JAM Chemical Kinetics — The Standard Question Set

By Aniket Bhardwaj · 5 September 2026 · IIT-JAM Chemistry

Kinetics is among the most predictable parts of physical chemistry for IIT-JAM. The questions come from a short, stable list: find the order, find k, find a time or concentration from an integrated rate law, find the activation energy from two rate constants, or derive a rate law from a mechanism. Each is arithmetic on top of a formula you can write from memory. This article sets out the formulas, then works every question type with the numbers computed.

Rate, order and molecularity

For aA + bB → products, the rate is defined so that it has a single value regardless of which species you follow:

rate = −(1/a)·d[A]/dt = −(1/b)·d[B]/dt = +(1/p)·d[P]/dt

rate = k[A]^m[B]ⁿ overall order = m + n

Two distinctions the paper tests directly. Order is experimental and may be zero, fractional or negative; molecularity counts species in one elementary step and must be a small positive integer. And m, n come from experiment, not from the coefficients a, b — they coincide only for an elementary reaction.

The integrated rate laws in one table

Order	Integrated form	Straight-line plot	Half-life	Units of k
0	[A] = [A]₀ − kt	[A] vs t	[A]₀ / 2k	mol dm⁻³ s⁻¹
1	ln[A] = ln[A]₀ − kt	ln[A] vs t	0.693 / k	s⁻¹
2	1/[A] = 1/[A]₀ + kt	1/[A] vs t	1 / k[A]₀	dm³ mol⁻¹ s⁻¹
n	—	—	∝ 1/[A]₀ⁿ⁻¹	mol¹⁻ⁿ dm³⁽ⁿ⁻¹⁾ s⁻¹

The half-life column alone answers a whole class of questions. First order is the only case where t½ is independent of the starting concentration, so "the half-life is unchanged on doubling [A]₀" means first order with no calculation at all. The units row works in reverse — give the units of k and the order is fixed.

Worked example 1 — first order, every standard sub-question. A first-order reaction has t½ = 20.0 min.

Rate constant: k = 0.693 / 20.0 = 0.03466 min⁻¹

Time for 75% completion: one quarter remains, which is two half-lives, so t = 2 × 20.0 = 40.0 min. By formula: t = (1/k)·ln(1/0.25) = 1.3863 / 0.03466 = 40.0 min ✓

Time for 90% completion: t = (1/k)·ln(1/0.10) = 2.3026 / 0.03466 = 66.4 min.
Cross-check: 90% done means 10% left, i.e. log₂10 = 3.32 half-lives, 3.32 × 20.0 = 66.4 min ✓

Fraction remaining after 50.0 min: [A]/[A]₀ = e^−kt = e^{−(0.03466 × 50.0)} = e^−1.733 = 0.177, i.e. 17.7%.

Worked example 2 — second order, and why its half-life keeps growing. A second-order reaction has k = 0.50 dm³ mol⁻¹ s⁻¹ and [A]₀ = 0.20 mol dm⁻³.

Time to fall to 0.050 mol dm⁻³:
1/[A] − 1/[A]₀ = kt → (1/0.050) − (1/0.20) = 20 − 5 = 15 dm³ mol⁻¹
t = 15 / 0.50 = 30 s

First half-life: t½ = 1/(k[A]₀) = 1/(0.50 × 0.20) = 1/0.10 = 10 s
Second half-life (from 0.10 to 0.050): 1/(0.50 × 0.10) = 20 s

Each successive half-life doubles. Contrast this with first order, where every half-life is identical — this is the fastest way to tell the two apart from a data table.

Finding the order from initial rates

Change one concentration at a time and see what the rate does. If doubling [A] multiplies the rate by 2^m, then m is the order in A.

Worked example 3. For A + B → products the following initial rates were measured, all concentrations in mol dm⁻³ and rates in mol dm⁻³ s⁻¹:

Run	[A]₀	[B]₀	Initial rate
1	0.10	0.10	2.0 × 10⁻⁴
2	0.20	0.10	8.0 × 10⁻⁴
3	0.10	0.20	4.0 × 10⁻⁴

Order in A (runs 1 → 2, [B] fixed): rate ratio = 8.0/2.0 = 4 when [A] doubles. 2^m = 4 → m = 2.
Order in B (runs 1 → 3, [A] fixed): rate ratio = 4.0/2.0 = 2 when [B] doubles. 2ⁿ = 2 → n = 1.
Overall order = 3, rate = k[A]²[B].

Value of k from run 1:
k = rate / ([A]²[B]) = 2.0 × 10⁻⁴ / ((0.10)² × 0.10) = 2.0 × 10⁻⁴ / (0.010 × 0.10)
= 2.0 × 10⁻⁴ / 1.0 × 10⁻³ = 0.20 dm⁶ mol⁻² s⁻¹

Unit check: for n = 3 the general form gives mol¹⁻³ dm³⁽³⁻¹⁾ s⁻¹ = mol⁻² dm⁶ s⁻¹ ✓. Second check: substitute into run 2 — 0.20 × (0.20)² × 0.10 = 0.20 × 0.040 × 0.10 = 8.0 × 10⁻⁴ ✓, which matches the table.

Temperature dependence — the Arrhenius equation

k = A·e^−Ea/RT ln k = ln A − Ea/RT

ln(k₂/k₁) = (Ea/R)·(1/T₁ − 1/T₂) R = 8.314 J K⁻¹ mol⁻¹

A plot of ln k against 1/T is a straight line of slope −Ea/R and intercept ln A — which answers most conceptual questions here: larger Ea, steeper line, rate more sensitive to temperature.

Worked example 4 — the classic "rate doubles per 10 K". The rate constant doubles from 300 K to 310 K. Find Ea.

ln(k₂/k₁) = ln 2 = 0.6931
1/T₁ − 1/T₂ = 1/300 − 1/310 = (310 − 300)/(300 × 310) = 10/93000 = 1.0753 × 10⁻⁴ K⁻¹
Ea = R × 0.6931 / 1.0753 × 10⁻⁴ = 8.314 × 6446 = 53 590 J mol⁻¹

Ea ≈ 53.6 kJ mol⁻¹

Now use it forward. Predict the rate ratio between 300 K and 320 K:
1/300 − 1/320 = 20/96000 = 2.0833 × 10⁻⁴
ln(k₃₂₀/k₃₀₀) = (53 590 / 8.314) × 2.0833 × 10⁻⁴ = 6446 × 2.0833 × 10⁻⁴ = 1.343
k₃₂₀/k₃₀₀ = e^1.343 = 3.83

Note that this is not 2 × 2 = 4. The "rate doubles every 10 degrees" rule of thumb is only approximate, because 1/T is not linear in T — a point worth making if a question asks you to comment.

Deriving a rate law from a mechanism

Two tools cover almost everything JAM asks. With a rate-determining step, the rate is that step's rate, and any intermediate in it must be replaced using the fast pre-equilibrium before it. With the steady-state approximation, set d[intermediate]/dt = 0 and solve.

Worked example 5 — a fractional order.

Step 1 (fast, at equilibrium): A₂ ⇌ 2A, equilibrium constant K
Step 2 (slow): A + B → P, rate constant k₂

rate = k₂[A][B]. But A is an intermediate, so eliminate it.
From step 1: K = [A]²/[A₂] → [A] = (K[A₂])^1/2
Substituting: rate = k₂·K^1/2·[A₂]^1/2[B]

Order = ½ in A₂, 1 in B, overall 1.5, and the observed rate constant is k_obs = k₂K^1/2. A half-order is the signature of a dissociative pre-equilibrium, and that is exactly what such a question is testing.

Three related items from the same block: pseudo-first order — flood one reactant and a second-order reaction behaves as first order; parallel reactions — the product ratio equals the ratio of the rate constants; reversible reactions — relaxation to equilibrium is governed by the sum k_f + k_r, while K is their ratio.

Where the marks go

Reading the order off the balanced equation. It is experimental unless the step is stated to be elementary.
Mixing time units. If k is in min⁻¹, the answer is in minutes. Convert before, not after.
Using 0.693/k for a non-first-order reaction. That expression is first order only.
Forgetting the stoichiometric factor in −(1/a)d[A]/dt when the question follows a species with a coefficient greater than one.
R in the wrong units. Use 8.314 J K⁻¹ mol⁻¹ and the answer comes out in joules per mole; divide by 1000 for kJ mol⁻¹.
Leaving an intermediate in the final rate law. A rate law may only contain species whose concentrations can be measured.
Celsius left unconverted. Every T in the Arrhenius equation is in kelvin.

How it appears in the paper

Question style	Tool to reach for
Order from a data table	Initial-rates comparison, or the half-life test
Time or concentration at time t	The correct integrated rate law
Identify order from units of k	mol¹⁻ⁿ dm³⁽ⁿ⁻¹⁾ s⁻¹
Ea, A, or k at a new temperature	Two-point Arrhenius, or the ln k vs 1/T slope
Rate law from a mechanism	Rate-determining step or steady state
Which plot is linear	[A], ln[A] or 1/[A] against t

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Check Ea and k without re-doing the logarithms by hand. The Arrhenius calculator takes two rate constants and two temperatures and returns the activation energy, or takes Ea and A and returns k at any temperature — exactly the two directions the exam asks for.

Open the Arrhenius Equation Calculator →

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