Unit 6: Chemical Energetics — Long Questions

Ans. Thermochemistry
Definition: Thermochemistry is one area of thermodynamics that is concerned with the study of the quantity of heat energy absorbed or evolved during physical or chemical changes. It is also known as energetics.
Basis of Thermochemistry

• Thermodynamics is the science of the relationship between heat and other forms of energy.
• Thermochemistry is largely based on the first law of thermodynamics which states that "Energy can neither be created nor destroyed, but it can be changed from one form to another".
• A special case of 1st law of thermodynamics is Hess's law. Applications of Thermochemistry
• Useful in calculating enthalpy changes of reactions that cannot be determined directly.
• Extensively used in analytical chemistry.
• Spontaneity of a reaction.
• Thermodynamics helps in predicting the feasibility or spontaneity of a chemical reaction under certain conditions.
• A spontaneous reaction is one that can occur naturally without external intervention.
• Thermodynamics has broad applications in Engineering and Physics.

ENTHALPY CHANGE

Enthalpy Change (ΔH)
Definition: The sum of all the possible potential and kinetic energies of a system is called its heat content or enthalpy denoted by H.
Explanation: Enthalpy is the heat content or energy possessed by a substance due to:

• Its structure (types of chemical bonds)
• Its physical state Every substance has a characteristic amount of enthalpy. This is the reason that total enthalpy of products (Hp) is never equal to that of reactants. During a reaction when reactants are converted into products, they show a change in enthalpy denoted by (ΔH). Mathematically ΔH = (Heat content of products, Hp) – (Heat content of reactants, HR). ΔH = Hp – HR

Ans. Energy Profile Diagrams
Bond Breaking and Making in Reactions: All chemical reactions involve breaking of bonds in reactant molecules. Followed by the formation of new bonds in product molecules.
Activation Energy
Definition: The minimum amount of energy that reactant molecules must possess for effective collisions to overcome the energy barrier and start a chemical reaction is called activation energy.
Symbol: Ea
Without enough activation energy, collisions between molecules will not lead to a reaction
Energy profile diagram

This diagram shows how energy changes during a chemical reaction.

• Reactant energy level
• Activation energy
• Peak or transition state
• Product energy level
• Enthalpy change (ΔH)

Exothermic Reactions

• Reactant have more energy than products
• ΔH is negative
• Heat is released
• Low activation energy give more stable products

Endothermic Reactions

• Products have more energy than reactants.
• ΔH is positive
• Heat is absorbed
• Ea is high less stable products are obtained.

Ans. Standard Enthalpy changes (ΔH°)
Factors affecting of enthalpy of substance
The enthalpy of a substance depends upon the following factors:

• Physical state (solid, liquid, gas)
• Pressure
• Temperature To make accurate comparisons between different reactions, we must define standard conditions. Conditions of standard enthalpy changes: (a) Temperature 25°C or 298K (b) Pressure 1atm or 101kPa When enthalpy is measured under these conditions, it is called standard enthalpy change, denoted as: ΔH°.

Did you Know?
S.Q. What is standard state of an element?
Ans. The standard state of an element is its most stable form at 298 K and I atm pressure. For example, the standard state of C is graphite not diamond. By definition, the standard enthalpy change of formation of any element in its standard state is zero.

(i) Enthalpy change of reaction ( ΔH°r )
Definition: The standard enthalpy of a reaction is the enthalpy change involved when stoichiometric amounts of reactants in their standard states react together completely to form products under standard conditions.
Example: The reaction between hydrogen and oxygen gases to form 1mole of water.
H₂(g) + 1/2 O₂(g) ——→ H₂O(l) ΔH°r = -286kJmol⁻¹

(ii) Enthalpy change of combustion ( ΔH°c )
Definition: The standard enthalpy change of combustion of a substance is the enthalpy change involved when one mole of the substance is completely burnt in excess of oxygen, under standard conditions.
It is always exothermic.
Example: Standard enthalpy of combustion ΔH°c of ethanol is -1368kJmol⁻¹.
Example:
C₂H₅OH(l) + 3O₂(g) ——→ 2CO₂(g) + 3H₂O(l) ΔH°c = -1368kJmol⁻¹
Enthalpy change of combustion is useful in calculating caloric content of foods and fuels.

(iii) Enthalpy change of formation ( ΔH°f )
Definition: Standard enthalpy change of formation of a compound is the enthalpy change involved when one mole of the compound is formed from its elements under standard conditions.
It can be éxothermic or endothermic.
Example: ΔH°f for methane is given below.
C(s) + 2H₂(g) ——→ CH₄(g) ΔH°f = -74.8 kJ mol⁻¹

(iv) Enthalpy change of atomization ( ΔH°at )
Definition: The standard enthalpy change of atomization of an element is the enthalpy change involved when one mole of gaseous atoms are formed from the element, under standard conditions.
Example: Standard enthalpy of atomization of H is given below:
H₂(g) ——→ 2H(g) EH-H = +436kJmol⁻¹
1/2 H₂(g) ——→ H(g) ΔH°at = +218kJmol⁻¹

(v) Enthalpy change of neutralization ( ΔH°n )
Definition: The standard enthalpy change of neutralization is the enthalpy change involved when one mole of water is formed by the reaction of an acid with an alkali under standard conditions.
It is always exothermic.
Explanation: The enthalpy of neutralization of NaOH by HCl is -57.1kJmol⁻¹. When these solutions are mixed together during the process of neutralization, the only change that occurs is the formation of water molecules leaving Na⁺ and Cl⁻ as free ions in solution.
Example: The enthalpy of neutralization is merely the heat of formation of one mole of liquid water from its ionic components and the actual neutralization reaction is:
H⁺(aq) + OH⁻(aq) ——→ H₂O(l) ΔH°n = -57.1kJmol⁻¹
For all strong acid-base reactions ΔH°n is always near – 57.1 kJmol⁻¹. When acid or base is weak, the value of enthalpy of neutralization will be less than 57.1 kJ/mol.

(vi) Electron Affinity ( ΔH°ea )
Definition: The first electron affinity is the enthalpy change involved when 1 mole of electron is added to 1 mole of gaseous atoms to form 1 mole of gaseous uni-negative ions under standard conditions.
Example: Electron affinity of chlorine atom
Cl(g) + e⁻ ——→ Cl⁻ ΔH = -348.8kJmol⁻¹
Since energy is released, so first electron affinity carries negative sign.

Summary Table of Enthalpy changes

Enthalpy change type | Definition | Exo/Endothermic | Example
Reaction ΔH°r | Stoichiometric amount of reactants react to form products | Exo or Endo | H₂(g) + 1/2 O₂(g) ——→ H₂O(l)
Combustion ΔH°c | 1 mole burned in excess O₂ | Always Exo | C₂H₅OH(l) + 3O₂(g) ——→ 2CO₂(g) + 3H₂O(l)
Formation ΔH°f | 1 mole compound formed from elements | Exo or Endo | C(s) + 2H₂(g) ——→ CH₄(g)
Atomization ΔH°at | 1 mole of gaseous atoms formed from element | Always Endo | 1/2 H₂(g) ——→ H(g)
Neutralization ΔH°n | 1 mole of H₂O formed from acid and base | Always Exo | H⁺(aq) + OH⁻(aq) ——→ H₂O(l)
Electron Affinity ΔH°ea | 1 mole of electrons added to gaseous atoms | Usually Exo | Cl⁻(g) + e⁻ ——→ Cl⁻(g)

Ans. Definition Bond energy (or Bond dissociation energy) is the average amount of energy required to break (dissociate) one mole of a particular bond in a substance.
Symbol: E
Units: kJmol⁻¹

• Breaking a bond requires energy (endothermic)
• Forming a bond releases energy (exothermic)
• Exact bond energy refers to a bond in a specific molecule (C-C in ethane)
• Average bond energy is the mean value taken over different compounds containing the same bond type. If bond energy of particular substance is determined, it is exact bond energy.

Bond energies vary even for the same type of bond depending on neighboring atoms and the molecular structure.

Examples of exact bond energies: C-C bonds usually have B.E values of approximately 350-380 kJ/ mol.
In ethane: [H₃C - CH₃] EC-C = 376kJmol⁻¹
In propane: [H₃C - CH₂ - CH₃] EC-C = 356kJmol⁻¹
In butane: [H₃C - CH₂ - CH₂ - CH₃] EC-C = 352kJmol⁻¹

In these examples EC-C for ethane, propane and butane are exact bond energies. When average of all EC-C energies is taken, it is average C-C bond energy from different compounds is 348 kJmol⁻¹

Table 6.1 Average bond energies of some important bonds (kJmol⁻¹)
[Table showing various bonds and their energies - H, C, N, O, S, Cl, Br, I columns]

Enthalpy change of reaction (ΔH°r) and chemical bonds

A chemical bond represents a form of energy known as chemical energy, which is interconvertible to all forms of energy.
A chemical reaction is a process in which:

• Old bonds are broken in the reactants
• New bonds are formed in the products

Bond Breaking
Requires energy → Endothermic → ΔH = +ve

Bond formation
Release energy → Exothermic → ΔH = – ve

Source of ΔH

• The energy required to break bonds in reactants.
• The energy released when new bonds form in products. ΔH°r = Total energy of bonds broken – Total energy of bonds formed

Difference between these deciding whether a reaction is exothermic or endothermic.

Condition | Result
Energy released > absorbed Exothermic | (ΔH is negative)
Energy absorbed > released Endothermic | (ΔH is positive)

Example Formation of water
Reaction
2H₂(g) + O₂(g) ——→ 2H₂O(g)

Break H-H and O=O bonds (energy absorbed)
Form: H-O bonds (energy released)
Since the energy released in forming H-O bonds is more than the energy absorbed in breaking H-H and O=O bonds, the reaction is exothermic. ΔH°r is negative.

Sample example 6.1
With the help of the following bond energy data; calculate the enthalpy change of the following reaction:
EH-H = 436kJmol⁻¹, EO-O = 495kJmol⁻¹ andEH-O = 463kJmol⁻¹
H₂(g) + 1/ 2O₂(g) ——→ H₂O(g)

Solution 436 + ½(495) 2 × (463)
(bond breaking energy) (bond forming energy)

∴ΔH°r = ∑ ER - ∑ EP
= [EH-H + EO-O] – [2EH-O]
= [436 + 247.5] – [2(463)]
ΔH°r = -242.5 K.kJmol⁻¹

Ans. Measurement of enthalpy change of a reaction

Calorimetry
Definition: Calorimetry is the measurement of heat evolved or absorbed during a physical or chemical process. The device used for this purpose is called a calorimeter.
Heat capacity: The amount of heat required to raise the temperature of a substance of mass 1kg though 1K (or 1°C) is known as specific heat capacity.
Example: Heat capacity of H₂O is 4.18 Jg⁻¹K⁻¹ (or Jg⁻¹ °C⁻¹) rounded of 4.2 Jg⁻¹k⁻¹ which is rounded off to 4.2 Jg¹K⁻¹.

Type of calorimeter

• Many types exist, but here we focus on the glass calorimeter, and phenomenon is called calorimetry.

Basic components of a simple calorimeter

• Insulated vessel: Prevents heat exchange with surroundings.
• Stirrer: Ensures uniform temperature distribution.
• Thermometer: Measures temperature change (ΔT).

GLASS CALORIMETER

Glass Calorimeter
Construction: A glass calorimeter is a simple, insulated vessel used to measure enthalpy changes (ΔH) for reactions in aqueous solutions.

• Beaker (reaction vessel)
• Thermometer
• Stirrer
• Loose-fitting lid
• Insulating material (cotton wool)

Limitations of glass calorimeter

• Reactions involving gases (gases may escape).
• Reactions producing very high temperatures (can damage glass or escape heat).
• Cotton wool insulation minimizes heat exchange with the surrounding air.

Working

• The reaction occurs inside the beaker.
• The temperature change (ΔT) is measured.
• Pressure is constant, the change in heat energy is equal to enthalpy change (ΔH).

Equations for heat calculations Heat energy absorbed or evolved. ΔH is calculated as follows:
(i) q = mwater × Cwater × ΔT
mwater = mass of water (or solution) in grams
Cwater = 4.18J/g°C
ΔT = Tf – Ti
(ii) Convert to kilojoules (kJ): By dividing with 1000.
(iii) Calculate ΔH for the reaction using relation. (enthalpy change per mole of reactant)

ΔH = - q/n (kJ / mol)
∴ q = mcΔT

ΔH = mcΔT/n (kJ / mol)

The solutions we are using here are so dilute that almost all of their mass consists of water therefore, we can simply use specific heat capacity of water. Such a calorimeter could be used to measure the heat of neutralization ( ΔH°n).

Sample Problem 6.2
Neutralization of 100cm³ of 0.5 moldm⁻³ NaOH at 25°C with 100 cm³ of 0.5 moldm⁻³ HCl at 25°C raised the temperature of the reaction mixture to 28.5°C. Find the enthalpy of neutralization. Specific heat of water = 4.2Jg⁻¹k⁻¹.

Solution
Density of H₂O is around 1gcm⁻³, so total volume of solution which is 200 cm³ ≈200g Rise in temperature, ΔT = 28.5 – 25.0 = 3.5°C = 3.5K
∴ Amount of total heat evolved, q = m × c × ΔT
= 200 × 4.2 × 3.5 = 2940J
= 2.94kJ

Calculation of number of moles of H₂O formed (n)
Using mole = concentration (mol/dm³)
n(HCl).n(NaOH) = 0.5×100/1000 = 0.05mol
Using equation, Number of moles of water formed, n(N2O) = 0.05mol
Heat evolved in the formation of 0.05 mole of water, q = -2.94 kJ

ΔH°n = q/n = -2.94 kJ/0.05mol

So, Enthalpy of neutralization, ΔH°n = -58.8 kJ mol⁻¹

Ans. Caloric content of food

The caloric content of food is a measure of the energy 'released' when the food is completely consumed in the body. This energy is typically expressed in units of kilocalories (k cal) or joules (J).

Food as a source of energy

When food is digested, chemical energy stored in food (also called calorie content) in it is released as heat energy.
This is similar to combustion, the enthalpy of combustion (ΔH°c) of food is equal to caloric content and is measured using a bomb calorimeter.

Caloric content is usually expressed as: Kilocalories (kcal), Joules (J) or kilojoules (kJ)
1kcal = 4.184 kJ.

Relation between enthalpy change and calorie content

The enthalpy of combustion of food (ΔH°c) is the calorie content of that food when it is translated or converted into kilocalories per gram.

Calorie content= ΔH° (kJ / g) / 4.184

Example Combustion of glucose.
Energy provided from glucose:
C₆H₁₂O₆(aq) + 6O₂(g) ——→ 6CO₂(g) + 6H₂O(l)

The calorie content of glucose can be calculated as follows:

Step by Step Find calorie content
ΔH per gram of glucose is measured.

• ΔH° = -2803 kJ/mol
• Molar mass of glucose = 180 g/mol 180.0g of glucose burns to produce energy – 2803 kJ

1. Convert ΔH per mole of per gram: -2803kJ / mol / 180 g / mol = -15.57kJ / g

2. Convert kJ/g to kcal/g using: 1 kcal = 4.184 kJ
Using the relation
ΔH (kJ/g) = – 4.184 × calorie content

Calorie content= ΔH°(kJ / g) / 4.184

3. Calorie content: Calorie content = -15.57kJ / g / 4.184 = -3.72kcal / g

Final result
Caloric content of glucose = 3.72 kcal/kg

Ans. Hess's law of constant heat summation

Statement "The total enthalpy change in a chemical reaction is independent of the route by which the chemical reaction takes place as long as the initial and final conditions are the same."

Enthalpies cannot be measured experimentally

German Henri Hess applied the law of conservation of energy to enthalpy changes. There are many reactions, for which ΔH cannot be measured directly by calorimetric method.

Example Tetrachloromethane CCl₄ cannot be prepared directly by combining carbon and chlorine. Hess's law helps us to calculate the enthalpy changes for such reaction or processes.

First law of thermodynamics (law of conservation of energy): "Energy can neither be created nor be destroyed, but can be converted from on form to another."

Energy Cycle Hess's law can be illustrated by drawing enthalpy cycles, often called energy cycles or Hess cycles.

Let's say a reaction goes from A → B:

• Direct route 1: A → B
• Indirect routes 2: A → intermediates → B
• The products formed in these routes (M, N and X, Y, Z) are called reaction intermediates.

By Hess's Law

For the indirect route 1, we can write
ΔH°r = ΔH1 + ΔH2

For the indirect route 2, we can write
ΔH°r = ΔH3 + ΔH4 + ΔH5

Hess's Law, which states "The total enthalpy change for a reaction is the same, no matter how it is carried out, provided the initial and final conditions are the same."

Figure 6.4 Hess's cycle

(i) Calculating enthalpy of formation (ΔH°f) using enthalpy of combustion (ΔH°c)

Sample problem 6.3
C(graphite) + O₂(g) ——→ CO₂(g) ΔH = -393.5kJmol⁻¹
CO(g) + ½O₂(g) ——→ CO₂(g) ΔH₂ = -283kJmol⁻¹
C(graphite) + ½O₂(g) ——→ CO(g) ΔH₁ = ?

Solution Applying Hess's law, we can write
ΔH° = ΔH1 + ΔH2
ΔH1 = ΔH – ΔH2
= – 393.5 – (– 283)
ΔH1 = – 110.5kJmol⁻¹

(ii) Calculating enthalpy change of reaction (ΔH°r) using enthalpies of formation (ΔH°f).

Sample Problem 6.4

Calculate the enthalpy change of reaction of
C₂H₄ + HCl ——→ C₂H₅Cl

Using Hess cycle with the help of following combustion data.
ΔH°f of C₂H₄ = +52.2 KJmol⁻¹
ΔH°f of HCl = -92.3 KJmol⁻¹
ΔH°f of C₂H₅Cl = -109 KJmol⁻¹
= -109[+52.2 + (92.3)]
ΔH°c = -68.9 KJmol⁻¹

Ans. Hess's law of constant heat summation

Statement "The total enthalpy change in a chemical reaction is independent of the route by which the chemical reaction takes place as long as the initial and final conditions are the same."

(i) Calculating enthalpy change of formation of a substance (ΔH°f) using Enthalpy of Combustion and Enthalpies of formation of other substances.

Sample problem 6.5

Propane (C₃H₈(g)) burns in oxygen according to the equation:
C₃H₈(g) + 5O₂(g) ——→ 3CO₂(g) + 4H₂O(l)

When 14.64g off propane is burned in an excess of oxygen in a calorimeter at 25°C and 1 atm pressure, 678.6 kJ of heat is evolved. Calculate the standard enthalpy of formation of propane. The standard enthalpies of formation of CO₂(g) and H₂O(l) are -393.51kJmol⁻¹ and -241.82kJmol⁻¹ respectively.

Solution The first step is to find the enthalpy change when one mole of propane is burned.
Molar mass of propane, C₃H₈ = 3(12.011) + 8(1.0079) = 36.033 + 8.0632 = 44.096
No. of moles of propane = 14.64/44.096 = 0.3320 mol
So, 0.3320 mol propane evolves heat = 678.6kJ
1.0 mol propane evolves heat = 678.6/0.3320 kJmol⁻¹ = -2044kJmol⁻¹

According to Hess' law
ΔH°c = [3× ΔH°f (CO₂) + 4× ΔH°f (H₂O)] – [1× ΔH°cf (C₃H₈) + 5× ΔH°cf (O₂)]
-2044kJ = [3× (-393.51) + 4× (-241.82)] – [1× ΔH°f (C₃H₈) + 5× (0)]
Taking the heat of combustion of O₂(g) to be zero and solving this equation for ΔH°f (C₃H₈) gives.
ΔH°f (C₃H₈) = -104kJmol⁻¹

(ii) Calculate the enthalpy change of reactions using bond energies.

Hess's Law statement
A special case of Hess's law allows us to calculate the enthalpy change of a reaction (ΔHr) using average bond energies (B.E.), particularly for gas-phase reactions.

Step 1 Breaking bonds in reactants

• All bonds in the reactant molecules are broken.
• This process requires energy input.
• The enthalpy change for this step is positive, since bond breaking is endothermic.
• Use the average bond enthalpies from Table 6.1 to calculate the energy required.

Step 2 Forming Bonds in Products

• Atoms recombine to form the product molecules.
• The enthalpy change for this step can again be estimated from the bond enthalpies of table 6.1.
• These values now be taken with minus sign because the bonds are being formed instead of being broken down.

In general, the heat of reaction for any gaseous chemical reaction can be calculated from average bond energies by use of the following version of Hess's law.
This formula gives an estimated enthalpy change using average bond energies.

ΔHr = ∑ Bond Energies of Reactants – ∑ Bond Energies of Products

Fig Hess's cycle; showing the relationship between bond energies and Hr. = ∑ ER – ∑ EP

Sample problem 6.6

In the case of formation of HCl(g) from H₂(g) and Cl₂(g), use B.E. data from Table 6.1 to estimate ΔH for the reaction:

H₂(g) + Cl₂(g) ——→ 2HCl(g)

And finally calculate the heat of formation of HCl, we replace this reaction by a hypothetical two-step process. The bonds in all the reactant molecules are first broken, and then the atoms are combined to make the products.

First Step H₂(g) + Cl₂(g) ——→ 2HCl(g)
EH-H = 436kJ mol⁻¹
∑ ER = (1 mol × 436 kJ/mol) + (1 mol × 242 kJ/mol)
= 436 kJ + 242 kJ
= 678 kJ

Second step

H₂(g) + Cl₂(g) ——→ 2HCl(g)

EH-Cl = 431kJmol⁻¹
Ep = 2mol × 431kJmol⁻¹ = 862kJ

The standard enthalpy change in the reaction is obtained by the following formula:
ΔH°r = ∑ ER – ∑ EP

ΔH°r = 678 – 862
ΔH°r = -184kJ (for 2mol of HCl)

Enthalpy of formation of HCl, = -184/2kJ = -92kJmol⁻¹

Quick Check 6.6

(a) The reaction for the Haber process is:
N₂(g) + 3H₂(g) ——→ 2NH₃(g)

The relevant bond energies are:
EN=N 945kJ mol⁻¹, EH-H = 436kJ mol⁻¹, EN-H = 391 kJ mol⁻¹

Calculate the enthalpy change of the above reaction.

Ans. Haber Process

Step 1 Bonds broken (Reactants)

• 1× N ≡ N = 945kJ / mol
• 3× H – H = 3× 436 = 1308kJ / mol Total energy to break bonds = 945 + 1308 = 2253 kJ/mol

Step 2 Bonds formed (Products)
In 2NH₃, each NH₃ has 3 N-H bonds → 2 × 3 = 6 N-H bonds
6 × N-H = 6 × 391 = 2346 kJ/mol

Step 3 Enthalpy change
ΔH = Bonds broken – Bonds formed
ΔH = 2253 – 2346 = -93 kJ/mol

Answer (a) ΔH = -93kJmol

(b) Calculate the enthalpy change for the following reaction
C₂H₅OH(l) + 3O₂(g) ——→ 2CO₂(g) + 3H₂O(l)

The bond energies of various bonds (in kJ mol⁻¹) are given below:
EC-C = 347, EC-H = +410, EC-O = +336, EC-O = +496, EC-O = +805, EO-H = +465

Ans. Combustion of Ethanol C₂H₅OH(l) + 3O₂(g) ——→ 2CO₂(g) + 3H₂O(l)

Step 1 Bonds broken (Reactants)
In C₂H₅OH

• 5 C-H bonds = 5 × 410 = 2050
• 1 C-C bond = 347
• 1 C-O bond = 336

Ans. Energetics of solution

Dissolution The process of dissolving a solute in a solvent is called dissolution. The formation of a solution takes place in three main steps.

Three steps in dissolution process
(i) Expanding the solvent
Overcoming intermolecular forces in the solvent to make space for the solute.

(ii) Expanding the solute
Breaking the solute into individual ions or molecules

(iii) Interaction between solute and solvent
Solute and solvent molecules interact to form a homogenous solution

Heat involved in dissolution
During dissolution, heat may be absorbed or released.
The total heat change is known as the standard enthalpy change of solution ( ΔH°ₛₒₗ )

Definition of ΔH°ₛₒₗ

Standard Enthalpy of Solution
"The standard enthalpy of solution is the amount of heat absorbed or evolved when one mole of a substance is dissolved in a solvent to give an infinitely dilute solution".

Symbol ΔH°ₛₒₗ

Unit kJmol⁻¹

Nature It may be exothermic (-ve) or endothermic (+ve)
Na₂CO₃(ₛ) + aq → 2Na₍ₐᵩ₎ + CO₃²⁻₍ₐᵩ₎ ΔH°ₛₒₗ = -25kJmol⁻¹
NH₄Cl(ₛ) + aq → NH₃₍ₐᵩ₎ + Cl⁻₍ₐᵩ₎ ΔH°ₛₒₗ = +16.2kJmol⁻¹

Ans. Hydration

Definition Hydration is the process when ionic compounds dissolve in water, they dissociate into individual ions. These ions are then surrounded by water molecules. This process is called hydration.
"The process in which water molecules surround and interact with the solute ions is called hydration".

Nature of Hydration forces
Ion-dipole forces are formed between:
• Ions from the solute
• Polar water molecules
These forces cause ions to become hydrated in aqueous solution.

Enthalpy of hydration

Definition The enthalpy change involved when one mole of a solute is dissolved in excess water to make an infinitely dilute solution under standard conditions is called enthalpy of hydration.

Symbol ΔH°ₕᵧd

This energy is always released (exothermic) because new ion-dipole bonds are formed.

Factors Affecting the magnitude of Hydration Energy
The heat of hydration depends on following factors:

(i) Charge on the ion
• Higher charge and stronger attraction to water molecules gives greater hydration energy.

(ii) Size of the ion
• Smaller ions and higher charge density gives stronger hydration energy.
Charge density is defined as charge per unit area.
• Greater charge density gives greater hydration energy.

Order of Hydration Energies Enthalpies of hydration of following ions are in the order:
Na⁺ < Mg²⁺ < Al³⁺

Mg²⁺ > Ca²⁺ > Sr²⁺

The smaller and more highly charged the ion, the greater is its hydration energy.

Hydration of Anions
Smaller more negatively charged anions (F⁻) have higher hydration energy than larger ones (I⁻).

Ans. Lattice energy

Definition The lattice energy of an ionic crystal is the enthalpy change involved when one mole of the ionic compound is formed gaseous ions under standard conditions.
Na₍(g) + Clₜ(g) → NaCl(ₛ) ΔH°ₗₐₜₜ = -787kJmol⁻¹

Factors affecting lattice energy

(i) Charge on the Ion
• Greater the ionic charge, greater is the lattice energy.
• This is because higher charged ions attract each other more strongly.
• Examples: ΔH_latt (MgO) = - 3923 kJmol⁻¹
ΔH_latt (LiF) = -1049 kJmol⁻¹
Mg²⁺ and O²⁻ have higher charges compared to Li⁺ and F⁻

(ii) Size of the Ion
• Smaller ions means closer packing and stronger attraction gives higher lattice energy.
• Lattice energy decreases with increasing size of either cation or anion.
• Example: Alkali Metal Halides
When we go down the group of alkali metals, lattice energy decreases.

Quick Check 6.6 (continued from previous page)
Factors affecting dissolution of ionic compounds

(i) Lattice Energy
• Energy required to break the ionic lattice.
• Higher lattice energy means more energy required to dissolve.

(ii) Hydration Energy
• Energy released when ions are surrounded by water molecules.
• Higher hydration energy favours dissolution.

Ans. Solubility trends of group 2 hydroxides and sulfates

(i) Group 2 Hydroxides: Increasing solubility of hydroxides of group 2.
• Both lattice energy and hydration enthalpy decrease down in the group 2.
• But lattice energy decreases faster, making ΔH°ₛₒₗ more exothermic.

(ii) Group 2 Sulfates: Decreasing solubility of sulfates of group 2.
Now look at the solubility of group 2 metal sulfates.
• SO₄²⁻ is large, so lattice energy doesn't decrease much.
• Hydration energy decreases significantly down the group.
• ΔH°ₛₒₗ becomes more endothermic, and solubility decreases down the group.

Calculating enthalpy change of solution ΔH°ₛₒₗ
• We can calculate the enthalpy change of solution or the enthalpy change of hydration by constructing an enthalpy cycle and using Hess's law
• From this enthalpy cycle:
ΔH°ₐₜₜ + ΔH°ₛₒₗ = ΔH°ₕᵧd

This energy cycle used to calculate enthalpy change of solution.

Sample problem 6.7
Determine the enthalpy change of solution ΔH°ₕᵧd of sodium fluoride (NaF) using the following data:
Lattice energy of sodium fluoride (NaF) = -902kJmol⁻¹
Heat of hydration of sodium ions (Na⁺) = -406kJmol⁻¹
Heat of hydration of fluoride ions (F⁻) = -506kJmol⁻¹

Solution Step 1: Draw the enthalpy cycle

NaF(s) → (ΔH°ₐₜₜ = 902 kJ mol⁻¹) → Na⁺(g) + F⁻(g)

ΔH°ₛₒₗ +aq H°ₕᵧd[Na⁺] = -406 kJ mol⁻¹

H°ₕᵧd[F⁻] = -506 kJ mol⁻¹

Na⁺(aq) + F⁻(aq)

Step 2 Rearrange the equation and substitute the values to find ΔH°ₛₒₗ.
ΔH°ₐₜₜ + ΔH°ₛₒₗ = ΔH°ₕᵧd

ΔH°ₛₒₗ = (-406) + (-506) – (-902)
ΔH°ₕᵧd[NaF] = +10 kJ mol⁻¹

Ans. Theoretical calculation of ΔH°ₗₐₜₜ
• It is impossible to determine the lattice energy of a compound by a single direct experiment.
• We can calculate the value for ΔH°ₗₐₜₜ using several experimental data and an energy cycle called the Born-Haber cycle.
• Born-Haber cycle is an application of Hess's law. It is used to measure lattice energy of binary ionic compounds M⁺ X⁻.

Example Consider the calculation of lattice energy of sodium Chloride using Hess's law and Born-Haber cycle.

ΔH°c: Standard enthalpy of formation of NaCl can be measured conveniently in a calorimeter.
ΔH°x: Total energy involved in changing sodium and chlorine from their standard physical states to gaseous ions. Applying Hess's law on the above energy cycle.
ΔH°f = ΔH°c + ΔH°ₐₜₜ
The above energy triangle has been extended to show the various stages involved in finding ΔH°c. The complete energy cycle is called the Born-Haber cycle and it is presented in Figure 6.10.

Na₍(g) + e⁻ + Cl₍(g)

ΔH_a of Cl₂(g) = +121 kJ mol⁻¹ ΔH_ea of Cl = -349 kJ mol⁻¹

Na⁺₍(g) + e⁻ + 1/2 Cl₂(g)
Na⁺₍(g) + Cl⁻₍(g)

ΔH_i of Na(g) = +496 kJ mol⁻¹

Na₍(g) + 1/2 Cl₂(g)

ΔH_a of Na(g) = +108 kJ mol⁻¹

Na₍(g) + 1/2 Cl₂(g)
ΔH_latt of Na⁺Cl⁻₍(s) = ?
(Lattice Energy of NaCl)

ΔH_f of Na⁺Cl⁻₍(s) = -411 kJ mol⁻¹

Na⁺Cl₍(s)

Fig Born-Haber's cycle of sodium chloride

Calculation of ΔH°c
This energy can be broken into several steps using Hess's Law.

Enthalpy Terms Involved

Process Symbol
Sublimation of Na (Na(s) → Na(g)) ΔH_at (Na)
Ionization of Na (Na → Na⁺ + e⁻) ΔH_i1 (Na)
Dissociation of Cl₂ (1/2 Cl₂ → Cl) ΔH_at(C₂)
Electron gain by Cl (Cl + e⁻ → Cl⁻) ΔH_ea1(C₂)

Using Hess's Law Calculation of ΔH°
From above cycle we have,
ΔH°x = ΔH_at(Na) + ΔH_i1(Na) + ΔH_at(Cl2) + ΔH_ea1(Cl2)
ΔH°x = 376 kJmol⁻¹
ΔH°_latt = ΔH°f - ΔH°x
ΔH°_latt = -411 – (+376)
ΔH°_latt = -787kJmol⁻¹

Conclusion • The Born – Haber cycle allows indirect calculation of lattice energy using other known enthalpy changes.
• It is a stepwise application of Hess's Law for ionic compounds.

Sample problem 6.8
Calculate the heat of formation of sodium fluoride which crystallizes in the sodium chloride lattice. The heat of atomization of Na(s) is 108kJ/mol, half the bond energy of F₂(g) is 79kJ/mol, the ionization energy of sodium atoms is 494 kJ/mol, the electron affinity of fluorine at4oms is -328 kJ/mol, and the lattice energy is -939 kJ/mol.

Solution Values are given for all the quantitates in the Born-Haber cycle, so we can apply Hess' law:

ΔH°f = ΔH°_latt + ΔH°_at(Na) + ΔH°_i1(Na) + ΔH°_at(F2) + ΔH°_ea(F)
= (-939+108 + 494 + 79 – 328)kJmol⁻¹
ΔH°f = -584kJmol⁻¹

(b) Calculate the heat of formation of lithium fluoride. The heat of atomization of Li(s) is 161 kJ/mol, half the bond energy of F₂(g) is 79 kJ/mol, the ionization energy of lithium atoms is 520 kJ/mol, the electron affinity of fluorine atoms is -328 kJ/mol, and the lattice energy is 1107 kJ/mol.

Ans.
Born-Haber Cycle for LiF(s):
Li(s) + 1/2 F₂(g) → (atomization of Li) ΔH = +161 kJ/mol
Li(g) + 1/2 F₂(g) → (bond dissociation of F₂) ΔH = +79 kJ/mol
Li(g) + F(g) → (ionization energy of Li) ΔH = +520 kJ/mol
Li⁺(g) + F(g) → (electron affinity of F) ΔH = -328 kJ/mol
Li⁺(g) + F⁻(g) → (lattice energy) ΔH = -1107 kJ/mol

LiF(s)

Calculation of enthalpy of formation (ΔH°f)

Step-by-step values Step Enthalpy Change (kJ/mol)
Atomization of Li +161
1/2 Bond energy of F₂ +79
Ionization energy of Li +520
Electron affinity of F -328
Lattice energy of LiF -1107

Apply Hess's Law ΔH°f = Sum of all steps above
ΔH°f = 161 + 79 + 520 – 328 – 1107
ΔH°f = -675kJ/mol

Answer (b) The standard enthalpy of formation of LiF(s) is -675kJ/mol.

Ans. Entropy

Definition Entropy is a measure of the number of ways energy can be distributed within a system at a specific temperature.

Explanation
Systems naturally evolve towards states of higher entropy. Entropy can also be thought of as a measure of the randomness or disorder of a system. The higher the randomness or disorder, the greater the entropy of the system.

Entropy, Diffusion and Number of Ways of arrangement

Diffusion and Entropy
• Diffusion is driven by random motion of particles and high probability of new arrangements.
• As particles spread out, the number of possible microstates increases, so entropy increases.

General formula
Number of arrangements = x^y
Where x = number of places, y = number f particles to be arranged

Factors Affecting Entropy

(i) Number of particles:
More particles, more microstates and higher entropy.

(ii) Hardness of substance:
Harder substances have lower entropy (due to restricted vibrations).
Example: Diamond < Graphite

(iii) Physical state:
Solids < Liquids < Gases (in terms of entropy).
Example: Water states near transition points.

Comparison of entropy values

Entropies of different substances can be compared on the basis of following factors.

(i) Number of particles: Small number of particles means low entropy and versa.
Example: CaCO₃ has higher standard entropy (92.9 JK⁻¹ mol⁻¹) than CaO (39.7 JK⁻¹ mol⁻¹).
This is because the number of possible arrangements is lower when the number of particles is smaller.

(ii) Physical properties: The entropy of substances having similar chemical nature is dictated by their hardness. Harder substances have lower entropy than softer ones.
Diamond has lower entropy than graphite because it is much harder. Stronger forces result in limited vibrations in harder substances which decrease the probability of disorder.

(iii) Physical states: A substance has lower entropy in solid state than in liquid and gaseous states. The entropy of ice near its melting point is 48.0 JK⁻¹mol⁻¹, for water, it is 69.9 JK⁻¹mol⁻¹, whereas water vapour just above the boiling point is 188.7 JK⁻¹mol⁻¹.

Quick Check 6.8
Explain the difference in the entropy of each of the following pairs of substances in terms of their states and structures.

Ans.
(i) Br₂(ℓ), S° = 151.6JK⁻¹mol⁻¹ and I₂(ₛ), S° = 116.8JK⁻¹mol⁻¹

Reason Br₂ is a liquid: particles are more mobile than in a solid (I₂), so entropy is higher.
I₂ is a solid, with tightly packed particles and low randomness → lower entropy.
Conclusion: Br₂(ℓ) has higher entropy due to the more disordered liquid state.

(ii) H₂(g), S° = 130.6JK⁻¹mol⁻¹ and CH₄(g), S° = 186.2JK⁻¹mol⁻¹

Reason • Both are gases, but CH₄ is a larger and more complex molecule with more vibrational and rotational motions → higher entropy.
• H₂ is a small diatomic molecule → fewer possible motions → lower entropy.
Conclusion: CH₄(g) has higher entropy due to greater molecular complexity.

(iii) Hg(ℓ), S° = 76.0JK⁻¹mol⁻¹ and Na(ₛ), S° = 51.2JK⁻¹mol⁻¹

Reason • Hg is a liquid, so its atoms are freer to move → more disorder.
• Na is a solid, so its atoms are arranged in a rigid lattice → less disorder.
Conclusion: Hg(ℓ) has higher entropy due to its liquid state.

(iv) SO₂(g), S° = 248.1JK⁻¹mol⁻¹ and SO₃(s), S° = 95.6JK⁻¹mol⁻¹

Reason • SO₂ is a gas → high particle randomness and motion → high entropy.
• SO₂ is a liquid → less freedom of motion than gas → lower entropy.
Conclusion: SO₂(g) has significantly higher entropy due to its gaseous state.

Ans. Entropy changes in reactions

In a chemical reaction, comparing the entropy of reactants and products helps determine whether the entropy change (ΔS°) is increased or decreased.

General rules to predict entropy change: The following rules must be followed to calculate change in entropy of a reaction.

(i) Solid → Liquid or Gas
• When a solid is converted to a liquid or a gas in the product, the entropy change is positive.
• Entropy increases (ΔS° is positive).
Example
H₂O(ₛ) → H₂O(ℓ) ΔS° > 0

(ii) Increase in Number of Moles (especially gases)
• If there is a change in the number of gaseous molecules in a reaction, due to high values of entropy are associated with gases.
• More the gas molecules, greater is the number of ways of arranging them and the higher the entropy.
• More moles in products than reactants shows increase in entropy.
• Especially important for gaseous molecules, due to their high entropy.

Example 1
CaCO₃(ₛ) → CaO(ₛ) + CO₂(g)

• Solid to solid or gas gives entropy increases (ΔS°>0)
• Spontaneous reaction

Example 2
2N₂O₄(g) → 4NO₂(g) + O₂(g)

• An increase of entropy of the system because there are a greater number of moles of gas molecules in the products (5 molecules) than in the reactants (2 molecules).
• Reactants: 2 gas molecules
• Products: 5 gas molecules
• ΔS° is positive, more disorder
• Reaction is spontaneous

Sign of Entropy Change (ΔS°) and spontaneity

ΔS° Sign Interpretation Spontaneous
Positive (+) Increased disorder: more ways to arrange Yes, usually spontaneous
Negative (–) Increased order: fewer arrangements possible No, may not be spontaneous

Entropy alone cannot always predict spontaneity. We need Gibbs Free Energy (ΔG) for final decision.

Calculating the entropy change of the system (reaction)

In order to calculate the entropy, change of the system we use the relationship:
ΔS°_system = S°_products - S°_reactants

Let us calculate the entropy change of the system for the reaction:
2Ca₍(s) + O₂(g) → 2CaO(ₛ)

The standard entropy values are:
S°_Ca(s) = 41.40JK⁻¹ mol⁻¹
S°_O₂(g) = 205.0JK⁻¹ mol⁻¹
S°_CaO(s) = 39.70JK⁻¹ mol⁻¹

As ΔS°_system = S°_products - S°_reactants

= 2× S°_CaO(s) – {2× S°_Ca(s) + S°_O₂(g)}
= 2× 39.70 – {2× 41.40 + 205.0}
= 79.40 – 287.8

ΔS°_system = -208.4 JK⁻¹mol⁻¹

Interpretation
• ΔS° system = -208.4 JK⁻¹ mol⁻¹ → Entropy decreases
• Even though entropy decreases, the reaction is spontaneous.
• This means we must also consider entropy of the surroundings or Gibbs Free Energy (ΔG) to fully judge spontaneity (feasibility).

Ans. Total entropy change, ΔS°_total

Definition ΔS°_total is sum of standard entropy change of system and standard entropy change of surrounding.

ΔS° for system can be found out by subtracting the standard entropies of reactants from the standard entropies of products (similar to the way you obtained ΔH°).

ΔS°_system = ∑ S°_(products) – ∑ S°_(reactants)

Respective number of moles of reactants and products in a balanced chemical equation are multiplied with entropy of reactants and products.

ΔS° for surrounding is calculated suing following equation:
ΔS°_surrounding = -ΔH°_r / T

Where,
• The value of ΔH°_r is in KJ/mol so must be multiplied with 1000. This is because entropy changes are measured in J/Kmol.
• The negative sign is part of equation and not the sign of ΔH°_r .

ΔS°_total is calculated using following equation:
ΔS°_total = ΔS°_system + ΔS°_surrounding

ΔS°_system = ∑ S°_(products) – ∑ S°_(reactants) + [-ΔH°_r / T]

It is useful to be able to predict about spontaneity of reaction using the sign of ΔS°_total

ΔS°_total is +ve ⟹ entropy increases ⟹ reaction is spontaneous
ΔS°_total is +ve ⟹ entropy decreases ⟹ reaction is nonspontaneous

Sample problem 6.9
2Ca₍(s) + O₂(g) → 2CaO(ₛ)

The standard entropy values are:
S°_[O₂(g)] = 205.0J K⁻¹ mol⁻¹
S°_[CaO(s)] = 39.70JK⁻¹ mol⁻¹

As S°_system = ∑ S°_(products) – ∑ S°_(reactants)

= 2× S_[CaO(s)] – {2× S°_[Ca(s)] + S°_[O₂(g)]}
= 2× 39.70 – {(2× 41.40) + 205.0}
= 79.40 – 287.8

S°_system = -208.4 J K⁻¹ mol⁻¹

Ans. The Free energy change, ΔG

In determining whether a chemical reaction is likely to be spontaneous, we use the new quantity Gibbs free energy change, ΔG. the Gibbs free energy change is given by the relationship.

ΔG = -TΔS_total

Gibb's Equation
Multiplying the ΔS°_total expression by – T, we can also get the expression without having to consider the entropy changes of the surroundings. This expression is called Gibb's equation.

–T × {ΔS°_total = ΔS°_system + [ΔH°_r / T]}

–TΔS°_total = TΔS°_system + ΔH°_r

ΔG° = ΔH°_r - TΔS°

Conclusion The following conclusion can be drawn from it.

(i) If ΔG < 0 (–ve), the given process may occur spontaneously.
(ii) If ΔG > 0 (+ve), the indicated process cannot occur spontaneously; instead, the reverse of it may occur.
(iii) If ΔG = 0 neither the indicated process nor reverse of it can occur spontaneously. The system is in a state of equilibrium.

Calculating ΔG° for a reaction

Sample Problem 6.12
For the reaction:
CaSO₄(ₛ) → Ca₍ₐᵩ₎ + SO₄²⁻₍ₐᵩ₎

Calculate ΔH°, ΔS° and ΔG° at 25°C using the following data; and discuss its spontaneity.

Heat of Formation ΔH°_f[CaSO₄(ₛ)] = -1432.7kJ , ΔH°_f[Ca²⁺₍ₐᵩ₎] = -543.0kJ, ΔH°_f[SO₄²⁻₍ₐᵩ₎] = -907.5kJ

Standard entropy S°[CaSO₄(ₛ)] = 106J / K, S°[Ca²⁺₍ₐᵩ₎] = -55.2J / K, S°[SO₄²⁻₍ₐᵩ₎] = +17.2J / K

Solution
ΔH° = ∑ ΔH_f (products) – ∑ ΔH_f (reactants) = [-543.0 – 907.5] – [-1432.7] = -17.8kJ

ΔS° = ∑ S° (products) – ∑ S° (reactants) = [(-55.2 + 17.2) – (106.7)]J / K = 1447J / K

ΔS° = -0.1447kJ / K

ΔG° = ΔH° – TΔS° = (-17.8kJ) – (298K) × (-0.1447kJ/K) = +25.3kJ

Result We conclude that this process is nonspontaneous at standard conditions at 25°C

Note that you substitute ΔS° in units of kJ/K in the formula of ΔG°.

Sample Problem 6.13
What is the standard free-energy, ΔG° for the following reaction (Haber's process) at 25°C?
N₂(g) + 3H₂(g) → 2NH₃(g)

Also discuss its spontaneity

Heat of formation ΔH°_f[N₂(g)] = 0kJ , ΔH°_f[H₂(g)] = 0kJ , ΔH°_f[NH₃(g)] = -45.9kJ

Standard entropy S°[N₂(g)] = 191.5J / K, S°[H₂(g)] = 130.6J / K, S°[NH₃(g)] = +193J / K

Solution
ΔH° = ∑ ΔH_f (products) – ∑ ΔH_f (reactants) = [2× (-45.9)] – [0 + 0] = -91.8kJ

ΔS° = ∑ nS° (products) – ∑ mS° (reactants) = [2× (193)] – [191.5 + 3× 130.6]J / K = -197J / K

ΔS° = -0.197kJ / K

ΔG° = ΔH° – TΔS° = (-91.88kJ) – (298K) × (-0.197kJ / K) = -33.1kJ

Result
Since ΔG° is a negative value thus it is concluded that Haber's process is spontaneous at standard conditions' i.e. at 25°C.

Ans. Spontaneity and temperature change

The free-energy for a reaction at 25°C, is calculated.
For the measurement of ΔG° at any temperature, following equation is used.
ΔG° = ΔH° – T ΔS°.

We get the value of ΔG at any temperature by substituting values of ΔH° and ΔS° at 25°C. But in this method, we assume that ΔH° and ΔS° are essentially constant w.r.t. temperature (this is only approximately true).

Remember that the superscript degree sign (°) refers to substances in standard states, which are substances at 1 atm and at the specified temperature. Although until new this was 25°C (298 K), we now consider other temperatures. Note that in general, ΔG depends strongly on temperature.

In table 6.3, all of the four possible choices of signs for ΔH° and ΔS°, give different temperature behaviors for ΔG° are being discussed. In all this discussion temperature is always positive as it is taken on Kelvin scale.

Table 6.3 Effect of temperature on the spontaneity of reactions

ΔH° ΔS° ΔG° Description
– + – Spontaneous at all T
+ – + Non– spontaneous at all T
– – + or – Spontaneous at low T; non-spontaneous at high T
+ + + or – Non–spontaneous at low T; spontaneous at high T

The term low temperature and high temperature are relative. For a particular reaction, high temperature could mean room temperature.

Sample Problem 6.14
Calculate ΔG° at 1000 °C for the following reaction and discuss its spontaneity.
CaCO₃(ₛ) → CaO(ₛ) + CO₂

Solution For this reaction, ΔH° = + 178.0 kJ and ΔS° = + 160.4 J/K = 0.1604 kJ/K

Substituting T = 1000 + 273 = 1273 K

ΔG = ΔH – T ΔS = (+178.0 kJ) – (1273 K) × (+0.1604 k J/K)
= + 178.0 kJ – 204.2 kJ
ΔG = – 26.2 kJ

We conclude that this reaction is spontaneous at 1000 °C and 1 atm, since ΔG has negative sign. In other words, at 1000 °C, limestone (CaCO₃) will decompose in an open container.