Which of the following is not the graphical representation of first-order kinetics?
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → plot of t1/2 against [A]0 drawn as a vertical line
For a first order reaction the integrated rate equation and its consequences are:
\( \displaystyle [A]_t = [A]_0\,e^{-kt} \qquad \ln\frac{[A]_0}{[A]_t} = kt \qquad \log[A]_t = \log[A]_0 - \frac{kt}{2.303} \qquad t_{1/2} = \frac{0.693}{k} \)
Option (1) log [A]t against t — A first order plot. It is a straight line with slope −k/2.303.
Option (2) [A]t against t — A first order plot. The concentration falls along an exponential decay curve.
Option (3) ln([A]0/[A]t) against t — A first order plot. It is a straight line through the origin with slope k.
Option (4) t1/2 against [A]0 drawn as a vertical line — Not a first order plot. \( t_{1/2} = 0.693/k \) does not depend on [A]0, so the plot must be a horizontal line parallel to the [A]0 axis. A vertical line would mean many half-lives for one initial concentration.
NCERT (2026-27 edition):
Half-Life of a Reaction – …
