The half-life of a reaction is 50 min at a particular initial concentration of the reactant. When the initial concentration is doubled, the half-life becomes 100 min. What is the order of the reaction?
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → Zero
The half-life t1/2 is the time in which the reactant concentration falls to half its initial value. How it changes with the initial concentration [R]0 shows the order:
Zero order: \( t_{1/2}=\frac{[\mathrm{R}]_0}{2k} \), so t1/2 is directly proportional to [R]0.
First order: \( t_{1/2}=\frac{0.693}{k} \), so t1/2 does not change with [R]0.
Here doubling [R]0 doubles the half-life (50 min → 100 min):
\( \displaystyle \frac{t_{1/2}'}{t_{1/2}}=\frac{100}{50}=2=\frac{2[\mathrm{R}]_0}{[\mathrm{R}]_0} \)
t1/2 ∝ [R]0, so the reaction is of zero order.
NCERT (2026-27 edition):
3.3.3 Half-Life of a Reaction – “For zero order reaction t1/2 ∝ [R]0. For first order reaction t1/2 is independent of [R]0.”