PHYSICAL CHEMISTRY FORM FIVE NOTES (New Syllabus)

Chapter One: Gases

Introduction

Many reactions performed in the laboratory or industries involve gases as reactants or products. This means that a chemist must be well equipped with the knowledge of gases to successfully handle such reactions. In this chapter, you will learn about the parameters of gases, gas laws, the kinetic theory of gases, the relative molecular masses of gases and the applications of gas laws in daily life. The competencies developed will enable you to manipulate, predict and associate the behaviour of gases to unlock and tackle various challenges that drive innovations in the manufacturing and transport industries.

1.1 Gas Parameters

Despite their wide differences in chemical properties, all gases more or less obey the gas laws. These laws show the relationship between macroscopic physical properties of gases, such as volume, pressure, temperature, and number of moles of a gas.

(a) Volume (V)

Gases have no definite volume as they expand to fill the volume of a container. Therefore, the volume of a gas is equal to the volume of a container occupied by the gas. The SI unit of volume is cubic metre (m³); however, other units such as litre (L), cubic decimetre (dm³), cubic centimetre (cm³) and millilitres (mL) are commonly used depending on the context and the precision required.

(b) Pressure (P)

The pressure of a gas is the force exerted by the impacts of its molecules per unit area of the wall of the container. The pressure of a gas in a container is measured by using a manometer. The pressure for gases in the atmosphere is measured by using a barometer.

The pressure can be calculated using the equation:

P = hρg

where h – height difference between the liquid levels in the U-tube, ρ is the density of the liquid in the manometer and g is the acceleration due to gravity.

The SI units of pressure are Pascal (Pa) and (Nm⁻²). The relationship among the units of pressure is:

1 atm = 760 Torr = 1.01 × 10⁵ Pa = 1.01 × 10⁵ N/m²

(c) Temperature (T)

The temperature of a gas is measured in Celsius degree or centigrade degree (°C). The SI unit of temperature is Kelvin (K). The Kelvin temperature (or absolute temperature) of a gas can be obtained using the following expression:

T(K) = T(°C) + 273

(d) Number of moles of a gas (n)

The number of moles of a gas in a container is equal to the mass of the gas (m) divided by its molecular/atomic mass (M).

Moles of a gas = mass of a gas / molecular mass or simply, n = m/M

1.2 The Gas Laws

The parameters needed to characterise a gas are inter-related. Any change in one parameter causes a change in one of the other parameters. The relationship between various properties of gases, such as pressure, temperature, volume and amount of gas molecules (moles) are described by the gas laws.

1.2.1 Boyle's Law

The Boyle's law was named after an Anglo-Irish philosopher and scientist, Robert Boyle (1627 – 1691), who published the law in 1662. The law states that at constant temperature, the volume of a fixed mass of a gas is inversely proportional to its pressure.

Mathematically: V ∝ 1/P (T and n are constant)

V = k/P ⇒ PV = k

For two sets of pressures and volumes:

P₁V₁ = P₂V₂

Example 1.1

A chemist observed that the air trapped in a U-tube occupies 25.5 cm³ at 1.15 atm. After adding a liquid to the tube, the pressure of the trapped air increased to 2.61 atm. What is the new volume of air in litres if the temperature remained constant?

Solution:

Given: V₁ = 25.5 cm³ = 0.0255 L, P₁ = 1.15 atm, P₂ = 2.61 atm

From Boyle's law: P₁V₁ = P₂V₂ ⇒ V₂ = P₁V₁/P₂

V₂ = (1.15 atm × 0.0255 L) / 2.61 atm = 0.0112 L

Therefore, the new volume of air is 0.0112 litres.

1.2.2 Charles' Law

The Charles' law was named after a French scientist, Jacques Alexander Cesar Charles (1746–1823). The law states that at constant pressure, the volume of a fixed mass of a gas is directly proportional to the absolute temperature.

Mathematically: V ∝ T (P and n are constant)

V = kT ⇒ V/T = k

For two sets of temperature and volume:

V₁/T₁ = V₂/T₂

Activity 1.1

Aim: To investigate the effect of temperature on the volume of a gas

Requirements: Balloon, balloon filler, two steady objects like small stones or bricks, and freezer or ice

Procedure:

1. Inflate the balloon with air using your mouth or the balloon filler and tie it.
2. Put the two steady objects side by side leaving the space in between, enough to fit the size of the filled balloon.
3. Fit the balloon between the two objects by allowing it to barely touch the objects on each side.
4. Remove the balloon and put it in a freezer or ice for about thirty minutes. Leave the steady objects on their position undisturbed.
5. Remove the balloon from the freezer or ice and place it between the two objects as in step 3. Record your observation.

Questions:

1. What variables were controlled in the experiment? Explain.
2. If you were to conduct the experiment again with a different gas, what would be the results? Explain.

1.2.3 Gay-Lussac's Law

The Gay-Lussac's law was named after the French chemist, Joseph Louis Gay-Lussac, who laid its foundation in 1802. The law states that the pressure of a gas is directly proportional to its absolute temperature provided that the volume and the amount of the gas are kept constant.

Mathematically: P ∝ T (V and n are constant)

P = kT ⇒ P/T = k

For two sets of pressure and temperature:

P₁/T₁ = P₂/T₂

1.2.4 Combined Gas Law

According to Boyle's law, pressure and volume of a gas are inversely related; and according to Charles' law, the absolute temperature and volume of a gas are directly related. These two relationships can be combined into a single equation known as the combined gas law.

V ∝ T/P (n is constant) ⇒ PV/T = k

For two sets of pressure, temperature and volume:

P₁V₁/T₁ = P₂V₂/T₂

Example 1.3

A 25.8 L quantity of gas has a pressure of 690 Torr and a temperature of 17 °C. Calculate the volume of the gas if the pressure is changed to 1.85 atm and the temperature to 345 K.

Solution:

Given: V₁ = 25.8 L, T₁ = 17 °C = 290 K, P₁ = 690 torr = 0.908 atm, P₂ = 1.85 atm, T₂ = 345 K

From combined gas law: P₁V₁/T₁ = P₂V₂/T₂ ⇒ V₂ = P₁V₁T₂/(P₂T₁)

V₂ = (0.908 atm × 25.8 L × 345 K) / (1.85 atm × 290 K) = 15.1 L

Therefore, the volume of the gas is 15.1 L.

1.2.5 Avogadro's Law

Avogadro's law was stated in 1811 by an Italian Chemist, Amedeo Avogadro (1776–1856). The law states that equal volume of gases at the same temperature and pressure contain equal number of particles.

Mathematically: V ∝ n (P and T are constant)

V = kn ⇒ V/n = k

For two sets of volume and number of moles at constant temperature and pressure:

V₁/n₁ = V₂/n₂

1.2.6 Ideal Gas Equation

A combination of Boyle's, Charles' and Avogadro's laws gives an equation known as ideal gas equation which indicates how the four variables are related to each other.

PV = nRT

where R is the universal gas constant.

Values of R:

• 0.0821 L atm mol⁻¹ K⁻¹
• 8.314 J mol⁻¹ K⁻¹
• 62.36 L Torr mol⁻¹ K⁻¹
• 8.314 m³ Pa mol⁻¹ K⁻¹

Example 1.5

What is the temperature in °C of a 9.65 g of oxygen gas in a 4560 mL container if the pressure is 895 Torr?

Solution:

Given: Mass of oxygen = 9.65 g, Volume = 4560 mL = 4.560 L, Pressure = 895 Torr = 1.18 atm

Number of moles (n) = mass/molar mass = 9.65 g / 32.0 g mol⁻¹ = 0.302 mol

From ideal gas equation: PV = nRT ⇒ T = PV/(nR)

T = (1.18 atm × 4.56 L) / (0.302 mol × 0.0821 L atm mol⁻¹ K⁻¹) = 217 K

T(°C) = T(K) - 273 = 217 K - 273 K = -56 °C

Therefore, the temperature of oxygen gas is -56 °C.

1.2.7 Dalton's Law of Partial Pressure

The concept of partial pressure was invented by an English chemist called John Dalton (1766–1844). Dalton proposed that in a mixture of gases which do not react chemically, each component exerts pressure as if it were the only component occupying the container.

Mathematically: P_T = P_A + P_B + P_C (T and V are constant)

The partial pressure of a gas is the product of its mole fraction and total pressure of the gases in the mixture.

P_A = χ_AP_T

The sum of the mole fractions of all gases in the mixture is equal to 1.

χ_A + χ_B + χ_C = 1

1.2.8 Graham's Laws of Diffusion and Effusion

Diffusion is a net movement of atoms or molecules from the region of high concentration to the region of low concentration. Effusion is the process by which gas molecules escape from their container through a tiny hole into an evacuated space.

The Graham's law of diffusion and effusion states that the rate of diffusion or effusion of a gaseous substance is inversely proportional to the square root of its molar mass.

Mathematically: R ∝ 1/√M

For two gases A and B:

R_A/R_B = √(M_B/M_A)

Since the rate is inversely proportional to time:

t_A/t_B = √(M_A/M_B)

Example 1.7

A flammable gas is found to effuse through a porous barrier in 1.50 min. Under the same temperature and pressure conditions, it takes 4.73 min for an equal volume of bromine vapour to effuse through the same barrier. Calculate the molar mass of the unknown gas.

Solution:

Given: t_A = 1.50 min, t_B = 4.73 min, M_B = 159.8 g/mol

From Graham's law: t_A/t_B = √(M_A/M_B)

1.50/4.73 = √(M_A/159.8) ⇒ M_A = 16.1 g/mol

Therefore, the molar mass of the unknown gas is 16.1 g/mol.

1.3 The Kinetic Theory of Gases

The kinetic theory was developed in the 19th century by James Clerk Maxwell (1831–1879) and Ludwig Boltzmann (1844–1906) who found that the physical properties of gases can be explained at the molecular level in terms of the motion of individual molecules as the pressure, volume, and temperature changes.

1.3.1 Relationships Among Volume, Pressure, and Temperature

The kinetic theory of gases explains how individual gas particles create a force on the surface, which is measured as gas pressure. According to Boyle's law, when gas pressure increases, the volume occupied by the gas decreases. The theory also explains what happens to the gas particles leading to the change in gas volume when pressure changes.

1.3.2 Assumptions of the Kinetic Theory

The following are the assumptions (postulates) of the kinetic theory of gases:

1. A gas consists of tiny particles called atoms or molecules, scattered throughout the container.
2. Gas molecules are in constant random motion and travel with uniform velocities and their direction changes on collision with other gas molecules or with the walls of the container.
3. All collisions between gas molecules are perfectly elastic, hence no loss of kinetic energy during collision.
4. The distance between gas molecules is very large and independent of the 'van der Waals' forces of attraction.
5. Gas pressure is caused by gas molecules colliding with the walls of the container.
6. The average kinetic energy (½mv²) of the gas molecules is directly proportional to the absolute temperature.

1.3.3 Relationship Between the Kinetic Gas Equation and the Gas Laws

The kinetic gas equation is:

PV = ⅓mNv²

where m is the mass of the individual gas molecule, N is the number of molecules, and v is the molecular speed.

This equation can be used to explain Boyle's law, Charles' law, Avogadro's law, and Graham's law of diffusion or effusion.

1.4 Relative Molecular Masses of Gases

The molecular mass is the sum of the atomic masses (in a.m.u, i.e., atomic mass unit) in the molecule. It is numerically equal to the molar mass (in grams) of a compound.

1.4.1 Determination of Molecular Mass by Dumas' Method

Dumas' method of determining the molecular mass was named after a French Chemist, John Dumas (1800 – 1884). This method is used to determine the molecular mass of gases. Since liquids are much easier to handle than gases, volatile liquids are used as sources of gases.

1.4.2 Determination of Molecular Mass by Victor-Meyer's Method

Victor-Meyer's method of determining the molecular mass was named after a German Chemist, Victor Meyer (1848–1897). This method is used for determining the molecular weights of volatile liquids.

1.4.3 Abnormal Vapour Density and Molecular Mass

The vapour density and molar mass found by vaporising volatile solid or liquid substances using Victor Meyer, or Duma's methods may be greater or less than expected due to association or dissociation of molecules.

Exercise 1.1

1. A sample of neon occupies a volume of 461 mL at STP. What will the volume of neon be if pressure is reduced to 93.3 kPa?
2. How does Boyle's law relate to the operation of a syringe or a bicycle pump?
3. A gas tank holds 2785 L of propane (C₃H₈) at 830 Torr. If the temperature is constant, calculate the volume of the propane at standard pressure.
4. Why is the understanding of Gay-Lussac's law important for ensuring the safe storage and transportation of compressed gases?
5. What is the density of helium gas at 298 K and 0.987 atm if its molar mass is 4.003 g/mol?
6. A 6.0 L sample at 25 °C and 2.00 atm contains 0.5 moles of a gas. If 0.25 moles of the gas at the same pressure and temperature are added, calculate the final total volume of the gas.
7. How does the application of Charles's law contribute to the design and operation of hot air balloons?
8. A container holds 50.0 mL of nitrogen gas at 25 °C at a constant pressure of 736 Torr. What will its volume be if the temperature increases by 35 °C?
9. Why are divers advised to ascend slowly as they return to the surface?

Chapter Two: Colligative Properties of Solutions

Introduction

Colligative properties of solutions are fascinating phenomena that occur when solute particles interact with solvent particles, leading to changes in the solution's physical properties. In this chapter, you will learn about the four types of colligative properties and their applications in water purification, dialysis, food preservation and antifreeze. The competencies developed will enable you to apply the colligative properties to understand the solution's responses to environmental changes.

2.1 Properties of Solutions Exhibiting Colligative Properties

Task 2.1
Use online simulation to explore colligative properties and their applications in various systems.

Colligative properties are the properties of solutions that depend on the number of solute particles dissolved and not on the nature of solute particles. For a solution to exhibit colligative properties, it must fulfil the following conditions:

(a) Dilute Solution

The solution should be dilute with a concentration less than or equal to 0.2 M. Colligative properties require that solute particles do not interact with each other, or at least that their interactions do not change with concentration.

(b) Non-electrolyte Solute

The solute should be non-electrolyte. Electrolyte solutes tend to form ions in a solution. Because colligative properties depend on the number of solute particles and not on the nature of particles, the formation of ions by a solute which is an electrolyte would increase the number of particles and hence the interaction between particles of the solute.

(c) Non-volatile Solute

The solute should be non-volatile. The solute is required to be non-volatile because the focus is on the effect of solute particles on the properties of the solvent. When a non-volatile solute is added to a solvent, it does not readily evaporate into the vapour phase.

2.2 Types of Colligative Properties

There are four colligative properties: vapour pressure lowering, boiling point elevation, freezing point depression, and osmotic pressure. These properties change in accordance with the amount of the solute added to the solvent to form a solution.

2.2.1 Vapour Pressure Lowering

Vapour pressure is the pressure exerted by the vapour on the liquid surface when an equilibrium is established between the liquid phase and the vapour phase, where the rate of evaporation is equal to the rate of condensation.

When a non-volatile solute is dissolved in a liquid solvent, the vapour pressure of the solvent is lowered. Unlike in pure solvents, the surface area in the solution is not completely available for a volatile solvent because it is partly occupied by a non-volatile solute.

ΔP = P° - P

where ΔP is the lowering of vapour pressure, P° is the vapour pressure of pure solvent and P is the vapour pressure of the solution.

Raoult's Law

Raoult's law states that the vapour pressure of a solvent in a solution (P_solvent) is equal to the vapour pressure of pure solvent (P°_solvent) times the mole fraction of the solvent (χ_solvent).

P_solvent = P°_solvent × χ_solvent

ΔP = P°_solvent × χ_solute

Example 2.1

Calculate the lowering of vapour pressure caused by the addition of 100 g of sucrose (molecular mass = 342 g mol⁻¹) to 1000 g of water if the vapour pressure of pure water at 25 °C is 23.8 Torr.

Solution:

Given: P°_solvent = 23.8 Torr, mass of sucrose = 100 g, mass of water = 1000 g

Moles of sucrose = 100/342 = 0.29 mol

Moles of water = 1000/18 = 55.56 mol

Relative lowering of vapour pressure = n/(n + N) = 0.29/(0.29 + 55.6)

ΔP = [0.29/(0.29 + 55.6)] × 23.8 Torr = 0.124 Torr

Therefore, the lowering of vapour pressure is 0.124 Torr.

2.2.2 Boiling Point Elevation

The boiling point of a liquid is the temperature at which the vapour pressure of the liquid is equal to the atmospheric pressure.

The vapour pressure of a solvent in a solution of non-volatile solute is always less than that of a pure solvent. Therefore, the solution will boil at a temperature higher than the boiling point of pure solvent.

ΔT_b = T - T° = K_bm

where ΔT_b is the elevation of boiling point, K_b is the molal boiling point elevation constant (ebullioscopic constant), and m is the molality.

Activity 2.1

Aim: To investigate the effect of the amount of solute on the boiling point of pure solvent

Requirements: Thermometer, glass beakers, glass rod, source of heat, measuring cylinder, analytical balance, tripod stand, wire gauze, sugar and water

Procedure:

1. Measure 20 mL of water and pour it into a 100-mL beaker.
2. Heat the water on the heat source and record the temperature at which the water starts to boil.
3. Dissolve 5 g of sugar in another 20 mL of water.
4. Heat the sugar solution and record the temperature at which the water boils.
5. Repeat steps 3 and 4 using 10 g of sugar.

Questions:

1. What is the boiling point elevation of the solution when 5 g of sugar was added?
2. What is the boiling point elevation when 10 g of sugar was added?
3. In what industries is the investigated phenomenon applicable?

Example 2.3

The boiling temperature of a solution prepared by dissolving 5.0 g of an organic solid in 100.0 g of benzene is 82.42 °C. If the boiling temperature of pure benzene is 80.10 °C, what is the molecular weight of the organic solid? (K_b = 2.53 °C/m).

Solution:

ΔT_b = 82.42 °C - 80.10 °C = 2.32 °C

m = ΔT_b/K_b = 2.32/2.53 = 0.917 mol kg⁻¹

Moles of solute = 0.917 × 0.1 = 0.0917 mol

Molar mass = 5.0/0.0917 = 54.5 g mol⁻¹

Therefore, the molecular weight of the organic solid is 54.5 g mol⁻¹.

2.2.3 Freezing Point Depression

The freezing point of a liquid is the temperature at which the vapour pressure of solid is equal to the vapour pressure of liquid.

A solution freezes at lower temperatures than pure solvent. This difference in the freezing point between pure solvent and a solution containing non-volatile solute is called freezing point depression.

ΔT_f = T° - T = K_fm

where ΔT_f is the depression of freezing point, K_f is the molal freezing point depression constant (cryoscopic constant), and m is the molality.

Example 2.5

Determine the freezing point depression of methyl alcohol (CH₃OH) in a solution made by dissolving 18.5 g methyl alcohol in 850 g of water (K_f for water = 1.86 °C kg mol⁻¹).

Solution:

Moles of methyl alcohol = 18.5/32.0 = 0.578 mol

Molality = 0.578/0.85 = 0.680 mol kg⁻¹

ΔT_f = K_fm = 1.86 × 0.68 = 1.26 °C

Therefore, freezing point depression of methyl alcohol solution is 1.26 °C.

2.2.4 Osmotic Pressure

Task 2.2
Use online resources to study the process of water purification.

Osmosis is the movement of solvent molecules from an area of the dilute solution to an area of concentrated solution through a semi-permeable membrane.

Osmotic pressure is the pressure which is applied on the side of a concentrated solution to stop the net flow of the solvent into the concentrated solution through a semi-permeable membrane.

Van't Hoff's Law of Osmotic Pressure

Van't Hoff's law states that the magnitude of osmotic pressure is proportional to the number of solute particles present in a given volume of solution.

πV = nRT

where π is the osmotic pressure, V is the volume, n is the number of moles, R is the gas constant, and T is the absolute temperature.

Example 2.7

A sugar solution with a concentration of 2.5 g dm⁻³ gave an osmotic pressure of 8.3 × 10⁻⁴ atm at 25 °C. Calculate the molar mass of the solute.

Solution:

πV = nRT = (m/M)RT

M = mRT/(πV) = (2.5 × 0.0821 × 298)/(8.3 × 10⁻⁴ × 1) = 7.37 × 10⁴ g mol⁻¹

Therefore, the molar mass of the solute is 7.37 × 10⁴ g mol⁻¹.

2.3 Abnormal Molecular Masses

For solutions of electrolytes, the values of the observed colligative properties can either be higher or lower than the theoretically expected value because the electrolyte solutes such as acid, base or salt can undergo dissociation or association when dissolved in a solvent.

Key Points:

• When a solute undergoes dissociation, the number of particles increases, increasing colligative properties
• When solutes associate, the effective number of particles decreases, decreasing colligative properties
• Dissociation lowers observed molecular mass
• Association increases observed molecular mass

2.3.1 Osmotic Pressure of Electrolytes

The observed osmotic pressure of the electrolyte solutes (π) in aqueous solution is higher than the osmotic pressure calculated using van't Hoff's equation π₀V = n₀RT.

i = π/π₀ = (moles of particles in solution)/(moles of particles in formula units dissolved)

where i is the van't Hoff's factor.

π = iMRT

Example 2.8

The osmotic pressure of equimolal solutions of CaCl₂ and sucrose at 298.15 K are 0.605 atm and 0.224 atm, respectively. Calculate the van't Hoff's factor and the degree of dissociation of CaCl₂.

Solution:

i = π/π₀ = 0.605/0.224 = 2.70

CaCl₂ ⇌ Ca²⁺ + 2Cl⁻

Total particles = 1 - α + α + 2α = 1 + 2α

i = (1 + 2α)/1 = 2.7

α = (2.7 - 1)/2 = 0.85 or 85%

Therefore, van't Hoff's factor is 2.7 and degree of dissociation is 85%.

2.3.2 Freezing Point Depression and Blagden's Law

Blagden's law states that the depression of freezing point of a dilute solution is directly proportional to the concentration of the dissolved solutes.

ΔT_f = K_fmi

Example 2.10

A solution of 20 g sodium chloride is added to 200 g of water to depress the freezing point. Calculate the freezing point depression of the solution; given that, K_f of water = 1.86 °C/m.

Solution:

Moles of NaCl = 20/58.5 = 0.342 mol

Molality = 0.342/0.2 = 1.71 mol/kg

NaCl dissociates to give 2 ions, so i = 2

ΔT_f = K_fmi = 1.86 × 1.71 × 2 = 6.36 °C

Therefore, the freezing point depression is 6.36 °C.

General Formulas for Colligative Properties of Ionic Solutions

Vapour pressure lowering (ΔP) = iχ_soluteP°_solvent

Freezing point depression (ΔT_f) = imK_f

Boiling point elevation (ΔT_b) = imK_b

Osmotic pressure (π) = iMRT

Task 2.3
Review literature (from various sources) on creating coolants for various household uses. Compile a report and share it with your peers. Submit the report to your subject teacher for evaluation.

Exercise 2.1

1. How does the vapour pressure lowering affect cooking processes?
2. A solution is made by dissolving 218 g of glucose (molar mass = 180.2 g/mol) in 460 mL of water at 30 °C. If the vapour pressure of pure water at 30 °C is 31.82 Torr, calculate:
   1. The vapour pressure of the solution
   2. The vapour pressure lowering of the solution
3. The vapour pressure of a 5% aqueous solution of a non-volatile organic substance at 373 K is 745 Torr. What is the molar mass of the organic substance? (The vapour pressure of pure water at 373 K is 760 Torr).

Exercise 2.2

1. Calculate the boiling point of a solution made by dissolving 45.0 g of a non-volatile and non-electrolyte solute with molar mass of 40 g/mol in 500 g of water (K_b for water = 0.512 °C/m).
2. Describe the practical application of boiling point elevation in automotive engineering.
3. Two solutions were prepared, one containing 105 g of sucrose (C₁₂H₂₂O₁₁) in 500 g of water, and the other containing 35.0 g of NaCl in 500 g of water. Which solution will have a higher boiling point? Comment on your answer (K_b = 0.512 °C/m).

Exercise 2.3

1. Ethanoic acid has a freezing point of 16.63 °C. On adding 2.5 g of an organic solute to 40 g of the acid, the freezing point was lowered to 15.48 °C. Calculate the relative molecular mass of the solute. (K_f for ethanoic acid = 3.9 °C/m)
2. Salts such as NaCl or CaCl₂ are spread on ice-covered roads or sidewalks during winter. Explain.
3. How many grams of ethylene glycol (C₂H₆O₂), a non-electrolyte anti-freeze, must be added to 4000 g of water to reduce the melting point to -40 °C? (K_f for water = 1.86 °C/m).

Exercise 2.4

1. How is the concept of osmotic pressure utilised in medical treatment such as dialysis?
2. Explain how osmotic pressure plays a role in the absorption of water by plant roots.
3. If a solution containing 30.0 g L⁻¹ of protein has a measured osmotic pressure of 9.4 Torr at 25 °C, calculate the molar mass of the protein.

Chapter Summary

Colligative properties depend only on the number of solute particles and not on their nature. The four main colligative properties are:

• Vapour pressure lowering: Described by Raoult's law
• Boiling point elevation: ΔT_b = K_bm
• Freezing point depression: ΔT_f = K_fm
• Osmotic pressure: π = MRT

For electrolyte solutions, the van't Hoff factor (i) must be included in the calculations to account for dissociation or association of solute particles.

Chapter Three: Two-Component Liquid Systems

Introduction

The preceding chapter focused on the properties of solutions that do not depend on the identity of particles but only on their relative concentrations. This chapter focuses on the properties of the liquid mixtures that depend on the identity of the individual components. In this chapter, you will learn about completely miscible liquids, immiscible liquids, and the distribution of solutes in immiscible solvents. The competencies developed will enable you to apply steam distillation, solvent extraction, and chromatography in the extraction, separation and purification of substances in everyday life.

Task 3.1
Use online resources such as scientific databases, research articles, educational websites, and interactive simulations to study the interactions in two-component liquid systems and the way they affect the system's behavior.

3.1 Completely Miscible Liquids

Miscible liquids dissolve completely in one another in all proportions to form a homogeneous solution. The resulting solution can be ideal or non-ideal depending on the strength of intermolecular forces of attraction between the components involved in forming the solution.

3.1.1 Formation of Miscible Liquid Mixture

Miscible liquids are formed between compounds which have similar types of intermolecular forces. For example, ethanol and water are miscible because they can form hydrogen bonds with each other. Likewise, ethane (C₂H₆) and propane (C₃H₈) are miscible because they both exhibit dispersion forces.

[Diagram: Formation of miscible liquid mixture showing two liquids combining to form a homogeneous solution]

3.1.2 Ideal Solutions

An ideal solution is one that is made from entirely miscible liquids whose intermolecular forces are the same as the intermolecular forces within the liquids involved in forming the solution.

Raoult's Law for Ideal Solutions:

P_A = χ_A × P°_A and P_B = χ_B × P°_B

where P_A and P_B are the partial vapour pressures, χ_A and χ_B are the mole fractions, and P°_A and P°_B are the vapour pressures of pure components.

According to Dalton's law of partial pressure, the total vapour pressure above the solution containing the volatile components A and B is:

P_T = P_A + P_B

3.1.3 Non-ideal Solution

A non-ideal solution is a solution that is made from completely miscible liquids whose intermolecular forces of attraction are different from the intermolecular forces of the liquid components that form the solution.

Positive Deviation from Raoult's Law

Positive deviation occurs when the intermolecular forces of attraction between particles in the solution (A–B) are less than those in pure components (A–A or B–B).

Properties of non-ideal solutions with positive deviation:

• Do not obey Raoult's law at any concentration
• Heat is absorbed during mixing (endothermic)
• Total volume of the solution increases
• P_A > χ_A × P°_A and P_B > χ_B × P°_B

Negative Deviation from Raoult's Law

Negative deviation occurs when the intermolecular forces of attraction between molecules in the solution (A–B) are stronger than those between pure components (A–A or B–B).

Properties of non-ideal solutions with negative deviation:

• Do not obey Raoult's law at any concentrations
• Heat is evolved during mixing (exothermic)
• Total volume of the solution decreases
• P_A < χ_A × P°_A and P_B < χ_B × P°_B

Activity 3.1

Aim: To investigate positive and negative deviation from Raoult's law in non-ideal solutions

Requirements: Acetone, ethanol, water, methanol, cyclohexane, chloroform, nitric acid, sulfuric acid, hydrochloric acid, thermometers, beakers and 50-mL measuring cylinders

Procedure:

1. Measure 10 mL of acetone, ethanol, water, methanol, cyclohexane, chloroform, hexane, sulfuric acid, and nitric acid, each in separate 100 mL beakers.
2. Record the temperature of each liquid.
3. Mix the liquids to form various solutions and record volume and temperature changes.

Questions:

1. Which solution(s) obey(s) the Raoult's law?
2. Which solution(s) deviate(s) from Raoult's law? Specify the kind of deviation.
3. How might the deviations impact industrial processes?

3.1.4 Separation of Miscible Liquid Mixture by Distillation

Task 3.2
Search on the internet or any other resources about useful products obtained by distillation and explain the principle behind their production.

Distillation is a method of separating and purifying liquid mixtures into their individual components based on their differences in boiling points.

Simple Distillation

Used when the difference in boiling points of the liquids is large. Involves heating the liquid mixture to form vapour and condensing the vapour to form liquid.

Fractional Distillation

Used for mixtures with closer boiling points. Involves a fractionating column where successive distillations occur.

[Diagram: Fractional distillation setup showing fractionating column, condenser, and collection flasks]

3.1.5 Applications of Fractional Distillation

Fractional distillation has numerous practical applications:

• Separation of crude oil into various components
• Chemical plants for large-scale manufacturing of chemicals
• Natural gas processing
• Purification of water
• Production of alcoholic beverages

3.2 Immiscible Liquids

Immiscible liquids are liquids which when mixed together in any proportions form two separate phases.

3.2.1 Formation of Immiscible Liquid Mixture

Liquids are immiscible with each other when the intermolecular forces are not alike. For example, water and butane are immiscible because water exhibits hydrogen bonding while butane exhibits only dispersion forces.

3.2.2 Properties of Immiscible Liquids

• Form two separate layers when mixed
• Lower layer is the liquid with highest density
• Boiling point of mixture is lower than that of either liquid
• Total vapour pressure P_t = P°_A + P°_B

3.2.3 Mixture of Partially Miscible Liquids

Partially miscible liquids dissolve in one another to a limited extent. Examples are mixtures of phenol and water, and ether and water.

3.2.4 Applications of the Knowledge of Immiscible Liquids

Steam Distillation

Steam distillation is the process whereby steam is passed through an impure liquid that is immiscible with water and in the presence of steam, where the liquid compound is made volatile and distils with steam.

[Diagram: Steam distillation setup showing steam generator, distillation flask, condenser, and collection flask]

Example 3.1

At a pressure of 760 Torr, a mixture of nitrobenzene (C₆H₅NO₂) and water boils at 99 °C. The vapour pressure of water at this temperature is 733 Torr. Find the ratio of water to nitrobenzene in the distillate obtained by steam distillation of impure nitrobenzene.

Solution:

Vapour pressure of nitrobenzene = 760 - 733 = 27 Torr

Molecular mass of nitrobenzene = 123 g mol⁻¹

Using distribution formula:

m_organic/m_water = (P_organic × M_organic)/(P_water × 18)

m_nitrobenzene/m_water = (27 × 123)/(733 × 18) = 0.25

Therefore, ratio of water to nitrobenzene = 4:1

3.3 Distribution of Solutes in Immiscible Solvents

When two immiscible liquids are mixed in the same container, they separate into two layers. When a solute soluble in both solvents is added, it distributes itself between the two solvents.

3.3.1 The Distribution Law

Distribution Law:

K_D = C₁/C₂

where K_D is the distribution coefficient, and C₁ and C₂ are concentrations of solute in solvents 1 and 2 respectively.

Example 3.3

A solid X is added to a mixture of benzene and water. After shaking well, 10 mL of benzene layer contains 0.13 g of X and 100 mL of water layer contains 0.22 g of X. Calculate the distribution coefficient.

Solution:

C_benzene = 0.13/10 = 0.013 g mL⁻¹

C_water = 0.22/100 = 0.0022 g mL⁻¹

K_D = C_benzene/C_water = 0.013/0.0022 = 5.9

Since K_D > 1, solute dissolves more in benzene than water.

3.3.2 Applications of the Distribution Law in Solvent Extraction

Solvent extraction is a method by which a compound is pulled out from one solvent to another where the two solvents are immiscible.

Multiple Extraction

Multiple extraction is more efficient than single extraction. Using several small portions of solvent extracts more solute than using one large portion.

Example 3.7

The distribution coefficient of Y between benzene and water is 10. Calculate the amount of Y extracted if 1 g dissolved in 100 mL of water is equilibrated with 100 mL of benzene.

Solution:

Let x be amount extracted

K_D = (x/100)/((1-x)/100) = x/(1-x) = 10

x = 10(1-x) ⇒ x = 10 - 10x ⇒ 11x = 10 ⇒ x = 10/11 = 0.909 g

Therefore, 0.909 g is extracted.

3.3.3 Applications of Distribution Law in Chromatography

Chromatography is a technique that separates the components of a mixture due to the difference in the affinity between the mobile phase and stationary phase.

Types of Chromatography:

• Column chromatography: Uses solid adsorbent as stationary phase
• Paper chromatography: Uses filter paper as stationary phase
• Thin layer chromatography (TLC): Uses thin layer of adsorbent

Exercise 3.1

1. How does the understanding of ideal solutions contribute to the development of pharmaceutical formulations?
2. What factors influence the miscibility of liquids?
3. Which of the following substances would be most soluble in CCl₄?
   1. CH₃CH₂OH
   2. H₂O
   3. NH₃
   4. C₁₀H₂₂
   5. NaCl

Exercise 3.2

1. With examples, explain any liquid mixtures that can be separated into pure components by fractional distillation.
2. How does fractional distillation contribute to producing purified chemicals?

Exercise 3.3

1. How practical is steam distillation in the pharmaceutical industry?
2. Why is steam distillation proper for thermally unstable compounds?
3. Calculate the molecular mass of compound B whose mixture with water distils at 95.2 °C with vapour pressures 119 Torr (B) and 641 Torr (water), and distillation ratio 1.62:1 (B:water).

Task 3.4
Evaluate the methods traditional healers and home remedies use to harness therapeutic properties of medicinal plants to create remedies tailored to specific health conditions.

Chapter Four: Energetics

Understanding Energy Changes in Chemical Reactions

Introduction

Energetics encompasses various principles and laws that govern the transformation, transfer and utilisation of energy in different systems. In this chapter, you will learn about the heat of the reaction, the enthalpy of the reaction, and the application of Hess's law in determining enthalpy changes for chemical reactions. The competencies developed are important in various aspects such as measuring the energy content of food and determining specific enthalpies of fuels, refrigeration and air conditioning.

Task 4.1
Explore online resources to learn about energetics and its application in food preparation, transport, health care, and environment.

4.1 Heat of Reaction

In chemical reactions, energy changes are involved in forming and breaking bonds. The study of chemical processes leading to energy changes is called energetics. A branch of chemistry that describes the energy changes that occur during chemical reactions is referred to as thermochemistry.

Heat is a form of energy that flows between two objects or two parts of an object because of the difference in temperature. Many chemical reactions involve the transfer of heat from the surroundings to the system or vice versa. The amount of heat absorbed or evolved when the reaction goes to completion is called the heat of the reaction.

Temperature vs. Heat

A physical property that quantitatively expresses the degree of hotness or coldness of an object is called temperature. When substances become warmer, it means that the average kinetic energy of their particles has increased.

Key Difference: Temperature is independent of the amount of substance present, while heat content depends on the amount of substance present.

Units of Heat Energy

• Joule (J): SI unit of energy (1 J = 1 N·m = 1 kg·m²/s²)
• Kilojoule (kJ): 1 kJ = 1000 J
• Calorie (cal): Amount of heat required to raise temperature of 1g water by 1°C (1 cal = 4.184 J)
• Kilocalorie (kcal): 1 kcal = 1000 cal = 4.184 kJ

4.1.1 Heat Capacity and Specific Heat Capacity

Heat capacity (C) is the amount of heat required to raise the temperature of a substance by one degree Celsius (1°C) or one Kelvin (1K).

Specific heat capacity (c) is the amount of heat required to raise the temperature of a unit mass of a substance by one degree Celsius (1°C) or one Kelvin (1K).

q = m × c × ΔT

where:
q = heat energy (J)
m = mass (kg)
c = specific heat capacity (J kg⁻¹ °C⁻¹)
ΔT = temperature change (°C or K)

Example 4.1

Calculate the amount of heat energy required to increase the temperature of 20 g of nickel from 50°C to 70°C given that the specific heat capacity of nickel is equal to 440 J kg⁻¹°C⁻¹.

Solution:

m = 20 g = 0.02 kg

ΔT = 70°C - 50°C = 20°C

c = 440 J kg⁻¹°C⁻¹

q = m × c × ΔT = 0.02 × 440 × 20 = 176 J

Therefore, the heat energy required is 176 J.

Example 4.2

Calculate the heat capacity of 80.0 g of water given that the specific heat capacity of water is 4.18 J g⁻¹°C⁻¹.

Solution:

C = m × c

C = 80 g × 4.18 J g⁻¹°C⁻¹ = 334.4 J°C⁻¹

Therefore, the heat capacity of 80 g of water is 334.4 J°C⁻¹.

4.1.2 Endothermic and Exothermic Reactions

Endothermic reaction: A chemical reaction which involves the transfer of heat from the surroundings to the reaction system (ΔH > 0).

Exothermic reaction: A chemical reaction which involves the transfer of heat from the reaction system to the surroundings (ΔH < 0).

Energy Profile Diagrams

[Energy profile diagram showing endothermic and exothermic reactions]
Endothermic: Products at higher energy than reactants
Exothermic: Products at lower energy than reactants

Examples:

Endothermic: NH₄NO₃(s) → NH₄NO₃(aq)   ΔH > 0

Exothermic: CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l)   ΔH < 0

4.2 Enthalpy of Reaction

Enthalpy (H) is the amount of heat absorbed or evolved at constant temperature and pressure. The corresponding amount of heat change during the reaction is called enthalpy change (ΔH).

ΔH = H_products - H_reactants

Thermochemical Equations

A thermochemical equation is a balanced stoichiometric chemical equation that includes the enthalpy change (ΔH).

Example: HNO₃(aq) + NaOH(aq) → NaNO₃(aq) + H₂O(l)   ΔH = -57.2 kJ mol⁻¹

Standard Conditions: ΔH° denotes standard enthalpy change where reactants and products are in their standard states (gases at 1 atm, solutions at unit concentration, substances in standard states).

4.2.1 Types of Enthalpy Changes

Type of Enthalpy	Definition	Example
Enthalpy of Formation (ΔHf)	Heat change when 1 mole of compound is formed from its elements	C(s) + O₂(g) → CO₂(g)
Enthalpy of Combustion (ΔHc)	Heat change when 1 mole of substance is completely burned in oxygen	CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l)
Enthalpy of Neutralization (ΔHn)	Heat change when 1 mole of water is formed from acid-base reaction	HCl(aq) + NaOH(aq) → NaCl(aq) + H₂O(l)
Enthalpy of Solution (ΔHsol)	Heat change when 1 mole of solute is dissolved in solvent	NH₄NO₃(s) → NH₄NO₃(aq)

4.2.2 Measurement of Heat of Reaction by Calorimetry

Calorimetry is the experimental method of measuring heat changes in chemical reactions using a calorimeter.

Calorimetry Experiment

Aim: To determine the enthalpy of neutralization of HCl and NaOH

Requirements: Calorimeter, thermometer, 1M HCl, 1M NaOH, measuring cylinder

Procedure:

1. Measure 50 mL of 1M HCl into the calorimeter
2. Record initial temperature
3. Add 50 mL of 1M NaOH quickly and stir
4. Record maximum temperature reached
5. Calculate heat evolved using q = m × c × ΔT

4.2.3 Hess's Law of Constant Heat Summation

Hess's Law: The total enthalpy change for a reaction is the same regardless of the route taken, provided the initial and final conditions are the same.

Application of Hess's Law

If: A → B   ΔH₁
B → C   ΔH₂
Then: A → C   ΔH = ΔH₁ + ΔH₂

4.2.4 Bond Energy

Bond energy is the average energy required to break one mole of a particular type of bond in gaseous molecules.

ΔH = Σ(bond energies of reactants) - Σ(bond energies of products)

4.2.5 Born-Haber Cycle

The Born-Haber cycle is an application of Hess's law used to calculate lattice energies of ionic compounds.

[Born-Haber cycle diagram for NaCl]
Showing the stepwise energy changes in formation of ionic compound

Exercise 4.1

1. Define the term "specific heat capacity" and give its SI units.
2. Calculate the heat required to raise the temperature of 250 g of water from 20°C to 80°C. (Specific heat capacity of water = 4.18 J g⁻¹°C⁻¹)
3. Distinguish between endothermic and exothermic reactions with suitable examples.

Exercise 4.2

1. What is meant by the term "enthalpy change"?
2. State Hess's law and explain its importance in thermochemistry.
3. Calculate the standard enthalpy of formation of methane (CH₄) given:
   • C(s) + O₂(g) → CO₂(g)   ΔH = -393.5 kJ mol⁻¹
   • H₂(g) + ½O₂(g) → H₂O(l)   ΔH = -285.8 kJ mol⁻¹
   • CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l)   ΔH = -890.3 kJ mol⁻¹

Exercise 4.3

1. Explain how calorimetry can be used to determine the heat of reaction.
2. What is bond energy and how is it related to the enthalpy of reaction?
3. Describe the Born-Haber cycle for the formation of sodium chloride.

Chapter Summary

• Energetics deals with energy changes in chemical reactions
• Heat capacity and specific heat capacity determine temperature changes
• Endothermic reactions absorb heat (ΔH > 0), exothermic reactions release heat (ΔH < 0)
• Enthalpy change (ΔH) is measured under constant pressure conditions
• Hess's law allows calculation of enthalpy changes for complex reactions
• Bond energies and Born-Haber cycles provide theoretical approaches to enthalpy calculations

Research Task
Investigate the energy content of common fuels and foods. Compare the enthalpy of combustion of different fuels (petrol, diesel, natural gas) and the calorific values of different foods. Present your findings in a report.

Chapter Five: Chemical Equilibrium

Understanding Dynamic Balance in Chemical Reactions

Introduction

Chemical equilibrium represents a fundamental concept in chemistry where opposing chemical reactions occur at equal rates, resulting in no net change in the concentrations of reactants and products. This chapter explores reversible reactions, equilibrium constants, and factors affecting chemical equilibrium. Understanding equilibrium is crucial for predicting reaction behavior and optimizing industrial chemical processes.

Task 5.1
Research industrial processes that utilize chemical equilibrium principles, such as the Haber process for ammonia synthesis or the Contact process for sulfuric acid production. Analyze how equilibrium concepts are applied to maximize product yield.

5.1 Reversible Reactions

A reversible reaction is a chemical reaction that can proceed in both forward and reverse directions. Such reactions are denoted by a double arrow (⇌) in chemical equations.

N₂(g) + 3H₂(g) ⇌ 2NH₃(g)

Forward Reaction
Reactants → Products

⇌

Reverse Reaction
Products → Reactants

5.1.1 Characteristics of Chemical Equilibria

Key Characteristics of Equilibrium:

• Dynamic process - reactions continue in both directions
• Constant concentrations of reactants and products
• Equal rates of forward and reverse reactions
• Can be approached from either direction
• Macroscopic properties remain constant

5.1.2 The Law of Mass Action

Law of Mass Action: The rate of a chemical reaction is proportional to the product of the concentrations of the reactants, each raised to the power of their stoichiometric coefficients.

aA + bB ⇌ cC + dD

5.2 Equilibrium Constants

5.2.1 Equilibrium Constant in Terms of Concentration (K_c)

K_c = [C]^c[D]^d / [A]^a[B]^b

Example 5.1

For the reaction: N₂(g) + 3H₂(g) ⇌ 2NH₃(g)
Write the expression for the equilibrium constant K_c.

Solution:

K_c = [NH₃]² / [N₂][H₂]³

5.2.2 Equilibrium Constant in Terms of Partial Pressure (K_p)

K_p = (P_C)^c(P_D)^d / (P_A)^a(P_B)^b

Example 5.2

For the reaction: 2SO₂(g) + O₂(g) ⇌ 2SO₃(g)
Write the expression for K_p.

Solution:

K_p = (P_SO₃)² / (P_SO₂)²(P_O₂)

5.2.3 Reaction Quotient (Q)

The reaction quotient (Q) has the same form as the equilibrium constant but uses initial concentrations rather than equilibrium concentrations.

Comparing Q and K:

• If Q < K: Reaction proceeds forward
• If Q = K: System at equilibrium
• If Q > K: Reaction proceeds reverse

5.2.4 Relationship Between K_c and K_p

K_p = K_c(RT)^Δn

where:
R = gas constant (0.0821 L·atm·mol⁻¹·K⁻¹)
T = temperature in Kelvin
Δn = (moles of gaseous products) - (moles of gaseous reactants)

Example 5.3

For the reaction: N₂(g) + 3H₂(g) ⇌ 2NH₃(g)
Calculate K_p at 298 K if K_c = 3.5 × 10⁸

Solution:

Δn = 2 - (1 + 3) = -2

K_p = K_c(RT)^Δn = 3.5 × 10⁸ × (0.0821 × 298)^-2

K_p = 3.5 × 10⁸ × (24.4658)^-2

K_p = 3.5 × 10⁸ × 0.00167 = 5.85 × 10⁵

5.3 Factors Affecting Chemical Equilibrium

Le Châtelier's Principle: If a system at equilibrium is subjected to a change in concentration, pressure, or temperature, the system will adjust to counteract that change and establish a new equilibrium.

5.3.1 Effect of Change in Concentration

Increasing reactant concentration: Equilibrium shifts toward products
Increasing product concentration: Equilibrium shifts toward reactants

5.3.2 Effect of Change in Pressure

Increasing pressure: Equilibrium shifts toward side with fewer moles of gas
Decreasing pressure: Equilibrium shifts toward side with more moles of gas

Example 5.4

Predict the effect of increasing pressure on the equilibrium:
N₂(g) + 3H₂(g) ⇌ 2NH₃(g)

Solution:

Left side: 1 + 3 = 4 moles of gas

Right side: 2 moles of gas

Increasing pressure favors the side with fewer moles

Equilibrium shifts to the RIGHT (toward NH₃)

5.3.3 Effect of Change in Temperature

Increasing temperature: Equilibrium shifts in the endothermic direction
Decreasing temperature: Equilibrium shifts in the exothermic direction

5.3.4 Effect of Catalyst

Catalysts do NOT affect the position of equilibrium. They only speed up the rate at which equilibrium is achieved by lowering the activation energy for both forward and reverse reactions.

5.4 Application of Le Châtelier's Principle in Industrial Processes

Case Study: Haber Process

Reaction: N₂(g) + 3H₂(g) ⇌ 2NH₃(g)   ΔH = -92 kJ/mol

Optimal Conditions:

• Temperature: 400-500°C (compromise between rate and yield)
• Pressure: 200-300 atm (favors product formation)
• Catalyst: Iron with promoters (speeds up equilibrium attainment)

Industrial Process	Reaction	Optimal Conditions	Equilibrium Principle Applied
Haber Process	N₂ + 3H₂ ⇌ 2NH₃	450°C, 200 atm	High pressure favors fewer moles
Contact Process	2SO₂ + O₂ ⇌ 2SO₃	450°C, 1-2 atm	Excess oxygen shifts equilibrium
Ostwald Process	4NH₃ + 5O₂ ⇌ 4NO + 6H₂O	850°C, 5-10 atm	High temperature for endothermic reaction

Exercise 5.1

1. Define chemical equilibrium and list three characteristics of systems at equilibrium.
2. Write the equilibrium constant expression (K_c) for the reaction:
   2NO(g) + O₂(g) ⇌ 2NO₂(g)
3. Explain why a catalyst does not affect the position of equilibrium.

Exercise 5.2

1. For the reaction: CO(g) + 2H₂(g) ⇌ CH₃OH(g)
   K_c = 14.5 at 500 K. Calculate K_p at this temperature.
2. Using Le Châtelier's principle, predict the effect of:
   1. Increasing temperature on an endothermic reaction
   2. Decreasing pressure on a reaction with more moles of gas on the product side
   3. Adding more reactant to a system at equilibrium

Exercise 5.3

1. The equilibrium constant for the reaction:
   H₂(g) + I₂(g) ⇌ 2HI(g) is 54.0 at 700 K.
   If initial concentrations are [H₂] = 0.100 M, [I₂] = 0.100 M, and [HI] = 0.300 M,
   calculate the reaction quotient Q and determine the direction the reaction will proceed.
2. Explain how the Haber process utilizes Le Châtelier's principle to maximize ammonia yield.
3. For the reaction: A + B ⇌ C + D, K_c = 1.0 × 10⁻² at 25°C.
   If initial concentrations are [A] = 0.100 M, [B] = 0.100 M, [C] = 0, [D] = 0,
   calculate the equilibrium concentrations of all species.

Chapter Summary

• Chemical equilibrium occurs when forward and reverse reaction rates are equal
• Equilibrium constants (K_c and K_p) quantify equilibrium positions
• Reaction quotient (Q) predicts direction of reaction shift
• Le Châtelier's principle describes how systems respond to disturbances
• Industrial processes optimize conditions using equilibrium principles
• Catalysts affect reaction rates but not equilibrium positions

Research Project
Investigate an industrial chemical process that relies on equilibrium principles (other than those mentioned in this chapter). Prepare a report explaining how Le Châtelier's principle is applied to optimize production, including specific temperature, pressure, and concentration conditions used.

Visualizing Equilibrium

[Equilibrium concentration vs. time graph]
Showing how reactant and product concentrations change over time
until reaching constant equilibrium concentrations

Chapter Six: Ionic Equilibria

Understanding Acid-Base Equilibria and Buffer Systems

Introduction

Ionic equilibria deals with the equilibrium established between ions and unionized molecules in electrolyte solutions. This chapter explores various acid-base concepts, pH calculations, buffer solutions, salt hydrolysis, and solubility equilibria. Understanding ionic equilibria is crucial for predicting solution behavior, controlling chemical processes, and applications in biological systems.

Task 6.1
Research the importance of pH control in biological systems, industrial processes, and environmental science. Prepare a report on how ionic equilibria principles are applied in maintaining optimal pH conditions.

6.1 Concept of Acids and Bases

6.1.1 Arrhenius Concept of Acids and Bases

Arrhenius Acid: A substance that dissociates in water to produce H⁺ ions
Arrhenius Base: A substance that dissociates in water to produce OH⁻ ions

HCl → H⁺ + Cl⁻     (Acid)
NaOH → Na⁺ + OH⁻     (Base)

6.1.2 Brønsted-Lowry Concept of Acids and Bases

Brønsted-Lowry Acid: Proton (H⁺) donor
Brønsted-Lowry Base: Proton (H⁺) acceptor

HCl + NH₃ ⇌ NH₄⁺ + Cl⁻
Acid    Base     Acid     Base

6.1.3 Lewis Concept of Acids and Bases

Lewis Acid: Electron pair acceptor
Lewis Base: Electron pair donor

Example 6.1

Identify the Lewis acid and base in the reaction: BF₃ + NH₃ → F₃B-NH₃

Solution:

BF₃ accepts an electron pair → Lewis Acid

NH₃ donates an electron pair → Lewis Base

6.2 Ionic Equilibria of Acids and Bases

6.2.1 Acid Dissociation Constant (K_a)

HA ⇌ H⁺ + A⁻     K_a = [H⁺][A⁻] / [HA]

6.2.2 Base Dissociation Constant (K_b)

B + H₂O ⇌ BH⁺ + OH⁻     K_b = [BH⁺][OH⁻] / [B]

6.2.3 Ionic Product of Water and pH

Ionic Product of Water: K_w = [H⁺][OH⁻] = 1.0 × 10⁻¹⁴ at 25°C

pH = -log[H⁺]     pOH = -log[OH⁻]     pH + pOH = 14

The pH Scale

0 3 6 7 8 11 14

Strong Acid Weak Acid Neutral Weak Base Strong Base

Example 6.2

Calculate the pH of a solution with [H⁺] = 3.5 × 10⁻⁵ M

Solution:

pH = -log[H⁺] = -log(3.5 × 10⁻⁵)

pH = -[log(3.5) + log(10⁻⁵)]

pH = -[0.544 - 5] = -[-4.456] = 4.456

6.3 Buffer Solutions

A buffer solution resists changes in pH when small amounts of acid or base are added. Buffers consist of a weak acid and its conjugate base, or a weak base and its conjugate acid.

6.3.1 Types of Buffer Solutions

Buffer Type	Components	Example	pH Range
Acidic Buffer	Weak acid + Salt of conjugate base	CH₃COOH + CH₃COONa	3-6
Basic Buffer	Weak base + Salt of conjugate acid	NH₃ + NH₄Cl	8-10

6.3.2 Henderson-Hasselbalch Equation

pH = pK_a + log([A⁻]/[HA])

Example 6.3

Calculate the pH of a buffer containing 0.1 M CH₃COOH and 0.2 M CH₃COONa (K_a = 1.8 × 10⁻⁵)

Solution:

pK_a = -log(1.8 × 10⁻⁵) = 4.745

pH = pK_a + log([A⁻]/[HA])

pH = 4.745 + log(0.2/0.1) = 4.745 + log(2)

pH = 4.745 + 0.301 = 5.046

6.3.3 Mechanism of Buffer Action

When acid is added: H⁺ + A⁻ → HA (conjugate base neutralizes added acid)
When base is added: OH⁻ + HA → A⁻ + H₂O (weak acid neutralizes added base)

6.4 Salt Hydrolysis

Salt hydrolysis is the reaction of salt ions with water to produce acidic or basic solutions.

6.4.1 Behavior of Salts in Water

Salt Type	Components	Hydrolysis	Solution pH
Strong acid + Strong base	NaCl, KNO₃	No hydrolysis	pH = 7 (Neutral)
Strong acid + Weak base	NH₄Cl, CuSO₄	Cation hydrolysis	pH < 7 (Acidic)
Weak acid + Strong base	CH₃COONa, KCN	Anion hydrolysis	pH > 7 (Basic)
Weak acid + Weak base	CH₃COONH₄	Both ions hydrolyze	Depends on Ka and Kb

6.5 Acid-Base Titration

Titration Experiment

Aim: To determine the concentration of an unknown acid using standard base solution

Requirements: Burette, pipette, conical flask, indicator, standard NaOH solution, unknown HCl solution

Procedure:

1. Pipette 25 mL of unknown acid into conical flask
2. Add 2-3 drops of phenolphthalein indicator
3. Titrate with standard NaOH solution until faint pink color persists
4. Record volume of NaOH used
5. Calculate acid concentration using M_aV_a = M_bV_b

6.6 Solubility Equilibria

6.6.1 Solubility Product Constant (K_sp)

A_xB_y(s) ⇌ xA^y+(aq) + yB^x-(aq)     K_sp = [A^y+]^x[B^x-]^y

Example 6.4

Write the K_sp expression for Ag₂CrO₄ and calculate its solubility if K_sp = 1.1 × 10⁻¹²

Solution:

Ag₂CrO₄(s) ⇌ 2Ag⁺(aq) + CrO₄²⁻(aq)

K_sp = [Ag⁺]²[CrO₄²⁻]

Let solubility = s, then [Ag⁺] = 2s, [CrO₄²⁻] = s

K_sp = (2s)²(s) = 4s³ = 1.1 × 10⁻¹²

s³ = 2.75 × 10⁻¹³ → s = 6.5 × 10⁻⁵ M

Exercise 6.1

1. Differentiate between Arrhenius, Brønsted-Lowry, and Lewis acid-base concepts.
2. Calculate the pH of 0.01 M HCl solution.
3. Identify conjugate acid-base pairs in the reaction: H₂PO₄⁻ + OH⁻ ⇌ HPO₄²⁻ + H₂O

Exercise 6.2

1. Explain how a buffer solution resists changes in pH.
2. Calculate the pH of a buffer containing 0.2 M NH₃ and 0.3 M NH₄Cl (K_b = 1.8 × 10⁻⁵)
3. Predict whether solutions of the following salts are acidic, basic, or neutral:
   1. NaCl
   2. NH₄NO₃
   3. KCN
   4. CH₃COONH₄

Exercise 6.3

1. The K_sp of BaSO₄ is 1.1 × 10⁻¹⁰. Calculate its solubility in mol/L.
2. 25.0 mL of 0.100 M HCl is titrated with 0.100 M NaOH. Calculate the pH:
   1. Before adding NaOH
   2. After adding 24.9 mL NaOH
   3. At equivalence point
   4. After adding 25.1 mL NaOH
3. Explain the common ion effect and its application in qualitative analysis.

Chapter Summary

• Acids and bases can be defined by Arrhenius, Brønsted-Lowry, and Lewis concepts
• pH scale quantifies acidity: pH = -log[H⁺]
• Buffer solutions resist pH changes using weak acid-base pairs
• Salt hydrolysis determines whether salt solutions are acidic, basic, or neutral
• Solubility equilibria are described by K_sp values
• Titrations are used to determine unknown concentrations

Research Project
Investigate the role of buffer systems in human blood (carbonate-bicarbonate buffer) and their importance in maintaining physiological pH. Explain what happens when this buffer system fails and the consequences of acidosis and alkalosis.

Chapter Seven: Chemical Kinetics

Studying Reaction Rates and Mechanisms

Introduction

Chemical kinetics is the branch of chemistry that deals with the rates of chemical reactions, the factors affecting these rates, and the mechanisms by which reactions occur. While thermodynamics tells us whether a reaction can occur, kinetics tells us how fast it will occur. This chapter explores reaction rates, rate laws, reaction orders, and the factors that influence reaction speeds.

Task 7.1
Investigate real-world applications of chemical kinetics in pharmaceutical drug design, food preservation, and industrial manufacturing. Prepare a report on how understanding reaction rates helps optimize these processes.

7.1 The Rate of Chemical Reactions

The reaction rate is defined as the change in concentration of a reactant or product per unit time. For a general reaction: aA + bB → cC + dD

Rate = -1/a Δ[A]/Δt = -1/b Δ[B]/Δt = 1/c Δ[C]/Δt = 1/d Δ[D]/Δt

Reactants
A + B

→

Products
C + D

Example 7.1

For the reaction: 2N₂O₅(g) → 4NO₂(g) + O₂(g)
If [N₂O₅] decreases from 0.100 M to 0.075 M in 5 minutes, calculate the average rate of reaction.

Solution:

Rate = -1/2 Δ[N₂O₅]/Δt

Δ[N₂O₅] = 0.075 - 0.100 = -0.025 M

Δt = 5 minutes

Rate = -1/2 × (-0.025)/5 = 0.0025 M/min

7.2 Rate Laws

The rate law expresses the relationship between the reaction rate and the concentrations of reactants. For a reaction: aA + bB → products

Rate = k[A]^m[B]ⁿ

where:
k = rate constant
m = order with respect to A
n = order with respect to B
Overall order = m + n

7.3 Order of Reactions

7.3.1 Zero-Order Reactions Order 0

Rate Law: Rate = k
Integrated Rate Law: [A] = [A]₀ - kt
Half-life: t_½ = [A]₀/2k

7.3.2 First-Order Reactions Order 1

Rate Law: Rate = k[A]
Integrated Rate Law: ln[A] = ln[A]₀ - kt
Half-life: t_½ = 0.693/k

Example 7.2

The decomposition of H₂O₂ is first-order with k = 7.30 × 10⁻⁴ s⁻¹. Calculate the time required for [H₂O₂] to decrease from 0.100 M to 0.025 M.

Solution:

ln[A] = ln[A]₀ - kt

ln(0.025) = ln(0.100) - (7.30 × 10⁻⁴)t

-3.689 = -2.303 - (7.30 × 10⁻⁴)t

t = (-3.689 + 2.303)/(-7.30 × 10⁻⁴) = 1900 seconds

7.3.3 Second-Order Reactions Order 2

Rate Law: Rate = k[A]²
Integrated Rate Law: 1/[A] = 1/[A]₀ + kt
Half-life: t_½ = 1/k[A]₀

Reaction Order	Rate Law	Integrated Rate Law	Half-life	Linear Plot
Zero	Rate = k	[A] = [A]₀ - kt	[A]₀/2k	[A] vs t
First	Rate = k[A]	ln[A] = ln[A]₀ - kt	0.693/k	ln[A] vs t
Second	Rate = k[A]²	1/[A] = 1/[A]₀ + kt	1/k[A]₀	1/[A] vs t

7.4 Experimental Determination of Rate Laws

Initial Rates Method

Aim: To determine the rate law for the reaction between iodine and acetone

Procedure:

1. Prepare solutions with different initial concentrations
2. Measure initial rates by monitoring color change
3. Compare rates to determine reaction orders
4. Calculate rate constant k

7.5 Factors Affecting the Rate of Chemical Reactions

7.5.1 Concentration of Reactants

Higher concentrations generally lead to faster reaction rates due to more frequent collisions.

7.5.2 Temperature

Arrhenius Equation: k = Ae^-Ea/RT
where A = frequency factor, Ea = activation energy, R = gas constant, T = temperature

Example 7.3

A reaction has Ea = 50 kJ/mol. By what factor does the rate increase when temperature rises from 25°C to 35°C?

Solution:

k₂/k₁ = e^{(-Ea/R)(1/T₂ - 1/T₁)}

T₁ = 298 K, T₂ = 308 K, Ea = 50,000 J/mol, R = 8.314 J/mol·K

k₂/k₁ = e^{(-50000/8.314)(1/308 - 1/298)}

k₂/k₁ = e^2.18 = 8.85

Rate increases by factor of 8.85

7.5.3 Surface Area

Increasing surface area of solids increases reaction rates by providing more collision sites.

7.5.4 Catalysts

A catalyst is a substance that increases the reaction rate without being consumed in the reaction. Catalysts work by providing an alternative reaction pathway with lower activation energy.

Energy Profile Diagram

[Energy profile showing uncatalyzed and catalyzed reaction pathways]
Catalyzed pathway has lower activation energy

7.6 Reaction Mechanisms

A reaction mechanism is the step-by-step sequence of elementary reactions by which overall chemical change occurs.

Example: Ozone Decomposition

Overall reaction: 2O₃ → 3O₂
Mechanism:
Step 1: O₃ → O₂ + O    (slow)
Step 2: O + O₃ → 2O₂    (fast)

Rate-Determining Step: The slowest step in a reaction mechanism determines the overall reaction rate.

Exercise 7.1

1. Define chemical kinetics and explain its importance.
2. For the reaction: 2HI(g) → H₂(g) + I₂(g)
   If [HI] decreases from 0.500 M to 0.250 M in 100 seconds, calculate the average rate.
3. Differentiate between rate law and rate constant.

Exercise 7.2

1. A first-order reaction has a half-life of 30 minutes. Calculate the rate constant.
2. For the reaction: A + B → products
   When [A] is doubled (keeping [B] constant), the rate doubles.
   When [B] is doubled (keeping [A] constant), the rate quadruples.
   Determine the rate law and overall order.
3. Explain how a catalyst affects the activation energy of a reaction.

Exercise 7.3

1. The decomposition of N₂O₅ follows first-order kinetics with k = 6.2 × 10⁻⁴ s⁻¹ at 45°C.
   Calculate the time required for 75% decomposition.
2. A reaction has Ea = 75 kJ/mol. By what factor does the rate increase when temperature increases from 300 K to 310 K?
3. Propose a two-step mechanism for the reaction: 2NO₂ + F₂ → 2NO₂F
   Given that the rate law is: Rate = k[NO₂][F₂]

Chapter Summary

• Chemical kinetics studies reaction rates and mechanisms
• Rate laws express the relationship between rate and reactant concentrations
• Reaction orders determine how concentration affects rate
• Temperature significantly affects reaction rates (Arrhenius equation)
• Catalysts lower activation energy and increase rates
• Reaction mechanisms describe the step-by-step process of reactions
• The rate-determining step controls the overall reaction rate

Research Project
Investigate enzyme catalysis in biological systems. Choose a specific enzyme (e.g., catalase, amylase, pepsin) and research its mechanism of action, including how it lowers activation energy and its importance in biological processes.

Chapter Eight: Electrochemistry

Exploring Chemical Reactions and Electrical Energy

Introduction

Electrochemistry is the branch of chemistry that deals with the relationship between electrical energy and chemical reactions. It encompasses the study of both spontaneous chemical reactions that produce electrical energy (galvanic cells) and non-spontaneous reactions that are driven by electrical energy (electrolytic cells). This chapter explores electrochemical cells, electrode potentials, conductivity, and practical applications of electrochemistry.

Task 8.1
Research the development of different types of batteries throughout history, from Volta's first battery to modern lithium-ion batteries. Analyze how electrochemical principles have been applied to improve energy storage technology.

8.1 Electrochemical Cells

An electrochemical cell is a device that converts chemical energy into electrical energy or vice versa through redox reactions.

ANODE
Oxidation occurs
Negative terminal

Salt Bridge

CATHODE
Reduction occurs
Positive terminal

8.1.1 Types of Electrochemical Cells

Cell Type	Energy Conversion	Spontaneity	Examples
Galvanic/Voltaic Cell	Chemical → Electrical	Spontaneous	Batteries, Fuel cells
Electrolytic Cell	Electrical → Chemical	Non-spontaneous	Electroplating, Electrolysis

8.1.2 The Salt Bridge

The salt bridge completes the electrical circuit in electrochemical cells by allowing ion flow between half-cells while preventing mixing of solutions. It maintains electrical neutrality.

8.2 Electrode Potential

Electrode potential is the tendency of an electrode to lose or gain electrons when in contact with its ions. The standard electrode potential (E°) is measured under standard conditions (1 M concentration, 1 atm pressure, 25°C).

E°_cell = E°_cathode - E°_anode

8.2.1 Applications of Electrochemical Series

Uses of Electrochemical Series:

• Predicting feasibility of redox reactions
• Determining relative oxidizing/reducing strengths
• Predicting corrosion tendencies
• Designing electrochemical cells

8.2.2 Standard Cell Potential

Example 8.1

Calculate the standard cell potential for the Zn-Cu electrochemical cell.
Given: E°(Zn²⁺/Zn) = -0.76 V, E°(Cu²⁺/Cu) = +0.34 V

Solution:

Zn is more negative → Anode: Zn → Zn²⁺ + 2e⁻

Cu is more positive → Cathode: Cu²⁺ + 2e⁻ → Cu

E°_cell = E°_cathode - E°_anode

E°_cell = 0.34 - (-0.76) = +1.10 V

8.2.3 Cell Diagram or Cell Notation

Zn(s) | Zn²⁺(aq) || Cu²⁺(aq) | Cu(s)

Single vertical line (|) = phase boundary
Double vertical line (||) = salt bridge
Left side = anode, Right side = cathode

8.3 Dependence of Cell Potential on Concentration

8.3.1 Nernst Equation

E = E° - (RT/nF) ln Q

At 25°C: E = E° - (0.059/n) log Q
where Q = reaction quotient, n = number of electrons, F = Faraday's constant

Example 8.2

Calculate the cell potential for the Zn-Cu cell when [Zn²⁺] = 0.1 M and [Cu²⁺] = 0.01 M
E°_cell = 1.10 V

Solution:

Reaction: Zn + Cu²⁺ → Zn²⁺ + Cu

Q = [Zn²⁺]/[Cu²⁺] = 0.1/0.01 = 10

n = 2 electrons

E = 1.10 - (0.059/2) log(10)

E = 1.10 - 0.0295 = 1.0705 V

8.3.2 Relationship Between Free Energy and Cell Potential

ΔG = -nFE_cell

8.3.3 Equilibrium Constants for Redox Reactions

E° = (RT/nF) ln K    or    E° = (0.059/n) log K (at 25°C)

8.4 Ionisation of Electrolytes in Solutions

8.4.1 Conductivity of Electrolytes

Conductivity (κ) is the ability of a solution to conduct electricity. Molar conductivity (Λ_m) is the conductivity of 1 mole of electrolyte in 1 m³ of solution.

Λ_m = κ / c

8.4.2 Ostwald Dilution Law

K_a = cα²/(1-α)    for weak electrolytes

where α = degree of dissociation, c = concentration

8.5 Applications of Electrochemistry

8.5.1 Electrolytic Extraction of Metals

Highly reactive metals like aluminum, sodium, and magnesium are extracted using electrolysis.

8.5.2 Purification of Metals

Electrorefining is used to purify metals like copper, where impure metal is the anode and pure metal deposits at the cathode.

8.5.3 Electroplating

Electroplating Experiment

Aim: To electroplate a copper object with silver

Requirements: Power supply, silver anode, copper cathode, silver nitrate solution

Procedure:

1. Clean the copper object thoroughly
2. Set up electrolytic cell with silver anode and copper cathode
3. Use AgNO₃ solution as electrolyte
4. Apply appropriate voltage
5. Observe silver deposition on copper

8.5.4 Corrosion Inhibition

Electrochemical principles are used to prevent corrosion through methods like galvanization, cathodic protection, and protective coatings.

8.5.5 Batteries Production

Battery Type	Anode Reaction	Cathode Reaction	Voltage	Applications
Zinc-Carbon	Zn → Zn²⁺ + 2e⁻	2MnO₂ + 2NH₄⁺ + 2e⁻ → Mn₂O₃ + 2NH₃ + H₂O	1.5 V	Flashlights, toys
Lead-Acid	Pb + SO₄²⁻ → PbSO₄ + 2e⁻	PbO₂ + 4H⁺ + SO₄²⁻ + 2e⁻ → PbSO₄ + 2H₂O	2.0 V	Automobiles
Lithium-ion	LiC₆ → C₆ + Li⁺ + e⁻	Li⁺ + CoO₂ + e⁻ → LiCoO₂	3.7 V	Electronics, EVs

Exercise 8.1

1. Differentiate between galvanic and electrolytic cells.
2. Write the cell notation for a cell with zinc anode and silver cathode.
3. Explain the function of a salt bridge in electrochemical cells.

Exercise 8.2

1. Calculate E°_cell for the Ag-Zn cell:
   E°(Ag⁺/Ag) = +0.80 V, E°(Zn²⁺/Zn) = -0.76 V
2. Using the Nernst equation, calculate the cell potential when:
   [Zn²⁺] = 0.001 M, [Ag⁺] = 0.1 M, E°_cell = 1.56 V
3. The standard cell potential for a reaction is +0.25 V. Calculate ΔG° for the reaction.

Exercise 8.3

1. Explain how electroplating works and give two practical applications.
2. Describe the working principle of a lead-acid battery.
3. A conductivity cell filled with 0.01 M KCl gives resistance 100 Ω at 25°C.
   The same cell with 0.01 M NaCl gives resistance 150 Ω.
   Calculate the molar conductivity of NaCl solution.

Chapter Summary

• Electrochemistry studies the interconversion of chemical and electrical energy
• Galvanic cells produce electricity from spontaneous reactions
• Electrolytic cells use electricity to drive non-spontaneous reactions
• Electrode potentials predict reaction spontaneity
• Nernst equation relates cell potential to concentration
• Conductivity measures electrolyte solution behavior
• Electrochemistry has wide applications in industry and technology

Research Project
Investigate fuel cell technology, focusing on hydrogen fuel cells. Research their working principles, advantages over conventional batteries, current applications, and future potential in sustainable energy systems.

Galvanic Cell Setup

[Detailed diagram of a galvanic cell showing:
- Zinc and copper electrodes in their respective solutions
- Salt bridge connecting the two half-cells
- Voltmeter measuring cell potential
- Electron flow through external circuit
- Ion flow through salt bridge]