Chemical Kinetics

Chemical kinetics deals with reaction rates and the factors that influence them. The rate of reaction for aA + bB → cC + dD is: rate = −(1/a)d[A]/dt = −(1/b)d[B]/dt = (1/c)d[C]/dt = (1/d)d[D]/dt. Factors affecting rate include concentration, temperature (a 10°C rise roughly doubles the rate), pressure (for gases), catalyst, surface area, and nature of reactants.

The rate law is experimentally determined: rate = k[A]^m[B]^n, where m and n are orders (NOT stoichiometric coefficients). Overall order = m + n. Units of k depend on order: for nth order, k has units (mol/L)^(1−n) × s⁻¹. So: zero order = mol/(L·s), first order = s⁻¹, second order = L/(mol·s).

Order vs Molecularity — critical distinction: Order is experimental, can be 0, fractional, or integer, applies to overall or elementary reactions, and can change with conditions. Molecularity is theoretical, always a positive integer (1, 2, or 3), applies only to elementary steps, and cannot change. The rate-determining step (slowest step) controls the overall rate.

Zero order kinetics: Integrated equation: [A] = [A]₀ − kt (linear decrease) Half-life: t₁/₂ = [A]₀/(2k) — depends on initial concentration Example: NH₃ decomposition on Pt surface at high pressure.

First order kinetics: Integrated equation: ln[A] = ln[A]₀ − kt, or equivalently k = (2.303/t) log([A]₀/[A]) Half-life derivation: At t₁/₂, [A] = [A]₀/2. Substituting: k × t₁/₂ = ln([A]₀/([A]₀/2)) = ln 2 = 0.693. Therefore t₁/₂ = 0.693/k — independent of initial concentration. Examples: radioactive decay, H₂O₂ decomposition, N₂O₅ decomposition.

Solved Example 1: First-order reaction with k = 6.93 × 10⁻³ s⁻¹. t₁/₂ = 0.693/k = 0.693/(6.93 × 10⁻³) = 100 s For 75% completion: [A] = 0.25[A]₀ t = (2.303/k) × log([A]₀/0.25[A]₀) = (2.$\frac{303}{6}$.93×10⁻³) × log 4 = 332.3 × 0.602 = 200 s = 2 × t₁/₂ ✓

Pseudo first-order reactions occur when one reactant is in large excess, making the reaction appear first order. Example: hydrolysis of ethyl acetate (SMILES: CC(=O)OCC) in excess water: CH₃COOC₂H₅ + H₂O → CH₃COOH + C₂H₅OH (SMILES products: CC(=O)O + CCO).

Arrhenius equation: k = Ae^(−Ea/RT), where A = frequency/pre-exponential factor, Ea = activation energy. Two-temperature form: log(k₂/k₁) = (Ea/2.303R)(1/T₁ − 1/T₂) Graph: ln k vs 1/T gives a straight line with slope = −Ea/R.

Solved Example 2: k at 300 K = 2.5 × 10⁻³ s⁻¹, k at 310 K = 5.0 × 10⁻³ s⁻¹. Find Ea. log(k₂/k₁) = (Ea/2.303R)(1/T₁ − 1/T₂) log(5.0×10⁻³/2.5×10⁻³) = (Ea/(2.303 × 8.314))($\frac{1}{300}$ − $\frac{1}{310}$) 0.301 = (Ea/19.147)($\frac{10}{93000}$) = Ea × 5.616 × 10⁻⁶ Ea = 0.$\frac{301}{5}$.616 × 10⁻⁶ = 53,596 J/mol ≈ 53.6 kJ/mol

Activation energy: Ea(forward) − Ea(backward) = ΔH. A catalyst lowers Ea by providing an alternative pathway but does NOT change ΔH or the equilibrium constant.

Solved Example 3: Zero-order reaction, k = 0.01 mol/(L·s), [A]₀ = 0.5 M. Time for complete decomposition: t = [A]₀/k = 0.$\frac{5}{0}$.01 = 50 s t₁/₂ = [A]₀/(2k) = 0.5/(2 × 0.01) = 25 s

Dimensional analysis for rate constant units: Rate = k[A]^n → mol/(L·s) = k × (mol/L)^n → k = (mol/L)^(1−n) × s⁻¹ ✓

The key testable concept is first-order half-life calculations (t₁/₂ = 0.693/k, independent of concentration) and Arrhenius equation activation energy determination.