Half life of the zero order reaction is:

The correct answer is Option (4) → Directly proportional to initial concentration of the reactant

The half-life \( t_{1/2} \) is the time in which the concentration of a reactant falls to half its initial value. For a zero order reaction, R → P, the rate does not depend on concentration, and the integrated rate equation is:

    \( \displaystyle k = \frac{[\mathrm{R}]_0 - [\mathrm{R}]}{t} \)

Putting \( [\mathrm{R}] = \frac{1}{2}[\mathrm{R}]_0 \) at \( t = t_{1/2} \) gives

    \( \displaystyle t_{1/2} = \frac{[\mathrm{R}]_0}{2k} \)

So the half-life of a zero order reaction is directly proportional to the initial concentration of the reactant (and inversely proportional to k): the more reactant there is, the longer it takes to use up half of it at a constant rate.

Why the other options are incorrect

• Option (1) Directly proportional to the square root of the initial concentration of the reactant — Incorrect. For an order n, \( t_{1/2} \propto [\mathrm{R}]_0^{\,1-n} \); a square-root dependence would mean an order of one-half, not zero.

• Option (2) Independent of initial concentration of the reactant — Incorrect. This holds for a first order reaction, \( t_{1/2} = 0.693/k \).

• Option (3) Inversely proportional to initial concentration of the reactant — Incorrect. This holds for a second order reaction, \( t_{1/2} = 1/(k[\mathrm{R}]_0) \).

NCERT (2026-27 edition):
Half-Life of a Reaction