Chemical Kinetics

1. Rate of Reaction and Factors Affecting It

• Rate is intrinsically positive: for a reactant, rate = −Δ[R]/Δt; for a product, rate = +Δ[P]/Δt. Instantaneous rate r_inst = lim_Δt→0(ΔC/Δt), i.e. the slope of the tangent to the concentration–time curve. Initial rate is the instantaneous rate at t = 0.
• Stoichiometric equivalence: for aA + bB → cC + dD, Rate = −(1/a)d[A]/dt = −(1/b)d[B]/dt = +(1/c)d[C]/dt = +(1/d)d[D]/dt. The rate of disappearance of a single species (say −d[A]/dt) is a directly measurable quantity; the "rate of reaction" divides it by that species' coefficient.
• Factors affecting rate: concentration (more collisions), temperature (more molecules cross E_a), physical state and particle size (gaseous > liquid > solid; smaller particles react faster), nature of the bonds being broken (stronger bonds ⇒ higher E_a ⇒ slower), catalyst (lowers E_a), and for ionic/enzymatic reactions, pH and the solvent's dielectric constant.

2. Rate Law, Order and Integrated Rate Laws

• Rate law: Rate = k[A]^p[B]^q; p and q are the partial orders (found experimentally, not from the balanced equation), and n = p + q is the overall order. Units of k = (mol L⁻¹)¹⁻ⁿ time⁻¹: mol L⁻¹s⁻¹ for zero order, s⁻¹ for first order, L mol⁻¹s⁻¹ for second order.
• Order can be zero, fractional or (rarely, w.r.t. one species) negative, but the overall order of a reaction has never been observed to be negative. Order is purely experimental; it cannot be predicted from stoichiometry except for elementary steps.
• Worked pattern 1 (order from k's units): if k has units L mol⁻¹min⁻¹, then 1−n = −1, so n = 2 (second order).
• Half-life patterns from the table above: zero order t_1/2 = [A]₀/2k (proportional to [A]₀); first order t_1/2 = 0.693/k (independent of [A]₀); second order t_1/2 = 1/(k[A]₀) (inversely proportional to [A]₀); nth order (n ≠ 1) t_1/2 ∝ 1/[A]₀ⁿ⁻¹.
• Worked pattern 2 (first-order time ratios): t = (2.303/k) log([A]₀/[A]_t). For 75% completion, [A]_t = 0.25[A]₀, so t_0.75 = (2.303/k) log 4 = 2 × t_1/2. For 99.9% completion, [A]_t = 0.001[A]₀, so t_0.999 = (2.303/k) log 1000 = 10 × t_1/2.

3. Pseudo-First-Order and Methods to Find Order

• Pseudo-first-order: when one reactant is in huge excess, its concentration stays effectively constant and the reaction behaves as first order in the other reactant. Acid hydrolysis of an ester, CH₃COOCH₃ + H₂O(excess) →^H+ CH₃COOH + CH₃OH, and inversion of cane sugar, C₁₂H₂₂O₁₁ + H₂O(excess) →^H+ glucose + fructose, are the two standard JEE examples. k_pseudo = k[H₂O], and Rate = k_pseudo[Ester].
• Initial rate method: hold all but one reactant's concentration fixed and compare two runs: r₁/r₂ = ([A]₁/[A]₂)^a, so a = log(r₁/r₂)/log([A]₁/[A]₂). This is the commonest way JEE gives a small concentration–rate data table and asks for the order.
• Integrated rate law (trial-and-error) method: substitute concentration–time data into each integrated equation in turn; the one that gives a constant k is the correct order.
• Half-life method: (t_1/2)₁/(t_1/2)₂ = ([A]_0,2/[A]_0,1)ⁿ⁻¹. Take log on both sides to solve for n directly from two half-lives measured at two different starting concentrations.
• Ostwald isolation method: flood all reactants except one with a large excess, so only that one reactant's order is seen; repeat, isolating each reactant in turn, and add up the partial orders.

4. Monitoring Progress: Gas Pressure, Titration, Polarimetry

• Gas-phase reactions (pressure monitoring), A(g) → nB(g): at time t, P_A = P₀ − x, P_B = nx, so total pressure P_t = P₀ + (n−1)x, giving x = (P_t − P₀)/(n−1) and P_A = [nP₀ − P_t]/(n−1). For a first-order gas reaction, k = (2.303/t) log[P₀(n−1)/(nP₀ − P_t)].
• Titration kinetics (H₂O₂ decomposition monitored by KMnO₄ titre): a ∝ V₀, (a−x) ∝ V_t, so k = (2.303/t) log(V₀/V_t).
• Ester hydrolysis monitored by NaOH titre: with V₀ (acid catalyst alone), V_t (catalyst + acid formed) and V_∞ (complete conversion): a ∝ (V_∞ − V₀), (a−x) ∝ (V_∞ − V_t), so k = (2.303/t) log[(V_∞ − V₀)/(V_∞ − V_t)].
• Optical rotation (inversion of cane sugar, polarimeter readings r₀, r_t, r_∞): a ∝ (r₀ − r_∞), (a−x) ∝ (r_t − r_∞), so k = (2.303/t) log[(r₀ − r_∞)/(r_t − r_∞)].

5. Arrhenius Theory and Catalysis

• Collision theory in brief: a reaction occurs only when colliding molecules have kinetic energy at or above a threshold and the correct spatial (steric) orientation — simple collision frequency alone overstates the rate, which is why an orientation/probability factor multiplies the collision number in the full collision-theory expression.
• Arrhenius equation: k = A e^−E_a/RT, so ln k = ln A − E_a/RT. A is the frequency (pre-exponential) factor; e^−E_a/RT is the fraction of molecules with energy ≥ E_a.
• Two-temperature form: log(k₂/k₁) = (E_a/2.303R)[(T₂−T₁)/(T₁T₂)]. This is the standard JEE numerical: given k at two temperatures (or a ratio and one E_a), solve for the missing quantity.
• Energetics: ΔH = E_a,forward − E_a,backward. For an exothermic reaction (ΔH < 0), E_a,forward can be as low as (but never below) 0. For an endothermic reaction (ΔH > 0), E_a,forward ≥ ΔH. A reaction with a larger E_a is more sensitive to a change in temperature (its k changes by a bigger factor for the same ΔT).
• Catalysis: a catalyst opens an alternative pathway of lower E_a, without being consumed. It speeds up k_f and k_b equally, so ΔH, ΔG and K_eq are unchanged — only the time to reach equilibrium falls. k_cat/k_uncat = e^{(E_a−E_a′)/RT}.

6. Molecularity vs Order; Complex Reactions

• Molecularity is the number of reacting species that collide simultaneously in one elementary step: always a positive integer, 1, 2 or 3 (never zero, fractional or negative), defined only for a single elementary step (not for the overall multi-step reaction), and independent of temperature and pressure.
• Order is the sum of the powers of concentration terms in the experimentally found rate law: can be zero, fractional, integer or (w.r.t. one reactant) negative; applies to elementary or complex reactions; and can itself vary with temperature, pressure and concentration for some real systems.
• Rate-determining step (RDS): the slowest step fixes the overall rate law. If the RDS involves only reactants/catalysts, write the rate law directly from it. If the RDS involves a reactive intermediate, use a fast prior equilibrium to substitute the intermediate's concentration.
• Worked pattern 3 (ozone decomposition, 2O₃ → 3O₂): Step 1 (fast equilibrium) O₃ ⇌ O₂ + O, K_eq = [O₂][O]/[O₃] ⇒ [O] = K_eq[O₃]/[O₂]. Step 2 (slow RDS) O + O₃ → 2O₂, Rate = k₃[O₃][O]. Substituting [O] gives Rate = k₃K_eq[O₃]²[O₂]⁻¹: order 2 in O₃, −1 in O₂, overall order 1.
• Steady-state approximation (SSA): for a short-lived, highly reactive intermediate, its rate of formation equals its rate of consumption, i.e. d[intermediate]/dt = 0. This is an alternative to using a fast pre-equilibrium and gives the same kind of rate law when applied correctly.

7. Parallel, Reversible and Consecutive Reactions

• Parallel (competing) first-order reactions: A decomposes simultaneously to B (k₁) and C (k₂). k_eff = k₁ + k₂, so 1/t_eff = 1/t₁ + 1/t₂ for any fixed fractional conversion, and [B]/[C] = k₁/k₂ at every instant (the product ratio is fixed, independent of time). Overall E_a,eff = (k₁E_a,1 + k₂E_a,2)/(k₁ + k₂).
• Reversible first-order reactions, A ⇌ B (forward k_f, backward k_b), starting from pure A = a: at equilibrium x_eq = a·k_f/(k_f+k_b), and (k_f+k_b) = (1/t) ln[x_eq/(x_eq−x)]. This behaves like a pseudo-first-order approach to equilibrium, with rate constant (k_f+k_b).
• Consecutive (sequential) first-order reactions, A →^k₁ B →^k₂ C, from [A]₀ = a, [B]₀ = [C]₀ = 0: [A]_t = ae^−k₁t; [B]_t = [ak₁/(k₂−k₁)](e^−k₁t − e^−k₂t); [C]_t = a − [A]_t − [B]_t.
• Worked pattern 4 (intermediate maximum): setting d[B]/dt = 0 gives the time at which B peaks, t_max = [1/(k₁−k₂)] ln(k₁/k₂) = [2.303/(k₁−k₂)] log(k₁/k₂), with [B]_max = a(k₂/k₁)^{k₂/(k₁−k₂)}. If B is the desired product (e.g. a drug intermediate), t_max is when to stop the reaction and isolate it.

8. How JEE Frames Questions

• Numerical: k from an integrated rate law or a half-life ratio, E_a from the two-temperature Arrhenius equation, order from an initial-rate data table, t_max and [B]_max for a consecutive reaction, k_eff and product ratio for a parallel reaction, k from pressure or titration data.
• Statements: effect of a catalyst on E_a/ΔH/K_eq; molecularity vs order comparisons; which straight-line plot matches which order; whether a stated reaction order is even possible.
• Data/graph questions: read a slope off a log k vs 1/T or 1/[A]_t vs t plot; identify the order from a given concentration–time table by testing which integrated law gives a constant k.

9. Quick Sheet and Checklist

Before the exam, check you can:

• Identify order from the units of k, and match [A]_t vs t / ln[A]_t vs t / 1/[A]_t vs t to zero, first and second order.
• Use t_0.75 = 2t_1/2 and t_0.999 = 10t_1/2 for first order without re-deriving them.
• Find order from an initial-rate table (log ratio method) or from a half-life ratio at two starting concentrations.
• Set up k from pressure, titre or optical-rotation data by identifying the correct "remaining reactant" proxy.
• Solve a two-temperature Arrhenius problem for E_a, A, k₂ or T₂, and state what a catalyst does and does not change.
• Handle parallel, reversible and consecutive first-order reactions, including t_max for a consecutive intermediate.