Physical Chemistry-4 (Electrochemistry, Quantum Chemistry and Characterization) BCH6T13

Syllabus topics (RTMNU)

• Rules of oxidation/reduction of ions based on half-cell potentials
• Chemical cells, reversible and irreversible cells with examples
• Electromotive force of a cell and its measurement
• Thermodynamic derivation of Nernst equation
• Standard electrode (reduction) potential and its application to different kinds of half-cells
• Reference electrodes
• Application of EMF measurements in determining (i) free energy, enthalpy and entropy of a cell reaction, (ii) equilibrium constants, and (iii) pH values, using hydrogen, quinone-hydroquinone and glass electrodes
• Concentration cells with and without transference, liquid junction potential
• Salt bridge & its functions
• Potentiometric Titration
• Numerical problems

Rules of Oxidation/Reduction of Ions Based on Half-cell Potentials

The standard reduction potential (E°) shows the tendency of a species to gain electrons. Arranging half-cells in order of E° gives the electrochemical series.

1. A species with a higher (more positive) E° is a stronger oxidising agent and gets reduced; one with a lower (more negative) E° is a stronger reducing agent and gets oxidised.
2. A metal with lower E° displaces a metal ion with higher E° from solution: Zn (E° = −0.76 V) displaces Cu²⁺ (E° = +0.34 V).
3. Metals with negative E° (below hydrogen) liberate H₂ from dilute acids; Cu, Ag, Au do not.
4. A redox reaction is spontaneous if E°_cell = E°_cathode − E°_anode is positive (ΔG° = −nFE° negative).

Chemical Cells; Reversible and Irreversible Cells

A chemical (galvanic) cell converts the free energy of a redox reaction into electrical energy, e.g. Daniell cell: Zn | Zn²⁺ || Cu²⁺ | Cu (anode on left, oxidation; cathode on right, reduction).

EMF of a Cell and its Measurement

EMF: the potential difference between the two electrodes when no current flows through the cell. It is measured with a potentiometer (Poggendorff compensation method), not a voltmeter, because a voltmeter draws current. The unknown cell is balanced against a driving cell along a wire; the balancing length is compared with that of a standard Weston cadmium cell (E = 1.0183 V at 25 °C): E_x = E_s × l_x/l_s. Today digital potentiometers/pH meters with high impedance are used.

Thermodynamic Derivation of the Nernst Equation

For a cell reaction aA + bB ⇌ cC + dD, the van't Hoff reaction isotherm gives ΔG = ΔG° + RT ln Q, where Q = [C]^c[D]^d/[A]^a[B]^b. Electrical work = maximum useful work: ΔG = −nFE and ΔG° = −nFE°. Substituting: −nFE = −nFE° + RT ln Q.

Example: Zn | Zn²⁺(0.01 M) || Cu²⁺(1 M) | Cu, E° = 1.10 V. E = 1.10 − (0.0591/2) log (0.01/1) = 1.10 + 0.0591 = 1.159 V.

Standard Electrode Potential and Types of Half-cells; Reference Electrodes

Standard electrode (reduction) potential: potential of an electrode in contact with its ions at unit activity (1 M), gases at 1 bar, 298 K, measured against the SHE (taken as 0.000 V).

• Metal–metal ion: Zn | Zn²⁺; E = E° + (0.0591/2) log[Zn²⁺].
• Gas electrode: Pt, H₂(1 bar) | H⁺; E = −0.0591 pH.
• Metal–insoluble salt–anion: Ag | AgCl | Cl⁻; Hg | Hg₂Cl₂ | Cl⁻ (calomel); E = E° − 0.0591 log[Cl⁻].
• Redox electrode: Pt | Fe³⁺, Fe²⁺; E = E° + 0.0591 log([Fe³⁺]/[Fe²⁺]).

Reference electrodes: (1) Standard hydrogen electrode (SHE) — primary reference, platinised Pt in 1 M H⁺ with H₂ at 1 bar; difficult to maintain. (2) Saturated calomel electrode (SCE) — Hg, Hg₂Cl₂ paste, saturated KCl; E ≈ +0.242 V at 25 °C. (3) Silver–silver chloride electrode — E ≈ +0.197 V (saturated KCl).

Applications of EMF Measurements: ΔG, ΔH, ΔS, Equilibrium Constant and pH

Example 1: A cell with n = 2 has E = 1.10 V at 298 K and (∂E/∂T)_P = −4.0 × 10⁻⁴ V K⁻¹. ΔG = −2 × 96485 × 1.10 = −212.3 kJ; ΔS = 2 × 96485 × (−4.0 × 10⁻⁴) = −77.2 J K⁻¹; ΔH = ΔG + TΔS = −212.3 + 298 × (−0.0772) = −235.3 kJ.

Example 2: Equilibrium constant of Zn + Cu²⁺ ⇌ Zn²⁺ + Cu (E° = 1.10 V): log K = 2 × 1.10/0.0591 = 37.23 → K ≈ 1.7 × 10³⁷ (reaction goes practically to completion).

pH measurement:

• Hydrogen electrode: cell Pt, H₂ | H⁺(unknown) || SCE. E_cell = 0.2422 + 0.0591 pH → pH = (E_cell − 0.2422)/0.0591. Example: E_cell = 0.500 V → pH = 0.2578/0.0591 = 4.36. Not usable with oxidising agents or poisons.
• Quinhydrone electrode: Pt in the solution saturated with quinhydrone (1:1 quinone–hydroquinone); Q + 2H⁺ + 2e⁻ → QH₂; E = 0.6996 − 0.0591 pH. Simple, but useless above pH 8 and with strong redox agents.
• Glass electrode: thin special-glass bulb with 0.1 M HCl and Ag/AgCl inside; potential depends on H⁺ across the membrane: E_G = E°_G − 0.0591 pH. Calibrated with buffers; used in pH meters; not affected by oxidising agents or poisons; high resistance needs electronic meter.

Concentration Cells with and without Transference; Liquid Junction Potential; Salt Bridge

A concentration cell has two identical electrodes in solutions of different concentrations; emf arises only from the concentration difference (E° = 0).

• Without transference: the two solutions are not in direct contact (joined through a salt bridge or a double cell), so no ions move between them. E.g. Ag | AgNO₃(c₁) || AgNO₃(c₂) | Ag: E = (0.0591/n) log(c₂/c₁) (c₂ > c₁).
• With transference: the solutions are in direct contact, ions move across the junction; E = 2t₋(0.0591/n) log(c₂/c₁) for a cation-reversible electrode, where t₋ is the anion's transport number.

Liquid junction potential: a small potential set up at the boundary of two different solutions because cations and anions diffuse at different speeds. Salt bridge (agar gel with saturated KCl, KNO₃ or NH₄NO₃, whose ions have nearly equal mobility): (1) completes the circuit; (2) keeps solutions electrically neutral; (3) minimises liquid junction potential; (4) prevents mixing of solutions.

Potentiometric Titration

The emf of a suitable indicator electrode (vs reference) is measured after each addition of titrant; the end point is where emf changes sharply.

• Acid–base: glass/quinhydrone electrode (pH changes).
• Redox: Pt electrode, e.g. Fe²⁺ vs K₂Cr₂O₇ or Ce⁴⁺.
• Precipitation: Ag electrode, e.g. Cl⁻ vs AgNO₃.

End point found from E vs V curve (steep part), first derivative ΔE/ΔV vs V (peak) or second derivative (zero crossing). Advantages: works for coloured/turbid solutions and dilute solutions, no indicator needed, can be automated.

Likely exam questions

1. (Long, 8 marks) Give the thermodynamic derivation of the Nernst equation. Calculate the emf of the cell Zn | Zn²⁺ (0.1 M) || Cu²⁺ (0.01 M) | Cu at 298 K (E° = 1.10 V).
2. (Long, 8 marks) Explain how EMF measurements are used to determine ΔG, ΔH, ΔS and equilibrium constant. Describe the determination of pH using hydrogen, quinhydrone and glass electrodes.
3. (Short, 4 marks) Distinguish between reversible and irreversible cells.
4. (Short, 4 marks) What is liquid junction potential? Give the functions of a salt bridge.
5. (Short, 4 marks) Explain potentiometric titration with one example.

Syllabus topics (RTMNU)

• Quantum Chemistry: Time-independent Schrödinger wave equation for the hydrogen atom, separation into three independent equations (radial and angular parts) - without mathematical derivation
• Total wave function for hydrogen atom in terms of radial and angular wave functions, energy of hydrogen atom (no derivation)
• Hydrogen-like wave functions (qualitative treatment), radial wave functions and angular wave functions - physical significance
• Interpretation and significance of quantum numbers
• Concept of atomic orbital and radial probability distribution curves for 1s, 2s, 2p, 3p and 3d orbitals
• Molecular orbital theory: Criteria for forming M.O. from A.O., LCAO-MO method for a molecule
• Qualitative expression for energy levels for bonding and antibonding wave functions
• Normalized wave functions for bonding and antibonding (no derivation)
• Physical interpretation of bonding and antibonding wave functions
• Introduction to M.O. theory for H2 molecule (Qualitative treatment, without derivation)
• Introduction to Valence bond theory for H2 molecule, Overlapping of atomic orbitals and bond formation
• Similarities and differences between valence bond and molecular orbital theories

Schrödinger Equation for the Hydrogen Atom and Separation of Variables

The hydrogen atom has one electron moving in the Coulomb field of the nucleus: V = −e²/(4πε₀r). Because V depends only on r, the equation is written in spherical polar coordinates (r, θ, φ):

The wave function can be written as a product (separation of variables, without derivation): ψ(r, θ, φ) = R(r) · Θ(θ) · Φ(φ) = R(r) · Y(θ, φ). This gives three independent equations:

• Φ equation (depends on φ) → gives the magnetic quantum number m.
• Θ equation (depends on θ) → gives the azimuthal quantum number l.
• R equation (radial, depends on r) → gives the principal quantum number n and energy.

R(r) is the radial wave function; Y(θ, φ) = Θ·Φ is the angular wave function (spherical harmonic).

Energy (no derivation): E_n = −μe⁴/(8ε₀²h²n²) = −13.6/n² eV — the same as Bohr's result, but obtained naturally from boundary conditions. For hydrogen-like ions: E_n = −13.6 Z²/n² eV.

Hydrogen-like Wave Functions: Radial and Angular Parts and their Significance

• Radial part R_n,l(r): decides the size of the orbital and the number of radial nodes (n − l − 1), where the probability is zero at a certain distance. Example: 1s has no radial node; 2s has one; 3s has two.
• Angular part Y_l,m(θ, φ): decides the shape and orientation of the orbital and the number of angular (planar) nodes (= l). s orbitals (l = 0) are spherical; p (l = 1) dumb-bell with one nodal plane; d (l = 2) cloverleaf with two nodal planes.
• Total nodes = n − 1.

Interpretation and Significance of Quantum Numbers

Concept of Atomic Orbital and Radial Probability Distribution Curves

An atomic orbital is a one-electron wave function ψ_n,l,m; ψ² gives the region of space where the probability of finding the electron is high (about 90–95% boundary surface).

Radial probability distribution function P(r) = 4πr²R² gives the probability of finding the electron in a thin spherical shell of radius r and thickness dr. Plots of 4πr²R² vs r:

• 1s: one maximum at r = 0.529 Å (Bohr radius a₀); no node.
• 2s: two maxima (small inner, large outer) with one radial node.
• 2p: one maximum, no radial node (n − l − 1 = 0); the maximum is closer than the main 2s maximum, but 2s has an inner peak (penetration).
• 3p: two maxima and one radial node.
• 3d: one maximum, no radial node, at about 9a₀.

Molecular Orbital Theory: Criteria for Forming MOs from AOs and the LCAO-MO Method

Criteria for effective combination of atomic orbitals:

1. The AOs should have comparable energies.
2. They should overlap to a considerable extent.
3. They must have the same symmetry about the molecular axis (e.g. 1s with 2p_z along z, not with 2p_x).

LCAO-MO method: molecular orbitals are formed by linear combination of atomic orbitals. For two atoms A and B: ψ_MO = c_Aψ_A ± c_Bψ_B. For a homonuclear molecule c_A = c_B:

• Bonding MO: ψ_b = ψ_A + ψ_B (constructive interference).
• Antibonding MO: ψ_a = ψ_A − ψ_B (destructive interference).

Energy Levels and Normalised Wave Functions for Bonding and Antibonding MOs

Qualitative energy expressions (with Coulomb integral α, resonance/exchange integral β (negative) and overlap integral S):

Physical interpretation: ψ_b² = ψ_A² + ψ_B² + 2ψ_Aψ_B — extra electron density (2ψ_Aψ_B) between the nuclei holds them together (bonding). ψ_a² = ψ_A² + ψ_B² − 2ψ_Aψ_B — electron density is reduced between nuclei with a nodal plane midway; nuclei repel (antibonding).

MO Theory for the H₂ Molecule (Qualitative)

Two 1s orbitals combine to form σ1s (bonding) and σ*1s (antibonding). H₂ has 2 electrons, both in σ1s with opposite spins: configuration (σ1s)². Bond order = ½(N_b − N_a) = ½(2 − 0) = 1; diamagnetic; stable. He₂ would be (σ1s)²(σ*1s)², bond order 0 — does not exist. H₂⁺ has bond order ½.

Valence Bond Theory for H₂; Overlap and Bond Formation

Heitler–London (1927): when two H atoms approach, each electron is attracted by both nuclei. Wave function ψ = ψ_A(1)ψ_B(2) ± ψ_A(2)ψ_B(1) (exchange of electrons). The symmetric combination with paired (opposite) spins gives an energy minimum at about 0.74 Å — a stable covalent bond; the other combination gives repulsion. Adding ionic terms (H⁺H⁻) improves the result.

Overlap and bond formation: a covalent bond forms when half-filled AOs with opposite spins overlap; greater overlap → stronger bond. Head-on (axial) overlap gives a σ bond (s–s, s–p, p–p); sideways overlap of parallel p orbitals gives a weaker π bond.

Similarities and Differences between VB and MO Theories

Similarities: both use overlap of atomic orbitals; both need comparable energy and same symmetry; both place electron density between nuclei in a bond; both predict bond formation and molecular shape (with hybridisation in VBT).

Likely exam questions

1. (Long, 8 marks) Write the Schrödinger equation for the hydrogen atom in spherical polar coordinates and explain its separation into radial and angular parts. Discuss radial probability distribution curves for 1s, 2s, 2p and 3d orbitals.
2. (Long, 8 marks) Explain the LCAO-MO method. Give the physical interpretation of bonding and antibonding wave functions and compare VB and MO theories.
3. (Short, 4 marks) Give the significance of the four quantum numbers.
4. (Short, 4 marks) State the criteria for combination of atomic orbitals to form MOs.
5. (Short, 4 marks) Explain the MO treatment of the H₂ molecule and its bond order.

Syllabus topics (RTMNU)

• Photochemistry: Interaction of radiation with matter, difference between thermal and photochemical reactions
• Laws governing absorption of light, laws of photochemistry
• Jablonski diagram depicting various processes
• Quantum yield, determination of quantum yield of reactions, reasons for low and high quantum yields
• Some examples of photochemical reactions (Photochemical decomposition of Hydrogen iodide, Photosynthesis of HBr from H2 and Br2 and photosynthesis of HCl from H2 and Cl2)
• Photosensitization, Photosensitized reactions, numerical problems
• Raman Spectroscopy: Raman effect, Explanation of Rayleigh lines, Stokes lines and antistokes lines
• Experimental set up of Raman spectrometer
• Pure rotational Raman spectra of diatomic molecules
• Rotational-vibrational Raman spectra of diatomic molecules
• Advantages of Raman spectroscopy over IR spectroscopy

Photochemistry: Interaction of Radiation with Matter; Thermal vs Photochemical Reactions

Photochemistry is the study of chemical reactions caused by the absorption of light (UV/visible, about 200–800 nm). When a molecule absorbs a photon (E = hν), it goes to an excited electronic state; it may then react (photochemical change), or lose energy as heat or light (photophysical processes).

Laws Governing Absorption of Light and Laws of Photochemistry

Lambert's law: the decrease in intensity with thickness is proportional to the intensity: −dI/dx = kI. Beer's law: also proportional to concentration. Combined Beer–Lambert law: log(I₀/I) = A = εcl.

Grotthuss–Draper law (first law): only the light that is absorbed by a substance can bring about a photochemical change. (Absorbed light does not always cause reaction.)

Stark–Einstein law of photochemical equivalence (second law): each molecule taking part in a photochemical reaction is activated by absorbing one quantum (photon) of light in the primary process. Energy absorbed per mole = one einstein = N_Ahν = N_Ahc/λ. At 400 nm, 1 einstein = 6.022 × 10²³ × 6.626 × 10⁻³⁴ × 2.998 × 10⁸ / 400 × 10⁻⁹ = 299 kJ.

Jablonski Diagram

A Jablonski diagram shows electronic states (S₀ ground singlet, S₁, S₂ excited singlets, T₁ triplet) with vibrational levels, and the processes after absorption:

1. Absorption S₀ → S₁/S₂ (about 10⁻¹⁵ s).
2. Vibrational relaxation to the lowest vibrational level (non-radiative).
3. Internal conversion (IC): S₂ → S₁, non-radiative, same multiplicity.
4. Fluorescence: S₁ → S₀ with emission of light; fast (about 10⁻⁹–10⁻⁶ s); stops as soon as light is removed.
5. Intersystem crossing (ISC): S₁ → T₁, non-radiative, change of spin.
6. Phosphorescence: T₁ → S₀ with emission; slow (10⁻³ s to seconds), continues after light is removed; at longer wavelength than fluorescence.

Quantum Yield: Definition, Determination, Reasons for Low and High Values

Determination: (1) moles reacted are found by chemical analysis (titration, spectrophotometry); (2) the energy absorbed is measured with a thermopile or chemical actinometer (uranyl oxalate or potassium ferrioxalate, whose φ is known) — reading with empty cell and with sample cell gives light absorbed. Number of photons = energy absorbed × λ/(hc).

Problem: 0.01 mol of product is formed when 5.0 × 10²¹ photons are absorbed. φ = (0.01 × 6.022 × 10²³)/(5.0 × 10²¹) = 1.20.

High quantum yield (φ > 1): the primary step starts a chain reaction or the excited product reacts further; e.g. H₂ + Cl₂ (φ up to 10⁴–10⁶). Low quantum yield (φ < 1): deactivation of excited molecules by collisions or fluorescence before reaction; recombination of fragments (primary recombination); reverse reaction; or the light absorbed is not enough for reaction; e.g. H₂ + Br₂ at ordinary temperature.

Examples of Photochemical Reactions

1. Decomposition of HI (φ = 2):
HI + hν → H + I (primary)
H + HI → H₂ + I
I + I → I₂
Net: 2HI → H₂ + I₂; one photon decomposes two HI molecules.

2. Formation of HBr from H₂ and Br₂ (φ very low, about 0.01 at room temperature):
Br₂ + hν → 2Br
Br + H₂ → HBr + H (endothermic, slow)
H + Br₂ → HBr + Br
H + HBr → H₂ + Br (inhibition)
Br + Br → Br₂
Most Br atoms recombine before the slow second step, so φ is low; it rises at higher temperature.

3. Formation of HCl from H₂ and Cl₂ (φ = 10⁴–10⁶):
Cl₂ + hν → 2Cl
Cl + H₂ → HCl + H
H + Cl₂ → HCl + Cl (chain propagates)
Chain ends at walls: 2Cl → Cl₂. Each Cl atom produces thousands of HCl molecules — the mixture may explode in sunlight.

Photosensitisation and Photosensitised Reactions

Photosensitisation: a substance that cannot absorb the light itself reacts when another substance (the photosensitiser) absorbs the light and transfers energy to it. The sensitiser is not consumed.

• Hg vapour + H₂: Hg + hν (253.7 nm) → Hg*; Hg* + H₂ → Hg + 2H (H₂ does not absorb this light itself).
• Photosynthesis: chlorophyll absorbs light and drives 6CO₂ + 6H₂O → C₆H₁₂O₆ + 6O₂.
• Decomposition of oxalic acid sensitised by uranyl ions (uranyl oxalate actinometer).

Raman Spectroscopy: Raman Effect; Rayleigh, Stokes and Anti-Stokes Lines

Raman effect (C.V. Raman, 1928; Nobel Prize 1930): when monochromatic light passes through a transparent substance, most scattered light has the same frequency (Rayleigh scattering), but a very small fraction has frequencies slightly lower and higher — Raman lines.

• Rayleigh line (ν₀): elastic collision; no energy exchange; most intense.
• Stokes lines (ν₀ − Δν): photon gives energy to the molecule (molecule goes to higher level); lower frequency; more intense.
• Anti-Stokes lines (ν₀ + Δν): molecule already in an excited level gives energy to the photon; higher frequency; weaker (fewer molecules in excited levels).

Raman shift Δν̄ is independent of exciting wavelength and characteristic of the molecule. Condition: the vibration/rotation must change the polarisability of the molecule. Example: with the Hg line at 4358 Å, CCl₄ gives a Stokes line at 4447 Å; Δν̄ = 1/(4358 × 10⁻⁸) − 1/(4447 × 10⁻⁸) = 459 cm⁻¹.

Experimental Set-up of a Raman Spectrometer

(1) Source: intense monochromatic light — a laser (He–Ne 632.8 nm, Ar⁺ 488/514.5 nm, Nd:YAG 1064 nm); earlier mercury arc. (2) Sample holder: glass/quartz tube or capillary; solids, liquids, gases and aqueous solutions. (3) Collection optics at 90° (or 180°) to the incident beam with a filter to remove the Rayleigh line. (4) Monochromator (grating). (5) Detector: photomultiplier or CCD, with computer recording.

Pure Rotational and Rotational–Vibrational Raman Spectra of Diatomic Molecules

Pure rotational Raman: selection rule ΔJ = 0, ±2. ΔJ = 0 gives the Rayleigh line; ΔJ = +2 gives Stokes, ΔJ = −2 anti-Stokes lines at Δν̄ = ±B(4J + 6) — the first lines at 6B from the exciting line, and then spacing 4B. Homonuclear molecules like H₂, N₂, O₂ (no dipole, no microwave spectrum) do show rotational Raman spectra, so B and bond length can be found.

Vibrational–rotational Raman: Δv = ±1 with ΔJ = 0, ±2, giving the Q branch (ΔJ = 0, intense central line at the vibrational shift), S branch (ΔJ = +2) and O branch (ΔJ = −2) on either side.

Advantages of Raman Spectroscopy over IR Spectroscopy

1. Homonuclear diatomic and symmetric vibrations (IR inactive) are Raman active.
2. Water is a weak Raman scatterer, so aqueous solutions can be studied (water absorbs strongly in IR).
3. Glass/quartz cells can be used (IR needs NaCl/KBr optics).
4. Covers both low and high frequency regions with one instrument; little sample preparation.
5. With IR, the mutual exclusion rule (for molecules with a centre of symmetry, a vibration active in IR is inactive in Raman and vice versa) helps find molecular structure, e.g. CO₂.

Likely exam questions

1. (Long, 8 marks) State the laws of photochemistry. Define quantum yield, explain its determination and give reasons for high and low quantum yields with examples of HCl and HBr formation.
2. (Long, 8 marks) What is the Raman effect? Explain Rayleigh, Stokes and anti-Stokes lines, the experimental set-up of a Raman spectrometer and the advantages of Raman over IR spectroscopy.
3. (Short, 4 marks) Draw and explain a Jablonski diagram.
4. (Short, 4 marks) What is photosensitisation? Give an example.
5. (Short, 4 marks) When 2.5 × 10²¹ photons are absorbed, 0.005 mol of a reactant decomposes. Calculate the quantum yield.

Syllabus topics (RTMNU)

• Colligative properties: Methods of expressing concentration of solutions
• Raoult's law, relative lowering of vapour pressure, molecular mass determination from relative lowering of vapour pressure
• Osmosis, osmotic pressure and its measurement by Berkeley and Hartley method, Determination of molecular mass from osmotic pressure
• Elevation of boiling point, determination of molecular mass from elevation of boiling point
• Depression of freezing point, Determination of molecular mass from depression of freezing point
• Van't Hoff factor, degree of dissociation and association of solute, Numerical problems
• Macromolecules: Macromolecules, classification of polymers
• Molar masses of polymers: number average and weight average molar masses, Polydispersity index (PDI)
• Determination of molar masses of macromolecules: viscometry, Osmometry and light scattering method
• Kinetics of polymerization, addition and condensation polymerization
• Electronically conducting polymers: poly(acetylene) and poly(aniline), applications of conducting polymers in Sensors and OLEDs, Numerical Problems

Methods of Expressing Concentration of Solutions

• Molarity (M): moles of solute per litre of solution.
• Molality (m): moles of solute per kg of solvent (independent of temperature — used for colligative properties).
• Normality (N): gram equivalents per litre of solution.
• Mole fraction (x): moles of a component ÷ total moles; x₁ + x₂ = 1.
• Mass percent (w/w), volume percent, and ppm for very dilute solutions.

Colligative properties depend only on the number of solute particles, not their nature: relative lowering of vapour pressure, osmotic pressure, elevation of boiling point and depression of freezing point.

Raoult's Law, Relative Lowering of Vapour Pressure and Molar Mass

Raoult's law: the vapour pressure of a solvent in a solution equals the vapour pressure of the pure solvent multiplied by its mole fraction: p = p⁰x₁. For a non-volatile solute:

Problem: 10 g of a non-volatile solute in 90 g of water lowers the vapour pressure from 23.76 to 22.97 mm Hg. M₂ = (10 × 18 × 23.76) ÷ (90 × 0.79) = 60.2 g mol⁻¹ (urea).

Osmosis, Osmotic Pressure, Berkeley–Hartley Method and Molar Mass

Osmosis: flow of solvent through a semi-permeable membrane from the pure solvent (or dilute solution) into the more concentrated solution. Osmotic pressure (π): the excess pressure that must be applied on the solution to just stop osmosis. van't Hoff equation: πV = nRT, or π = CRT.

Berkeley and Hartley method: a porous porcelain tube with a copper ferrocyanide membrane deposited in its walls is placed inside a gun-metal vessel. The tube contains solvent (with a capillary indicator); the outer vessel contains the solution and is fitted with a piston and pressure gauge. Pressure is applied on the solution until the solvent level in the capillary stops falling — this pressure is the osmotic pressure. Advantages: quick, no change in concentration, no strain on the membrane.

Elevation of Boiling Point and Molar Mass

A solution of a non-volatile solute has lower vapour pressure, so it boils at a higher temperature. ΔT_b = K_bm, where K_b is the molal elevation (ebullioscopic) constant (water 0.52 K kg mol⁻¹).

Measured by Landsberger's or Cottrell's method.

Depression of Freezing Point and Molar Mass

ΔT_f = K_fm; K_f for water = 1.86 K kg mol⁻¹ (cryoscopic constant). Measured by Beckmann's method or Rast's camphor method (camphor has a very large K_f, about 40).

van't Hoff Factor; Degree of Dissociation and Association

Electrolytes dissociate and some solutes associate, so observed colligative properties differ from calculated values. van't Hoff factor i = observed colligative property ÷ calculated (normal) value = normal molar mass ÷ observed molar mass.

• Dissociation (A_xB_y → n ions): i = 1 + (n − 1)α → α = (i − 1)/(n − 1).
• Association (n A → A_n, e.g. benzoic acid dimer in benzene): i = 1 − α(1 − 1/n) → α = (1 − i)n/(n − 1).

Problem: 0.1 m NaCl solution freezes at −0.348 °C. Calculated ΔT_f = 1.86 × 0.1 = 0.186 K. i = 0.348/0.186 = 1.87; α = (1.87 − 1)/(2 − 1) = 0.87 (87% dissociated). Modified equations: ΔT_f = iK_fm, π = iCRT.

Macromolecules and Classification of Polymers

Macromolecules are very large molecules (molar mass 10³–10⁷ g mol⁻¹). Polymers are macromolecules made of repeating units (monomers).

• By source: natural (starch, cellulose, proteins, natural rubber), semi-synthetic (cellulose acetate, vulcanised rubber), synthetic (polythene, nylon, PVC).
• By structure: linear (HDPE), branched (LDPE), cross-linked/network (bakelite).
• By mode of polymerisation: addition (polythene, PVC, polystyrene) and condensation (nylon-6,6, terylene, bakelite).
• By molecular forces: elastomers (rubber), fibres (nylon), thermoplastics (PVC), thermosetting (bakelite).

Molar Masses of Polymers: Number Average, Weight Average and PDI

A polymer sample contains chains of different lengths, so average molar masses are used.

Problem: A sample has 3 mol of chains of M = 10,000 and 2 mol of M = 20,000. M̄_n = (3 × 10,000 + 2 × 20,000)/5 = 14,000. M̄_w = (3 × 10,000² + 2 × 20,000²)/(3 × 10,000 + 2 × 20,000) = 1.1 × 10⁹/70,000 = 15,714. PDI = 15,714/14,000 = 1.12.

Determination of Molar Masses: Viscometry, Osmometry and Light Scattering

• Viscometry: flow times of solvent (t₀) and polymer solutions (t) are measured in an Ostwald/Ubbelohde viscometer. Relative viscosity η_r = t/t₀; specific viscosity η_sp = η_r − 1; η_sp/C is plotted against C and extrapolated to C = 0 to get intrinsic viscosity [η]. Mark–Houwink equation: [η] = K M^a (K and a are constants for a polymer–solvent pair) gives the viscosity-average molar mass. Simple and cheap.
• Osmometry (membrane osmometer): π/C is plotted against C; extrapolation to C = 0 gives RT/M̄_n → number-average molar mass.
• Light scattering: polymer molecules scatter light; the intensity of scattered light (turbidity) depends on molar mass and concentration (Debye equation). Plot of Hc/τ vs c gives 1/M̄_w as intercept → weight-average molar mass; also gives size of molecules.

Kinetics of Polymerisation: Addition and Condensation

Addition (chain-growth) polymerisation — free radical mechanism: (1) Initiation: initiator I → 2R• (rate R_i = 2fk_d[I]); R• + M → RM•. (2) Propagation: RM_n• + M → RM_n+1• (k_p). (3) Termination: by combination or disproportionation of two radicals (k_t). Using the steady-state approximation for radicals: rate of polymerisation R_p = k_p[M](fk_d[I]/k_t)^1/2 — first order in monomer, half order in initiator. High polymer forms early; monomer remains till the end.

Condensation (step-growth) polymerisation: functional groups react step by step with loss of small molecules (H₂O, HCl), e.g. diacid + diol → polyester. For a self-catalysed or acid-catalysed reaction, the rate is second (or third) order in functional groups. The degree of polymerisation follows Carothers equation: X̄_n = 1/(1 − p), where p is the extent of reaction — high molar mass only at very high conversion (p = 0.99 gives X̄_n = 100).

Electronically Conducting Polymers: Polyacetylene and Polyaniline; Applications in Sensors and OLEDs

Conducting polymers have a conjugated backbone (alternate single and double bonds) whose conductivity rises enormously on doping (oxidation = p-doping with I₂; reduction = n-doping with Na). Heeger, MacDiarmid and Shirakawa received the Nobel Prize in Chemistry (2000) for this discovery.

• Polyacetylene (–CH=CH–)_n: made by Ziegler–Natta polymerisation of acetylene; trans form; iodine doping raises conductivity close to metals. Unstable in air.
• Polyaniline (PANI): made by oxidative polymerisation of aniline with ammonium persulphate in acid; the emeraldine salt form is conducting; stable, cheap, colour changes with pH/redox state.

Applications: Sensors — conductivity changes on exposure to gases (NH₃, NO₂), humidity or pH, so PANI and polypyrrole films are used in gas sensors and biosensors (enzyme electrodes). OLEDs — conjugated polymers (e.g. PPV) emit light when current passes (electroluminescence); PEDOT:PSS and PANI serve as hole-injection layers. Also in batteries, anti-static coatings, electromagnetic shielding and corrosion protection.

Likely exam questions

1. (Long, 8 marks) Define osmotic pressure. Describe the Berkeley–Hartley method and show how molar mass is found from osmotic pressure. 1.2 g of a solute in 200 mL of solution gives an osmotic pressure of 0.82 atm at 300 K; find its molar mass.
2. (Long, 8 marks) Define number-average and weight-average molar masses and PDI. Describe determination of molar mass of polymers by viscometry and light scattering, and add a note on conducting polymers.
3. (Short, 4 marks) What is van't Hoff factor? How is degree of dissociation calculated from it?
4. (Short, 4 marks) Explain the kinetics of free-radical addition polymerisation.
5. (Short, 4 marks) 2.5 g of a solute in 50 g of water lowers the freezing point by 0.93 K (Kf = 1.86). Find its molar mass.