CSIR-NET Chemical Kinetics — The Question Patterns Worth Drilling
Chemical kinetics is one of the most reliably examinable areas of physical chemistry, because the questions can be made numerical, mechanistic or conceptual from the same small set of results. Part B questions usually ask you to apply an integrated rate law or an Arrhenius relation directly. Part C questions change one condition — a limiting concentration, a competing pathway, a temperature range — and ask what the rate expression collapses to. This article walks through the patterns that recur, with the arithmetic done in full.
Pattern 1 — Integrated rate laws and half-lives
Know these three as a block, and know the shape of the linear plot each one produces, because many questions give you a graph instead of numbers.
First order: ln([A]0/[A]) = kt · linear in ln[A] vs t · t½ = ln2 ÷ k = 0.693 ÷ k
Second order (one reactant): 1/[A] − 1/[A]0 = kt · linear in 1/[A] vs t · t½ = 1 ÷ (k[A]0)
The diagnostic that examiners love: only for a first-order reaction is the half-life independent of the starting concentration. For zero order t½ is proportional to [A]0; for second order it is inversely proportional. The general statement is t½ ∝ [A]0(1−n) for order n ≠ 1, which gives a direct route to the order from two experiments.
Worked example 1 — order from half-life data. With [A]0 = 0.10 M the half-life is 100 s; with [A]0 = 0.20 M it is 50 s. Find the order.
t½,1 ÷ t½,2 = 100 ÷ 50 = 2, and [A]0,2 ÷ [A]0,1 = 0.20 ÷ 0.10 = 2.
n = 1 + log 2 ÷ log 2 = 1 + 1 = 2, second order.
Cross-check with the second-order half-life formula: k = 1 ÷ (t½[A]0) = 1 ÷ (100 × 0.10) = 0.10 M⁻¹ s⁻¹, and for the second run 1 ÷ (50 × 0.20) = 0.10 M⁻¹ s⁻¹ — the same k, which confirms the assignment.
Pattern 2 — Activation energy from two temperatures
This is the single most common numerical in the whole topic. Take the Arrhenius equation, write it at two temperatures and subtract:
Worked example 2 — the "rate doubles per 10 K" rule. A reaction's rate constant doubles when the temperature rises from 300 K to 310 K. Find Ea (R = 8.314 J K⁻¹ mol⁻¹).
ln 2 = (Ea ÷ 8.314) × (310 − 300) ÷ (300 × 310)
0.6931 = (Ea ÷ 8.314) × 10 ÷ 93 000
0.6931 = Ea × 10 ÷ (8.314 × 93 000) = Ea × 10 ÷ 773 202 = Ea × 1.2933 × 10⁻⁵
Ea = 0.6931 ÷ (1.2933 × 10⁻⁵) = 53 590 J/mol ≈ 53.6 kJ/mol
Note the point the question is really testing: the popular "rate doubles for every 10 °C" statement is not a law. It happens to hold near room temperature only for activation energies around 50 kJ/mol. Repeat the same calculation between 400 K and 410 K and you get Ea = 0.6931 × 8.314 × 164 000 ÷ 10 = 94 500 J/mol ≈ 94.5 kJ/mol — a completely different answer.
Related results worth carrying on your formula sheet: the Arrhenius plot of ln k against 1/T has slope −Ea/R; and from transition state theory the Eyring equation k = (kBT/h) eΔS‡/R e−ΔH‡/RT connects the two, with Ea = ΔH‡ + RT for a unimolecular gas-phase or solution reaction. A large negative ΔS‡ signals an ordered, associative transition state — a standard one-line Part C inference.
Pattern 3 — Steady-state approximation on a two-step mechanism
Part C almost always includes one mechanism where you must apply the steady-state approximation to a reactive intermediate and then examine the limiting cases. The template:
I + B → P (k2)
Worked example 3 — derive and take limits.
Set d[I]/dt = 0: k1[A] = k−1[I] + k2[I][B]
So [I] = k1[A] ÷ (k−1 + k2[B])
Rate = k2[I][B] = k1k2[A][B] ÷ (k−1 + k2[B])
Limit 1 — low [B] (k2[B] ≪ k−1): Rate = (k1k2/k−1)[A][B], overall second order, and the pre-equilibrium is maintained.
Limit 2 — high [B] (k2[B] ≫ k−1): Rate = k1[A], first order in A and zero order in B — the first step has become rate-determining and adding more B changes nothing.
This saturation shape is the same algebra as the Lindemann–Hinshelwood mechanism for unimolecular gas reactions and as Michaelis–Menten enzyme kinetics. Recognise the form Rate = a[X] ÷ (b + c[X]) and you have already solved half the question.
Pattern 4 — Parallel and consecutive reactions
For competing first-order pathways A → B (k1) and A → C (k2), A decays with the sum of the constants and the products appear in a fixed ratio:
For the consecutive scheme A → B → C with rate constants k1 and k2, the intermediate B rises then falls, and its maximum occurs at
Worked example 4. For A → B → C with k1 = 0.10 min⁻¹ and k2 = 0.40 min⁻¹, when is [B] greatest?
tmax = ln(0.40 ÷ 0.10) ÷ (0.40 − 0.10) = ln 4 ÷ 0.30 = 1.3863 ÷ 0.30 = 4.62 min
Because k2 > k1, B never accumulates much — this is exactly the condition under which the steady-state approximation on B is justified. If instead k1 ≫ k2, B piles up and the second step is rate-determining. Being able to say which approximation is valid, and why, is what Part C is checking.
Mistakes that cost marks in kinetics questions
- Assuming order equals molecularity. Order is experimental and can be fractional or negative; molecularity applies only to an elementary step and is a small positive integer.
- Mixing up 1/T1 − 1/T2 with 1/T2 − 1/T1. Check the sign at the end: if k rose with T, Ea must come out positive.
- Using Celsius in Arrhenius. T must be absolute, every time.
- Applying the steady-state approximation to a species that accumulates. It is valid only for a low-concentration, highly reactive intermediate.
- Forgetting the stoichiometric coefficient. For 2A → P the rate of reaction is −(1/2)d[A]/dt, so k from an [A] plot differs from the reaction rate constant by a factor of 2.
- Treating a pseudo-first-order constant as the true one. If kobs = k[H₂O] because water is in vast excess, kobs carries different units and a different value.
Summary — what each pattern tests
| Pattern | Key result | Typical trap |
|---|---|---|
| Integrated rate law | Which plot is linear | Reading the graph axis wrongly |
| Half-life method | t½ ∝ [A]0(1−n) | Assuming t½ is always constant |
| Arrhenius two-point | ln(k₂/k₁) = (Ea/R)(T₂−T₁)/T₁T₂ | Celsius temperatures; sign errors |
| Steady state | Rate = a[X]/(b + c[X]) saturation form | Not testing both limits |
| Parallel paths | kobs = Σk; yields fixed by k ratio | Thinking the ratio changes with time |
| Consecutive | tmax = ln(k₂/k₁)/(k₂−k₁) | Applying steady state when k₁ ≫ k₂ |
| Transition state theory | Ea = ΔH‡ + RT; ΔS‡ sign | Confusing Ea with ΔH of reaction |
Check your activation-energy arithmetic in one step. The Arrhenius calculator takes two rate constants and two temperatures and returns Ea and the pre-exponential factor A — ideal for verifying the two-point numericals above before you trust your own working under exam pressure.
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