Integrated Rate Equations & Half-Life Why integrate the rate law? Differential rate laws tell how fast concentration changes at an instant; integrated rate equations tell how concentration changes over time. For NEET questions, integrated forms help you compute concentration at any time t, half-life, or the rate constant k from experimental plots. An equation expressing instantaneous rate as a function of concentration(s), e.g., rate = k[A] m[B] n. Rate law (differential) Integrated rate equation Time-dependent relation between concentration and t obtained by integrating the differential rate law. Time required for reactant concentration to fall to half its initial value. Its dependence on [A]0 is diagnostic of reaction order. Half-life (t1/2) Differential forms describe instantaneous change (slope at a point). Integrated forms relate concentration and time over a finite interval and are used to calculate [A] at time t, t1/2, and to linearize data to find k. Differential and integrated rate equations can be applied interchangeably for any time interval. Zero-order kinetics In zero-order reactions, the rate is constant (independent of [A]). Concentration decreases linearly with time. Typical examples include photochemical reactions at high light intensity and surface-catalyzed reactions when the catalyst surface is saturated. Linear decrease of [A] with slope −k and intercept [A]0. Zero-order integrated rate law Half-life is directly proportional to initial concentration for zero order. Zero-order half-life Zero-order: [A] falls along a straight line with constant slope −k; think of a steady outflow lowering liquid level at a constant rate. First-order kinetics In first-order reactions, rate ∝ [A]. The concentration decays exponentially. Many decompositions and radioactive decay follow first order. Common examples at NEET level: decomposition of dinitrogen pentoxide (N2O5), inversion of sucrose (acid-catalyzed), and acid-catalyzed hydrolysis of methyl ethanoate (methyl acetate, SMILES: CC(=O)OC) in dilute HCl (effectively first-order in ester, see pseudo-first-order below). First-order integrated (natural log form) First-order integrated (base-10 log form) Equivalent to the log forms; shows exponential decay. First-order exponential form Signature feature: half-life is independent of [A]0. First-order half-life Spot the order fast: if t1/2 is constant when [A]0 changes, it is first order. If t1/2 increases with [A]0 → zero order; if t1/2 decreases with [A]0 → second order. neet-alert 2026-05-26T17:04:45.760Z Straight-line plot for first-order kinetics: ln[A] (y-axis) vs time t (x-axis), clear line with negative slope. Mark slope = −k and intercept = ln[A]0. Include inset showing curved [A] vs t exponential decay. Clean 2D vector style, axes labeled. First-order linearization: plot ln[A] vs t to get a straight line with slope −k and intercept ln[A]0. gpt-image-2 Integrated rate equations guide real decisions: zero-order in some surface/photochemical setups; first-order often models drug breakdown (shelf-life) and radioactive decay. Pseudo-first-order reactions When one reactant is in large excess, its concentration is effectively constant. Its factor merges into the observed rate constant, making the reaction appear first order in the limiting reactant. Classic NEET example: acid-catalyzed hydrolysis of an ester in excess water. Example: Methyl ethanoate (methyl acetate, SMILES: CC(=O)OC) hydrolysis in dilute HCl with water in large excess behaves as rate = kobs[ester], where kobs = k[H2O] (and includes the catalytic H+ dependence when kept constant). Ester hydrolysis in very dilute aqueous acid (water in huge excess) Inversion of sucrose in strong excess of water and fixed [H+] Any bimolecular reaction A + B when [B] ≫ [A] and [B] is held effectively constant Where pseudo-first-order applies Second-order kinetics Two common types: 2A → products (rate = k[A] 2) and A + B → products (rate = k[A][B]). NEET focuses on the 2A case for integrated form and half-life. A well-known laboratory example of overall second order is base hydrolysis (saponification) of ethyl ethanoate (ethyl acetate, SMILES: CC(=O)OCC) with NaOH: rate ∝ [ester][OH−]. Second-order integrated (2A → products) Half-life is inversely proportional to initial concentration. Second-order half-life (2A case) Second-order linearization: plot 1/[A] vs t to get a straight line; slope = k, intercept = 1/[A]0. gpt-image-2 2026-05-26T17:04:45.940Z Straight-line plot for second-order kinetics: 1/[A] (y-axis) vs time t (x-axis). Show positive slope = k and intercept = 1/[A]0. Add small inset of [A] vs t curve that flattens over time (hyperbolic decay). Clean vector style. Units of k: zero order → mol L⁻1 s⁻1; first order → s⁻1; second order → L mol⁻1 s⁻1. Check units to cross-verify order in numericals. remember Units of k depend on overall order: zero order (mol L⁻1 s⁻1), first order (s⁻1), second order (L mol⁻1 s⁻1). Dimension analysis helps catch mistakes. The rate constant k has the same units for all reaction orders. Compare zero, first, and second order at a glance Zero [A]t = [A]0 − kt t1/2 = [A]0/(2k) mol L⁻1 s⁻1 Straight line (downward) [A] vs t (slope = −k) First ln[A]t = ln[A]0 − kt or [A]t = [A]0 e (−kt) t1/2 = 0.693/k s⁻1 Exponential decay ln[A] vs t (slope = −k) or log[A] vs t Second (2A) 1/[A]t = 1/[A]0 + kt t1/2 = 1/(k[A]0) L mol⁻1 s⁻1 Curved (hyperbolic-like) 1/[A] vs t (slope = k) Case Order summary — integrated forms, half-life, k units, and useful plots Order Integrated form Half-life Units of k [A] vs t shape Linear plot for k (slope) 2026-05-26T17:04:46.099Z Three small panels in one figure showing concentration [A] (y) vs time t (x): Panel 1 linear decline (zero order), Panel 2 exponential decay (first order), Panel 3 hyperbolic-like decay (second order). Label each panel clearly. Textbook vector style. Side-by-side [A] vs t curves: zero-order (straight line), first-order (exponential), second-order (flattening curve). gpt-image-2 Order t1/2 dependence on [A]0 How t1/2 depends on [A]0 Zero Directly proportional (↑[A]0 → ↑t1/2) First Independent of [A]0 (constant) Second Inversely proportional (↑[A]0 → ↓t1/2) Half-life vs [A]0: 0–Up, 1–Flat, 2–Down (with [A]0). Radioactive decay and dating Radioactivity follows first-order kinetics. If N is the number of undecayed nuclei at time t, N decreases exponentially, and half-life is constant. This property powers dating methods in archaeology and geology. First-order decay law (radioactivity) Here, is the decay constant (s⁻1). Half-life in decay Fraction remaining after n half-lives Carbon-14 dating: 14C has t1/2 = 5730 years (NCERT). If a sample retains 25% of its original 14C, it has passed two half-lives → age ≈ 2 × 5730 = 11,460 years. Uranium-238 and K–Ar systems date much older rocks in geology. Radioactive decay curve: equal time steps reduce the amount by half each time (50%, 25%, 12.5%, ...). clinical Pharmacokinetics: Many drugs show near first-order elimination. Knowing t1/2 helps set dosing intervals to keep plasma levels therapeutic while limiting side effects. Quick numericals (with shortcuts) Worked examples First-order: Given [A]0 = 0.100 M and [A]t = 0.0250 M at t = 138.6 s. Compute k. Use k = (2.303/t) log([A]0/[A]t) = (2.303/138.6) log(4.00) ≈ 0.01662 × 0.60206 ≈ 1.00 × 10⁻2 s⁻1. Then t1/2 = 0.693/k ≈ 69.3 s. Zero-order: [A]0 = 0.50 M, k = 1.0 × 10⁻3 mol L⁻1 s⁻1. t1/2 = [A]0/(2k) = 0.50/(2 × 10⁻3) = 250 s. Second-order (2A): [A]0 = 0.100 M, k = 0.50 L mol⁻1 s⁻1. t1/2 = 1/(k[A]0) = 1/(0.50 × 0.100) = 20 s. Fraction from half-lives: After n half-lives, remaining fraction = (1/2) n. After 3 half-lives → 1/8 = 12.5% remains, so 87.5% has reacted. Carbon-14 dating: A sample has 25% of original 14C. That is 2 half-lives → age ≈ 2 × 5730 = 11,460 years. Trap: “All reactions have constant half-life.” False — only first-order reactions have t1/2 independent of [A]0. neet-alert Extra: rearranged forms you’ll use in calculations Slope form to get k (first order, base-10 logs) How to read slopes: annotate slopes and intercepts on [A] vs t, ln[A] vs t, and 1/[A] vs t plots. gpt-image-2 2026-05-26T17:04:46.128Z Composite figure with three mini-plots: [A] vs t (slope −k, zero order), ln[A] vs t (slope −k, first order), 1/[A] vs t (slope k, second order). Use red arrows for slopes, black labels for axes and intercepts. Clean vector style. t1/2 = 0.693/k applies only to first-order processes (including radioactive decay). For zero order and second order, t1/2 depends on [A]0. The half-life formula t1/2 = 0.693/k applies to every reaction.