Chemical Kinetics: How Fast Do Reactions Go?

Q: For a first-order reaction, which statement is correct?

For a first-order reaction, t½ = 0.693/k, which contains only k — so the half-life is completely independent of the initial concentration (option 3 is correct). The unit of k for first order is s⁻¹, not mol L⁻¹ s⁻¹, so option 2 is wrong. Since Rate = k[R], doubling [R] doubles the rate, so option 4 is wrong. And option 1 contradicts the very formula t½ = 0.693/k.

A from-scratch guide to reaction rates for CBSE Class 12: what rate means, how to write a rate law, the difference between order and molecularity, the rate constant and its units, and the first-order integrated equation with half-life — complete with worked numericals.

By the PadhoDost Team · 📖 7 min read · Updated 4 August 2026

Part of Class 12 (CBSE) prep

🧠 Two reactions, two very different speeds

Think of an iron gate slowly rusting over months, and a Diwali phatka bursting in a fraction of a second. Both are chemical reactions, but their speeds are worlds apart. Chemical Kinetics is the branch of chemistry that answers one simple question: HOW FAST does a reaction happen, and what controls that speed? Padho, dost — once you can measure and predict reaction speed, a big chunk of physical chemistry suddenly clicks into place.

What Do We Mean by 'Rate of Reaction'?

The rate of a reaction is the change in concentration of a reactant or product per unit time. As a reaction proceeds, reactants get used up (their concentration falls) and products form (their concentration rises). The average rate is measured over a time interval, while the instantaneous rate is the rate at one particular instant — the slope of the concentration-versus-time curve at that exact moment. Because a reactant's concentration decreases, we put a minus sign in front of its change so that the rate always comes out positive.

For aA + bB → cC + dD : Rate = −(1/a)·d[A]/dt = −(1/b)·d[B]/dt = +(1/c)·d[C]/dt = +(1/d)·d[D]/dt

📝 Average rate — a quick numerical

In a reaction, the concentration of reactant R falls from 0.50 mol/L to 0.40 mol/L in 10 minutes. Find the average rate.

Average rate = −Δ[R] / Δt

= −(0.40 − 0.50) / 10

= 0.10 / 10 = 0.01 mol L⁻¹ min⁻¹

So R is being consumed at 1 × 10⁻² mol L⁻¹ per minute.

ℹ️ Units of rate are always concentration ÷ time, i.e. mol L⁻¹ s⁻¹ (or mol L⁻¹ min⁻¹). Do not confuse this with the unit of the rate constant k, which changes with the order of the reaction — more on that just below.

Rate Law, Order and Molecularity

The rate law (or rate expression) tells us how the rate depends on the concentration of the reactants. The single most important idea in this chapter: the rate law is found by EXPERIMENT, not by reading off the balanced equation. The powers of concentration in the rate law need not match the stoichiometric coefficients.

Rate = k [A]ˣ [B]ʸ → Order of reaction = x + y (k = rate constant)

Order is the sum of the powers of the concentration terms in the experimentally observed rate law. It can be 0, 1, 2, or even a fraction. Molecularity, on the other hand, is the number of reacting species that come together in a single elementary step. Molecularity is a theoretical idea, is always a whole number (1, 2 or 3), and is defined only for elementary reactions — while order is experimental and can be zero or fractional.

Order	Rate law	Unit of k	Half-life (t½)
Zero	Rate = k	mol L⁻¹ s⁻¹	[R]₀ / 2k (depends on [R]₀)
First	Rate = k[R]	s⁻¹	0.693 / k (independent of [R]₀)
Second	Rate = k[R]²	mol⁻¹ L s⁻¹	1 / (k[R]₀) (depends on [R]₀)

⚠️ Common mistake: reading the rate law off the balanced equation. For 2N₂O₅ → 4NO₂ + O₂ the order is 1, NOT 2. Remember order and molecularity are different: order is experimental (can be 0 or fractional), while molecularity is theoretical (a whole number, defined only for elementary reactions).

First-Order Reactions and Half-Life

Integrated first-order equation: k = (2.303 / t) · log([R]₀ / [R]) and Half-life: t½ = 0.693 / k

📝 Worked numerical — first-order reaction

A first-order reaction has a half-life of 60 minutes. Find k, and the time for the reaction to be 75% complete.

Step 1 — Rate constant: k = 0.693 / t½ = 0.693 / 60 = 0.01155 min⁻¹

Step 2 — 75% complete means 25% of R is left, so take [R]₀ = 100 and [R] = 25 → [R]₀/[R] = 4

Step 3 — Use k = (2.303/t) log([R]₀/[R]) → t = (2.303 / k) · log 4

t = (2.303 / 0.01155) × 0.6021 = 199.4 × 0.6021 ≈ 120 minutes

Cross-check: 75% done = 2 half-lives = 2 × 60 = 120 minutes ✓

Chapter in a nutshell

✓Rate = change in concentration per unit time; unit mol L⁻¹ s⁻¹. Instantaneous rate = slope of the concentration–time graph.
✓Rate law is found by experiment: Rate = k[A]ˣ[B]ʸ; order = x + y.
✓Order can be 0, 1, 2 or fractional (experimental); molecularity is 1, 2 or 3 (theoretical, elementary steps only).
✓Rate constant k is independent of concentration; its unit depends on the order.
✓First order: k = (2.303/t) log([R]₀/[R]); half-life t½ = 0.693/k is constant and independent of initial concentration.

💡 Exam tip: 'independent of initial concentration' is a dead giveaway for FIRST order. And if a numerical says a reaction is 50% / 75% / 87.5% complete, that is simply 1, 2 or 3 half-lives — you can often answer in one line without touching the log formula.

⚡ Quick check

For a first-order reaction, which statement is correct?

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