+6 \times 80-(-1100)] \mathrm{kJ}=1855 \mathrm{~kJ}\)

Thus 1855 kj energy is required to break 6 mol of S — F bonds. Hence, the energy required to break lmol of S —F bond \(=\left(\frac{1}{6} \times 1855\right)=309.166 \mathrm{~kJ} \text {. }\)

So, the bond energy of the S—F bond = 309.166 kj.mol^-1

Question 3. Determine the standard enthalpy of formation of isoprene(g) at 298 K temperature. Given:

⇒ \(\Delta H^0(\mathrm{C}-\mathrm{H})=413 \mathrm{~kJ} \cdot \mathrm{mol}^{-1}\)

⇒\(\Delta H^0(\mathrm{H}-\mathrm{H})=436 \mathrm{~kJ} \cdot \mathrm{mol}^{-1}\)

⇒ \(\Delta H^0(\mathrm{C}-\mathrm{C})=346 \mathrm{~kJ} \cdot \mathrm{mol}^{-1}\)

⇒ \(\Delta H^0(\mathrm{C}=\mathrm{C})=611 \mathrm{~kJ} \cdot \mathrm{mol}^{-1}\)

⇒ \(\mathrm{C}(\text { graphite }, s) \rightarrow \mathrm{C}(\text { graphite }, \mathrm{g}) ; \)

⇒ \(\Delta H^0=717 \mathrm{~kJ} \cdot \mathrm{mol}^{-1}\)

Answer: The formation reaction of isoprene [C4Hg(g)]:

⇒ \(5 \mathrm{C} \text { (graphite, } s)+4 \mathrm{H}_2(g) \rightarrow \mathrm{H}_2 \mathrm{C}=\mathrm{C}-\mathrm{CH}=\mathrm{CH}_2(g) \quad \cdots[1]\)

This reaction can be considered as the sum of the following two reactions:

According to Hess’s law, the total change in enthalpy of these two reactions is equal to the change in enthalpy of the reaction [1],

Therefore; \(\Delta H^0=\Delta H_1^0+\Delta H_2^0\)

⇒ \(\Delta H_1^0\) = Enthalpy of atomization of 5 mol C(g) + Bond energy of 4 mol H —H bonds

= (5 × 717 + 4 × 436) = 5329 kJ

⇒ \(\Delta H_2^0=(-)\) Bond energy of 8 mol (C—H) + Bond energy of 2mol (C=C) bonds + Bond energy of 2 mol (C —C) bonds]

[Here, the – ve sign indicates the bond is formed in step (2)] =-[8 × 413 + 2 × 611 + 2 ×  346]kJ

=-5218 kJ

Therefore \(\Delta H^0=(5329-5218) \mathrm{kJ}=111 \mathrm{~kJ} .\)

Determination Of The Value Of Δu And ΔH: Calorimetry

The process of measuring the amount of heat transferred during any physical or chemical transformation Is called calorimetry. The device by which the amount of heat transferred is measured is called a calorimeter. The heat change at constant pressure {qp) and that at constant volume (qv) are determined by calorimetry.

The heat change at constant pressure is equal to the change In enthalpy (AH) of the system, and the heat change at constant volume is equal to the change in internal energy (AH) of the system, Among the different types of calorimeters we will discuss here only the bomb calorimeter. The heat of the reaction at constant volume can be measured by using a bomb calorimeter.

Bomb calorimeter:

In general, a bomb calorimeter is used to determine the heat of combustion of a reaction at a constant volume.

• The bomb is a rigid closed steel container that can resist high pressures and the inside portion of the container is coated with platinum metal.
• A known amount of a substance (whose heat of combustion is to be determined) is taken in a platinum crucible and placed inside the bomb.
• The bomb is then filled with excess O₂ by passing pure O₂ at 20-25 atm pressure through a valve.
• The bomb is now immersed in an insulated, water-filled container fitted with a mechanical stirrer and thermometer. The sample present in the crucible is then ignited electrically in the presence of oxygen.
• During combustion, heat is evolved. As the calorimeter is insulated, heat evolved during combustion cannot escape.

Evolved heat is absorbed by the bomb, water, and other parts of the calorimeter. As a result, the temperature of the calorimeter increases, which is recorded from a thermometer.

Calculation If the heat capacity of the calorimeter is cal and the Increase in temperature due to absorption of heat by the calorimeter is ΔT,

Then the beat of reaction,

⇒ \(q_{\text {reaction }}=(-) \text { heat absorbed by the calorimeter }\)

=\(-q_{\text {calorimeter }}=-C_{c a l} \times \Delta T\)

Determination of Using die same bomb calorimeter and die same amount of water, a substance of known heat of combustion is burnt In the calorimeter, and AT is measured. From the amount of the substance of known heat of combustion, we calculate the reaction Putting the values of reaction end ΔT into the above relation, we get the value of C cal.

As the wall of the bomb calorimeter is very rigid, the volume of the die system remains unchanged (ΔV = 0) during a reaction, ffence a/ [=- PΔ P =0 and according to the first law of thermodynamics, All = \(\Delta U=q+w=q+0=q_V.\) Thus, the heat of reaction for a reaction occurring In a bomb calorimeter ~ changes in internal energy of the reaction system.

Numerical Examples

Question 1. When 1.0 g of a compound (molecular weight = 28) Is burnt In a bomb calorimeter, the temperature of the calorimeter rises from 25′-C to 25.45’T). Calculate the heat that evolves when 1 mol of this compound is completely burnt (Ccal = 2.5 kj – K-1 ).
Answer:

ΔT = [(273 + 2545)-(273 + 25)] = 0.45 K

Therefore, heat evolved due to the combustion of 1 g of the compound

= \(C_{c a l} \times \Delta T=2.5 \mathrm{~kJ} \cdot \mathrm{K}^{-1} \times 0.45 \mathrm{~K}=1.125 \mathrm{~kJ}.\)

∴ Feat produced due to combustion of1 moJ or 28 g of that compound J.125 × 28 = 31.5 kj.

Question 2. At 25-C, the heat of combustion at a constant volume of 1 mol of a compound Js 5150 of. The temperature of a bomb calorimeter rises from 25- C to 30.5°C when a certain amount of the compound Is burnt In It. If the heat capacity of the calorimeter Is 9.76 kJ .K^-1 then how much of the compound was taken for combustion? (Molar mass of the substance = 128
Answer:

As given in the question, the grain-molecular weight of that compound = 128 g- mol^-1

Therefore, the amount of heat that evolves in the combustion of 128 g ofthe substance = 5150 kJ.

Again \(\text { Again, } \Delta T=[(273+30.5)-(273+25)] \mathrm{K}=5.5 \mathrm{~K} \text { and }\)

Thus, die amount of heat evolves when a certain amount of compound is burnt = Ccal × AT = 9.76 × 5.5 = 53.68 kJ

Now 5150 kj of heat is liberated due to the combustion of the 128g compound. Hence, the amount ofthe compound required for the evolution of53.68 kJ of heat

⇒  \(\frac{128}{5150} \times 53.68=1.334 \mathrm{~g} .\)

The Second Law Of Thermodynamics

The second law of thermodynamics can be stated in various ways. Some common statements ofthe law are given below:

Clausisus statement:

It is impossible to construct a device, operating in a cycle, that will produce no effect other than the transfer of heat from a lower-temperature reservoir to a higher-temperature reservoir.

Planck -Kelvin statement:

It is impossible to construct an engine, operating in a cycle, that will produce no effect other than the extraction of heat from a reservoir and the performance of an equivalent amount of work.

Spontaneous And Non-Spontaneous Processes

A process that occurs on its own accord under a given set of conditions, without any outside assistance, is called a spontaneous process.

Examples:

1. The flow of heat from a hotter body to a colder body.
2. Evaporation of water kept in an open container at normal temperature.
3. Conversion of water to ice at a temperature below 0°C.
4. Rusting of iron.

Many processes require to be initiated by some external assistance. However, once initiated, they continue to occur under the prevailing conditions without any assistance from outside.

Examples:

1. No reaction occurs when a piece of coal is kept in an open atmosphere. However, once it is ignited, its combustion starts and continues spontaneously till it is completely burnt out with the formation of CO₂ and H₂O.
2. A candle when enkindled in air bums spontaneously till the end. Here, the combustion of candles (hydrocarbon) produces CO₂ and H₂O.
3. When an electric spark is created for once in a mixture of hydrogen and oxygen, the reaction between these two gases starts and goes on spontaneously at room temperature to produce water.

Non-spontaneous processes

A process that needs external assistance for its occurrence is called a non-spontaneous process.

Examples:

1. Underground water is lifted to the roof with the help of an engine. As long as the engine is in operation, water continuously goes up. But the upward flow of water ceases at the moment when the engine is stopped.
2. O₂ gas is produced when KClO₃ is heated with MnO₂. However, the evolution of oxygen ceases when the source of heat is removed.
3. Energy must be supplied from the external source to recoil a spring.

Spontaneity Of Physical Processes And Chemical Reactions

There are many physical processes or chemical reactions which occur spontaneously in preferred directions under certain conditions. For this type of process, energy is not required from any external agency.

A large number of processes, both exothermic and endothermic, are found to occur spontaneously. The question that comes into our mind is: ‘What is the driving force that makes such processes spontaneous

Factors affecting the spontaneity of a process

The tendency of a system to attain stability through lowering its energy or enthalpy:

Our experience shows that every system has a natural tendency to attain stability by lowering its intrinsic energy.

Examples:

1. When water falls from a high region, its potential energy goes on decreasing gradually and when it reaches the earth’s surface, its potential energy gets converted completely into a different form of energy.
2. Heat always flows from a hot body to a cold body. As a result, the internal energy of the hot body decreases.
3. From the above examples, we see that a process occurring spontaneously is accompanied by a decrease in the energy of the system.
4. In the case of chemical reactions, the decrease or increase in energy is usually in terms ofthe change in enthalpy (AH) in the reaction.
5. In a reaction, if the total enthalpy of the products Is less than that of the reactants, then heat is evolved, and hence AH = -ve, indicating that the reaction is exothermic. It has also been observed in actual experiments that most of the exothermic reactions occur spontaneously. For example

⇒ \(\mathrm{C}(\mathrm{s})+\mathrm{O}_2(\mathrm{~g}) \rightarrow \mathrm{CO}_2(\mathrm{~g})\)

⇒  \(\Delta H=-393.5 \mathrm{~kJ}\)

⇒  \(\mathrm{H}_2(\mathrm{~g})+\frac{1}{2} \mathrm{O}_2(\mathrm{~g}) \rightarrow \mathrm{H}_2 \mathrm{O}(l)\)

⇒  \(\Delta H–285.83 \mathrm{~kJ}\)

The fact that the decrease in energy of the reaction system in an exothermic reaction led the chemists to believe that only exothermic reactions would occur spontaneously. This means that the condition of spontaneity Is ΔH = -ve.

To put it in another way the decrease in enthalpy in a reaction is the driving force for the reaction to occur spontaneously. However, there are quite some reactions that do not occur spontaneously even though ΔH is negative for these reactions.

Spontaneous processes in which energy (enthalpy) of the system increases:

The endothermic reactions are associated with the increase in energy of the reaction system. It is probably not wrong to believe that such reactions would be nonspontaneous.

But, there are some processes, both physical and chemical, which occur spontaneously even though they are associated with the increase in enthalpy. Here are a few examples.

Although the evaporation of water at room temperature Is an endothermic process, It occurs spontaneously

⇒ \(\mathrm{H}_2 \mathrm{O}(l) \rightarrow \mathrm{H}_2 \mathrm{O}(\mathrm{g}); \Delta H=+44.1 \mathrm{~kJ} \cdot \mathrm{mol}^{-1}\)

The dissolution of NH4Cl in water Is a spontaneous process although this process is endothermic.

⇒ \(\mathrm{NH}_4 \mathrm{Cl}(s) \rightarrow \mathrm{NH}_4^{+}(a q)+\mathrm{Cl}^{-}(a q) ; \Delta H=+15.1 \mathrm{~kJ} \cdot \mathrm{mol}^{-1}\)

The reaction of Ba(OH)₂. 8H₂O(s) with NH4NO3(s) is so endothermic that it decreases the temperature of the reaction system to a very low value.

⇒ \(\mathrm{Ba}(\mathrm{OH})_2 \cdot 8 \mathrm{H}_2 \mathrm{O}(s)+2 \mathrm{NH}_4 \mathrm{NO}_3(s)\)

⇒ \(\mathrm{Ba}\left(\mathrm{NO}_3\right)_2(a q)+2 \mathrm{NH}_3(a q)+10 \mathrm{H}_2 \mathrm{O}(l)\)

Spontaneous process and tendency to increase the randomness of a system:

Our experience tells us that the natural tendency of any spontaneous process is that it tends to occur in a direction in which the system associated with the process moves from an order to a disordered state.

Examples:

Some red marbles are kept on one side of a tray, and on the other side the same number of blue marbles, identical in mass and size, are kept perfectly in order.

If the tray is shaken properly, the marbles of both types will be mixed (a disordered condition), and the system will be in a state of disorderliness, which is considered to be a more stable state of the system.

Because if that tray is shaken several times, the marbles will never return to their previous orderly arrangement i.e., the natural tendency of the system is to achieve a state of randomness from a well-ordered state. In ice, water molecules exist in an ordered state as their motions are restricted.

But, in liquid water, molecules have more freedom of motion as intermolecular forces of attraction in water are not as strong as in ice. Thus, the melting of ice leads to an increase in molecular disorder in the system.

⇒ \(\mathrm{H}_2 \mathrm{O}(s) \rightarrow \mathrm{H}_2 \mathrm{O}(l) ; \Delta H=+v e
(ice) (water)\)

⇒ \(\mathrm{H}_2 \mathrm{O}(l) \rightarrow \mathrm{H}_2 \mathrm{O}(\mathrm{g}) ; \Delta H=+v e(water) (vapour)\)

When salts such as NH₄Cl, NaCl, etc., are added to water, they undergo dissolution spontaneously. In the crystal structures of the salts, ions are held in an ordered arrangement.

On dissolution of the salts, the ordered arrangement of the ions is lost as the ions get dispersed throughout the solution in a disordered way. Thus, the dissolution ofa salt in water leads to a disordered state.

Thus from the above examples, it is clear that spontaneous processes occur with an increase in the randomness of the system. Hence we can say, that the primary condition of the spontaneity of any process is the increase in randomness in the system.

Spontaneous processes involving a decrease in randomness:

It is true that in most of the spontaneous processes, the degree of randomness of the system gets increased. However many spontaneous processes are accompanied by the decrease in randomness in the system.

Examples:

Water molecules in the clouds exist in extremely disordered states. But, when they fall in the form of rain, their freedom of movement decreases.

Molecules in water are in a state of randomness. However, when water undergoes freezing, the motion of the molecules becomes restricted, leading to a decrease In disorder in the system. Thus, the increase in randomness of the constituent particles of a system is not an essential condition for the spontaneity ofthe process that the system undergoes.

The driving force in a spontaneous process

From the above discussion, we see that neither the decrease in energy alone nor the increase in randomness alone can determine the spontaneity of a process. This is because many processes occur with an increase in energy, and any processes occur with a decrease in randomness.

Thus, we can conclude the effects of both of these factors have a role in determining the spontaneity of a process. To put it in another way we can say that the combined effect of these two factors is the driving for a process to occur spontaneously.

In this regard, it is important to note that these factors are not dependent on each other, and they may work in the same or opposite direction. Hence, we may have the following combinations of these two factors, the results of which may be favorable or unfavorable about spontaneity.