CBSE Class 12 Chemical Kinetics — The Numericals That Always Come

By Aniket Bhardwaj · 1 September 2026 · CBSE / ICSE

Chemical Kinetics is one of the most dependable scoring chapters in Class 12 Chemistry, because the numerical questions repeat in shape year after year. There are really only three question types you must be able to do without thinking: a first order calculation, a zero order calculation, and finding activation energy from two rate constants at two temperatures. Everything else in the chapter is theory built around those three.

This guide gives you the exact formulas, tells you what every symbol means, and works through each type with the arithmetic shown line by line — the same way you should write it in the answer sheet.

First, the vocabulary that decides the formula

The rate law is found by experiment, never from the balanced equation. For a reaction A → products, if rate = k[A]ⁿ, then n is the order and k is the rate constant. The order decides which integrated rate law you use, so read the question for the words "first order" or "zero order" before you write anything.

A useful cross-check: the units of k depend on the order.

Units of k = mol⁽¹⁻ⁿ⁾ L⁽ⁿ⁻¹⁾ s⁻¹
Zero order (n = 0): mol L⁻¹ s⁻¹ · First order (n = 1): s⁻¹ · Second order (n = 2): L mol⁻¹ s⁻¹

If a question gives you k in s⁻¹, the reaction is first order even if nobody said so. That single observation solves many "identify the order" questions instantly.

Zero order — the straight-line case

In a zero order reaction the rate does not depend on concentration at all. Concentration therefore falls in a straight line with time.

[A] = [A]₀ − kt ⇒ k = ([A]₀ − [A]) / t
Half-life: t_1/2 = [A]₀ / 2k

Note carefully: for zero order the half-life does depend on the starting concentration. Double the starting amount and the half-life doubles.

Worked example 1. The decomposition of ammonia on a hot platinum surface is zero order with k = 2.5 × 10⁻⁴ mol L⁻¹ s⁻¹. If the initial concentration is 0.10 mol L⁻¹, how long until it falls to 0.04 mol L⁻¹, and what is the half-life?

[A] = [A]₀ − kt
0.04 = 0.10 − (2.5 × 10⁻⁴) t
(2.5 × 10⁻⁴) t = 0.10 − 0.04 = 0.06
t = 0.06 ÷ (2.5 × 10⁻⁴) = 240 s

t_1/2 = [A]₀ / 2k = 0.10 ÷ (2 × 2.5 × 10⁻⁴) = 0.10 ÷ (5.0 × 10⁻⁴) = 200 s

First order — the logarithm case

This is the one that appears most often. In the base-10 form used in Indian textbooks:

k = (2.303 / t) log ( [A]₀ / [A] )
Equivalently: log [A] = log [A]₀ − (k / 2.303) t (straight line, slope −k/2.303)
Half-life: t_1/2 = 0.693 / k

The half-life of a first order reaction is independent of concentration. That is the single most examined fact of the chapter, and it is why radioactive decay and many drug-elimination processes are described as first order.

Because the formula uses a ratio of concentrations, you may put in any quantity proportional to concentration — moles, pressure, percentage remaining, even mass. This is why "30% of the reactant decomposes" questions are solvable without knowing the actual concentration.

Worked example 2. A first order reaction is 30% complete in 40 minutes. Find k and the half-life.

If 30% has reacted, 70% remains. Take [A]₀ = 100 and [A] = 70.

k = (2.303 / 40) × log (100 / 70)
log (100/70) = log 1.4286 = 0.1549
2.303 / 40 = 0.05758
k = 0.05758 × 0.1549 = 8.92 × 10⁻³ min⁻¹

t_1/2 = 0.693 / k = 0.693 ÷ (8.92 × 10⁻³) = 77.7 minutes

Sanity check: 30% gone in 40 min, so half gone should take a bit under twice that. 77.7 min is sensible.

Activation energy from two rate constants

The Arrhenius equation links the rate constant to temperature:

k = A e^−E_a/RT ⇒ log k = log A − E_a / (2.303 R T)
Two-temperature form: log (k₂ / k₁) = [ E_a / (2.303 R) ] × [ (T₂ − T₁) / (T₁T₂) ]

Here A is the frequency (pre-exponential) factor, E_a the activation energy, R = 8.314 J K⁻¹ mol⁻¹, and T is in kelvin. A plot of log k against 1/T is a straight line of slope −E_a/2.303R — the graph question of this chapter.

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

k₂/k₁ = 2, so log 2 = 0.3010

(T₂ − T₁) / (T₁T₂) = 10 ÷ (300 × 310) = 10 ÷ 93000 = 1.0753 × 10⁻⁴ K⁻¹

2.303 R = 2.303 × 8.314 = 19.147 J K⁻¹ mol⁻¹

E_a = (0.3010 × 19.147) ÷ (1.0753 × 10⁻⁴) = 5.763 ÷ (1.0753 × 10⁻⁴) = 53 598 J mol⁻¹

E_a ≈ 53.6 kJ mol⁻¹

The same rearranged equation lets you go the other way: given E_a and k at one temperature, predict k at another. Practise both directions, because the paper can ask either.

Mistakes that quietly cost marks

Dropping the 2.303. It converts natural log to log base 10. If you use ln, there is no 2.303; if you use log, there always is. Never mix the two in one line.
Temperature in °C. Every T in Arrhenius work must be kelvin. 27 °C is 300 K, not 27.
Wrong R. Use R = 8.314 J K⁻¹ mol⁻¹ and your E_a comes out in joules per mole — then divide by 1000 for kJ mol⁻¹. State the unit in the final answer.
Using the wrong half-life formula. 0.693/k is first order only. For zero order it is [A]₀/2k.
Reading order from the balanced equation. Order is experimental. Molecularity comes from a mechanism step; order can even be fractional or zero.
Using amount reacted instead of amount left. In the first order formula the denominator is what remains. "30% complete" means [A] = 70, not 30.

Quick revision table

Quantity	Zero order	First order
Rate law	rate = k	rate = k[A]
Integrated form	[A] = [A]₀ − kt	k = (2.303/t) log([A]₀/[A])
Straight-line graph	[A] vs t, slope = −k	log[A] vs t, slope = −k/2.303
Half-life	[A]₀ / 2k	0.693 / k
Depends on [A]₀?	Yes	No
Units of k	mol L⁻¹ s⁻¹	s⁻¹

Memorise this table and nearly every kinetics numerical becomes a matter of choosing the correct column and substituting. Write the formula first, then substitute, then compute — an examiner can award method marks only for working that is visible.

Check your activation energy answers in seconds. The free Arrhenius Equation calculator takes two rate constants and two temperatures and returns E_a, or takes E_a and predicts k at a new temperature — perfect for verifying a whole exercise after you have solved it by hand.

Open the Arrhenius Equation Calculator →

Want kinetics taught properly, with question practice graded every week? ABC Chemistry runs Class 11–12 chemistry coaching from its Gurugram centre and online across India — see abcchemistry.in. Families in Delhi-NCR who prefer one-to-one teaching at home can also arrange home tuition.