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MHT-CET Chemistry · Chemical Kinetics

Zero-Order Kinetics

A zero-order reaction runs at a constant rate k regardless of concentration: [A]ₜ = [A]₀ − kt, k has the unit of a rate (mol dm⁻³ s⁻¹), and the half-life [A]₀/2k grows with the initial concentration.

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Why this matters

11 PYQs, none HARD — one of the shortest pages in the chapter. Rate constant from a concentration drop over a time, percent unreacted after a time, the half-life formula and its proportionality to [A]₀, and the one statement question (rate independent of concentration). It is worth five minutes because the contrast with first order — half-life dependent on [A]₀ here, independent there — is a favourite comparison stem.

Concept 1 of 2: [A]ₜ = [A]₀ − kt: Constant Rate, k in Concentration per Time

With r=k[A]0=kr = k[A]^0 = k the concentration falls in a straight line: the drop divided by the time IS the rate constant. Its unit is the unit of a rate, mol dm−3^{-3} s−1^{-1}.

Definition

  • k=[A]0−[A]ttk = \dfrac{[A]_0 - [A]_t}{t}: 1.2→0.41.2 \to 0.4 in 240240 s gives 0.8240\dfrac{0.8}{240} mol dm−3^{-3} s−1^{-1} =0.2= 0.2 mol dm−3^{-3} min−1^{-1}; 0.8→0.20.8 \to 0.2 in 66 min gives 0.10.1 mol dm−3^{-3} min−1^{-1}.
  • Percent unreacted: with k=1k = 1 mol dm−3^{-3} s−1^{-1} and [A]0[A]_0 taken as 100100, after 9090 s [A]=100−90=10%[A] = 100 - 90 = 10\%.
  • The rate is independent of the reactant's concentration; the rate constant's unit is mol dm−3^{-3} s−1^{-1}, NOT s−1^{-1}.
  • Examples: decomposition of NH3_3 on a hot platinum surface, photochemical H2_2 + Cl2_2, enzyme reactions at saturation.

Zero order

[A]t=[A]0−kt,k=[A]0−[A]tt  (mol dm−3 s−1)[A]_t = [A]_0 - kt,\qquad k = \frac{[A]_0 - [A]_t}{t}\ \ (\text{mol dm}^{-3}\,\text{s}^{-1})

Worked example

A zero-order reaction has k=0.05k = 0.05 mol dm−3^{-3} min−1^{-1}. Starting from 1.51.5 mol dm−3^{-3}, what is the concentration after 1212 minutes?
Practise this conceptself-check · 4 quick reps

The same idea in a real exam question:

MHT-CET · 2023 · 10th May Shift 1 · Q55Easy

Example 1 · Chemical Kinetics · Zero-Order Kinetics

For a zero-order reaction, A⟶\text{A}\longrightarrow product, concentration of A decreases from 1.2 mol dm−31.2\,\text{mol dm}^{-3} to 0.4 mol dm−30.4\,\text{mol dm}^{-3} in 240 second. What is rate constant of the reaction?

Forgetting the seconds-to-minutes conversion

0.8240=0.0033\dfrac{0.8}{240} = 0.0033 per second is 0.20.2 per minute; the options are per minute. Convert before matching.
Drill 3 more on [a]ₜ = [a]₀ − kt: constant rate, k in concentration per time

Concept 2 of 2: Zero-Order Half-Life: t½ = [A]₀/2k, Proportional to the Initial Concentration

Half of [A]0[A]_0 must be removed at the constant rate kk, so t1/2=[A]0/2kt_{1/2} = \dfrac{[A]_0/2}{k}. Doubling the starting amount doubles the time — the opposite of first order, where the half-life is fixed.

Definition

  • t1/2=[A]02kt_{1/2} = \dfrac{[A]_0}{2k}; with [A]0=a[A]_0 = a: a2k\dfrac{a}{2k}.
  • kk from a half-life: t1/2=0.2t_{1/2} = 0.2 min with [A]0=0.2[A]_0 = 0.2 gives k=0.22×0.2=0.5k = \dfrac{0.2}{2 \times 0.2} = 0.5 mol dm−3^{-3} min−1^{-1}.
  • Half-life is directly proportional to the initial concentration and inversely to kk; it does not depend on the amount of product or on temperature except through kk.

Zero-order half-life

t1/2=[A]02kt_{1/2} = \frac{[A]_0}{2k}

Worked example

A zero-order reaction with k=0.04k = 0.04 mol dm−3^{-3} s−1^{-1} starts at 0.80.8 mol dm−3^{-3}. Find the half-life, and the half-life if the start were 1.61.6.
Practise this conceptself-check · 4 quick reps

The same idea in a real exam question:

MHT-CET · 2025 · 21 April Shift II · Q94Easy

Example 2 · Chemical Kinetics · Zero-Order Kinetics

For a reaction, A⟶BA \longrightarrow B, rate equation is r=k[A]∘r = k\lbrack A\rbrack^{\circ}. If initial concentration of reactant is 'a'mol dm−3dm^{- 3} find half life time of reaction.

Using 0.693/k

0.693/k0.693/k is the FIRST-order half-life. For zero order the half-life carries [A]0[A]_0; option (B) ak\dfrac{a}{k} forgets the 22.
Drill 2 more on zero-order half-life: t½ = [a]₀/2k, proportional to the initial concentration

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (2)

Watch out for (2)

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