Physical Chemistry-2 BCH3T05

Syllabus topics (official RTMNU)

Thermodynamics

(A) Second law of thermodynamics

• Need for second law of thermodynamics, statements of second law of thermodynamics, Carnot’s cycle and its efficiency, thermodynamic scale of temperature, concept of entropy, entropy as a state function of V & T, P & T, entropy change in phase change for ideal gas, entropy as criteria of spontaneity & equilibrium.

(B) Free energy functions and Systems of variable composition

• Helmholtz free energy (A) & Gibb’s free energy (G) and their properties, standard free energies, effect of temperature on free energy (Gibb’s – Helmholtz equation) & its applications, A & G as criteria for thermodynamic equilibrium.
• Partial molar quantities, chemical potential, Van’t-Hoffs reaction isotherm, relation between standard free energy change & equilibrium constant (thermodynamic derivation of law of mass action), effect of temperature on equilibrium constant (reaction isochore).

(A) Second law of thermodynamics

Need for the second law

• The first law (ΔU = q + w) only says energy is conserved. It does not tell us the direction of a process or whether it will happen on its own.
• Example: heat flows from hot to cold body by itself, never the reverse, though both obey the first law.
• The second law gives the direction of spontaneous change and the limit of conversion of heat into work.

Statements of the second law

• Kelvin–Planck: It is impossible to construct an engine working in a cycle that takes heat from a single reservoir and converts it completely into work.
• Clausius: Heat cannot flow by itself from a colder body to a hotter body without external work.
• Entropy form: In a spontaneous process the entropy of the universe increases (ΔS_univ > 0).

Carnot’s cycle and its efficiency

An ideal reversible engine using 1 mol ideal gas between a hot source (T₂) and cold sink (T₁). Four reversible steps:

1. Isothermal expansion at T₂ (V₁ → V₂): heat absorbed q₂ = RT₂ ln(V₂/V₁).
2. Adiabatic expansion (V₂ → V₃): temperature falls T₂ → T₁, q = 0.
3. Isothermal compression at T₁ (V₃ → V₄): heat rejected q₁ = RT₁ ln(V₄/V₃) (negative).
4. Adiabatic compression (V₄ → V₁): temperature rises T₁ → T₂, q = 0.

For the two adiabatic steps V₃/V₄ = V₂/V₁. Net work = q₂ + q₁ = R(T₂ − T₁) ln(V₂/V₁).

Diagram (P–V): a closed loop of two isotherms (gentle curves) joined by two adiabats (steeper curves); the area enclosed equals the net work.

Example: engine between 500 K and 300 K: η = (500 − 300)/500 = 0.40 (40 %).

Thermodynamic (Kelvin) scale of temperature

• For a reversible engine q₂/|q₁| = T₂/T₁; the ratio depends only on temperatures, not on the working substance.
• Kelvin defined temperature using this heat ratio, so the scale is independent of any material property. Zero on this scale is the temperature of a sink at which η = 1 (absolute zero).

Concept of entropy

• From the Carnot cycle: q₂/T₂ + q₁/T₁ = 0, i.e. Σ q_rev/T = 0 for a cycle. So dq_rev/T is an exact differential: dS = dq_rev/T.
• Entropy (S) is a measure of disorder/randomness. It is a state function and an extensive property. Unit: J K⁻¹ mol⁻¹.

Entropy as a function of V and T, and of P and T (ideal gas)

• dS = C_V dT/T + R dV/V → ΔS = C_V ln(T₂/T₁) + R ln(V₂/V₁)
• dS = C_P dT/T − R dP/P → ΔS = C_P ln(T₂/T₁) − R ln(P₂/P₁)
• Isothermal expansion: ΔS = R ln(V₂/V₁) = R ln(P₁/P₂). For 1 mol doubling volume: ΔS = 8.314 × 0.693 = 5.76 J K⁻¹.

Entropy change in phase change

At the transition temperature, phase change is reversible at constant P: ΔS = ΔH_trans/T.

• Fusion: ΔS_f = ΔH_f/T_f. Ice: 6010/273 ≈ 22.0 J K⁻¹ mol⁻¹.
• Vaporisation: ΔS_v = ΔH_v/T_b. Water: 40700/373 ≈ 109 J K⁻¹ mol⁻¹. (Trouton’s rule: ≈ 85–88 J K⁻¹ mol⁻¹ for normal liquids.)

Entropy as criterion of spontaneity and equilibrium

• For an isolated system (or universe): ΔS > 0 → spontaneous; ΔS = 0 → equilibrium; ΔS < 0 → non-spontaneous.
• Clausius inequality: dS ≥ dq/T (equal sign for reversible).

(B) Free energy functions and systems of variable composition

Helmholtz and Gibbs free energy

• Helmholtz free energy: A = U − TS. At constant T, −ΔA = maximum work (w_max). Hence also called work function.
• Gibbs free energy: G = H − TS. At constant T and P, −ΔG = maximum useful (non-PV) work.
• Both are state functions and extensive. Relation: G = A + PV.
• dG = V dP − S dT, so (∂G/∂P)_T = V and (∂G/∂T)_P = −S. Similarly dA = −P dV − S dT.

Standard free energy

ΔG° is the free energy change when reactants and products are in standard states (1 bar, specified T). ΔG°_f of elements in standard states = 0. ΔG° = ΔH° − TΔS°.

Gibbs–Helmholtz equation

1. G = H − TS and S = −(∂G/∂T)_P.
2. Substitute: G = H + T(∂G/∂T)_P; for a change: ΔG = ΔH + T[∂(ΔG)/∂T]_P.
3. Equivalent form: [∂(ΔG/T)/∂T]_P = −ΔH/T².

Applications: calculating ΔH from emf of a cell and its temperature coefficient (ΔH = −nF[E − T(∂E/∂T)_P]); finding ΔG at another temperature.

A and G as criteria of equilibrium

Partial molar quantities and chemical potential

• Partial molar quantity of component i: X̄_i = (∂X/∂n_i)_{T, P, nj} (change in property when 1 mol of i is added to a large amount of mixture).
• Chemical potential μ_i = (∂G/∂n_i)_{T, P, nj} = partial molar free energy.
• G = Σ n_iμ_i; Gibbs–Duhem equation: Σ n_i dμ_i = 0 (constant T, P).
• For an ideal gas: μ = μ° + RT ln p. Matter flows from higher to lower μ.

Van’t Hoff reaction isotherm

For aA + bB ⇌ cC + dD (ideal gases), using μ_i = μ°_i + RT ln p_i:

ΔG = ΔG° + RT ln Q_p, where Q_p = (p_C^c p_D^d)/(p_A^a p_B^b).

At equilibrium ΔG = 0 and Q = K_p, so ΔG° = −RT ln K_p. Since ΔG° is constant at fixed T, K_p must be constant: this is the thermodynamic derivation of the law of mass action.

• K > 1 → ΔG° negative; K < 1 → ΔG° positive.

Effect of temperature on K (reaction isochore)

Combining ΔG° = −RT ln K with Gibbs–Helmholtz: d ln K_p/dT = ΔH°/RT². Integrated: log(K₂/K₁) = (ΔH°/2.303R)(T₂ − T₁)/(T₁T₂).

• Endothermic (ΔH° > 0): K increases with T. Exothermic: K decreases with T (agrees with Le Chatelier).

Likely exam questions

Long answer

1. Describe Carnot’s cycle and derive the expression for its efficiency. How does it lead to the concept of entropy?
2. Derive Van’t Hoff reaction isotherm and show that ΔG° = −RT ln K_p. Hence derive the reaction isochore.

Short answer

1. State Kelvin–Planck and Clausius statements of the second law.
2. Derive the Gibbs–Helmholtz equation and give one application.
3. What is chemical potential? Show that it is a partial molar free energy.

Syllabus topics (official RTMNU)

Phase Equilibria

(A) Phase rule

• Statement of phase rule, definition of phase, component and degree of freedom, Derivation of phase rule, Applications of phase rule to one component system i) water system, ii) Sulphur system , and two component system i) Pb-Ag system. ii)Bi-Cd system ,Pattinson’s process.

(B) Liquid-Liquid mixtures

• Ideal liquid mixtures, Raoults law of ideal solutions, Henry’s law, non- ideal systems, azeotropes : HCl-H2O & ethanol- water system.
• Partial miscible liquids : phenol-water system, trimethylamine- water, nicotine-water system, lower & upper con solute temperature, effect of impurity., Nernst distribution law, limitations, deviations & applications.

(A) Phase rule

Statement (Gibbs, 1876)

For a heterogeneous system in equilibrium, not affected by gravity, electric or magnetic forces, the number of degrees of freedom is:

Definitions

• Phase (P): a homogeneous, physically distinct and mechanically separable part of a system. All gases form one phase; each solid is a separate phase; miscible liquids form one phase.
• Component (C): the least number of independent chemical species needed to express the composition of every phase. Water system: C = 1. CaCO₃ ⇌ CaO + CO₂: C = 2.
• Degree of freedom (F): the least number of variables (T, P, concentration) that must be fixed to define the system completely.

Derivation of phase rule

1. System with C components distributed in P phases. Composition of each phase needs (C − 1) concentration terms, so total = P(C − 1). Add T and P: total variables = P(C − 1) + 2.
2. At equilibrium, chemical potential of each component is the same in all phases: μ₁ (phase 1) = μ₁ (phase 2) = … gives (P − 1) equations per component, total C(P − 1) equations.
3. F = variables − equations = P(C − 1) + 2 − C(P − 1) = C − P + 2.

One-component system: water

• Phases: ice, water, vapour. C = 1, so F = 3 − P.
• Areas (one phase, F = 2, bivariant): ice, liquid, vapour.
• Curves (two phases, F = 1, univariant): OA vaporisation curve (water ⇌ vapour), ends at critical point 374 °C, 218 atm; OB sublimation curve (ice ⇌ vapour); OC fusion curve (ice ⇌ water), slopes to the left (negative slope) because ice is less dense than water, so pressure lowers the melting point; OA′ metastable curve of supercooled water.
• Triple point O (three phases, F = 0, invariant): 0.0098 °C (273.16 K) and 4.58 mm Hg.
• Diagram: P on y-axis, T on x-axis; three curves meet at O like a “Y”, with OC nearly vertical and leaning left.

One-component system: sulphur

• Four phases: rhombic (S_R), monoclinic (S_M), liquid (S_L), vapour (S_V). Only three can coexist (F cannot be negative).
• S_R ⇌ S_M at 95.6 °C: transition temperature (enantiotropy).
• Stable triple points: B (S_R, S_M, V) at 95.6 °C; C (S_M, L, V) at about 120 °C; E (S_R, S_M, L) at about 151 °C and 1290 atm.
• Metastable triple point G (S_R, L, V) at about 114.5 °C, reached when rhombic sulphur is heated quickly.
• Monoclinic sulphur is stable only within the triangle BCE.

Two-component system: Pb–Ag (simple eutectic)

• Solid–liquid system at constant pressure, so the reduced phase rule F′ = C − P + 1 is used.
• Curve AO: melting point of Ag (961 °C) falls as Pb is added; Ag separates. Curve BO: melting point of Pb (327 °C) falls as Ag is added; Pb separates.
• Eutectic point O: 303 °C, 2.6 % Ag and 97.4 % Pb. Solid Ag, solid Pb and melt coexist; F′ = 2 − 3 + 1 = 0 (invariant).
• Pattinson’s process (desilverisation of lead): argentiferous lead (less than 2.6 % Ag) is melted and cooled. Pure lead crystallises out and is removed, so the melt gets richer in silver until it reaches the eutectic composition (2.6 % Ag). The Ag is then recovered by cupellation.

Two-component system: Bi–Cd

• Bi (m.p. 271 °C) and Cd (m.p. 321 °C) are completely miscible as liquids but do not form a compound or solid solution.
• Simple eutectic at about 140 °C, about 40 % Cd and 60 % Bi.
• Diagram: a “V” shape; the two freezing curves meet at the eutectic point; below the eutectic line the solid is a mixture of Bi and Cd.

(B) Liquid–liquid mixtures

Ideal solutions and Raoult’s law

• Raoult’s law: partial vapour pressure of a component = its mole fraction × vapour pressure of the pure component: p_A = x_Ap°_A.
• Ideal solution: obeys Raoult’s law at all compositions; ΔH_mix = 0 and ΔV_mix = 0. A–B forces equal A–A and B–B forces. Example: benzene–toluene, n-hexane–n-heptane.
• Henry’s law: for a dilute solute (often a gas), p = K_Hx (partial pressure proportional to mole fraction); K_H differs from p°. In a dilute (ideal-dilute) solution the solvent follows Raoult’s law and the solute follows Henry’s law.

Non-ideal systems and azeotropes

Azeotrope: a constant-boiling mixture whose vapour has the same composition as the liquid, so it cannot be separated by simple fractional distillation. Hence absolute alcohol cannot be obtained by fractional distillation of dilute ethanol.

Partially miscible liquids

• Phenol–water: two layers below 66 °C (water-rich and phenol-rich). Mutual solubility rises with temperature; above the upper consolute (critical solution) temperature, about 66 °C at about 34 % phenol, they mix completely.
• Triethylamine–water (written “trimethylamine” in the syllabus): miscible below about 18.5 °C, two layers above it — a lower consolute temperature. Reason: hydrogen bonding between amine and water breaks on heating.
• Nicotine–water: closed solubility loop with both a lower CST (about 61 °C) and an upper CST (about 210 °C); between them the liquids form two layers.
• Effect of impurity: an impurity soluble in only one liquid raises the UCST (e.g. NaCl or KCl in phenol–water); an impurity soluble in both liquids lowers the UCST. This is used to test purity of phenol.

Nernst distribution law

• Statement: at constant temperature, a solute distributes itself between two immiscible solvents so that the ratio of its concentrations is constant, provided it has the same molecular state in both: C₁/C₂ = K_D.
• Limitations: constant temperature; dilute solutions; the two solvents must be immiscible and not react with the solute; equilibrium must be reached.
• Deviations: association in one solvent (benzoic acid dimerises in benzene): C₁/√C₂ = K. Dissociation in one solvent: C₁(1 − α)/C₂ = K.
• Applications: solvent extraction (many small extractions remove more than one large extraction); Parkes process (Ag extracted from molten Pb by zinc); finding association or dissociation of a solute; finding equilibrium constant of KI + I₂ ⇌ KI₃; distribution indicators.

Likely exam questions

Long answer

1. State and derive the phase rule. Explain the phase diagram of the water system.
2. Discuss the Pb–Ag system with a neat labelled diagram and explain Pattinson’s process.

Short answer

1. What are azeotropes? Explain with HCl–water and ethanol–water systems.
2. Explain the phenol–water system. What is the effect of impurity on its critical solution temperature?
3. State Nernst distribution law. Give its limitations and two applications.

Syllabus topics (official RTMNU)

Electrochemistry

(A) Electrical transport

• Conductance in metals (electronic) & in electrolyte solutions (ionic conductance), conductivity of electrodes, specific, equivalent & molar conductance, measurement of equivalent conductance, variation of equivalent & specific conductance with dilution, mobility of ions & Kohlrausch’s law, Arrhenius theory of electrolyte dissociation & its limitation, Debye-Huckel theory (elementary treatment)

(B)

• Migration of ions, velocity of ions & change in concentration around electrode, transport number: definition & determination by Hittorfs method & moving boundary method, factors affecting transport number of ions, relation between transport number & ionic conductance.
• Application of Kohlrausch’s law & conductance for the determination of degree of dissociation, dissociation constant of acids, solubility of sparingly soluble salt, conductometric titrations (Acid-Base & precipitation) .

(A) Electrical transport

Metallic and electrolytic conductance

Conductivity of electrolytes: specific, equivalent and molar conductance

• Resistance R = ρ(l/a); conductance G = 1/R (unit S or Ω⁻¹).
• Specific conductance (conductivity) κ = 1/ρ = G × (l/a). l/a is the cell constant (cm⁻¹). Unit of κ: S cm⁻¹.
• Equivalent conductance Λ_eq = κ × 1000/N (N = normality). Unit: S cm² eq⁻¹.
• Molar conductance Λ_m = κ × 1000/M (M = molarity). Unit: S cm² mol⁻¹.

Measurement of equivalent conductance

1. The solution is placed in a conductivity cell with platinised platinum electrodes.
2. Its resistance is measured with a Wheatstone bridge using AC (about 1000 Hz) to avoid electrolysis and polarisation; a headphone or magic-eye detector shows the null point.
3. Cell constant is found using standard KCl solution of known κ: cell constant = κ_KCl × R_KCl.
4. Then κ = cell constant/R and Λ_eq = κ × 1000/N.

Variation with dilution

• Specific conductance decreases on dilution, because the number of ions per cm³ decreases.
• Equivalent (molar) conductance increases on dilution and reaches a limiting value Λ° at infinite dilution.
• Strong electrolytes (KCl): small linear rise; Λ = Λ° − A√C, so Λ° is got by extrapolating the Λ vs √C straight line to C = 0.
• Weak electrolytes (CH₃COOH): very steep rise at low concentration (degree of ionisation increases); Λ° cannot be found by extrapolation.
• Diagram: Λ on y-axis, √C on x-axis; KCl is a gently sloping straight line, CH₃COOH a curve shooting upward near the y-axis.

Mobility of ions and Kohlrausch’s law

• Ionic mobility: speed of an ion under a potential gradient of 1 V cm⁻¹; λ° = F × u (u = mobility).
• Kohlrausch’s law: at infinite dilution each ion migrates independently and contributes a definite amount to the conductance: Λ° = λ°₊ + λ°₋ (for 1:1 electrolyte).
• Example: Λ°(CH₃COOH) = Λ°(CH₃COONa) + Λ°(HCl) − Λ°(NaCl).

Arrhenius theory of electrolytic dissociation

• Electrolytes in water split into ions spontaneously; ions are in equilibrium with undissociated molecules.
• Degree of dissociation α = Λ_c/Λ°. Conductance depends on the number of ions.
• Limitations: strong electrolytes do not obey Ostwald’s dilution law; ignores interionic attraction; ignores role of solvent; strong electrolytes are fully ionised even in solid (NaCl crystal is ionic).

Debye–Hückel theory (elementary)

• Strong electrolytes are completely ionised; Λ is less than Λ° due to interionic attraction.
• Each ion is surrounded by an ionic atmosphere of opposite charge.
• Asymmetry (relaxation) effect: the moving ion’s atmosphere lags behind, pulling the ion back.
• Electrophoretic effect: the ionic atmosphere with its solvent moves the opposite way, slowing the ion.
• Debye–Hückel–Onsager equation: Λ = Λ° − (A + BΛ°)√C.

(B) Migration of ions and transport number

Velocity of ions and concentration change near electrodes

During electrolysis ions move at different speeds, so concentration falls unequally around the electrodes. Hittorf’s rule: (fall in concentration around anode)/(fall around cathode) = (speed of cation)/(speed of anion).

Transport number

• Definition: fraction of total current carried by an ion: t₊ = u₊/(u₊ + u₋), t₋ = u₋/(u₊ + u₋); t₊ + t₋ = 1.
• Relation with ionic conductance: λ₊ = t₊Λ and λ₋ = t₋Λ (at infinite dilution λ°₊ = t₊Λ°).

Hittorf’s method

1. A Hittorf cell (anode, middle and cathode compartments) is filled with AgNO₃ solution; silver electrodes; a silver coulometer in series.
2. A small current is passed for a few hours; the middle compartment concentration must stay unchanged.
3. Anode solution is analysed. With Ag anode: Ag dissolved (from coulometer) = w; increase in Ag at anode = x. Ag⁺ migrated out = w − x.
4. t_Ag⁺ = (w − x)/w; t_NO₃⁻ = 1 − t_Ag⁺.

Moving boundary method

• Two electrolytes with a common ion (e.g. HCl over CdCl₂ as indicator) form a sharp boundary; the indicator ion must be slower.
• When Q coulombs pass, the boundary moves a distance l in a tube of area a: volume V = la.
• t₊ = V × C × F/Q (C in mol cm⁻³).

Factors affecting transport number

• Temperature: values tend towards 0.5 as T rises.
• Concentration: changes slightly with concentration.
• Nature of the ion and its hydration: Li⁺ is small but heavily hydrated, so it moves slowly (low t).
• Nature of the accompanying ion (t of Cl⁻ differs in HCl and NaCl).

Applications of Kohlrausch’s law and conductance

• Degree of dissociation: α = Λ_c/Λ°.
• Dissociation constant of weak acid: K_a = Cα²/(1 − α).
• Solubility of sparingly soluble salt (AgCl, BaSO₄): κ_salt = κ_solution − κ_water; solubility S = κ × 1000/Λ° (mol L⁻¹); K_sp = S² for AgCl.

Worked example: κ(AgCl) = 1.86 × 10⁻⁶ S cm⁻¹, Λ° = 138.3 S cm² mol⁻¹. S = 1.86 × 10⁻⁶ × 1000/138.3 = 1.345 × 10⁻⁵ mol L⁻¹. K_sp = (1.345 × 10⁻⁵)² = 1.81 × 10⁻¹⁰.

Conductometric titrations

Conductance is plotted against volume of titrant; the end point is the intersection of two lines.

• Strong acid vs strong base (HCl + NaOH): conductance falls (fast H⁺ replaced by slow Na⁺), then rises after the end point (excess fast OH⁻). V-shaped graph.
• Weak acid vs strong base (CH₃COOH + NaOH): low start, slow rise as CH₃COONa forms, steep rise after the end point.
• Strong acid vs weak base (HCl + NH₄OH): falls till the end point, then almost constant.
• Precipitation (KCl + AgNO₃): Cl⁻ is replaced by NO₃⁻ of similar mobility, so conductance stays nearly constant; after the end point excess Ag⁺ and NO₃⁻ make it rise.
• Advantages: works for coloured or dilute solutions and weak acids, where indicators fail.

Likely exam questions

Long answer

1. Define specific, equivalent and molar conductance. How is equivalent conductance measured? Explain its variation with dilution.
2. What is transport number? Describe Hittorf’s method for its determination.

Short answer

1. State Kohlrausch’s law and give its applications.
2. Explain the conductometric titration of strong acid against strong base with a graph.
3. Write a note on Debye–Hückel theory of strong electrolytes.

Syllabus topics (official RTMNU)

Chemical Kinetics

(A)

• Concept of reaction rate, factors affecting the rate of a reaction – concentration., temperature, pressure, solvent, light, catalyst.
• Mathematical expression for rate constant of first and second order reactions, their characteristics.
• Pseudo order reactions.
• Half life and mean life of reactions with examples.
• Methods of determination of order of reaction –integration differential method, method of half life period and isolation method Experimental methods based on conductometry, polarimetry etc.
• Effect of temperature on rate of reaction.
• Arrhenius equation, activation energy.

(B)

• Collision theory of bimolecular reactions (hard sphere model).
• Transition state theory, expression for rate constant based on equilibrium constant and thermodynamic aspects.
• Lindmann’s theory of unimolecular reactions

(A) Reaction rate, order and its determination

Concept of reaction rate

• Rate = change in concentration per unit time: for A → B, rate = −d[A]/dt = +d[B]/dt. Unit: mol L⁻¹ s⁻¹.
• Rate law: rate = k[A]^m[B]ⁿ; k = rate constant; order = m + n (found by experiment; may be zero or fractional).

Factors affecting the rate

• Concentration: more molecules, more collisions, higher rate.
• Temperature: rate usually becomes 2–3 times for every 10 °C rise (temperature coefficient).
• Pressure: for gaseous reactions, higher pressure means higher concentration.
• Solvent: polarity and viscosity of the solvent change the rate (polar solvents speed up reactions with charged transition states).
• Light: photochemical reactions (H₂ + Cl₂ → 2HCl) need light.
• Catalyst: provides a path of lower activation energy; does not change equilibrium.

First-order reactions

1. −d[A]/dt = k[A]. With initial concentration a and x reacted at time t: dx/dt = k(a − x).
2. Integrating: k = (2.303/t) log[a/(a − x)].
3. Half-life (x = a/2): t_½ = 0.693/k, independent of initial concentration.

• Unit of k: s⁻¹. Plot of log(a − x) vs t is a straight line with slope −k/2.303.
• Examples: decomposition of N₂O₅, radioactive decay, decomposition of H₂O₂.

Second-order reactions

• Equal initial concentrations (a = b): dx/dt = k(a − x)²; k = (1/t)·x/[a(a − x)].
• Half-life: t_½ = 1/(ka), inversely proportional to initial concentration.
• Unequal concentrations: k = [2.303/t(a − b)] log[b(a − x)/a(b − x)].
• Unit of k: L mol⁻¹ s⁻¹. Plot of 1/(a − x) vs t is a straight line.
• Example: saponification of ethyl acetate by NaOH.

Pseudo-order reactions

A reaction of higher order that behaves as a lower order because one reactant is in large excess (its concentration stays practically constant).

• Acid hydrolysis of ethyl acetate: CH₃COOC₂H₅ + H₂O → CH₃COOH + C₂H₅OH (water in excess): pseudo-first order.
• Inversion of cane sugar: C₁₂H₂₂O₁₁ + H₂O → C₆H₁₂O₆ (glucose) + C₆H₁₂O₆ (fructose): pseudo-first order.

Half-life and mean life

• Half-life: time for half the reactant to react. General: t_½ ∝ 1/aⁿ⁻¹.
• Mean (average) life τ of a first-order reaction = 1/k = 1.44 t_½. Example: radioactive decay.

Methods of determining order

1. Integration (hit and trial) method: substitute data into first-, second-order equations; the equation giving constant k gives the order.
2. Differential method (van’t Hoff): rate = kCⁿ; measure rates r₁, r₂ at C₁, C₂: n = (log r₁ − log r₂)/(log C₁ − log C₂).
3. Half-life method: n = 1 + log(t₁/t₂)/log(a₂/a₁), where t₁, t₂ are half-lives at initial concentrations a₁, a₂.
4. Isolation method (Ostwald): all reactants except one are taken in large excess; order with respect to the isolated one is found; repeat for each and add.

Experimental methods

• Conductometry: saponification of ethyl acetate; fast OH⁻ is replaced by slower CH₃COO⁻, so conductance falls with time.
• Polarimetry: inversion of cane sugar; optical rotation changes from + (sucrose) to − (invert sugar). k = (2.303/t) log[(r₀ − r_∞)/(r_t − r_∞)].
• Others: titration (hydrolysis of ester with NaOH), measuring gas volume (decomposition of H₂O₂), spectrophotometry.

Effect of temperature: Arrhenius equation

• k = A e^−Ea/RT; A = frequency (pre-exponential) factor, Ea = activation energy.
• log k = log A − Ea/(2.303RT): plot of log k vs 1/T is a straight line, slope = −Ea/2.303R.
• Two temperatures: log(k₂/k₁) = [Ea/2.303R]·[(T₂ − T₁)/(T₁T₂)].
• Activation energy: minimum extra energy reactants need to form the activated complex.

Worked example: rate doubles from 300 K to 310 K. Ea = 2.303 × 8.314 × log 2 × (300 × 310)/10 = 2.303 × 8.314 × 0.301 × 9300 ≈ 53.6 kJ mol⁻¹.

(B) Theories of reaction rates

Collision theory (hard sphere model)

• Molecules are hard spheres; reaction occurs only when they collide.
• Collision frequency Z_AB = n_An_Bσ_AB²(8πk_BT/μ)^½ (σ = collision diameter, μ = reduced mass).
• Only collisions with energy ≥ Ea are effective: fraction = e^−Ea/RT. Rate = Z_ABe^−Ea/RT.
• Correct orientation is also needed: k = P Z e^−Ea/RT, P = steric (probability) factor.
• Limitation: P cannot be calculated; works only for simple gas molecules.

Transition state theory (activated complex theory)

• Reactants form an activated complex (X‡) at the top of the energy barrier, in quasi-equilibrium: A + B ⇌ X‡ → products.
• K‡ = [X‡]/[A][B]; the complex breaks with frequency k_BT/h.
• k = (k_BT/h)K‡ (rate constant from equilibrium constant).
• Thermodynamic form: ΔG‡ = −RT ln K‡ and ΔG‡ = ΔH‡ − TΔS‡, so k = (k_BT/h) e^ΔS‡/R e^−ΔH‡/RT (Eyring equation).
• e^ΔS‡/R explains the steric factor of collision theory.

Lindemann’s theory of unimolecular reactions

1. Activation by collision: A + A ⇌ A* + A (k₁ forward, k₋₁ deactivation).
2. Reaction: A* → products (k₂). There is a time lag between activation and reaction.
3. Steady state for A*: [A*] = k₁[A]²/(k₋₁[A] + k₂). Rate = k₂[A*] = k₁k₂[A]²/(k₋₁[A] + k₂).
4. High pressure (k₋₁[A] ≫ k₂): rate = (k₁k₂/k₋₁)[A], first order.
5. Low pressure (k₂ ≫ k₋₁[A]): rate = k₁[A]², second order.

This explains why unimolecular gas reactions change from first to second order at low pressure.

Likely exam questions

Long answer

1. Derive the rate constant expression for a first-order reaction and show that its half-life is independent of initial concentration.
2. Explain Lindemann’s theory of unimolecular reactions. Discuss the transition state theory of reaction rates.

Short answer

1. What are pseudo-order reactions? Give two examples.
2. Explain the half-life method for determining the order of a reaction.
3. Write the Arrhenius equation. How is activation energy determined from it?