GATE Chemical Kinetics — Integrated Rate Laws and Half-Lives

By Aniket Bhardwaj · 30 August 2026 · GATE Chemistry

Chemical kinetics is the friendliest part of physical chemistry for a GATE candidate: the results are closed-form, the numericals are self-checking, and a handful of relations covers nearly everything that gets asked. The difficulty is never the algebra. It is correctly identifying the order, then keeping the stoichiometric factor straight. This article works through the integrated rate laws, the half-life relations, the Arrhenius treatment and consecutive reactions, with every number actually computed.

The three integrated rate laws

For a reaction whose rate of consumption of A is −d[A]/dt = k[A]ⁿ:

Zero order (n = 0): [A] = [A]₀ − kt · t_½ = [A]₀ / 2k · units of k: M s⁻¹
First order (n = 1): ln[A] = ln[A]₀ − kt · t_½ = ln2 / k · units of k: s⁻¹
Second order (n = 2): 1/[A] = 1/[A]₀ + kt · t_½ = 1 / (k[A]₀) · units of k: M⁻¹ s⁻¹

[A]₀ is the initial concentration, [A] the concentration at time t, k the rate constant. The units of k are a free diagnostic: if a question hands you k in M⁻¹ s⁻¹, the reaction is second order overall and you need not read further to know which formula applies.

The half-life fingerprint

The general result for order n (n ≠ 1) is t_½ ∝ [A]₀¹⁻ⁿ. This gives you a one-line way to determine order from data:

Order	Effect of doubling [A]₀ on t_½	Linear plot that gives k
Zero	Doubles	[A] vs t, slope = −k
First	No change	ln[A] vs t, slope = −k
Second	Halves	1/[A] vs t, slope = +k
Third	Falls to one quarter	1/[A]² vs t, slope = +2k

The concentration-independence of the first-order half-life is the property that makes radioactive decay so clean, and it is also the property GATE most often tests obliquely.

Worked example 1 — first order. A first-order reaction has a half-life of 30.0 minutes. How long does it take to reach 90% completion?

k = ln 2 / t_½ = 0.69315 / 30.0 = 0.023105 min⁻¹

At 90% completion, [A]/[A]₀ = 0.10, so kt = ln([A]₀/[A]) = ln 10 = 2.302585

t = 2.302585 / 0.023105 = 99.66 minutes

Cross-check: 90% completion means 10% remains, which is between three half-lives (12.5% remaining) and four (6.25%). Precisely, log₂(10) = 3.322 half-lives × 30.0 min = 99.66 min. ✔

Worked example 2 — second order. For 2A → products with −d[A]/dt = k[A]², k = 0.050 M⁻¹ s⁻¹ and [A]₀ = 0.20 M. Find (a) t_½ and (b) the time needed for [A] to fall to 0.050 M.

(a) t_½ = 1 / (k[A]₀) = 1 / (0.050 × 0.20) = 1 / 0.010 = 100 s

(b) 1/[A] − 1/[A]₀ = kt
1/0.050 − 1/0.20 = 20.0 − 5.0 = 15.0 M⁻¹
t = 15.0 / 0.050 = 300 s

Cross-check: 0.050 M is one quarter of 0.20 M, so it takes one half-life to reach 0.10 M (100 s) plus a second half-life at the new starting concentration 1/(0.050 × 0.10) = 200 s. Total 300 s. ✔ Notice the second half-life is twice the first — the second-order fingerprint.

The Arrhenius equation

k = A e^−E_a/RT · ln k = ln A − E_a/RT · ln(k₂/k₁) = −(E_a/R)(1/T₂ − 1/T₁)

A is the pre-exponential factor (same units as k), E_a the activation energy in J mol⁻¹, R = 8.314 J K⁻¹ mol⁻¹. A plot of ln k against 1/T is linear with slope −E_a/R and intercept ln A — that plot is itself a standard GATE question.

Worked example 3 — activation energy. The rate constant of a reaction doubles when the temperature is raised from 300 K to 310 K. Find E_a.

ln(k₂/k₁) = ln 2 = 0.69315

1/T₁ − 1/T₂ = 1/300 − 1/310 = 0.00333333 − 0.00322581 = 1.07527 × 10⁻⁴ K⁻¹

E_a = R ln(k₂/k₁) / (1/T₁ − 1/T₂) = (8.314 × 0.69315) / 1.07527 × 10⁻⁴

= 5.7630 / 1.07527 × 10⁻⁴ = 53 597 J mol⁻¹ ≈ 53.6 kJ mol⁻¹

The old rule of thumb that "rate doubles for every 10 K rise" is therefore only true for reactions with E_a near 50 kJ mol⁻¹ around room temperature — a fact GATE has used as a trap in conceptual questions.

Consecutive reactions and the intermediate maximum

For A →^k₁ B →^k₂ C with only A present initially:

[B](t) = [A]₀ · k₁/(k₂ − k₁) · (e^−k₁t − e^−k₂t)
t_max = ln(k₂/k₁) / (k₂ − k₁) · [B]_max/[A]₀ = (k₁/k₂)^{k₂/(k₂−k₁)}

Worked example 4. k₁ = 0.10 min⁻¹, k₂ = 0.20 min⁻¹. Find t_max and the maximum fraction of B.

t_max = ln(0.20/0.10) / (0.20 − 0.10) = 0.69315 / 0.10 = 6.93 min

[B]_max/[A]₀ = (0.5)^0.20/0.10 = (0.5)² = 0.25

Direct verification. At t = 6.93 min: k₁t = 0.6931 and k₂t = 1.3863.
[B]/[A]₀ = (0.10/0.10)(e^−0.6931 − e^−1.3863) = 1 × (0.500 − 0.250) = 0.250. ✔

Approximation methods you are expected to apply

Steady-state approximation: set d[intermediate]/dt = 0, solve for the intermediate concentration, substitute back. This is the standard route through the Lindemann–Hinshelwood mechanism and enzyme kinetics.
Pre-equilibrium approximation: valid when the first step equilibrates much faster than the slow step consumes the intermediate. Gives a rate law containing the equilibrium constant of the fast step.
Michaelis–Menten: v = v_max[S] / (K_M + [S]), with K_M equal to the substrate concentration at half of v_max. The Lineweaver–Burk double reciprocal plot linearises it: 1/v against 1/[S] gives slope K_M/v_max and intercept 1/v_max.
Transition-state theory: the Eyring equation k = (k_BT/h) e^−ΔG‡/RT, from which ΔH‡ and ΔS‡ come out of an Eyring plot of ln(k/T) against 1/T.

The traps that cost kinetics marks

The stoichiometric factor. For 2A → P, the rate of reaction is −½ d[A]/dt, not −d[A]/dt. If a question defines rate one way and gives you k measured the other way, your answer is out by a factor of two. Read the definition before you substitute — this is the single most common error in the topic.
Assuming order equals molecularity. Molecularity is a property of an elementary step and is a small whole number; order is experimental and can be fractional, zero, or even negative.
Using E_a in kJ inside the exponent. R is in J K⁻¹ mol⁻¹, so E_a must be in J mol⁻¹. A factor of 1000 here produces an answer that is not even the right order of magnitude.
Forgetting that a catalyst does not change ΔG or K. It lowers E_a for both the forward and reverse directions equally, so it changes how fast equilibrium is reached, never where equilibrium lies.
Applying t_½ = ln2/k to a non-first-order reaction. That relation is a special property of first order and nothing else.

Check your activation-energy work instantly. The Arrhenius calculator takes any three of k₁, k₂, T₁, T₂ and E_a and returns the fourth, showing the substitution so you can see exactly where a hand calculation went wrong.

Open the Arrhenius Equation Calculator →

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