Kinetics Graphical Methods
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Direct answer
Graphical analysis pins down reaction order without calculus: a zero-order reaction plots [A] linearly against time (slope −k); a first-order reaction plots ln[A] linearly (slope −k); a second-order reaction plots 1/[A] linearly (slope +k). Half-lives diagnose order just as fast — constant t1/2 means first order (0.693/k), t1/2 proportional to [A]0 means zero order, and t1/2 inversely proportional to [A]0 means second order. Indian JEE papers add a signature twist: first-order gas decompositions tracked by total pressure, resolved with k = (2.303/t) log(p0/(2p0 − pt)) for A → 2B-type stoichiometry.
What you must remember
- Integrated equations: zero order [A] = [A]0 − kt; first order ln[A] = ln[A]0 − kt; second order 1/[A] = 1/[A]0 + kt — match the linear variable to the order.
- Slope signs: ln[A] versus t has slope −k; 1/[A] versus t has slope +k; [A] versus t has slope −k — the sign slip is a classic mark-loser.
- Half-life set: zero order t1/2 = [A]0/2k; first order t1/2 = 0.693/k, independent of concentration; second order t1/2 = 1/(k[A]0).
- Order from two half-lives: n = 1 + log(t1/2 ratio)/log(concentration ratio reversed); a fourfold [A]0 increase that cuts t1/2 fourfold means n = 2.
- Units of k betray order: mol L−1 s−1 for zero, s−1 for first, L mol−1 s−1 for second, from the general (mol L−1)^(1−n) s−1 — a one-mark regular straight out of NCERT.
- Gas-phase first order: for A(g) → 2B(g) with total pressure pt and initial p0, the surviving reactant pressure is pA = 2p0 − pt, and k = (2.303/t) log(p0/(2p0 − pt)).
- Initial-rate method: double one reactant; rate doubles → first order in it, quadruples → second order, unchanged → zero order.
- Ostwald isolation: hold all but one reactant in large excess to reduce the rate law to pseudo-order in the limiting species — ester saponification with excess alkali is the NCERT staple.
Reading data like a kineticist
Suppose an experimenter reports t1/2 = 40 min at [A]0 = 0.1 M and t1/2 = 10 min at [A]0 = 0.4 M. The half-life shrank fourfold while concentration rose fourfold, so t1/2 ∝ 1/[A]0 — second order — and k = 1/(t1/2 × [A]0) = 1/(40 × 0.1) = 0.25 L mol−1 min−1. Had the same 40-minute half-life persisted at both concentrations, k = 0.693/40 = 0.0173 min−1 and the reaction is first order. Now the pressure variant: for A(g) → 2B(g) starting at p0 = 100 mm with total pressure 120 mm after 20 minutes, pA = 2(100) − 120 = 80 mm and k = (2.303/20) log(100/80) = 0.1151 × 0.0969 ≈ 0.011 min−1. One caution travels with the shortcut: it is stoichiometry-specific. For A → 3B, the material balance gives pA = (3p0 − pt)/2, and blind reuse of 2p0 − pt silently corrupts the answer.
What the graphs hide
JEE Main hands you a labelled plot and asks for order and k — remember that 1/[A] versus t rises through a positive intercept 1/[A]0, while ln[A] versus t falls through ln[A]0. Advanced prefers raw data tables and the pressure trick, or a pre-built axis of ln(p0/(2p0 − pt)) whose slope already equals −k. Two traps recur: quoting k from a second-order plot without its L mol−1 s−1 units, and recomputing a first-order half-life after dilution when the defining property is that it never changes — a candidate who recalculates 0.693/k for the new concentration has misunderstood the entire graph family. Pseudo-order wording also bites: saponification with hydroxide in excess is first order in ester, but the true molecularity and order of the elementary reaction remain two.
Frequently asked questions
Which plot is linear for a second-order reaction?
1/[A] against time, with slope k and intercept 1/[A]0, provided the rate law is −d[A]/dt = k[A]^2.
How is order determined from half-life data?
Compare t1/2 at two initial concentrations: constant means first order, proportional to [A]0 means zero order, inversely proportional means second order — formalised as n = 1 + log(t1/2 ratio)/log([A]0 reversed ratio).
What are the units of a third-order rate constant?
(mol L−1)^(1−3) s−1 = L^2 mol−2 s−1, from the general unit pattern (mol L−1)^(1−n) s−1.
How is total pressure used for a first-order gas decomposition?
Convert pt to the surviving reactant pressure through stoichiometry — pA = 2p0 − pt for A → 2B — then apply k = (2.303/t) log(p0/pA).
Why does ester saponification appear first order?
With sodium hydroxide in large excess, its concentration is effectively constant, collapsing the second-order law to pseudo-first order in ester — the isolation method at work.